ELECTRICAL
ENGINEERING
THOMÄLEN
TRANSLATED BY
G.W.O. HOWE
TK
145
T453
LONGMANS & CO
B 429369
?་

ARTES
1817
SCIENTIA
VERITAS
LIBRARY
OF THE
UNIVERSITY OF MICHIGAN
EL FLURITIUS UMUNT
TUEBOR
SI QUÆRIS-PENINSULAM ·AMŒNAM ·
CIRCUMSPICE
THE GIFT OF
Mrs. G. W. Patterson
A TEXT-BOOK
OF
ELECTRICAL ENGINEERING
TRANSLATED FROM THE GERMAN OF
DR ADOLF THOMÄLEN
BY
GEORGE W. O. HOWE,
M.Sc., WHIT. SCH., A.M.I.E.E.
LECTURER IN ELECTRICAL ENGINEERING AT THE CENTRAL TECHNICAL COLLEGE,
SOUTH KENSINGTON
Um
NEW YORK
LONGMANS, GREEN & CO.
LONDON: EDWARD ARNOLD
1907
[All Rights reserved]
ཏྟཾ,
Bew. Lib
CIFT
MRS. C. W. PATTERSON
12-11-1931
PREFACE.
GR
REAT difficulty has always been experienced in recommending a
text-book to the second and third year students of electrical
engineering at the Central Technical College. There appeared to be
nothing to bridge the gap between the elementary text-books and the
specialised works on various branches of electrical engineering. I have
no doubt that the same difficulty has been experienced by all who have
lectured to the more advanced students in our Universities and Technical
Colleges.
<<
It was while contemplating the preparation of a suitable text-book
that my attention was drawn to Dr Thomälen's work, written with the
same object and covering much the same ground as I had contemplated.
The present work is a translation of the second edition of this Kurze
Lehrbuch der Elektrotechnik," but includes additional matter which it is
intended to introduce into the third German edition now in preparation.
Generally speaking, it is a close translation of the original, but I have
not hesitated to use other methods where they appeared preferable. This
applies especially to Section 101, which differs entirely from the corre-
sponding Section in the German edition. A Section has been added to
Chapter XX, dealing with the Cascade Converter.
The book is concerned almost exclusively with principles and does not
enter into details of the practical construction of apparatus and machines.
It is not intended to take the place of the standard works on the design
of dynamo machinery, both direct and alternating, but rather to lay a
thorough foundation which shall make the study of such works more
profitable.
+
4
iv
Preface
In the preface to the first German edition, Dr Thomälen expresses his
desire to lead the student to enter mentally into the various phenomena
and to give him a physical conception of the underlying principles. This
“elektrotechnisches Denken" will enable the student to predetermine with
confidence, either graphically or mathematically, the behaviour of various
electrical machines and apparatus.
The higher mathematics employed never goes beyond the simplest
elements.
CENTRAL TECHNICAL COLLEGE,
SOUTH KENSINGTON.
June, 1907.
G. W. O. HOWE.
CONTENTS.
CHAPTER I.
1.
The electric current
2.
Electromotive force
PAGE
3. Ohm's law
4. The relations between the resistance of a conductor and its material, cross-
section, length and temperature
5. Kirchhoff's rules
6. Resistances in parallel
•
7. Series and parallel arrangement of cells
8. Wheatstone's bridge
9. Measurement of electromotive force by the potentiometer method
10. Joule's law, electrical energy and electrical power
11.
Potential difference
12. Energy lost in heating conductor
CHAPTER II.
1
24
8
10
13
14
16
18
19
21
23
#
13. Electrolysis
14. Quantitative laws of electrolysis
15. Polarisation
16.
Accumulators
17. Primary cells
18. The voltameter
26
28
31
34
38
40
CHAPTER III.
19. Strength of magnet pole
42
20. Strength of magnetic field
21. Lines of force
45
22. Magnetic potential
47
23. Iron in a magnetic field
49
24. The earth's field
*****8
43
50
CHAPTER IV.
25. Magnetic effect of a straight conductor
26. Magnetic effect of a single-turn coil
27. Magnetic field of a solenoid
28. Magnetisation curves
29. Ohm's law and the magnetic circuit
30. The lifting power of an electromagnet
31. Hysteresis
32. Dynamic effect of parallel currents
33.
Induced electromotive force
34. The laws of mutual induction
35.
Self-induction
36. Eddy currents
51
54
58
60
64
70
72
76
77
82
84
88
vi
Contents
*
CHAPTER V.
37. The units of length, mass, and time in the absolute system
38. Dimensions and units of velocity, acceleration and force
39. Dimensions and units of pole-strength, field-strength and magnetic flux
40. Dimensions and units of electromotive force, current, quantity of electricity
and resistance
41. Dimensions and units of energy, heat and power
42. Dimensions and units of the coefficient of self-induction and of capacity.
CHAPTER VI.
43. Bipolar ring-winding
44. Bipolar drum-winding
45. Multiple circuit ring-winding
46. Multiple circuit drum-winding
47. Two-circuit ring-winding
48. Two-circuit drum-winding
49. Series-parallel ring-winding
50. Series-parallel drum-winding
•
51. The excitation of dynamos
52. The field magnets
53.
Position of the brushes
54. Armature reaction
55. Sparkless commutation.
56. Three-wire dynamos
CHAPTER VII.
CHAPTER VIII.
57. Characteristics of separately-excited dynamo
58. Characteristics of series dynamo .
59. Characteristics of shunt dynamo
60. Dynamo and battery in parallel
61. The efficiency of dynamos
•
•
PAGE
91
92
94
95
98
99
102
107
112
115
120
123
127
131
134
137
142
144
149
156
•
160
168
169
173
180
CHAPTER IX.
62. Direction of rotation of motors
187
•
63. Torque, speed and output of a D.C. motor .
64. Motor with constant excitation
190
194
65. The starting and regulation of a shunt motor
199
66. Principles of the series motor
205
67. Example
208
68. The regulation of series motors
211
CHAPTER X.
69. The instantaneous value of the induced E.M.F.
70. The electrolytic mean or average value of an alternating current
71. Alternating current power and root-mean-square current
72. Vector diagrams
73. The electromotive force of self-induction
217
221
222
226
228
Contents
vii
PAGE
75.
74. Ohm's law for alternating currents
Resistance and inductance in series
232
236
76.
Resistance and inductance in parallel.
239
4:
77.
Effect of phase difference on A. c. power
239
78. Effect of capacity.
245
79. Resistance and capacity in series
248
t
80.
Circuit containing resistance, inductance and capacity
249
81. Self-induction and capacity in parallel
251
CHAPTER XI.
83. The magnetising current
82. The electromotive forces induced in a transformer
84. The hysteresis current .
Transformer on non-inductive load
253
254
255
258
264
267
88. Types of alternating current generators
89. The mean E. M. F. of a generator
85.
86.
Transformer on inductive load
87. The effect of magnetic leakage
CHAPTER XII.
272
278
279
of a single-slot winding
of a double-slot winding
·
of a treble-slot winding
281
282
284
285
287
289
291
90. The effective E. M. F. with sinusoidal field
91. The E. M. F.
92. The E. M.F.
93. The E. M. F.
94. The E.M.F. of a distributed winding
95. The alternating E.M. F. of a closed D.C. winding.
96. The E. M.F. of a creeping bar or wave-winding
97. The E. M.F. of a creeping coil-winding .
CHAPTER XIII.
98. The E. M. F. diagram for an A. C. generator
99. The ampere-turn diagram
100.
•
•
Calculation of armature reaction
101. Experimental determination of armature reaction and armature leakage.
102. Predetermination of exciting current and pressure regulation
103. Effect of polar leakage .
•
293
297
298
302
305
307
CHAPTER XIV.
104. Alternator with constant excitation and constant terminal pressure
105. Synchronising power of armature
106. The parallel connection of alternators.
107. The effect of field regulation on alternators in parallel
108. Phase-swinging of alternators
•
CHAPTER XV.
109. The principle of the synchronous motor
110. Synchronous motor with constant P.D. and excitation
111. Synchronous motor with constant load and variable excitation
310
316
318
•
321
326
332
334
337
viii
Contents
CHAPTER XVI
PAGE
112. The rotary field in a two-phase motor
340
113. The rotary field in a three-phase motor
343
114. Mesh or delta connection
115. Star connection
116. The measurement of power in polyphase circuits
344
347
350
353
117. General principles of the rotor
CHAPTER XVII.
118. Distributed windings and the electromotive forces induced in them.
119. The magnetic flux of an induction motor
357
362
120. The effect of the iron in the magnetic path
367
121. The torque of an induction motor
372
122. Calculation of slip
374
CHAPTER XVIII.
123. Rotor current, torque, and output in relation to the slip, neglecting leakage.
124. The circle-diagram, neglecting primary losses
377
380
125. Output, torque and slip from the circle-diagram
383
126. Normal load, starting torque and maximum torque
386
127. The circle-diagram, corrected for primary copper loss
388
128. The corrected values of output, rotor current and slip
129. The most convenient form of the circle-diagram
130. Practical example.
393
395
399
131. The leakage factor
403
CHAPTER XIX.
132. Resolution of the primary M.M.F. of the single-phase motor into two
constant rotating components .
133. The E. M. F. induced in the actual stator
134. The circle-diagram of the single-phase motor
135. Single-phase commutator motors .
•
410
413
415
417
CHAPTER XX,
136. Relation between direct and alternating currents in rotary converters
137. Armature copper loss in rotary converters .
138. Comparison between rotary converters and direct-current generators with
regard to armature copper loss
139. La Cour's cascade converter.
APPENDIX
LIST OF SYMBOLS USED
INDEX
429
431
433
435
438
·
•
•
•
•
449
451
CHAPTER I.
1. The electric current.-2. Electromotive force.-3. Ohm's law.-4. Resistance, dependent on
material, cross-section, length and temperature.-5. Kirchhoff's rules.-6. Resistances in
parallel.-7. Series and parallel arrangement of cells.-8. Wheatstone's bridge.-9. Poten-
tiometer method of measuring electromotive force.-10. Joule's law, electrical energy and
electrical power.-11. Potential difference.-12. Energy lost in heating conductor.
1. The Electric Current.
The heating of a glow lamp filament, the power exerted by an electric
motor, the magnetisation of the iron in an electromagnet, the decomposition
of liquids and many other phenomena, are said to be due to the effect of the
electric current. The question arises: why do we ascribe these phenomena
to a current, i.e. a continuous movement of electricity?
The electricity with which we are here concerned is essentially the same
as static or frictional electricity. If a glass rod is rubbed with silk it becomes
electrified and attracts small light bodies. In the same way a rod of resin,
when rubbed with a woollen cloth, becomes electrified, but in the opposite
sense to the glass rod. This difference of sense or sign is based on the fact
that the electrical charges of the glass and resin mutually neutralise each
other. Because of this, the electricity of the glass is called positive, and
that of the resin negative. The equalising exchanges between positive and
negative electricity follow in exactly the same way as those between heat
and cold, high pressure and low pressure. As water flows from the higher
to the lower level, or heat from the hot to the cold body, so electricity flows
from the higher positive to the lower negative level. This equalising flow
of electricity is what we call the electric current. There is no essential
difference between the flow of electricity between the terminals of a cell,
or of a dynamo, and that between two metal balls charged with electricity
of opposite sign.
Although the same in principle, there is a great difference as regards
quantity between the frictional electricity with which we are familiar and
the current of electricity from a cell or dynamo. The quantity of electricity
stored in a Leyden jar, which flows through the spark on discharge, is
very small compared with the quantity flowing in a short time through a
Daniell cell, and quite negligible compared with the quantity flowing in a
day through the supply mains of a town. The pressure of the frictional
electricity is, on the other hand, very much higher than that of the ordinary
dynamo. This can be seen from the ease with which frictional electricity
T. E.
1
9
#
2
Electrical Engineering
sparks across large air gaps. Moreover, in the case of frictional electricity,
the equalising flow is, as a rule, a sudden momentary rush known as the
discharge, and can hardly be called an electric current.
In all that follows, we assume that the electric current is due to the
movement of positive electricity only. The strength of the current is then
defined as the quantity of electricity that flows through the cross-section of
the conductor in one second. To measure the current we make use of its
electrolytic or of its magnetic effect. We say that one current is twice as
strong as another when it deposits twice as much silver from a silver solution
in the same time, or when it exerts double the force on a magnetic needle
under exactly the same conditions.
Now, the absolute determination of a current from its magnetic effect is
a very laborious procedure, whereas the electrolytic method is simple, involv-
ing merely the accurate weighing of a silver deposit. For this reason the
legal definition of the unit of current is based on its electrolytic effect*.
According to this definition, that current has unit strength which deposits
1·118 milligrammes of silver per second from a silver solution.
current is called an Ampere.
This
For practical purposes the strength of a current is more conveniently
measured by means of its magnetic effect, a current-carrying coil acting on
a magnet or piece of iron with resulting attraction or change of direction.
Such instruments are called Ammeters. If intended for the measurement
of very small currents they are called Galvanometers.
The quantity of electricity that passes through the cross-section of
a conductor per second, when the current strength is one ampere, is
called a Coulomb. If we assume, for example, that the current passing
through a glow lamp is 0.5 ampere, then 0-5 coulomb of electricity passes per
second through any cross-section of the conductor. The quantity of electri-
city passing in an hour is 0.5.3,600 or 1,800 coulombs. In general, if
Q = the quantity of electricity in coulombs,
and
then
i = the current in amperes,
t=the time in seconds,
Q=i.t coulombs
We see therefore that an ampere-hour is equal to 3600 coulombs.
2. Electromotive Force.
………..(1).
We have already mentioned the idea of electricity flowing from a higher
to a lower level, the word level being used in a special sense. This difference
of electrical level can be produced by machines in which windings are con-
tinually passed before the poles of magnets, or by galvanic cells consisting of
two chemically different plates in a liquid. The Bichromate cell, for example,
consists of zinc and carbon plates dipping into a mixture of chromic acid
* This is not strictly true for England, as the ampere is legally defined as the current which
gives a certain reading on an ampere balance at the Board of Trade. This balance was, how-
ever, standardised by deposition of silver and not from its dimensions.
Electromotive Force
3
Zn
C
+
(CrO,) and sulphuric acid (H₂SO4). If the cell be examined with the aid
of an electrometer, the upper end of the carbon
is found to be positively charged, while the upper
end of the zinc is charged negatively. Hence, a
difference of electrical level exists between the
upper ends of the carbon and zinc or between
the terminals of the cell, and if these terminals
be joined by means of a metallic or liquid con-
ductor, electricity will flow from the higher to
the lower level, that is, an electric current will
flow through the conductor. Positive electricity
will flow from the carbon C through the external
conductor to the zinc Zn (Fig. 1).

Croз H2SO4
Стоз
Fig. 1.
It is found, however, that a steady flow of electricity continues through
the external conductor so long as it connects the terminals. Hence, we
must assume that the electricity, which flows externally from the carbon
to the zinc, passes back in the cell from the zinc to the carbon, being lifted,
as it were, from the lower to the higher level. In a similar manner the
water which flows from the mountains to the sea is raised again to the higher
level by evaporation due to the heat of the sun. We see, therefore, that
there is something in the cell which causes a movement of electricity and
is able to maintain a steady flow or current round the circuit, through the
internal resistance of the cell itself, and through the resistance of the external
conductor. The cause of this movement of electricity is called the
electromotive force of the cell. It is analogous to a pump which raises
the water from a lower to a higher level. The electromotive force of the
cell is equal to the difference of level or pressure measured between its
terminals when on open circuit, i.e. with the external circuit broken.
The question arises: How can we determine the magnitude of this
difference of pressure, or compare it with any other pressure? We could
use a gold-leaf electroscope, and determine the pressure from the deflection
of the gold leaves. As a rule, however, we determine the pressure from
the strength of the current it produces. If, for example, two vessels com-
municate with each other by means of a pipe having a known constant
resistance, the quantity of water flowing through it per second is proportional
to the difference of pressure or of level in the two vessels. In the same way
we connect one and the same conductor, viz. the coils of a galvanometer,
successively between the terminals of the various cells which are to be
compared, or between the two points between which the pressure difference
is to be measured. The strength of the current in the galvanometer is
then a measure of the magnitude of the electrical pressure difference, i.e.
of the electromotive force of the cell.
In this way we find that the electromotive force of a given type of cell
is an essentially constant magnitude, quite independent of the size of the
cell. In this respect a cell resembles a high-level reservoir. The difference
of level, and thereby also the pressure is fixed, once for all, by the difference
in altitude between the reservoir and the point at which the water power
1-2
4
Electrical Engineering
is used; it is quite immaterial whether the reservoir is large or small, so
long as it is kept full. The current or flow of water is entirely in the hands
of the consumers, depending only on the number of cocks opened. Hence,
we see that it is very misleading to speak about the current strength of
a cell; it suggests the idea of a certain fixed current in the cell, which must
always flow when the cell is in use. This idea is entirely wrong, for the
cell has only one essentially characteristic property and that is its electro-
motive force, which is present whether the cell is in use or standing idle.
The strength of the current is, however, entirely in our hands, depending,
as it does, on the convenience of the path open to it, that is, on the
magnitude of the external resistance connected between its terminals.
We are now at liberty to choose any pressure as the unit with which
to compare all others. The unit used in practice is roughly half as large
as the electromotive force of the above-mentioned Bichromate cell, and
is called a volt.
The following table gives particulars of the cells commonly employed.
Type of Cell
Bichromate
Daniell
Bunsen
Leclanché
Clark Standard
Weston Standard
Composition of Cell
Zinc and Carbon in sulphuric and
chromic acid
Zinc in dilute sulphuric acid, copper
in solution of copper sulphate
Zinc in dilute sulphuric acid, carbon
in strong nitric acid
Zinc and carbon in solution of sal-
ammoniac, the carbon being sur-
rounded with manganese dioxide
Zinc in zinc sulphate solution, mercury
in solution of mercurous sulphate
Cadmium amalgam in solution of cad-
mium sulphate, mercury in mercurous
sulphate
Electromotive
force in volts
2
1.07
1.8 to 1.9
1.4
1.433 (15° C.)
1.019 (15° C.)
The zinc, or the cadmium, which is chemically very similar to zinc, forms
the negative pole in all these cells, and the current flows from the positive.
pole to the zinc through the external circuit.
3. Ohm's Law.
We have already seen that if the same external conductor be connected
successively across cells having different electromotive forces, the strength
of the current, in each case, is proportional to the electromotive force. We
go now a step further and connect the terminals of the same cell successively
by means of wires of different materials, lengths, and cross-sections. We
observe that the current differs in each case, and conclude therefore that
the various wires offer different resistances to the current. If the current
is small, we conclude that the wire has a large resistance, whereas, if the
I
1
Ohm's Law
5
current is large, the resistance of the wire must be small. Hence, the
strength of the current is inversely proportional to the resistance of the
path. Similarly, the quantity of water, forced per second through a narrow
pipe by means of a given pressure, is smaller, the longer the pipe, the
greater the friction against the walls of the pipe, and the smaller the bore.
In other words, the quantity of water per second is inversely proportional
to the total resistance of the pipe. We come thus to the conclusion that
the current is proportional to the electromotive force and inversely
proportional to the electrical resistance. Hence, if
E=the electromotive force in volts,
i=the current in amperes,
and
R= the resistance of the whole circuit,
then
E
R
.(2).
This fundamental law of electrical technology is known as Ohm's law. At
first sight, it might appear necessary to introduce a coefficient after the sign
of equality; this, however, is unnecessary, or rather it becomes unity if the
resistance be expressed in terms of a suitable unit. Inversely, the unit of
electrical resistance is determined by Ohm's law as expressed in (2), and
cannot now be arbitrarily chosen. If we put i=1 and E = 1, then from (2)
we have R= 1. Hence, the unit of resistance is that resistance through
which a pressure of 1 volt produces a current of 1 ampere. This unit
of resistance is called the Ohm. It is found experimentally that a column
of mercury 106.3 cms. long and 1 sq. mm. in cross-section has a resistance
of 1 ohm.
To take an example, let the electromotive force of a dynamo be 115 volts,
its resistance 0.05 ohm and the external resistance 11 ohms; then
E = 115 volts,
R = 0·05 + 1·1 = 1·15 ohms,
i
E 115
R 1.15
100 amperes.
I
The legal definition of the electrical units follows, however, a different
procedure from that which we have followed. We have defined the units of
pressure and current and have derived the unit of resistance from them.
Legally, however, the ampere is defined as that current which deposits
1.118 mgs. of silver per second, and the ohm as the resistance of a column of
mercury, having a cross-section of 1 sq. mm. and a length of 106·3 cms.
From these two legally defined units, it follows that a volt is that pressure
which produces a current of 1 ampere in a resistance of 1 ohm*.
Equation (2) can be written thus,
R
E
i
..(3).
* Considerable confusion and difference of opinion exists at the present time on the subject
of legal definition of electrical quantities. It is now being considered by an international
committee.
T
6
Electrical Engineering
This equation expresses mathematically the fact which we have already
mentioned, viz. that if, in an experiment, we find the current to be small, in
spite of a large electromotive force, we can conclude that the resistance is
large. Hence, we see that resistance can only be accurately defined as the
ratio of electromotive force to strength of current. One must be very careful,
however, not to regard resistance as a back pressure. The back pressure is
only obtained when we multiply the resistance by the current flowing through
it. Thus, equation (2) can be written thus,
E = i . R ....
(4).
The left-hand side of this equation is the pressure or electromotive force
of the source of supply, while the right-hand side is the pressure necessary to
send the current i through the resistance R. They are both equal to one
another.
E
Ohm's law, however, is not only applicable to the circuit as a whole, but
to each individual part of the circuit. Sup-
pose, for example, that the current i flows
successively through the resistances R₁, R₂
and R. (Fig. 2). If the differences of po-
tential between the terminals of the resist-
ances R₁, R, and R, be denoted by e₁, e2
and e, respectively, it is found by experi-
ment that

e₁ = i. R₁,
R₁
=
e₂ = i. R₂,
R₂ = 22,
es = i . R₂,
R₂ =
2
від я
R₁
R2
Kez
Fig. 2.
R
We see, therefore, that although we may make the current as large as we
like, by using, say, a stronger cell or more of them, the ratio of the terminal
pressure e̟ to the current i has always a constant value. This is on the
assumption that neither dimensions, nature or temperature of the material
undergoes any change. This constant ratio we call the resistance R₁. It is
really nothing more than a coefficient, by which the proportionality of the
drop of pressure in a conductor to the current can be expressed, as in the
equation e̟=i. R₁.
If, for example, a resistance of 1 ohm is placed in circuit with two arc
lamps connected in series, and the current is 15 amperes, then the drop of
pressure in the resistance is 1.15 = 15 volts. If the supply pressure is
110 volts there remains only 110-15 95 volts for the two lamps.
In general, if e denote the potential difference between the terminals of a
resistance R, then
e=i. R
……..(5).
This equation enables us to understand the difference between ammeters
or current strength meters and voltmeters or potential difference meters.
Generally speaking, both classes of instruments depend on the magnetic
↓
1
Ohm's Law
7
E
Ø
effect of the current. They differ, however, in details of construction and in
the method of connection. Ammeters are put in series with the main current
and have as a rule few turns of thick
wire. The heat developed, and the
pressure drop in the instrument, are
consequently very small, even with
heavy currents. If, on the other hand,
an instrument is to be used as a volt-
meter, it is wound with many turns of
fine wire and connected, in series, if
necessary, with a high constant resist-
ance, across the points between which
the potential difference is to be mea-
sured, e.g. the dynamo terminals in Fig. 3. The voltmeter is thus connected
in parallel with the resistance across which the terminal pressure is being
measured.

e
* * * * * * *
Fig. 3.
The current that flows through the voltmeter is a loss to the supply. The
large resistance of the instrument is for the object of keeping this current as
small as possible. This is also necessary in order that the supply pressure
and current may not be affected by the connection of the voltmeter. To
obtain a powerful magnetic effect it is necessary to employ a large number of
turns in its coils, and we are compelled, from considerations of economy and
space, to use fine wire. The instrument acts really as an ammeter in that
its deflection is caused by the current flowing through it. If, however, this
current be multiplied by the resistance of the voltmeter, we obtain the
potential difference between its terminals, and the scale can therefore be
marked in volts. That ammeters and voltmeters are the same in principle
can best be seen from the fact that there are many instruments which can be
used for both purposes. If, for example, the sensibility of an ammeter be
such that a deflection of one degree is produced by a thousandth of an
ampere, and external resistance be added bringing the total resistance up to


R
****
Logo
Fig. 4.
Fig. 5.
1000 ohms, then a deflection of one degree will correspond to a potential
difference of 100 10001 volt.
•
From the above it is evident that, to compare two or more ammeters, we
must connect them in series (Fig. 4). If, on the other hand, we wish to
compare several voltmeters, we must connect them in parallel, each with its
terminals connected to the ends of the resistance R, the pressure across
8
Electrical Engineering
which is to be measured. Such an arrangement is shown in Fig. 5, in which
the lamps enable us to regulate the pressure across the resistance R by
dropping a larger or smaller fraction of the terminal pressure of the dynamo.
Above all, when comparing two voltmeters, one must be careful not to
connect them in series, as, owing to their unequal resistances, the total
pressure is distributed unequally between them.
Having thus explained the principles underlying the measurement of
current and pressure, mention must be made of their combined use to deter-
mine resistance. This indirect method of resistance measurement by observing
the pressure and current is specially suitable for very small or very large
resistances. If, for example, a current of 10 amperes be passed through the
armature of a stationary electric motor and the potential difference between
the brushes be measured and found to be 2 volts, then the armature resist-
ance Ra is given by the equation
Ra = 2;
e
2
10
=0·2 ohm.
The advantage of this method is that the resistance is measured exclusive
of the resistance of any connecting leads. We have assumed that the resist-
ance to be measured does not include an electromotive force; if it does, the
calculation of the resistance is not so simple.
4. The relations between the resistance of a conductor and
its material, cross-section, length and temperature.
The importance of Ohm's law lies largely in the fact that the resistance
of a conductor, that is, the constant ratio of the potential difference between
its ends to the current flowing through it, stands in a simple relation to its
cross-section and its length; it depends largely on the material and to a far
lesser degree on the temperature. Thus, if
1 = length of conductor in metres,
A =
cross-section in sq. mms.,
p = a coefficient depending on the material,
it is found experimentally that
Ꭱ
R=PA
p
………….(6).
Hence, the resistance of a wire is proportional to its length and inversely
proportional to its cross-section. The coefficient p is very different for
different materials. Its meaning can be seen by putting = 1 and A=1
in equation (6). R is then equal to p, so that the coefficient p is the
resistance of a wire of the given material 1 metre long and 1 sq. mm.
cross-section. This is called the specific resistance of the material*.
To determine its value for a given wire, the values of 1, A and R are
* It is customary in England to define the specific resistance as the resistance of a wire
1 cm. long and 1 sq. cm. cross-section. In this case the values of p given in the table must be
multiplied by 10→4.
Specific Resistance
9
found from experiment and p calculated from them. In this way we find
the following values:
Copper at 15° C.
p=0·017,
1
p=0·94 =
=
1.063'
>
Mercury
German Silver
Carbon....
p=0·2-0·4,
p = 100-1000,
Sulphuric Acid (25-30 °%) p = 14,000.
Resistances for the purpose of weakening the current or for causing a
drop in the pressure are made, therefore, of German silver, whereas the
windings of machines and the leads are made of the best conducting material,
viz. copper, to avoid all unnecessary pressure losses. Even then the pressure
losses are not inconsiderable. Consider, for example, the supply of
60 amperes to a point 20 metres distant, through wires with a cross-
section of 50 sq. mms. The total length of wire, go and return, is
l=2.20=40 metres,
and we have
7
0.017.40
R=p
0.0136 ohm.
A
50
The pressure drop e, in the leads is then found as follows:
e₁ = i. R= 60 . 0·0136 = 0·8 volt.
The specific resistance is, however, not constant, but increases in all
metals with increasing temperature, the increase of resistance being directly
proportional to the rise of temperature. The temperature coefficient of
a material is defined as the increase in resistance for a temperature
rise of 1 degree centigrade, the increase being expressed in terms of
the resistance at 0° C. Thus, when we say that experiment gives a mean
temperature coefficient for copper of 0·004, we mean that for every degree
rise of temperature the resistance increases by 0·004, that is 0.4 per cent. of
its value at 0° C. If, then,
and
it follows that
and
therefore
R₁ be the resistance at the initial temperature T₁,
R₂ the resistance at the final temperature T₂,
R, the resistance at 0° C.,
▲ the temperature coefficient,
R₁ = R。 (1+A. T₁),
R₂ = R。 (1+A. T₂),
R₂ — R₁ = R。 (T₂ — T₁) A;
·
2
R₂- R₁
Ꭱ
A
~ Ro (T₂- T₁)*
In practice it is usually quite accurate enough to assume that
Ꭱ
R₂- R₁
A
2
R₁ (T₂- T₁)
·(7).
10
Electrical Engineering
Under working conditions the temperature of electrical machines rises
about 50° C., which corresponds to an increase in resistance of 0·4. 50 = 20 per
cent. The specific resistance of warm copper is, for this reason, usually taken
as 0.02 instead of 0·017.
With the help of the temperature coefficient 0·004, we are able to deter-
mine the temperature rise of parts of a machine which are quite inaccessible
to a thermometer. Suppose, for example, that the resistance of the field
winding at 15° C. is 50 ohms, that the field current after running for several
hours is 2 amperes, and that the pressure across the terminals of the field
winding is 114 volts. The resistance of the field winding under working
conditions is therefore 11457 ohms, and we have
R₁ = 50, R₂ 57, T₁ = 15, and A=0.004.
From equation (7) we have
whence
=
0.004
57-50
50 (T₂ — 15) '
T₂ = 50° C.
The temperature rise is therefore 35° C.
As the temperature coefficient of all single metals, with the exception
of mercury, is relatively large, they cannot be used for standard resist-
ances. For such purposes we use an alloy, such as German silver, which, in
addition to the advantage of a high specific resistance, possesses a very small
temperature coefficient, viz. 00002 to 0.0004. The resistance of manganin,
an alloy of copper and manganese, is practically independent of the tem-
perature.
In contrast to the metals, the resistance of carbon and of liquid conductors
decreases with rising temperature, that is, their temperature coefficient is
negative. The temperature changes to which a glow lamp filament is
subjected are so great that its resistance must necessarily be determined
while burning, i.e. by the ammeter and voltmeter method.
Calculations are sometimes simplified by the use of the conductance, that
is, the reciprocal of the resistance of a conductor. The specific conductance
or the conductivity of copper, for example, is = 59 or roughly 60. It
1
0.017
was formerly the custom to define the conductivity of liquids in another
way, viz. by comparison with mercury; modern electro-chemical works, how-
ever, follow the above method and use the reciprocal of the ohm.
5. Kirchhoff's Rules.
(a) KIRCHHOFF'S FIRST RULE.
The sum of the currents flowing to any point is equal to the sum of
the currents flowing away from that point. Calling the currents flowing
to the point positive and those flowing away negative, we have for any point
in a wire or network,
Σi=0 ..
.......(8).
Kirchhoff's Rules
11
This rule is of fundamental importance. The idea of many beginners
that the electricity is gradually used up as it flows through the circuit
from the positive to the negative terminal is entirely wrong. The whole
quantity of electricity that leaves the positive terminal flows through the
circuit to the negative terminal and passes through the battery or dynamo
to the positive terminal whence it started. Hence, there is no continual
generation of electricity but simply a setting in motion around the
whole circuit of a quantity of electricity, the rate of flow, that is the
current, being the same at every point in the circuit. Similarly, water
supplied from a reservoir to a house is not used up in the sense of ceasing
to exist; the whole quantity flows from the higher level to the lower level,
to be raised again, if by no other means, by evaporation, to the higher
level. When we speak in practice of a loss of current, we simply mean
that a part of the current, by taking a bypath, has prevented its useful
employment, and represents therefore a monetary loss. It is, therefore, not
the current, or a quantity of electricity, which is used up, but pressure.
By means of Kirchhoff's first rule, we are enabled, from a knowledge of
two or more currents at a point in a network, to calculate the other current.
For example, let the terminal pressure e of
a shunt dynamo be 220 volts, the current i
in the external circuit (Fig. 6) 100 amperes,
and the resistance Rm of the shunt field
winding 50 ohms. We wish to find the
armature current ia. By Ohm's law,

e 220
im
=44 amperes.
Rm 50
i
Ra
**** *
ia Jim
Rm
Fig. 6.
By Kirchhoff's first rule
ia = i + im = 100 +44 1044 amperes.
=
(b) KIRCHHOFF'S SECOND RULE.
In every closed circuit, the sum of the products of current and
resistance taken round the whole circuit is equal to the sum of the
electromotive forces in the same circuit. Hence, for any closed circuit
Σ(iR) = ΣE
ΣΕ...
..(9).
In applying this rule to any circuit, or closed mesh forming part of a
network, we can carry out the summation in either direction, all currents or
electromotive forces in the opposite direction being given the negative sign.
If the direction of any current or electromotive force is unknown it must be
assumed. If the calculation then gives a negative result for this quantity
it means that its direction is opposite to that which had been temporarily
assumed.
As an example, suppose a dynamo giving an E. M. F. (electromotive force)
of E = 116 volts to be connected up in opposition to a battery of 50 accumu-
lators, each having an E.M.F. of 2 volts, in order to charge it (Fig. 7). Let
12
Electrical Engineering
e
the dynamo resistance Ra be 01 ohm, the battery resistance R 0·18 ohm,
and the resistance of the leads R 0·12
ohm. We are required to find the current
i and the terminal pressure across both
dynamo and battery.

The back E.M.F. of the battery
E=50.2=100 volts.
If now we go round the circuit in a clock-
wise direction, as indicated by the arrow,
and carry out the summations of equa-
tion (9), we get
or
iRa+iR¿+iRɩ = E — Eʊ,
i:
E-E
116-100
Ra+Ro+ R₂ ~¯¯¯ 0·1 + 0·18 + 0·12
Rb
E
Ra
B
Eb
Fig. 7.
40 amperes.
The various ohmic pressure losses, or drops, are then as follows:
i. Ra=40.01 = 4 volts.
i. Rr 40.0·18=
=
i. Rr = 40.0·12=
7.2
4.8
""
adding the back E. M. F. = 100
""
Total = 116 volts.
We see that the E.M. F. of 116 volts is exactly used up in overcoming the
back E. M.F. of the battery and in supplying the pressure losses, or drops, in
the various resistances.
In finding the terminal pressure of the dynamo, which is also the pressure
across the external circuit, we have to remember that a part of the electro-
motive force of the dynamo is used up in the dynamo itself in driving the
current through its internal resistance. The greater part, however, remains
available for the external circuit. Similarly, in a water supply the full
pressure corresponding to the difference of level is only obtained when all
cocks are closed. Directly water is drawn, a pressure loss occurs due to
friction between the water and the inner surface of the pipes, and the
pressure available for the consumer drops below that corresponding to the
difference of level. To find the terminal pressure e we must therefore sub-
tract the internal drop from the electromotive force, thus,
e = E-iRa=116-4=112 volts.
For the battery, which is being charged, the relations are however quite
different. The terminal pressure es across the battery has a double duty,
viz. to overcome the back E.M.F. E, and to drive the current through the
ohmic resistance of the battery. Hence
eq = Ev+i. Rr 100+7.2=107.2 volts.
=
t
Resistances in Parallel
6. Resistances in Parallel.
13
It is self-evident that the resistance of several conductors connected in
series is equal to the sum of their individual resistances. If, on the other
R₁
ww
R
www
R2
Fig. 8.
iz
w
Fig. 9.
hand, two resistances R₁ and R₂ are connected in parallel, as shown in Fig. 8,
the combined resistance is evidently less than that of either, for the current
has a more convenient path than would be offered by either resistance alone.
The conductance of the loop is, as can be shown experimentally, equal to the
sum of the conductances of the parallel branches, that is,
1
1 1
+
RR₁ R₂
Ꭱ Ꭲ
From this we have
R₁R₂
Ꭱ
R₁+ R₂
.(10).
In making very low resistances, for example, it is almost impossible to
cut the wire to the exact length required. It is therefore cut so as to have
a resistance R₁, slightly higher than that required, which is then brought
down to the correct value by connecting another resistance R, in parallel
with it. Suppose that R, has been made 0·102 ohm. The question is: What
resistance must be put in parallel with R, so that the total resistance R may
be exactly 0.1 ohm? From equation (10) it follows that
0.1=
0.102. R₂
0.102 +R₂'
.. R₂=5·1 ohms.
Suppose now that an error of 2 per cent. is made in the adjustment of this
5.1 ohms, so that R₂ is really only 5 ohms. The total resistance will then be
2
R=
R₁R₂ 0.102.5
R₁+ R₂¯¯¯ 0·102 +5
an error of only 0.04 per cent.
= 0.09996 ohm,
If more than two resistances be connected in parallel, we find in the
same way that
1 1 1 1
+ + +...
R R R2 R3
…………….(11).
If the individual resistances are equal, the calculation is much simpler.
Suppose, for example, that the armature of a 4-pole machine is parallel
wound, so that there are four equal parallel paths through it (Fig. 9). If
the total length of wire on the armature is 200 metres and its cross-section
+


14
Electrical Engineering
10 sq. mm., the resistance of each branch, taking the specific resistance of
warm copper as 0·02, is equal to
P. A
= 0·02 200/4
ohm.
10
The combined resistance of the four equal parallel branches is only a quarter
of this, so that
Ra
0.02.200
42.10
= 0·025 ohm.
To find the ratio of the currents in the two branches in Fig. 8 we observe
that both resistances R, and R₂ have the same terminal pressure e. If & be
the current in the resistance R₁, and i₂ that in the resistance R₂, then we
have
wherefore
e = i₂ R₂ = i₂ R₂,
in R₂
22
R₁
.(12).
Hence, when a current divides between two or more parallel paths, the
currents in the various branches are inversely proportional to their resistances.
It is on this principle that large currents are
measured by means of sensitive galvanometers
carrying only very small currents. The gal-
vanometer is put in parallel with a known low
resistance, and is said to be shunted. The
main current i will divide, a small portion ig
flowing through the galvanometer of resist-
ance Rg, while the remainder i, flows through
the shunt R.. We have then (Fig. 10),

Rg
о
ig
ww
is
Rs
ig __ Rs
Rg
Fig. 10.
ig
or
ig + is
i = ig
but igisi, the main current; therefore
..(13).
To simplify calculation the resistance of the shunt is usually made equal
to, or of the resistance of the galvanometer. If, for example, the
galvanometer has a resistance of 100 ohms, and is shunted with 488 ohms,
we have
Rg
Rs + Rg'
R₂+ Rg
Rs
i=ig
1000+100
100
= 1000 ig.
Hence the main current is 1000 times the galvanometer current.
7. Series and Parallel arrangement of cells.
When cells are connected in series, the positive terminal of one cell is
connected to the negative of the next, and so on (Fig. 11). The same current
flows through all the cells one after another, and all the electromotive forces
Series and Parallel arrangement of cells
15
act in one direction. Hence, not only must the electromotive forces of all
the cells be added together, but also their internal resistances. The series
arrangement is used, therefore, where a large current is required in spite
of a high external resistance. The increased battery resistance due to the
series arrangement is then negligible, whereas the great electromotive force
thus obtained is important. To connect cells in parallel, all their positive


m
Fig. 11.
m
Fig. 12.
terminals are joined together, on the one hand, and all their negatives on
the other. By this means the battery resistance is made very small, but
the pressure across the external circuit is only that due to a single cell.
We can no more add parallel connected electromotive forces than parallel
connected hydraulic pressures. If, for example, two pipes come down from
two equally high reservoirs, the pressure obtained is no greater than if a
single reservoir were used, if the resistance of the pipes be neglected. Cells
are therefore connected in parallel to obtain a large current in an external
circuit of low resistance, that is, when a large current can be obtained
without a high pressure. The parallel arrangement has here the advantage
of keeping the battery resistance, which is now considerable compared with
the low external resistance, as small as possible, thus enabling a larger current
to be obtained.
A combination of these two methods leads to the mixed arrangement
shown in Fig. 13, in which several cells are connected in series, and then
a number of such rows connected in
parallel. The question arises: How
should a given number of cells be
arranged so as to give the maximum
current? Let
R be the external resistance,
R₂ the battery resistance,
R; the internal resistance of one cell,
> the number of cells in series,
E the E.M.F. of each cell,

HHH
Fig. 13.
z the total number of cells.
The number of parallel paths is, and the resistance of each branch R¡~.
The battery resistance is therefore
OC
ᎡᏘ
R₂ x R₂ x²
Ro
z/x 2
16
Electrical Engineering
Since the E.M.F. of the battery is equal to that of a single branch, that
is E. x, we have, by Ohm's law, for the current i through the whole battery,
E.x
E
i =
R₂x²¯¯¯ R R¿x'
R+
+
2
OC
z
For a maximum value of this expression the denominator must be a
minimum. The denominator is a function of the variable x, which fact can
be expressed thus:-
f(x):
Ꭱ ᎡᏆ
+
OC
Differentiating with regard to x, we have
2
ƒ' (x) = − Rx¯² +
Ri
2
The function f(x) will have a minimum value when f'(x) = 0, that is when
R₂ x²
R
2
The left-hand side of this equation represents the external resistance, while
the right-hand side is equal to the internal resistance of the whole battery.
Hence, to obtain the maximum current through a given external resistance
with a given number of cells, these should be so arranged that the internal
battery resistance is, as nearly as possible, equal to the external resistance.
8. Wheatstone's Bridge.
The measurement of resistance with the Wheatstone bridge is an im-
portant and instructive application of Kirchhoff's rules.
The cell E (Fig. 14), whose E.M. F. need neither be known nor constant, is
connected across the ends of a bare straight wire AB.
scale by means of which the exact
position of the sliding contact C
can be read. Connected in parallel
with this wire is a branch consisting
of the unknown resistance x, the
value of which is to be determined,
and a known resistance R. The
resistance of the connecting wires
is assumed to be negligible. A
galvanometer is connected between
the sliding contact C and the con-
necting point between x and R.
The sliding contact is moved until
no current flows through the gal-
A
Under the wire is a

E
b
B
a
in
www.
iz
iz
Fig. 14.
vanometer. The current & then flows uniformly through the resistances a
and b, and the current i, through the resistances x and R. We now apply
Kirchhoff's second rule to the left-hand lower mesh, going round it in a
Wheatstone's Bridge
17
-25
clockwise direction as indicated by the dotted arrow. This circuit contains
no E.M.F. and the current in the galvanometer is zero; therefore
Similarly in the right-hand mesh
Hence it follows that
i₁a +0―i₂x=0.
ibi₂ R+0=0.
α
b- R
.(14).
The ratio a b in this equation refers to the resistances of the two portions.
of the slide wire; these are, however, proportional to their lengths, so that
the ratio ab can be read off directly on the scale.
It is instructive to deduce this result, without using Kirchhoff's rules,
from the analogy of a water main which divides into two parallel vertical
mains. If a horizontal pipe joins the two mains at any level, no water will
flow through this connecting pipe because there is no difference of level
or pressure between its ends. This is exactly analogous to the Wheatstone
bridge. When no current flows through the galvanometer there is no
difference of pressure between the points C and D, and the pressure drop
from A to C is equal to that from A to D. Writing this in the form of
an equation, we have
and
a
i₁.α=i₂.x,
i.bi₂. R.
Equation (14) follows directly from this.
This form of Wheatstone's bridge is known as the slide-wire bridge or
metre-wire bridge, the wire AB generally being a metre long.
The lengths of wire a and b can be replaced by accurately known re-
sistances, and it is then more convenient
to obtain a balance by altering the value
of R. The well-known Post Office box
is of this type.
The method outlined above is suit-
able for resistances of medium size.

x
x
C
F
+
a
b
R₁
R₂
D
W
d
Fig. 15.
In the case of very small resistances,
the connecting wires would introduce
considerable error. To avoid this, the
cell could be connected directly to the
ends of the resistances ≈ and R, while a
and b could be represented by accurate
resistances of such values that the end
connections would be quite negligible. In the ordinary form of bridge,
however, it is impossible to eliminate the effect of the connection between
≈ and R. If, in Fig. 14, the galvanometer be connected to the end of x, R
will be increased by the resistance of the connection between x and R,
whereas, if the galvanometer be connected to R, a will appear too large.
In Thomson's (Kelvin's) double bridge this difficulty is surmounted by
connecting the galvanometer through resistances R, and R₂ to both x and R
T. E.
2
18
Electrical Engineering
(Fig. 15). The former direct connection between x and R, shown at the
bottom of Fig. 15, still remains. We see that there are now two leads
connected to each end of both x and R. The resistances R, and R, are so
chosen that their ratio R: R₂ is equal to the ratio a b. If, for example,
a: b=1: 10 then R: R₂ must be made 1:10. The point D will then have
the same potential as the point F which divides the direct connection be-
tween x and R in the ratio cd=1:10. We can therefore imagine the
galvanometer directly connected to F instead of to D, and when, by altering
the value of R, the current in the galvanometer has been reduced to zero,
we have
In our case, where a : b=1: 10,
α
b
x + c
R+d'
1
10'
x + c
R+d
and as c d is also in the ratio 1 : 10, it follows that a: R=1:10. In this
way the resistance of the connecting leads is eliminated and it is therefore
possible to measure very low resistances, such as the armature resistance
of a large dynamo, with great accuracy*.
9. Measurement of Electromotive Force by the
Potentiometer Method.
A
E.
E
B
A cell of constant but unknown E. M. F. is connected across the ends of a
calibrated wire AB (Fig. 16). The P. D. (potential difference) between A and
B must be greater than the E. M. F. x
which has to be measured, and also
greater than the E. M. F. E。 of the
standard cell. A Bunsen cell, for ex-
ample, could be used for E, and a
Daniell cell for E。. A galvanometer,
the standard cell and a high resistance
are connected in series between the end
A of the calibrated wire, and the sliding
contact, the cells E and E, having simi-
lar poles connected to A. The sliding
contact is moved along the wire until a
point C is found where no current flows
through the galvanometer, the final ad-
justment being made with the high resistance short-circuited. The cell to
be tested is now put in the place of the standard cell so that the electromotive
forces E and x oppose each other as before. The sliding contact is again
adjusted until no current flows through the galvanometer, its new position
being C'.

Fig. 16.
In order to find the relation between the electromotive forces E, and x
* For further information on the subject of measurement of resistance, see Price's "Measure-
ment of Resistance," Kempe's "Handbook of electrical testing," or Chapter XXVI in Watson's
"Practical Physics."
Measurement of Electromotive Force
19
from the measured lengths AC and AC', we observe that the terminal
pressure of the Bunsen cell E is used up along the whole length of the wire
AB. Across half the wire, for example, there is half the pressure, for, with
the same current, pressure differences are proportional to resistances. The
ratio of the P. D. between A and C to the P. D. between A and C' will there-
fore be equal to the ratio of the lengths of wire, that is AC: AC'. We have
seen, however, that the P. D. between A and C exactly balances the E. M. F. E。,
while the P. D. between A and C'exactly balances the E. M. F. ∞.
If now we
replace potential differences by corresponding lengths of wire, we have
AC E.
AC' XC
.(15).
This method, which brings home very clearly to the beginner the fall of
potential along a resistance, is one of the most convenient methods of
calibrating instruments, and it may be said that almost all instruments are
now calibrated in this way. The calibrated slide wire is replaced, either
entirely or for the most part, by accurate resistance coils, an accumulator
takes the place of E, and a Weston or Clark cell is used as the standard.
In a similar way it can be shown that the pressure y across the points A
and B is such that
AB y
AC E
It must be carefully noticed, however, that y is not the E. M. F. of the
upper cell but only the P. D. between its terminals. The cell E differs from
the cells and E, in that it is carrying a current, and its terminal pressure is
therefore less than its E. M. F. by an amount equal to its internal pressure
drop.
XC
10. Joule's Law, Electrical Energy and Electrical Power.
A conductor through which an electric current is flowing becomes heated.
The amount of heat so developed was measured by the English physicist
Joule, who determined the relation between the heat and the current, pressure
and time. The quantity of heat taken as the unit in Electrical Engineering
is the gramme-calorie, which is the amount of heat required to raise the
temperature of 1 gramme of water from 0° to 1° C., or, what is practically the
same thing, the heat required to raise 1 gramme of water through 1°C. If
Qh represents the quantity of heat in gramme-calories, e the pressure across
the terminals of the conductor in volts, i the current in amperes, and t the
time in seconds, then it is found experimentally that
Qh = 0·24.e.i.
...(16).
The experiment can easily be repeated by immersing a platinum spiral in
a measured quantity of water, the spiral being soldered to two thick copper
leading-in wires of negligible resistance (Fig. 17). The inner glass vessel,
which contains the water, rests on cork knife-edges and is separated by an
air space from the external vessel to avoid as far as possible any loss of heat.
A current is passed through the spiral and readings are taken of the pressure
2-2
20
Electrical Engineering
across the terminals, the current and the resulting temperature rise T₂ — T₁
in a certain time. The amount of heat is determined by multiplying the
weight of water W in grammes by the tempera-
ture rise T₂- T₁. It is found that this quantity

2
of heat is proportional to the product e.i.t.
Qr=W (T₂- T₁₂)=c.e.i.t.
| | || | ||
With a little care the value 0.24 will be ob-
tained for c. It is a good plan to start the
experiment with the water a few degrees below
the temperature of the room and to continue it
until the water is an equal number of degrees
above the room temperature. The amount of heat
radiated out in the second half of the experiment
will be equal to the heat received from the sur-
roundings during the first half, thus correcting any
radiation error. It must be remembered moreover
that the value of W is made up of the actual
weight of the water and the water equivalent of
the inner glass vessel. The latter is found by multiplying the weight of
the empty glass vessel in grammes by the specific heat of glass, i.e. 0·19.
www
Fig. 17.
Joule's experiment is as fundamentally important in Electrical Engineering
as his determination of the mechanical equivalent of heat, according to which
a kilogramme-calorie is equivalent to 427 metre-kilogrammes or 1 British
Thermal unit (Fahrenheit) to 778 foot-lbs. According to Joule's law the
product e.i.t is proportional to a quantity of heat. Now, as heat is nothing
more than a special form of energy, the product e.i.t must represent an
amount of energy, and it is therefore a measure of the electrical energy
or work.
The unit of electrical energy is the work done in one second when a
current of 1 ampere flows under a pressure of 1 volt. This amount of
energy is called a joule or a watt-second. The electrical work J is given
therefore by the equation
J=e.i.t joules
.(17).
We have now to find the relation between the electrical work in joules
and the mechanical work in metre-kilogrammes or in foot-pounds. Putting
e, i and t equal to 1 in equations (16) and (17) we find that the electrical
energy is one joule while the heat Qh is 0-24. Hence 1 joule is equivalent to
0·24 gramme-calorie. According to the mechanical equivalent of heat a
gramme-calorie is equivalent to 0427 metre-kg. and therefore
or
1 joule = 0·24 gr.-cal. = 0·24. 0·427 met.-kg. = 0·102 met.-kg.
1 met.-kg. = 9.81 joules
Similarly, we could show that
1 foot-pound = 1.356 joules.
...(18).
It is no mere coincidence that we get the figure 9.81, the acceleration due
to gravity; it follows necessarily from the units employed.
Joule's Law, Electrical Energy and Electrical Power 21
Let us consider, as an example, a dynamo giving a current of 50 amperes
at a pressure of 220 volts for 10 hours, i.e. for 36,000 seconds; the electrical
work done or energy given out is
J = 220.50.36,000 = 396. 10º joules,
which corresponds to 405.10 metre-kgs. or 292.10 foot-lbs. Again, if a
Daniell cell gives a current of 0.55 ampere for 1 hour at a terminal pressure
of 1 volt, the energy given out is
J=1.0·55.3,600 = 1,980 joules.
This is the same amount of work as is required to raise
one metre.
1,980
= 200 kgs.
9.81
From the unit of work we can obtain the unit of electrical power, i.e. the
rate at which work is done. If the product e.i.t represents the electrical
work, the product e.i must be the electrical power. Hence, the unit of
electrical power is given by one ampere under a pressure of 1 volt.
This unit of electrical power is called a watt. If the electrical power in
watts be represented by P, we have
Pe. i watts
..(19).
From equation (18) we obtain the relation between the watt and a
metre-kg. per second. One metre-kg. per second is equal to 9·81 joules per
second, that is, to 981 watts. Hence
=
1 H. P. = 75 met.-kgs. per second = 75.9.81 watts 736 watts.
This is the continental or metric horse-power, which is about 14 per cent.
smaller than the British horse-power.
1 British H. P. = 550 ft.-lbs. per second = 550.1·356=746 watts.
The power taken by a 10 H. P. motor, having an overall efficiency of 0·85,
will thus be
P =
10.736
0.85
8,700 watts.
At a pressure of 220 volts, the current taken will be
i:
8,700
220
≈ 40 amperes.
The above units of energy and power are often, in practice, found incon-
veniently small. Larger units are therefore used, as follows:
1 hectowatt
1 kilowatt
= 100 watts (rarely used).
= 1,000 watts.
1 watt-hour
= 3,600
watt-seconds or joules.
1 kilowatt-hour-3.6.106
11.
""
Potential Difference.
""
In the preceding section we based our consideration of electrical energy
on the experimental evidence of Joule, as it appeared advisable to do this
before giving a theoretical proof. We have repeatedly made use of the
analogy between the electric current and a current of water. The analogy is
22
Electrical Engineering
also very close with regard to electrical work. The work done by a waterfall
in a given time is obtained in metre-kilogrammes by multiplying the weight
of water which has fallen in the given time by the height of the fall in metres.
We proceed in a similar manner to calculate the electrical work by forming
the product e.i.t. For e is the pressure or difference of level, while i. t is
the quantity of electricity which has passed during the time t from the higher
to the lower level or potential. From this we obtain a more accurate definition
of that which we have, up to the present, represented by e and designated
difference of pressure or difference of level. Thus, if in the equation
J=e.i.t
we make the product i.t=1, that is, we make the quantity one coulomb, we
have J=e. In other words, the difference of pressure e in volts is the
work, measured in joules, which is done when one coulomb flows from
the higher to the lower electrical level. While this coulomb of electricity
was at the higher level it possessed potential energy or capacity for doing
work, just as a raised weight possesses potential energy, which is greater the
greater the difference of level. We speak therefore of an electrical difference
of potential, and define it as the work done when a unit of positive
electricity flows from the higher to the lower electrical level.
Inversely, we could define potential difference as the work in joules which
we must do to raise one coulomb of electricity from the lower to the higher
potential. Referring to Section 2 we see that this raising of the electricity
from the lower to the higher electrical level or potential is the special
function of the electromotive force. Electromotive force and potential
difference or pressure are thus, in a sense, identical and are measured in
the same units.
The above definition of potential difference is identical with the well
known definition met with in frictional or static electricity. Fig. 18 repre-
sents a positively charged insu-

lated metal sphere. Imagine a
small freely moving body with
unit charge of positive electricity
to be placed on the surface of the
sphere. Since like charges repel,
this small body will be repelled
to an infinite distance by the
large sphere. By this means work
will be done equal to the summa-
+
Fig. 18.
•B
tion over the whole distance of the product of force and distance, or this
amount of kinetic energy will be imparted to the body. Then the potential
at the surface of the sphere is the work done by the electric forces in
repelling the unit of positive electricity to an infinite distance.
When the small body is at A, it possesses with respect to the point B a
capacity for doing work, i.e. it possesses potential energy, analogous to that
of a raised weight. The potential at the point A is, therefore, higher than
that at B, and there is a difference of potential, or of electrical level, between
Potential Difference
23
the two points. This difference of potential is equal to the work done on a
unit positive charge in repelling it from A to B.
In principle it is immaterial whether the unit of electricity which is
driven from A to B is a charge on a small body, moving with this body, or
a quantity of electricity moving along a stationary conductor as in an electric
current. One point to which attention must be specially drawn is this,
that if, in the case represented in Fig. 18, the coulomb be taken as the unit
charge on the small body and the work be measured in joules, then the
potential difference will be obtained directly in volts.
12. Energy lost in heating conductor.
Substituting for e its value i. R, the formula for the electrical power can
be written as follows:-
P=e.i=¿². R……………..
..(20).
The power expended in the useful resistance, such as lamps, for example,
is proportional to the square of the current and to the resistance of that
portion of the circuit, while the same holds true for the leads. This fact has
had considerable influence on the development of electrical engineering. To
make this clear let us consider the case in which 10,000 H.P. has to be
transmitted over a distance of 30 kilometres (about 19 miles). We can
transmit this power by means of a large current at a low pressure or by
means of a small current at a high pressure. In either case, to transmit
10,000 H.P. or 7,460,000 watts, we have
i
P 7,460,000
e
e
If, now, we try successively pressures of 100, 1,000 and 10,000 volts, we
obtain as the corresponding currents 74,600, 7,460 and 746 amperes.
Allowing 10 per cent., i.e. 746,000 watts for the loss in the leads or
cables, the resistance of which is R, we have
or
¿². R₁ = 746,000,
746,000
R₁ =
The total length of cable is 2.30 kms. = 60,000 metres, and using equa-
tion (6) on page 8, the cross-section can be found as follows:-
A
0.017. 60,000
746,000
2
2
= 13.7.10.¿².
Hence, the cross-section of the copper is directly proportional to the
square of the current, or inversely proportional to the square of the pressure
employed. The above equations give the following values:
A = 13.7.10-4.2.
760.104 sq. mm.
e
100
i
74,600
1,000
7,460
10,000
746
760
760.10%
"
1
24
Electrical Engineering
#
It is evident that the last case is the only one of the three which is at all
practicable.
We shall now endeavour to make the effect of a higher pressure on the
loss in the mains, or on the cross-section of the conductor, clearer by a
consideration of the 3-wire system. Imagine a glow lamp to be constructed
so as to burn normally with a current of 0.5 ampere and to have a resistance,
when so burning, of 220 ohms. To maintain a current of 0.5 ampere the
pressure across the lamp terminals must be 0.5.220110 volts. We shall
assume that 200 such lamps have to be supplied at a distance of 2 kilometres,
with an allowable loss in the mains of 10 per cent. of the power transmitted,
and we shall calculate the necessary cross-section of cable, using pressures of
110 and 220 volts respectively.
1st case. Pressure 110 volts. All lamps in parallel (Fig. 19).
The total current taken by 200 lamps at 0.5 ampere each is
i=0.5.200=100 amperes.
The total power transmitted is
P=e.i= 110. 100 = 11,000 watts.
With 10 per cent. loss 1,100 watts
will be dissipated as heat in the
mains. If R be the resistance of
the mains, then
1002. R₂ = 1,100
i2. Rr
1,100
and
R₂:
= 0·11 ohm.
1002
え
​
The total length is 2.2,000 = 4,000 metres.
Z
R=p · A
we have
Fig. 19.
From the equation
A
p. l ___ 0·017. 4,000
=
R
0.11
620 sq. mm.
2nd case.
Pressure 220 volts.
To obtain this pressure we connect two 110 volt dynamos in series
(Fig. 20). If, now, we connect the lamps 2 in series, the pressure across
i

48 17
Fig. 20.
each lamp is 110 volts. We have, then, for 200 lamps 100 parallel paths
each taking 0.5 amperes, the total current being i 100. 0·5 = 50 amperes.
The power transmitted is therefore
P=e.i=220.50 11,000 watts.
Energy lost in heating conductor
25
This is the same power as in the first case.
The loss in the mains shall,
of course, remain the same as before, viz. 1,100 watts. We have therefore
i2. Rr502. R₂ = 1,100
=
and
wherefore
R₂ =
Rr
1,100
502
= 0·44,
A==
Rr
= 155 sq. mm.
0.44
p. l __ 0·017. 4,000
Thus, by doubling the pressure on the mains, the cross-section of the
conductor has been reduced to a quarter of its previous value. To enable
each lamp in the second case to be switched on and off independently of the
others, a balancing wire or neutral wire connects the middle point of the
dynamos to the middle point of every two lamps. This conductor is usually
made of half the cross-section of each of the outer conductors. There is a
great saving in copper in spite of the three wires. It is evident, moreover,
that, with the same weight of copper in the mains, doubling the working
pressure enables nearly four times the power to be transmitted with the
same percentage loss in the mains.
The above example has been worked out in detail because of the difficulty
experienced by the average beginner in forming clear and tangible ideas on
this question of power transmission at high or low pressures. It is evident
that in both the above cases the same number of lamps were supplied, each
lamp having the same current and burning, therefore, with the same brilliancy.
The result obtained is therefore exactly the same in both cases.
CHAPTER II.
13. Electrolysis.—14. Quantitative laws of electrolysis.-15. Polarisation.-16. Accumulators.—
17. Primary cells.-18. The Voltameter.
13. Electrolysis.
Conductors of the first class are those through which the electric
current passes without producing any chemical change. The metals and
carbon belong to this class. Second class conductors, on the other hand,
undergo chemical changes or decomposition when a current passes through
them. This decomposition is called electrolysis and such conductors are
known as electrolytes. To this class belong the bases, acids and salts, either
in solution or in the molten state.
The vessel or apparatus, in which electrolysis is carried on, is called an
electrolytic cell, and the immersed conductors, by means of which the current
is led in and out of the liquid, are called the electrodes. The positive electrode,
by which the current enters the liquid, is called the anode, while the negative
electrode, by which the current leaves the cell, is called the cathode.
The constituents into which the liquid is split up appear at the electrodes,
one constituent wandering with the current to the cathode, while the other
makes its way against the current to the anode. These constituents are
therefore called ions, i.e. wanderers. According to modern views some of the
ions are charged with positive electricity. These so-called cations move in
the positive direction of the current and give up their positive charges to the
cathode. The other ions, called anions, are charged with negative electricity
and move against the current to the anode, to which they give up their
negative charges. It is perhaps difficult for the beginner to reconcile these
views with the general idea that an electric current consists only in a move-
ment of positive electricity. It is probable, however, that the latter is correct
when applied to first class conductors, except that it is the negative electricity
which moves and not the positive. The student will do well to let both
ideas exist peacefully side by side.
To obtain a better understanding of electrolysis we will now turn our
attention to the difference between metals and non-metals. Metals, such as
potassium, magnesium, iron and gold, have a characteristic lustre; they are
good conductors of heat and electricity, and their compounds with hydrogen
and oxygen are bases, e.g. caustic soda (NaOH), caustic potash (KOH), slaked
lime (Ca [OH]). The basic character of the metallic oxides is shown by the
facts that they turn red litmus blue, act as caustics, have an alkaline taste
F
Electrolysis
27
and neutralise acids. The basic nature of many metallic oxides is not very
pronounced, being only shown by their ability to neutralise acids.
In addition to the above characteristics, all metals show a remarkable
peculiarity when present in a solution through which an electric current is
flowing, in that their ions always move with the current, towards the cathode.
The metals are therefore electro-positive, that is, their ions are the carriers
or transporters of positive electricity (cations). Since hydrogen also moves
with the current towards the cathode, it is also to be classed as a metal, more
especially as it can be chemically replaced in its compounds by metals.
To the non-metals belong chlorine, bromine, iodine, nitrogen, oxygen,
sulphur, etc., which are without metallic lustre, and which, in so far as they
occur in the solid state, are bad conductors of heat and electricity. Finally,
they combine with hydrogen and oxygen to form acids, e.g. hydrochloric acid
(HCl), sulphuric acid (H,SO), nitric acid (HNO,) and phosphoric acid
(H,PO₁). Acids are characterised by a sour taste, their ability to turn blue
litmus red, to dissolve metals and to neutralise bases with the formation of
salts.
Of the non-metals, chlorine, bromine, iodine and fluorine move against
the electric current and belong, therefore, obviously, to the anions; moreover,
the hydroxyl group (OH) among the bases, as well as the acid radicals SO,
PO, NO,, etc., are anions.
We will now endeavour to make the processes of electrolysis clearer by
means of a few characteristic examples.
1. ELECTROLYSIS OF BASES.
When a solution of caustic potash (KOH) is electrolysed, the metal
potassium moves with the current and causes hydrogen to be liberated at
the cathode, in accordance with the equation
2K+ 2H,O=2KOH + H₂.
The hydroxyl group OH, on the other hand, moves against the current and
causes oxygen to be liberated at the anode in accordance with the equation
40H = 2H₂O + O2.
From the result it would appear that water only was being decomposed.
The decomposition of the water, however, is purely chemical and takes no
part in the passage of the current, which is due entirely to the decomposition
of the caustic potash. Pure water is a perfect insulator.
2. ELECTROLYSIS OF ACIDS.
The simplest case is that of hydrochloric acid (HCl). Hydrogen is liberated
at the cathode and chlorine at the anode. We have in this case a simple
decomposition without secondary reactions. In the electrolysis of sulphuric
acid (H,SO₁) hydrogen is likewise liberated at the cathode. The acid radical
SO₁, on the other hand, is not liberated at the anode, but if the anode
consists of carbon or platinum, oxygen is liberated, the chemical reaction
being as follows:
2SO₁+ 2H₂O = 2H₂SO₁+ O₂.
The result is therefore the same as if water alone were decomposed.
28
Electrical Engineering
If, however, the anode consists of copper, the radical SO, attacks it,
forming copper sulphate in accordance with the equation
SO,+ Cu = CuSO..
3. ELECTROLYSIS OF SALTS.
When a solution of potassium chloride (KCl) is electrolysed, the potassium
ions wander to the cathode, where caustic potash is formed and hydrogen
liberated, as above; the chlorine goes to the anode. If the two electrodes
are separated by a porous partition, e.g. an unglazed earthenware pot, the
final products of electrolysis will be caustic potash and chlorine. In the
absence of such a partition, a tertiary reaction ensues between the chlorine
and caustic potash with the formation of potassium chlorite (KClO) in
accordance with the equation
2KOH + 2C1 = KClO + KCl+H,O.
If, however, the solution be hot, potassium chlorate is formed, thus:
6KOH + 6C1 = KClO3 + 5KCl + 3H₂O.
As a further example of the decomposition of a salt solution, we may take
the electrolysis of copper sulphate (CuSO). The copper wanders with the
current and electroplates the cathode. The group SO, moves, as before,
to the anode and, if the latter consists of carbon or platinum, causes a
liberation of oxygen. If the anode is of copper it will be eaten away, with
the formation of copper sulphate, thus:
Cu + SO, = CuSO4.
14. Quantitative laws of Electrolysis.
The amount of an element or of a chemical compound deposited on an
electrode was found by Faraday to be proportional to the current and to the
time, or, in other words, proportional to the quantity of electricity which has
passed. Thus, if
and
m be the weight of deposit in milligrammes,
i the current strength in amperes,
t the time in seconds,
c a coefficient,
then it is found experimentally that
m = c.i.t.
The coefficient c is different for various ions. Faraday found that it was
proportional to the atomic weight and inversely proportional to the valency.
Thus, if
and
a be the atomic weight,
k the valency,
then Faraday found experimentally that
a
m = 0·0103862.i.t mgs.
Tc
….(21).
Quantitative laws of Electrolysis
29
To understand this equation it is necessary to define the terms atomic
weight and valency. The atomic weight of an element is the weight of its
atom, relatively to the weight of the hydrogen atom. An atom is defined as
the smallest particle of matter which can take part in chemical action, or
enter into the chemical structure of a compound. The atomic weight of
chlorine, for example, is 354, since hydrochloric acid (HCl) contains 35-4
parts of chlorine to one part of hydrogen. Water, on the other hand, contains
8 parts of oxygen to one part of hydrogen. At first sight it would appear,
therefore, that the atomic weight of oxygen was 8, but on examining
the relative volumes in which hydrogen and oxygen combine to form water,
we find that its formula must be written H₂O. One atom of oxygen is there-
fore combined with two atoms of hydrogen, and the atomic weight of the
former must be 16, in accordance with the ratio 1:8 2:16. From the
atomic weights of the elements and the formula for the compound, the
percentage of any element in a compound can be calculated. It is evident
that the atomic weights are of prime importance in calculating the amounts
of various elements deposited by electrolysis.
Another term used in equation (21) is valency. By this is understood
the number of atoms of hydrogen which can combine with, or be replaced by,
one atom of the element. Chlorine, for example, is univalent because it
combines with a single atom of hydrogen to form hydrochloric acid (HCI).
Similarly, potassium is univalent since it combines with a single atom of the
univalent element chlorine to form potassium chloride (KCl), or, in other
words, it replaces the hydrogen atom of hydrochloric acid. Oxygen, on the
other hand, is divalent since one of its atoms combines with two hydrogen
atoms, forming water (H2O). It is best to look upon the valency as the
number of arms or bonds by which the atom is linked with the arms of other
atoms. Copper, for example, is divalent in most of its compounds, and links
its two arms with the two arms of the divalent oxygen atom, forming copper
oxide, as shown by the equation
Cu +0 = CuO.
Nitrogen, on the other hand, is trivalent and links its three arms with the
single arms of three hydrogen atoms, forming ammonia (NH), a molecule
of which is therefore built up in the following manner:
H
N-H
H
We are now in a position to illustrate Faraday's law by means of a simple
experiment. By means of platinum electrodes the same current is passed
through a row of electrolytic cells containing, respectively, sulphuric acid
(H,SO), hydrochloric acid (HCI), copper sulphate (CuSO4), cupric chloride
(CuCl₂), and cuprous chloride (CuCl) (Fig. 21). We will assume that the
experiment is carried on until 2 mgs. of hydrogen have been liberated in the
first cell. The amounts liberated or deposited in the same time in the various
cells will be found to be as follows, the atomic weights being given in brackets:
30
Electrical Engineering
t
In the 1st cell (H₂SO,):
2 mgs. hydrogen (H=1), 16 mgs. oxygen (0 = 16).
In the 2nd cell (HCl):
2 mgs. hydrogen (H=1), 70.8 mgs. chlorine (Cl = 35·4).
In the 3rd cell (CuSO₁):
63.2 mgs. copper (Cu = 63·2), 16 mgs. oxygen (0 = 16).
In the 4th cell (CuCl½):
63.2 mgs. copper (Cu = 63.2), 70.8 mgs. chlorine (Cl = 35·4).
1
2
3
4
5

TELU
H₂SO
нсг
CuSO4
Fig. 21.
Cu Cl₂
Cu Cl
So far, everything is quite regular, and it appears that a given current
flowing for a given time liberates the same weight of a given element in
whatever combination it may exist in the electrolyte. The actual weight of
any element liberated is seen to depend primarily on its atomic weight. We
obtain in each cell either atomic weights or multiples of them. From this
fact we could have calculated from the weight of hydrogen liberated in
the first cell, the results obtained in the next three cells. We meet a
difficulty, however, when we come to the fifth cell. We cannot say before-
hand if we shall obtain the same weight of copper as in the third and fourth
cells, viz. 63.2 mgs. If so, the electrolyte being CuCl, the amount of chlorine
liberated would only be 35'4 mgs. If, however, the chlorine liberated is the
same as in the second and fourth cells, viz. 70.8 mgs., the amount of copper
deposited must be 126'4 mgs. Experiment proves this latter to be the case.
For the same current flowing for the same time, twice as much copper is
deposited from a cuprous chloride solution as from a cupric chloride solution.
In cuprous chloride one atom of copper is bonded with one atom of
chlorine; the copper is therefore univalent. Assume for a moment that an
atom of hydrogen weighs 1 mg., then in the first cell we have separated
2 atoms of hydrogen from the group SO,, that is, we have broken two bonds
or valencies. Similarly, we find that two bonds have been broken in each
cell, leading us to the simplest expression of Faraday's law: The same
current flowing for the same time breaks up always the same number
of bonds or valencies,
We can, however, express this law in another form. The number obtained
by dividing the atomic weight by the valency is called the chemical equiva-
lent of the element. The weights of various elements liberated by a given
quantity of electricity are therefore proportional to the chemical equivalents,
i.e. the quantities are chemically equivalent. Now, according to modern
views of electrolysis, the electric pressure applied to the cell causes those
electrically charged ions which exist in a free state in the solution to wander
in or against the direction of the applied pressure. These ions give up their
I
Quantitative laws of Electrolysis
31
ہے
charges on arrival at the electrodes, and act therefore as the carriers of the
electricity through the liquid. We see, then, that equivalent weights of
various elements carry or transport equal quantities of electricity, or, in
other words, equivalent weights of different ions carry equal electrical
charges.
0.010386.63·2
2
The weight in milligrammes of an element which is deposited in
one second by a current of one ampere is called its electro-chemical
equivalent. This can be calculated from equation (21) if we know the
atomic weight and valency of the element. For example, silver is univalent
and has an atomic weight of 107·6, whence its electro-chemical equivalent is
0.010386.107.6
= 1·118; for copper (cupric) we get
= 0·328.
1
In the same way the weight of hydrogen and oxygen liberated by an ampere
in one second can be calculated. From a knowledge of its specific gravity
or density the volume of the gas liberated can also be found. In this way
we find that an ampere liberates in one second 0·174 cubic centimetre of the
explosive mixture of hydrogen and oxygen, measured in a dry state at 0° C.
and at a pressure of 760 mm. of mercury.
The relative volumes of hydrogen and oxygen evolved by the electrolysis
are proportional, according to Avogadro's law, to the number of molecules of
each gas. Since the molecules of both gases contain two atoms, the volumes
are proportional to the number of atoms, that is, from the formula H₂O, in
the ratio of 2 to 1.
15. Polarisation.
E
e
If dilute sulphuric acid be electrolysed in a cell with platinum electrodes,
there is found to be a potential difference of two or three volts between the
terminals of the cell. If a smaller
pressure than this be applied the
electrolytic action ceases, while
the application of a larger pres-
sure of, say, 10 or 12 volts to the
cell and an external resistance R
in series simply causes such a
current to flow that the drop in
the external resistance leaves a
pressure of 2 or 3 volts across
the cell.

R
Fig. 22.
The current is then given by the equation
E-e
R
>
Ri
where E is the applied P.D. or the E. M. F. of the generator in Fig. 22, e the
P.D. between the terminals of the cell and R the external resistance, including
that of the generator in the figure. It is remarkable that, although the P.D.
e is applied externally to the cell and it is this P.D. which drives the current
through the cell, yet the magnitude of the P.D. is determined entirely by the
cell. It appears to be a function of the cell which we cannot arbitrarily
32
Electrical Engineering
?
alter, which remains practically constant, for example, when we so alter E
and R that the current varies over a wide range. It even remains very
nearly constant when the resistance of the liquid is considerably changed by
varying the distance between the electrodes. We see, then, that the P.D.
across an electrolytic cell cannot be found, as in the case of a metal resistance,
by multiplying the internal resistance R by the current i. The P.D. e is
very much larger than this product.
The reason for this is obvious when we notice that, on breaking the
circuit, the electrolytic cell sends a current through the voltmeter. It acts
now as a generator or source of electrical energy, the current coming out
from that terminal of the cell at which it previously entered. This current
is called the polarisation current and the electromotive force producing it is
the E.M.F. of polarisation. Its effects are only noticeable so long as the
electrodes are covered with bubbles of the liberated gases. We have thus in
the electrolytic cell a galvanic cell made up of hydrogen, sulphuric acid, and
oxygen, the electromotive force of which is opposed to the original current.
This galvanic cell was, however, active during the passage of the original
current. The terminal pressure e had therefore a double function to perform,
viz. to overcome the electromotive force of polarisation E₁, and to supply the
pressure drop due to the internal ohmic resistance of the cell. We have then
for the terminal pressure of the cell the equation
e = E₁+i. Ri.
Hence, we define the E. M.F. of polarisation, or the back E. M.F., as the
electromotive force of the new element or galvanic cell produced by the
chemical changes of the electrodes. If, as is usual, the polarisation is very
large compared with the ohmic resistance, it follows that the terminal
pressure will depend almost entirely on the value of E₁, and very little on
the current or internal resistance.
The phenomenon of polarisation will be made clearer by considering it in
the light of the principle of the conservation of energy. By multiplication
with i, the above equation becomes
e. i = E₁. i + i¿ª. Ri.
The product e. i is the total power supplied to the cell, while 2. R is the
power dissipated as heat in the liquid. The power E. i must therefore be
used in decomposing the water. This is obviously so, for the mixture of
gases produced possesses a store of potential energy. We can explode the
mixture at any moment and obtain in the form of heat or of mechanical work
the energy used in its formation.
Now, the energy stored in the gas is proportional to the quantity of gas,
which, as we have seen, is proportional to the quantity of electricity which
liberated it. The energy stored in the gas is therefore
J = c.i.t,
where c is a constant for the gas concerned, viz. water-gas. On the other
hand, the energy stored must be
J=E,.i.t.
Polarisation
33
From this it follows that E₁ = c, that is, that the E.M.F. of polarisation for
a given electrolytic process has a constant value c. We assume, of course,
that the liquid and electrodes are similar in each case, so that the final
products are of exactly the same chemical composition. For other types
of cells the E.M.F. of polarisation is different, depending, as it does, on the
energy
of chemical composition, or calorific value, as it were, of the electro-
lytic products. When copper sulphate is decomposed with platinum electrodes
the back E.M.F. will be quite different to that found for sulphuric acid. In
the first case, the products of electrolysis are copper and oxygen. When
copper and oxygen combine to form copper oxide the heat evolved is quite
different to the heat produced by the explosion of equivalent amounts of
hydrogen and oxygen.
It is very interesting to calculate the back E.M.F. from a knowledge of the
chemical data. We know, for example, that when 18 grammes of water-gas
are exploded to form 18 grammes of water, 68,000 calories of heat are evolved.
The electrical energy required for the decomposition of 18 grammes of water
must be equal to this; it is given by the equation
68,000 = 0·24 . E₁ . i . t.
From equation (21) on page 28, 1 ampere flowing for 1 second liberates
0.010386.1
1
mg. hydrogen and
0.010386.16
2
mg. oxygen,
or, together,
mg. = 9.35. 10-5 grammes of water-gas.
0.010386.18
2
The quantity of electricity necessary for the liberation of 18 grs. of water-
gas is therefore
i. t =
18
9:35.10-5
= 1.92.105 coulombs.
Putting this value of i. t in the above equation we get
E₁ =
68,000
0.24.1.92.105
= 1.47 volts.
We see, then, that it would be quite impossible to electrolyse acidulated
water by means of a Daniell cell. It is important to remember, however,
that the actual back E.M.F. is much higher than the calculated theoretical
value. The foregoing considerations were merely for the purpose of illustrating
the main principles underlying the phenomenon of polarisation.
We will now consider a case in which there is practically no polarisation,
because the nature of the electrodes is not changed by the electrolysis and
no energy is stored in the final products. If copper sulphate solution be
decomposed with a plate of pure copper as anode, chemically pure copper
will be deposited on the cathode, while an equal weight of copper is dissolved
from the anode. Both electrodes are therefore always of the same chemical
nature and cannot form a polarising element or cell with the liquid. The
back E.M.F. is therefore zero and e is equal to i. R. We come to the same
conclusion when we consider that the energy used in depositing copper on
T. E.
3
;
34
Electrical Engineering
the cathode is equal to the energy liberated at the anode where copper is
dissolved, i.e. oxydised. The decomposition requires therefore no supply of
energy, except to overcome the internal ohmic resistance.
16. Accumulators.
The first accumulators or secondary cells were made by Planté, who
electrolysed dilute sulphuric acid between electrodes of lead, the surfaces of
which became thereby chemically changed, or formed, as the change is
technically termed. To increase the capacity Faure made the plates with
deep grooves or holes into which pastes of various oxides of lead, or finely
divided metallic lead, were introduced. The positive plates are now largely
made of deeply grooved lead plates having an enormous surface (Fig. 23).
These plates are "formed" in the works by electrolysis of sulphuric acid, their
surface being thus converted into lead peroxide. The negative plates (Fig. 24)
consist of lead grids into the meshes of which lead oxide paste is forced. In
this unfinished condition they are delivered and, after fitting up the battery,
it is necessary to charge it continuously for about 40 hours, thus converting
the paste in the negative plates into metallic lead. A cell consists of a
number of positive and negative plates placed alternately, resting on projecting
lugs on the sides of the glass vessel. All the positive plates in a cell are
connected together by a lead bar running along the top, and, similarly, the
negative plates by another bar (Fig. 25).



Fig. 23.
Fig. 24.
Fig. 25.
To understand the working of the cell we may assume that we have two
pure lead plates and a vessel of sulphuric acid from which we can construct
an accumulator by the original method devised by Planté.
On passing a
current through our electrolytic cell, hydrogen will be liberated at the
cathode, which will therefore be unchanged. At the positive plate or anode,
however, oxygen will be evolved which will attack the lead and produce a
coating of the brown peroxide of lead (PbO₂). On stopping the current we
have now a cell consisting of lead, sulphuric acid and lead peroxide, and
having an electromotive force of 2 volts. This E.M.F. was also acting during
the passage of the current, that is when charging, its direction being opposed
to the current. If now the terminals of the cell be joined through a resistance,
the cell discharges and a current flows through the resistance, leaving the
cell by the terminal at which the charging current entered. The terminal
which was positive while charging is also positive when discharging, for in
Accumulators
35
an electrolytic cell that terminal is called positive at which the current
enters, while in any source of electrical energy the positive terminal is that
by which the current leaves.
When discharging the changes in the cell are as follows:
+
Before discharging:
PbO2
2H₂SO₁
Pb
Direction of current in cell :
Movement of the ions:
H₂◄
SO₁
Reactions at the electrodes:
PbO2 + H₂+ H₂SO
Pb+SO₁
=PbSO₁+2H₂O
= PbSO₁
Fully discharged:
PbSO
4
PbSO 2H₂O
Both plates thus become converted into lead sulphate; the positive
through reduction of the peroxide by means of the liberated hydrogen, the
negative through oxidation of the metallic lead by the oxygen, the lead
sulphate being considered as a compound of lead oxide (PbO) and SO,,
obtained by displacing water from sulphuric acid. The result is that the
excess of oxygen in the peroxide plate is transferred to the lead plate, the
oxidation of which is the source of the electrical energy, just as the oxidation
of coal in a furnace is the source of heat energy. When, finally, the positive
plate has given up all its excess oxygen, and the negative plate has been
thereby coated with sulphate, the store of potential energy is used up and
the cell is discharged. This is evident from the fact that the plates are now
similar and can therefore no longer make up a voltaic cell, one essential of
which is the presence of two chemically different electrodes. We must not
forget to mention that the concentration of the acid gradually decreases as
the cell becomes discharged, its specific gravity indicating to a certain degree
the extent to which the cell is discharged. This change of concentration
must be taken into account, in addition to the reduction and oxidation, when
considering the energy supplied by the cell.
The discharged accumulator is treated again as an electrolytic cell, that
is, it is charged. The following changes then take place :
Before charging:
Direction of current in cell:
+
PbSO
SO₁
H₂SO4
PbSO₁
4
→H₂
Movement of the ions:
Chemical reactions :
Fully charged:
PbSO₁+ SO₁ + 2H₂O
4
= PbO₂+ 2H,SO,
2
4
PbSO4 + H2
= H₂SO₁+ Pb
PbO2 3H₂SO Pb
4
The positive plate is again converted into peroxide and the negative plate
to metallic lead, and the electrolytic cell has become once more capable of
being used as a source of electrical energy.
It is evident that the process of charging an accumulator is quite different
from the charging of a condenser, for in the latter case the energy is stored
in an electrical form, whereas in the accumulator the storing consists of
chemical changes in the active materials of the cell. When we speak of the
3-2
36
Electrical Engineering
capacity of a cell, we do not mean the same thing as we do when we speak
of the capacity of a condenser. The capacity of a condenser is defined as the
quantity of electricity which flows into it at the positive terminal when a
P.D. of one volt is applied to its terminals. It is important to note that the
same quantity flows out at the negative terminal, there being no more
electricity in the condenser when charged than when uncharged; its
distribution is, however, different in the two conditions. The capacity of
an accumulator, on the other hand, is the total quantity of electricity which
is set in motion by its discharge.
When an accumulator is said to have a certain discharge capacity in
ampere-hours, it is theoretically immaterial whether a large current be taken
for a short time or a small current for a long time and also whether the
previous charging current was small or large. Like a Daniell cell, for
example, an accumulator has a certain electromotive force but no fixed
current. The makers, however, specify a certain discharge current as the
normal, or as the maximum, for a given size and type of cell. This is not
the current which the cell is bound to give, neither is it the largest current
the cell is capable of giving, but merely the largest current which the cell
should be called upon to give without running the danger of loosening the
active material, which may come away from the plates and fall to the bottom
of the vessel, or of buckling the plates, which may bend so much under bad
treatment as to cause positive and negative plates to come into contact. We
can therefore choose any discharging current within practical limits, and will
obtain corresponding times of discharge. Theoretically, the capacity should
be the same in every case, since, according to Faraday's law, the product i.
is proportional to the quantity of active material decomposed.
In practice, however, it is found that the capacity is considerably decreased
when the cell is discharged with a heavy current. This is due to the action
being largely superficial in such a case, only a portion of the active material
taking part in the chemical changes.
t
The ampere-hour efficiency is the ratio of the ampere-hours in the
discharge to the ampere-hours in the charge. Theoretically, this ratio should
be equal to 1, since charge and discharge consists in the chemical conversion
and reconversion of the active material, and the quantity of electricity set in
motion is proportional to the quantity of converted active material. Owing
to various causes, however, the accumulator, especially if left standing for a
long time, discharges itself. The causes of this are as follows: faulty
insulation, local currents between parts of the same plate, due to impurities,
imperfect conversion of the active material during charge, or to unequal
concentration of the acid. Finally, a part of the charging ampere-hours is
used in the useless production of gas. This occurs mainly towards the end
of the charge when the active material is almost all converted and the
hydrogen and oxygen, failing to find any material with which to combine,
is given off and leads to so-called "gassing." For these reasons the ampere-
hour efficiency is less than 1. It is, however, very high, generally exceeding
0·9, and by suitably arranging the experiment it can be made very nearly
equal to 1.
Accumulators
37
In determining the ampere-hour efficiency it is, of course, very important
to cease charging or discharging at the correct time. The charge is generally
carried on until the active material on the surface of the plates is nearly all
converted, and the cell "gasses," the pressure across the cell then rising to 2·6
or 2.7 volts owing to the gas bubbles which adhere to the electrodes. The
discharge, on the other hand, is continued until the pressure drops to 18
volts, the drop being to a large extent due to increasing internal resistance
as the result of the formation of lead sulphate on the plates. This final
value of 1.8 volts should be read on the voltmeter during the discharge with
normal current and not after opening the circuit, for the E.M.F. of the cell
may then read 2 volts. We see then that the cell is not entirely discharged,
but it is found that a further discharge does great harm to the plates,
besides being of little use owing to the rapidly dropping pressure.
The watt-hour efficiency is, however, of greater practical importance than
the ampere-hour efficiency. To determine this, we observe the terminal
pressure at frequent intervals during charge and discharge, and plot the
values observed as ordinates with the time as abscissae (Figs. 26 and 27).
We see that the average pressure during charge is much higher than that
during discharge. During charge the terminal pressure is given by the
equation
while during discharge the equation is
e = E₁+i. Ri
e = E₂-i. Ri
.(22),
.(23).

3

25
Charging
2
1.5
05
3
2.5
Discharging
12
1.5
10.5
Time
Fig. 26.
Time
Fig. 27.
The average terminal pressure during charge is therefore greater than
that during discharge by twice the internal drop, quite apart from the fact
that the electromotive force during discharge is less than that during charge,
for in the latter case the E. M. F. is increased by the presence of the gases
evolved. The watt-hour capacity is therefore less than the ampere-hour
efficiency, and lies in favourable cases between 0·8 and 0·9 since we have, in
addition to the losses already mentioned, the loss through ohmic heating*.
* For further information on accumulators, see "The Storage Battery" by Treadwell, "Theory
of the Lead Accumulator" by Dolezalek, or “Storage Battery Engineering" by Lyndon.
38
Electrical Engineering
17. Primary Cells.
The simplest voltaic cell consists of plates of copper and zinc in dilute
sulphuric acid, the copper terminal exhibiting a positive charge and the zinc
a negative one. If the terminals be joined by means of a conductor, a current
flows through it from copper to zinc, the current flowing from zinc to copper
in the cell. The sulphuric acid is thereby decomposed and the hydrogen,
going with the current, adheres to the copper plate in the form of small
bubbles. We have, then, a new cell consisting of hydrogen, sulphuric acid,
and zinc, the electromotive force of which opposes that of the original cell.
As a consequence, the E.M.F. of the cell gradually dies away, and the cell is
said to polarise. As the cell, for practical purposes, should have a more or
less constant E. M. F., steps must be taken to prevent the polarisation, that is,
to hinder the formation of free hydrogen. Either a metal must be deposited
on the cathode instead of hydrogen, or the hydrogen must be combined with
some other substance at the moment of its evolution.
In the Daniell cell the difficulty is overcome by the use of two liquids,
copper sulphate and dilute sulphuric acid, separated by a porous earthenware
pot. The constituents of the cell may be represented as follows:
+ Cu—CuSO₁ || H„SO.—Zn — .
The positive pole consists of a plate of copper dipping into copper
sulphate, the negative pole of a zinc plate in dilute sulphuric acid. The
zinc is amalgamated to prevent its consumption by the acid when the cell is
not being used. The current flows through the cell from zinc to copper and
the hydrogen moving in the direction of the current passes through the
pores of the earthenware pot and throws down the copper out of the copper
sulphate solution, as shown by the formula
H₂+ CuSO₁ = H₂SO₁ + Cu.
4
4
The zinc is dissolved by the group SO,, forming zinc sulphate (ZnSO₁),
and this, being an oxidisation, supplies the energy corresponding to the
electrical energy supplied by the cell. This includes the work done in the
external circuit, the heating in the cell itself and the energy consumed in
precipitating the copper. To put it simply, the total electrical energy
developed by the cell is equal to the difference between the energy evolved
in the solution of the zinc and the energy consumed in the precipitation of
the copper. Since the electrodes remain unchanged, there is no polarisation,
and the electromotive force remains constant at 107 volts, assuming that we
are dealing with pure materials. The current varies, of course, according to
the external resistance. The internal resistance depends on the size of cell,
but assuming it to be 0.5 ohm the current which would flow through the cell,
1.07
were it short-circuited, is
2.14 amperes.
0.5
The Bunsen cell consists of carbon in strong nitric acid and zinc in
dilute sulphuric acid, the liquids being separated, as before, by a porous
diaphragm.
+ C-HNO, || H₂SO,-Zn -
Primary Cells
39
The carbon is the positive, the zinc the negative pole. In the sulphuric
acid the action is exactly the same as before. The hydrogen passes through
the pores and combines with the nitric acid, as shown in the formula :
3H₂+ 2HNO, = 2NO + 4H₂O.
The nitrous oxide (NO) thus liberated is further oxidised by the air to
nitric oxide (NO2) in the form of brown fumes with an objectionable odour.
The E.M.F. of a Bunsen cell is 1·8—1·9 volts. Its internal resistance is less
than that of a Daniell cell.
The bichromate cell consists of carbon and zinc in a mixture of dilute
sulphuric acid and potassium bichromate (K₂CrO4, CrO3):
4
+C—H₂SO4, K₂CrO4, Cro¸‚—Zn
There is no porous partition. The carbon is again the positive pole.
The radical SO, dissolves the zinc as before, while the hydrogen is oxidised
by the chromic acid (CrO,), with the formation of water and chromium oxide
(Cr₂O₁), thus
2CrO3 + 3H2 = Cr₂O3 + 3H₂O.
The chromium oxide (Cr₂O₂) combines with the sulphuric acid, giving a
greenish coloration, which gradually replaces the red tint of the chromic acid
(Cro). At the same time, the electromotive force, which was primarily
2 volts, falls, so that with continuous heavy use the cell has not a constant
E.M.F.
The Leclanché cell consists of carbon and zinc in a solution of sal am-
moniac:
+ C—NH¸Cl—Zn —
With the decomposition of the sal ammoniac, the chlorine goes to the zinc,
which it attacks, forming zinc chloride. Although not oxidation, this process
is very similar to it, and supplies the energy of the cell. The group NH
behaves very like a metal and goes with the current to the carbon, where it
splits up into ammonia and hydrogen:
2NH₁ = 2NH, + H₂.
2.
The hydrogen, thus produced, would polarise the cell, were the carbon
not surrounded with manganese dioxide, which combines with the hydrogen,
giving up a part of its oxygen, in which it is very rich, and producing water
and a lower oxide of manganese:
2MnO₂+ H₂ = Mn₂O3 + H₂O.
This action is not very rapid and the depolarisation is not instantaneous
in its action as in the case of nitric or chromic acid. Although the E.M.F. of
the cell is 14 volts when carrying no current, it rapidly falls when in use,
and the cell is therefore only suitable for intermittent work.
The Weston Standard cadmium cell consists in its best form of an
H-shaped glass vessel (Fig. 28) into which two platinum wires are fused.
The platinum wires pass through the bottom of each limb into the liquid
electrodes, the positive being pure mercury and the negative cadmium,
rendered liquid by being amalgamated with mercury (12 to 13 per cent.
of cadmium). The electrolyte is a saturated solution of cadmium sulphate
1
40
Electrical Engineering
"
H
(CdSO,), its saturation being assured by the addition of some crystals of
the salt. When current passes, the cadmium in the electrolyte moves
towards the mercury electrode and would
alloy itself with it, thus making the two elec-
trodes more and more alike, and gradually
reducing the E.M. F. of the cell. This is pre-
vented by placing above the mercury a paste
consisting of mercurous sulphate (Hg₂SO),
and cadmium sulphate (CdSO), and metallic
mercury. The cadmium ions displace the
mercury from the mercurous sulphate form-
ing cadmium sulphate and metallic mercury,
thus:

Ca
Cd + Hg₂SO₁ = CdSO₁ +2Hg.
4
caso,
Ag₂SO
Bg
Fig. 28.
The positive electrode is unaffected and remains pure metallic mercury,
polarisation being thus avoided. At the negative electrode cadmium is, of
course, dissolved by the SO, group.
The electromotive force is very constant at 1019 volts, and is hardly
affected by small changes of temperature. It must only be allowed, however,
to give very small currents and is therefore only used in the potentiometer
method. It is now largely used for the calibration of instruments by this
method, and has, to a large extent, replaced the Clark cell, which was formerly
used for the same purpose, but which is much more sensitive to changes of
temperature*.
18. The Voltameter.
The legal definition of the ampere is based on the silver voltameter. A
small platinum bowl is used as the cathode, and contains a solution of silver
nitrate (AgNO3) (20 to 40 parts by weight to 100 parts of water). The
anode consists of pure silver. The solution is only used until 3 grammes of
silver have been deposited per 100 c.cms. of solution, and the deposit on the
bowl must not exceed 0.1 gramme of silver per square centimetre. The
current density on the anode surface must not exceed 0.2 ampere per sq. cm.
while on the cathode it must not exceed 0.02 ampere. The bowl is weighed
before the experiment, and after the experiment it is washed with distilled
water, free from chlorine, until the washings give no cloudiness on the
addition of hydrochloric acid. This proves the perfect removal of every
trace of silver nitrate, which would otherwise form a precipitate of silver
chloride when the acid was added, in accordance with the formula:
HCl + AgNO3 = AgCl + HNO3.
The bowl is then washed for 10 minutes with hot distilled water
(70-90° C.) and again rinsed with cold water until the washings remain
perfectly clear on the addition of hydrochloric acid. It is then dried in warm
air, cooled and weighed.
Great care must be taken not to touch the inside of the bowl with the
* For further information on primary batteries see "Primary Batteries " by W. R. Cooper.
The Voltameter
41
fingers either before or during the experiment. It is a good thing to heat
the bowl to redness in the tip of a Bunsen flame to destroy any organic
matter, before commencing the experiment. This should not be done if the
bowl contains any silver, as an easily fusible alloy of silver and platinum may
be formed. The cold or luminous part of the Bunsen flame must be avoided
or the bowl may become very brittle owing to the formation of platinum
carbide.
If we represent by
m₁, the weight of bowl before experiment,
M₂
""
""
t, the time in seconds,
after experiment,
then, since 1 ampere deposits 1·118 mgs. of silver in 1 second, we have
i
M₂ - Mr
1.118.t
amperes
.(24).
Copper voltameters and water voltameters are suitable for heavier
currents but do not give such reliable results as the silver voltameter. They
are very useful, however, for general laboratory work, from an educational
point of view.
علیہ
ہے
}
1
CHAPTER III.
19. Strength of magnet pole.-20. Strength of magnetic field.-21. Lines of force.-
22. Magnetic potential.-23. Iron in magnetic field.-24. The earth's field.
19. Strength of Magnet Pole.
N
•A
S
ល
Fig. 29.
A magnet is a piece of steel which possesses the property of attracting
pieces of iron to itself. The name is supposed to have originated in the
discovery near the town of Mag-
nesia of pieces of iron ore possess-
ing this property. If a bar of
steel possessing this property be
plunged into iron filings, they are
found to cling to it, especially at
its ends (Fig. 29). These places, where the force of attraction seems strongest,
are called the poles of the magnet. In very long and thin magnets it can be
assumed that the attractive force is directed towards a single point near
each end. These points are not at the extreme ends, but at the points N
and S in Fig. 29. The line joining these points is called the magnetic axis.
If the magnet is free to rotate about a vertical axis A, one of its poles
will point approximately towards the geographical north. This pole is called
the north pole of the magnet, the other being called the south pole. If two
magnets be brought near together, it is observed that their like poles repel
each other, while their unlike poles attract each other. It follows from this
that the geographical north pole of the earth possesses physical south polarity,
and vice versa.
Coulomb was the first to measure the force between two magnet poles.
He suspended a long thin horizontal magnet by means of a fine metallic
wire. By turning the upper end of this wire he moved the north pole
of the suspended magnet away from the south pole of another vertically
placed magnet. Since the turning-moment or torque exerted by a twisted
wire is proportional to the fangle of twist, he was able to determine the
relation between the attractive force between the poles and the distance
between them. He found the force to be inversely proportional to the square
of the distance. He found, moreover, that the force was doubled by putting

Strength of Magnet Pole
43
another similar magnet with one of the two already in use, that is, by
doubling the strength of one of the poles. If then
and
we have
f = the force exerted on each pole by the other,
m₁ = the strength of one pole,
m₂ = the strength of the other pole,
r = the distance in centimetres,
f=
Mr. M₂
p2
(25).
If we choose the units of force and length, this equation gives us the
definition of unit pole strength. As the unit of length we take the centi-
metre, and as the unit of force, that force which, acting on the mass
of 1 cubic centimetre of water, produces an acceleration of 1 cm. per
second per second. This force is called a dyne. Since the kilogramme
weight is a force which produces an acceleration of 981 cms. per second per
second in the mass of a kilogramme or 1000 c.cms. of water, a kilogramme
weight is equal to 981,000 dynes. Hence
1 dyne
=
1
981,000
kg.* = 1.02 milligrammes.*
If we wish to express Coulomb's law in the simple form of equation (25)
we cannot now choose arbitrarily the unit of pole strength, but must obtain
the unit from the equation. If in equation (25) we put m₁ = 1, m₁ = 1 and
r=1, then we have ƒ=1. Hence that pole has unit strength which
exerts a force of 1 dyne on a similar pole at a distance of 1 cm. The
strength of a pole or a quantity of magnetism is therefore measured by the
force it exerts under certain conditions, and we define the pole strength m as
the force exerted by the pole on a pole of unit strength at a distance of
1 centimetre.
20. Strength of Magnetic Field.
S
The space surrounding a magnet, or any space where magnetic effects
can be observed, say, by means of a compass needle, is called a magnetic
field. The magnetic effect at any point
near a magnet can be determined from a
consideration of the combined effects of
the two poles. At the point A (Fig. 30),
in the neighbourhood of a magnet,
imagine a freely moveable north pole,
the corresponding south pole being so
far away as to be negligible. This north
pole will be repelled in the direction AB

Fig. 30.
by the north pole N, and attracted in the direction AC by the south pole S.
The forces are inversely proportional to the squares of the distances. In
* To indicate that forces and not masses are referred to, a star will be added to the units of
mass when used as forces.
44
Electrical Engineering
Fig. 30, for example, the distances are 2:1 and the forces 1:4. By the
parallelogram of forces the resultant of AB and AC is AD. This is there-
fore the direction of the magnetic force at this point. If a compass needle
be placed at A the magnetic forces due to the bar magnet will act on its
poles and turn the needle until its magnetic axis lies along AD. Thus
the magnetic axis of a magnetic needle, free to move in any direction, in-
dicates the direction of the magnetic field at any point.
To find the strength of the magnetic field at any point, we place a pole
of strength m at the point and determine the force acting on it. This force
is proportional both to the strength of the pole m on which the field is acting
and to the strength of the field. If H represents the strength of the field,
we have
f=m.H...
……….(26).
If, in this equation, we put m=1, then ƒ=H. The field strength H is
therefore equal to the force in dynes exerted on unit pole at the given
point. Hence that field has unit strength which exerts a force of 1 dyne on
unit pole.
The force on a pole at A in Fig. 30 was due to the two poles N and S.
We shall consider now the case represented in Fig. 31, where one pole of
the magnet is very far away. As its
effect decreases as the square of the
distance, this pole can be neglected. If
now the pole strength of the magnet
be m and a unit pole be placed at the
point A, r cms. from the pole m, the force
on the unit pole, by Coulomb's law, is
솜
​H =
m.1
m
2.2
зав
r
Fig. 31.
……..(27).
Since we are dealing with the force on unit pole, i.e. with strength of
field, we have put H in the above equation in the place of f.
F2
We imagine now a magnet needle of pole strength m and having a
magnetic axis of length 7, placed in a uniform field of strength H. The
axis of the needle is at right
angles to the direction of the
magnetic force, which in Fig. 32
is supposed to act downwards.
On the north pole there acts a
force of fi=m. H with an arm
or leverage of and on the south

2'
pole is a force ƒ½=m. H with an
2
equal arm. The total turning-
moment is therefore
N
S
Ն
Fig. 32.
1
ž(fi+f2)=2mH
· H.ml.
2
Since the force is in dynes and the arm or length in cms., the turning-
Strength of Magnetic Field
45
?
moment is in centimetre-dynes. The turning-moment is therefore the
product of two quantities, of which the first is the field strength and the
second a magnetic characteristic of the needle, viz. the product of its pole
strength and length. This product m. is called the magnetic moment of
the needle.
21. Lines of Force.
So far, we have based our assumptions on the now obsolete view ac-
cording to which a certain quantity of magnetism is concentrated at the
polar points, from which effects are caused at a distance without reference
to the intervening space. As a matter of fact, poles are never concentrated
at points, and modern physics no longer recognises action at a distance.
The older view is, however, of great value inasmuch as it explains in a
simple manner the results of many magnetic experiments, and enables us
to express them mathematically. For a deeper insight into magnetic phe-
nomena we must turn to Faraday's ingenious conception of lines of force,
which enables us to form a mental picture of the processes underlying the
various phenomena. Faraday did not believe in action at a distance, and
to him the force in the neighbourhood of a magnet did not come into
existence only when another pole was placed there on which the force could
act. The space around a magnet is in a peculiar state. It is filled with
magnetic forces which flow, as it were, from pole to pole. This view explains
magnetic phenomena as well as, or even better than, the old theory, and
the agreement between the results obtained mathematically from the one
and the results obtained by a consideration of the other strengthens our
belief in the correctness of both theories, or, at least, of the results obtained
from them.
n
-E-
Faraday's conception is based on a simple experiment which every student
should repeat for himself. If iron filings are shaken through a sieve on to a
magnet and the board or table on
which it lies is lightly tapped, the
filings arrange themselves in charac-
teristic curves (Fig. 33). The iron
filings become magnetised by their
proximity to the magnet, and acting
as magnetic needles arrange them-
selves with their axes along the
direction of magnetic force. The
north pole of one filing attracts the
south pole of the next with the result
that curves are produced which indi-
cate the direction of the magnetic
forces much more plainly than the
geometrical construction of Fig. 30.

N
S
Fig. 33.
These curves are called "lines of force" and are conventionally assumed to
proceed from the north pole, through the surrounding medium, to the south
pole. A compass needle sets itself with its axis tangential to the line of
46
Electrical Engineering
*
force, and its north pole pointing along the positive direction of the line. A
freely moving north pole, whose corresponding south pole is far removed, will
move along the line of force from N to S (Fig. 33). Hence we can define
a line of force as the path of a freely moving north pole. It is essentially
nothing more than the direction of the magnetic force. When dealing with
magnetic problems it is usual to refer to the lines of force as if they had
a real existence, somewhat similar to the visible lines of the iron filings
experiment.
The line of force is, however, far more than a convenient indication of
the direction of the magnetic force; it enables us to express, in a very con-
venient way, the magnitude of the force, that is, the strength of the magnetic
field. To this end we observe that the lines of force radiating from a pole do
not lie in one plane, but radiate out into space in all directions. Imagine
a number of concentric spheres with the pole as centre; radiating from the
pole the lines of force will pass through the surfaces of successive spheres.
Since the area of a sphere increases as the square of its radius, the number
of lines of force per unit area must decrease in the same ratio. We are now
able to see why, in Coulomb's law, the force ƒ varied inversely as the square
of the distance, and why, in equation (27) on page 44, the strength of the
field in the neighbourhood of a pole decreased as the square of the distance.
We can now define the strength of a magnetic field very simply as the
number of lines of force per square centimetre.
m=1
If now the strength of field is to be measured both by the force on unit
pole and by the number of lines per square cm., both methods giving the
same result, the unit for the line of force
cannot be arbitrarily chosen. In a field
of unit strength there must be one line
per sq. cm. To make this clearer imagine
a unit pole at the centre of a sphere of
1 cm. radius (Fig. 34). If now another
unit pole be placed on the surface of the
sphere, the force acting on it, according to
Coulomb's law, is

H
1.1
12
m=1
Fig. 34.
On the above assumption the number
of lines per sq. cm. must be the same figure,
viz. 1. Hence a line of force represents the total flux of force, so to say,
which passes through 1 sq. cm. of the surface of a sphere of 1 cm. radius,
described about a unit pole. The surface of a sphere is 4πr², that is, in our
case, 4π sq. cms. Since there is one line per sq. cm. the total number of lines
radiating from unit pole is 4π. A line of force is thus 1/4π of the total flux
of force radiating from unit pole.
The total number of lines N radiating from a pole m can be found from
the formula:
N = 4πm
.(28),
Lines of Force
47
and the total number of lines crossing a surface of A sq. cms. over which
there is a normal uniform field strength is
N = H . A ..
.(29).
If, for example, the field strength in the air-gap between the poles and
armature of a dynamo is 7,000, this means that 7,000 lines of force leave.
every sq. cm. of the north polar surface. If the surface of the pole facing the
armature is 400 sq. cms. the total number of lines entering the armature
from the north pole is
N=H.A=7,000. 400 = 2.8. 10° lines.
22. Magnetic Potential.
S
N
E
Fig. 35.
We have already seen that a freely moving north pole in a magnetic
field moves along the lines of force. If now we form the product from point
to point of the force acting on the
pole and the distance moved through
by it, and take the sum of the pro-
ducts so obtained, we find the work
done by the magnetic field in moving
the pole. This work is either used in
overcoming mechanical resistance, or
in storing kinetic energy in the moving
body. If we, on the other hand, move
a north pole n (Fig. 35) against the
direction of the line of force from S
to N, we have to overcome the oppos-
ing force of the magnetic field, which
tends to drive the north pole from N
to S. We must therefore do an
amount of work equal to the summation of the product of force and
distance along the whole path traversed. This work is stored as potential
energy, to be liberated once more when the north pole, under the influence.
of the field, is driven back. This is quite analogous to the case in which we
lift a weight, thus overcoming the opposing force of gravity. The work done
is stored as potential energy in the raised weight, to be liberated when the
weight falls from the higher to the lower level. We see, therefore, that
there is a difference of magnetic level or magnetic potential between
the different points on a line of force. As in Section 11, we define this
magnetic potential difference as the work done in moving the positive
unit quantity of magnetism, that is, a unit north pole, from the lower
to the higher potential. Inversely, it could be defined as the work done
by the field when a unit north pole is moved from the higher to the lower
potential. We have here assumed that the potential of the north pole is
positive and that of the south pole negative. The potential at the middle
point of a line of force will be zero.
In moving the north pole n from C to D in Fig. 35 it is immaterial

48
Electrical Engineering
whether we follow the line of force or go any other way. This follows from
the conservation of energy. The energy used in moving the north pole n
from C to D is constant, by whatever path we go, for it is stored as potential
energy, its amount depending only on the initial and final positions C and D.
The same amount of energy will be restored in every case when, under the
action of the field, the pole n is driven back from D to C.
On the other hand, to move the pole n from C to E requires no expendi-
ture of energy, since the motion is at right angles to the lines of force. There
is no difference of potential between the points C and E. Such points do
not, of course, lie wholly in one plane, but exist in the surrounding space.
They lie on a so-called equipotential surface or plane, which is normal to the
direction of the field at every point, and on which a pole can be moved
without work being done.
We will now consider a homogeneous field, in which all the lines of force
are parallel. If a pole be moved against the direction of the field (Fig. 36)
the force on the pole is constant over
the whole distance l. If the pole have
unit strength, the potential difference
between the points A and B separated
by a distance will be equal to the
product of the force and the distance,
i.e. H. l.


N
Fig. 36.
S
Up to the present we have spoken
only of potential differences, but we will now go a step further and define
magnetic potential itself. We will imagine a north pole m, concentrated
at a point from which the south pole is so far removed as to be negligible.
The lines of force radiate into space, and under their influence a north pole
at A (Fig. 37) would be driven away to an infinite distance. The force
名
​Fig. 37.
acting on it would decrease as the square of the distance, and would finally
vanish. The total amount of work done in driving unit pole from A to
infinity is called the potential at A, or the potential of m at A, and we thus
assume that at an infinite distance the potential is zero.
It may appear strange that the potential at an infinitely distant point is
the same as that at the middle point of a line of force, which we found above
to be also zero. It would, however, be possible to bring a north pole from
an infinite distance, along the path indicated by the dotted arrow in Fig. 35,
to the centre of any line of force without any expenditure of energy, for the
Magnetic Potential
49
north pole would be equally attracted by the south pole of the magnet and
repelled by its north pole, and we see that the motion is always at right
angles to the lines of force.
If the idea of potential proves somewhat of a stumbling-block to the
beginner, he will do well to think always of a difference of level and remember
that, as a rule, we are concerned with differences of potential, and not with
the absolute value of potential.
23. Iron in a magnetic field.
If iron filings be sprinkled on a magnet near which is a piece of iron,
we notice that the iron draws the lines of force within itself (Fig. 38). As a
result, the lines of force entering the end
of the piece of iron nearest the magnet
pole are very numerous, while, in other
places, the field is weakened.
This was
formerly explained as a case of induction,
the magnet pole inducing a pole of oppo-
site sign to itself on the side of the iron
nearest to it. This difference in sign of
the poles of magnet and iron explained
the well-known fact that the iron is
attracted by the magnet.

N
n
Fig. 38.
The theory of the lines of force leads to the same conclusion. We
suppose the iron to have a special conductivity or permeability for the lines
of force, so that they exhibit a great tendency to pass through iron, if
possible. The attraction of the iron by the magnet pole leads us then to
the assumption that the lines of force strive to shorten themselves as much
as possible, in much the same way as a stretched elastic band tends to shorten
itself. We must also imagine the lines to exert a lateral pressure tending to
force themselves apart, since they would otherwise collect in a dense mass
along the shortest possible path. They must therefore repel each other
mutually in a direction at right angles to their own direction, as do two
parallel bar magnets, placed side by side with their similar poles together.
If the lines of force have a great tendency to pass through iron, owing to
its superior conductivity compared with the surrounding air, it is evident
that in Fig. 38 a great number of lines will enter the side of the iron nearest
the pole N of the magnet. This side of the iron will therefore be a south
pole, since, on our former assumptions, south polarity is exhibited where lines
of force enter the material. The further extremity of the iron will be a
north pole, as the lines leave the iron at that end. The magnetisation of
the iron is stronger, the better its conductivity for the lines of force. It is
stronger, for example, in wrought iron or mild steel than in hard cast iron or
hardened steel under exactly similar conditions.
This fact leads us to the assumption that the process involved in the
magnetisation of a piece of iron is similar to the action of a magnetic field on
a number of iron filings. We assume that the smallest particles, the so-
T. E.
4
50
Electrical Engineering
called molecules of the iron or steel, are naturally magnetic, but point in all
directions without law or order (Fig. 39). They have therefore no external
effect and the iron appears unmagnetised. When a magnet is brought near,
the molecules are turned, like compass needles, to point in the same direction.
The ends of the piece of iron will now exhibit "free" magnetism, while
towards the middle of the piece the molecules neutralise each other. It is
evident that the molecules of wrought iron are more readily turned than
those of hard steel.


·S'.
n
KZW1
n
Fig. 39.
S
n
N
2
$ — N
s-n
S
n
S
n
S
n
n
Fig. 40.
This molecular theory of magnetism is supported by the fact that wrought
iron is only temporarily magnetised, while steel becomes a permanent magnet.
When the magnetising force is removed, the molecules of the elastic wrought
iron return almost entirely to their former chaotic arrangement, leaving only
a small trace of remanent magnetism. With hard steel, on the other hand,
the molecules, when once turned, remain in their new position and we have
a permanent magnet. It is now clear why a steel magnet exhibits no
magnetism at the middle of its axis, and yet, when broken in two, the two
rough ends are found to be poles of opposite sign.
The strongest point in favour of the molecular theory is, however, the
development of heat in a piece of iron subjected to rapidly reversing mag-
netism. A reversal of magnetism means a rotation of the molecules, and
this is opposed by their friction, the molecules tending to remain in their
previous condition and opposing any change. This phenomenon is therefore
called hysteresis, i.e. a lagging behind. The work done in reversing the
magnetism appears as heat, the energy expended per second being propor-
tional to the frequency of the reversal and to the volume of the iron. It is
also, of course, dependent on the hardness of the iron.
24. The earth's field.
In
As the magnetic poles of the earth do not correspond with the geo-
graphical poles, the axis of a magnetic needle makes an angle with the
geographical meridian, which is known as the angle of declination.
England, the declination is to the west and in London at the present time
is 16°. If the needle be freely movable in the vertical as well as the
horizontal plane, it will form an angle with the horizontal, known as the
inclination. In England the north pole will be below the centre and the
inclination in London is 67°. With the ordinary magnetic needle, moving
about a vertical axis, we are only concerned with the horizontal component
of the earth's field. In London, at present, this is equal to 0·185, that is, the
horizontal force on a unit pole is 0·185 dyne.
CHAPTER IV.
25. Magnetic effect of a straight conductor.-26. Magnetic effect of a single-turn coil.-
27. Solenoid.-28. Magnetisation curves.-29. Ohm's law and the magnetic circuit.
30. Lifting power of an electromagnet.-31. Hysteresis.-32. Dynamic effect of parallel
currents.-33. Induced electromotive force.-34. Laws of mutual induetion.-35. Self-
induction.-36. Eddy currents.
1
25.
Magnetic effect of a straight conductor.
It is found by experiment that a magnetic needle is deflected out of its
north and south direction when near a conductor carrying a current. Hence
the electric current produces a field in its neighbourhood. The direction of
the lines of force of this field can be found by moving a compass needle con-
tinually in the direction indicated by it, or by sprinkling iron filings on a
plane through which the conductor passes normally (Fig. 41).
The filings arrange themselves in concentric circles around the conductor
as a centre. The lines of force produced by the current are therefore closed
circles and a freely moving north pole-
its corresponding south pole being far
removed-would rotate about the con-
ductor in the direction of the lines of
force. This direction is always normal
to the plane through conductor and pole.
It can be found by Ampère's swimming
rule: If a person swims with the cur-
rent and looks to the magnet needle,
its north pole will be deflected to his
left hand. Another rule, requiring less
imaginative effort, is the corkscrew rule:
The directions of current and field
are related in the same way as the
directions of translation and rotation of a right-handed screw.

Fig. 41.
The action between conductor and pole is naturally mutual. If the pole
is fixed and the conductor is movable, the latter will move in a direction at
right angles to a plane through conductor and pole. The motion will thus be
at right angles to the plane through the conductor and the lines of force
meeting it, due to the pole alone. The direction of motion can be found by
Ampère's rule, in this case also, if the rule be expressed as follows: If a
4-2
52
Electrical Engineering
1
person swims with the current and looks toward the north pole,
whence the lines of force proceed, the north pole tends to move to the
left. If the pole is fixed, the conductor tends to move to the right. In
this form, the rule is applicable to motors, in which a uniform field proceeds
from a large pole face. In Fig. 42 a conductor, carrying a current in the
direction indicated, is in front of a north pole, from which the lines of force
come out normally to the paper. If we imagine ourselves swimming with the
current and facing the north pole, we see that the conductor will move to the
right, as shown by the dotted arrow. Similarly, in Fig. 43 a current-carrying
conductor is in front of a south pole into which the lines of force enter. If
we imagine a swimmer with his head to the right and feet to the left, looking
out from the surface of the paper, his right hand will point in the direction
of the dotted arrow and the conductor will therefore move towards the top of
the page.


Fig. 42.
Fig. 43.
As a further example, imagine a horizontal conductor carrying a current
in a magnetic field, the direction of which is from right to left, perpendicular
to the vertical plane through the conductor (Fig. 44). Swimming with the

S
N
Fig. 44.
current and looking to the right, where we may imagine a north pole to be
situated, our left hand points downwards, and the north pole would, if
possible, move in this direction. If the magnetic field is stationary the
conductor will, if possible, move in the direction of our right hand, that is,
vertically upwards as shown by the dotted arrow.
We must now determine the force exerted on each other by a pole
and a small element of current-carrying conductor. In Fig. 45 let
I= the current, in units which we have yet to determine,
dl= the length of the element of conductor in centimetres,
Magnetic effect of a straight conductor
53
:
m=the strength of the pole, in units already defined,
r=the distance between pole and element of conductor in cms.,
= the angle between r and dl,
df = the force in dynes between pole and element of current.
It is found by experiment that the force is proportional to the current, to
the pole strength, and to the length of the element of conductor. If is not
a right angle, the length dl must be replaced by dl sin o, as the number of
the lines radiating from the pole m which meet dl is equal to the number
crossing AB = dl sin 4. Finally, the force is inversely proportional to the
square of the distance, since the pole m radiates lines of force equally in all
directions and their density must therefore decrease with the square of the
distance. We arrive, therefore, at the experimental result, that
df=
=
m.I.dl.sino
зав
.(30).

ab
B
Fig. 45.
m
Although sometimes called Biot-Savart's law, this equation was originally
established by Laplace. From equation (27) on page 44 we see that p2
is equal to the field strength H produced by the pole m at the point occupied
by the element of conductor. Equation (30) can therefore be written thus:
...(31).
df = H.I. dl sin
& now represents the angle between the conductor and the lines of force
cutting it. If now the field is uniform and the conductor is straight and
7 cms. long and perpendicular to the lines of force, the total force acting on
the conductor is
ƒ=H.I.l dynes
..(32).
This equation is applicable to the cases represented in Figs. 42, 43 and
44. We must, however, determine the unit in which I must be expressed,
for, having chosen the units for ƒ, H and 1, the current cannot be expressed
arbitrarily in any unit, such as the ampere.
If in equation (32) we put ƒ= 1, H = 1, and l=1, then I=1. The unit
of current in equation (32) is therefore that current which exerts a
force of 1 dyne on 1 cm. of conductor through which it flows, in a field
of unit strength, the conductor and field being at right angles. This
unit is called the absolute unit of current. It appeared, formerly, to be too
large for practical purposes and a tenth part of it was therefore adopted as
the practical unit and called an ampere. Very accurate experiments were
then made to determine the amount of silver deposited in a given time by
this practical unit of current, the result being 1·118 milligrams per second.
!
54
Electrical Engineering
If now
while
I represents the strength of current in absolute units,
""
""
""
amperes,
i
we see that the number of amperes will always be ten times as large as the
number of absolute units, i.e.
i = 101,
i
or
I =
10
Substituting in equation (32) we have
i
ƒ = H.
]
10
.(33).
..(34).
N
The importance of this result is evident at once, if we apply it to electric
motors. Suppose, for example, that 944 wires each 24 cms. long lie upon the
periphery of an iron drum (Fig. 46).
Suppose, further, that two-thirds of the
total number of wires lie under the poles
and that each wire carries a current of
i= 10 amperes.
We assume that the
wires are so connected that the turning-
moments produced under each pole are
in the same direction and can therefore
be added. Let the strength of the field
or the number of lines of force per sq. cm.
be 7,000, and the diameter of the arma-
ture 40 cms. We are required to find
the turning-moment or torque of the
motor in metre-kilogrammes.
The total length of wire under the
poles is

From equation (34) we have
N
Fig. 46.
2
944.24 = 15,000 cms.
3
1 = 7,000. 10
10
•
15,000 = 105. 10° dynes.
1
kg.
*
981,000
ƒ = H. 10.1 = 7,000.
Now from page 43 we know that 1 dyne
The force acting on the periphery is therefore:
S
F
105.106
981,000
=107 kgs.*
*
This force acts at a radius of 20 cms. = 0.2 metre, giving a turning-moment of
M₁ = 107.0·2 = 21.4 metre-kilogrammes.
Magnetic effect of a single-turn coil
55
26. Magnetic effect of a single-turn coil.
If either the swimming rule or the corkscrew rule be applied to a single-
turn coil (Fig. 47), we find that all the lines of force produced by the current
enter the coil from one side and come out at the other, closing
on themselves by passing round outside the coil. The coil is
thus a magnetic shell or disc, having a south pole on one
face, where the lines of force enter, and a north pole on the
other, where they leave. If then we face the disc or coil, and
the current is flowing in a clockwise direction, we are facing
a south pole; whereas, if the current is flowing in an anti-
clockwise direction, we are facing a north pole.
In order to determine the magnitude of the force exerted
on a pole m, situated at the centre of the coil, we shall con-
sider, in the first place, a small element dl of the coil. The
Fig. 47.
length of the element is at right angles to the radius, that is, to the line
joining the element of conductor to the pole at the centre. The value of
sino is therefore 1. If the radius of the coil be r cms., we have from equation
(30) on page 53

m.I.dl
df =
p2
The total length of all such elements of conductor is 27r. The total force
is therefore
f=
m.1.2πr
p.2
m.1.2π
..(35).
Now the strength of field H is defined as the number of lines of force per
sq. cm. or as the force on unit pole. Putting m=1 in equation (35) we
obtain the field strength at the centre as
2π. I
H:
r
…………….(36).
We will now make use of this result to explain the principle of a certain
class of practical measuring instruments. A large number of such instruments
depend for their action on the effect of a current-carrying coil or bobbin on a
magnet needle. The movement, however, by which the effect is measured,
alters the relative position of coil and needle, thus destroying the proportion-
ality between the current in the coil and its force on the needle. Such
instruments must therefore be calibrated empirically.
In other instruments, however, a definite relation between the current
and the reading of the instrument is maintained. If, for example, the
deflected needle is always brought back to its neutral or zero position by
twisting one end of a spiral spring, the other end of which is attached to the
needle, the turning-moment exerted on the needle by the current is exactly
equal to that exerted by the spring, which, we know, is proportional to the
angular twist applied to it. The current in the coil is therefore proportional
to the angle through which the torsion head has to be turned.
56
Electrical Engineering
Another method, by which proportionality may be maintained within
certain limits, is to work with very small deflections which can be read by
means of a beam of light, reflected on to a scale from a small mirror attached
to the magnet needle. In such mirror galvanometers the relative movement
between coil and needle is so small that proportionality is maintained over a
certain range.
f
α
1
LB I
Finally, the diameter of the coil can be made so large, compared with the
length of the needle at its centre, that, whether deflected or not, the poles of
the needle can be assumed to be at
the centre of the coil. The instrument
is then known as a tangent galvano-
meter. For the sake of clearness, the
needle in Fig. 48 is shown much
larger, compared with the diameter
of the coil, than it actually is. The
tangent galvanometer is set up so
that the plane of the coil is vertical
and in the magnetic meridian. When
no current is flowing, the axis of the
needle will lie in the plane of the coil.
In the plan of the galvanometer (Fig.
48) the coil will be represented by a
straight line parallel to the lines of
force of the horizontal component h of
the earth's magnetism. When a cur-
rent flows in the coil, each pole m of

A
α
1
1
Fig. 48.
ih
the needle will be subjected to a force ƒ perpendicular to the plane of the
coil; from equation (35) we have
m.I.2π
f=
r
The force on each pole due to the earth's field is equal to m.h and acts
parallel to the plane of the coil. Under the action of these two forces the
needle comes to rest at an angle a to the plane of the coil. The component
of the force due to the earth, acting at right angles to the needle, is
AB=m.h.sin a,
while the opposing component of the force due to the current in the coil is
m. 1.2π
AC=ƒ.cos a=
cos a.
Since the needle is in equilibrium, AB = AC, or
m.1.2π
m.h.sina =
cos α.
r
Cancelling m out from each side of the equation, we see that the reading of
the galvanometer is quite independent of the strength or size of the magnetic
needle. Solving the above equation for I, we have
I=
h.r
2π
tan a....
…………..(37).
Magnetic effect of a single-turn coil
57
Since the number of amperes is 10 times the number of absolute units,
we have
10.h.r
2π
tan a
..(38).
If the horizontal component of the earth's field at any place is known, the
strength of current can thus be calculated from the values of r and a. Con-
versely, a standardised tangent galvanometer can be used to determine the
horizontal component.
We will now consider the case in which the pole m lies on the axis AA
of the coil, but at a distance from it (Fig. 49). If the line a joining the pole
to the circumference of the coil makes an angle a with the axis, then
А
r
α
sin a
Now the element of conductor at B is perpendicular to the plane of the
B
a
B
A
α
α
D
Fig. 49.
paper and is therefore at right angles to the line a which lies in that plane.
Putting sin = 1 in equation (30) on page 53, we get
df =
=
m.I.dl m.I.dl.sin² a
a²
p2
This force acts perpendicularly to the plane containing the pole and the
element of conductor, that is, in the direction CD. The component acting
along the axis is CE=df. sin a. Adding up the forces due to all the
elements of conductor constituting the coil, we find that all the components
such as ED, at right angles to the axis, neutralise one another, while the
components along the axis must be added. The total force acting on the
pole m is therefore along the axis AA and has a magnitude
ƒ = Σdƒ. sin a =
m. 1. sin³ a
.Σdl.
p2
Since Σdl is equal to 2πr, this reduces to
2πm. I
f=
sin³ a
go
(39).
4

58
Electrical Engineering
To find the field strength at point C, that is, the force on a unit pole at
this point, we must put m = 1 in equation (39), giving
2π. I
H
sin³ a
γ
.(40).
This equation will prove of great assistance to us in the next section in
calculating the field of a solenoid.
27. Magnetic field of a solenoid.
The lines of force due to two parallel conductors carrying currents in the
same direction combine to form lines of force which encircle both conductors.
Two conductors are shown, for example, in Fig. 50, passing vertically through
the paper, and carrying currents which flow towards us. The points in the
cross-sections of the wires may be considered as representing the points of
arrows indicating the direction of the current. If now we draw the lines of
force due to each individual conductor, we find that they are in opposite
directions in the space between the conductors, and therefore mutually
1
O
Fig. 50.
Fig. 51.
neutralise one another, more or less. At points such as A and B, on the
other hand, the forces due to both conductors are in the same direction and
the density of the lines of force is greater than that due to either conductor
alone. We could calculate in this way the strength and direction of the
field at every point, but it is simpler to make use of iron filings and obtain a
diagram such as Fig. 51. We shall thus see clearly that the lines embrace
both conductors.
If the wire be wound into a solenoid, that is, a long cylindrical coil of
small diameter, the lines of force pass through each individual turn or coil as
seen in the previous section. They combine now to form much longer lines
threading more or less through the whole solenoid, inside which they run
approximately parallel to the axis. Some of the lines such as 1 and 1' in
Fig. 52 are produced by the two middle turns only, others such as 2 and 2′
by the four middle turns and so on, while others such as 4, 4′ and 5 are due
to the combined action of all the turns. We see, therefore, that a current-
carrying solenoid is very similar to a bar magnet, for its lines of force do not
enter and leave entirely at the ends, but, to a large extent, emerge from the
sides of the north polar half of the magnet and doubling back re-enter the
sides of the south polar half.
The polarity of the solenoid can be determined by either of the rules
}
Magnetic field of a solenoid
59
already given. Swimming with the current and looking into the solenoid,
the left hand indicates the north polar end and also the positive direction of
the lines of force inside the solenoid.
2
It can be seen that every line of force goes through the solenoid at its
mid-point. The field strength, that is, the number of lines of force per
square centimetre inside the solenoid, is a maximum at this mid-section, and
falls off on either side towards the ends. This is also the reason why an iron
core is drawn into such a solenoid. Iron always tends to move into the
strongest part of the magnetic field,
so that the number of lines passing
through it from end to end is a
maximum. If we imagine a rod
of iron half in the solenoid and
half out, we see that the lines
would have a more convenient
path and therefore increase in
number in the iron rod, were it
further inside the solenoid; it will
therefore be drawn in. The seat
of the force on the rod can be seen
at once when we consider that the

5€
Fig. 52.
lines enter or leave that end of the rod which is inside the solenoid in
parallel paths, whereas these same lines leave or enter that part of the rod
which is outside the solenoid in all directions. A great number of galvano-
meters, ammeters and voltmeters depend for their action on this principle;
they are generally known as soft-iron instruments.
We will now investigate the field strength at the centre of the solenoid.
We will represent by
I the strength of current in absolute units,
S the number of turns or windings,
7 the length of solenoid in centimetres,
r the radius of solenoid in centimetres.
The S turns carrying the current I are equivalent to a belt 7 cms. wide
carrying a current S. I.
The
current which in Fig. 53 flows
in the strip of width do is there-
fore equal to
dx
S.I.
ī
T'
Imagine a unit pole to be
α
Zx

B
а
placed at the centre of the solen-
oid. The force exerted on it by the
band do can be found by equa-
oool
Ն
Fig. 53.
tion (40) on page 58. For I in the equation we must substitute S.I.
this gives us
ан
¿H =
2π.S.1 dx
Τ
sin³ α.
·
dx
Τ
60
Electrical Engineering
Now x=r.cotan a and do
↑
da.
sin² a
Putting this value for da in the equation for dĦ and integrating between
the limits π- a₁ and +αa₁, we get
H = - [
+a 2π.S.I
sin a. da
=
2π.S.I
1
· [
tar
4.S.I
cos a
cos a₁
………….(41).
4π.S.I
H=
.(42).
For very long solenoids cos a, is practically equal to 1, so that
If the current be expressed in amperes, this formula becomes
H =
0.4π.S.I
l
………….(43).
If the unit pole be placed at the centre of one of the ends of the solenoid,
π
the integration must be carried out between the limits and a, (Fig. 54).
We obtain then
2
H = - ["
а 2π.S.I
2π.S.I
1
2π.S.I
sin a. da =
cos a
cos α1
し
​π/2
#/2
.(44).
01
a
x
da

Ն
Fig. 54.
As before, if the solenoid is very long, a, is very small and cos a₁ can be
taken as unity. The field strength at the ends is thus half that at the
middle, and of the total number of lines passing through the middle of the
solenoid a half passes through the ends while the other half leaves by the
cylindrical surface.
28. Magnetisation Curves.
Under similar conditions, the number of lines of force per square centi-
metre is greatly increased by the presence of iron within the coil. The
solenoid then becomes an electromagnet. It is as if the lines of force which
would be present were the space occupied by air produced or induced a much
greater number of lines in the iron. The field strength or number of lines
of force per square centimetre in the air, which is represented by the
letter H, is therefore also called the magnetising force. The number
of lines per square centimetre in the iron is called the magnetic induc-
tion and is represented by the letter B. It is to be noticed that the
induction B includes the lines which were present before the iron was intro-
duced as well as the new lines due to its introduction.
Magnetisation Curves
61
We account for the greatly increased number of lines in iron by assuming
that its magnetic conductivity or permeability is much greater than that of
air. The ratio of this permeability to that of the air is represented by the
Greek letter μ. The permeability μ gives, therefore, the magnetic
conductivity of the iron compared with air; other things being equal,
the number of lines produced in the iron is μ times the number which
would have been produced in the air. From this it follows that
or
μ
B=μ.H
B
H
For air the permeability is evidently 1, and we have B =H.
.(45),
...(46).
The value of μ depends above all on the quality of the iron. The
permeability of annealed armature stampings or of mild cast magnet steel
may sometimes exceed 3,000. The permeability of a given piece of iron
or steel varies moreover with the degree to which it is magnetised.
The experimental determination of the permeability is carried out in the
way indicated in Fig. 55. The rod of iron to be tested is made in two parts,

www
S
E
S₁
a
ал
Ն
Fig. 55.
9
and passes through holes a and a, in a massive iron yoke and also through
the magnetising coils S and S, and the small secondary coil E. A current i
is passed through the coils S and S₁, and from a knowledge of the number
of turns in these coils, together with the length l of the iron rod, the value of
H can be found from the equation
H
0.4π. S. i
The length 7 does not include the parts of the rod inside the holes ɑ and a₁,
as these carry very little of the magnetic flux, which leaves the rod in all
directions as soon as the rod enters the yoke. The latter being very massive
offers very little resistance to the magnetic lines and can therefore be neglected.
The left-hand half of the rod is now pulled out very suddenly, whereupon the
secondary coil E is ejected by means of a spring. The magnetic flux through
62
Electrical Engineering
the coil is thus suddenly removed, with the result that an electromotive force
is induced in the coil, causing a momentary current to flow through the so-called
ballistic galvanometer g, which is connected up in series with the coil E.
From the deflection or swing of the ballistic galvanometer, the number of
lines passing through the iron rod, and therefore cut by the coil E, can be
calculated. By dividing this number of lines by the cross-section of the
rod we obtain the induction density B. The experiment can then be repeated
with another value of the magnetising current, and in this way a set of
B
14000
12000
10000
ast Steel
B


Stampings
26000
24000
22000
00
8000
6000
20000
4000
18000
2000
16000
3 6
9 12
12
15
15 18 21 24
400 800 1200 1600 2000
H
H
Fig. 57.
Fig. 56.
corresponding values of H and B obtained. Tests made on a sample of cast
steel gave the following results (compare Fig. 56):
B
=
H=0·9
B = 1,130
μ =
= 1,260
H
1.55
5,200
3,350
2.7
8,160
3,020
3.75
9,480
2,530
8.55
12,440
1,460
18.1
14,510
· 800
34.5
15,710
460
82.7
17,150
210
145.3
18,200
130
A peculiar point in the above table is the low permeability possessed by
the steel when weakly magnetised. The molecules are evidently but slightly
affected by small values of the magnetising force. With stronger fields,
covering the range of inductions from 5,000 to 9,000, B is roughly proportional
to H. Finally, we see that beyond a certain point a continual increase in
the magnetising force has little effect on the magnetic induction. The iron
is then said to be saturated. It is impossible, however, to say at what point
saturation commenced.
If now we plot these results on squared paper, with the magnetising force
Has abscissae and the induction B as ordinates, we obtain the so-called
Magnetisation Curves
63
20000
18000
16000
14000
12000
10000
Flux density in lines per square Centimetre.
stampings is given, which is of special interest on account of the fact that
the teeth of modern armatures are very highly saturated in order to improve
the commutation. This curve gives the following values*:
H=100
300
500
B = 18,700
μπ 187
21,300
71.
22,500
45
700
23,100
33
* From the E. T. Z. (Elektrotechnische Zeitschrift), 1901, page 769.
0008
6000
4000
2000
Ampere-turns per Centimetre.
1.0
20
30
40
50
60
70
80
90
100
110 120 130 140 150 160 NO 180 190 200
Ingot-Iron
Cast Steel
Stampi
wrought
Iron
Stampings
Ingot
Armature
ob
2
3
4
5
Wrought Iron
Cast Iron
Cast-Iran
magnetisation curve. This is shown for this specimen of steel in Fig. 56.
Beside it in Fig. 57 the upper part of the magnetisation curve for armature

9
7
8
6
10
12
11
13
14
15
16
17
18
19 20
Ampere- turns per Centimetre.
Fig. 58.
(From Kapp's "Dynamo machines.")
64
Electrical Engineering
If the magnetising force be still further increased, the value of B increases
uniformly by an amount 2.5 times the increase of H. The upper part of the
curve in Fig. 57 is thus a straight line. If this line be produced backwards
it cuts the ordinate axis in the point B = 21,350. For very high saturation
we have therefore
B = 21,350 + 2·5H.
For practical purposes the magnetisation curves as drawn above are not
convenient, as we generally require to know the necessary number of ampere-
turns per centimetre length of path for a given value of B. This cannot be
read off directly from the curve but must be obtained from equation (43) on
page 60 by dividing the corresponding value of H by 04π. Thus, if the
number of ampere-turns be X, the ampere-turns per cm. will be
X
ī
X S.i H
ī = ī 0.4π
= 0·8H
…………..(47).
The calculations are simplified if the magnetisation curves are drawn with
for abscissae instead of H, as is done in Fig. 58. For any value of the
flux density B we can then read off directly the necessary number of ampere-
turns per centimetre of path.
From Fig. 58 we see that wrought iron, cast steel, and ingot iron require
very small magnetising forces up to B = 14,000. Cast iron, on the other
hand, is much less satisfactory and it is for this reason that the magnet
frames of large machines are now very often made of cast steel notwithstanding
its higher price.
29. Ohm's law and the magnetic circuit.
For a closed magnetic circuit such as, for example, an iron anchor ring,
it is practically immaterial whether the ampere-turns are distributed over
the whole circumference or are wound all together
at one part, as shown in Fig. 59. In either case
we can apply equation (43) on page 60 and get

В=µH :
μ.0·4π.S.i
1
μ. 0.4π
X
ī ...(48),
where l applies no longer to the length of the
coil, but to the mean length of the path of the
magnetic lines.
If now A represents the cross-sectional area
of the iron measured perpendicularly to the lines,
Ն
Fig. 59.
then the total number of lines or the total magnetic flux is given by the
equation
N=B.A=µ.H.A.0.4π. X. A
or
,
0.4π. X
N
μ.A
Α
.(49).
Ohm's law and the magnetic circuit
65
The denominator of the right-hand side of this equation has a similar
form to the expression for electrical resistance in the equation
R = p·
P
1
α
The similarity becomes plainer when we remember that the conductivity
is the reciprocal of the specific resistance. The permeability μ in equation (49)
can be looked upon as the magnetic conductivity and its reciprocal then
corresponds to the specific resistance p.
]
—>
μ
μ.A represents a magnetic
From this point of view the expression
resistance. It is proportional to the length and inversely proportional to
the conductivity and to the cross-section of the magnetic path. Through
this magnetic resistance we may imagine the lines of force to be driven in
much the same way that the electric current is driven through an electrical
resistance. It is because of this analogy that we speak of the total number
of lines as the magnetic flux. The force that drives this magnetic flux
through the magnetic circuit is produced by the ampere-turns X. The
ampere-turns may therefore be compared to the electromotive force which
drives the electric current through the resistance of the electric circuit. This
is, perhaps, made more evident by a consideration of the equation
H.1=0·4π. X,
which follows from equation (43) on page 60. Since H is the force on a unit
pole, the product H.1 represents the work done in moving the unit pole in
air against the lines of force over the whole distance l. This is, however, what
we considered in Section 22 under the name of magnetic potential difference.
We can therefore look upon the magnitude 0'47. X as the magnetic pressure
or force acting round the magnetic circuit, and it is for this reason called the
magnetomotive force. Equation (49) can now be written in the form:
magnetic flux =
magnetomotive force
magnetic resistance
The magnetic resistance of the whole or any part of a magnetic circuit is
generally referred to as its reluctance.
Although this Ohm's law for magnetism, as it has been called, has been,
and still is, of enormous importance in the design of electrical machinery, we
must be very careful lest we be misled by this electrical analogy. The
similarities between the electric and the magnetic circuit are, after all,
mainly only apparent. Although we speak of the total number of lines N
as a magnetic flux, it is merely a name which enables us to follow the
electrical analogy more readily; once the lines of force are produced they
remain at rest and there is no flow of magnetism around the circuit.
A great and fundamentally important difference between an electric
current and a magnetic flux lies in the fact that work must be done to
drive a current through a resistance, the energy thus expended appearing
as heat in the resistance. The maintenance of a magnetic field or of
T. E.
5
1
66
Electrical Engineering
magnetic induction involves, on the other hand, no expenditure of energy.
None of the energy expended in the field windings of a dynamo, for example,
is transformed into magnetism, but every trace of it is transformed into heat
in the resistance of the field winding, and can be calculated from the product
im Rm. The consumption of energy would be just the same if the current
remained the same, although the space within the coil were filled with air
instead of iron, thus reducing the number of lines to a very small value.
For this reason there is no direct loss of energy in a dynamo due to a
part of the magnetic flux leaking from pole to pole without passing through
the armature.
Above all, however, attention must be drawn to the fact that electrical
resistance depends simply on the length and cross-section of the conductor,
whereas magnetic reluctance is also dependent on the degree of saturation of
the iron.
We will now illustrate, by means of an example, the application of this mag-
netic Ohm's law to the calculation of a dynamo. The magnetic circuit consists
of several parts differing in length, cross-section and material (air, wrought iron,
cast iron, etc.). A part of the total magnetic flux is lost, so far as the arma-
ture is concerned, by leakage through the air, so that some parts of the
magnetic circuit carry a larger number of lines of force than other parts.
We have then to calculate the number of ampere-turns required on the poles
to produce a given magnetic flux through the armature. To do this we
divide the flux or total number of lines in each part of the magnetic circuit
by the cross-section of that part. In this way we obtain the induction or
flux density in lines per square centimetre at each point, thus
N₁ 1
B₁
A₁
.(50).
We then find from the magnetisation curve of the material of which any
part is constructed, the value of H₁, that is, the magnetising force necessary
to produce the induction B, in the material concerned. Unfortunately the
relation between B and H is no simple one, but depends in a complicated
way on the saturation of the iron; we are therefore compelled to use mag-
netisation curves, which have been found experimentally for the materials
which we are using. From H, and equation (47) on page 64 we have
X₁ = 0·8H₁. l₁
1
..(51),
where ↳ is the length of path in the part under consideration. This calcula-
tion must be made for each portion of the magnetic circuit, and the results
added together, giving
1
ΣX = X₁ + X₂ + X¸ . . . = 0·8H₂ . l½ + 0·8H₂ . l½ + 0·8H3 . lg …………..(52).
X
If the magnetisation curves are drawn with, i.e. ampere-turns per cm.
for abscissae, instead of H, the calculation for the path in iron is simpler,
while for the air-gap we still have the equation
Xg=0·8Bg. lg = 0·8Hg.lg.
.
X
We then have ΣΧ (= (4), 4+ (4) ·
•
· l₂ + ... + 0·8 Hg. lg,
2
1
Ohm's law and the magnetic circuit
67
where the numbers 1, 2, etc. refer to various parts of the iron circuit and
g refers to the air-gaps.
As an example we take a machine with the following main dimensions:
Diameter of armature.....
D
= 20 cms.,
shaft
ds
3
وو
""
""
Axial length of armature
L
20
"
Angle subtended by pole
B
= 120°,
Length of each air-gap
0.4 cm.,
Cross-section of poles and yoke ......
Am=400 sq. cms.,
Im
110 cms.
Length of path in poles and yoke...
The armature is built up of laminations stamped out of wrought iron
sheet and separated from each other by thin paper, the space lost in this
way being 15 per cent. of the whole.
We shall assume that the whole
magnet is of cast iron. As indi-
cated in Fig. 60, a part of the
magnetic flux produced in the field
magnets is lost by leakage.
shall assume that a sixth part of
the flux produced is lost in this
way, so that, if N be the flux in
the armature and Nm that in the
magnet,

Nm = 1.2. N.
We
We will now determine the
ampere-turns required to produce
a flux N in the armature of 2.5.106
lines.
We must first find the length
-Um
Fig. 60.
and cross-section of each part of the magnetic circuit.
The cross-section
of the armature perpendicular to the lines of force is found by multiplying
the difference D-d, by the axial length L, and also by 0.85 to allow for the
paper insulation. Hence,
Aa=L(D-ds). 0.85 = 290 sq. cms.
The mean value of la is evidently about 20 cms.
To find the cross-section of the path in the air-gaps we must reduce the
cylindrical surface of the armature in the ratio ẞ: 360. We have then
Ag= D.π.L.
B
360
420 sq. cms.
To find the length of the path in air we must multiply the length of the
gap between armature and pole by 2, since the lines have to cross the gap
both on entering and on leaving the armature. We have, therefore,
lg
l, = 2.0·4 = 0·8 cm.
5-2
68
Electrical Engineering
For the field magnet we have given that
7
Am
= 400 sq. cms.,
lm=
= 110 cms.
We know also that
Nm = 1·2 . N = 1·2 . 2·5 . 10° = 3 . 10″.
Bringing all the above results together, we have
N=25.10,
Aa=290, A,= 420,
la = 20, lg = 0·8,
From these figures it follows that
Nm = 3.10°,
Am = 400,
lm = 110.
N
N
m
N
Ba
8,600, Bg
=5,960, Bm=
= 7,500.
A a
Ag
Am
From the magnetisation curve for armature stampings in Fig. 58 on
page 63 we see that the ampere-turns per cm. required to produce an
induction B of 8,600 are
(†).-2
(주​)
=
α
α
This is the number of ampere-turns required to drive the induction Ba
through one centimetre of armature iron. The total number of ampere-turns
necessary to drive the flux through the armature is therefore
X
(14) · la = 40.
Xa=(²
α
a
Similarly we find that the ampere-turns per cm. for an induction
Bm7,500 in cast iron are
X
1) = 40.
m
From this it follows that to drive the flux at an induction Bm7,500
through a length lm 110 cms. of cast iron requires a number of ampere-turns
=
X
Xm= (†)__ . Im = 40. 110 = 4,400.
1 m
Finally, the number of ampere-turns X, required to drive the flux across
the air-gap is found from the equation
We have now
Xg=0·8Bg. lg = 3,820.
ΣX=Xa+Xm+Xg = 8,260.
It is immaterial whether, for example, we use 8,260 turns carrying
1 ampere, or 4,130 turns carrying 2 amperes, neglecting, for the present,
all considerations of efficiency, etc. In many cases the magnetic circuit is
more complicated than the above, including armature teeth and pole cores.
The necessary ampere-turns for these parts are, however, calculated in exactly
the same way as we have already indicated.
By means of this magnetic Ohm's law it is possible, under certain con-
ditions, to calculate the leakage from the dimensions of the machine. This
calculation makes the analogy between magnetic and electric potential
Ohm's law and the magnetic circuit
69
α
t
difference very plain. As electrical potential differences are expressed in
volts, so can magnetic potential differences be expressed in ampere-turns.
If a resistance lies in parallel with a lamp across which a certain pressure is
maintained, a part of the total current flows through the resistance. If this
parallel resistance is due to an earth, that is, to a fault in the cable, the
current flowing by this by-path is a loss. Similarly, there is a magnetic
potential difference or pressure between the pole-shoes of a machine, which
drives the magnetic flux along two parallel paths, the useful flux going
through the armature and the leakage flux through the air. This magnetic
pressure is due to the ampere-turns Xa+X+Xg, which are necessary to
drive the useful flux through the armature core, the teeth and the air-gaps.
In the multipolar alternator shown in Fig. 61, assume that the sum
Xa+Xt+Xg= 6,200. This same magnetic pressure acts across the air
space between two adjacent pole-
shoes. Assume that the axial length
of the armature is 30 cms., the radial
depth of the pole-shoe 2.5 cms., and
the distance lz, between the adjacent
pole-shoes 6.5 cms. The cross-section
of the leakage path between the pole-
shoes is, therefore, 30.2575 sq.
cms. Now, the lines of force pass
from the north pole to the south
pole on each side of it, so that the total cross-section is doubled, making it
150 sq. cms. Assuming that the lines of force spread out as shown in the
figure, and that their maximum cross-section is double that at the ends, we
get for a mean value of the cross-section

A₁₁ =
Al
150+300
2
225 sq. cms.
Fig. 61.
N
Putting these values of , and A, in equation (49) on page 64, and
remembering that
μ for air is 1, we get for the leakage from the pole-shoe
0'4π. 6,200
Nu
0.27.106 lines.
6.5
225
04π.S.i
The fundamental formula H=
would have led to the same
result.
To this leakage from the pole-shoe we must add the leakage from the
pole-core. We shall assume that the mean distance between two adjacent
pole-core faces is 10 cms., and the radial length of the pole-core 16 cms.
With an axial length of 30 cms. the cross-section of this leakage flux taken
on each side of the pole is
A₁ = 2.30.16=960 sq. cms.
At the top of the pole the magnetic potential difference is due to
70
Electrical Engineering
6,200 ampere-turns while at the root it is 0. As a mean we can therefore
take 3,100 ampere-turns, which gives a leakage flux of
0·4πX
0·4π. 3,100
Ny=
10
Ab₂
960
0.37.10º.
The total leakage flux from the pole is therefore
N=N+N= 027 .10 +037.10%=064.10
We have neglected the leakage between the flanks or end surfaces of the
poles, which would make the total leakage slightly more than the above
value.
If we assume that the useful flux through the armature is 3.10 lines
per pole, the flux at the root of the pole will be
Nm=N+N₁ = 3.64. 106,
which gives a leakage coefficient of
Nm
3.64.106
λ=
= 1·2.
N 3.10€
30. The lifting power of an electromagnet.
Although, as pointed out in the last section, no work is done in main-
taining a magnetic field, work must be done to establish the field. In
Fig. 62 is shown an iron ring which we shall suppose to be uniformly wound
over its whole periphery with S turns. Some of these turns are shown,

Ն
Fig. 62.
b

'H'
Fig. 63.
on the right in elevation, on the left in section. By the dotted lines of force
we do not mean to suggest that a part of the flux produced leaks across
the interior space, although, were the ring wound as shown in the figure
this would be the case. The dotted lines show how the lines of force, due
to the establishment of a current in the turns shown in section, spread out,
commencing as small circles round the individual wires and then combining
with the lines due to other wires to form lines encircling several wires. This
spreading out goes on until each line of force lies entirely in the iron, it
having cut all the S wires passing through the interior of the ring. As
soon as the current ceases to grow and becomes steady, the lines of force
become stationary.
The lifting power of an electromagnet
71
In order to calculate the work done in thus establishing the magnetic
field, we will consider a straight wire of length 7 cms. carrying a current
of I absolute units, in a field of strength H perpendicular to the wire
(Fig. 63). The direction of the field is assumed to be away from the reader
and perpendicular to the paper, as indicated by the dots. We will assume
that the conductor is moved a distance s in a direction at right angles both
to its own length and to the direction of the field, the direction of motion
being such that the force exercised by the field on the conductor has to be
overcome. The work done is found by multiplying the force in dynes by
the distance in centimetres, thus
J=f.s=H.I.l.s ergs.
Now l.s is the area swept out by the motion of the wire, and as H is the
number of lines of force per sq. cm., H. l.s is equal to the total number of
lines N cut by the wire; hence
J=H.I.l.s=N. I ergs
..(53).
Now, if the current I in the case represented in Fig. 62 is increased
by an amount dĨ, the induction or flux density increases by an amount dB,
no matter whether the ring contains iron or air. If the cross-sectional area
of the ring be A sq. cms., the number of newly created lines will be
A.dB=dN. These newly created lines cut, as we have seen, the S turns
composing the ring winding, which carries a current I. The work done in
this change, according to equation (53), is
dJ = S. dN. I = S. dB. A. I ergs.
From equation (42) on page 60, we have
S.I
H.l
477
and substituting this in the above equation, we have
A.l
dJ
.H. dB ergs
477
..(54).
If the coil be wound on a ring of wood or other non-magnetic material,
B = H, and the total amount of work done in raising the magnetic induction
from 0 to B will be
BA.l
J==
4π
0
.B.dB
B2. A.l
8π
ergs
.(55).
This amount of energy has to be given to the coil, as we shall see in
Section 35, in the form of electrical energy. The volume of the magnetic
field is A. l, so that each cubic centimetre of the magnetic field contains
B2
8π
a store of potential energy equal to ergs. We may picture the ether
in a state of strain, somewhat like a twisted steel spring.
We will now make use of this result, in order to find the lifting power
of an electromagnet. Fig. 64 represents an iron ring, uniformly wound so
as to obviate magnetic leakage, and divided into two similar halves. We
have to calculate the force required to separate the two halves of the ring.
..
72
Electrical Engineering
Let this force be ƒ dynes, and imagine that the two halves have been
separated by an infinitely small distance dl. If this distance be small
enough, we can neglect the change of flux
produced by introducing this air-gap into
the magnetic circuit, and the flux-density
B is unaltered. The work done in causing
this separation is equal to f. dl ergs, and,
since the flux-density in the iron is un-
altered, the energy stored in the magnetic
field in the iron is unchanged, so that this
supply of energy f. dl must be stored in
the air-gap. Now, let the total cross-sec-
tion of the two air-gaps be A sq. cms., so
that the volume of the magnetic field in air

Fig. 64.
is equal to A. dl cubic centimetres. The flux-density B is the same in the
air-gap as in the iron, and we have found above that the energy stored
in each cubic centimetre of the field in air is equal to ergs.
B2
8π
Hence:
B2
energy stored in air-gaps
A .dl ergs,
8πT
and this must be equal to the work f. dl done in producing the air-gaps, or
and
B². A
f.dl=
dl,
8π
B². A
f=
dynes
8π
F
B2. A
8π.981,000
4B². A. 10-8 kgs.
If F be the force in kilogrammes,
.(56).
.(57).
In a horse-shoe magnet both poles are effective, and the double cross-
section must therefore be included in A. If, for example, the cross-section of
each pole is 10 sq. cms. and the induction B = 18,000, we have
F= 4. 18,0002.2. 10.10-260 kilogrammes.
The lifting power of a magnet is often largely affected by magnetic
leakage, which we have here neglected.
31. Hysteresis.
In determining the magnetisation curve, we start with the iron in an
unmagnetic state and gradually increase the magnetising current, thus
increasing the value of H, and with it the value of the induction B. Plotting
0.4π. S. i
the values thus obtained with the magnetising forces H =
1
abscissae and the inductions B as ordinates, we obtain the curve OA in
Fig. 65.
as
If, after reaching any arbitrary value of the induction, such as AG = Bmax;
Hysteresis
73
B
B
we decrease the magnetising current, we find that, for any given value of the
magnetising force H, the induction is greater than it was with increasing
magnetisation. Plotting these values of B and H for decreasing values of H
we obtain the curve AB. It appears
as if a certain amount of the mag-
netic induction still remains in the
iron after the magnetising force has
been removed. This phenomenon
is therefore known as hysteresis,
i.e. a lagging behind. For a mag-
netising force H=0 the value of
the induction B is OB. This is
therefore the induction or flux
density of the remanent magnet-
ism. Hysteresis and remanent
magnetism are therefore, in a
sense, one and the same thing.
We explain both phenomena by the tendency of the molecules, in virtue of a
kind of molecular friction, to remain in any position they have once taken up.

D
Fig. 65.
&
H
If, now, the direction of the current be reversed, a certain value of the
magnetising force will be required before the remanent magnetism is
destroyed. This force is represented in Fig. 65 by the length OC, and is
called the coercive force. It is evident that the iron molecules oppose the
reversal of magnetism, and it is only when the magnetising force exceeds OC
that an induction is obtained in the reverse direction. A further increase in
the magnetising current gives the curve CD and the succeeding decrease in
the current, the curve DE. The ordinate OE is, as before, the remanent
magnetism, etc.
An experiment with the specimen of cast steel already mentioned
(Fig. 56) gave the following results :
Decreasing Induction
H
B
145.3
18,250
62.7
16,800
24.2
15,590
3.2
13,080
0
10,200
1.25
1.5
6,110
0
Increasing Induction
H
B
2.2
- 6,240
5.9
- 11,060
11.9
— 13,460
34.2
- 15,710
- 61.6
- 16,680
- 145.3
- 18,250
We see from this table that the remanent magnetism has a value
B = 10,200, while the coercive force is 15. Both of these figures are
dependent on the value of Bmax reached in the experiment, in the above
case 18,250; for high values of the induction, however, the variation is very
small.
If the above values be plotted on squared paper we get a curve like
ABCD in Fig. 65. The other side of the loop DEA can be drawn symmetrical
74
Electrical Engineering
with the first side. If the area of the loop is now measured, the abscissae
being taken to the scale of H, and the ordinates to the scale of B, it is found
to be 170,000. It can now be shown that this area is closely related to the
amount of energy dissipated during the complete cycle through the inter-
molecular friction. From equation (54) on page 71 we know that the work
done when the induction B is increased by the amount dB is
dJ
A.l
4π
•
H.dB.
A. l is equal to the volume V of the iron ring in c.cms. If we divide by
V and integrate between the limits 0 and B we obtain the work done per
cubic centimetre, thus
J
1 B
ΣH.dB
√
4π O
.(58).
Now H.dB is represented by one of the narrow strips in Fig. 66, and
ΣH. dB by the whole shaded area in the same figure. The work done in



B
B
B
B
Fig. 66.
H
H
H
H
Fig. 67.
Fig. 68.
Fig. 69.
magnetising the iron is thus obtained by dividing the shaded area in Fig. 66
by 4π. If, now, the lines of force decrease, dB is negative and the work
done is also negative, that is, energy is restored. The amount of energy thus
restored is represented by the shaded area in Fig. 67. If the magnetising
current be now reversed, the magnetising force H becomes negative, and the
product H. dB will again be positive, so that work must be done during
this period corresponding to the shaded area in Fig. 68. With decreasing
magnetisation a part of the energy is again restored as shown in Fig. 69.
The actual energy dissipated in a complete period is the resultant of the
positive and negative amounts of energy. If we go continually through the
cycle represented by the loop ABCDEA in Fig. 65, the energy dissipated
per cubic centimetre of iron in each cycle is, from equation (58), equal to
the area of the loop divided by 47. For the specimen of cast steel above
mentioned the area was 170,000, whence the loss per cubic centimetre per
cycle is
A ΣH.dB
477
170,000
12.5
= 13,600 ergs per cubic centimetre.
Hysteresis
75
Although this method of finding the hysteresis loss by drawing and
measuring the loop is of great importance in enabling us to understand the
question of hysteresis, it is rarely adopted in practice on account of the time
required to determine experimentally with a ballistic galvanometer the points
on the hysteresis loop. It is now a rule to test the iron, as far as possible,
under the conditions obtaining in actual practice. The magnetism is sub-
jected to rapid reversals by means of an alternating current, and the energy
consumed is measured on a wattmeter, the deflection of which is caused by
the electrodynamic forces between a coil carrying the current and another
coil connected across the mains and therefore carrying a current proportional
to the potential difference between the mains.
It was found by Steinmetz that the hysteresis loss per cubic centimetre
per cycle is approximately proportional to the 16th power of the maximum
induction.
J
nh.
B16
max ergs.
...(59).
The coefficient is practically constant for a given specimen of iron, at
least up to Bmax = 7,000. For the types of iron generally used it varies from
0.001 to 0.004. For the specimen of cast steel which we have considered, we
found for Bmax = 18,250 a value of
J
13,600, whence
J
Nh
V. Bmax
13,600
= 0.0028.
18,25016
This value is, however, of little interest in the present case, as hysteresis
need only be considered when dealing with alternating currents, and massive
iron or steel is never used where it would be subjected to a rapidly alternating
magnetisation.
Knowing 7, Steinmetz's formula enables us to calculate the loss of power
in watts due to hysteresis. If W is the weight of iron in kilogrammes,
assuming a specific gravity of 77, we have for the volume in cubic centimetres,
W. 1,000
V =
7.7
If the complete cycle is gone through
dissipated will be
~ times per second* the energy
2.
1'6
W. 1,000
max 7.7
❤ergs per second.
ባ
Now, in Section 41 we shall see that
1 erg= 1 centimetre-dyne
1
1
100981,000
metre-kilogramme,
wherefore (see page 21),
1
1
1 erg per sec. =
9.81.107
met.-kgs. per sec. = watt.
107
* The sign ~, roughly symbolising a sine-curve, is often used to denote the frequency in
cycles per second.
76
Electrical Engineering
To obtain the loss in watts we have therefore to divide the loss in ergs
per second by 107, hence
Ph
B16
Nh
max
=
W . ~ . 10-4
7.7
watts
..(60).
If, for example, r = 0·002, W= 100 kgs., ~=50 and Bmax = 7,000, we have
0.002. 7,00016. 100.50.10-4
Ph=
7.7
184 watts.
Lately, however, Steinmetz's coefficient ŋ has been used much less than
formerly. When, as is now usual in practice, the hysteresis loss is determined
by the wattmeter method, using alternating current, a new difficulty crops up,
for the reading on the wattmeter includes not only the loss due to hysteresis
but another loss due to the so-called Foucault currents or eddy currents. In
Germany the loss is now always specified as being so many watts per kilo-
gramme at a frequency of 50 ~ per second, the maximum induction being
10,000. This loss-coefficient, as it is called, lies between 3 and 4 for the
iron generally used for stampings.
32. Dynamic effect of parallel currents.
Two parallel wires are shown in Fig. 70, carrying currents in the same
direction. Looking along the wires from A and B we shall see the feathers
of the arrows, which are represented by crosses in the cross-sections of the
wires. The lines of force due to the individual wires combine, as we have
already seen in Section 27, to form longer lines encircling both wires. We
A
X
B
(X)
Fig. 70.
have already likened the lines of force to stretched elastic bands, and have
calculated, for one special case, the force with which they oppose any attempt
to lengthen them. The lines will accordingly tend to contract in the case
shown in Fig. 70 and thereby draw the conductors together. The attraction
will be the greater because of the lines of force between the wires being in
opposite directions, the resultant field vanishing at a point between the
wires. We arrive therefore at the conclusion that: Parallel wires attract
each other when carrying currents in the same direction.
If, however, the currents flow in opposite directions (Fig. 71), we shall
see from A the point, and from B the feather of the arrows representing the
direction of current. Drawing the lines of force due to each wire separately,
we see that they are in the same direction between the wires and can there-
fore be added, whereas, in the space outside, they counteract each other.
Dynamic effect of parallel currents
77
We have seen that lines of force in the same direction exert a lateral pressure,
tending to mutually repel one another, and in the present case this will lead
to the wires being mutually repelled. Hence, parallel wires repel each
other when carrying current in opposite directions.
-A
Fig. 71.
In each case we observe that the tendency is to increase the total number
of lines produced.
The application of this dynamic effect to the construction of wattmeters
was mentioned in the previous section. It is also employed in the con-
struction of ammeters and voltmeters of the so-called dynamometer type.
33. Induced Electromotive Force.
Imagine a conductor of length 7 cms. perpendicular to the lines of force
to be moved in a direction perpendicular both to the field of strength H,
and to its own length. If the distance moved through in the time dt be ds,
the velocity is
ds
dt
cms. per second.
In Fig. 72 the lines of force pass normally through the paper from front
to back, and are represented by points. The conductor 7 is moved along the
metallic rails in the plane of the paper and it is
found experimentally that an electromotive force
is induced in it. This electromotive force only
exists while is in motion and vanishes directly
7 comes to rest. If the slide-rails are metallically
connected, as shown in the figure, then the elec-
tromotive force produces a current, the direction
of which is indicated by the dotted arrow, when
7 is moved downwards. During the time dt a
certain amount of work is done by the electric
current, which, from Section 10, is equal to the product of electromotive
force, current, and time. We will now express the current in absolute units
and the work also in absolute units, that is, in centimetre-dynes or ergs. We
have therefore

dJE.I.dt ergs,
Fig. 72.
where E is the electromotive force in some units not yet defined, but certainly
not volts. This work dJ is, naturally, only obtained by the expenditure of an
78
Electrical Engineering
equal amount of mechanical work. We know from page 53 that the current
in l, and the magnetic field, exert a certain force ƒ, which must be overcome
when moving the conductor. It follows from Lenz's law that the induced
current is always in such a direction that it opposes the motion inducing it.
For the magnitude of this opposing force, we have from equation (32) on
page 53,
f=H.I.l dynes.
The mechanical work is the product of the force and the distance,
so that
dJ=H.I.l.ds ergs.
Equating the mechanical energy expended and the electrical energy
developed, we have
whence
T
*
H.I.l.ds,
E=H.l.
ds
dt
H.l.v
.(61).
Hence, the electromotive force induced in a conductor by its movement
in a magnetic field is proportional to the field strength, to the length of the
conductor, and to the velocity with which it moves. From equation (61) we
can obtain the definition of the absolute unit of electromotive force. If we
put H=1,l=1 and v= 1, then we get E-1. Hence, the absolute unit of
electromotive force is induced when a conductor 1 cm. long moves with
a velocity of 1 cm. per second perpendicular to its length and to the
field of unit strength.
We are now in a position to calculate in a simple manner the electro-
motive force of a machine. Suppose, for example, we require to find the
maximum value of the E. M. F. of an alternator, that is, the E. M. F. at the
moment when the armature wires are under the middle of the poles. Let
the total number of armature wires be 400, each having an active induced
length of 30 cms. and let the field strength at the middle of the pole be
5,000. Then, at a peripheral speed of 20 metres per second, i.e. 2,000 cms.
per second, the electromotive force will be
E = 5,000.30.400. 2,000 =1,200. 10° absolute units.
The absolute unit is far too small for practical purposes and a multiple of
it is therefore used, viz. 108 absolute units, and is called a volt. This is the
practical unit of pressure which was introduced in Section 2, but could not
then be accurately defined. If E represents the electromotive force in volts,
we have
E=H.l.v.10-8
.(62).
In the example calculated above the momentary electromotive force was
therefore 1,200 volts.
If, instead of being at right angles, the conductor is at an angle to the
lines of force, it follows from equation (31) on page 53 that
E = H.l.v.sin &. 10–8
.(63).
If, further, the motion is not perpendicular to the field, but at some angle,
then v must be put equal to the component at right angles to the field.
Induced Electromotive Force
79
1.
The equations which we have thus established can be still further
simplified. The product .ds in equation (61) represents the area swept
out by the conductor, and the product H.l.dS the total number of lines dN
cut by the moving conductor. We write dN because we are dealing with an
infinitesimal distance ds, the number of lines cut being therefore also infinitely
small. For a single conductor, we get from equation (61)
E
dN
dt
…..(64).
The electromotive force is thus obtained in absolute units by dividing the
number of lines cut by the time taken to cut them, or, in other words, the
electromotive force is equal to the rate at which the lines are cut.
If, in the time t, the flux N cuts S turns, it is equivalent to NS lines
cutting a single wire, and the average value of the induced E. M. F. in volts is
E =
N.S
t
.10-8
The product of lines and turns linked by the lines is sometimes called the
number of linkages. In the above example the linkages are N.S. The
electromotive force is therefore equal to the rate at which linkages are made
or broken.
The absolute unit of electromotive force is induced when one line of
force is cut per second. One volt is induced by the cutting of 108 lines
per second.
Ն
Having determined the magnitude of the induced E. M. F. we must now
consider its direction. This can be determined by means of a swimming rule,
due to Faraday, which is as follows:
To a person swimming along the
lines of force, that is, from north
to south pole, and looking in the
direction along which the conduc-
tor is moving, the induced electro-
motive force is from left to right.
In Fig. 73 the lines of force are in the
plane of the paper from right to left.
They pass normally through a vertical
plane in which the horizontal conduc-
tor moves downwards. If we swim
from right to left with the face down-
wards our right hand will point in the
direction of the dotted arrow, which is therefore the direction of the induced
electromotive force. That this direction for the induced current is necessarily
correct follows from a comparison of the two swimming rules of Ampère and
Faraday respectively. If we swim with the current along the dotted arrow
and look towards the north pole whence the lines of force proceed, the north
pole would, according to Ampère's rule, tend to move towards our left, that
is, downwards. If the pole is fixed the conductor tends to move upwards.
Hence, the conductor tends to oppose the downward movement, as indeed it

Fig. 73.
80
Electrical Engineering
must, to conform to Lenz's law or to the principle of the conservation of energy.
The direction found for the induced E. M. F. is therefore the correct one.
We can apply this rule directly to the armatures of dynamos. In Fig. 74
is shown a hollow iron cylindrical or ring armature, with a closed winding of
insulated copper wire, which is rotated in a clockwise direction between the
poles N and S of an electromagnet. The lines of force leave the north pole
and pass right and left through the armature to the south pole, leaving the
air space inside the armature almost free from lines of force. It is therefore
only the wires on the outside cylindrical surface that cut the lines of force.
Applying Faraday's swimming rule we find that in the wires under the north
pole the induced electromotive force is from front to back, while in those

N

A
N
+
H
S
Fig. 74.
Fig. 75.
under the south pole it is from back to front. When current flows in the
armature its direction will be as indicated by the arrows, viz. away from the
south pole towards the north in the end connections facing us in the figure.
We see that the current flows to the right both in the upper and lower half
of the armature, meeting at a point halfway between the poles, where the
currents from each half join and leave the armature together by means of the
positive brush. After flowing through the external circuit the current enters.
the armature again by the negative brush, and divides between the upper
and lower halves of the winding. For the sake of simplicity the brushes are
shown rubbing on the external armature wires, from which the insulation
would then have to be removed.
If the armature remains stationary while the poles rotate, the swimming
rule must be applied to the relative motion. If, for example, in Fig. 75,
which represents the usual arrangement of alternators, i. e. dynamos which
generate alternating current, the north pole moves to the right, the effect is
the same as if the pole were fixed and the armature wires moved to the left.
If we imagine ourselves to be swimming with the field, that is, from N to A
and looking in the direction of the relative motion of the conductors, that is,
towards the left, the direction of the induced E. M. F. will be indicated by our
right hand, that is, downwards through the paper. The current in the wires
Induced Electromotive Force
81
at A will therefore be represented by crosses, while in a similar manner we
find that the current in the wires under the south pole is coming towards us
and must be represented by dots.
Another simple rule for finding the direction of the induced E. M. F. is the
right-hand rule. If the thumb and first two fingers of the right hand point
mutually at right angles, like the three edges at a corner of a cube, and the
first finger point along the lines of force, while the thumb points in the
direction of motion, the second finger will give the direction of the induced
electromotive force.
We must be careful to notice, however, that, although we speak some-
times of the direction of the induced current, it is electromotive force and not
current which is induced. In Fig. 61, for example, electromotive force was
induced in even when its ends were not metallically connected, and this
electromotive force could be shown and measured on an electroscope. We
must, for this reason, accustom ourselves to think of the electromotive force
as the thing induced, its magnitude depending, as we have seen, on the field,
on the length of conductor, and on its velocity, whereas the current is quite
an arbitrary thing depending on the external resistance, that is, on the load
taken by the customers of a station. The fact that the E. M. F. of a dynamo is
not constant, but varies because of the effect of the current on the magnetic
field, need not prevent us from seeing in the electromotive force the cause,
and in the current, the effect.
In the foregoing, we have always looked upon the electromotive force as
being caused by the cutting of lines of force. It is, however, often convenient
to look upon the phenomenon from another point of view, according to which
electromotive force is due to an increase or a decrease in the number of lines
of force passing through a coil. In Fig. 72, for example, the number of lines
passing through the coil, made up of the conductor l and the slide-rails, is
decreased by the motion of l. A current is thereby induced which flows
round the circuit in a clockwise direction. This current. produces lines of
force which pass down through the paper inside the circuit, and which are
therefore in the same direction as those of the original field. From this it
follows that, if the number of lines through a circuit be decreased, a
current is induced which tends to keep up the number of lines, whereas,
if the lines be increased, a current is induced which tends to weaken
the field. Lenz's law may thus be expressed in the following general
form: The current induced by a change in the magnetic field is such as
to oppose the change.
For this reason, equation (64) must be re-written thus:
E
dN
dt
...(65).
dN represents an increase in the number of lines. If this is positive, the
electromotive force will be negative, that is, it will produce a current in such
a direction as to weaken the field. If the field passes through more than one
turn of the coil or circuit, N will represent the number of linkages and dN
the change in the number of linkages.
T. E.
6
82
Electrical Engineering
34. The Laws of Mutual Induction.
So far, we have considered the electromotive force induced when a con-
ductor cuts the lines of force produced by a magnet. We will now consider
the effect of one conductor cutting the lines of force produced by another
conductor. The induction in such a case is called mutual induction. In the
primary wire I in Fig. 76 a current flows from right to left, producing lines
of force as shown in the figure, viz. coming out from the paper at A, B and C.
If the secondary conductor II is moved towards I, it will cut the lines of
force at A, B, C, etc. and an electromotive force will be induced in it, the
direction of which is found by the swimming rule or by the right-hand rule
to be from left to right as shown by the dotted arrow.

0 0 0
A
B
Fig. 76.
I
ㅍ
​Here again we can check our result by means of Lenz's law, which is only
an application of the conservation of energy. If the ends of conductor II
are externally joined by means of a conductor, a current will flow in the
opposite direction to that in I. We have seen in Section 32 that the con-
ductors will in this case repel one another, and therefore oppose the motion.
of II which is towards I. The work done in the latter case in forcibly
bringing the wires nearer appears as heat in the secondary wire. We come
therefore to the conclusion that, on decreasing the distance between the
wires, the induced secondary current is opposed to the primary,
whereas on increasing the distance the secondary current is in the
same direction as the primary current.
It is, however, not necessary to move the wires mechanically nearer
together, for an electromotive force is induced in the secondary whenever
the primary current is increased or decreased, since its lines of force must.
thereby cut through the secondary conductor.
We are thus led to the most important conception, already used in
Section 30, according to which the lines of force do not suddenly appear
or disappear throughout a space, but, leaving the conductor, gradually
spread out from it, to be followed by others, in much the same way as the
ripples spread out from the point where a stone has been thrown into a lake.
As the current in I increases the lines of force will be represented successively
by Figs. 77 and 78, in which we see that they cut the secondary conductor in
a downward direction. The result is the same as if the field were stationary
and the conductor II were moved upwards towards I. From this it follows
that the induced secondary current is opposed to the growing primary
current, while it is in the same direction as the decreasing primary
current.
The Laws of Mutual Induction
83
It is possible to utilise this mutual induction for the transformation of
direct into alternating current. Fig. 79 represents diagrammatically a so-
called induction coil. It consists of a primary winding I supplied with
direct current, one end of the winding being joined directly to the battery,
while the other end is connected to the fulcrum D of the spring J. The
circuit is completed by means of a contact point touching the spring and
connected to the other pole of the battery. Directly the circuit is closed
the coil acts as an electromagnet and attracts a piece of iron attached to the

I
I
Л
I
I
I
н


Fig. 77.
Fig. 78.
Fig. 79.
end of the spring, and thereby pulls the spring away from the contact point.
The current being now broken, the coil loses its magnetism and allows the
spring to go back and close the circuit once more. The current in the
primary coil is thus intermittent. Wound over this primary coil is a
secondary coil, which, for the sake of clearness, is shown beside the primary
coil in Fig. 79. The primary lines of force grow and die away again in rapid
succession, and thus cut through the secondary windings first in one direction
and then in the other, thereby inducing in them an electromotive force of
rapidly alternating direction. If the secondary terminals are connected by
means of a conductor, an alternating current will flow in the secondary
circuit. This current increases from zero to a maximum, dies away again
to zero and then reverses and goes through the same changes in the other
direction. The current changes therefore both in magnitude and in direction.
By increasing the number of secondary turns and making the primary
current break as rapidly as possible, the secondary electromotive force can
be made so large that the spark will jump across a considerable air-gap. In
this way induction coils are now made to give a spark through the air over a
yard long.
Of even greater importance from a technical point of view is the applica-
tion of this principle of mutual induction to the construction of transformers
which transform the high pressure alternating current transmitted over the
line down to alternating current of a low pressure suitable for ordinary use.
We have seen that a very high pressure is essential for long-distance trans-
mission in order to keep the losses small, or, if the losses are fixed, in order
to keep the size of wire within practicable limits. To use such high pressures
a great number of lamps would have to be put in series, in which case we
would lose the independence of the separate lamps. In addition to this, the
insulation for such high pressures would be extremely difficult, and pressures
above 500 volts are very likely to give a fatal shock to a person touching the
6-2

84
Electrical Engineering
live wires. From the results of experiments it is known that the human
body can, generally speaking, stand a current of about a hundredth of an
ampere for a short time without any serious damage. If the resistance of
the body, when contact is lightly made with the finger-tips, be estimated at
50,000 ohms, we see that the dangerous pressure begins at about 500 volts,
thus,
e=i. R=0.50,000 = 500 volts.
We see therefore that it is necessary to reduce the pressure to a value
more suitable for the consumer, and this reduction of pressure is accomplished
very simply by means of an alternating current
transformer (Fig. 80). It consists of an iron core
built up of insulated stampings, which carries a
primary and a secondary winding. In our case the
transformer is a step-down transformer, and the
primary winding will therefore be connected to
the high pressure supply. The primary winding
must, for this reason, consist of a great number
of turns of fine wire. The lines of force produced
by this winding will cut the secondary winding,
which must consist of a few turns of thick wire, so
that the electromotive force induced in it may be
small. Now, the lines of force produced by the
primary winding will, as they alternately grow and die away, cut the primary
turns themselves, and induce in them an electromotive force proportional to
the number of primary turns. We shall see later that this E.M.F. is almost
exactly equal and opposite to that applied to its terminals. We see, there-
fore, that the primary and secondary pressures are proportional to the number
of turns in the corresponding windings. The principles of the transformer
are very easy to follow until current is taken from the secondary side, when
the question becomes more complicated owing to the lines of force being now
due to the combined action of both primary and secondary currents. We shall
postpone a full investigation of the transformer until Chapter XI.
35. Self-Induction.
Fig. 80.
Towards the end of the previous section it was mentioned that the lines
of force produced by a coil, in growing and dying away, cut, not only neigh-
bouring wires, but also the turns of the coil itself. In Fig. 81 the dotted
lines represent the growth of the lines of force due to a single turn of the
coil. Each line of force emerges from the wire of the turn to which it is due,
and gradually grows until it finally lies entirely in the iron ring, having cut,
on its way, every turn in the coil.
turn in the coil. We will now calculate the magnitude of
the electromotive force induced in the coil in this way.
and
Let
S be the number of turns in the coil,
A, the cross-section of the iron,
1, the length of the magnetic path in the ring.
Self-Induction
85
We will assume that the permeability μ is constant, which is approxi-
mately true for low values of the induction, for which B is roughly propor-
tional to H. If, now, the current

of I absolute units increases by an
amount dI in the time dt, the in-
crease in the number of lines of
force per sq. cm. will be (see equa-
tion (42) on page 60)
4π.S.dl.μ
dB =
1
and the increase in the total flux
will be
HHH
4π.S.dI.μA.
dN
7
Fig. 81.
These lines cut S turns in the time dt. The electromotive force of
self-induction thereby produced was seen in Section 33 to oppose
whatever change of current produced it. Since dN represents, as before,
an increase in the number of lines, we must put a negative sign before the
right-hand side of the E.M.F. equation. Substituting this value of dN in
equation (65) on page 81,
we have
E =
E = – S.
dN
dt
>
4π. S².μ. A dI
し
​absolute units
dt
....(66).
If, now, the current is expressed in amperes, the right-hand side becomes
10 times too big, and must therefore be multiplied by 10-¹. To get the
electromotive force in volts the right-hand side must be further multiplied
by 10-8. We get, then,
E=
4π. S².µ. A
10-9 volts
7
di
dt
...(67).
The expression
4π. S².µ. A
.10- gives us the coefficient of self-
induction of the coil in practical units, that is, in henries. Denoting this
by the letter L, we have
E=-L volts
di
dt
The coefficient of self-induction can also be expressed thus:
0.4π. S.μ. A
1
.S.10-8
..(68).
..(68a).
0·4π. S.μ. A
Now
is the flux produced by a current of one ampere, and
multiplying this by S gives the total number of linkages produced by one
ampere. It follows from this that the coefficient of self-induction in
henries is the number of linkages produced by a current of 1 ampere,
multiplied by 10-8.
The henry is thus the unit of self-induction and that coil has unit coeffi-
86
Electrical Engineering
cient of self-induction for which the above expression works out to 1. The
henry is a practical unit and can be introduced directly into calculations
involving amperes and volts.
From equation (68) it is seen that the electromotive force of self-induction
depends not only on the construction of the coil, but also on the rate at
which the current is varying.
The expression
0·4π.μ. A
represents the lines produced by one ampere-
turn. This can be estimated in many cases, e.g. dynamo armatures, with
sufficient accuracy. The flux produced by a current i, flowing in S turns,
will be
N
0.4π. μ. A
.S.i.
If, now, the current in a coil is changed from +i toi in a time T, due
to the commutator segments to which it is connected passing under a brush,
the number of linkages cut in the time T will be 2N. S. The average value
of the electromotive force of self-induction will therefore be
Es mean
2NS
T
10-8=2.
0·4π.µ. А. §¹². i
1. T
.10-8
04π.μ. Α
A
or since
L
S2.10-8,
Es mean
2Li
T·
To find L we have simply to multiply the flux per ampere-turn, which, as
we have already said, can often be estimated from past experience, by the
square of the number of turns and by 10-8.
The effect of self-induction is seen in the gradual growth of current in
a coil to which a certain fixed P.D. is applied. It also causes the current to
die away slowly when the circuit is broken. It acts therefore as a sort of
inertia, opposing any change in the current. Its effect is very considerable
when the circuit of an electromagnet is suddenly broken, for the large
number of lines in the magnet suddenly collapses, cutting all the turns of the
magnet-winding at a high speed. Hence, in the equation
E
N.S
t
. 10-8 volts,
which we established on page 79, both N and S are large, while t is very
small. The electromotive force may be so great in such a case that the
insulation of the coil is punctured. In any case, the spark produced on
breaking such a circuit is very vicious owing to the electromotive force of
self-induction trying to keep the current flowing, even across the air-gap
of the switch.
For these reasons the field current of dynamos and motors is often greatly
reduced by inserting resistance before breaking the circuit. In addition to
this a resistance R (Fig. 82) is often put in parallel with the field winding
Rm. When the circuit is broken, the current in the electromagnet, and the
Self-Induction
87
field, do not die away instantly, but gradually. As the field dies away it
induces an E.M.F. of self-induction in the magnet winding, in the same
direction as the decreasing current. This E.M. F. drives the current round the
circuit made up of R and Rm, and the current gradually falls to zero.
Ꭱ


Fig. 82.
Rm
Fig. 83.
It is sometimes necessary to make a coil which shall have no self-
induction. In one arrangement of the Wheatstone bridge the current is
supplied from the secondary of an induction coil, while the galvanometer
is replaced by a telephone receiver. The resistance to be measured must.
contain no self-induction, since the readings would then be dependent on
the self-induction as well as on the resistance. This method can therefore
only be applied to resistances having negligibly small coefficients of self-
induction, e.g. glow lamps and straight wires. The other resistances making
up the bridge must naturally also be free from self-induction. This is
arranged by doubling the wire on itself, and then winding the two parallel
halves side by side as shown in Fig. 83. The magnetic effect of one wire
is then neutralised by that of the neighbouring wire.
It was mentioned in Section 30 that, although no energy was expended
in maintaining a magnetic field, yet a certain amount of work was done in
building up the field, which remained stored in it until it collapsed, when
the energy was restored. We are now in a better position to understand
this point. We saw that the current in the ring-shaped coil was such as
to oppose the movement of the gradually widening lines of force. We
calculated the work done in overcoming this opposition, by multiplying the
number of linkages by the current, and we asked ourselves the question
as to how this energy was given to the coil. This question we can now
answer. As the current increases, the electromotive force of self-induction
is opposed to it, and must be overcome by the applied terminal pressure.
If i is the current at any moment and R is the resistance of the coil, then, at
that moment,
e=E,+i. R;
multiplying both sides by i. dt, we have
‚
e.i.dt E. i. dt +i. R. dt.
Of these three terms, e.i.dt is the total energy supplied to the coil in
the time dt, while 2. R. dt is the energy transformed into heat in over-
coming ohmic resistance. The remainder E,.i.dt is used in forcing the
lines of force through the inner wires of the ring winding. This energy
is given back in the same form, viz. electrical energy, when the current
decreases and the lines of force shrink up like a stretched elastic band.
88
Electrical Engineering.
While the field is maintained, this energy is stored up in it, like the potential
energy of a stretched spring. On breaking the circuit, the E.M. F. of self-
induction keeps the current flowing just long enough to give back this store
of energy. As a rule it is all transformed into heat, a large part of which
is often evident in the spark. If the coil contains iron, only a part of the
expended energy is given back, the remainder being converted into heat
by the hysteresis of the iron.
36. Eddy Currents.
The name of eddy or Foucault currents is given to those currents which
do not flow along a particular prescribed path, but, following the line of
least resistance, flow where they will in the material. Such currents are
induced, for example, when a massive conductor is cut by a magnetic field,
but their exact path and intensity cannot generally be accurately determined.
We know, however, that they flow, in a general way, at right angles both
to the magnetic field and to the direction of motion. If, for example, a turn
of rectangular copper strip be wound on a massive iron cylinder, as shown
in Fig. 84, and the cylinder be rotated so that the upper part in the figure
comes out from the paper, the direction of the induced E. M. F. will be as


N
S
Fig. 84.
Fig. 85.
shown by the arrow. An equal E. M.F. will, however, be induced in the iron
below the copper strip. In consequence of the large cross-section and there-
fore the small resistance of the iron, the current produced by this latter
E.M.F. will be very considerable. The heat, thus generated, will cause a
prohibitive temperature rise in the iron, besides representing a great waste
of power. The iron in Fig. 84 may be looked upon as a short-circuited
dynamo and work must be done in turning it, because the induced currents
oppose the motion to which they are due.
The energy consumed by eddy currents can be demonstrated in a very
simple manner by means of a pendulum with a bob of sheet copper, so
arranged that the latter can swing freely through the narrow gap between
the poles of an electromagnet. On closing the exciting circuit of the magnet,
the pendulum is brought rapidly to rest, as if the bob were immersed in
a viscous fluid. The energy of the pendulum is used up in overcoming
the opposing forces due to the induced eddy currents in the copper sheet,
Eddy Currents
89
and the energy is converted into heat in accordance with Joule's law. It
is evident that eddy currents represent, under all circumstances, a loss of
energy, and to avoid this loss the armatures of machines are always built up
of sheet iron stampings or discs, as shown in the lower half of Fig. 84. The
plates are insulated from one another by means of varnish or thin paper, or
sometimes by nothing more than the oxide on the surface. The iron must
evidently be laminated in a plane perpendicular to the armature conductors,
so as to break up the path along which the eddy currents would flow.
When the armature winding is placed in open slots it is often necessary
to laminate the pole-shoes, as the lines of force go mainly through the teeth,
and cross to the pole-face in tufts as shown in Fig. 151. The flux-density
in the pole-shoe is greater opposite a tooth than opposite a slot, so that,
as the armature rotates, each point of the pole-face is subjected to a rapidly
varying field. In this way, electromotive forces are induced in the pole-shoe,
necessitating its lamination.
t
Eddy currents can also occur in the copper bars or wires of the armature
winding itself; especially if these be very massive, as shown, in a very
exaggerated way, in Fig. 85, which may be regarded as a section through
Fig. 84. In the position shown, one edge of the bar is moving in a strong
field while the other edge is outside the pole and therefore in a weak field.
The electromotive force induced in the part under the pole will drive current
principally through the end connections, and thus through the external
circuit, but, at the same time, it will drive current back along the other
edge as shown by the dotted lines in the small side-view of the conductor.
These eddy currents are diminished by rounding off the edges of the poles
or by making the air-gap larger towards the edges than at the centre, thus
causing a more gradual change in the field strength. They are, however,
almost entirely avoided by using toothed armatures, in which the lines pass
almost entirely through the teeth, as shown in Fig. 151. When a tooth
leaves the pole, its tuft of lines is dragged a little way with it, and finally
snaps back to the following tooth at an enormous speed. Every part of the
conductor is cut at the same moment in a similar manner, and eddy currents
are thus prevented.
Although, in all the cases above mentioned, eddy currents were positively
harmful, there are cases in which they are turned to account with great
advantage. In one form of tramcar brake, an iron disc, keyed to the axle,
revolves between the poles of an electromagnet. When the latter is excited
the eddy currents induced in the disc oppose its motion, and the kinetic
energy of the moving car is converted into heat in the disc.
Eddy currents are also used for damping galvanometers, the needles of
which move in a hollow block of copper. The instrument is thus made
aperiodic, that is, it takes up its new position without swinging about. If
the instrument is of the moving coil type, as so many mirror galvanometers
are, the coil is rapidly brought to rest by short-circuiting it. The currents
induced in it, as it swings between the poles of the permanent magnet,
oppose the motion producing them; these currents are not eddy currents,
however, as they flow in the windings of the coil.
90
Electrical Engineering
A class of instrument which we may well consider before leaving the
question of eddy currents is the "shaded-pole" type of alternating current
measuring instrument, as made by the Allgemeine Elektricitäts Gesellschaft
of Berlin. In Figs. 86 and 87, a metal disc rotates about a centre A between
the poles M of an electromagnet. Two metal plates T are fixed to the poles
so as to cover, or shade, a portion of the pole-face. When an alternating
current flows through the coil of the magnet, the lines of force pass, first
in one direction and then in the other, through the disc and plates T. The
M
T
M
M
Fig. 86.
Fig. 87.
currents induced will flow round the metal plates and also round the lower
half of the disc in the same direction. The current flowing along the left-
hand vertical edge of the plate T in Fig. 86 will therefore be in the same.
direction as that in the left-hand lower side of the disc, and will exert an
attractive force on it. The disc will thus rotate in the direction of the arrow.
The right-hand sides of the plates T in Fig. 86 not only cover the disc,
as shown, but are also bent out away from it, thus screening the right-hand
side of the disc from forces similar to those on the left.
1


}
CHAPTER V.
37. The units of length, mass, and time in the absolute system.-38. Dimensions and units of
velocity, acceleration, and force.-39. Dimensions and units of pole-strength, field-strength,
and magnetic flux.-40. Dimensions and units of pressure, current strength, quantity of
electricity, and resistance.—41. Dimensions and units of energy, heat, and power.—42. Di-
mensions and units of the coefficient of self-induction, and of capacity.
37. The units of length, mass, and time in the absolute
system.
The unit of length in the absolute system is the centimetre. This is
defined as the hundredth part of the standard metre preserved in Paris. The
standard metre is approximately a ten-millionth part of the length of a
quadrant of the earth, taken through Paris.
The absolute unit of mass is the gramme. This is defined as the mass of
a cubic centimetre of water at 4° C., or better, as the mass which weighs as
much as 1 cubic centimetre of water. Hence, the weight of a body in
grammes gives its mass directly in absolute units. The student must be
careful to distinguish between mass and weight. The mass of a cubic centi-
metre of water represents a certain quantity which is the same at any point
on the earth's surface, or in space. Its weight, on the other hand, will vary
from point to point, and will be appreciably different at the poles to what it
is on the equator. The weight of a given mass is the force with which it is
attracted by the earth. Since this force of gravity is found experimentally to
be strictly proportional to the mass, for any given point on the earth's surface,
the mass of a body is generally determined from its weight and expressed in
the same units. We shall represent masses in grammes by the letter M.
The second is chosen as the absolute unit of time. It is defined as the
86,400th part of the mean solar day. A time expressed in seconds is repre-
sented by the letter t.
Most of the other quantities with which we shall be concerned, such as
velocity, energy, etc. can be expressed as functions, or dimensions, of length,
mass, and time. As the expression "dimension,” used in this way, is apt to
prove a difficulty to beginners, we will make its meaning plainer by means of
a few simple examples. Area or surface is the second dimension of length,
while volume is the third dimension of length. Area and volume are thus
dimensions or functions of length, that is, they are quantities which can be
calculated from measurements of length. Velocity is similarly a function or
dimension of length and time, since its magnitude can be determined by
dividing a length by a time. In this way, the majority of our physical con-
ceptions are capable of expression in terms of length (L), mass (M), and
92
Electrical Engineering
time (T). The dimensions of area, for example, are Ľ³, of volume L³, and of
velocity LT-
It follows that the units for these various quantities in the absolute
system of units can no longer be chosen at random, but are necessarily
determined from the three fundamental units, viz. the centimetre, the gramme
and the second. Thus, the unit of area is necessarily the square centimetre,
that of volume the cubic centimetre, while the unit of velocity must be one
centimetre per second. Units derived in this way from these three funda-
mental units are known as C.G.S. units from the initial letters of the three
units chosen. We proceed now to determine the dimensions of the various
quantities, together with their absolute units, and to compare the latter with
the practical units.
38. Dimensions and units of velocity, acceleration,
and force.
(a) VELOCITY.
Velocity is defined as the ratio of the distance travelled to the time taken,
or as the distance moved through in unit time. We have therefore,
L
T
Dimensions of velocity: L.T
If v be the velocity in absolute units, we have
7 cms.
t sec.
……..(69).
The absolute unit of velocity is one centimetre per second, or 1 cm./sec. The
word "per" signifies as much as "divided by," so that the term "centimetre
per second" indicates at once the dimensions of velocity and the method of
calculating the velocity from the length and the time.
Example: The armature of an alternator has a diameter of 1.6 metres and
runs at 300 revolutions per minute. What is its peripheral speed in absolute
units?
We have
circumference of armature = 1·6.π = 5 metres,
distance per minute
=
= 5.300 1,500 metres,
or, expressing length in centimetres and time in seconds,
7 = 1,500.100 150,000 cms.
The peripheral speed is therefore
t=60 seconds.
7
150,000
cms.
=2,500
= 2,500 c.G.s. units.
t
60
sec.
(b) ACCELERATION.
Acceleration is the increase of velocity in unit time, or the ratio of
increase of velocity to time:
Acceleration:
Increase of velocity
time
Dimensions and units of velocity, etc.
93
The dimensions of an increase in velocity are the same as those of
velocity, viz. LT-1. We get therefore
Dimensions of acceleration:
L.T-1
T
=L.T-2.
If v₁ be the initial velocity and v₂ the final velocity, we have for the
acceleration, if uniform, in absolute units,
α =
V2 — V1
t
C.G.S. units
....(70).
A body has unit acceleration when its speed increases in one second by
the absolute unit, i.e. one centimetre per sec.
Example: A body starts from rest and attains in 3 seconds a speed of
29.43 metres per second. What is its acceleration in absolute units?
We have
Wherefore
V₁ = 0,
V2
v₂ = 29.43.100
cms.
sec.
t = 3.
α=
t
V2 — V1 2,943
3
981
cms.
sec.2
981 C.G.S. units.
We see, at once, that this is an example of a body falling freely under the
action of gravity. The result is obtained in absolute units by putting each
individual quantity in the calculation in absolute units. It is evident, more-
over, that acceleration, whether due to gravity or not, can be expressed
neither in cms. nor in cms. per second, but only in cms. per sec. per sec.
An
acceleration can no more be expressed in units of velocity than can an area in
centimetres. Similarly a horse-power is not equal to 76 metre-kilogrammes
or 550 foot-pounds, but to 550 foot-pounds per second. Carelessness in the
expression of units leads to great confusion. Only those quantities can be
compared which have the same dimensions and an error in calculation has
often been discovered through a want of agreement between the dimensions
of the two sides of an equation. It is therefore very important to master the
principles of dimensions, and to gain that confidence in their application,
which is the result of frequent use.
(c) FORCE.
Force is defined in mechanics as the product of mass and acceleration:
Force = Mass. Acceleration.
We have therefore
Dimensions of force: M.L.T-L.M. T-2*.
* The mechanical idea of force differs from the astronomical. In its simplest form the law
of gravitation is expressed by the equation
m.m
f=
r.2
where ƒ is the force, m the mass, and r the distance. The dimensions of force in astronomical
units are therefore M². L-2. Dimensions of force in the two different systems cannot be equated,
resembling in this respect the dimensions of quantity of electricity in absolute and in electro-
static units. The difference in the dimensions of force proves that the dimensions of a quantity
are not based a priori on the nature of the quantity or, at any rate, only represent one point of
view.
94
Electrical Engineering
If ƒ be the force, and M the mass in absolute units, then
f=M.a
·(71).
Hence, the absolute unit of force causes the absolute unit of mass, that is
a cubic centimetre of water, to move with an acceleration of 1 cm./sec. This
force is called a dyne.
Example: With what force does the earth attract a 1 kilogramme
weight?
We have
Wherefore
and
mass = 1 kilogramme = 1,000 grammes,
acceleration = 9.81 met./sec. 981 cms./second.
M = 1,000,
2 =
a = 981,
ƒ = 1,000.981 = 981,000 C.G.s. units or dynes.
Thus, the kilogramme weight, which is the practical metric unit of force,
is equal to 981,000 dynes. If we represent a weight or force of 1 kilogramme
by kg.* we get
Similarly
1 kg.* =981,000 dynes.
1
1 dyne =
981,000 kg.* = 1·02 milligrammes
1 pound* = 453'6 grammes* = 445,000 dynes.
*
(72).
39. Dimensions and units of pole-strength, field-strength,
and magnetic flux.
(a) POLE-STRENGTH.
The strength of a pole, or the quantity of "free magnetism" on a pole,
is measured by the force exerted by the pole under certain conditions.
According to Coulomb's law the force exerted by one pole on another is
given by the equation
ƒ=
mr. m2
p2
where m, and m₂ are the strengths of the two poles and r is the distance
between them. If we are not concerned with numerical values, we can omit
the indices and write
f
=
m.m
p2
or
m = r√f.
Hence, to find the dimensions of pole-strength we must take the square
root of the dimensions of force and multiply the result by the dimensions of
the distance r, i.e. by a length. We have therefore
Dimensions of pole-strength: L.√L.M.T~² — Lª.M¹‚T~,
From Coulomb's law it follows that that pole has unit strength which exerts
a force of 1 dyne on a similar pole placed 1 cm. away from it. It has been
proposed to call this unit a Weber, but the proposal has not been generally
adopted.
Dimensions and units of field strength, etc.
95
(b) FIELD-STRENGTH.
The force exerted on a pole by a magnetic field is, according to equation
(26) on page 44, proportional to the strength of the pole and to the strength
of the field, or
Hence
f=m.H.
f
H=±.
H:
m
The student should accustom himself to reading equations, such as this, in
words. The field-strength H is the force per unit pole, or the force exerted
on a pole of strength 1. Its dimensions are therefore obtained by dividing
the dimensions of force by those of pole-strength. We have therefore
Dimensions of field-strength:
L.M. T-2
MT-1
=L.M. T-
That field has unit strength which exerts a force of 1 dyne on a pole of
strength 1.
Example: The force on a
field is found to be 20 dynes.
We have
north pole of 100 absolute units in a magnetic
What is the strength of the field?
H
f
20
0.2 C.G.S. units.
m
100
(c) MAGNETIC FLUX.
We have seen in Section 21 that the number of lines of force per square
centimetre is equal to the strength of the field.
The total number of lines,
or the magnetic flux, is therefore obtained by multiplying field-strength by
area :
N=H.A.
The dimensions of magnetic flux can thus be found by multiplying the
dimensions of field-strength by those of area. We have therefore
Dimensions of magnetic flux: L. L.M. T-L. M¹‚T¬¹.
It is rather striking that the dimensions of magnetic flux should be the same
as those of pole-strength. This is, however, in accordance with equation (28)
on page 46, that is, with the fact that the total magnetic flux from a pole is
found by multiplying the strength of the pole by 4π:
N = 4πm.
If the values of pole-strength and magnetic flux only differ by a constant
factor 4π, their dimensions must be the same. A line of force can be defined
as the 47th part of the flux emanating from a unit pole.
40. Dimensions and units of electromotive force, current,
quantity of electricity, and resistance.
(a) ELECTROMOTIVE FORCE.
Electromotive force can be defined as the number of lines, or the magnetic
flux, cut per second. We have then
Dimensions of electromotive force: L.M. T-2.
96
Electrical Engineering
The absolute unit of electromotive force is induced when one line of force is
cut per second. The volt is 108 absolute units.
1 volt = 108 lines cut per second
10º C.G.S. units.
If E be the electromotive force in volts induced in a coil of S turns by the
introduction of dN lines in the time dt, we have
E
or, from equation (62) on page 78,
d.N.S
.10-8
dt
E=H.l.v.10-8.
(b) CURRENT.
The force ƒ experienced by a conductor of length 7, which carries a
current I in a field H, is given by the equation
wherefore
f=H.I.l,
f
I =
H.1
The dimensions of current can therefore be found from those of force,
field-strength and length, thus
Dimensions of current:
L.M. T-2
L-.M². T¬¹. L
Lª.M². T-¹.
That current has unit strength, which experiences a force of 1 dyne
per cm. of length in a field of unit strength. The tenth part of this unit
has been chosen as the practical unit and called an ampere, wherefore
1 ampere = C.G.S. unit.
The number of amperes is therefore always 10 times the number of
absolute units in the same current.
The legal definition of the ampere is based on the fact that it deposits
1.118 mgs. of silver per second from a solution of silver nitrate.
(c) QUANTITY OF ELECTRICITY.
Since the current is defined as the quantity of electricity that flows
through any cross-section of the conductor in one second, quantity of electri-
city must be the product of current and time. Hence, we have
Dimensions of quantity of electricity: L.M. T.T Lª.M¹.
-
An absolute unit of electricity flows in one second through any cross-
section of a conductor carrying unit current. The tenth part of this quantity
is the practical unit, corresponding to the ampere, and is called a coulomb.
1 coulomb C.G.S. unit of quantity.
If Qe represents the quantity of electricity in coulombs, then
Other units of quantity of electricity are as follows:
1
1 microcoulomb
coulomb = 10 coulomb,
106
1
1 ampere-hour = 3600 coulombs.
1
Dimensions and unit of resistance
97
The dimensions of quantity of electricity in the electrostatic system are
not the same as in the above electromagnetic system. According to Cou-
lomb's law for electricity, the force between two charges, or quantities of
electricity, is given by the formula
f=
Mr. M₂
p2
where m, and m, are the quantities expressed in electrostatic units. In this
system, the unit is that quantity which exerts a force of 1 dyne on a similar
quantity at a distance of 1 cm. This unit is very small, a coulomb being
3.10º times as large, while an absolute electromagnetic unit of quantity
is equal to 3.1010 electrostatic units. The dimensions of quantity of elec-
tricity in the electrostatic system are the same as those of pole-strength,
since both are derived from the formula, m =r√ƒ.
The units of quantity of electricity in the electrostatic and electromagnetic
systems do not differ merely by the factor 3.10¹º, but also in their dimensions
by the factor L. T-, i.e. by the dimensions of velocity.
The importance of this is shown by an experiment due to Rowland. If a
ring with a static charge of 1 electrostatic unit per centimetre of periphery
is rotated with a velocity of 3.1010
cms.
sec.
=
kms.
sec.
300,000 the magnetic effect
is exactly the same as if the ring were at rest, but carried a current of
1 absolute electromagnetic unit. It is worthy of note that 300,000 kms. per
second is the velocity of light and of the electromagnetic waves used in
wireless telegraphy.
(d) RESISTANCE.
Resistance is defined as the ratio of electromotive force to current.
E
R
From this we have
Dimensions of resistance:
L.M. T-2
L.M.T
L.T-1.
Resistance has therefore the same dimensions as velocity, and, strange
though it may seem, the absolute unit of resistance is equal to 1 centimetre
per second. This is the resistance through which the very small absolute
unit of electromotive force sends the relatively large absolute unit of current.
The practical unit of resistance, the ohm, is 10 times as large as this
absolute unit. This is the resistance through which 1 volt sends a current
of 1 ampere.
1 ohm =
10º C.G.S. units of resistance.
Legally, the ohm is the resistance of a thread of mercury 1 sq. mm. in
cross-section and 106.3 cms. long.
The legal volt is that pressure which produces a current of 1 ampere
in a resistance of 1 ohm, or inversely, a volt is the potential difference between
the ends of 1 ohm resistance when a current of 1 ampere is passed through it.
T. E.
7
98
Electrical Engineering
41. Dimensions and units of energy, heat and power,
(a) ENERGY AND WORK.
Mechanical work is defined as the product of force and distance. Hence,
Dimensions of work: L2. M.T.
The absolute unit of work is done when a force of 1 dyne is overcome
through a distance of 1 cm. This unit of work or energy is called a centi-
metre-dyne or an erg. If J represents the work done, in ergs, in overcoming
a force of ƒ dynes through a distance of s centimetres, we have
J = f.s
.(73).
Example: How many ergs of work are done in lifting a kilogramme a
height of 1 metre?
We have
ƒ= 981,000,
1 kg.* = 981,000 dynes,
1 metre = 100 cms.
s = 100.
J=ƒ.s=981,000 . 100 = 9·81.10' ergs.
Hence, the practical unit of work
gramme, is equal to 9.81. 10' ergs.
is equal to 1.36.10' ergs.
in the metric system, viz. a metre-kilo-
Similarly the British unit, a foot-pound,
Since we saw in Section 10 that the product E.i.t represents the
electrical work, its dimensions must be the same as those found above for
mechanical work. By multiplying together the dimensions of electromotive
force, current and time, we find that this is so. The absolute unit of electrical
energy is also, of course, the erg.
Now, a volt is 108 c.G.S. units, and an ampere is 10-1 c.G.s. units, from
which it follows that the joule, which is the product of 1 volt. 1 ampere. 1
second must be equal to 108. 10-¹= 107 absolute units of energy, i.e. 107 ergs.
1 joule = 10' ergs.
We saw above that
and
Therefore
and
10–¹
=
1 metre-kg. 9.81.10' ergs,
1 foot-pound = 1.36.10' ergs.
1 metre-kilog. = 9.81 joules,
1 foot-pound = 1·36 joules.
(b) HEAT.
Since heat is a form of energy its dimensions are necessarily the same
as those of energy. Seeing, however, that the scale of the thermometer has
been chosen quite arbitrarily, it is but natural that a numerical coefficient
has to be introduced into the expression of Joule's law, whereas, by a suitable
choice of units, such coefficients have been eliminated from the laws of Ohm,
Coulomb, and Laplace. In choosing as the unit of heat the quantity of
heat necessary to raise the temperature of 1 gramme of water through 1º C.,
we are taking a step quite outside the absolute system of units. This unit
Dimensions and units of energy, heat and power
99
of heat is called the gramme-calorie. Its relation to the unit of mechanical
work is what is known as the mechanical equivalent of heat, and was first
determined by Joule.
1 gramme-calorie = 0·427 metre-kilog.
Similarly, 1 British thermal unit (F°) = 778 foot-pounds.
Since 1 metre-kilog. = 9.81 joules, we have
1
1 gramme-calorie = 0·427 . 9·81:
=
0.24 joules,
or
1 joule = 0·24 gramme-calories.
This is only another way of expressing Joule's law, according to which
the quantity of heat in gramme-calories is given by the equation:
Qr= 0.24 E. i.t.
(c) POWER.
Power is the rate of doing work, i.e. the work done in a given time.
Power =
work
time*
We have therefore
Dimensions of power: L2. M. T▬³.
The absolute unit of power is 1 erg 'per second. This is an exceedingly
small power, and a practical unit 107 times as large is used; it is called
a watt.
1 watt = 10' ergs per second = 1 joule per second.
Since the electrical work in joules is equal to the product E.i.t, the
electrical power in joules per second, or in watts, must be equal to E.i. If
P represents the power in watts, we have
PE.i.
Example: How many watts are equivalent to one horse-power?
We know that
and
therefore
1 horse-power = 550 foot-pounds per sec.,
1 foot-pound per second = 1.36 joules per second;
1 horse-power = 550. 1·36 = 746 watts.
42. Dimensions and units of the coefficient of self-induction
and of capacity.
(a) THE COEFFICIENT OF SELF-INDUCTION.
In Section 35 we saw that the coefficient of self-induction in absolute
units was equal to
S2
4π. S². μ. A
し
​As 4π, S, and μ are mere numbers, we have
Dimensions of coefficient of self-induction
L2
:
I
= L.
է
7-2
Un
100
Electrical Engineering
Hence, the dimension is simply a length, and the unit is a centimetre. If,
however, in calculating the self-induction, the pressure be expressed in volts
and the current in amperes, we get from equation (67), on page 85,
where
4π. S². μ. A
di
di
E =
10-9.
L.
dt
dt'
L =
4π. S².μ
A
•
10-9
し
​..(74).
This gives the coefficient of self-induction in practical units or henries.
Since the number of henries is 109 times as small as the number expressing
the same self-induction in absolute units, the henry must be 109 times as
large as the absolute unit.
1 henry = 10º C.G.S. units=109 cms.
Now 109 cms. or 10,000 kilometres is the length of a quadrant of the
earth. For this reason the practical unit of self-induction was formerly
called a quadrant. According to equation (74) that coil has a self-induction
4π. S². μ. A
of 1 henry for which the value of
10-⁹ is equal to 1, or in which
an electromotive force of 1 volt is induced when the current increases steadily
at the rate of 1 ampere per second.
(b) CAPACITY.
A condenser consists of two metal plates very near together, but yet
separated by a thin sheet of insulating material known as the dielectric.
When the two plates are connected up to the terminals of any source of
electricity, the condenser becomes charged by positive electricity flowing into
one plate and negative into the other. This flow of electricity continues
until the back pressure of the condenser exactly balances the terminal
pressure of the supply. The quantity of electricity passing into, say, the
positive plate of the condenser is greater, the greater the pressure applied,
and also the greater the capacity of the condenser. This capacity is pro-
portional to the area of the plates and inversely proportional to the distance
between them. It is also dependent on the nature of the dielectric.
We have therefore
Quantity of electricity = pressure. capacity.
The dimensions of capacity are therefore obtained by dividing the
dimensions of quantity of electricity by those of electrical pressure or E.M. F.
L. M
Dimensions of capacity: 1. MT-2
L-¹. T².
A condenser has unit capacity when a pressure of 1 absolute unit charges
it with an absolute unit of quantity, or when unit quantity of electricity
charges it to a pressure of one absolute unit. We have seen that the
absolute unit of pressure is exceedingly small, viz. a hundred-millionth part
of a volt, while the absolute unit of quantity of electricity is very large, viz.
10 coulombs. A condenser with a capacity of one absolute unit would there-
Dimensions and unit of capacity
101
$
fore have enormous dimensions, in order that such a small pressure might
produce such a large charge. For this reason the absolute unit of capacity
is not used in practice, but is replaced by a practical unit called a farad.
This is the capacity of a condenser which is charged with 1 coulomb by
a pressure of 1 volt, or, in which a charge of 1 coulomb causes a back-pressure
of 1 volt. If the capacity in farads be represented by K, and the quantity
of electricity in coulombs by Qe, then
QeK. E coulombs,
or
Qe
K:
farads
E
.(75).
The relation between the farad and the absolute unit of capacity can be
found as follows:
1 farad =
1 coulomb
1 volt
10-1 C.G.S. unit
108 c.G.S. units
10-9 C.G.S. unit.
The farad is thus the thousand-millionth part of the absolute unit.
Even the farad is too large a unit for ordinary practice, and a microfarad,
i.e. a millionth of a farad, is used.
1
1 microfarad =
farad = 10— farad = 10-15 C.G.S. unit.
106
CHAPTER VI.
43. Bipolar ring-winding.-44. Bipolar drum-winding.-45. Multiple-circuit ring-winding.—
46. Multiple-circuit drum-winding.-47. Two-circuit ring-winding.-48. Two-circuit drum-
winding.-49. Series-parallel ring-winding.-50. Series-parallel drum-winding.
43. Bipolar ring-winding*.
The principles of electromagnetic induction, and especially the induction
of electromotive force by the movement of a conductor in a magnetic field,
were first established by Faraday and published in 1831 and 1832 in his
celebrated Experimental Researches. The first machines which were made
consisted of a coil of insulated copper wire wound in two diametrically
opposite slots in an iron cylinder. The ends of the wire were connected to
two insulated slip-rings carried on the armature spindle. For the sake of


N
N
A
D
S
A
Х
X
D
II
Fig. 88.
+
S
Fig. 89.
I
B
clearness, these rings are shown one over the other in Fig. 88, instead of side
by side. Two fixed brushes are held by springs against the slip-rings and
maintain the connection between the coil and the external circuit.
On rotating the iron cylinder or armature between the poles of a magnet,
the wires in the slots cut through the lines of force and induce electromotive
force. In Fig. 88 the bunch of wires constituting a coil-side is shown under
the middle of the pole. With a clockwise direction of rotation, an applica-
* For a fuller discussion of armature windings see E. Arnold's "Armature Windings.”
Bipolar ring-winding
103
tion of either the swimming or the right-hand rule shows the induced electro-
motive force in the wires under the north pole to be directed away from the
observer. Since the electromotive force in the wires under the south pole is,
at the same moment, directed towards the observer, the electromotive forces
in the two sides of the coil can be added together. In a similar manner, the
electromotive forces induced in each individual turn of the coil are added
together. If the brushes are joined by means of an external resistance,
a current flows in the direction indicated in the figure. This current leaves
the machine and enters the external circuit by the outer slip-ring and brush,
which is therefore marked positive. After flowing through the external
circuit, the current re-enters the machine through the inner brush and slip-
ring, which can therefore be called the negative terminal.
The lines of force leaving the north pole cross the gap normally or radially
into the iron of the armature and the strength of the field H is practically
constant over the whole gap. So long as the side of the coil moves under
the pole, the electromotive force induced in it can be found from equation (62)
on page 78:
E=H.l.v. 10-8 volts.
7 here represents the total length of wire under both poles, in centimetres,
while v is the peripheral speed in cms. per second. The wire not actually
under the poles, such as the end connections of the coil, is inactive and must
not be included in l.
As soon as the coil-side I leaves the pole-tip B, the electromotive force
drops rapidly to zero and remains there as long as the coil-side moves between
the pole-tips B and C. Directly it passes, however, under the south pole at
C (Fig. 89) an electromotive force is induced in it towards the observer, and
the direction of the current is reversed. Since the polarity of the brushes,

A
B
C
D
A
B
1
Fig. 90.
and the direction of the current in the external circuit are also reversed at
the same time, the machine gives an alternating current. The changes in
the current can best be represented by plotting the circumference ABCD as
abscissae and the electromotive forces induced, when the coil-side is at these
various points, as ordinates (Fig. 90). The current at any moment can be
found by dividing the electromotive force at that moment by the total
resistance of the circuit. It is evident that the machine gives current which
104
Electrical Engineering
not only reverses its direction, but which is also intermittent, that is, it
ceases altogether for certain intervals.
The transition from the above machine to the modern dynamo is shown
in Fig. 91, where the ends of the coil are connected to the two insulated


A
N
XX
B
N
+
I
S
Fig. 91.
s
Fig. 92.
halves of a single slip-ring. In Fig. 91 the direction of the current in the
wires is such that the right-hand brush is positive, the current leaving the
machine by this brush. It remains positive, however, when the coil-side I
has passed from the north to the south pole, and the current in the coil has
been reversed (Fig. 92). To obtain this result the brushes must be placed in
the neutral zone, that is, they must touch the commutator at the extremities
of a diameter perpendicular to the magnetic field. We have assumed that
the insulation, which divides the slip-ring into two parts, is in the plane
of the coil. On passing through the neutral zone, the direction of the current
ДДД

B
Fig. 93.
D
A
B
in the coil is reversed, as before, but its connections with the brushes and,
therefore, with the terminals of the external circuit are reversed at the same
time. The current in the external circuit flows now in one direction only, and
instead of an alternating current we have an intermittent direct current
(Fig. 93).
In order that the external current may remain constant in magnitude as
well as direction, the slip-ring must be divided into a greater number of
parts, becoming thus transformed into the ordinary commutator with many
segments. This arrangement was invented by Pacinotti in 1860, but was
Bipolar ring-winding
105
N
little used until taken up by Gramme. The Gramme ring armature consists
of a hollow iron cylinder, through which a continuous spiral of insulated
copper wire is wound. We must not imagine
that the axial length of the armature is small,
for it is quite considerable, and the armature
is far more a hollow cylinder than a ring. In
Fig. 94 the winding consists of 8 coils each of
2 turns. In an actual machine, however, the
number of coils is much greater, the whole
ring being closely covered with winding.
Every spiral must be wound in the same
direction.
The collector or commutator is mounted
on the armature spindle. This is divided by
planes passing through the axis into as many
insulated copper segments as there are coils
on the armature. As in Figs. 91 and 92, the
insulation between the segments is shown
opposite the coils.

S
Fig. 94.
The coils are now joined and a tap taken from each connecting wire down
to a commutator segment (Fig. 95). The figure, thus obtained, is very
simple and brings out clearly the continuous endless spiral. In actual
practice, however, the number of soldered T-joints in the armature winding


A
N
N
+
B
E
S
Fig. 95.
S
Fig. 96.
would be disadvantageous, and it is preferable to bring the end of each coil
down to the commutator segment and to connect it there to the beginning
of the next coil. The commutator segment itself constitutes, then, the con-
nection between two adjacent coils.
If the ring is rotated between the poles of a powerful electromagnet, the
wires on the periphery of the armature cut through the lines of force.
Electromotive forces are induced in them, the direction of which can be
found by either of the rules already given. Since the lines of force pass
106
Electrical Engineering
from the north to the south pole through the iron of the armature, and
leave the internal air space almost free from magnetic field, only those wires.
cut through the lines of force which are situated on the external periphery
of the ring. For a clockwise rotation the induced electromotive force at A
in Fig. 95 is found to be away from the observer. Instead of indicating the
direction of E.M. F. or current by means of dots and crosses we can put corre-
sponding arrows on the end connections. We see, therefore, that, for clock-
wise rotation, the current in the end connections facing the observer
flows away from the south pole towards the north pole.
This current can only flow, however, if suitable arrangements are made
whereby it can leave the armature and flow through the external circuit.
We see from the arrows in Fig. 95 that in both the upper and lower halves
of the armature the electromotive forces act towards the point B. It is as if
two equal forces opposed each other at B and mutually neutralised each
other. If, however, brushes be placed on the commutator at B and C, and
connected through an external resistance, current will leave the machine
at B, flow through the external circuit, and enter the machine again at C,
where it will divide between the two parallel halves of the armature winding
and flow back to B. B is therefore the positive and C the negative brush.
The brushes lie, as before, on a diameter at right angles to the magnetic
field, that is, they are in the neutral zone. As before, we assume here that
the insulation between two commutator segments is opposite the correspond-
ing armature coil.
As in the previous example, the electromotive force of the machine can
be found from the equation
E=H.l.v. 10-8 volts.
Since the electromotive forces in the two halves of the armature cannot
be added, but are in parallel, I must represent the length of active conductor
under one pole. If the number of conductors on the periphery is very large,
the number under one pole will be constant and the electromotive force will
be practically constant. The advantages possessed by the Gramme machine
with its commutator of many segments, compared with the earlier machines,
is self-evident.
The electromotive force is not decreased when, in consequence of the
rotation, the brush makes contact simultaneously with two adjacent seg-
ments (Fig. 96). At such a time the two coils lying in the neutral zone
are cut out of the circuit since each one is short-circuited by a brush. The
current flows, for example, directly from the points D and E in Fig. 96 to
the positive brush. The electromotive force is not decreased since the short-
circuited coils are cutting absolutely no lines of force, assuming that the
brushes are in their theoretically correct position. If the number of coils
or commutator segments is very large, the change of armature resistance
produced by short-circuiting a coil at the brush is quite negligible.
If
D= diameter of armature in centimetres,
L = axial length
""
"
""
B = angle subtended by pole,
Bipolar ring-winding
107
then we have
N = total flux leaving one north pole,
z = total number of wires on external periphery,
n = revolutions per minute,
n
v = D.π
60*
2.B
The number of conductors under one pole is 360
The length l of active wire under one pole is therefore
z.B
L.
360
Substituting these values for v and 7 in the equation for E, we get the
electromotive force of the machine
E=H.
z. B. L
360
N
.D.T
10-8 volts.
•
60
In this equation
D.T.B
360
D.T.B
is the length of the polar arc, and
L
360
is therefore the area of a pole, which, when multiplied by the flux-density H;
must give the flux N leaving one north pole. Introducing the term Ñ in
the above equation, we have
E = N. 2.10-8 volts
n
60
If, for example, N=3. 10º, n=1,100 and z = 200, then
(76).
E = 3.106.
1,100.200
60
.10-8-110 volts.
To find the armature resistance we must remember that the two halves
of the winding are in parallel. If
and
7= total length of wire on armature in metres,
A cross-section of wire in square millimetres,
=
then the resistance of each half of the winding is
the two halves in parallel will be a half of this, viz.
Ra
p.1/2 _p.l
2A ᏎᎪ
1/2
ρ A
The resistance of
..(77).
The specific resistance p of warm copper can be taken approximately
as 0·02.
44. Bipolar drum-winding.
The ring armature, already described, possesses the advantage of a simple
winding, in which a single coil can be repaired without unwinding the whole
armature. Since, moreover, the pressure between two adjacent wires is only
a small fraction of the terminal pressure of the machine, the insulation can
be easily and efficiently provided for. On the other hand, the ring armature
has the disadvantage of requiring a large amount of inactive winding. The
armature resistance and the weight of wire required are therefore both rela-
tively large.
*
108
Electrical Engineering
1
We have now to examine the drum armature of Hefner-Alteneck, which
is an improvement on the ring armature so far as the proportion of active
armature wire is concerned. Here, however, two wires having a large
difference of potential may be very close together. In the drum winding,
after a wire has passed along the external cylindrical surface under a north
pole, it passes right across the end surface to a point diametrically opposite,
where it then passes along the armature under the south pole. The two
diametrically opposite wires constitute the sides of an armature coil. The
end of the first coil is connected to the beginning of the next coil, care being
taken that the winding is distributed uniformly over the whole periphery.
To ensure this we divide the armature periphery into a suitable number
of parts (in Fig. 97 we have taken 8), and number the points successively
1, 2, 3, etc. These are the starting points of the armature coils. To wind
the first coil we start at 1, pass down the armature, that is, away from the
8
2
5'
3
N


6'
2
5
3
8
73
6
Fig. 97.
6
Fig. 98.
observer, and should then pass across the back to a point diametrically
opposite. This point is, however, already taken up by the beginning of
coil 5, and we therefore take the back end-connection of coil 1, as shown
by the dotted line, to a point l' lying near point 5. The wire is then
brought up to the front along the surface of the armature at 1'. The wires
1-1' together with their end-connections constitute one turn of the first
coil. From I' we pass across the front, back to 1, and wind another turn
1-1', repeating this until the first coil is completely wound. For the sake
of simplicity the coils in Fig. 97 consist of a single turn. There is, however,
no reason why each turn in the figure should not represent a coil of many
turns.
After completely winding the first coil, we pass across the front to the
beginning of the second coil at the point 2. After passing down the arma-
ture at point 2 we should cross the back end to point 6, but as this is
already occupied we go to the neighbouring point 2', as shown by the dotted
Bipolar drum-winding
109
line. The coil 2-2' is then wound in exactly the same way as coil 1-1',
and when it is finished we pass to coil 3, and so on round the whole
armature.
The winding can be represented in tabular form in the following manner:
8'
1-1'
1
2-2'
3-3'
4-4'
4'
5-5'
6-6'
7_7
8-8'
The horizontal lines represent connections at the back of the armature,
such as 1—1′, while the slanting lines represent connections at the front or
commutator end, such as l'—2.
When completely wound in this way there will be sixteen wires or coil-
sides on the periphery, dividing it into sixteen equal parts, of which each
coil spans 7. In winding coil 1, for example, we started at the point 1 and
stepped forward over 7 divisions to the point l'. The forward step, which
we shall call y₁, is therefore equal to 7, or
Y₁ = 7.
From 1' we step back over 5 divisions to the point 2, whence the back-
ward step y, is equal to 5, or
Y₂ = 5.
If the sixteen wires had been consecutively numbered from 1 to 16 as in
Fig. 99, the steps y, and y½ would still have been 7 and 5, and starting from
1, the first coil would have gone to 1+7=8 and have stepped back to
8-5=3. The winding table for Fig. 99 would be as follows:
6
1-8
3-10
14
9—16
11-2
5-12
7-14
13-4
15-6
If the coils have many turns the end-connections of the coil last wound
will lie over the end-connections of the coils already wound, and the re-
sistances of the various coils will be unequal. This can be avoided by
winding a half of the first coil, a half of the second and a half of the third,
then all the fourth coil, and finally the remaining halves of the third, second
and first coils. In modern armatures, however, the coils are usually all alike,
being wound on formers in such a way that they can be slipped one over the
other.
After completing the last coil, its end must be connected to the beginning
of the first coil, thus completely closing the winding, as shown in Fig. 98 and
the following figures. We have then to connect the wires joining successive
110
Electrical Engineering
coils to the commutator segments. It is simplest to draw the insulation
between the segments opposite to the corners of the regular octagon formed
by the front end-connections. Each front end-connection is then joined to
the nearest segment.
It is interesting to notice that the drum-winding is identically the same
in principle as the ring-winding. The end of each coil is connected to the
beginning of the next or adjacent coil. In the drum-winding, however, space
is left between the beginnings of successive coils for the ends of other coils.
The agreement becomes plainer when we notice that, except for the negli-
gibly small lack of symmetry, the coil-sides 1', 2', etc. in Fig. 98 are exactly
equivalent to the coil-sides 1, 2, 3, etc. They have the same position relatively
to the poles and have the same electromotive force induced in them at any
moment. In considering the drum-winding we could therefore neglect the
coil-sides 1', 2′, 3', etc., and in their place imagine the number of wires in the
coil-sides 1, 2, 3, etc. to be doubled. In this way we obtain a number of
coil-sides around the periphery of the armature, connected successively in
series, as in the ring armature.
In order to trace out the path of the current in the drum-winding we
will consider in the first place the position of the armature shown in Fig. 99.

12
01
13
N
15
16
14
11
10
+
10
N

14
15
13
16
12
"
3
8
S
7
Fig. 99.
7
6
S
Fig. 100.
We saw in the previous section that for clockwise rotation of the armature
the current in the front end-connections flows from the south pole towards
the north. If arrows are drawn in accordance with this rule, as in the figure,
it is evident that the electromotive forces in the turns of any coil, and also
in the various coils, act, in general, in the same direction and can be added.
In the end-connection 5-10, however, the two forces oppose each other, and
a brush must therefore be placed on the corresponding segment. When the
external circuit is closed, current will leave the machine by this brush, which
must therefore be marked +. Similarly, the negative brush must be placed
on the segment connected to the end-connection 2-13.
Bipolar drum-winding
111
The current enters the machine at the negative brush, divides and flows
through two parallel paths to the positive brush. For the moment repre-
sented in Fig. 99 the path of the current through the armature can be shown
in the following manner:
2 11 16 9 14 7 12 5
+
13 4 15 6 1 8 3 10
We will now consider the case when each brush touches two segments
simultaneously (Fig. 100). At this moment the coils 2-11 and 3—10 are
short-circuited by the negative and positive brushes respectively, and are
thereby cut out of the circuit. No arrows are drawn, therefore, to indicate
the current in the end-connections of the wires 2, 11, 3 and 10. The path of
the current through the armature is as follows:
16 9 14 7 12 5
+
13 4 15 6 1 8
As before, we see that the short-circuited coils lie in the neutral zone, if
the brushes are in their correct position.

G
F
E
D
C
C
3 2
16/15 14 13 12 11.
10
8
5
B
A
of
4
G
E
D
Ꮳ
C
3
B
A
Fig. 101.
The path of the current is clearly shown by developing or unwinding the
periphery of the armature, and spreading it out flat, as is done in Fig. 101
for the moment represented in Fig. 99 when each brush touches one segment
only. We must imagine the winding to move in the direction of the upper
112
Electrical Engineering
arrow in front of the stationary poles N and S, while the commutator slides
along the fixed brushes. A sixteenth of a second later each brush will touch
two segments and short-circuit two coils lying in the neutral zone.
The electromotive force of a drum armature is naturally the same as that
of a ring armature with the same number of external wires. We see the
advantage of introducing the number of external wires, instead of the
number of turns or coils, into the formula for the electromotive force. We
have then for the bipolar drum armature
n
E = N . z. 10-8 volts.
60
The formula for the armature resistance is also exactly the same as for
the ring armature.
45. Multiple-circuit ring-winding.
We have previously mentioned that the electromotive force is a funda-
mental characteristic of the ordinary machine, depending on its construction
and the speed of rotation. The current, on the other hand, is dependent on
the external resistance, which is entirely in the hands of the consumers. A
limit is set to the current, however, by the amount of heat allowable in the
armature without endangering the insulation. Machines for large currents
must therefore have a low armature resistance and a sufficiently large cooling
surface. This leads to conductors of large cross-section, and copper strips or
bars are used instead of round wire.
The losses due to eddy currents in massive copper bars would be very
considerable. Moreover, in machines of large size the bipolar magnets would
be very clumsy and unfavourable to ventilation. Finally, if the current in
each armature coil is too large, the short-circuiting of the coils at the brushes
leads to sparking. Large machines are therefore made multipolar and the
armature winding has as many parallel paths as there are poles. The
simplest form of such a winding is the continuous spiral ring-winding. It is
exactly the same as for the bipolar machine. The magnet system is arranged
so that the poles are alternately north and south. If the armature is turned
in the clockwise direction, we know that the current in the front end-connec-
tions flows towards the north poles and away from the south poles. We see
that the current flows from both sides towards the points A and B, and from
here, by means of the commutator, to the positive brushes. Both positive
brushes are connected together and to the positive terminal of the external
circuit. In the same way the two negative brushes are connected together
and to the negative terminal of the external circuit.
The armature winding is thus divided into four parallel paths, so that if
1 = total length of wire on armature in metres,
p = number of pairs of poles, and
A = cross-section of wire in sq. mm.,
the resistance of one path between two brushes of opposite sign will be
:
Multiple-circuit ring-winding
113
P.L
2p. A
Since there are 2p such paths in parallel the resistance of the
whole armature will be 2p times as small, that is,
Ra=
ρ.ι
4p². A
..(78).
This multiple-circuit or parallel winding has, thus, the advantage of
giving a very small armature resistance. Since the external current divides
in 2p parts, the current in each coil is relatively small, viz. ia/2p. Let us

N
B
ти

A
www
N
Fig. 102.
www
www
*
*
consider, as an example, a 4-pole machine for 110 volts and 100 amperes
armature current. If the total length of wire on the armature is 200 metres
and its cross-section 10 sq. mm. we have, since p= 2,
Ra
p.l
4p². A
0.02.200
4.4.10
=
0.025 ohm.
The heating or so-called copper loss is therefore
ia²Ra= 100². 0.025 = 250 watts.
This is about 2.5 per cent. of the output of the dynamo. The pressure
drop in the armature is
ia Ra= 100.0·025 = 2.5 volts,
and the current density in the armature winding
ia 100
4.10
2p. A
2.5 amps. per sq. mm.
To calculate the electromotive force of armatures with parallel winding
we observe that the number of lines of force cut by a conductor in one
revolution is p times as great as in a bipolar machine having the same
number of lines per pole. On the other hand, the number of wires connected
in series is p times as small as in a bipolar machine with the same total
T. E.
8
114
Electrical Engineering
number of wires. The electromotive force of a multipolar dynamo with
parallel winding and N lines leaving each north pole is therefore
E = N. z. 10-8 volts
as in the bipolar machine.
N
60
With regard to the number of brushes we see that there is a brush in
each neutral zone, which gives a total of 2p brushes. This number can be
reduced to 2 by interconnecting all segments separated by an angle of
360
p
degrees. These connections can easily be placed on the end of the
commutator towards the armature. They are shown in Fig. 103 for the
commutator of a six-pole parallel wound machine (p=3). In this case every
three segments separated by 360 120° are connected together, the connec-
tion thus made replacing that which would otherwise be made between the
=
S
N


4
Fig. 103.
N
Fig. 104.
three brushes of like sign. If the connecting strips are cut from sheet copper
and bent, so that the parts shown by heavy lines lie in a different plane to
those shown by light lines, contact between the various strips is easily
avoided.
In this way we have not only connected the three segments on which the
brushes would lie at any moment, but also all the other segments in groups
of three. Since, however, all interconnected points have always the same
potential, their connection will have no effect on the distribution of current
in the armature, and no current will flow through any connection except
those of the segments under the brushes at the given moment. This is,
perhaps, more evident from Fig. 104, in which the equipotential connections
are made on the wires leading from the armature to the commutator seg-
ments. These connections were first introduced by Mordey.
A reduction in the number of brushes is, of course, only permissible when
the current density under the brushes is small. Equipotential connections
Multiple-circuit ring-winding
115
are used to a large extent in modern dynamos, not for the purpose of reducing
the number of brushes, but for equalising the distribution of current in the
armature, and between the various brushes, so as to prevent sparking at the
commutator due to any one brush being heavily overloaded. It is for this
reason that they are generally known as equalising connections.
46. Multiple-circuit drum-winding. (Lap-winding.)
In a multipolar lap-winding, the wire, after passing under a north pole,
returns under the adjacent south pole, passes across the end to the point
from which it started and repeats the process until the first coil is completely
wound. The wire then passes to a point very near the starting point of the
first coil, and the second coil is wound in exactly the same way, its position
being but slightly displaced from that of the first coil. If the return wires
under the south pole be neglected the similarity, in principle, of the lap- and
ring-winding can be seen, and it is evident that, like the ring-winding, the
lap-winding must lead to multiple-circuit or parallel connection. We shall
first consider:
(a) LAP-WINDING WITH LONG COILS.
In this winding each coil spans, at least, the pole pitch. Assume, for
example, that there are 4 poles and 8 armature coils. As shown in Fig. 105
the periphery is divided into 8 parts and the points at which the 8 coil-sides

6
8
5
N
1

N
16
2
15
14
OP
3
உ
13
S
12
3
4
+
S
5
6
5
N
2'
Fig. 105.
10
9
N
8
Fig. 106.
will commence are numbered 1, 2, 3, etc. A wire is passed from front to
back at 1, bent round, and brought up from back to front at a corresponding
point under the south pole. As this point 3 is already taken for the
beginning of the third coil, the return side of the first coil must come up at a
point near it such as 1'. When the coil 1—1′ is completely wound we pass
across the front from the end l' to the beginning of the second coil at 2.
Proceeding in this way the successive coils are all in series and finally the
end of the eighth coil is joined up to the beginning of the first. The wires
8-2
116
Electrical Engineering
joining successive coils, such as 1-2, are connected to the commutator
segments. Since there are 8 coils, there are 8 segments.
The step from 1 to 1' is equal to 5, while that from 1' back to 2 is equal
to 3. If, as before, y, represents the forward step and y, the backward step,
then
y₁=5 and y=3.
In general, if there are 2p poles, the forward step should theoretically be
equal to the 2pth part of the periphery, but practically 1 division greater or
less than this. Hence, if there are S divisions or coil-sides on the periphery,
for a lap-winding with long coils we have
S
Уг
+1,
2p
Y2
=
S
- 1.
2p
S must evidently be divisible by 2p in order that y, and y, may be whole
numbers, and, in addition to this, y, and y₂ must be odd numbers. The
necessity for this last condition is evident if we number the coil-sides succes-
sively from 1 to 16 as shown in Fig. 106 and write down the corresponding
winding table, thus:
4
12
1-6
9-14
3-8
11-16
5-10
13-2
7-12
15-4
Had the step been even, commencing at 1, we should have arrived at
uneven numbers only, and the winding would have closed without using the
evenly numbered coil-sides.
As before, it is easily seen that for clockwise rotation the currents in the
front end-connections flow towards the north poles and away from the south
poles. The path of the current in Fig. 106 can be represented as follows:
7 12 9 14
10
5 8 3
16 11
2 13
15
4 1 6
+
Figures 105 and 106 represent smooth armatures or toothed armatures
with a single coil-side per slot. Smooth armatures are now little used, since
the number of wires which can be placed on the periphery is very limited.
In addition to this, it is much cheaper to wind the coils on formers and lay
them in the slots than to wind the smooth-cored armatures. Another
advantage is the reduction of loss due to eddy currents as pointed out in
Section 36. Modern machines have therefore, almost exclusively, toothed
armatures and several coil-sides in each slot. There is no reason, however,
why Fig. 106 and the winding-step which we have given for it should not
represent such an armature. We have merely to imagine the coil-side 2
Multiple-circuit drum-winding
117
placed below coil-side 1 instead of beside it, and in finding the step to count
both upper and lower coil-sides in each slot.
It is simpler, however, to count the step by the number of slots and not
by the number of coil-sides. We assume, as is practically always the case,
that, of the two sides 1 and 1' of a coil, one occupies the top of a slot and
the other the bottom of another slot (Fig. 107). With the winding arranged
in two layers, the step with respect to coil-sides is therefore necessarily odd.

N
S
11
12
X
X

+
+d
X
X
X
X
7
کے
6
ވ
ハ
​4
S

N
Fig. 107.
We see then that for toothed armatures we need not calculate the step at all
but need only make the coils equal to the pole pitch in width. After
finishing coil 1-1' we go back to slot 2, to the coil occupying the same
position in this slot as the previous coil occupied in slot 1. This is shown in
Fig. 107 for a 4-pole machine with 12 slots. Comparison with Fig. 105 will
show that both are the same in principle and in the method of numbering.
For the sake of clearness the back end-connections are drawn outside the
armature.
118
Electrical Engineering
(b) SHORT-CHORD LAP-WINDING.
The width of the armature coils need not be so great as the pole pitch.
By making them considerably narrower we obtain the short-chord winding of
Swinburne. For a smooth armature the winding-step with respect to the
consecutively numbered coil-sides can be found at once from the following
formulae:
Yı
8-b
+ 1,
2p
Y2
s - b
2p
1,
where b is any even number giving odd whole numbers for Y1
and y2.
The
larger we choose b the narrower are the coils. If, for example, s=22, p=2
and b6 we get the winding shown in Fig. 108, for which we have
Y₁ = 5,
Y2=3.

S
3
7
78
+
12
13
S
20
16
19
18
17
N
14
1
Fig. 108.
The position of the brushes can be determined by putting arrows on the
front end-connections to indicate the direction of the current. We see that
current flows from the wires 11 and 14 towards the right-hand commutator
segment and we therefore place the positive brush on this segment and
Multiple-circuit drum-winding
119
arrange the other brushes every degrees. The left-hand brush short-
360
2p
circuits the coil 21-4, which lies in the neutral zone. The path of the
current through the armature is as follows:
5 10 7 12 9
8
3 6 1
20 15 18 13 16
17 22 19 2
+
In this case, only one coil is short-circuited at the same time. This is
always so if the number of segments is not divisible by the pairs of poles.
N

SY
|
X
ハ
​X
+
X
+
+
+
X
+
+
X
Λ
"
محمد
N
Fig. 109.
Generally speaking, this would be an advantage from the point of view of
sparkless commutation, but a disadvantage if the number of coils in the
different armature branches is thereby made to differ considerably. The
number of segments should preferably be divisible by the number of pairs of
poles; if still indivisible by the number of poles, all the positive brushes will
short-circuit coils at the same time, and all the negative brushes at another
time.
The advantage of the short-chord winding lies in the fact that adjacent
wires in the neutral zone carry currents in opposite directions and thus have
120
Electrical Engineering
no demagnetising effect (see Section 54). This is seen very clearly in the
slot-winding shown in Fig. 109, which also shows the simplicity of this type
of winding. No calculation of the winding-step is required and it is only
necessary to stipulate that, of the two sides of a coil, one shall occupy the
upper and the other the lower position in a slot. For a total number of slots
Z=20 and 4 poles we have a slot-step of 5 forwards and 4 backwards. If,
however, we choose a short-chord winding with slot-steps of Y₁=3 and
Y₂ = 2, we obtain the winding shown in Fig. 109. The wires in the neutral
zone are seen to carry currents in opposite direction. The shortening of the
coils is here, however, carried too far and would cause a decrease in the
electromotive force.
2
47. Two-circuit ring-winding.
In multipolar machines with two-circuit or series winding the current
passes through the armature by two parallel paths as in bipolar machines.
If N represents, as before, the magnetic flux leaving one north pole, the
induced electromotive force for the same number of wires z and the same
speed n will be p times as great as that of a two-pole machine. We have
thus for two-circuit winding,
n
E=p.N. 2.10-8.
60
It is convenient to have, if possible, the same formula for the electromotive
force for both series and parallel windings. If now we let a represent the
number of pairs of parallel paths through the armature, we have in both
cases,
E=2.N.·
z.10-8
α
60
.(79).
The resistance of a series-wound multipolar armature is, of course, the
same as if it were bipolar, viz.
Ra
=
p.1
4A
where is the total length of wire in metres, and A is its cross-section in
sq. mms.
It is easy to see from the formulae for electromotive force and
resistance that the series winding is specially suitable for machines with a
high pressure and small current.
The principle of the series ring-winding is as follows: A coil under a
north pole is connected in series with the coil similarly situated under the
next north pole and this again with the similar coil under the next north pole,
and so on. After making p steps, we have been once round the armature and
must come to a coil next to that from which we started. Thus, if there are
s coils on the armature and each step is equal to y, we have,
or
y
py=8±1,
s±1
p
In order that the winding may close on itself after the whole armature
Two-circuit ring-winding
121
is wound, y and s must have no common factor. The number of coils s must
always be odd.
In the bipolar winding each coil was connected to its neighbour. This is
also true for the multipolar winding, except that the corresponding coils
under the other similar poles are connected between them. The consider-
ation of the multipolar winding can thus be referred to that of the bipolar
winding. In every complete circuit round the armature the winding creeps.
forward or backward by one coil. Hence, there are p coils connected between
two adjacent commutator segments. If, for example, p= 2 and s
13, then
y must be 7 or 6. In Fig. 110, we have chosen y = 6. Commencing at any

12
13
11
10
2
I
N
3
5
6
Ꮽ
N
Fig. 110.
segment, the beginning of a coil passes across the front of the armature,
while the end of the coil is brought up to the front through the interior
of the ring and connected, at the commutator segment, to the beginning of
the corresponding coil under the next similar pole. The winding can be
tabulated as follows:
1-7
11-4
13-6
10-3
12-5
9_2
11
8-1
122
Electrical Engineering
We now draw the arrows on the front end-connections in the usual
manner, omitting those on coils 1 and 4 in which no electromotive force is
induced. There can be no doubt as to the direction of the current in the
other coils, even coil 7 being safely considered as under a south pole.
The direction of the current in coil 11 shows that the negative brush
must be placed on segment I, while from coil 8 we see that the positive
brush must be on segment II. The brushes are thus not diametrically
360
2p
opposite, but subtend an angle of 90° or degrees. The direction of the
current in coils 1 and 4 can now be seen, and the path of the current through
the armature may be represented as follows:
11 5 12 6
12 6 13
13 7 1
4 10 3 9 2 8
+
The small want of symmetry in the two halves of the armature is of no
practical importance.
It is easily seen that the brushes might have been placed on segments
diametrically opposite to those on which we have placed them. The current
in the coils 1 and 4, which lie in the neutral zone, would thereby have been
reversed. In general, there are p possible positions for each brush, separated
360
by an angle of degrees. We can, however, go a step further and place
brushes on the commutator at each of these points and thus have p positive
and p negative brushes. The coils lying in the neutral zone will then be
short-circuited through pairs of similar brushes. With series windings, how-
ever, multipolar machines can be, and often are, run with two brushes only.
p
A moment later than that represented in Fig. 110, the positive brush
will make contact with two segments and thereby short-circuit the coils 1
and 7 connected in series. Since, from the nature of the series ring-winding,
we start at any segment and pass through p coils to get round the armature
to the adjacent segment, it follows that p coils in series must be short-
circuited whenever a brush touches two segments simultaneously.
If we wish the brush to short-circuit one coil at a time, instead of p coils
in series, we must make the number of segments p times as large as the
number of coils. The number of coils can then be kept relatively small.
The beginning and the end of every coil are connected to separate, but
adjacent, segments. The connection of the coils in series is effected by the
interconnection of the commutator segments, every set of p equally-spaced
segments being connected together. The result is the same as if the coils
had been connected together in accordance with the formula
s+1
Y =
Ρ
In Fig. 111, for example, s = 7 and p=3. The commutator has, there-
fore, 21 segments, which are connected together in sets of three, separated
from each other by 120 degrees. Only a few of these connections are shown
in the figure, the remainder being indicated by letters.
At the moment shown in the figure the coil 6 is exactly in the neutral
Two-circuit ring-winding
123
zone and is short-circuited by the positive brush. The negative brush lies
60° away from it. The path of the current is, then, as follows:
-1
7 2 4
5 3 1
+
This shows very clearly that p positive and p negative brushes can be
used, since the p segments are interconnected and equivalent to one large
segment.

N
d
e
*
g
N
1
a
9
b
C
a
e
e
d
b
g
a
2
3
ર
N
S
Fig. 111.
4
It is important to notice the positions, with regard to the poles, of the
coils constituting one of the two paths through the armature. Coil 4, for
example, lies partly under the beginning of a south pole, coil 2 lies wholly
under a south pole, while coil 7 lies partly under the end of a south pole.
The three coils are therefore equivalent to a single coil with three times
as many turns, spread out over a whole pole-pitch.
48. Two-circuit drum-winding.
In the series or two-circuit drum-winding, a coil consists of a coil-side
under a north pole and a coil-side in a somewhat similar position under the
adjacent south pole. The end of the coil is not brought back to a point
near the beginning, but is connected to the beginning of a coil occupying
a corresponding position under the next pair of poles. After proceeding
round the armature in this way and commencing to continue round a second
124
Electrical Engineering
time, the coil-sides are found to lie next but one to the corresponding coil-
sides of the first round. If each coil-side consists of a single wire or bar,
the complete coils disappear and the winding becomes a simple wave-winding
(Fig. 112).

12
11
10
13
14
N
I.
==
+
7
2
3
5
៩
8
N
Fig. 112.
From these considerations the formula for a series drum-winding
should be
or
p (Y₁+ Y2) = s ± 2,
Y₁ + Y 2
8 ± 2
p
Both winding-steps are continuously in one direction and there is no
backward-step. Both y, and y, must be odd since, otherwise, beginning with
coil-side 1 we should either come to odd coil-sides only, or find the winding
impossible. As in all drum-windings s must be an even number. If the
winding-steps y, and y, be made equal, we have
y
where s is even and y is odd.
=
8 +2
or 2py=s±2,
2p
The choice of unequal values for y, and y, may be due to two causes.
If the number of wires or coil-sides is fixed, it may be necessary to make y₁
and y, unequal in order to make them odd. If, for example, s = 214 and
p=6, we have
Y₁ + Y₂
214+2
6
= 36.
Two-circuit drum-winding
125
Now we cannot take y₁ = y₂ = 18 as the winding would then close on
itself before the armature was completely wound. We should therefore make
y₁ = 19 and y₁ = 17. By making y₁ and y, widely different we obtain a similar
result to that obtained by Swinburne's short-chord winding, viz. a reduced
armature reaction.
A series drum-winding is shown in Fig. 112 in which s=14 and p = 2.
We have
s2 14+2
Y₁ + Y₂
p
= 8 or 6.
2
We choose y₁+ y₂ = 6 and make y₁ = y₂ = 3.
The winding scheme is then simply 1-4-7-10-13-2, etc. Starting
from the front at 1 we pass down the armature under the north pole, across
the back-end to 4, up to the front and then across the front to 7, making
connection with the commutator on the way. The back end-connections are
drawn dotted in the figure.
We put in the arrows to indicate the direction of current in the front
end-connections, omitting, for the present, those for wires 13 and 6, which
lie in the neutral zone. Because of the direction of the current in wire 9, we
place the negative brush on segment I, and the positive brush, because of
the current in wire 10, we place on segment II. This settles the current in
wires 13 and 6, and the current path through the armature is as follows:
9 12 1 4 7 10
6 3 14 11 8 5 2 13
+
The odd number of coils causes, at times, a slight dissymmetry between
the two armature paths.
Here again 4 brushes could be used, or, in general, 2p brushes separated
360
by
degrees. In Fig. 112, for example, two more brushes are shown
2p
dotted, lying diametrically opposite to those on segments I and II. Since
the brushes of like sign will be connected together, the coils 3-6 and 2—13,
lying more or less in the neutral zone, will be short-circuited at the moment
represented in the figure. The path of the current through the armature
will then be as follows:
-1
9 12 1 4 7 10
14 11
11 8 5
+
As in the series ring-winding p coils are connected in series between two
neighbouring segments. This can be obviated, with a view to improving the
commutation, by making the number of segments p times the number of
360
p
coils. Those segments which are separated by degrees are then inter-
connected. In Fig. 113, for example, s = 16 and p=3, whence
y
16+2
6
3.
The winding proceeds as follows: 1-4-7-10, etc. Of the connections
within the commutator, only those are shown which are of importance at the
126
Electrical Engineering
moment. The others are clearly indicated by means of the letters on the
segments.
The arrows are now drawn on the front end-connections. Assuming that
the coils 1-4 and 9-12, lying in the neutral zone, are short-circuited at
the moment chosen, we get the position for the brushes shown in the figure.
The path of the current is then as follows:
-1
6 3 16 13 10
13 10 7
15 2 5
8 11 14
+
Naturally, as the segments are connected in groups of p, the number
of brushes can be increased to 2p.

S
T
4
e
3
7
る
​8
2
с
d
e
h
a
10
N
11
a
+
hg
fe
d
13
8.
16
N
Fig. 113.
As before, the slot winding with two coil-sides per slot is extremely
simple. After going once round the armature, that is, after 2p steps, we
arrive at the slot next to that from which we started. The result is the
same as when, in a smooth-cored armature, we arrive at the coil-side next
but one to that from which we started. If Y, and Y₂ represent the winding-
steps with regard to the slots, and the number of slots be S, then the above
consideration leads to the formula
1
p. (Y₁+ Y₂) = S ± 1,
2
or
S+1
1
Y₁ + Y₂ =
Ρ
Two-circuit drum-winding
127
If, for example S 15 and p = 2, we have
=
Y₁+ Y₂ =
15 ± 1
2
8 or 7.
We choose Y₁ = Y₂ = 4 and obtain the winding shown in Fig. 114.
2
-

N
S
"
X
✓ 13
172
114
15
X
X
X
Χ
11
X
+
1/

4
6
S

N
Fig. 114.
49. Series-parallel ring-winding.
Up to the present we have divided armature windings into parallel or
multiple-circuit windings and series or two-circuit windings. Of these, the
former are suitable for machines for large currents, and the latter for machines
for high pressures. As a general rule, that type of winding is chosen which
gives a suitable current in the armature wires. If, for example, the armature
current be 300 amperes, a series winding would give a current of 150 amperes
in each armature path. Experience shows this to be a suitable value, re-
quiring a copper strip of large cross-section and wasting little space in
insulation. A series winding would therefore be chosen in such a case. If,
on the other hand, the armature current be 600 amperes, the current per
armature wire would be too heavy with a series winding, and it would be
necessary to adopt a parallel winding.
128
Electrical Engineering
E,
M
८,
ig
A difficulty which occurs with all the parallel windings already described
is due to the fact that each branch of the armature winding lies entirely
under one pair of poles. If now the
various pairs of pole have different num-
bers of lines of force, the electromotive
forces induced in the different branches
will be different. This may cause great
differences between the currents in the
various branches. As an example, let us
assume that an electromotive force E, of
115 volts is induced in the upper half,
consisting of two parallel branches, in
Fig. 115. As the result of inequality

E2
m
Fig. 115.
What
in the magnetic flux, let the electromotive force E, in the lower half be
114 volts. Let the resistance R' of each double branch be 0.05 ohm.
will be the current in each branch when the total current is 100 amperes ?
Since the terminal pressure is the same for each branch, we have
or
Wherefore
E₁ — ¿¿R' — E‚ — i¿R′ = e,
—
2
115 – ¿½. 0·05 = 114 — 2. 0·05.
ii 20.
=
As the total current is 100 amperes, we have
¿= 60 amperes,
i₂ = 40 amperes.
The greater load and therefore the greater currents under some pole-pairs
may lead to destructive sparking at the commutator. It is therefore best to
provide such armatures with equalising connections by means of which these
differences between the currents in the different branches are equalised within
the armature itself, and the same current is carried by each brush.
The difficulty can, however, be overcome to a large extent by using
Arnold's series parallel windings. Each branch of the armature winding is
distributed around the whole armature so that any inequality between the
poles acts in a similar manner on every branch. The winding progresses
in one direction only, and, after finishing a coil, passes to the corresponding
coil under the next pair of poles.
Up to this point the winding does not differ from the series windings
already described. These latter were, however, two-circuit windings, whereas
Arnold's winding has more than two parallel paths. This result is obtained
by so arranging the winding that, after going once round the armature, we
arrive at the coil next but one, or next but two, as the case may be, to
the coil from which we started; in the two-circuit winding we arrived at
the next coil.
To make this plain we shall consider a bipolar machine and examine the
effect of winding successively every other coil, so that it is necessary to
go twice round the armature to complete the winding. This is shown in
Fig. 116, where the step is 2, and the number of coils 15. s and y have no
Series-parallel ring-winding
129
common factor and the winding is therefore simply closed on itself, although
electrically it is divided into four parallel paths. The brushes must be wide
enough to cover more than one segment. In general, a winding-step of a
gives a winding with 2a parallel paths.
In this way we are led to the following rule for multipolar machines with
ring-windings: If, after passing once round the armature, we come to the
ath coil from that at which we started, the winding will be divided into
2a parallel paths. It is easy to see that the series ring-winding previously
described was merely a special case of the series-parallel winding, in which a
was made equal to 1.

N
+
S
Fig. 116.
We shall confine ourselves to the more important case in which the
number of parallel paths is equal to the number of poles, or pa. In a ring
armature with series-parallel winding, after going once round the armature,
that is, after making p steps each equal to y coils, we must arrive at the
pth coil from that at which we started. We have, therefore,
or
pys±p,
s± p
S
y
± 1.
p
p
In this formula s may be odd or even according to circumstances. If the
winding is to be singly closed, s and y must have no common factor. In
Fig. 117, for example, s = 16 and p = 2. That gives y=9 or 7.
9 or 7. We have
chosen 7, and the winding scheme is simply 1-8-15—6—etc.
T. E.
9
130
Electrical Engineering
The connections between the coils themselves and with the commutator
are made at the front of the armature by means of connectors in two
planes.
Putting in arrows to show the direction of the current in the front end-
connections of the coils, we see that it flows towards the point joining coils 8
and 15, and also towards the point joining coils 16 and 7. This determines
the position of the positive brushes, and gives us the following path for the
armature current:
14
13
15
12
16
11
2
9 16
4 13
6 15
12
5
14 7
3
10
1
8
N
2
+

3
4
5
+
S
10
N
Fig. 117.
9
A moment later each of the four brushes will make simultaneous contact
with two segments. The negative brushes, for example, will short-circuit the
coils 3 and 11, the short-circuited path being as follows:
I, II, 3, III, IV, 11, I.
Thus, the coils 3 and 11, lying in the neutral zone, are short-circuited at
the given moment by the negative brushes. In general; both the positive
and negative brushes short-circuit p coils in series. Since a closed circuit can
be broken at one point without destroying the electrical connection between
all the parts of the circuit, it is possible to remove one of the positive and one
Series-parallel ring-winding
131
of the negative brushes. The remaining brushes must be wider than one
segment. Each brush is connected to the segment, from which the other
similar brush has been removed, through the coil lying in the neutral zone.
50. Series-parallel drum-winding.
We have already drawn attention to the similarity in principle between
ring- and drum-windings, and we will now use this fact to find a formula for
the winding-step of series-parallel drum-windings. We saw in the last section
that, for a ring-winding to have 2a parallel paths, it is necessary that a com-
plete circuit round the armature should bring us to the ath coil from that at

N
22
S
N
20
21
19
18
24
23
15
2
I
I
11
6
S
N
14
S
13
Fig. 118.
which we started. Now the space occupied by a coils on the ring is repre-
sented by 2a coil-sides on the drum. In a complete circuit of the drum
armature we make p double steps of y₁+y, coil-sides. For a pure parallel
winding (pa) we have, therefore,
1
p (y₁ + y2) = s ± 2p.
Since the two sides of a coil always embrace an even number of other
coil-sides, the steps y, and y₂ must both be odd. The sum y₁+y, is therefore
even and y₁+y, and s have always the common factor 2. Except for this
they must have no common factor if the winding is to be singly closed.
9-2
132
Electrical Engineering
If the winding-step is the same both back and front, we have
S
y = 20 ± 1.
2p
Here s and y must have no common factor. If, for example, s = 24 and
p=3, we get y = 5 or 3. The latter value, i.e. y = 3, would not give a singly-
closed winding since s and y would have a common factor 3. Starting at 1
we would only make eight steps before arriving again at 1, and the winding
would thus be closed after only a third of the armature had been wound. The
complete armature winding would thus be triply closed, that is, it would
consist of 3 distinct windings. If, on the other hand, we choose the value
y=5, we get the winding 1-6-11-16—etc., which is singly closed.
This winding is shown in Fig. 118, in which the back end-connections are
shown dotted, while those at the front are arranged in two planes in the usual
way. The number of segments is half the number of coil-sides. Putting in
the arrows to show the direction of the current in the front connections, we
find the position of the brushes. Positive brushes are placed where the
currents flow together from both sides, while negative brushes are placed
where the current flows away in both directions. The paths followed by the
current are as follows:
11
6 1 20
16 21 2 7
8
13
13 18 23
3
22
22 17 12
24
5 10 15
4
+
19 14 9
A moment later the brushes will make simultaneous contact with two
segments, and the negative brushes, for example, will shorten the following
circuit and cut it out of the main armature circuit:
I, II, 14, 19, III, IV, 22, 3, V, VI, 6, 11, I.
.
Three coils will be short-circuited at the same time by the positive
brushes. The coils short-circuited in this way lie in the neutral zone. In
general, p coils or 2p coil-sides in series are short-circuited by both the
positive and negative brushes. What was said concerning the leaving off of
some of the brushes in the ring armature holds equally well in the case of
drum armatures.
As in other cases, this winding is very simple when applied to toothed
armatures with the slots numbered consecutively and each containing two
coil-sides. The rule for a series-parallel winding is then: If, after going once
round the armature, we come to the ath slot from that at which we started,
the armature will have 2a parallel paths. If Y₁ and Y, are, as before, the
winding-steps with regard to the slots, we have, for a pure parallel winding
(p=a), the formula
p (Y₁+Y₂)=S±P,
S
1
or
Y₁+Y₂ == ± 1.
p
Series-parallel drum-winding
133
If, for example, as in Fig. 119, the number of coil-sides is 24, so that the
number of slots S is 12 and the pairs of poles p = 2, then we have
1
Y₁+Y₂ = 12±1=7 or 5.
2
=2
We have chosen Y₁ =3, Y₂-2 and obtain the winding shown in Fig. 119.
It is found in practice, however, that sparkless commutation is difficult to
obtain with series-parallel windings, especially when the winding is such that,
N

+
Χ
X
S
+
十二
​ハ
​Y
1/
X
X
|1
N
Fig. 119.
commencing on the neutral zone, and passing once round the armature, we
arrive at a coil-side under a pole-tip. To ensure successful operation, the use
of equalising connections is to be recommended. If the number of segments
is divisible by the number of pairs of poles, which is desirable for the sake of
symmetry and sparkless commutation, there can be no difficulty in determining
which segments should be interconnected. It is by no means necessary that
all the segments should be so connected, since a small fraction of the total
number of possible connections is generally found to suffice.
CHAPTER VII.
51. The excitation of dynamos.-52. The field magnets.-53. Position of the brushes.—
54. Armature reaction.-55. Sparkless commutation.-56. Three-wire dynamos.
51. The excitation of dynamos.
The earliest machines in which electromotive force was induced by the
motion of a wire in a magnetic field were provided with permanent magnets
of hardened steel. They were built up of laminations which gave much
stronger magnets than could be obtained from massive steel. Even these
compound magnets, however, gave a relatively small magnetic flux, so that
the electromotive force and current of a machine of reasonable dimensions
were small.

пиши
$
N
N₁
S₁
Fig. 120.
It was a great advance, therefore, when Dr Wilde of Manchester passed
the current obtained from the two-part commutator of a small permanent-
magnet dynamo around the wrought-iron poles of a second dynamo. Since
wrought-iron has a high permeability, it follows that the magnetic flux in
this latter machine, and therefore also the electromotive force induced, were
very large. An arrangement similar to that of Wilde is shown in Fig. 120,
in which N₁ and S₁ represent the steel magnet of the exciting machine. The
current taken from this machine is passed through the field coils of a larger
dynamo with magnets of wrought-iron. The coils must be wound in such a
way that one pole is north and the other south. When, for example, the
bottom end of the left-hand coil lies in front, the bottom end of the right-
hand coil, to which it is connected, must be behind. The substitution of
electromagnets for permanent magnets had been proposed and carried out
1
$
The excitation of dynamos
135
by Wheatstone and Cooke in 1845, the exciting current being obtained from
batteries.
One of the greatest factors in the enormous progress made by electrical
engineering during the latter years of the nineteenth century was the dis-
covery of the self-exciting power of a dynamo with wrought-iron magnets.
This discovery was made almost simultaneously by several workers in different
countries, and the name of dynamo was coined by Werner von Siemens in
1866 in his paper describing such a machine. This discovery, together with
that of the commutator with a large number of segments, laid the foundation
of modern electrical engineering.
The word dynamo is now used in a broader sense to signify all machines
for generating electromotive force by electromagnetic induction, whether
these machines be self-exciting or not. It is usually confined, however, to
direct-current generators and is rarely applied to alternators.
The process of self-excitation in a direct-current dynamo is as follows.
On starting, the armature conductors cut the lines of force due to the


N
X
S
оос
1
Fig. 121.
remanent magnetism and thereby induce a small electromotive force. If the
circuit is closed as in Fig. 121 this produces a small current, which, passing
round the field coils, strengthens the remanent magnetism. The electro-
motive force is thereby increased and with it the current, and so on. In this
way the machine gradually builds up its field. The strength of field and
electromotive force finally reached depend on the construction of the machine
and on the conditions under which it is working. It might appear at first
sight that the magnetic field would increase until the iron was completely
saturated. This is, however, not the case, quite apart from the fact that the
saturation of the iron is a very indefinite idea. As a matter of fact, the
current increases until the product of current and resistance of circuit is equal
to the electromotive force induced by the corresponding flux.
This increases the difficulty of understanding the action of the dynamo,
since the flux, which produces the electromotive force, is itself dependent on
the current. We are thus led to the conclusion that although the current, in
accordance with Ohm's law, is dependent on the electromotive force and the
resistance, yet, on the other hand, the electromotive force itself is dependent,
on the current and therefore on the resistance. We must therefore give up
136
Electrical Engineering
our previous view, according to which the electromotive force and the resist-
ance of the circuit are the given quantities, from which we can calculate the
current. The electromotive force of a dynamo is not merely a function
depending on constants of the machine such as flux, speed, or number of
wires on the armature. Over and above all these, it is dependent on the
conditions under which it is run, that is, on the resistance between its
terminals, or the current taken from it.


******
+
N
S
www
*
Fig. 122.
There are several ways in which the magnet winding of a self-excited
dynamo may be arranged. In the series dynamo (Fig. 121) it consists of a
few turns of thick wire, through which the main current passes. Hence the
armature, field, and external circuit are all connected in series.


N
+
******
احد
*****
N
S


ww
Fig. 123.
*****
ww
* * * * X
W
Fig. 124.
In the shunt dynamo (Fig. 122) the field winding consists of a large
number of turns of fine wire. This winding is connected directly to the
brushes and is thus in parallel with the external circuit. Since the resistance
The excitation of dynamos
137
of the field winding is large, the current through it is very small compared
with the normal current in the external circuit.
In the compound dynamo (Figs. 123 and 124) the field winding consists
of a combination of shunt and series windings. It is really a shunt dynamo,
the magnetic field of which is strengthened by means of a few series turns.
The coils must be wound in such a way that the currents in the two windings
are in the same direction. In Fig. 123 the shunt winding is connected
directly to the brushes, and is thus in parallel with the series winding and
the external circuit. This is known as the short-shunt arrangement. In
Fig. 124 the shunt winding is connected to the terminals of the machine, and
is simply in parallel with the external circuit. This is known as the long-
shunt arrangement. The object of the compound winding is to maintain a
constant or even rising pressure at the terminals of the machine in spite of
the increasing drop in the armature with increasing current.
It is of special interest to determine the conditions under which a
machine may refuse to excite. They are as follows:
I. If the initially induced electromotive force is too small, i.e.
(1) with too little remanent magnetism,
(2) at too low a speed.
II. If the magnetising current produced is too small, i.e.
(1) with large brush-contact resistance,
(2) with large external resistance on series dynamos,
(3) with large field-circuit resistance on shunt dynamos,
(4) with very small external resistance on shunt dynamos, such as a
short-circuit, which would cause the terminal pressure to sink to
zero and the magnetising current to vanish.
III. If the flux produced either does not strengthen the remanent
magnetism at all or not sufficiently, i.e.
(1) with large air-gaps between poles and armature,
(2) with wrong field connections for a given direction of rotation,
(3) with a wrong direction of rotation for given field connections.
A reversal of the polarity of a machine does not prevent its excitation,
but simply reverses the polarity of the brushes.
52. The Field Magnets.
As a rule the field magnets consist of a number of parts of different
material bolted together. The materials used are stampings, wrought-iron,
cast-iron, cast-steel, and ingot-iron.
Wrought-iron could be used throughout for the magnets of separately
excited dynamos or for motors and would be preferable in some cases on
account of its high permeability. It could not be used throughout in a self-
exciting dynamo on account of its small remanent magnetism which would
probably be insufficient to enable the field to build up. Its use might there-
138
Electrical Engineering
fore be confined to the pole-cores, which must be kept as small as possible,
not only to save iron but to make the length of each turn of the field coil as
short as possible and thus save copper.
Stampings of sheet-iron are used for the pole-shoes in order to diminish
the eddy losses caused in them by the armature teeth.
Cast-iron was formerly largely used for the field magnets of dynamos and
motors. On account of its low permeability it is now little used except for
small machines, in which the magnet frame and bed plate are cast in one. In
this case the cheapness of construction and the low price of the cast-iron more
than compensate for the increased weight of copper and iron due to the low
permeability.
In large modern machines, however, cast-iron has been replaced by cast-
steel. It possesses the twofold advantage of a high remanent magnetism and
a high permeability, hardly differing from that of wrought-iron. The flux
density can be carried much higher than in cast-iron while requiring relatively


Fig. 125.
Fig. 126.
few ampere-turns on the poles. Whereas it was formerly common to find a
half of the total number of ampere-turns used up in driving the flux through
the cast-iron field magnet, it is now common in large machines for the whole
path in iron, including armature core and teeth, to require no more than a
quarter of the total ampere-turns*. The weight of iron and of copper on the
field magnets are thus both reduced by the use of cast-steel. This makes the
machine lighter and less clumsy in appearance and improves the ventilation
Moreover, the higher flux-densities which can be used have an appreciable
influence on the commutation, as we shall see later on. Against these
advantages we have to place the higher price of cast-steel and the probable
increase in cost of manufacture due to the magnet frame being no longer in
one piece with the bed-plate.
Turning now to the shape of the field magnets we find that the early
machines were of the horse-shoe type. It was soon seen that the long pole-
cores, first introduced, were quite unnecessary and only lengthened the path
* This depends, however, on the purpose for which the machine is designed. In machines
which have to light glow lamps without a battery connected in parallel, a half of the total
ampere-turns may be required for the iron and for armature reaction. (See Section 59.)
The Field Magnets
139
of the magnetic lines. It was further found that the best form from a mag-
netic point of view was a well rounded type following the path of the lines of
force. So long as a line of force remains in the same medium it never forms
a sharp corner and the sharp edges in Fig. 125 only add to the weight of the
machine. The edges should be rounded off as shown in Fig. 126.
The Manchester type shown in Fig. 127 is of special historical interest.
The two field coils are connected so that their magnetomotive forces act in
parallel. The lines of force flow upward in both cores and produce at N a
so-called consequent pole. In calculating the necessary ampere-turns it is
best to divide the machine into two parts by a vertical plane through the
spindle and calculate the ampere-turns for each half separately. Each core
A


N
ΠΟΠ
S
Fig. 127.
Fig. 128.
carries a half of the armature flux. The ampere-turns so calculated must be
placed on each core since the magnetic effects of the two cores are in parallel
and cannot be added.
The arrangement shown in Fig. 128 is commonly known as the Lahmeyer
type and is now very largely used. The winding is completely enclosed by
the iron. Practically all multipolar machines are built on this principle, each

N
S
S
N
Fig. 129.
pole carrying a field coil (see Fig. 150). In Fig. 129 is shown an arrangement
whereby a four-pole machine can have two field coils. This arrangement has
been used in traction motors where space in one direction is very limited.
The unsymmetrical arrangement may, however, lead to sparking at the com-
mutator.
A great number of machines were made in Germany with the poles inside
140
Electrical Engineering
the armature. Mechanical difficulties with the rotating external ring armature
caused this type to be abandoned. For alternating current generators, how-
ever, this arrangement is the better, except that the armature is stationary
and the internal poles rotate.
We must now compare these various types of machines with respect to
magnetic leakage.
By the term leakage flux we mean those lines of force which pass from
pole to pole through the air instead of going through the armature. As a
result of this leakage, the flux through the magnet cores is much greater than
that through the armature and it is only this latter which is usefully employed.
The uselessly increased flux through the poles and yoke necessitates a greater
number of ampere-turns in the field coils.
If
then
N = the flux entering the armature from a north pole,
N₁ = the flux leaking from the pole on all sides,
Nm=N+N₁ = the total flux in the pole core,
λ=
Nm
N
= the leakage coefficient
.(80).
To determine this leakage coefficient experimentally a small secondary
coil is wound round the pole-core as shown at the bottom of the right-hand
core in Fig. 125. The ends of this coil are connected to a ballistic galvano-
meter. When the exciting circuit is broken the field collapses and the lines
of force cut through this secondary coil. The swing of the ballistic galvano-
meter is a measure of the flux which has collapsed in this way. The experiment
is repeated with the secondary coil wound round the armature so as to enclose
the armature flux. The ratio of the swings obtained in the two cases gives
the leakage coefficient X.
As average values we find that
λ = 1.5 for the Manchester type,
λ = 1.36 for the horse-shoe type,
λ = 1.1 to 1.2 for the Lahmeyer and multipolar types.
The leakage is greatest in the Manchester type, because the field coils are
on the part of the magnetic circuit most remote from the armature, and the
top and bottom masses of iron are of large extent and have the full difference
of magnetic potential between them. It has little effect to cut away the
metal at A (Fig. 127) and so follow a little more closely the natural path of
the lines.
The leakage of the horse-shoe type is considerable, since the magnet cores
run close together and the polar surfaces are large and cause a lot of external
leakage. Rounding the poles, as shown in Fig. 126, reduces the leakage by
causing the leakage paths to be longer and less convenient.
The Lahmeyer type (Fig. 128) is more favourable since the field coils are
close to the armature and the leakage paths have consequently a small cross-
section.
We must remember, however, that the values given above for λ are only
rough approximations, and that the leakage coefficient for machines of the
The Field Magnets
141
same type may vary considerably. The leakage will be greater, for example,
with rectangular poles than with round poles under the same conditions, and
greater when unlike poles are near together than when they are far apart.
The leakage coefficient is greatly influenced by the magnetic reluctance of
the path of the useful flux. If, for example, the air-gap between poles and
armature is specially large, a great number of ampere-turns will be required
to drive the main flux through this reluctance. This causes a correspondingly
large leakage flux and therefore a high leakage coefficient.
This becomes clearer if we look upon the leakage and useful fluxes as two
parallel currents of strengths inversely proportional to the magnetic resist-
ances of the paths. The ratio of the leakage to the useful flux is thus-equal
to the ratio of the reluctance of the main path to that of the leakage path.
It is evident that machines of the same type may have very different leakage
coefficients.
On the other hand the value of λ may vary for the same machine with
the conditions of load, etc., since the flux through the armature is not always
the same.
For the same reason the saturation of the teeth and therefore
their reluctance varies with the load on the machine. In addition to this,
the ampere-turns on the armature itself represent a magnetic back pressure
equal to X, ampere-turns, as we shall see in Section 54. These back ampere-
turns increase with the load and necessitate an increased number of ampere-
turns on the field, which leads, as we have already seen, to an increased
leakage.
This can be taken into account by calculating the magnetic reluctance of
the leakage path, if the shape of the machine lends itself to simple geo-
metrical measurement. If A, and ly are the cross-section and length of the
leakage path, we have seen in Section 29 that its reluctance is
=
Ն
Ri=
Δι
(81).
If Xa+Xt+ X, ampere-turns are necessary to overcome the reluctance
of core, teeth and gap and X ampere-turns are required to overcome the
back ampere-turns of the armature, then the magnetomotive force between
the poles will be
M.M.F. = 0·4π (Xa+Xt+ Xg+Xo).
From the so-called Ohm's law of the magnetic circuit we have for the
leakage flux
N₂ =
Ni
M.M.F.
Xa+ Xt + X g + Xv
= 0·4π
Rɩ
Rr
..(82).
The total flux in the pole-cores is then Nm = N+ N₁. An example of
such a calculation of the leakage is given in Section 29. Too much reliance,
however, must not be placed on such calculations, since they are only roughly
approximate.
When the machine does not lend itself to geometrical measurement the
matter is more complicated and the results even less trustworthy. In multi-
polar machines it is safe to assume that the polar arc is 70 per cent. of the
pole-pitch, so that the interpolar space is 30 per cent. of the pitch. The
142
Electrical Engineering
TD
2p
length l of the leakage path is accordingly proportional to the pitch
The cross-section A, of the leakage path is certainly proportional to the axial
length L and roughly proportional to the radial length of the pole. This
latter may
be taken as approximately proportional to the pole-pitch, so that the
Dπ
cross-section of the leakage path is roughly proportional to L .
In the
2p
1
same way as we find the electrical resistance from the formula
the magnetic reluctance to be
P
we get
Α
Dπ/2р c
R₁ = C₁· L. Dπ/2pL
c₁ is here a factor having different values depending on the type of
machine. It might be called the specific reluctance of the leakage path.
Kapp uses the geometrical mean of the armature length and the pole-pitch
instead of the armature length L. This alters the above formula to the
following:
C
R₂ =
L
P
.(83),
where c is a constant for which Kapp gives the following values:
Manchester type
c = 0.15,
Horse-shoe type
c = 0.36,
Multipolar or Lahmeyer type
...
c = 0·44—0·69.
If, for example, the diameter of the armature is 60 cms., its length
40 cms., the pairs of poles 2, and the value of c be estimated at 0.44, we get
C
Ri
D
L
p
0.44
√1200
= 0·013.
If the number of ampere-turns required to drive the main flux from pole
to pole be 10,000, we get for the leakage flux
N₁ = 0·4π
10,000
0.013
= 1.106 lines.
We must remember here again that such calculations of the leakage are
only rough approximations, but are generally sufficiently accurate for ordinary
design.
53. Position of the Brushes.
We saw in Section 43 that the current in each division of the armature
flowed towards the neutral zone. The brushes must therefore be placed in
the neutral zone, or rather, since the commutator connections are often bent,
in such a position that a coil passes from one division of the armature to
another while it is in the neutral zone. If the brushes are placed in the
wrong position the individual coils of a branch of the armature winding lie
Position of the Brushes
143
under opposite poles and are therefore the seat of opposing electromotive
forces, which neutralise each other and lower the electromotive force of the
machine. The correct position of the brushes is on the diameter at right-
angles to the lines of force through the armature. This is called the neutral
axis.
If the brushes are displaced, not only is there a loss of electromotive
force, but the coil, or coils, short-circuited by the brush are moving in a
magnetic field and are therefore the seat of an induced E. M. F. The resistance
of the short-circuited coil is very small and the current produced may
may be
very
large. When the short-circuit is broken by one of the segments leaving the
brush there is a vicious spark which rapidly destroys the commutator.
Hence, to obtain sparkless commutation, it is necessary that the brushes be
placed more or less in the neutral axis.

N
SI
Fig. 130.
1 N₂

B
N
N,
C
D
Fig. 131.
1
We find, however, that when the armature carries a current, the magnetic
flux through it is distorted and twisted round in the direction of rotation.
The neutral axis found at no-load, which we may call the geometrical neutral
axis, since it is geometrically symmetrical, is no longer the magnetic neutral
axis or the point where no E.M.F. is induced in the moving armature coil.
This true magnetic neutral axis is twisted round with the field in the direc-
tion of rotation. This is shown in Fig. 130; it can be shown experimentally
by means of iron-filings.
The direction of the magnetic neutral axis can be found by considering
that the armature itself is an electromagnet tending to drive lines of force
through its interior in the direction of the diameter through the brushes.
The resultant flux N is compounded of the flux N₂ produced by the armature
and the flux N, produced by the field-magnets.
2
2
By applying either the swimming or the screw rule the direction of N, is
found to be BA, while that of N, is evidently BD (Fig. 131). Since, for the
correct position of the brushes, the brush-axis is perpendicular to the resultant
flux N, and N, acts necessarily along the brush-axis, it follows that the angle
ABC is a right-angle.
2
144
Electrical Engineering
If a represents the double displacement angle (Fig. 130), the angle
α
through which the brush is moved is and we have
α
N2
sin 2-N
N₁°
1
N
With the brushes in this position the machine gives its highest terminal
pressure for the given armature current. It is important to notice that the
reluctance of the path of the lines generated by
the armature is very different to that of the lines
generated by the field-magnets. A calculation of
the theoretically correct position of the brushes by
means of the above formula is practically impossible.
In practice, however, the brushes are pushed forward
until they are almost under the pole-tip. The
current in the coil under the brush has to reverse
its direction during the time of short-circuit. This
reversal is retarded by the self-induction of the
coil (see Section 35). The brushes of a dynamo
are therefore moved beyond the magnetic neutral
axis (Fig. 132) so that an E. M. F. is induced in the
short-circuited coil in such a direction as to neutralise the effect of self-
induction and make the commutation sparkless. The small loss of E. M. F.
thus caused is quite negligible.

54. Armature Reaction.
S
Fig. 132.
If the brushes are placed in the neighbourhood of the pole-tip, the whole
armature winding may be looked upon as consisting of two parts of which the
first lies within the angle a (Fig. 130) and the second within the remaining
angle 180-a. The windings constituting the first part lie within the
neutral zone while those of the second part lie mainly under the poles. As
can be seen from Fig. 133 the windings in the neutral zone are directly
opposed to the field magnets and tend to produce a flux in the direction of
the dotted arrows. This can also be seen from Fig. 131 where the armature
flux N₂ has a component acting in the opposite direction to the magnet flux
N₁. The actual resultant flux, which passes from pole to pole through the
armature, is due to the difference between the field ampere-turns and these
back ampere-turns X of the armature. If z is the total number of wires on
z α
the armature, the number of wires included in the angle a is (Fig. 133).
360
This is therefore the number of back armature turns. If there are 2a parallel
ia
paths through the armature, the current in each armature wire is
2
•
and for
2a'
the back ampere-turns we have
ia
Xo=
2a
·
z.a
360
.(84).
This equation is equally applicable to series or parallel windings, drum or
Armature Reaction
145
ring armatures. In the ring armature shown in Fig. 134 the coils in each
neutral zone embrace the single cross-section of the ring and a pair of
opposite coils could be replaced by a single coil embracing the whole arma-
ture as in a drum-winding. In either case two external wires represent one
turn opposing the field magnets. The effect of the armature back ampere-
turns is to decrease the flux as the load comes on the dynamo, and thus cause
a decrease in the electromotive force. In calculating the necessary ampere-
turns for the field-magnets when designing a machine, it is necessary to add
the back ampere-turns X to those required to overcome the reluctance of the
iron and air-gap. As a rule X is from 10 to 15 per cent. of the total excita-
tion, but as modern dynamos work with highly saturated iron, that is, on the
flat part of the magnetisation curve (Fig. 149), a large number of back
ampere-turns causes but a small drop in the electromotive force. The
demagnetising effect is largely reduced by the use of short-chord windings
(p. 118).


N
४
α
S
Fig. 133.
S
Fig. 134.
We turn now to the remaining windings lying within the angle 180° — a
and practically all under the poles. These produce what are known as cross
ampere-turns, since they tend to produce a flux through the armature at
right-angles to the main flux. These lines of force may be looked upon as
leakage; they follow paths through the air-gap and pole as shown in Fig. 135.
They have their greatest density under the pole-tips, since the lines, which
cross the gap at this point, link all the coils lying under the pole. At the
point where the armature passes under the pole, they oppose the main flux,
while at the point where the armature leaves the pole they act in the same
direction as the main flux. The cross ampere-turns do not therefore directly
either strengthen or weaken the armature flux, but merely distort it as
shown in Fig. 130. Now the pole-tip at which the armature passes under
the pole is of special importance, since it is in this neighbourhood that the
short-circuited coil has to find a suitable external field to reverse its current
and give sparkless commutation (see Section 55). If, then, the cross-mag-
netisation is too strong, the field under this pole-tip will be too weak, or may
even become negative, making sparkless commutation impossible. As a
rough approximation it may be assumed that a flux density of 1,500 lines
per sq. cm. is necessary in the case of drums and 2,500 in the case of ring-
armatures to ensure sparkless commutation.
T. E.
10
146
Electrical Engineering
The fact that a stronger field is required for the commutation of a ring-
armature does not indicate that the cross-magnetisation is greater in this
case. The reason is to be found in the greater self-induction of the ring
winding due to the inside of the ring not being entirely free from lines of
force. To overcome this greater self-induction during the interval of short-
circuit the ring-armature requires a stronger commutating field than the
drum-armature.
N


·S--
Fig. 135.
o . . . . . . . . . . . . . ......
Fig. 136.
If B, represents the mean flux-density in the air-gap and B. the flux-
density under the pole-tips due to the cross-magnetisation, then the resultant
flux-density under the commutating tip is
B = B。 — Bc.
Now the ampere-turns of cross-magnetisation are
X₁ = 2.ß ia
с 360 * 2a
(85).
The principal reluctance in the path of the cross ampere-turns is that
due to the air-gap, and, neglecting the reluctance of the iron, we have from
equation (43) on page 60
where l is double the air-gap in cms.
Combining this with the equation
0·4π. Xc
Bc=
lg
0·4π . X。
Bg
lg
g
we have for the actual strength of the commutating field
B = B,― B₁ = 0·4π .
g
9
lg
The empirical conditions for sparkless commutation, which were men-
tioned above, may therefore be expressed as follows:
Xg - Xe
0.4π.
≥ 1,500 for drum-armatures
…..(86),
lg
0.4π.
Xg - Xe ≥
2,500 for ring-armatures....
.(87).
lg
$
Armature Reaction
147
We see from these equations that the influence of the cross ampere-turns
increases with the armature current and that we have here a limit to the
overload capacity of the dynamo. This limit is not merely due to the fact
that very large loads would cause a prohibitive temperature rise in the
armature, but is fixed by the number of cross ampere-turns beyond which
sparkless commutation is impossible. In designing a dynamo it is necessary
above all things to keep the cross-magnetisation small, and to this end the
following means are adopted:
1. From equation (85) it can be seen that a small polar-arc ß is an
advantage. As a rule the polar-arc is about of the pole-pitch.
2. It is an advantage to have a small number z of wires on the armature.
To obtain the necessary electromotive force, this will necessitate a large
magnetic flux.
3. The high flux-density in the air-gap thereby rendered necessary
(8,000—10,000) will lead to a large value of X,, which, again, will make it
easier to fulfil the equations (86) and (87).
4. The value of X, can be increased by increasing the air-gap, the two
having a constant ratio. Since X. is thereby unaffected the result is to
increase the left-hand side of equations (86) and (87). It must be borne in
mind, however, that large fluxes and long air-gaps necessitate a great number
of field ampere-turns.
•
5. By making the teeth narrow, so that their flux-density is very high
(18,000—24,000), their permeability is largely reduced. This is practically
equivalent to an increase in the air-gap and its effect is the same as in
number 4*. The common idea that the highly saturated teeth reduce the
self-induction of the short-circuited coil is entirely wrong, since, when the
coil is short-circuited, the teeth on either side of it are not saturated.
6. Attempts have been made to reduce the cross-magnetisation by
making a slit across the pole as shown in Fig. 136. These attempts have
not proved successful even when the slit was extended so as to entirely
divide the magnet system. The failure is possibly due to the large area and
narrow width of the slit, whereby its reluctance to the cross-flux is very
small.
7. The cross-magnetisation is neutralised in a very effective manner by
means of auxiliary poles placed in the centre of the neutral zone. These
poles are excited by the main current and produce a commutating field, the
strength of which is proportional to the armature current, that is, to the
current which has to be reversed in the short-circuited coil. The brushes
can therefore be fixed in the geometrical neutral axis. This idea, originally
suggested by Menges, has been developed by Swinburne, Fischer-Hinnen
and, latterly, by many others †. In Fig. 137 the cross-magnetisation acts from
right to left, while the magnetomotive force of the auxiliary poles acts from
left to right. Machines of this type can be very heavily overloaded without
causing sparking at the brushes and the maximum output is limited solely
by the heating of the armature. The use of auxiliary poles is extending
* See Fischer-Hinnen, "Gleichstrommaschinen." 5th edit., 1904.
+ See several articles in the "Elektrotechnische Zeitschrift," 1905–1906.
10-2
148
Electrical Engineering
very rapidly, since the designer can then economise in many "directions
without fear as to the commutation, and may thus more than compensate for
the cost of the auxiliary poles.

X
X
N
S
Fig. 137.
Another method, due to Déri, is theoretically more perfect than the fore-
going, but is more difficult to carry out. The field-magnet has no salient

X
N
+
X
X
X
X
8
X
X
+
X
+
+
+
S
Fig. 138.
१.
+
poles, but, like the stator of an induction motor, is a ring carrying a drum-
winding (Fig. 138). This drum-winding is represented by the outer ring in
the figure and produces a field running vertically through the armature from
Armature Reaction
149
top to bottom. In the same slots is the compensating winding, which carries
the main current and which would produce a field running horizontally from
left to right. This field is, however, exactly equal and opposite to that
which would be produced by the cross-magnetisation. The latter is therefore
completely neutralised. By making the axis of the compensating winding
slightly inclined to the horizontal in the direction of rotation, it can be made
to strengthen the main field, and thus to counteract the drop of pressure
with increasing load. The dynamo is therefore compounded*.
55. Sparkless Commutation.
In order that we may clearly understand the process of commutation and
find the conditions for sparkless running, we shall confine ourselves to the
simplest case, in which a brush is never in contact simultaneously with more
than two segments. We shall also neglect the influence of neighbouring
wires which may be undergoing commutation at the same time under
another brush. We shall consider
1. the reversal of the current in the short-circuited coil under
the influence of the brush contact resistance, that is, neglecting self-
induction;
2. the effect of self-induction;
3. the conditions for sparkless commutation in the magnetic
neutral axis;
4. the neutralisation of the effect of self-induction by shifting
the brushes.
1. The coil is assumed to have no self-induction and commutation is
assumed to occur in the magnetic neutral axis. The width of the brush is
taken to be that of one segment and the resistance of the coil and com-
mutator connections is neglected because of its smallness. We shall make
use of the following symbols:
I the current in one branch of the armature,
t the time, reckoned from the moment of short-circuit,
T the duration of short-circuit,
i the short-circuit current at the moment t,
the current in the segment coming under the brush,
2 the current in the segment leaving the brush,
R the contact resistance of the whole brush,
r₁ the resistance between brush and segment coming under it,
1
r₂ the resistance between brush and segment leaving it.
2
If we neglect the variation of contact resistance with current density, the
resistances r₁ and r₂ are inversely proportional to the contact surfaces. We
have then
T
r₁ = R .
2
R.T
r₂ = T - t'
* See Arnold, "Die Gleichstrommaschine," 1902. Vol. 1., p. 405.
150
Electrical Engineering
The current 21 flows from the external circuit to the brush and there
divides into two unequal portions. The current flowing through the resist-
ance r, is seen from Fig. 139 to be
made up of the current I of the
right-hand half of the armature
and the current i of the short-cir-
cuited coil. We have then
i=I-i and i₂=I+i.
These equations will still remain
true when the current i has been
reversed and become negative.
Applying Kirchhoff's second rule
to the short-circuit, and neglecting
the small resistance of the coil itself,
we have
—¿½rı+¿½r₂ = 0.
1
21
↓

R₁
R2
Fig. 139.
Substituting the values found above for i, i, r, and r,, we have
R.T
t
R.T
− ( I − i ) ¹³ ; ¹² + (1 + i) T — t
By solving this equation for i, we find that
= 0.
t
i = 1 — 21.
T
A

G
t=7
0 t=0
t=
iz
E
D
t
*
T-t
Fig. 140.
In Fig. 140 the time t is plotted along the horizontal axis OG, while the
corresponding values of the short-circuit current i are set up as ordinates.
From equation (a) we see that, when
t = 0, i=I,
T
t=½, i=0,
2
t=T, i=- I.
We thus get the straight diagonal line shown in the figure. At any time
t from the commencement of short-circuit, represented by OC, we have
i = BC, i=I-i- AB, i₂ = I + i = BD.
Sparkless Commutation
151
The ordinates measured between the horizontal line EF and the diagonal
represent the current i, in the trailing tip of the brush. Now the area of
contact of the trailing tip is proportional to T-t. The current density
under this brush tip is therefore proportional to
i₂
T-t'
which we see from the
21
figure is a constant ratio equal to We see then that the current density
T
under the brush is constant at every stage of the commutation. There is
therefore no sparking, since the current in the trailing tip decreases pro-
portionately with the decreasing area, until as the brush leaves the segment
the current has fallen to zero. At the same moment the current in the coil
is - I, so that the coil passes into the new branch of the winding without
undergoing any change.
5
H
A
K
4
M
B
3
2. As a matter of fact the conditions are not nearly so favourable, as the
short-circuited coil produces a flux, which, on the collapse of the old current
and the establishment of the new
reverse current, cuts the coil itself
and induces in it an E.M.F. opposing
the change of current. This re-
tardation of the process of com-
mutation, due to the self-induction
of the coil, causes the current to
depart from the straight line, or
uniform rate of change, and follow
a curve as shown in Fig. 141. The
short-circuit current no longer
passes through its zero value at

T
2'
the time but much later.
To find the shape of the above
curve we apply Kirchhoff's 2nd
law to the short-circuit, taking
into account its self-induction.
2
C
2
3
4
5 6
7
8
9
20
1
2
3
4
t
D
8
Fig. 141.
T-t
We know that the E.M.F. of self-induction E, is in the direction of the
decreasing current, that is, at the moment shown in Fig. 139, in the direction
of i. We have then
8
E¿=¿½r₂ — ¿½rı.
di
2
dt
Substituting for E, the equivalent – L. from equation (68) on page 85,
and putting in the values found for r₁ and r, on page 149, we have
1
di
- L.
ջ
dt
R.T
T—t
R.T
i.
t
It follows from this that
di
R.Ti
(b),
dt
L
T-t t
where i₂=I+i and I-i.
The solution of this differential equation has
152
Electrical Engineering
been given by Arnold. We shall obtain the short-circuit curve, however, by
a graphical method*. For the sake of simplicity, we shall assume that
so that
T=10, I=5, L=1, R= ÷
R.T
1.
I
Ꮮ
R=†
In the first place we shall determine the slope of the curve at its
beginning, i.e. for t=0. Assuming that this part of the curve is straight,
we see from Fig. 141 that
or
di i
dt t
Now, when t=0, i₂ = 21 and T-t-T, so that equation (b) becomes
21
t
t
11
R.T (21 21
LT
(홀​)
21.R
t
L (1 + R₂T)
L
This equation gives us the slope of the curve for the moment t=0. In
our case we have
21
t
10.to
1+.10
= 1.
ì
We draw then from the point H in Fig. 141, a straight line having this
inclination to the horizontal and cutting the ordinate t=1 in the point K.
HK is the first part of the curve, and we must now determine the next
part. When t=1, we have
i=0·5, 95, t=1, T-t-9.
From equation (b) we have
di
dt
1
1 (9,5 - 0,5) = -0.555.
This, then, is the trigonometrical tangent of the angle made with the
horizontal by the geometrical tangent to the curve at the point K. We draw
the line KM having this slope, and have now, with sufficient approximation,
got the curve from H to M. At the point M we have
i=1055, i₂ = 8945, t = 2, T-t-8.
di
dt
From these values we can calculate and draw the next portion of the
curve, and so on, until the whole curve is obtained. At any given moment
we have AB=i₁, BC=i, BD=12. The value of i finally becomes negative.
It is evident that the current changes very rapidly towards the end of the
short-circuit.
3. To determine the conditions for sparkless commutation in the mag-
netic neutral axis we shall find the current density under the trailing
brush-tip. We have already mentioned that the current density is inversely
proportional to the contact area, that is, to T—t.
* See Rothert, E. T. Z., 1902, p. 865.
Sparkless Commutation
153
We have then, neglecting a constant coefficient,
where c is the current density.
v₂
C
T-t'
During the last moments of commutation, when t is nearly equal to T,
the value of ¿ will approach 21, since practically the whole current passes
through the new segment. If we assume that the end of the curve is a
straight line making the angle a with the horizontal, we have
di
i2
tan α(t=T)
dt
T-t'
V2
di
At this final moment
T-t
dt
>
i 21
or, using equation (b) and putting
t
T›
C
R₂T (0-27)
T
21. R
Solving for c, we have
C
R.T-L
.(88).
It follows from this equation that with commutation in the neutral zone
the current density under the trailing brush-tip can only then have a positive
R.T
L Ꮮ
finite value, when is greater than 1. This result is found to agree very
well with the conditions for sparkless commutation derived from actual prac-
tice when carbon brushes are employed. With copper brushes, however, the
resistance R is very small and the above condition cannot be fulfilled, but
yet the machines run sparklessly. This may be due to a large increase in
the contact resistance with a slight and inappreciable amount of sparking.
From the condition
R.T
>1,
L
we see, in the first place, that, to obtain sparkless running, it is advantageous
to have large brush contact resistance, and perhaps even to increase the
resistance of the short-circuited coil by the use of high-resistance connectors
between armature and commutator. Carbon brushes are consequently used
in almost every case in preference to those of copper. In electrolytic
dynamos, however, where the currents are large and the pressures very
small, carbon brushes cannot be used on account of the pressure drop they
introduce.
We see further the unfavourable effect of high speed in reducing the
time available for commutation. This can be overcome to some extent by
making the brush wider and thus lengthening the period of commutation.
The fact that several neighbouring coils will then be short-circuited at the
same time considerably complicates the problem.
It is of prime importance to reduce the self-induction of the coils as much
as possible. To this end the armature slots are open and coils of a single
turn used wherever possible.
154
Electrical Engineering
To calculate the self-induction of the coil it is necessary to multiply the
flux produced per ampere by the number of turns in the coil and by the
number of such coils short-circuited in series. Very often another conductor
in the same slot and carrying current in the same direction is short-circuited
at the same time under a brush of opposite polarity. This has the effect of
doubling the inductance of the part of the coil within the slot. The calcula-
tion is greatly simplified by making an assumption due to Hobart, who
showed that the flux produced per cm. per ampere was about 4 lines within
the slot and about 0·8 line in the end connections.
The effect on the above relations of making the brush wider than a single
segment can hardly be followed mathematically. Hobart assumes that the
self-induction increases proportionately with the number of coils simul-
taneously short-circuited. This, however, hardly appears compatible with
the good results obtained by using wide brushes. These good results are
due to the lengthening of the time T and to the effect of a neighbouring
short-circuited coil in reducing the self-induction of the active coil. This
latter effect is due to the currents induced in the neighbouring short-circuited
coil, which by Lenz's law oppose any rapid change of flux.
R.T
If the inequality>1 be multiplied by I, we get the condition
21.R>
21.L
T
21, 1.
The left-hand side represents the ohmic pressure-drop in the contact
resistance of the brush when the current is uniformly distributed, while the
right-hand side is equal to the average electromotive force of self-induction
of the short-circuited coil (see page 86). The condition for sparkless com-
mutation is therefore that the ohmic drop is greater than the mean E.M.F. of
self-induction.
It is customary, however, to discuss the problem from the point of view
of another E.M. F., the so-called reactance voltage. This is determined by
means of a formula, which we shall use in connection with alternating
currents, viz.
E,=w.L.I.
I is the current in one branch of the armature winding, and w = 2π~ is
the angular velocity of the alternating current vector. ~ is the number
of complete periods per second, and since in our case a half period occupies
the time T, the number of periods per second is given by the equation
1
2T*
The reactance voltage, calculated in this way, should not exceed 1 volt
for sparkless commutation in the neutral zone. There is, however, much
uncertainty in such predictions as to whether a machine will run sparklessly
or not, and we shall not enter into an actual numerical example.
4. The conditions just established for sparkless running refer to machines
with their brushes in the neutral zone, such, for example, as reversible motors.
The brushes of generators are displaced in the direction of rotation so that
Sparkless Commutation
155
commutation takes place in a field of suitable strength (Fig. 132). We have
already seen that this commutating field induces an electromotive force Ex,
which opposes and counteracts the electromotive force E, of self-induction.
To show the effect of the commutating field we shall consider two specially
simple cases. We saw above under (1) that a constant current density under
the brush, that is, a straight line short-circuit curve, is advantageous. We
shall assume that, under the joint action of the electromotive forces E, and
Ex, the current curve becomes linear. We see then from Fig. 140 that the
di 21
constant value of is and we have, at every moment,
dt T
E₂ = - L.di
21
- L.
T
k
Hence, to obtain a straight line short-circuit curve and constant current
density under the brush, the externally induced electromotive force Ex must
be constant, that is, the short-circuited coil must move in a field of constant
strength during the whole time of commutation. If this is so it is evident
that the self-induction may be as large as we please. As a matter of fact,
however, such a field cannot be obtained, and the strength of the field, in
which commutation occurs, varies with the load, owing to the distortion pro-
duced by armature reaction.
It would evidently be advantageous if the current density under the
trailing brush-tip were already reduced to zero at the moment of leaving the
segment. To find the condition for this we apply Kirchhoff's second law:
Eş - Ek = i2. r½ — i₁ . r₁.
8
2
We make the following substitutions from our previous considerations :
E‚=-L.di and i̟,r,—ir,= R . T. (동​).
dt
2
1
At the last moment of short-circuit we have
T t
i2
di
21 21
and
T-t
dt
t T
Our expression of Kirchhoff's law may therefore be written thus:
in
L.
T-t
Solving for
i
T-t'
in
Ex=R.T.
T.(
(1-2).
T-t T
which is the current density s, we get
21.R-Ex
S
R.T-L
.(89).
ᎡᎢ
I
This equation shows that the current density is infinitely large when
= 1, even when commutating in an external field. There is, however, an
exception, and that is when Ex=2IR.
For RT
L
> 1 the current density under the trailing edge of the brush is
zero, if the externally induced electromotive force E becomes equal to 2IR,
that is, exactly equal to the pressure drop in the brushes. Whether this
156
Electrical Engineering
ᎡᎢ
condition can also be obtained when is less than one is not quite certain,
I
although equation (89) certainly points to this conclusion*.
The whole question depends, therefore, on the production of a commutat-
ing field of suitable strength. The most obvious way of obtaining this is to
suitably shape the pole shoes. The passing of a coil-side from the neutral
zone into the active field must be gradual. Bad sparking has often been
cured simply by rounding the edges of the poles. Skewing the sides of the
pole shoes so that they no longer lie parallel to the armature slots appears to
facilitate the correct placing of the brushes. Dolivo-Dobrowolsky's idea of
connecting up all the pole shoes by means of an iron ring encircling the
armature appears to be obsolete. By this means the gradation of the field
was made very gradual, but the commutating field was very sensitive to
armature reaction. It was therefore very satisfactory when the brushes could
be shifted with changing load.
The requirements of modern specifications, whereby machines must run
sparklessly from no-load up to 25 per cent. overload without shifting the
brushes, have greatly increased the difficulties. As a fact, most machines
will work sparklessly at half-load, but at no-load and full load, with the
brushes in the same position, the commutation depends largely on the effect
of the cross-magnetisation (see page 145).
56. Three-wire Dynamos.
The advantages of the three-wire system of distribution were pointed out
on page 25. At the same time we mentioned the earliest and simplest
method of obtaining the intermediate pressure, viz. by connecting two
dynamos in series. Since, however, two small machines are dearer than
a single large one, a number of other methods have been tried and are in
use to-day.
It is simple to obtain an intermediate pressure when a battery works
in parallel with the dynamo. The middle or neutral wire is then joined
to the mid-point of the battery. This arrangement has, however, the dis-
advantage of unequally working the two halves of the battery.
The method most commonly adopted is to connect two similar motors,
which are rigidly coupled together, in series and to connect the neutral wire
to their common terminal. Such a set is called a balancer (Fig. 142). When
both sides of the system are equally loaded, the two machines run light, and
with similar construction and equal excitation the pressure across the mains
will be equally shared between them. If now the two sides of the system
are unequally loaded, as shown in the figure, the machine I will act as a
generator and deliver the extra current to the positive, overloaded side of
the system. The motor II will take enough current to drive the generator I
and to cover its own losses. To make the action of the set as clear as possible
we shall assume that the ohmic resistance of the machines is negligible.
* The validity and meaning of the above equations have been warmly discussed in the
E. T. Z., 1906.
A
Three-wire Dynamos
157
If the no-load current of each motor is i, and the pressure across each half
of the mains e volts, then the no-load power taken by the balancer set is
2e.io. If now the machine I acts as a generator and supplies a current x,
the load on it is e.x. The power required by the motor II is therefore
2e.i+e.x. The current through it will be 2i +x. From Kirchhoff's first
law we know then that the current in the middle wire is 2+2x. This
must be the difference between the currents & and i₂ in the outers. We
have then
ii₂ = 2io + 2x,
or
XC =
i1 — iz
2
io.
If, for example, i₁ = 200, i, = 150, ∞ = 5, then the current in the generator I
will be
X =
200-150
2
5 = 20.
The machine II will take the current
x+2i20 + 10 = 30,
and the current in the main dynamo (Fig. 143) will be 180 amperes.
+

1
I
T
e
in
zij 2x
I
Ziotx
i 2
Fig. 142.
*
+

180
20
180
200
50
30
150
Fig. 143.
As we shall see in Section 63, a back E.M.F. is induced in the motor,
which, if we neglect the armature resistance, is equal and opposite to the
terminal pressure. Since the machines I and II are similar, and have the
same excitation and speed, it follows that their electromotive forces are
the same.
Hence, their terminal pressures are equal and the balance of
pressure is perfect.
As a matter of fact, however, the result will be affected by the armature
resistance, which is not negligible. The terminal pressure of the generator
will be given by the formula
e = E - ia. Ra,
and that of the motor by the formula
e=E+ia. Ra.
As the generator is connected across the more heavily loaded side, this
side will have the lower pressure. The difference can be diminished by
exciting each machine of the balancer set off the opposite side of the system.
The field coils of the generator would then be supplied from the pressure
of the lightly loaded side, thus increasing its E. M. F., while the E.M.F. of the
158
Electrical Engineering
motor would be decreased. The same result can be obtained by passing
the middle-wire current round a few series-turns on the fields of each
machine.
A simple and interesting method of obtaining a middle point is that due
to Dolivo-Dobrowolsky (Fig. 144). Two points of the winding, separated
from each other by a pole-pitch, are connected to two slip-rings. By means

N
S
Fig. 144.
*
*
*
*
*
*
*
N₂
N₁
of brushes these rings are connected to the terminals of a so-called choking
coil, a coil of large self-induction wound on an iron core. An alternating
current flows through this coil and produces lines of force, which cut through
the turns of the coil and induce electromotive forces opposing the current.
The alternating current passing through the coil is therefore only large
enough to produce this flux, and is prac-
tically negligible. The coil has a very
small resistance to continuous currents.
If now the middle wire is connected to
the middle point of the coil, the pressure
difference between it and each of the outer
wires is a half of the terminal pressure of
the dynamo. A more perfect arrangement
was proposed by Sengel, who connected
three points of the winding, separated by
of the pole-pitch, to three slip-rings.
The three brushes were connected to the
terminals of three choking coils, the other
terminals of which were connected together

Sz
S₁
Fig. 145.
and to the middle wire (compare three-phase star connection).
Dettmar solved the problem in another manner by connecting the middle
wire to an auxiliary brush placed on the commutator half-way between the
main brushes (Fig. 145)*. As this auxiliary brush short-circuits two seg-
ments whenever it bridges the insulation between them, it is necessary that
the coil, thus short-circuited, should be moving in a weak field. This is
* See E. T. Z., 1897, pp. 55 and 230.
Three-wire Dynamos
159
2
arranged by leaving a gap in the middle of the pole. The figure represents
really a two-pole machine, the north pole of which is split into two parts N₁
and N₂ with a gap between them. In order that the pressure of either side
of the system may be regulated without interfering with the other, the field
coils of the poles N₁ and S, must be connected in series and controlled
separately from the field coils of the poles N, and S₁, which are likewise
in series, since diametrically opposite wires on the armature are connected in
series.
1
2
It can be easily seen that the cross-magnetisation of the armature will
lower the pressure of one half of the system and raise that of the other. As
in the case of the balancer it is advantageous to excite the field magnets of
one side from the pressure of the other side of the system.
A
CHAPTER VIII.
57. Characteristics of separately-excited dynamo.-58. Characteristics of series dynamo.-
59. Characteristics of shunt dynamo.-60. Dynamo and battery in parallel.-61. Efficiency
of dynamos.
57. Characteristics of separately-excited dynamo.
In this machine the field magnets are excited from an external in-
dependent source (Fig. 146). In the first place we shall consider the effect
of varying the speed when the machine is running light, i.e. unloaded, with
constant excitation. From equation (79), on page 120, the electromotive
force is known to be
n
E=P.N. z. 10-8 volts.
α
60
Since, under the assumed conditions, n is the only variable on the right-
hand side of the equation, the electromotive force must be proportional to

HHHHHHF - -
|-----

www
О
в
n
e
Fig. 146.
Fig. 147.
the speed. If then we plot the observed values of the speed as abscissae and
the corresponding values of the electromotive force as ordinates, we shall
obtain a straight line through the origin as shown in Fig. 147.
If now we vary the experiment by keeping the speed constant and varying
the exciting current by means of a resistance in series with the field winding
(Fig. 148), and plot the observed electromotive forces against the corre-
sponding values of the exciting current, we obtain the curve shown in
Fig. 149. This curve is called the no-load characteristic, or magnetisation
The electromotive force increases proportionately with the magne-
tising current up to a certain point, beyond which, however, the electromotive
force increases less rapidly owing to the increasing saturation of the iron.
curve.
Characteristics of separately-excited dynamo
161
The values of the ordinates are obtained by connecting a voltmeter across
the dynamo terminals. At no-load the P.D. between the terminals is equal
to the E.M.F. induced in the armature, which is proportional to the flux N.
The ordinates in Fig. 149 therefore represent to a certain scale the values
of N corresponding to the various values of the exciting current. Hence
we are quite justified in calling the curve the magnetisation curve of the
dynamo.
There is, however, an important difference between the magnetisation
curves considered in a previous section and this no-load characteristic. There
the abscissae represented the ampere-turns per cm. and the ordinates the
flux density or lines per sq. cm., whereas here the abscissae represent the
ampere-turns for the whole path, partly in iron and partly in air, and
the ordinates represent the total flux.

HHHHHH------

le
TUM
гип
به
im
e
Fig. 148.
Fig. 149.
We will now predetermine the no-load characteristic of a given machine,
shown quarter-size in Fig. 150. The armature teeth and the air-gap are
shown full-size in Fig. 151. We must first determine the cross-section of
the magnetic path in the armature core, teeth, air-gap, pole-cores and yoke.
We have,
diameter of armature...
internal diameter of armature (in this case the
diameter of the shaft)
axial length of armature
depth of slot
breadth of slot
number of slots
distance from pole to armature....
angle subtended by pole
D= 15 cms.
D₁ = 3.5 cms.
L= 11 cms.
d = 2 cms.
b = 0·5 cm.
space lost in paper insulation between laminations...
cross-section of magnet yoke ring…………..
S = 36
= 0.2 cm.
B = 70°
=
15%
37 sq. cms.
For the cross-section of the armature-core normal to the flux, we have
Aa
=
0·85 (D — D。 – 2d) . L
2d). L = 70 sq. cms.
To find the effective cross-section of the teeth we must subtract the sum
of all the slot-widths from the periphery measured half-way down the slots,
T. E
11
162
Electrical Engineering
and multiply the remainder by
LB
360*
Taking into account the factor 0.85 for
the paper insulation, we have
At = 0·85 [(D − d) π – S. b]
L.B
360
= 41.5 sq. cms.

.--Ն
Նյ
Fig. 150.
If we make the simple but approximate assumption that the lines of
force cross the air-gap in the manner shown in Fig. 151, the cross-section
of the gap A, is the mean of the area of the pole-face and the total area of
the tops of the teeth facing the pole. If the double air-gap has a length
lg = 0'4 cm.,
the area of the pole-face is
(D+lg) π
·
B.L
360
= 104 sq. cms.
The tops of the teeth have an area
L.B
(Dπ – 36b) = 62 sq. cms.
360
As the mean of these two values, we get
104 +62
Ag=
83 sq. cms.
2
Characteristics of separately-excited dynamo
163
The cross-section of the round pole-cores can be seen from the drawing
to be
π
A₂ = 7·42.
Ap
43 sq. cms.
4
Since the flux of each pole divides into two equal parts in the yoke, the
effective cross-section of the yoke is twice that of the actual magnet ring,
that is,
Ay=2.3774 sq. cms.

Fig. 151.
The lengths of the paths in the different portions of the magnetic circuit
can be found from the drawing. It is important to notice that the air-path l
is twice the gap between armature and pole, while the path in the teeth
is twice the depth of a tooth. Similarly lp is twice the length of a pole
since the flux passes through both a north and a south pole. For ly, how-
ever, we put the simple length through the yoke from pole to pole. The
pole-shoes need not be separately considered. The values so obtained are
collected in the following table:

Armature core Armature teeth
(stampings)
(stampings)
Air-gap
Pole-core
Yoke
(W. I.)
(C. L.)
Aa=70
At=41·5
la= 7
4 = 4
A,= 83
lg=0·4
A₂ = 43
A₁ = 74
lp = 12.5
ly = 22
We have now to determine the number of ampere-turns required to drive
the flux through each individual part of the magnetic circuit. For this
purpose we shall assume four different values of the flux, and work them
out side by side. By dividing the flux by the cross-section of the part under
consideration, we get the flux-density B, and by reference to the curves in
Fig. 58, the necessary ampere-turns per cm.
This must be multiplied
X
by the length of the path in order to get the total number of ampere-turns
for the part of the magnetic circuit under consideration.
11-2
164
Electrical Engineering
*
1. ARMATURE CORE.
N= 0.2
0.4
0.5
0.6.106
N N
Ba
2850 5700
7150
8550
Aa 70
from the curve for stampings =
0.5
1.1
1.4
1.8
α
X₂ = L (†). -" (†).-
Xa la 7 <=7
α
2. ARMATURE TEETH.
3.5
7.7
9.8
12.6
a
N = 0.2
0.4
0.5
0.6.106
N
N
Bt
4800 9650
12000
14400
At 41.5
from the curve for stampings =
0.8
2.3
4
9
t
Xt = lt
t
(+). = 4 (-4). -
X
3.2 9.2
16
36
t
3. AIR-GAP.
For the air-gap, we have the fundamental equation (43) on p. 60,
0·4π. Xg
Bg= Hg
lg
or
Xg=
Bg.bg
= 0·8 Bg.lg.
0.4π
We have, therefore,
N
= 0.2
0.4
0.5
0.6.106
N N
Bg
2400 4820
6020
7220
Ag 83
Xg=0·8Bg • lg = 0·32B, 770 1540
1930
2310
=
4. POLE-CORES.
In calculating the ampere-turns for the poles and yoke we must take
into account the fact that the flux in them is larger than the armature flux,
on account of leakage. Assuming that the leakage coefficient
Nm
we have
(+1)
P
λ
=
1.2,
N
Nm = 1.2N.
N
0.2
0.4
Nm = 1.2N =
0.24
0.48
0.5 0.6.10°
0.6 0.72.106
Nm Nm
Bp
5600
11200
14000
16800
Ap 43
from the curve for wrought iron =
1.6
5
14
100
Xp=lp
X, - 4 (†), - 12·5 (7),
=
20
62.5
175 1250
Characteristics of separately-excited dynamo
165
!
5. YOKE.
N=
0.2
0.4
0.5
0.6.106
Nm = 1·2N=
0.24
0.48
0.6
0.72.10°
N
N
m
m
By
3250
6500
8100
9720
Ay
74
from the curve for cast iron
4.1.
19
53
107
Y
Xy
X, = (†), . ly = 22
2 (7),
90
418
1160
2350
N

0-6×106
0.5
A
B
04
0.3
0.2
0.1
P
Xy
D
Xg
E
ΣΧ
120
100
80
60
40
20
1000
2000
3000
4000
5000
6000 X
Fig. 152.
The total ampere-turns can now be found by adding up the five com-
ponents, thus
N= 0.2
0.4
0.5
0.6.10€
Xa=
3.5
7.7
9.8
12.6
Xt= 3.2
9.2
16
36
Xg
X, 770
1540
1930
2310
Xp
X₂ =
20
62.5
175
1250
Xy
90
418
1160
2350

EX=X+X+X+X+ Xy
== 890
2040
3290
5960
α
α
In Fig. 152 the values of armature flux N are plotted as ordinates, and the
values of X, Xy, Xg, and EX as abscissae. The curve X, is naturally a straight
line. The values of X and Xt are so small that they have been omitted in
the figure. It is easy to see from the figure how the total ampere-turns EX
are used up in the various parts of the machine. To produce the flux OA,
for example, we have to employ an excitation AE. Of this, a part AB is
166
Electrical Engineering
required for the poles, a part AC for the yoke and the larger part AD for
the air-gap, and AE=AB+AC+AD.
Near the origin the curve for EX coincides very nearly with the straight
line X,, while as the saturation of the iron increases it becomes gradually
flatter.
The electromotive force can be calculated directly from the flux N by
means of equation (79) on p. 120,
E=2.N.
α
n
60
z. 10-8 volts.
If, for example, the speed is 1665 revs. per minute and the number
of wires on the armature periphery is 720, we have for a parallel wound
armature
E = N.
1665
60
.720.10-8 = 200. N. 10 volts.
The ordinates of the curve EX give the electromotive force directly on
the scale shown on the right-hand side of the figure.
It must be particularly noticed that EX is the number of ampere-turns
per pair of poles, since the calculation was made for a complete magnetic
circuit linking both a north and a south pole. The total number of field
ampere-turns on the whole machine is p. EX.
Having thus fully investigated the effect on a separately excited dynamo
of varying its field current, we must now consider the effect of loading the
machine, that is, taking current from it. We assume that both its speed
and excitation are kept constant while its terminals are connected by means
of an external resistance such as a bank of glow-lamps. The greater the
number of lamps connected in parallel across the terminals the bigger is
the load on the dynamo. This increase in the current taken from it causes
a decrease in the P.D. between the terminals. When running on open circuit
the terminal pressure is equal to the electromotive force, but when loaded it
is less by the pressure drop due to the armature resistance. If the resistance
of the armature be Ra, the terminal pressure will be given by the formula
e = E - i. Ra.
If we neglect the effect of armature reaction, the electromotive force will
be constant since the exciting current is constant, and it can be represented
by the horizontal line E in Fig. 153. The pressure drop in the armature
is proportional to the current, that is, to the abscissae, so that the terminal
pressure is given by the straight line e drawn at an angle a below the line E,
where
i. Ra
tan a =
=
2
Ra
When setting out the angle a it must be carefully noticed that the scales
for the ordinates and abscissae are probably different, so that we cannot draw
simply the geometrical tangent. The difference between the ordinates of
the curves E and e is equal to the ohmic pressure drop for a certain value
of the current, while the ordinates of the curve e give us the corresponding
terminal pressure. Experimentally the process is of course reversed, since we
•
Characteristics of separately-excited dynamo
167
observe the terminal pressure and add to it the calculated armature drop to
find the electromotive force.
The results can be plotted in another manner, namely, by setting out
the external resistance R, i.e. the quotient, as abscissae and plotting both e
and i as ordinates (Fig. 154). Since the product i (R+ Ra) is equal to the
constant electromotive force, the curve representing i must be a rectangular
hyperbola. Its asymptotes are the horizontal axis and a vertical line through
a point representing - Ra. The curve i cuts the vertical axis in the point A.
This corresponds to no external resistance, or, in other words, the dynamo
E
is short-circuited and OA is the short-circuit current. This point lies,
Ra
naturally, quite outside the working limits and could only be found by
plotting the curve for a very small value of the excitation.

Fig. 153.
E
α
e
Ra
A

i
E
e
R
Fig. 154.
The limits of the curve e in Fig. 154 can be found from the consideration
that the terminal pressure is equal both to the product iR and to the
difference E-i. Ra. On short-circuit it must be zero, since the terminals
are electrically equivalent to a single point and can have no difference of
potential. The whole electromotive force is then used up in driving the
enormous short-circuit current through the armature resistance. On the
other hand, the terminal pressure attains its maximum value and becomes
equal to the electromotive force E when the external resistance becomes
infinitely great, i.e. on open-circuit.
As we have already indicated, the curves would be modified in an actual
case by the demagnetising effect of armature reaction. Moreover, the
armature resistance is not constant, but increases with its temperature
and therefore with the load. It is hardly possible, however, to take these
points into consideration in a simple manner. Another disturbing influence,
which we have neglected and which we shall continue to neglect, is the effect
of remanent magnetism.
168
Electrical Engineering
58. Characteristics of series dynamo.
Both the consideration and the actual working of a separately excited
dynamo are extremely simple, since with constant speed and constant excita-
tion the electromotive force is independent of changes in the external circuit.
The behaviour of self-excited dynamos is, however, far more complicated,
since every change in the external circuit affects the magnetising current,
and also, therefore, the magnetic flux and the electromotive force.
The relations are, perhaps, simplest in the series dynamo since the arma-
ture current is here identical with the magnetising current and the external
current. The series-connection of armature, field and external circuit was
that which first suggested itself. It has, however, become almost obsolete as
far as generators are concerned. It was used for supplying a number of arc
lamps connected in series, but is quite unsuitable when the lamps, as is now
general, are connected in parallel between the mains. As motors, however,
series machines are of enormous importance.
Rm
E Ra
e
R
Fig. 155.
i (Ra+Rm)
区
​

Fig. 156.
B.
We shall now consider the effect of varying the external resistance of a
series dynamo running at a constant speed (Fig. 155). We take correspond-
ing readings of the current and the terminal pressure, and plot a curve with
abscissae equal to the current i and ordinates equal to the pressure e
(Fig. 156). We then draw a straight line through the origin making an
angle a with the base line, where
tan a = Ra + Rm,
Rum being the resistance of the field winding. The ordinates of this straight
line represent the pressure drop in the dynamo itself, and by adding this to
the corresponding ordinate of the curve e we obtain the curve E of the
electromotive force. This is the characteristic curve of the series dynamo
and represents the relation between electromotive force and magnetising
current. If, instead of the current i, we had plotted the product of current
and the number of turns on the field, we should have obtained the same
magnetisation curve as that found in the previous section (Fig. 152).
Characteristics of series dynamo
169
We see from the curve that the electromotive force vanishes on open-
circuit, since when the magnetising current ceases the magnetic flux dis-
appears. If the terminals be joined by means of a reasonably small resistance,
the machine will excite and produce an electromotive force which will drive
a current round the circuit. The smaller we make the external resistance,
the larger, by Ohm's law, will be the current and therefore also the magnetic
flux and the electromotive force. This reaches its limit when the machine is
short-circuited. The terminal pressure then vanishes and the whole electro-
motive force AB is used up in driving the short-circuit current OB through
the internal resistance. This occurs at the point A, in which the straight
line cuts the curve E.
With regard to the terminal pressure we see that the curve reaches a
maximum with a certain load and then decreases as the load is further
increased. That this must be so is evident, for, as the iron becomes saturated,
a point will be reached at which the increase of electromotive force, due to a
given increase in the current, is not enough to balance the increased drop
thereby produced.
This peculiarity was turned to useful account in the machines designed
for lighting a string of arc lamps in series. It was necessary to make the
increase of current, due to the short-circuiting of a lamp, as small as possible.
The cross-section of the iron was therefore made so small that it was highly
saturated at the normal working current. The machine worked beyond the
knee of the characteristic curve E, where an increase in the current causes
but a slight increase in the E.M.F. By making the armature reaction, which
we have neglected, strong enough, an increase of current might even cause a
decrease in the E.M.F. On short-circuiting a lamp the current increased
slightly, but not so much as it would have done, had the electromotive force
increased with the current.
The peculiar behaviour of a self-exciting dynamo, in that the electromotive
force depends on the current taken, is easier to comprehend if we fix our
attention on the external circuit. The question arises as to whether the
current is produced by the electromotive force, or the electromotive force by
the current, through the medium of the magnetic flux. Both views must be
looked upon as correct, and we then have a complete cycle, without being
able to say which is cause and which effect. In practice, however, it is the
external resistance which is arbitrarily varied, and both the electromotive
force and the current are thereby caused to change simultaneously.
59. Characteristics of shunt dynamo.
It is customary, in connection with shunt dynamos, to distinguish between
an internal and an external characteristic. The internal or static character-
istic is obtained by running the machine light at a constant speed and
varying the resistance of the field circuit by means of a rheostat or variable
resistance. By plotting the values of the magnetising current im as abscissae
and the values of the terminal pressure e as ordinates we obtain a curve
exactly similar to the characteristic of a series dynamo or to the no-load
{
£
170
Electrical Engineering
characteristic of a separately excited machine and there is therefore no need
to examine it further. This curve is called the internal or static characteristic
of the shunt dynamo.
We must now consider the behaviour of a shunt dynamo when its load or
external current is varied. In the series dynamo the field winding and the
external circuit are connected in series, with the result that the electro-
motive force is largely dependent upon the external current. In the shunt
dynamo, on the other hand, the field winding is in parallel with the external
circuit, that is, it is connected directly to the brushes (Fig. 157). As the
result of this arrangement variations in the external current do not react
directly on the electromotive force. The external current has, in fact, only a
small secondary effect on the excitation and E. M. F., and, within the practical
working limits of the machine, we are dealing with a relatively constant
electromotive force. The behaviour of a shunt dynamo in actual operation is

E

ми
Ø
e
W
e
.
Fig. 157.
Fig. 158.
consequently simple, but a clear understanding of its behaviour is usually a
matter of some difficulty to the beginner.
In the first place, an increase in the external current causes an increase
in the armature current. The armature current is the sum of the external
and the shunt currents, thus
ia = i + iş.
In consequence of this increase the pressure drop in the armature is
increased and the terminal pressure falls, in accordance with the equation
e=E―ia Ra
The terminal pressure is therefore greatest at no-load and gradually falls
with increasing load. Since the field winding is connected directly across
the brushes, the decrease in the terminal pressure causes a corresponding
decrease in the shunt current. This leads to a diminished flux and a pro-
portionately smaller, E. M.F., in consequence of which the terminal pressure
decreases still further. The variation of pressure with load is therefore
greater in a shunt dynamo than in a similar separately excited machine. It
Characteristics of shunt dynamo
171
is, however, very much smaller than the pressure variation of a series
dynamo.
It is possible, however, to design the shunt dynamo so that the drop of
pressure from no-load to full load is almost as small as that of a separately
excited machine. This result is obtained by working some part of the
magnetic circuit at a high saturation. The magnetic flux is then but little
affected by considerable variation in the magnetising current, and the drop is
almost entirely due to the ohmic resistance. Such machines are specially
suitable for electric light stations, where constancy of pressure is an absolute
necessity in the interests of steady illumination. The small unavoidable
changes in the pressure which still occur are counteracted by means of
variable resistances, or rheostats, in the shunt circuit.
Machines which are to be used for charging accumulators should, on the
other hand, be designed with low flux-densities in the iron. When beginning
to charge, resistance is put into the field circuit, which, as the charging
proceeds, is gradually cut out. In this way the flux and E. M.F. are gradually
increased, and the current maintained at a constant strength in spite of the
increasing counter-electromotive force of the battery.
If now we determine the external characteristic of a given machine
experimentally by switching lamps, one after the other, in parallel across the
mains and so varying the external current, we find that the P.D. between the
terminals gradually falls. This is exactly what we should expect from our
above considerations and is shown graphically by the upper branch of the
curve in Fig. 158. If, however, we increase the number of lamps, and there-
fore decrease the external resistance, beyond a certain point, the current does
not continue to increase, as we should expect, but decreases. The terminal
pressure continues to decrease until the machine is short-circuited, when it
naturally falls to zero. The magnetising current ceases at the same time
and with it the magnetic flux which it had produced. Were it not for
remanent magnetism, there would be no E.M. F. and consequently no armature
current. The lower part of the curve in Fig. 158 would then pass through
the origin, whereas the remanent magnetism causes it to meet the base line
a little to the right of the origin.
The action of the shunt dynamo will probably be made much clearer if
we evolve the external or dynamic characteristic from the internal or static
characteristic. We shall neglect the effect of armature reaction. We plot,
in the first place, the magnetising current as abscissae and the electromotive
force E as ordinates. This is the internal characteristic and is shown in
Fig. 159. Since, in our present experiment, we vary only the external
current and leave the resistance of the field circuit unaltered, the terminal
pressure must be proportional to the magnetising current, in accordance with
the equation
e = is Rs.
For the curve of terminal pressure e we therefore get the straight line in
Fig. 159. Now the difference between the ordinates of the two curves E
and e must be equal to the pressure drop ia . Ra = (i+i). R。 in the armature,
172
Electrical Engineering
and is therefore proportional to the armature current. By subtracting the
shunt magnetising current from the armature current found in this way, we
get the external current, which, plotted as ordinates, gives us the curve i in
the figure. As the curve shows, there is one special load at which the
current is a maximum.
Since the abscissae is in Fig. 159 are proportional to the terminal pressure,
the curve i in Fig. 159 is identical, except as regards scale, with the curve e
in Fig. 158. It is, however, turned through 90° so that the left-hand part of
the curve i in Fig. 159 corresponds to the lower curve in Fig. 158, where,
owing to the large number of lamps connected in parallel, the machine is
nearly short-circuited. The right-hand part of Fig. 159 corresponds to the
upper curve in Fig. 158, that is, the working portion of the characteristic
where considerable variation in the load causes but slight changes in the
pressure.
i
Compound dynamos may be regarded as shunt dynamos, the fields of
which are strengthened when the load comes on by means of a few series
turns carrying the main current. In this
way the drop due to armature resistance
and armature reaction is neutralised and
the terminal pressure maintained constant
at all loads. If over-compounded, that is,
provided with a greater number of series
turns, the terminal pressure increases with
the load, rising, for example, in many trac-
tion generators from 500 volts at no-load
to 550 volts at full load, thus making it
possible to maintain a pressure of 500 volts
at distant points of the system.

E
e
im
Fig. 159.
Now the maintenance of a constant
pressure in spite of variations of the load
is of the utmost importance for the supply
of power to glow-lamps, since these are extremely sensitive to changes of
pressure and the effect of such changes on the eye is very disagreeable.
The compound dynamo appears therefore to be specially suitable for
electric lighting. As a matter of fact, however, it is rarely used for this
purpose and the majority of lighting stations will be found to contain shunt
dynamos only. This is due to the fact that sudden overloads cause a
momentary drop of pressure even on a compound dynamo, since the engine
governor cannot act instantaneously and the speed is consequently reduced.
This causes the E. M.F. and the terminal pressure to decrease, and this leads
to a decrease in the shunt current and a further fall of pressure. A lighting
station, moreover, is not subjected to the sudden, and often enormous,
variations of load experienced in a traction station. The changes are usually
gradual and allow the switchboard attendant ample time to counteract the
drop by means of the field rheostat. The added complication of compound
dynamos is therefore unnecessary in such a station.
Compound machines, like series machines, are quite unsuitable for charging
Characteristics of shunt dynamo
173
accumulators. If they be used for this purpose and the speed of the engine
falls on any account, the back E. M.F. of the battery may overcome the
forward E.M.F. of the dynamo and drive a reverse current round the circuit.
This current will pass through the series field winding and reverse the
polarity of the magnets. The electromotive forces of dynamo and battery
are no longer in opposition, but acting together in a circuit of very
low resistance. The current will therefore rapidly rise to a dangerous
value.
If, however, a shunt dynamo is used, the positive pole of the battery is
always connected to the positive pole of the dynamo, and the shunt current
in the dynamo field-coils is always in the same direction, even when the main
current is reversed. The electromotive forces of the battery and dynamo are
always in opposition. When the main current reverses, the battery drives
the machine as a motor and the current cannot reach a dangerous limit.
Trouble might be caused, however, by sparking at the commutator, since the
correct position of the brushes is different in the case of a dynamo and that
of a motor. This can be prevented by putting a minimum cut-out in the
main circuit. This consists of an electromagnet excited by the main current
and holding up an armature so long as the current exceeds a certain value;
if, however, the current falls below this value, the armature is released and
automatically opens a switch in the main circuit.
60. Dynamo and Battery in parallel.
The practically constant supply pressure which we saw in the last
section to be necessary for electric lighting can be obtained by the use of a
battery of accumulators connected in parallel with the dynamo. The battery,
moreover, constitutes a very desirable reserve, as the dynamo need only be
large enough to deal with the average load instead of the maximum load,
and may, at times of very light load, or in case of necessity, be shut down
without interrupting the supply. Finally, the battery enables the dynamo
to work at constant full load in spite of the variable load on the station
(buffer batteries in traction stations). The steam consumption is thereby
made uniform and considerable economy effected.
These advantages depend on two characteristic properties of the accumu-
lator, viz. its constancy of E. M. F. and its low internal resistance. As a result
of the latter the drop of pressure in the battery may be almost neglected
and the terminal pressure is practically equal to the electromotive force.
The shunt dynamo may be looked upon as separately excited, since the field
winding is virtually connected across the constant terminal pressure of the
battery. We shall consider the effect of
1. variation of load,
2. variation of engine speed,
3. variation of excitation.
174
Electrical Engineering
1. EFFECT OF VARIATION IN THE EXTERNAL LOAD.
If the speed and excitation be constant and the electromotive force be E,
then the armature current will also be constant and will be given by the
equation
ia
E-e
Ra
where e is the terminal pressure, maintained constant, as we have seen, by
the battery. We see then that neither the terminal pressure nor the dynamo
current are affected by changes in the external load. If, for example, the
battery is discharging (Fig. 160) and more glow-lamps are switched into the
external circuit, the extra current required is supplied almost entirely by the
battery discharging at a higher rate. If, on the other hand, the battery is
being charged (Fig. 161) when the extra lamps are switched on, the current
required will be drawn from the battery by taking a part of its charging

Discharging
Rm
wwwwwwww
Ra
E
ia
|ir
i
Charging

HHHHHHHHHK
Rm
wwwwwwww
Ra
E
La
i
e
е
* * * * *
Fig. 160.
*****
Fig. 161.
current. We have here, therefore, an exceptional case in which the generator
gives practically a constant current, which we divide, as we choose, between
the battery and the external circuit.
The battery will pass from a condition of charging to one of discharging
simply by changing the load in the external circuit. If, for example, the
external current is smaller than the armature current, the excess current will
pass into the battery and charge it. If, now, the external current increases
beyond the armature current, the battery will give a discharging current. If
the dynamo gives the average current, the battery will be continually
changing from charge to discharge and vice versa. This change is only
possible if the terminal pressure of the battery changes. When charging the
terminal pressure must be greater than the E. M. F. in accordance with the
equation
e = Ef+iz. Rь.
When discharging, on the other hand, the terminal pressure must be
smaller than the E.M.F. of the battery, as shown by the equation
e= Ez-iz. Rь.
These changes are small and we can therefore neglect them in our con-
sideration of the subject.
Dynamo and Battery in parallel
175
2. EFFECT OF SPEED VARIATION.
We have assumed in the foregoing that the electromotive force of the
dynamo is constant, but we shall now consider the effect of speed variation
on terminal pressure and armature current. We shall assume the external
load to be constant. When running without a battery in parallel a change
of speed has a large effect on the terminal pressure, but in the present case
the terminal pressure is maintained constant by the battery. A change in
the E.M.F. of the dynamo will, however, cause the armature current to
change, in accordance with the formula
ia
E-e
Ra
If, for example, the speed suddenly drops, the E. M. F. decreases and with
it the armature current. Since the current in the external circuit is con-
stant, the battery, if discharging, must give a greater current, the smaller
that given by the dynamo. If charging, a decrease in the dynamo current
causes a decrease in the current passing into the battery. Hence, speed
variations of the engine are converted into variations of battery current.
3. EFFECT OF VARIATION OF EXCITATION.
A change in the rheostat in the field circuit has, naturally, a similar effect
to a change of speed. If, for example, we wish to charge the battery, we
adjust the field current until the terminal pressure of the machine is slightly
above that of the battery. The switch is then closed so that similar poles of
the machine and battery are connected. Since the electromotive forces
oppose each other, the current is practically nil, but can be brought up to
the desired value by adjusting the field rheostat (compare the above formula
for ia). The reading on the voltmeter is only slightly changed, since the
terminal pressures of both dynamo and battery are identical.
To stop charging, the field current is weakened until the armature current
falls to the value of the external current. The battery is then carrying no
current, but can remain across the bus-bars or terminals to act as a pressure
regulator.
At times of exceptionally heavy load the shunt rheostat is used to
distribute the load in a suitable manner between battery and machine. The
excitation can be adjusted until the machine is working at full load, so that
the battery is left to deal with the variable overload.
We have assumed that the battery pressure is constant, and we are justified
in making this assumption, in so far as the small effect of the variable current
is concerned. We have already seen, however, that the terminal pressure of
the battery gradually changes as charging or discharging proceeds (Figs. 26
and 27). To maintain a constant pressure, battery switches are often em-
ployed, by means of which cells can be added to or removed from the battery.
In order that the current may not be interrupted when switching another cell
in or out, an arrangement is shown in Fig. 162, in which auxiliary contacts
are fitted between the main contacts and connected to them by means of
176
Electrical Engineering
resistances. The moving contact is wide enough to span across the insulation
between the segments. In the position shown the lamps are connected across
the main cells only, and the first regulating cell is short-circuited through the
resistance between the contacts. By turning the switch-arm to the right it
is made to lie on the auxiliary contact only. The first regulating cell is then
in series with the battery across the mains and by suitably choosing the
auxiliary resistance, which is now in the main circuit, the supply pressure
may be made to rise about 1 volt, depending on the strength of the current.
On moving the arm over on to the main contact the first regulating cell is
connected directly to the mains and the pressure increased by another volt.
The pressure can thus be adjusted by steps of two volts, the auxiliary contacts
being only used to pass from one step to the next.
Another device is illustrated in Fig. 163. Here the switch-arm consists
of two parts rigidly fixed together but insulated from each other. In the
position shown the first regulating cell is short-circuited through the two


AHHHHHHHHH
HHHHHHHHHP
M
Fig. 162.
* * * * *
Fig. 163.
arms and the auxiliary resistance. On turning the switch further to the right,
the main arm leaves the end contact and thereby puts the first regulating
cell in series with the battery across the mains. The auxiliary resistance is
in circuit, however, and the increase of pressure is less than two volts by an
amount depending on the current. On turning the switch still further to the
right, both arms will lie simultaneously on the second contact block, and
finally the auxiliary contact will leave the block and lie between it and the
next block. The pressure on the mains has then been increased by two volts.
The use of these simple battery switches is to be recommended where the
charging is done when there is no external load (Fig. 164). The switch is
then used during the morning as a charging switch, beginning with it on the
right-hand stop and gradually moving it over to the left as the cells become
charged. The right-hand regulating cells were probably only used towards
the end of the previous evening and are consequently fully charged long before
the main battery. During the process of charging, therefore, the switch-arm
is gradually moved from right to left. Since the battery pressure during the
charge is greater than that for which the lamps are designed, it is necessary
Dynamo and Battery in parallel
177
to keep the switch A open. When the cells are all fully charged the switch
A₁ is also opened.
When lamps are to be lighted, the switch A is put on to such a contact
that the volts across the mains on closing the switch A will be correct. The
switch A is then closed and the dynamo run up and put in parallel as already
described. The field current is adjusted until the dynamo is on full load, if
possible, and the battery is not called upon to share the load on the station
until it exceeds the full load of the dynamo. During the night the dynamo
can be shut down and the battery left to take charge of the all-night load.
To maintain the pressure on the lamps, the switch-arm will have to be moved
over gradually from left to right.
The number of regulating cells required with this arrangement can be
found from the fact that the pressure of a cell at the commencement of dis-
charge is about 2 volts, while after discharging as far as permissible it is only
1.8 volts. If the supply pressure is 110 volts the number of newly charged
cells must be 110/255. At the end of the discharge the number must be

E

HHHHH----
40
www
HHHH-----
*
Fig. 164.
Fig. 165.
increased to 110/1.860. We see, therefore, that about 10 per cent. of the
battery must be connected up to the battery switch.
The above plan of disconnecting the external circuit while charging the
battery, will hardly be practicable in the majority of cases. If we have to
maintain the supply during the time that the battery is being charged, we
must use our single battery switch as a discharging switch to maintain
constant supply pressure (Fig. 165). The dynamo must be connected to the
last cell by means of the two-way switch EL. The dynamo current divides
between the battery and the external circuit, except that the regulating cells
to the right of the switch-arm carry the whole current. Hence, the very cells
which need it least get the greatest charging current. When the charging is
complete the dynamo must be disconnected and its pressure reduced by means
of the field rheostat to that of the mains. The two-way switch is then put on
to E, and the battery used simply as a pressure regulator, until the load
increases to such an extent that battery and dynamo must work in parallel.
The disadvantages of this arrangement are evident. The regulating cells
are not only a source of waste, but are rapidly destroyed through being con-
T. E.
12
178
Electrical Engineering
tinually overcharged at a high current density. Another disadvantage is that
the dynamo must give the full charging pressure of the whole battery, which,
with a supply pressure of 110 volts and the 60 cells calculated above, would
be 60.2·7=162 volts. If, however, the end cells could be cut out as they
became charged, we should only require a terminal pressure of 150 volts.
The cost of the machine would therefore be unnecessarily high, since such a
machine is always designed for the maximum pressure, i.e. for the corre-
sponding flux in the field magnets. When the machine is working in parallel
with the battery and giving a terminal pressure of 110 volts, the flux is
reduced by means of the field rheostat and the machine is larger than actually
required for working under these conditions.
In order that the cells may be charged without interrupting the supply,
the regulating cells must be connected up to a double battery-switch (Fig. 166).
The supply mains are then permanently connected to the discharging side of
the switch, which serves as a pressure regulator both when charging and when

*
Fig. 166.
discharging. The charging switch-arm lies always to the right of the dis-
charging arm and should never be brought inside it. Both arms will be
gradually moved from right to left as the charging proceeds until finally both
are on the left-hand contacts, and dynamo, battery and mains have the same
pressure. Although the end cells are switched out as they become charged
and are consequently saved from the continual overcharging which they
suffered in the previous arrangement, the cells lying between the two switch-
arms are still charged with a current equal to the sum of the battery current
and that supplied to the lamps.
The two-way switch is now moved over to the upper contact and the
dynamo and battery are in parallel across the mains. By suitably adjusting
the field rheostat the load can be subdivided as desired. As the battery dis-
charges the lower switch-arm is gradually moved to the right, sometimes
carrying the upper one with it by means of a stop, so as to ensure that the
latter is not to the left of the former when charging is commenced.
As can be seen from the figure, the corresponding contacts of the charging
and discharging switches are metallically connected and only a single row of
contacts is actually necessary over which two switch-arms can be independently
Dynamo and Battery in parallel
179
moved. The arrangement shown in Fig. 166 is, however, more easily under-
stood.
Although the use of a double battery-switch makes it possible to charge
without interrupting the constant pressure supply, and also to switch off the
regulating cells as they become fully charged, it does not obviate the necessity
for making the dynamo capable of giving a pressure far in excess of the
normal supply pressure. The disadvantage of requiring the dynamo to work
normally at one pressure and yet be capable of giving another, much higher,
pressure when necessary, is specially noticeable in larger stations. The
machines are compelled to run normally at an output far below their maxi-
mum and the capital outlay is unnecessarily large. This disadvantage is
completely removed by the use of a so-called booster, which is a comparatively
small dynamo connected in series with the main dynamo. The latter is then
designed for the normal supply pressure, without any provision for a large
increase of pressure. The iron of the machine can be highly saturated, thus
reducing both the size and cost of the machine. A further saving is effected
in the field rheostat which need not be nearly so large.
·

----
my
Lis
Fig. 167.
An arrangement including both a double battery-switch and a booster is
shown in Fig. 167. The main dynamo is connected permanently to the
discharge switch-arm, while the booster is connected between the two switch-
arms and charges the cells lying between them. If the excitation of the
booster is adjusted until its current is equal to the charging current through
the main battery, no current will flow through the discharge switch-arm.
The latter will then resemble the galvanometer connection in the Wheatstone
bridge in that it connects two points at the same potential. The battery will
then act simply as a pressure regulator.
To determine the number of regulating cells required in this arrangement,
we have, in the first place, to be able to obtain the supply pressure of, say,
110 volts towards the end of the charging when each cell has a pressure of
about 2.7 volts. The main battery must therefore contain no more than
12-2
180
Electrical Engineering
110
2.7
= 40 cells. In the second place, when the cells are almost discharged,
the pressure of each will drop to 1.8 volts and the number then required will
be
110
1.8
= 60. The whole battery must therefore consist of 60 cells of which
20, i.e. about 30 per cent., are connected to the battery-regulating switch.
In many modern stations battery-regulating switches are more or less
eliminated by the use of automatic boosters, in which the field is produced
by the resultant effect of several distinct windings. One winding will cause
the E.M.F. of the booster to vary with the load, while another will take into
account the state of the battery. In this way the pressure of supply is main-
tained practically constant and the battery caused to charge or discharge
according as the load on the station is below or above the normal*.
61. The efficiency of dynamos.
In considering the losses occurring in dynamos we shall first consider the
no-load losses. In the first place we have mechanical friction at the bearings
and brushes. To this must be added the air-friction, or so-called windage,
which is very considerable in modern well-ventilated armatures. These
frictional losses are independent of the excitation. We have also, however,
the losses due to hysteresis and eddy currents, which depend on the excitation.
The hysteresis loss occurs in the iron of the armature and is approximately
proportional to the 16th power of the flux-density. The eddy current loss
occurs in the armature iron, the armature conductors and the pole-shoes.
This loss is proportional to the square of the flux-density, since an increase in
the number of lines causes an increase not only of the electromotive force but
also of the eddy currents. The no-load loss for a given excitation can generally
be looked upon as a constant loss.
To this no-load loss we must add:
the copper loss in the armature ia. Ra,
2
α
2
the copper loss in the series field winding im². Rm,
the copper loss in the shunt field winding i. e.
The last one of these losses is dependent on the excitation and will there-
fore vary somewhat with the load under which the machine is running. The
variation is small, however, and this loss may be considered constant. We
are assuming here that the terminal pressure is maintained constant.
The overall efficiency of the machine is the ratio of the useful output e.i
to the total power supplied to the machine. The latter is made up of the
no-load losses and the total electrical power E. ia generated in the armature.
We have then for the efficiency
n
e.i
E. ia + Po
* See "Storage Battery Engineering" by Lamar Lyndon.
.(90),
The efficiency of dynamos
181
where P, is the power taken when running light, excluding the loss in the
field windings, or, if we introduce the individual losses,
n =
e.i
e. i + ia². Ra+im². Rm + e. is + Po
.(91).
This formula applies both to series, shunt and compound dynamos. If
the compound machine has the short-shunt arrangement, that is, with the
shunt field across the brushes instead of the terminals, the pressure across the
shunt will be less than e, but the difference is so small that we may safely
neglect it. In the simple shunt dynamo im = 0 and iɑ = i + is.
We have then for a simple shunt dynamo
n =
or
n
i³.
e.i
e. i + ¿². Ra + 2i. iş. Ra + iş². Ra + e. is + P。
e
P¸+i².Ra
Po+i. Ra+e. is
2
e+2ig. Ra+i.Ra+
..(92),
The efficiency will be a maximum when the denominator is a minimum.
We must therefore differentiate this expression with respect to i, and put the
result equal to 0. We have seen above that both e and i, may be considered
constant. We will also neglect the product is. Ra on account of its smallness.
2
We get then
Pote.is
Ra
=
0,
or
¿². Ra=P₂+e.ig.
Now the external current is practically equal to the armature current, so
that the left-hand side of this equation represents the armature copper
The right-hand side represents the losses which are practically constant, viz.
the friction loss, the iron loss, and the shunt field loss. Hence, the efficiency
is a maximum when the variable losses are equal to the constant losses.
Moreover, it is specially noteworthy that the efficiency curve, plotted to a
base of variable output, is more or less parallel to the base line over a large
range on either side of the maximum point. A large variation of load has
consequently little effect on the efficiency.
As an example we shall take a machine for 110 volts and 100 amperes.
Its no-load loss, that is, the power required to drive it when excited, but
on open circuit, is 750 watts. Its full load efficiency is 0.88. The total
power transmitted from the engine to the dynamo shaft is therefore
110.100
0.88
=
12,500 watts. The total losses are
12
100
12,500 1,500 watts.
The dynamo has then its maximum efficiency at full load. The no-load loss
of 750 watts is distributed between the field coils, which take 250 watts, and
the friction and iron loss, which amount to 500 watts.
We have then
e. is
i8 = 250
=250,
= 2.3,
ia = i + is=100 + 2·3 = 102·3,
ia². Ra
Ra
=
750 750
2
га 102.32
= 750,
=
= 0.071
182
Electrical Engineering
We now assume that the dynamo is run first on half-load and then on
100 per cent. overload. Neglecting any small change in the field current,
we have
i
load
100 per cent. overload
50 and
200 respectively,
ia = i + is = 50+ 2·3 = 52-3
ia². Ra=52·32.0·071 = 194
""
33
39
202.3
2,900
250
22,000
""
e. is
e.i=output
Po
250
= 5,500
500
Input = 6,450
5,500
n =
0.85
6,450
35
""
500
25,650
22,000
= 0.86.
25,650
Hence we see that the efficiency is very constant in the neighbourhood of
its maximum, even for large variations in the load. The designer has a
certain amount of freedom in the distribution of the total losses between
constant losses and armature copper loss. The necessity for keeping down
the pressure drop in the armature makes it impossible, as a rule, to design
the machine so as to have its maximum efficiency at the normal load. In
the above example the armature drop ia. Ra=100.0·0717 volts, the
E.M.F. is 110+7=117 volts and the percentage drop in the armature is
7/1176 per cent. Moreover, if the armature resistance be made too great,
the heat generated may be more than the armature surface can safely get rid
of without reaching a dangerous temperature. In fact, the exact point of
maximum efficiency is hardly considered when designing a dynamo, and the
losses are distributed from entirely different practical considerations.
rule the armature copper losses are less than the constant losses under normal
running conditions so that the efficiency increases with the load, even up to
considerable overloads.
We shall now consider the experimental determination of the various
individual losses which go to make up the no-load loss of a dynamo. The
loss in the field windings is easily found from the ohmic resistance and the
field current, so that we need consider only the remaining losses P。. If we
have another suitable machine which can be used as a motor to drive the
dynamo we are testing, the two are coupled together and the power measured
which the motor takes to run the dynamo at the normal speed with its field
magnets unexcited. The amount by which this exceeds the power taken by
the motor when uncoupled is almost exactly equal to the friction loss in the
dynamo. If the experiment be repeated with the dynamo fields excited, the
power taken by the motor will increase by an amount Pr+ Pe equal to the
hysteresis and eddy current loss in the dynamo.
Having found in this way the loss due to friction and that due to hysteresis
and eddy currents, the experiment is repeated at another speed, the excitation
of the dynamo being the same as before.
The loss Ph+ Pe is divided by the speed n to which it corresponds
and the quotient plotted vertically on a base representing the speed n
(Fig. 168). Now the hysteresis loss is directly proportional to the speed,
t
The efficiency of dynamos
183
while the eddy current loss varies as the square of the speed, so that we
may write
Pr+ Pe=c₂n + C₂n²
..(93),
where c₁ and c₂ are constants.
From this it follows that
Pr+ Pe
C₁ + c₂n.
n
Ph + Pe
This shows that the points representing
in Fig. 168 should lie
n
Ph + Pe
on a straight line which cuts the vertical axis at a height c equal to the
hysteresis loss at a speed of one revolution
per minute. Multiplying this by the speed
in revolutions per minute, we get the actual
hysteresis loss. We have thus separated the
loss P. into its three components, friction,
hysteresis and eddy current loss.

C1
Fig. 168.
n.
In the absence of a second machine, the
losses can be separated by the so-called retar-
dation method. The machine is run light as a
motor with a constant field excitation, but with
an adjustable P.D. applied to its armature terminals. In this way the speed
is varied and the power P, supplied to the armature is observed for each
speed. In Fig. 169 the values of n are plotted as ordinates and the corre-
sponding values of P, as abscissae.

n
a
n
a
→Pu
Fig. 169.
B
0
C
-t
0
B1
C1
Fig. 170.
Fig. 171.
a
>t
The armature circuit is then opened suddenly without altering the field
excitation, so that the armature gradually slows down and comes to rest.
The speed is observed at frequent intervals during the slowing down and
plotted to a base representing the time (Fig. 170). It is sometimes more
convenient to read the armature P.D. on a voltmeter than the speed, to
which it is proportional. The decrease in the kinetic energy of the armature
during any interval is necessarily equal to the work done in the same time
in friction, hysteresis and eddies. At a certain moment t seconds after
opening the armature circuit, the speed n= AC, and the kinetic energy J in
1
184
Electrical Engineering
Energy is being dissi-
joules is decreasing at the rate of dJ in the time dt.
dJ
dt
pated, therefore, at the rate of joules per second. The kinetic energy
at any moment is proportional to the square of the speed. We have then
J=k.n²,
dJ=k. 2n. dn,
or
dJ
dn
2k.n
=
•
dt
dt'
where k is a coefficient depending on the moment of inertia of the armature.
We have just seen, however, that the energy dissipated per second is equal
dJ
to
and since the energy dissipated per second must be equal to the
dt
sum of the losses P., we have
P₁ = – 2k. n.
dn
dt
(94).
If we draw both the tangent and the normal to the curve at the point A,
we have (see Fig. 170)
and since AC=n,
Hence
dn BC
tan a=-
dt AC'
dn
n.
= BC.
0
dt
P₁ = 2k. BC.
We have already found P, for this speed in Fig. 169, and we can thus
calculate 2k. It is important to notice, however, that BC must be measured
from the figure in seconds and multiplied by the square of the ratio of the
ordinate scale to the abscissa scale, before it can be used in the above
equation. Thus, if 1 cm. vertically represent n' revs. per minute, while 1 cm.
horizontally represent t' seconds, the time in seconds represented by BC in
Fig. 170 must be multiplied by (n'/t') before substituting it for BC in the
above equation. The reason for this will be seen after a little consideration;
it is also interesting to note that, P, being a power and k a moment of inertia
multiplied by a simple number, the dimensions of BC must be T-3 or T. T−4.
As read from the curve BC is a simple time, but by multiplying it by the
square of the ratio above mentioned, the dimensions of which are T-2, the
dimensions of each side of the equation are brought into perfect agreement.
This calculation is repeated for different speeds and the average taken of
the values of 2k thus obtained.
The experiment is then repeated with the field magnets unexcited, in
which case the time taken to slow down is much longer, since there is no
hysteresis or eddy loss. In this way the curve in Fig. 171 is obtained, which
has a subtangent B'C' at the chosen speed n. In the same way as before, we
could show that the power wasted in friction is given by the equation
Pƒ= 2k. B'C',
B'C' being corrected for the scale as above.
The efficiency of dynamos
185
We can avoid the determination of 2k, and all trouble as to the scale, by
simply using the ratio
Pƒ
B'C'
Po
P. BC
Up to the present we have tacitly assumed that for a constant speed and
excitation, the hysteresis and eddy current losses are the same at full load as
at no-load. This assumption is not strictly correct, so that an accurate
determination of the efficiency is only possible when the machine is actually
running on full load. The supply of power required to test a large machine
in this way would be very considerable. A great saving can be effected if
we have two similar machines, by coupling them mechanically together and
running one of them as a motor from an external source. This motor drives
the other machine as a generator and the latter supplies current to the
motor. The external source of supply and the generator are thus connected
in parallel to supply the motor. The power supplied from the external
source is, of course, merely the total loss in the two machines, and can be
easily measured. The excitation of each machine must be so adjusted that
the speed and armature currents of the two machines correspond to the
normal running conditions. This method of testing two similar machines
was devised by Hopkinson and is known as the Hopkinson test. At first
sight, the exact conditions under which the two machines are running appear
rather complicated. We must remember, however, that both machines have
the same speed and the same terminal pressure. The E.M. F. of the generator
is greater than the terminal pressure, whereas in the motor, as we shall see
in Section 63, the terminal pressure exceeds the E. M. F. induced in the arma-
ture. Since the two machines have exactly the same speed, terminal pressure
and number of wires on the armature, it follows that the generator must be
more strongly excited than the motor, or, to put it more correctly, the
machine with the weaker excitation will run as a motor and drive the other
machine as a generator. If
and
N₁ is the flux per pole in the motor,
1
N₂, the flux per pole in the generator,
i, the armature current of the motor,
i2, the armature current of the generator,
iii, the current supplied from external source,
E₁, the back E. M. F. of the motor,
E₂, the E.M.F. of the generator,
then for the motor (see Section 63) we have
e=E₁ + i₁. Ra
P
a
N
N₁. 2.10-8+i. Rɑ,
60
Ra›
and for the generator
e = E₂ — i₂, R₂ = 2. N₂.
n
Ra
2.10-8 — i₂. Ra.
a
60
186
Electrical Engineering
Adding these two equations, we get
2e
- P. (N₁ + N₂) .
N
a
60·2.10~+i. Ra·
If we neglect the very small quantity i。. Ra, we have
p n
a 60
2e
2.10-8=
N₁+N₁°
·
The speed is thus inversely proportional to the sum of the fluxes N
and N₂. If we substitute this value of P.n
p
α'60
.z.10-8 in the above equation
for the generator, we get
e=
N₂.2e
N₁+N₂
i₂. Ra,
N2
N₂-N₁ e
1
i₂
or
1
N₁ + N₂` Ra
Hence, while the speed is proportional to
is proportional to
N,- N₁
N₁+N₂*
1
N₁+ N₂'
,
the generator current
Both the exact speed and the exact current
strength at which we wish to test the machines can therefore be obtained
by suitably adjusting the excitation of the two machines. The product e.io,
obtained from simple observations, is equal to the total losses occurring in
the two machines.
CHAPTER IX.
62. Direction of rotation of motors.-63. Torque, speed and output of a D. C. motor.-64. Motor
with constant excitation.-65. The starting and regulation of a shunt motor.-66. Principle
of series motor.-67. Example.-68. The regulation of a series motor.
62. Direction of rotation of motors.
Direct-current motors are identical in construction with direct-current
generators. Since the same machine may act at one moment as a motor and
at the next as a generator, much of what has been said concerning generators
can be applied to motors. We must consider, however, the mechanical
characteristics of motors, such as the direction of rotation, the speed, the
torque and the output, and we have specially to investigate the manner in
which these mechanical characteristics depend on the electrical and magnetic
relations.


N
X
N
X
Dynamo
Fig. 172.
Motor
Fig. 173.
To determine the direction of rotation we have only to remember that, to
rotate the armature of a dynamo delivering current, it is necessary from the
principle of the conservation of energy that mechanical work is done. The
current induced by the motion tends to stop the motion, owing to the opposing
force exerted on the conductor by the magnetic field. If the dynamo repre-
sented in Fig. 172 be rotated in a clockwise direction, the current will flow
from front to back under a north pole and in the opposite direction under
a south pole. The magnetic field exerts a force on the conductor, the direc-
tion of which can be found by the left-hand rule. Pointing the first finger of
the left hand in the direction of the magnetic field and the second finger
along the current, the thumb indicates the direction of the motoring force on
the wire. In this case, however, this is unnecessary, as we know from Lenz's
law that the force will be opposed to the motion which induced the current.
This opposing force is shown by a dotted arrow. We may imagine that
188
Electrical Engineering
the machine is continually seeking to act as a motor, but never succeeds, as
the driving force of the prime mover overcomes the motoring force.
This latter force has full play, however, if we shut down the engine and
supply the armature current from some external source. We assume that
the current in both field and armature is still maintained in the original
direction (Fig. 173). It is evident, therefore, that the motor shown in
Fig. 173 will rotate in an anti-clockwise direction as indicated by the arrow.
We conclude, therefore, that with similar magnetic polarity and similar
direction of current in the armature, a machine runs as a motor in
the opposite direction to that in which it was driven as a generator.
The same result is obtained by simultaneously reversing both the magnetic
polarity and the armature current.
The student must be careful, however, not to make the mistake of
assuming that any machine running as a generator will run, as a matter
of course, in the opposite direction as a motor. We were careful to point
out that both the magnetic polarity and the armature current were un-
changed in our example, in which the rotation was reversed. We must now
99


Dynamo
Fig. 174.
Motor
Fig. 175.
turn to the special cases of shunt and series motors and see whether this
supposition is correct.
In the series machine field windings and armature are connected in series.
We assume that the series machine shown in Fig. 174 is driven in a clock-
wise direction as a generator and that the right-hand brush is positive. We
now use the same machine as a motor without changing the connections in
any way. From what has been said above it is evidently immaterial, as far
as the direction of rotation is concerned, how we connect the motor terminals
to the mains. The simpler plan, of course, is to leave the connections as
they were (Fig. 175). The motor current is then reversed both in field and
armature. The result is the same as if the current had remained unchanged
in both. Hence, if the connections of a series generator remain un-
changed, it will run as a motor in the reverse direction, i.e. against
the brushes.
If, however, we wish the machine to run in the same direction, we must
reverse the field connections. The direction of the current in the field
windings will then be the same both as a generator and as a motor, and the
magnetic polarity will remain unchanged.
We shall now consider the case in which a machine which has been
Direction of rotation of motors
189
running as a series motor is driven in the same direction and used as a series
generator. This case is of practical importance, since series motors are
braked by being cut off from the supply mains and loaded as generators
by means of a resistance connected between their terminals. The kinetic


и
о
м
Dynamo
Fig. 176.
Motor
Fig. 177.
energy of the tramcar, for example, is thereby converted into electrical energy
and dissipated as heat in the resistance, and the car is rapidly retarded.
Now the E.M.F. induced in the motor armature is opposed in direction to the
motor current (Section 63). When used as a generator the direction of
rotation is unchanged and the field is the remanent magnetism of the motor
field, and therefore in the same direction. The E.M.F. in the generator is
therefore in the same direction as in the motor and produces a current in the
opposite direction to the motor current. This current would weaken the
remanent magnetism instead of strengthening it and the machine would
refuse to excite. Hence, it is necessary to reverse the field connections when
using the motor as a brake.
In the shunt motor, however, the relations are very different. A glance
at Figs. 176 and 177 shows us that the current in the field windings is in


+
+
т
ми
Fig. 178.
ww
Fig. 179.
the same direction, whether the machine be used as a motor or as a generator,
if the connections remain unchanged. The armature current, on the other
hand, is reversed. Had the current remained the same in both armature
and field, the direction of rotation would have been changed, but, as it has
only reversed in the armature, the direction of rotation of the motor will be
190
Electrical Engineering
the same as that of the generator. Hence, a shunt dynamo will run as a
motor in the same direction, i.e. with the brushes; similarly a shunt motor
can be run as a generator in the same direction without altering the con-
nections.
We have finally to consider the reversal of the direction of rotation of
motors. Were we merely to change the connections with the supply mains,
we should reverse the current in both armature and field windings, and the
direction of rotation would be the same as before. We must evidently reverse
the connections of either the field or armature, but not of both. As a rule it
is the armature current which is reversed (Figs. 178, 179).
63. Torque, speed and output of a D. C. motor.
An example was worked out in Section 25, showing how the turning-
moment or torque could be calculated from the flux, the number of armature
wires and the armature current. We must now find a general formula for
the torque of a motor. Let
and
H be the strength of the field in the air-gap,
L the length of the armature in cms.,
D the diameter
z the number of wires on the armature periphery,
B the angle subtended by the pole,
I the current in each wire in absolute units,
ia the external armature current in amperes.
Then, from equation (32) on page 53 the force exerted on each armature
wire under a pole by the magnetic field is
H.I.l dynes.
The total number of wires under the 2p poles is
z.2p.B
360
The total force is therefore
f = H. 1, 2 p . B . z. L
360
dynes
.(95).
To convert this into kilogrammes we must divide it by 981,000. The
radius of the armature is
r =
Ꭰ
2.100
metres.
M₁ = H. I. 2p . B .z. L
Ꭰ
For the torque in metre-kilogrammes we have, therefore,
1
360 *2.100 981,000
Now the total flux per pole N is equal to the product of the strength of
the field in the gap and the area of the pole face, or
N =
D.T.B.L
360
. H
...
.(96).
Torque, speed and output of a D. C. motor
191
Substituting this value, we have
Mt
M₁ = P. N.z. I
10-7
π.9.81
..(97).
•
This equation applies equally to all machines, bipolar or multipolar, series
or parallel wound. If, as before, there are 2a parallel paths through the
armature, we have
Equation (97) then becomes
ia
I
10.2a'
p N.z.ia
Mt=
.10-8 met.-kgs.
a 2π.9.81
.(98).
For series winding a = 1 and for parallel winding a=p. This formula
gives the total torque exerted on the armature. The useful torque is less
than this by an amount necessary to overcome the friction.
The full meaning of these formulae for the turning-moment only becomes
apparent when we exchange the two sides of the equation. In their above
form the equations state that the torque exerted by the motor is proportional
to the product of the magnetic flux and the armature current.
When a
motor is working and has reached a steady condition, it exerts a torque
exactly equal to the resisting torque due to the load on the motor.
We are
thus led to the important conclusion, that the product N. i automatically
adjusts itself to the load on the motor, so as to exert the requisite torque.
We do not send a current of an arbitrarily fixed strength through the motor
and allow the motor to exert a corresponding torque, whatever it may be.
This is only the case before the motor moves, when the current is not sufficient
to exert the necessary starting torque. As soon as the motor reaches a uni-
form steady speed, equation (98) must be satisfied and the current must adjust
its strength to suit the torque given by the load.
We have yet to answer the question as to how it comes about that the
current in the armature is just that corresponding to the load, and neither
more nor less. There is no apparent regulator by means of which the current
can be varied, since the armature has a negligible resistance. We have,
however, in the motor a back electromotive force which, like the governor
of a steam engine, is dependent on the speed. It is immaterial whether
the machine is rotated as a generator by means of external forces, or as
a motor by means of forces within the machine itself. In both cases the
armature wires cut through the lines of force, and thereby induce electro-
motive forces in themselves. A motor is therefore not only similar to a
generator in construction, enabling the same machine to be used first as
the one and then as the other, but, while running as a motor, the machine is
continually generating an E.M.F. and tending to act as a generator. By
applying the right-hand rule it can be seen that this induced E. M. F. is in
the opposite direction to the terminal pressure and current, and it is for
this reason that it is often referred to as the back E.M. F. of the motor.
The back E.M.F. is, of course, proportional to the speed, and it is this
fact which governs the adjustment of the current to the exact strength
192
Electrical Engineering
necessary for the torque. If, when the motor is running, its load is suddenly
increased, the torque which it has been exerting is no longer sufficient to
overcome the load. The speed of the motor drops, with the result that
the lines of force are not cut so rapidly and the back E.M.F. is therefore
decreased. The terminal P.D. is then able to send a larger current through
the armature. This decrease of speed and consequent increase of current
goes on until the current is large enough to exert a torque corresponding
to the new load.
Inversely, if the load be decreased, the torque exerted by the motor is
too large. The motor therefore speeds up and increases the back E. M.F.
until the current is choked down to a value corresponding to the torque
required for the new load. When the motor has settled down again to a
steady speed, the torque exerted by it is exactly equal to the resisting
torque of the load, and not slightly larger than it, as many beginners are
apt to imagine. Due allowance must be made, however, for the torque
necessary to overcome friction at the motor bearings, etc.
The speed can be found from equation (79) on page 120, according to
which
n
E P.N z. 10-8 volts.
α
60
The back electromotive force E is determined by the armature current,
for the applied terminal pressure e must exceed the back E. M. F. by an amount.
necessary to drive the current through the resistance of the armature, that
is, by ia. Ra volts. We have then
From these two equations, we have
N
e=E+ia. Ra
…………..(99).
E.60.108__ (e — ia · Ra). 60. 10º
p/a.N.z
p/a. N.z
…………….(100).
It is evident from this formula that a change of load, and therefore of
armature current, does not cause merely a momentary change of speed, but a
permanent change. The variation of speed with change of load is, however,
very small in many cases.
power of the motor,
We have purposely
torque, although in
We must now find the formula for the output or
that is, the rate at which it does mechanical work.
postponed this until after the consideration of the
practice the capability of a motor is always expressed in terms of its output
and not of its torque. To get a clear idea, however, of the action of
a motor it is absolutely necessary to consider the torque first, since the
output depends on the product of two variables, viz. the torque and the
speed.
Now, if Mt be the turning-moment in metre-kgs. and
w = 2π.
n
60
be the angular velocity, we know from mechanics that mechanical work is
being done at the rate of Mt. w metre-kgs. per second. To obtain the power
Torque, speed and output of a D. C. motor
193
"
in watts we must multiply this by 9.81 (see Section 41); in this way we get
n
P = Mt. ∞.9·81 = Mť . 2π . 9.81 watts .........(101).
60
If Mt be expressed in foot-pounds, the figure 9.81 becomes 1.36.
It is important to find the output as a function of the electrical quantities.
For this purpose, we multiply equation (99) by ia and get
e.ia = E. ia '+ ia². Ra
.(102).
The product e.ia is the total power transmitted to the armature, while
ia. Ra is the power dissipated as heat due to the armature resistance. Hence,
the remainder E. ia must represent the power transformed into a mechanical
form. This will include both the useful output and the power wasted in
the bearings and in the iron of the armature.
It is possible for us now to compare the two expressions found for the
output. Equating them, we have
E. ia Mt. 2π .
n
60
.
9.81..
.(103).
If we substitute for E and M, the values found for them in equations (79)
and (98), we shall find that both sides of our equation are identical.
When we consider the relation between the output and the armature
current, we see at once that for a very small current, and consequently a very

ia Ra
E
Fig. 180.
ia Ra
E

Fig. 181.
small torque, the output will be small. With increasing load the output
increases, until with very large loads the speed drops so much that the
output is decreased. To find when the output is a maximum, we must
differentiate the equation
P = E. iae. ia-ia Ra,
and equate the result to 0. We have thus
dP
dang
dia
e 2ia. Ra = 0,
e
or
ia. Ra = 2.
2'
For this case we have also ia. Ra = E.
The same result is obtained from Figs. 180 and 181, in which the side
of a square is made to represent the terminal pressure e, and is divided into
two parts, one equal to the ohmic drop ia. Ra and the other to the back
13
T. E.
194
Electrical Engineering
α
E.M.F. Ra being constant, the area of the shaded square is proportional
to the power i Ra wasted in heating, while the area of the shaded rectangle
is proportional to ia . E and represents the mechanical output. This shaded
rectangle is small both for small (Fig. 180) and for large (Fig. 181) values
of ia. It has its maximum area when it is a square, i.e. when ia. Ra = E.
This is therefore the condition for maximum output, as we have already
found. On account of the large armature current, this case lies far outside
the limits of practical working. It would evidently be very bad practice
to waste half of the energy in heating the armature wires.
64. Motor with constant excitation.
We shall consider in this section a motor in which the field winding
is connected directly to the constant pressure supply mains. In the first
place we shall also assume that the armature is connected directly across the
mains (Fig. 177).
The total torque exerted by the motor is given by equation (98) on
page 191 as
Mt
=
p N.z.ia
a' 2π.9·81
•
10-8 metre-kgs.


₂met-kgs.
2
1.5-
0.5-
Mt
Mu
+
+
1000
n
800
600
400
200
ያ 4 6
8 10 12 14 16 18 20 Amps.
Fig. 182.
2 4
6
8 10 12
Amp
Fig. 183.
Since the magnetising current, and therefore the flux N, is constant in
the present case, the total torque must be proportional to the armature
current i. Plotting the armature current horizontally and the torque M
vertically, we get a straight line passing through the origin (Fig. 182).
The useful turning-moment Mu is less than Mt by an amount necessary
to overcome the resisting torque due to friction, hysteresis and eddy currents.
If the no-load current in the armature is io, the useful torque is proportional
to the difference i-io. If we plot the useful torque in Fig. 182, we get
another straight line parallel to the first and cutting the horizontal axis
at the point i。.
For the speed, we have from equation (100) on page 192
n =
(e-ia. Ra). 60. 108
p/a.N.z
revolutions minute ......(104).
per
Since the flux is still constant, the speed is proportional to the back
E. M.F. (e-ia. Ra). The drop of speed from no-load is therefore proportional
Motor with constant excitation
195
to the pressure drop ia. Ra, and therefore also to the current ia. If then we
plot the armature current horizontally and the speed vertically, we get the
slightly falling straight line in Fig. 183. Since the ohmic pressure drop
in the armature is always small, the motor with constant excitation has
almost a constant speed from no-load to full load.
The output in H.P. on the pulley, the so-called brake horse-power (B.H.P.),
can either be found from the useful torque and the angular velocity, in which
Mu. 2π.n
or it can be found from the back E.M.F. and the
76.60
case we get
useful current iɑ — i., in which case we get E. (ia — io).
If i is the sum of the armature current ia and the field current i, the
total power supplied to the motor is i.e watts, and the overall efficiency is
E. (ia — i) __ (e — ia . Ra) (ia — is)
η
e .i
e. (ia + is)
(105).
It is evident that at no-load, that is, when iaio, the efficiency must
be nil. This is also the case when the load on the motor is so large that
it cannot start even with the largest possible current
e
ia
Ra®
The fact that such a current would burn out the armature does not
concern us at the present moment. The important point to notice is that
there must be some intermediate load for which the efficiency is a maximum.
To find the most efficient armature current we multiply out the numerator
of equation (105) and get
2
e. ia-ia². Ra - e . io+ia. i, . Ra
e. (ia + is)
We now differentiate this with regard to ia and put the result equal
to 0. For the sake of simplicity we neglect the constant factor e in the
denominator.
dn (e
(e – 2ia. Ra + io. Ra) (ia + is) — (e. ia-ia². Ra - e. i,+ia.io. Ra)
(ia + is)²
dia
This expression will vanish when the numerator vanishes. If we multiply
the numerator out and put it equal to 0, we get
ia². Ra-e.is - e. i + is. Ra. (2ia — is) = 0.
The meaning of this equation becomes clearer when we remember that,
owing to the smallness of the magnetising current, the right-hand term is
so small as to be practically negligible. We get then as the condition for
maximum efficiency
2
ia². Ra = e. is + e. i = e (is + i)
.(106).
Now is is the total motor current at no-load. Hence, the efficiency
is a maximum when the ohmic heating of the armature conductors is equal
to the constant losses both in armature and field, including the friction. In
other words, the efficiency is a maximum when the losses dependent on the
load become equal to those independent of the load. The decrease of friction
due to the small decrease of speed with increasing load is quite negligible.
13—2
196
Electrical Engineering
11 1
In order to make the above considerations clearer, we shall apply them
to a numerical example.
We shall consider a bipolar motor (p = 1, a = 1) having the following
constants:
e = 110,
Ra = 0·3,
N= 2.10,
is = 1,
z = 300,
i。 = 2.
We have then
Mt=
p N.z.ia
•
•
10-8 = 0·1.ia,
α
а˚ 2π.9.81
น
Mu=
= 0·1iα-0·2,
= 1100-3ia.
n =
0·1 (ia — io)
(e - ia. Ra). 60. 10º
p/a. N.z
If the normal current for which the armature is designed, is 10 amperes, we
obtain a total full load turning-moment of 1 metre-kg. of which 0.8 metre-kg.
is available at the pulley. The speed at full load is 1070. The drop of speed
from full to no-load is 30 in 1100, or roughly 3 per cent.
For an armature current of 10 amperes, the overall efficiency will be
η
(e – ia. Ra). (ia – io)
e. (ia + is)
107.8
110.11
= 0·7.
To find the conditions at maximum efficiency, we must first find the
constant losses. These are
e. i + e. i; = 330 watts.
For maximum efficiency, the armature copper loss must also be equal
to 330 watts, that is,
ia². Ra = 330 or ia
330
V 0.3
33 amperes.
Adding the field current of 1 ampere, we get a total current of 34 amperes
and a motor input of
e i = 110.34 = 3740 watts.
If we subtract from this input the total losses which are 2.330 watts,
we obtain the output 3740-6603080 watts, and the overall efficiency
is therefore
nmax
3080
3740
0.825.
The output on the pulley is then
3080
746
= 4.1 B.H.P. The percentage drop
of speed from no-load is equal to the percentage pressure drop in the arma-
ture, that is, 0·3. 33 in 110 or 9 per cent. (see page 194).
It is interesting to determine how the efficiency varies when the load
is varied over a large range. If we assume that the load is reduced to about
half of that corresponding to maximum efficiency, so that the armature current
is 17 amperes, then we have
E=e-ia. Ra = 110 17.03 105 volts,
-
=
i = ia + is = 17 + 1 = 18 amperes,
iai 17-2 = 15 amperes.
=
Motor with constant excitation
197
The overall efficiency will then be
η:
n =
E. (ia) 105.15
e. i
110.18
0.8.
In the same way, if the load is about doubled, so that the armature
current is 66 amperes, we have
E = e-ia. Ra = 11066.03 = 90 volts,
i=ia + is = 66+1=67 amperes,
ia - io 662= = 64 amperes.
Under this heavy load, the efficiency will be
η
E. (ia — is) 90.64
e
•
г
110.67
0.78.
Hence, we see that a large variation in the load has very little effect.
on the efficiency. It is therefore unimportant to design a motor to have
its maximum efficiency at full load. To reduce the drop of speed as much
as possible the normal load should be below the point of maximum efficiency,
while a motor which is to work, as a rule, on very light loads, and only have
its full load at intervals, should be designed to have its maximum efficiency
at a small load.
We pass now to the consideration of a motor the field winding of which
is connected directly across the constant pressure supply, but the armature of
which has an adjustable resistance in series

м
му
with it (Fig. 184). We make the striking
discovery that, whether loaded or running
light, the current is in no way affected by
a change in the adjustable resistance. If,
however, we turn to equation (98), on page
191, we see that the armature current must
be determined entirely by the load, since
the magnetic flux is constant. If the load
on the motor remains constant, the arma-
ture current will remain constant, however
much the series resistance may be varied.
On suddenly decreasing the resistance there
is, it is true, a momentary increase of the
armature current. This causes the motor to exert a greater torque than
that required to overcome the load, with the result that the armature is
accelerated and runs permanently at a higher speed. This increase of speed,
however, causes an increase in the back E. M. F., which reduces the armature
current to its former value corresponding to the load. The only way to vary
the armature current of a constantly excited motor is to vary the load.
Fig. 184.
The speed, on the other hand, is very dependent on the resistance R
in series with the armature. The applied terminal pressure e has now to
overcome the back E.M. F. and cover the pressure drop in both the armature
and the resistance R. We have then
or
e=E+ia. Ra+ia. R,
Ee-ia. Ra-ia. R.
198
Electrical Engineering
In this equation e and Ra are constant, and ia is also constant so long as
the load remains unchanged. It follows, therefore, that a change in the
value of R will produce a large effect on E and consequently on the speed.
This becomes much clearer if we neglect the small drop of pressure in
the armature, and assume that the pressure across the brushes is exactly
equal and opposite to the back E.M.F. On this assumption, a voltmeter
connected across the brushes will indicate directly the back E.M.F. We find
then that, on varying the value of R, the speed is almost directly proportional
to the pressure across the brushes. The accuracy is greatest when the
pressure drop in the armature is as small as possible, that is, at no-load.
In the numerical example which we have worked out above, the speed, when
running light with a P.D. of 110 volts between the brushes, is found from
equation (104), on page 194, to be
N
(110 - 03. 2). 60.108
2.106.300
=
1094.
If a resistance of 27.5 ohms be connected now in series with the armature,
the no-load current of 2 amperes will be unaltered. The drop of pressure in
this resistance will therefore be 27.5.255 volts and the pressure read on
the voltmeter across the brushes will be reduced to 110 − 55=55 volts. The
speed will therefore be
n =
--
(55 03.2). 60.108
2.106.300
544.
The speed is thus almost exactly a half of what it was when the P.D.
between the brushes was 110 volts. The speed is therefore practically pro-
portional to the P.D. between the brushes when the motor is running light.
The same relation is approximately true when the motor is loaded. We are
thus led to the interesting conclusion that, in a motor with constant excita-
tion, the current depends only on the load, while the speed depends only on
the P.D. between the brushes. Hence the speed can be regulated to any
desired extent by altering the P.D. applied to the brushes.
There are, however, two errors which we must be careful to avoid. In
the first place, it is very important to notice that in the above experiments
the adjustable resistance was in the armature circuit only and had no effect
on the field. If the armature and field windings are connected directly to
the same terminals, and the resistance connected in the common external
circuit, an alteration in the resistance may, under some circumstances, have
very little effect on the speed. If, for example, we double the terminal
pressure of both armature and field, the magnetic flux may be nearly doubled,
if the iron be only slightly saturated. In the formula
n
E.60.108
p/a.N.z
both the value of E in the numerator and the value of N in the denominator
would then be doubled, leaving the speed n unchanged. If, on the other
hand, the magnets are highly saturated, the motor will behave more like a
constantly excited one and doubling the terminal pressure will cause the
speed to increase nearly 100 per cent.
Motor with constant excitation
199
The other point on which we must be quite clear is that speed regulation
by means of a series resistance is only possible when an attendant can stand
by the regulating switch, as on a tramcar, for example. Apart from the
large loss of power in such a resistance, there is the disadvantage that every
change of load causes a change of current and therefore a change in the
amount of pressure lost in the resistance. The result is that the pressure
across the armature terminals and the speed of the motor are subjected to
large variations. It is possible, for instance, for a motor to be running light
with so much resistance in series with its armature that it comes to rest
when the load is put on.
65. The starting and regulation of a shunt motor.
The shunt motor is really no more than a constantly excited motor. It
possesses, therefore, all the fundamental characteristics which we have found
for this type of motor in the previous section. The current is proportional
to the load and the speed is practically constant from no-load up to full load.
This latter property makes the shunt motor specially suitable for driving
lines of shafting and machine tools, where constancy of speed is required.
As we have already studied the principles of the motor, we need only consider
here the more important points in connection with their starting and regu-
lation.
The primary object of the starting resistance is to protect the armature
from a dangerously large current. At the first moment of starting, the
armature is at rest and there is consequently no back E.M.F. If the armature,
with its comparatively small resistance, were connected directly across the
full pressure of the supply mains there would be an enormous rush of current.
A starting resistance must therefore be put in series with the armature, of
such a value that the current is limited to a permissible strength with regard
to its heating effect on the armature. There are also mechanical reasons for
limiting the turning-moment exerted at starting, especially when there are
heavy masses to set in motion, A further reason is the large pressure drop
and consequent unsteadiness of the lamps in the neighbourhood, when the
supply mains are suddenly called upon to carry a large current.
When the motor has been set in motion, the back E. M. F. increases with
the speed and limits the current to a value corresponding to the load. The
starting resistance can therefore be gradually cut out as the speed increases.
If it is required to regulate or vary the speed of a shunt motor, it must
be done by altering the resistance of the shunt field circuit. This is clearly
seen by applying equation (104) on page 194 to the case of a motor running
on no-load. We have then
n =
e. 60.108
p/a.N.z
If resistance is put in the field circuit, and the magnetising current and
the flux N thereby reduced, we see from this equation that the speed n must
increase. This is also evident when we consider that the motor must run
faster to produce the same back E.M.F. in the weaker field as it formerly did
in the stronger field.
200
Electrical Engineering
If the load remains the same, this weakening of the field will naturally
affect the armature current. Since the load or torque is proportional to the
product N. ia, a weakening of the field must cause a corresponding increase
in the armature current. To make this clearer we shall go back to our
example in which the terminal pressure e = 110 volts, the armature resist-
ance Ra= 0.3 ohm, the number of wires z=300 and the flux N=2.10
lines. For an armature current ia of 10 amperes, we have for a bipolar
motor, from equation (104) on page 194,
n
(e-ia. Ra). 60.108 ___ (110 – 10.0·3). 60. 10º
p/a.N.z
2.106.300
1070.
If now, without changing the load, we reduce the magnetic flux 20 per
cent., the armature current must increase in the same proportion, viz. 0·8 : 1.
We shall then have
N=0.8.2.106 1.6.10°,
10
ia
0.8
=
=
12.5 amperes.
The product N. ia has remained unchanged but the armature current has
increased. As this will cause an increase of pressure drop in the armature,
the back E. M. F. will decrease. This will naturally affect the speed and make
the actual increase less than we should otherwise expect. In the above
example, for instance, we have weakened the field by 20 per cent. and should
therefore expect the speed to increase in the ratio 0.8: 1, that is from 1070 to
1070
1338. As an actual fact, the new speed attained on weakening the
0.8
field is
n =
(110-12.5.0.3). 60.108
1.6.106.300
1328.
This indicates at the same time that there is a limit to the speed
obtained by weakening the field. The effect of the pressure drop in the
armature may become so great that, on further weakening the field, the
speed is decreased. This is, of course, an absolute necessity, since we should
otherwise be led to the conclusion that the loaded motor would run at an
infinite speed when the field circuit was broken and the flux reduced to zero.
As a matter of fact, the motor remains stationary in such a case, since one of
the two factors that produce the torque has vanished. Hence, there must be
a certain value of N, or of ia, for which the speed at a given load is a
maximum. We can find this critical value of a from equation (103) on
page 193, in which we equated the mechanical and the electrical power. We
have then
n
E. ia= e. ia-ia². Ra=Mt. 2π.
.9.81.
60
As one of the conditions of our experiment is that the load be kept
constant, Mt is a constant, and we can put
n = c.(e. ia-ia²Ra),
where c is merely a coefficient.
The starting and regulation of a shunt motor 201
dn
For n to be a maximum must vanish, or
dia
dn
dia
=c.(e-2ia. Ra) = 0.
e
Hence
ia. Ra = 12.
The maximum value of the speed is therefore reached when the field is
so far weakened that the pressure drop in the armature is equal to a half of
the terminal pressure. The electrical efficiency of the armature would then
be 50 per cent., but the conditions, if for no other reason than the enormous
armature current, are far beyond the normal working range.
With regard to the arrangement of starting resistance in the armature
circuit and field rheostat in the field circuit, care must be taken that the
field circuit is not suddenly broken when the motor is being switched off.
This would not only cause sparking at the point where the circuit was broken,
but might break down the insulation of the field coils, owing to the large

Fig. 185.
Fig. 186.
electromotive force induced by the sudden cessation of the current through
the highly inductive circuit. The motor can be switched off sparklessly by
opening the external circuit and leaving the field connected across the arma-
ture (Fig. 185). The motor will slow down slowly because of its kinetic
energy, and will act as a dynamo sending a gradually decreasing current
through the field windings in the same direction as before. The field will
gradually diminish as the motor slows down.
It is, however, very dangerous to adopt the plan, shown in Fig. 185, of
switching off in the external circuit, since the moving arm of the starting
resistance is left in such a position that a dangerous short-circuit would
result on again closing the switch. It is impossible to place the starting
resistance in the external circuit, as shown in Fig. 186, since the P.D. between
the brushes at the moment of starting would only be iɑ. Ra, which is so
small that there would be practically no field current and no magnetic flux.
The motor would therefore not start.
The unbroken connection of field and armature, which we have seen to be
necessary for sparkless switching, can, however, be arranged by connecting the
field across the free ends of the armature and starting resistance as shown in
202
Electrical Engineering
Fig. 187. When the arm is in the middle of the row of contacts, the right-
hand half of the resistance is acting as starting resistance while the left-hand
half is in series with the field winding and is weakening the field. This
arrangement has the great advantage of the permanent connection of field,
armature and starting resistance in series, thereby making it possible to
switch off the motor without any dangerous sparking, especially if the switch-
arm is moved rapidly from the running position, right across the contacts
and off at the left-hand side before the motor slows down appreciably. We
have the disadvantage, however, of the starting resistance being normally in
series with the field winding and causing a small but unnecessary loss due to
the heat continually being generated in the resistance.
This loss can be diminished by connecting the field to some intermediate
point of the resistance, as shown in Fig. 188. This brings with it, however, the
disadvantage of a weak field at the first moment of starting and is really a
compromise between Figs. 186 and 187. Since the self-induction of the field
winding causes the growth of the field to be gradual, it is an advantage to
apply the full P.D. of the mains at the first moment so as to hasten the
growth of the flux as much as possible.



Fig. 187.
Fig. 188.
Fig. 189.
These disadvantages are overcome in the arrangement shown in Fig. 189,
in which the field circuit is connected at once to the full P.D. of the mains
and remains so connected by means of a short contact arm which moves over
a segment. The longer contact arm, which is virtually a prolongation of the
short one, moves over the contacts of the starting resistance.
This arrange-
ment does not give quite so sparkless a break as that shown in Fig. 187,
since, at the moment of switching off, the field current is supplied by the
armature and, having to pass through the starting resistance, is suddenly
reduced. This sudden reduction of the current through the inductive field
windings causes a spark.
This is even more pronounced in the arrangement shown in Fig. 190.
The "off" position of the switch-arm is on the extreme left; on moving the
arm to the right, the field is switched on to the full pressure of the mains,
and the resistance w₁ is in series with the armature. As the motor runs up
to speed the switch-arm is moved to the right, until it finally stands on the
right-hand contact of w, and on the right-hand end of the left-hand segment.
If now we wish to increase the speed, the arm is moved still further to the
right, putting resistance w, into the field circuit and supplying the armature
directly through the right-hand segment.
The starting and regulation of a shunt motor 203
The spark on breaking is diminished by connecting the left-hand end of
the starting resistance with the right-hand end of the regulating resistance.
When switched off, the motor will slow
down while acting as a generator and
sending current through armature, fields,
starter w, and field regulator w₂, all con-
nected in series.

будут
музициу
Considerable difficulties arise when
starting apparatus has to be designed
for very large motors, and the amount
of energy lost as heat in the starting
resistance is by no means negligible
when motors are continually being
started, more especially when large
masses have to be set in motion. For
this reason some electrically operated
hoisting plants for mines have been
Fig. 190.
started by means of a battery of accumulators connected across the supply
mains. By means of a battery switch, the battery is divided into groups and
only a fraction of the total pressure applied to the motor armature at starting.
As the speed increases the back E. M.F. of the motor also increases, and the
applied P.D. is gradually increased by means of the battery switch. The
battery acts also as a buffer and equalises the load on the supply station to a
certain extent.
Another method of starting a motor and regulating its speed conveniently
over a wide range is to employ a motor-generator set. In the Ward-Leonard
arrangement, for instance (Fig. 191), an auxiliary shunt-motor is driven off
the mains, and coupled to it is a separately excited generator. This set runs
at a practically constant speed, but the excitation of the generator can be
varied over a wide range. It is made very weak when starting the main
motor, the armature of which is supplied directly from the generator. By
gradually increasing the excitation of the generator, the motor is brought up
to full speed and can then be switched over on to the mains.
power
If, however, the supply is alternating, the main motor must be a direct-
current motor and cannot then be switched over on to the supply mains. In
this case the motor-generator must be capable of transmitting the full
continuously and not merely during starting. Although this transformation
from A.C. to D.C. appears to complicate the plant, it is often necessary from
other considerations, since a low-speed hoisting motor is less difficult to
construct as a D.C. than as an A.C. motor.
It is specially advantageous to fix a heavy flywheel between the auxiliary
motor and the generator in the above arrangement. This is generally known
as the Ilgner system and has been largely adopted within the last two or
three years. When the load is heavy, as when starting to hoist, the flywheel
gives up a part of the energy stored in it, whereas at light loads or no-load
energy is restored to the wheel by its acceleration. The flywheel, therefore,
acts as a buffer and equalises the load on the station.
204
Electrical Engineering
The auxiliary motor must, of course, be so constructed that the flywheel
can act in this way. Its speed is therefore made to drop considerably as the
load increases. In an induction motor this can be attained by increasing the
rotor resistance, and in a direct-current motor, by compounding it with a few
series turns so arranged that they strengthen the shunt field (see Section 66
for the effect of the series turns on the speed).
One method of starting large motors, which is of considerable interest,
depends on the use of an auxiliary generator, the E. M.F. of which is at first
opposed to the supply pressure, but afterwards acts with and is added to it.
The machine I in Fig. 192 is the hoisting motor, constructed for a pressure
of 1,000 volts. Machines II and III constitute a motor-generator set, both
machines are for 500 volts and the set runs at an approximately constant
speed. The supply pressure is also 500 volts. We shall confine ourselves to
a description of the method of starting the hoisting motor; a complete
understanding of the various changes involved can only be obtained by
applying Kirchhoff's rules to the circuit.
Before starting, motor I is at rest and merely acts as a supply lead to
motor II. The latter drives the machine III as an unloaded dynamo, since


Auxiliary
Motor.
Main Motor
+
でっ
​III
I
Dynamo.
Fig. 191.
Fig. 192.
the field of III is adjusted until its E. M.F. exactly counterbalances the pressure
of the supply mains. By slightly weakening the field of motor II its speed
is increased by a small amount. This increase of speed increases the E. M. F.
of machine III and enables it to act as a loaded generator and supply a
current . This increases the load on motor II and causes its current to
increase to i₂. This is exactly similar to the Hopkinson method of testing
two similar machines, which was described on pages 185 and 186. The
current flows round to motor II and the mains supply the losses.
The field of motor II is weakened until the current i, reaches such a
value that the main motor I, which is separately excited, starts. For a
given value of the excitation of II there is a fixed speed of motor I, corre-
sponding to a certain distribution of the supply pressure between the two
machines I and II. We can also see that, by further weakening the field of
II, its speed may decrease, since the decrease in its P.D. may more than com-
pensate for its weakened field. This is what actually happens, with the
result that the E. M. F. of generator III diminishes and with it the current i.
Finally, the field of motor II is reduced to zero. Its back E. M.F. has
L
The starting and regulation of a shunt motor
205
disappeared and the full P.D. of 500 volts is applied to the motor I. The
speed of the motor-generator set falls so low that the pressure of the mains
overcomes the E. M.F. of III and reverses the current i. Machine III is now
a motor driving the other machine II. We now reverse the field current of
II so that it acts as a generator. Its E. M. F. is added to that of the mains,
and its field is gradually strengthened until its E.M.F. is 500 volts. In this
way the P.D. applied to the main motor I is gradually increased to 1,000 volts.
It has then reached its full speed and one half of its input is supplied
directly from the supply mains and the other half indirectly through the
motor III and the generator II. The whole power must, of course, come
ultimately from the supply mains.
66. Principle of the series motor.
The behaviour of the series motor is not nearly so easy to understand as
that of the shunt motor since the magnetic field of the former is not constant
but varies continually with the load. The field winding is in series with the
armature, and the armature current, which in conjunction with the field pro-
duces the torque, is at the same time the magnetising current which produces
the field. If the load on a series motor be increased both the armature and
field current is increased. The larger torque is seen from equation (98) on
page 191 to be due to the growth of both of the active factors N and ia.
If the motor is working below the knee of the magnetisation curve, that
is, with unsaturated iron, doubling the current causes the field to be doubled
and the torque, in accordance with the above equation, to be quadrupled.
Inversely, if the load be quadrupled the current is only doubled.
The speed of a series motor is given by equation (104) on page 194, viz.
(e – ia. Ra). 60.108
n =
P.N.z
α
To avoid complicating the formula, we shall let Ra represent here the
combined resistance of armature and field. If the load and therefore also the
armature current be increased, the numerator of the above expression will be
but slightly altered, since the combined resistance of armature and field is
always very small. The denominator, on the other hand, will be considerably
changed, since the flux varies with the current. The speed will therefore
decrease as the load is increased, since with the strengthened field a smaller
speed will suffice to produce a back E.M.F. nearly equal to the supply
pressure.
The large torque with a relatively small current and the variable speed
make the series motor specially suitable for cranes and traction purposes.
When starting, the torque exerted by the motor must exceed the resisting
torque due to the apparent load, in order to accelerate the masses which have
to be brought up to speed. It is the ability of the series motor to provide
this heavy starting torque with a comparatively small current that makes it
so invaluable for the above purposes.
206
Electrical Engineering
The series motor possesses, moreover, special advantages in respect of the
variations of load which the continually changing slope of the track puts upon
the tramcar motors. The large torque required when running up an incline
is produced with a relatively small demand on the power station.
This is only possible, of course, because of the corresponding decrease of
speed mentioned above. The matter is quite clear, apart from the above
considerations, if looked at from a mechanical point of view. Power is the
product of force and speed. If then the series motor exerts a large pull with
a small current and consequently with a small supply of power, it is evident
that its speed must be small. The generators and motors may therefore be
constructed for a reasonably small output and the load on the station will not
be so irregular as it would be with shunt motors on the cars. The variable
speed of a series motor must therefore be considered a great advantage and
it is an important point in the interests of the power station that a heavily
loaded car goes very slowly uphill.
N
N

Mt

500 5-10°
6-10°
Eo
400 4-106
4.100
N
300 3-106
200 2.106
3.10°
20
.15
N
2-106
10
100
Mt
5
10
15
20
29
5
10
15
20
25
Amp
Amp
Fig. 193.
Fig. 194.
We must now investigate more fully the behaviour of a series motor
under various conditions. For this purpose we first determine the magnet-
isation curve by running the motor as a generator by means of another motor.
If the speed be maintained at a constant value n, and the generator be loaded
by means of different external resistances, we shall obtain a set of corresponding
values of the current i, and the terminal pressure e。. From these we can cal-
culate the E. M.F. of the series generator by means of the equation
E。=e+ia. Ra,
where Ra includes both armature and field. As in Section 58, the E. M.F. is
plotted as ordinates and the external current, which is, at the same time, both
armature and field current, as abscissae. This curve is the characteristic
of the series machine. By dividing the values of E. by the constant
p.no.z.10-8
we get the magnetic flux N which is produced by the cor-
a.60
responding magnetising current ia. Curve N only differs from curve E, in
the matter of scale (Fig. 193), but is, unlike curve E, independent of speed.
If it is impossible to make the experiment in this manner by running the
Principle of the series motor
207
motor as a generator, it can be run as a motor at a constant terminal pressure
with variable mechanical load on the brake. The flux can be calculated from
the observed speed and current by means of equation (104) on page 194:
N
(e-ia. Ra). 60.108
P
α
z.n
When we have found the flux corresponding to each value of the current
ia, the two are multiplied together and the product N. ia set up as ordinates.
In this way we obtain the curve Mt in Fig. 194, which represents the torque
to some scale not yet determined. At the beginning the curve Mt is convex
to the base, showing that the torque increases more rapidly than the current.
As saturation sets in, however, the curve straightens out and gradually
approximates to a straight line through the origin, corresponding to an
increase of torque proportional to the increase of current.
The curve of magnetic flux N enables us, moreover, to find the speed
corresponding to any condition of load at any terminal pressure. For this
E
N

n
Pu
Mu 7

500 5.106
400 4-10€
300 3.106
2002-106
100
N
น
1000
10000
E
800
8000
D
0,8
600
6000
15
400
4000
Mu
0,6
10
0,4
200
2000
.5
0,2
10
15
20
25
Amp.
Fig. 195.
5
10
15
20
25
Amp
Fig. 196.
purpose we draw a horizontal line at a height above the base line equal to
the terminal pressure e. In Fig. 195 this pressure is 500 volts. Through
the point where this line cuts the vertical axis we draw a straight line sloping
down from the horizontal at an angle a, where
tan α =
Ra.
In measuring the tangent we must, of course, measure the ordinate on the
scale of volts and the abscissa on the scale of amperes, and we must also
remember that Ra includes both armature and field winding. The difference
between the horizontal and the inclined line at any point is equal to iɑ. tan a,
that is to ia. Ra, which is the ohmic pressure drop in the motor. The
ordinates of the inclined line are therefore equal to the back E. M. F., which we
denote by E. If now we divide each value of E by the corresponding value
of N, we obtain a figure proportional to the speed (see equation (104) on
page 194). In this way we obtain the curve n in Fig. 195.
It is evident from this curve that a series motor will race if the load be
removed. The speed, in fact, may reach such a value that the motor will fly
208
Electrical Engineering
เ
to pieces owing to excessive centrifugal forces. The speed drops rapidly as
the load is increased so long as the iron is comparatively unsaturated, but as
saturation sets in the speed approaches an approximately constant value.
To find the total efficiency we must subtract the losses due to heating of
conductors, hysteresis, eddies and friction from the motor input and divide
the remainder, which is the output, by the input. In the shunt motor the
losses, other than copper loss, were determined in a simple manner by finding
the power required to run the motor at no-load and assuming that these
losses were the same at all loads. This was justified to a certain extent
because of the constant excitation and speed. In the series motor, however,
such an assumption would be palpably incorrect since both magnetisation and
speed are variable.
The efficiency can be determined approximately by assuming that the
torque M, required to overcome the friction and iron losses is constant at all
loads. This assumption is not very incorrect if we consider the efficiency of
the motor alone and exclude any reduction gear. By subtracting this constant
torque M, from the values of Mt we obtain the dotted curve M, of useful
torque in Fig. 196.
น
The output Pu is given to a certain scale by the product of the useful
torque and the speed. To obtain the efficiency this output must be divided
by the input e.ia. In this way we obtain the curve n, the ordinates of which
are equal to the efficiency. It is seen that the efficiency is small at low loads,
but increases with increasing load until it reaches a maximum, beyond which
it decreases owing to the large copper loss in armature and field windings.
67. Example.
We shall now determine the speed, torque and efficiency of an actual
motor, the details of which are given in Kapp's well-known "Dynamo Con-
struction." The motor is a 4-pole tramway motor made at Oerlikon in
Switzerland. The armature has a series or 2-circuit winding arranged for a
terminal pressure of 500 volts. The internal resistance of the motor is
2.75 ohms and the number of wires on the armature is 944, so that Ra = 2·75
and z = 944. The motor was driven as a series dynamo at a constant speed
no of 450, and the characteristic so obtained is given by the first two columns
in the following table.
The magnetic flux corresponding to each value of the current is found
from equation (79) on page 120, as follows:
N
E..60.108 E..60.108
7070 E..
p/a.no.z 2.450.944
From these values of the flux and the corresponding values of the current,
the torque exerted on the armature can be found from equation (98) on
page 191, and we get
M₁ =
Mt
p.N.ia.z
10-8
а.2π.9.81
2.N.ia.944
2π.9.81
10-0.307.10. N. ia mkg.
Example
209
To calculate the speed of the motor we must know its back electromotive
force E. This is given by the formula
E=e-ia. Ra
=
500 -2.75.ia.
The speed can then be found from equation (100) on page 192, viz.
E.60.108
n
P
E.60.108
2. N.944
E
3.18.106.
'N'
.N.z
a
The results of these calculations are given very clearly in the following
table and they are represented graphically by the curves in the previous
section.
Generator
Motor and Generator
ia
E。
N = 7070 E.
t
M₁ = 0·307 . 10−6. N. ia
5 225
1.59.106
2:45
10
362
2.56.106
7.85
15 450
3.18.106
14.6
20 490
3:46.106
21.3
25
505
3.56.106
27.3
Motor
ia
E-500-275.ia
5
486.2
10
472.5
15
458.7
20
445
25
431.2
E
n = 3·18.106.
N
970
590
460
410
390
By means of these figures we can determine the speed at which a tramcar
of given weight will ascend a given incline, as well as the current taken in
doing so.
Let
W be the weight of the car in metric tons (1 metric ton = 1,000 kgs.*),
f the tractive coefficient, i.e. the force in kgs. required to move one ton
at a steady speed on the level,
and
then the pull required on the level will be W.f. On the slope, the weight
AB (Fig. 197) acting vertically downwards can be resolved into two com-
ponents one normal to the surface and the other AO, which has to be directly
overcome by the tractive pull. We have from the figure
AO AB. sin d.
s the slope or gradient in parts per thousand,
S
If we substitute for sin 9 its value
kilogrammes, we have
and for AB the weight of the car in
1,000
$
AO = 1,000 W
W.s.
1,000
T. E.
14
210
Electrical Engineering
The pull on the car must therefore be given by the formula
F=W.ƒ+W.s = W (ƒ+s) kg.*
If the radius of the car-wheels be r metre, the torque on the axles
must be
F.r=W(f+s). r met.-kgs.
If the gear is such that the motor runs k times as fast as the car axles, the
torque of the motor, by the conservation of energy, must be a kth part of that
on the axles. If we assume, further, that the gearing has an efficiency ŋ, we
obtain the following formula for the total torque of all the motors on the car:
F.r_W.r.(f+8)
k.n
ΣMt=
ΣΜ k.n
In our present example the radius of the wheels was 0.39 metre and the
ratio of the gearing 49. If we assume that the weight of the car is 8 tons
and the tractive coefficient 12 kgs. per ton, we have
ΣΜ
ΣM₁ =
8.0·39. (12 + s) ___ 0·635 . (12 + s)
4.9.7
Պ

1000
A
Ꮽ
Я
B
५
Fig. 197.
In our case, as is almost universal, each car was provided with two motors,
so that each motor had to produce half the required torque. Hence, for each
motor, we have
or
ΣMt 0·318 (12 + s)
Mt
2
պ
s=3·14.n.Mť – 12.
This gives us the slope up which the car can run steadily for the given
torque on each motor.
η
As a rule the efficiency of the gearing, including the friction and iron
losses of the motor, can be taken as 08, but in the present case it was
accurately determined in each case and is given in the table below. In the
same table we have the calculated slopes up which the car can run with the
given strengths of current. The torque exerted by the motor has been taken
from the table on page 209. A negative value of the slope denotes running
down hill.
We have yet to calculate the speed at which the car runs in any given
To obtain the revolutions per hour of the car wheels we must multiply
the revolutions of the motor per minute by 60 and divide by the gear ratio k.
case.
Example
211
If the revolutions per hour made by the car wheels be multiplied by the cir-
cumference of the wheels in metres and divided by 1,000, we get the speed of
the car in kilometres per hour. Calling this speed S, we have
n.60 2r.π
S
k 1,000
kms. hour.
per
If we assume the radius of the car wheels to be 0.39 metre and the gear ratio
k to be 4.9, the above equation becomes
S=0.03. n kms. per hour.
This formula has been employed to calculate the last column but one in the
following table. The values of n have been taken from the table on page 209.
The current i in the last column is the total current taken by the car, which,
as the two motors are in parallel, is equal to twice ia.
s = 3·14. n . Mť – 12
-6.2%
- 6·2 °。。
ia
η
5
0.75
S = 0·03.n
29 kms. per hour
i=2ia
10
10
0.775
+7
17.7
20
""
""
""
15
0.805
+25
13.8
30
""
""
""
20
0.835
+44
12.3
40
"
""
25
0.85
+61
11.7
50
""
i
kmns. per hour

60+ 30
50+25
40
30+15
20+10
i
Speed
10+5
1
+
2
3
4
5
Percentage Gradient.
Fig. 198.
In Fig. 198 the total current taken by the car and the speed at which it
runs, are plotted as ordinates on a base representing the slope or gradient.
It is more customary in England to refer to a slope of 50% or 5 per cent. as
a rise of 1 in 20.
68. The regulation of series motors.
The speed of a motor is given by equation (100) on page 192, viz.
n =
(e-ia. Ra). 60.108
p/a.N.z
revolutions per minute.
In this equation e represents the terminal pressure of the motor. A
method of regulating the speed of a series motor, which suggests itself at once,
is by regulating the terminal pressure by means of the starting resistance.
The current corresponding to the load on the motor causes a fall of potential
14-2
212
Electrical Engineering
in the resistance, thus reducing the P.D. between the armature terminals.
We may look at the matter from another point of view and let e represent
the supply pressure of the mains. Putting in an external resistance R
increases the total resistance from Ra to Ra+R₁, and the above equation
becomes
n = {e — ia (Ra + R₁)}. 60. 10ª
p/a.N.z
It is quite immaterial which of these two ideas we adopt. In the one
case we imagine the positive term in the numerator to be decreased, while in
the other case we imagine the negative term to be increased. Both lead to
the same result, viz. a fall of speed.
Although this method of regulating the speed is the one generally
adopted, it is very wasteful, since the power saved at the motor by de-
creasing its speed is all dissipated in the regulating resistance. Another

wwwww
my
Fig. 199.
method which is occasionally resorted to is that of altering the excitation.
As we have already seen when considering the shunt motor, the speed is
increased by decreasing the value of N in the denominator of the above
equation. We cannot here, however, weaken the field by putting resistance
in series with the field winding, since such resistance would be in series with
the armature and would merely lower the P.D. between the brushes. By
putting a resistance in parallel with the field winding, as shown in Fig. 199,.
a part of the armature current is shunted and takes no part in the ex-
citation.
If we assume, for the sake of simplicity, that the resistance connected in
parallel with the field winding has the same resistance as the field winding,
and that the magnetic flux is proportional to the magnetising current, it
appears at first sight that the magnetic flux has been halved, and the speed
consequently doubled. This, however, is not the case, for we have seen, when
considering the shunt motor, that with a constant load a weakening of the
field leads to an increase in the armature current. To determine the strength
of the armature current that flows when the field winding is shunted, we
must first draw the characteristic curves of the motor without the resistance
(Fig. 200). The ordinates of the curve Mt are equal to the product of the
The regulation of series motors
213
N
t
corresponding values of the flux N and the current ia. In Fig. 201 we have
the same curves for the motor when the field winding is shunted. The flux
curve is exactly the same as before, but the numbers along the base have
been doubled, since the same flux is still produced by the same magnetising
current, and therefore by double the armature current. The curve Mť in
Fig. 201 is obtained, as before, by multiplying together the flux N and the
current ia. We can then see from these two figures the values of the flux
and current for the same turning-moment or load. From Fig. 200, for
example, we find that a torque of 14 met.-kgs. requires a current of
15 amperes and a flux of 3·1.106 lines. For the same load, after the field
has been weakened, Fig. 201 shows a current of about 19 amperes and a flux
of 2.5.10 lines. If we neglect the small ohmic pressure drop, the speed is
inversely proportional to the flux. In the present case, therefore, the speed
has been increased in the ratio of 3·1 to 2·5, or 24 per cent., by putting an
equivalent resistance in parallel with the field winding.


Mi
5.10 25
4.10 20
2.10 15
2.10 10
1.1095
5
N
10
15
Armature Carrent
Fig. 200.
'W
N
5.10
4.10
N
3.10
2.10€
1.10€
20
25
10
20
30
Armature Current
40
50
Fig. 201.
This method of varying the speed by shunting the field current necessi-
tates a certain loss of energy in the shunt. This loss is small, but can be
avoided by adopting a method due to Sprague, in which the field coils are all
connected in series at starting. At full speed they are connected in parallel,
so that each magnet coil carries only a part of the armature current. The
speed can be economically varied over a considerable range by this method.
Two steps in such an arrangement are shown in Figs. 202 and 203.
The regulation of the speed of series motors by weakening the field is
now seldom resorted to. From what we saw in Section 54 concerning the
effect of cross-magnetisation, it is evident that an undue weakening of the
main field must give rise to sparking at the commutator. Moreover, there is
the danger of the magnetic flux not being the same in each of the several
motors on a car. This may result from differences between the magnetie
reluctances of the motors, or from faulty contacts due to the continual
operation of the switching gear, which would cause the current to divide
unequally between the various parallel paths. The effect of such differences
214
Electrical Engineering
can be seen clearly from the following example. A tramcar has two motors
connected in parallel, the pressure e across the brushes is 500 volts and the
armature resistance of each motor is 1 ohm. If the current in one motor is
15 amperes, the back electromotive force can be found as follows:
E-500-15.1 = 485 volts.
We shall now assume that the magnetic flux of the second motor is 5 per
cent. less than that of the first motor. As the two motors are on the same
car they must necessarily run at the same speed. The back E.M.F. of the
second motor is therefore 5 per cent. smaller than that of the first, and we
have in this case
E=485.0·95 = 460 volts.
To find the current in the second motor, we have
e - E 500 – 460
ia
40 amperes.
Ra
1


т
m
m
Fig. 202.
Fig. 203.
The total current taken by the car is 40+15=55 amperes, and of this
total only 15 amperes flow through the more strongly magnetised motor,
while 40 amperes pass through the weakly magnetised motor. The latter
does almost three times as much work as the former. It is not difficult to
calculate the conditions under which the back E.M.F. of one motor becomes
equal to the terminal pressure, so that its current and input are both zero.
It is possible, however, to go still further, and find a speed at which the
E.M. F. of the strongly excited machine is greater than the supply pressure,
and the machine works as a generator supplying power to the other motor.
The latter has then not only to drive the car, but also to drive the other
machine as a loaded generator. The result is that the armature of the under-
excited motor becomes over-heated and is rapidly destroyed.
It may also occur that the armature resistances are unequal. This would
also be very disadvantageous in the Sprague arrangement.
The regulation of series motors
215
+
(
As a result of these difficulties, speed regulation of series motors is rarely
carried out by shunting or re-arranging the field windings. The general
method now adopted is the simple resistance in series with the motor. The
armature is always in series with its own field winding, so that the armature
current of any motor is identical with the magnetising current of that motor.
At starting, both the current and power taken by the car can be greatly
reduced by connecting the motors in series. Each motor then acts as a
resistance in series with the other (Fig. 204) and the pressure across each
motor is only half that of the mains. The current taken by the car is only
that of a single motor; this is a most important point, since the current
taken to start a car is very great. When a certain speed has been reached,
the motors are connected in parallel (Fig. 205). It would appear at first
sight as if we had again encountered the difficulty of an unequal distribution
of current and work, if by any chance the internal resistances or field strengths
were not exactly equal. If, for example, one of the armatures in Fig. 205 had
a larger resistance than the other, its current
e - E
ia
Ra
would apparently be smaller than that of the other motor. Since, however,
its flux and back E.M.F. would be thereby simultaneously reduced, the in-
•


aa aa
Fig. 204.
Fig. 205.
equality is almost entirely neutralised. Suppose, for example, that the
following figures refer to one of the motors:
e=500, N=3.10°, ia = 20, Ra = 2,
while the resistance of the other is so different that its current is 18 amperes
and the magnetic flux, found from the characteristic for a magnetising current
of 18 amperes, is 2.9.106 lines. We shall now calculate the resistance ≈ of
the second motor. Since both motors run at the same speed, we have from
equation (100) on page 192:
n =
From this we get
(500 – 20 . 2) .60.108 (500 – 18 . 2) . 60 . 108
P.3.10.z
α
X 3:05 ohms.
P. 2·9.106.z
α
Hence we see that an increase in the resistance of 50 per cent. merely causes
the current to decrease from 20 to 18 amperes. Similarly, a difference in
the magnetic flux will cause a very small difference in current (Figs. 206 and
207). We may assume that the gap between armature and poles is greater
in the second motor, thus causing its characteristic curve, which shows the
216
Electrical Engineering
relation between magnetising current and flux, to lie lower than the corre-
sponding curve of the first motor. For the sake of simplicity the values of
the flux in Fig. 207 are taken 10 per cent. less than those in Fig. 206. If
now we draw the speed characteristics under the assumption that the terminal
pressure is the same and the internal resistance is negligible, we find that it
lies higher in the second case. For any given speed, such as that shown by
the dotted horizontal line, we see that the second motor takes a larger


n
n
N
N
Amp
Fig. 206.
Атри
Fig. 207.
current. The difference is, however, very small and never reaches the values
obtained above for the Sprague arrangement. The motor having the larger
magnetic reluctance develops a smaller E.M.F. and thereby enables a larger
current to flow through it. This larger current, however, passes round the
magnets of the same motor and makes up to some extent for its greater
reluctance, so that the difference in the flux cannot become very con-
siderable.
CHAPTER X.
69. The instantaneous value of the induced E.M.F.-70. The electrolytic mean or average value
of an alternating current.-71. Alternating current power and R.M.S. current.-72. Vector
diagrams.-73. The E. M. F. of self-induction.-74. Ohm's law for alternating currents.—
75. Resistance and inductance in series.-76. Resistance and inductance in parallel.
77. Effect of phase difference on a.c. power.-78. Effect of capacity.—79. Resistance and
capacity in series.-80. Circuit containing resistance, inductance and capacity.-81. Self-
induction and capacity in parallel.
69. The instantaneous value of the induced E.M.F.
When a coil of wire is rotated about a diameter in a uniform magnetic
field, electromotive forces are induced in it which change periodically both
in strength and direction. The axis about which the coil rotates should be
at right angles to the direction of the field, as shown in Figs. 208 and 209,
where a rectangular coil is wound on a non-magnetic cylinder. The dotted
lines represent the lines of force, the large circle the non-magnetic cylinder,
↑ ↑ Not ↑ ↑ ↑ ↑


↑ ↑ ↑
↑
1
1
1
Fig. 208.
Fig. 209.
and the smaller circles the sides of the coil in section. In Fig. 208 the coil
is at right angles to the lines and the magnetic flux threading the coil is a
maximum. The motion of the conductors on the cylindrical surface is, at
the given moment, parallel to the lines of force and no E.M.F. is induced in
them. We shall take this position, in which there is no induced E.M.F., as
our starting point or origin. In Fig. 209 the coil has moved through an
angle a and is cutting obliquely through the lines of force, so that an E.M.F.
is being induced in the coil. If we assume that the lines of force pass up
218
Electrical Engineering
•
the page and that the cylinder is rotated in a clockwise direction, an applica-
tion of the right-hand rule shows the direction of the induced E.M.F. to be as
indicated in the figure.
The E.M.F. induced in the coil is greater the more rapidly the lines of
force are cut, that is, the less obliquely the coil cuts through them. The
E.M.F. will therefore reach its maximum value when the lines are cut at
right angles, that is, when the coil is vertical and is threaded by no line
of force. On rotating the coil still further the instantaneous value of the
E.M.F. decreases and reaches zero when the coil is once more horizontal.
this moment the direction of the induced E.M.F. in the coil is reversed.
long as the wire is on the upper half of the drum it cuts the lines from left
to right, but when it passes over to the under side of the drum it cuts them
from right to left. Hence, the induced E.M.F. varies both in magnitude and
in direction.
and
If H be the strength of field in lines per sq. cm.,
N the maximum flux through the coil,
v, the constant peripheral speed in cms. per sec.,
n the revolutions per minute,
D the diameter of the drum in cms.,
L the induced length of the coil-side,
At
So
the number of wires connected in series (if the coil consist of a
single loop z' = 2),
then for the instantaneous value of the induced E.M.F. we have equation (62)
on page 78, according to which
E=H.l.v. 10-8 volts.
For l, which is the total length of induced wire, we must put L.z'. With
regard to the speed v with which the lines are cut, it is evident from Fig. 209
that it is merely the horizontal component of the peripheral speed v。. The
same lines would be cut if the conductor moved horizontally with a speed v
equal to v, sin a. Now
V₂ = π.D.
n
60°
Making these substitutions in the above equation, we have
n
E=H.L.z.D.π. sin a. 10-8.
60
The product D. L is equal to the area of the coil in square centimetres
and H is the number of lines per sq. cm. Hence the product H.D.L is
equal to the total flux N passing through the coil when it is perpendicular
to the field.
We have, therefore,
n
E=T.N. z'. 10-8. sin a
60
.(107).
The instantaneous value of the E.M. F. is therefore proportional to the
sine of the angle which the plane of the coil makes with the initial position.
When a=0, E=0, and when a 90°, sin a=1 and the E.M.F. has its
The instantaneous value of the induced E. M.F. 219
N
=π.N. .z′.10-8
maximum value, as we have already seen.
E.M.F. is seen from equation (107) to be
The value of this maximum
Emax
.....(108).
E = Emax sin a
.(109).
60
Equation (107) can now be written in the form
=
For values of a between 0 and 180°, sin a is positive, whereas for values
between 180° and 360° it is negative. This is in agreement with the fact
already mentioned, that the E.M.F. changes its direction in the horizontal
position of the coil, corresponding to a 180°. If a curve be drawn, having
the values of the angle a for abscissae and the corresponding instantaneous
values of the induced E. M.F. for ordinates, the well-known sine curve is
obtained. We can, however, introduce the time t as the variable and ex-
press the E.M.F. as a function of the time. If w be the angular velocity

E
t
to
Fig. 210.
of the coil and t the time taken to reach its present momentary position,
the angle w. t is identical with the angle a, and we may write equation (109)
in the form
E = Emax. sin (w. t)
..(110).
If the abscissae of our curve be made to represent the time t instead
of the angle a, we merely alter the scale and obtain the curve shown in
Fig. 210. This shows clearly how the E. M. F. varies from moment to moment.
The time t, of a complete revolution corresponds to the angle 27. During
this time the E.M. F. passes through both a positive and a negative maximum,
and returns to its initial value ready to pass again through the same cycle.
The time thus taken to go through a complete cycle is called the period,
or the periodic time. A common value of the periodic time in alternating
current practice is a fiftieth of a second, so that the current or E.M. F. passes
through 50 periods or cycles per second. It is then said to have a frequency
or periodicity of 50. Some American engineers call a half-period an alter-
nation, and speak of a current with a frequency of 50 as having 6,000
alternations per minute.
If the drum on which the coil is wound be made of iron and it be rotated
220
Electrical Engineering
between the poles of an electromagnet in the usual way (Fig. 211), the
induced E.M.F. will not be a sine function of the time. The lines of force
now pass perpendicularly across the gap, so that they are no longer cut
obliquely according to a fixed trigonometrical rule. It is possible, however,
by suitably shaping the pole-tips, to make the gradation from the neutral
zone to the strong field under the pole very gradual, so that we may assume
that the field strength is distributed round the armature according to the
sine law. Its maximum value will be under the centre of the pole and
it will gradually decrease on either side towards the neutral zone, in the
centre of which it will be zero. The induced E.M.F. will again be a sine
function of the time, and will be given by equation (107). As before, N
represents the maximum flux passing through the coil. This is here the
same thing as the flux leaving the north pole of the field magnet.
For multipolar machines with p pairs of poles, the equation must be
slightly modified. As in direct current armatures each coil will span a pole-
pitch, the wire which passes under the centre of a north pole crossing at the
N
N


S
S
S
Fig. 211.
N
Fig. 212.
end and coming up under the next south pole (Fig. 212). The number of
lines of force cut in one revolution will be p times as many as in a bipolar
machine with the same value of N. The E.M.F. induced in the coil will
be p times as great, and we get the following general formula
n.
The product P·60
n
E=T.N.P·60
.z'. 10-8. sin a.
is equal to the number of cycles per second. This is
generally represented by the sign ~, which is derived from the sine wave.
If we substitute the frequency ~ for p in the above equation, we get
Similarly
N
60
E
= T.N.~. z . 10–8 . sin a
Emax = T.N.~.2.10-8
……..(111).
…....(112).
These formulae are equally applicable to two-pole or multipolar machines.
The angle a has not, however, the same meaning in both cases. In the
four-pole machine in Fig. 212, a movement of the wire from the neutral axis
to the middle of the pole represents a geometrical angle of 45°. The E.M.F.
The instantaneous value of the induced E. M.F.
221
has increased, however, from 0 to its maximum value, so that sin a must
have its maximum value 1 for the position under the middle of the pole.
It is evident that we cannot, in such a case, put sin a equal to sin 45°.
The difficulty could be avoided by putting sin (p. a) instead of sin a in the
equation, in which case a would always represent the geometrical angle.
This is, however, an unnecessary complication. We shall assume in all that
follows that a degree is not the 360th part of a complete revolution, but
the 360th part of a complete period or cycle. A complete cycle thus cor-
responds to an angle 2π. The angle π represents half a period, that is,
the time taken for a wire to pass from the middle of a north pole to the
middle of the adjacent south pole. A rigid adherence to this assumption
will make a mistake in applying the above, or similar, equations impossible.
70. The electrolytic mean or average value of an alternating
current.
We have seen in the previous section that the induced electromotive
force varies as a sine function of the time. If we assume that the circuit
contains only ohmic resistance and is free from self-induction, we can deter-
mine the current at any moment by dividing the electromotive force at that
moment by the constant resistance. Hence, by simply altering the ordinate
scale, the sine curve of E. M. F. may represent the alternating current. The
instantaneous value of the current is therefore proportional to the sine of
the angle a which the plane of the coil makes with its initial position, and
we may write
i = imax sin a.
In order to determine the mean value of the current experimentally, the
ends of the coil are connected to a two-part commutator, in the manner
already described in Section 43. The current in the external circuit is then
pulsating, but always in one direction, and we are said to have rectified the
alternating current. If this current is passed through a copper voltameter,
the quantity of copper deposited in a given time is a measure of the quantity
of electricity which has passed through the circuit. To find the mean value
of the current, we must divide the mass m of the deposit in milligrammes
by the time t and by the electrochemical equivalent of copper 0-328. We
thus have
mean.
m
0.328.t'
This average value of the current may thus be called the electrolytic
To determine it graphically, we plot the time t as abscissa and the
current i as ordinate. After a time t, reckoned from the initial position,
the value of the current is given by the equation
i = imax sin a imax sin (w. t).
=
(w.t).
For an infinitely short interval of time dt, the current may be assumed
to be constant. Since the current is defined as the quantity of electricity
passing per second, the product i.dt, which is represented by the narrow
222
Electrical Engineering
vertical strip in Fig. 213, is equal to the quantity of electricity passing in
the time dt. The total quantity passing during the whole period t, is equal
to the sum of all such elemental strips, that is, to the shaded area in the
figure. The average value of the current is found by dividing this area.
by the time t₁. If a rectangle of equal area be drawn on the same base,
the height of this rectangle gives the required mean current.
An approximate value can be obtained by finding the mean of the values
of sin 0, sin 10°, sin 20°, and so on up to sin 90°, and multiplying it by max-
To obtain an accurate value of the mean current, we must determine the
area of the shaded area (Fig. 214) corresponding to half a period. The
integration is simplified by plotting angles as abscissae. The area of the
narrow vertical strip in Fig. 214 is
To find the whole
the limits 0 and T.
at
жK
i. da imax. sin a. da.
=
area we must integrate, or add up all the strips, between
This gives us
ži. da = imax ["
π
sin a. da = 2imax.
0
da

t
Fig. 213.
π
Fig. 214.
By dividing this area by the base π we get the average height or average
current. Hence
2
imax=0.636. imax
π
....(113).
Hence, on the assumption that the current is a simple sine function of
the time, its average value is to the maximum value as 2: π or as 7: 11.
Similarly for the electromotive force, we have
E2T
2
π
Emax=0.636. Emax
..(114).
These average values of the alternating current and pressure are, however,
little used in electrical engineering, since they cannot be used to calculate
the power. The reason for this will be clear after reading the next section.
71. Alternating current power and root-mean-square current.
As we have just mentioned, the power cannot be obtained by multiplying
together the mean values of the current and pressure. We must find the
momentary value of the power by multiplying the momentary value of the
1-
Alternating current power and root-mean-square current 223
current by the value of the pressure at the same instant. The same result
is obtained by squaring the momentary value of the current and multiplying
by the resistance. Hence, for the power at any instant, we have
P = emax. sin a. imax. sin a = (imax. sin a)². R.
These values of P are found for various moments during the period, and
plotted as ordinates in Fig. 215. As power is the work done per second,
the area of the small strip P. dt must represent the work done during the
interval dt. The whole shaded area must represent the total work done
during a period. If this be divided by the time t in which the work is
done, we get the mean work per second, that is, the mean power. This
is given by the height of a rectangle of the same area and on the same base
as the shaded figures. This height is found to be exactly half of the total
height of the curves.
The same result is obtained by calculation.
by plotting the angle a as abscissa (Fig. 216).
at
Як
This is simplified, as before,
The area of the narrow strip

da
t₁
Fig. 215.
Л
Fig. 216.
in Fig. 216 is equal to P. da. Substituting the value of P found above, we
find the total area of the figure as follows:
but
Π
["P. da = ['""'x. R. sin² a. da,
0
["sin' a. da-
0
max •
hence, the shaded area in Fig. 216 is equal to
ㅋ
​α
1
π
sin a. cos a
2
2
2'
0
2ºmax. R["
π
Ꭱ sin² a. da = i²,
=2 R
max
2
0
To obtain the mean power we must divide this area by the base π.
We thus get
23³max · R
P =
2
If the circuit contain no self-induction, we can put
Emaximar. R,
.(115).
and equation (115) becomes
P =
emax. Imax
2
..(116).
It is interesting to obtain this result in another way, which requires
no higher mathematics, but which, at the same time, does not give one such
224
Electrical Engineering
an insight into the reasons for the result, as is obtained from a careful study
of the above. We assume that a two-pole machine has a drum containing
two independent windings at right angles to each other. Both windings are
connected to equal external resistances. When the plane of one coil makes
an angle a with the zero position, the other will make an angle 90 + a with
the same position, and the momentary value of the total power of the two
windings together is
Emax. max. [sin² a + sin² (90° + a)] = Emax · imax · (sin² a + cos² a).
Since sin² a + cos² a is always equal to 1, it follows that the power of
the two coils together is always constant and equal to Emax. max. This is
the mean power of the two coils, and that of each will therefore be a half
of this, viz.
P =
Emax. imax
2
as found above. Had we calculated the power by multiplying together the
average values of the current and pressure, the result would have been
2
π
Emar •
2
π
imax 0405 Emax. imax.
=
This is 20 per cent. smaller than the real average power.
We must now define and calculate the values of current and pressure,
which can be applied directly to the calculation of the power. For this
purpose we write equation (115) in the following form:
P:
=
imax Imax
√2 √2
max R.
imax
√2
The expression in this equation represents a certain strength
of current which, if squared and multiplied by the resistance, gives
the mean power. This strength of current is called the effective or
root-mean-square value of the alternating current. This is the current
strength which is meant when speaking generally of the strength of an
alternating current. We shall represent it by the letter i. Although this
letter also represents the instantaneous value of the current, confusion be-
tween the two is almost impossible. We have therefore
i
2m8x
= 0·707. imax
√2
.(117).
We can therefore define the effective value of an alternating current
in the three following ways:
(1) as that value which, squared and multiplied by the resistance, gives
the mean power;
(2) as the root of the mean value of the squares of the momentary
values of the current;
(3) in true sine waves, as 0.707 of the maximum value of the current.
Since we found the electrolytic mean to be
effective value is to the average value as 11: 10,
of the maximum, the
or more accurately as
Alternating current power and root-mean-square current 225
T: 2√2. If, therefore, we wish to calibrate an ammeter for alternating
currents by rectifying the alternating current and passing it through a copper
voltameter, it will be necessary to multiply the average current, as calculated
from the deposit in the voltameter, by 11 in order to obtain the correct
value of the effective current which should be indicated on the scale of
the ammeter. Those instruments, however, in which the deflecting force
is proportional to the square of the current at any moment, will give the
same deflection with direct current as with an alternating current of the
same effective value.
They take up a position corresponding to the mean
value of the squares of the alternating current, and, if calibrated with direct
current, indicate correctly the effective value of the alternating current.
To this class belong those instruments in which a current-carrying coil
attracts or deflects a small piece of soft iron. The force exerted at any
moment is proportional to the product of the momentary value of the current
and that of the magnetism induced in the soft iron. For low values of the
saturation the force will be therefore proportional to the square of the current,
and we should expect the instrument to read correctly both direct and alter-
nating currents. As, however, no iron is entirely free from hysteresis and
remanent magnetism, these instruments are not suitable for very accurate
work even with direct currents. With alternating currents the hysteresis
causes the reading to be lower than that obtained with a direct current
of the same effective value.
Hot-wire instruments, on the other hand, have exactly the same scale
for all currents. In this type of instrument a platinum or platinum-silver
wire is heated by the passage of the current and thereby caused to elongate.
This elongation is magnified, and transmitted to the pointer of the instru-
ment. Since the wire possesses a certain thermal capacity and the pointer,
etc., a certain inertia, the instrument is unable to follow the rapidly pulsating
current, and consequently takes up a position corresponding to the mean
power dissipated as heat in the wire. The readings on the scale correspond
to effective currents, that is, the currents which, if passed steadily through
the wire, would produce the same heating as the alternating current. It
follows that the instrument must be equally applicable to both direct and
alternating currents.
The same holds true for dynamometers, in which the current flows
through a fixed and a moving coil connected in series. It was seen in
Section 32 that currents flowing in the same direction are attracted, while
currents in opposite directions are repelled from each other. Since the
current is reversed simultaneously in both coils, the turning moment exerted
by the fixed on the moving coil will act always in one direction. The
deflection, being proportional to the product of the currents in the two coils,
is proportional to the square of the current. In consequence of the inertia
of the moving coil, the pointer takes up a position corresponding to the
mean torque, and therefore to the mean of the squares of the current. The
scale can be marked so as to read directly the effective current. This is,
as before, that continuous current which gives the same deflection as the
T. E.
15
226
Electrical Engineering
periodically varying current. The same scale can evidently be used both for
direct and alternating currents. The same is naturally true for volt meters.
72. Vector diagrams.
The sine curve represents very clearly the changes occurring in the
current or in the electromotive force during a period, and we shall often
make use of it in our further studies of alternating current phenomena.
The vector diagram is to be preferred, however, when we wish to determine
rapidly the relations between the various currents and electromotive forces
in an alternating current circuit. In Fig. 218 the line OE represents Emax
to a certain scale, and rotates about the point 0 with a constant angular
velocity w = 2π~. If we are considering a bipolar machine, the vector will
rotate at the same rate as the armature. The direction of rotation is im-
material, but we shall always assume it to be clockwise. In its initial
position the vector will lie horizontally to the left of O. The angle a which

F
A
N
S
α
B
0
C
D
π
Fig. 217.
Fig. 218.
it makes with its initial position at any moment is equal to the angle made
by the plane of the coil in Fig. 217 with its zero position. The projection of
OE on the vertical axis is evidently equal to OE. sin a, that is, to Emax. sin a.
Hence, the instantaneous value of the electromotive force is equal at
every moment to the projection of the radius vector upon the vertical
axis.
To make the matter still clearer, a sine curve representing half a period
has been added in Fig. 218 to show the connection between it and the vector
diagram. The radius of the circle described about point 0 as centre is
equal to the maximum ordinate of the sine curve. The instantaneous value
OA in the vector diagram is equal to the ordinate BD`in the sine curve.
Finally, the arc subtending the angle a at unit radius in the vector diagram
is equal to the abscissa CD in the sine curve.
Electromotive forces of different magnitude and of different phase can be
very conveniently compounded and their resultant determined by means of
the vector diagram. When we speak of electromotive forces being of different
phase we mean that they are out of step, as it were, and do not attain their
Vector diagrams
227
maximum values at the same moment.
On the armature in Fig. 219, for
example, there are two coils consisting of a different number of turns and
displaced by a certain angle from each other. Coil I possesses the greater
number of turns and will therefore have a greater maximum E.M.F. than
coil II. Two circles are drawn about the centre 0 (Fig. 220), one having a
radius OE₁ = Emax and the other a radius OE₂ = Eamax, and the vectors
OE, and OE, are drawn in directions corresponding to the positions of the
coils I and II on the armature. As in the case of forces, we find the
resultant OR by completing the parallelogram OE, RE₂. At the given
moment
OA is the instantaneous value of the E.M.F. in coil I,
OB
""
""
""
""
""
II,
and OC is the projection on the vertical of the resultant OR. From the
similarity and equality of the shaded triangles it follows that
OC=OA + OB.

I
"
II
A
E
B
E₂
2
R
S
Fig. 219.
Fig. 220.
Now, OC is the instantaneous value of an E.M. F. represented both in
magnitude and phase by the vector OR. Thus the sum of the instantaneous
values of the electromotive forces in the two coils is equal at every moment
to the instantaneous value of the resultant. It follows that electromotive
forces can be compounded by the parallelogram law in exactly the same way
as mechanical forces.
To make this important result still clearer, we have represented in
Fig. 221 the moment when the resultant E. M. F. has its maximum value. Its
vector lies vertically, while the vector of E, lies a little to the left and that
of E, a little to the right of the vertical. If we consider the wires con-
stituting the adjacent coil-sides on one side of the armature, we can see that
the maximum resultant E.M. F. will be induced when the wires as a whole lie
under the middle of the pole. In this position, however, the wires of coil I
lie just before the pole middle, while the wires of coil II lie just beyond it,
exactly as we have found from the vector diagram.
In a similar manner we may consider the moment when the coils lie in
the neutral zone and the resultant E. M.F. is nil (Fig. 222). One coil-side
15-2
228
Electrical Engineering
is under the influence of the north pole while the other is under the influence
of the south pole, and the opposing electromotive forces induced in the two
coils exactly neutralise each other. We arrive at the same conclusion from
a consideration of the vector diagram in which the resultant OR is horizontal
and its projection on the vertical axis is O. The vector OE, lies above the
abscissa axis while the vector OE, lies below it. The projection of OE, is
therefore positive and the projection of OE₂ negative, and since they oppose
each other, their resultant vanishes. Here again, the directions of the vectors
correspond with the geometrical positions of the coils. Such considerations
as these diminish, to a certain extent, the difficulty experienced by the
student in grasping the idea of difference of phase and in the application of
the parallelogram of forces to electromotive forces.
R
2

I
E

E2
I
E
I
E₂
Fig. 221.
Fig. 222.
We can, however, go a step further and substitute effective values of the
electromotive forces instead of maximum values. This will simply alter the
scale of the vector diagram without altering the angles. The length OR will
then represent the effective value of the resultant. In a similar way we can
find the resultant of two currents flowing in parallel paths, but differing in
magnitude and phase. Such a compounding of electromotive forces or
currents is called a vectorial or geometrical addition.
73. The electromotive force of self-induction.
It was seen in Section 35 that the electromotive force E,, induced by a
change of current in a coil of self-induction L, is given by the equation
where
di
-L.
dt
volts,
0·4π. S².µ. A
L =
10-8 henries.
し
​For the angular velocity of the radius vector we have
w=2π.~,
so that
i = imax. sin a = imax. sin (wt).
Differentiating, we have
di
dt
i'max. COS (wt). w = imax. w. cos a.
The electromotive force of self-induction
229
Introducing this into the above formula for E,, we get
- Lw. imax⋅ cos a
Es
..(118).
The E.M.F. of self-induction is thus a cosine function and by plotting it as
ordinates with either the angle a or the time t as abscissae we obtain a cosine
curve. This is identical in form with the sine curve, but is displaced 90°
from it. The relations found for the sine function can therefore be applied
here, so that the effective value of the E.M. F. of self-induction is 0.707 of the
maximum value.
From equation (118) it can be seen that the maximum value is reached
when cos a
– 1, that is, when a = 180°. Its value will therefore be
Esmax = Lw.imax·
If both sides of this equation be multiplied by 0·707, the maximum values
of E.M.F. and current will be converted into effective values, and we can write
E₁ = Lw.i.
8
Hence, the effective value of the E.M.F. of self-induction is given by the
product of the coefficient of self-induction, the angular velocity of the vector
and the effective current.
A coil purposely made so as to have a large self-induction is called a
choking coil. Such a coil is shown in Fig. 227, and consists of an iron core
bent into a ring with a small air-gap and wound with insulated copper wire.
If the magnetic flux density is kept small, the reluctance of the iron may be
neglected and only the air-gap need be considered in calculating the self-
induction.
We shall assume the following data for the choking coil:
Length of magnetic path in air-gap
Cross-section of path normal to the flux......
Number of turns in coil
Effective value of current.
Frequency
1 = 0.6 cm.
A = 12 sq. cms.
S
= 200.
i = 10 amps.
~=
Then, since the permeability of the air is 1, we have
50.
Ꮮ
Į =
0·4π. S². μ. A
1
10-8=1·25
2002.1.12
0.6
.10-0.01 henry.
We have also
w=2π ~=314.
Hence, for the electromotive force of self-induction, we have
E₁ = Lw.i=001. 314. 10-31-4 volts.
We shall see presently that a voltmeter connected across the terminals of
the choking coil will read almost exactly 31-4 volts, showing that the coil
acts like a resistance in causing this drop of pressure.
In order to design a choking coil, however, the formula needs to be modi-
fied, as we might otherwise choose a value of cross-section for the iron or a
number of turns for the coil which would lead to a very unsuitable flux-
density in the core. In Section 29 we found the following relation between
the maximum current and the maximum flux-density :
B Вmax = μ. Hmax
0.4π. S. imax. μl
1
230
Electrical Engineering
If we introduce the effective current by putting imax = √2i, and solve the
equation for i, we get
Bmax. 1
1.78.S.μ
..(119).
This equation is of great importance in the calculation of the magnetising
current of a transformer. Here, however, we shall merely use it to get the
equation for the induced E. M. F. into a convenient form. If we substitute for
L, w and i in the equation E.-L.w.i the values just found for them,
we get
E = 444. Bmax. A.~. S. 10–º.
Since the product Bmax. A gives the total flux N embraced by the coil,
we may write
E.=444 N.~. S. 10-8
8
.(1·20).
From the two equations (119) and (120) a choking coil can be designed
when we have chosen a suitable value for the magnetic flux-density. If, for
example, an E.M.F. of 30 volts is to be induced by a current of 10 amperes at
a frequency of 50, we may proceed in the following manner. The flux-
density is taken at 5,000 lines per sq. cm. and the path in the iron is
neglected, so that we need only consider the length of the air-path. The
cross-section of the iron is assumed to be 15 sq. cms. We have then
N = Bmax • A = 5,000. 15 = 75,000.
From equation (120) we have
S
E,.108
30.108
4:44. N.~ 4-44.75,000. 50
=
180.
Putting the number of turns S equal to 180 in equation (119) we get
1.78.S.μ.i_178.180.1.10 = 0.64 cm.
Bmax
5,000
This is the length of the air-gap. The cross-section of the path across
the air-gap has been taken as equal to the cross-section of the iron. As a
matter of fact, however, the lines of force do not pass straight across the gap,
but curve round from pole to pole, being driven outwards by the lateral
pressure which we have seen to exist between the lines. The cross-section
of the path across the gap is therefore considerably greater than the cross-
section of the iron. The presence of the paper or varnish insulation between
the sheet-iron stampings of which the core is built up will still further
increase this effect. The above coil will consequently cause a greater drop
than 30 volts when carrying 10 amperes.
Having considered the magnitude of this E. M. F. of self-induction, we must
now turn to its phase. We saw above that the curve is a cosine curve and is
consequently displaced 90° from the sine curve of the current. This will be
made plainer if we determine the magnitude and direction of the E.M.F. of
self-induction at several critical moments during the period.
The current curve is represented by the thick black line in Fig. 223.
When a = 0, sin a=
0, sin a=0, but cos a = 1. This is the moment when the current
passes through zero and the E. M. F. of self-induction has its negative value, as
The electromotive force of self-induction
231
seen from equation (118) on page 229. The current curve is steepest at this
point, which means that, although there is no current at the moment, yet the
rate at which the current, and with it the flux, is changing is then a maxi-
mum. That the E.M.F. induced in the coil is negative follows at once from
the fact that it opposes the growing current and tends to keep up the dying
current.
When the angle a is equal to 90° we have sin a = 1 and cos a=0. The
current is then a maximum and the induced E.M.F. is equal to 0. At the
moment when the current is a maximum, the current curve in Fig. 223 is
parallel to the base, and for an instant both current and flux are constant.
There is therefore no cutting of lines at this moment and the dotted curve
must pass through the base line at the point representing a = 90°.
8
By finding the various points on the curve E, in this manner, the whole
curve can be drawn, with the result that the E.M.F. of self-induction is found
to lag 90° behind the current. This point should be carefully noticed, since
one might easily come to the conclusion that, as the curve E, in Fig. 223 is
8

i
Es
0
元
​Fig. 223.
further to the right than the curve i, the E.M.F. is therefore ahead of the
current. This conclusion is quite wrong, for we see that the curve E, reaches
its maximum value at a time when the curve i has already passed its maxi-
Hence the E.M. F. of self-induction lags behind the current by
quarter of a period or by an angle of 90°.
mum.
This is made clearer by a consideration of the vector diagram in combi-
nation with the laws of self-induction already found in Section 35. The
diagram is shown in Figs. 224, 225 and 226 for three successive moments.
When the current vector passes through the horizontal position (Fig. 225),
its projection on the vertical, and therefore also the instantaneous value of
the current, is equal to 0. The E. M. F. of self-induction is a maximum at this
moment and its vector must consequently be vertical; moreover, since it
lags 90° behind the current, it must be vertically downwards.
We shall now consider the relations a moment earlier and a moment
later. In Fig. 224 the current is decreasing and its vector lies below the
horizontal so that its projection is in the same direction as that of the
232
Electrical Engineering
electromotive force E. The E. M. F. is thus in the same direction as the
decreasing current, which agrees exactly with what we found in Section 35.
In Fig. 226 the current has passed through its zero value and is increas-
ing. The projection of its vector is now above the horizontal and therefore
in the opposite direction to the projection of the E. M. F. which has hardly
changed. The self-induced E. M. F. is thus opposed to the growing current, as
was found in Section 35.

i
000
Fig. 224.
Fig. 225.
Es
Fig. 226.
If now we agree to consider the magnetic flux as having the same sign as
the current producing it, there will be no difference of phase between the
magnetising current and the flux. The current curve will then represent to
a certain scale the number of lines threading the winding at any moment.
Hence, the self-induced E. M. F. lags not only 90° behind the current, but also
90° behind the magnetic flux, through the variation of which the E.M. F. is
induced.
74. Ohm's law for alternating currents.
We shall now consider the circuit represented in Fig. 227, in which an
ohmic resistance R is connected in series with a choking coil, the resistance
of which is so small compared with its self-induction that it can be neglected.
e

R
iR
ως
Es
Fig. 227.
We have to determine the magnitude of the current which flows round the
circuit under the joint influence of the terminal pressure e of the generator
and the self-induced E.M.F. of the choking coil. In addition to the magnitude
of the current, its phase relation with regard to the terminal pressure should
be considered,
To simplify the problem we shall first consider it in the reverse form,
viz. what terminal pressure will be required to drive a given current
Ohm's law for alternating currents
233
through the external circuit? In the first place we determine the mo-
mentary value of the ohmic pressure drop by multiplying the momentary
value of the current by the resistance. This ohmic pressure drop is plotted
as ordinates and gives the heavy black curve in Fig. 228. Since it must
have its maximum value at the same moment as the current, it must be in
phase with it. The curve i. R* differs from the current curve merely in the
matter of scale.
The ohmic pressure drop must be provided by the combined action or
resultant of the terminal pressure, which is at present unknown, and the
electromotive force E, the curve of which lags 90° behind the curve i. R.
We proceed now to determine the value of the terminal pressure for a few
characteristic points during the period.

G
160
120
H
80 F
40
0
10
A B
C
-40
-80
-120
-160
Es
Fig. 228.
At the origin O there is momentarily no current, although the E.M. F. of
self-induction has at this very moment its maximum negative value. It is
evident, therefore, that its effect is exactly neutralised by an equal and
opposite terminal pressure OF. The current can only vanish when forward
and back pressures exactly neutralise each other.
At the moment A, on the other hand, the terminal pressure has not only
to overcome a considerable E.M.F. of self-induction, but has now to cover a
large ohmic pressure drop. The terminal pressure must therefore be equal to
the sum of these two, and will be represented by GA.
At the moment B there is no E. M. F. of self-induction, and the terminal
pressure has merely to provide for the ohmic pressure drop HB. The curves
of terminal pressure and ohmic drop will therefore cross at the point H.
Finally, at the moment C, the self-induced E. M. F. is just sufficient to drive
the diminishing current through the resistance without assistance from the
generator, the terminal pressure of which is therefore zero. The curve e cuts
* In a few of the figures the ohmic resistance is denoted by the letter n instead of R.
234
Electrical Engineering
the horizontal axis at the point C. In this way we obtain the thin full-line
curve in Fig. 228, the data for which are as follows:
i = 10 amperes, R= 10 ohms, wL=6.28.
The maximum value of the ohmic pressure drop is found as follows:
imar. R√2.10.10 1414 volts,
=
while for the maximum value of the self-induced E. M. F. we have
Esmax=wL. imax = 6.28.√/2.10 = 88.8 volts.
The maximum value of the terminal pressure of the generator is found
from Fig. 228 to be 167 volts.
An examination of the figure leads to the following important con-
clusions:
1. The curve of ohmic pressure drop does not reach such a high value as
the terminal pressure curve, that is, the current is smaller than one would
expect from the terminal pressure and the given resistance. The formula
e=i.R is therefore not applicable to alternating current circuits containing
self-induction.
2. The sum of the maximum values of the ohmic pressure drop and the
self-induced E. M. F. is greater than the maximum value of the terminal
pressure. The sum of the first two is 230-2 volts, whereas the latter is given
by the figure as 167 volts. This fact is illustrated in a striking manner by
connecting three voltmeters as shown in Fig. 227. One is across the machine
terminals, one across the resistance, and the other across the choking coil.
The following effective values will be read on the three instruments:
e=
emax 167
√2 √/2
i. R= 10.10 = 100,
118
E¸=wL.i= 6·28. 10 = 62·8.
Thus, the sum of the two separate pressures i. R and E, is greater than
the total pressure e. This result seems quite contradictory to the ideas which
we have formed in connection with direct currents. It must be carefully
borne in mind, however, that this is only true of the effective or maximum
values, and not of the instantaneous values at any moment. For the latter
the total pressure is the algebraic sum of the separate pressures.
3. The curve representing the ohmic pressure drop or the current lags
behind the curve of terminal pressure by the angle & which is represented in
Fig. 228 by the distance CD. The current reaches its maximum value a
short time after the pressure has passed through its maximum value. The
angle corresponding to this interval is known as the angle of lag. When the
current is ahead of the pressure, that is, when the current reaches its maximum
a short time before the pressure, the current is said to lead by a certain angle
which we call the angle of lead.
The striking phenomena occurring in a circuit containing self-induction
are brought out very clearly in the vector diagram. Instead of the current
vector we may consider the vector of the ohmic pressure drop i. R, which is
in phase with, and proportional to, the current. In Fig. 229 it is passing
Ohm's law for alternating currents
235
e
-Es
through the zero position and the vector of the E.M.F. of self-induction points
vertically downwards. Since the current is equal to 0 at this moment, the
E.M.F. of self-induction must be exactly neutralised by an
equal and opposite component of the applied terminal
pressure. This component -E, is therefore drawn vertically
upwards from 0. The resultant terminal pressure e is
given by the hypotenuse of a right-angled triangle, the
sides of which are equal to the E.M.F. of self-induction
and the ohmic pressure drop respectively.

i.R 6 10
Fig. 229.
Es
It is seen that the terminal pressure has a double
duty to perform, viz., to overcome the ohmic resistance
on the one hand, and counterbalance the E.M.F. of self-induction on the other.
Although the E. M.F. of self-induction is sometimes acting with the current
and sometimes against it, we must always look upon it as a back E.M. F.
which must be overcome by a component of the applied pressure. This is
specially evident in the case shown in the figure, for the whole terminal
pressure is momentarily engaged in counterbalancing this back E.M.F. of
self-induction.
8
Several of the characteristic points noticed in connection with Fig. 228
are even more evident in the vector diagram. The terminal pressure e is
greater than the pressure i. R required to overcome the ohmic resistance,
while the sum of the component pressures i. R and E, exceeds the total
pressure e. As before the current lags behind the terminal pressure by an
angle . It is to be observed that this angle must be measured between the
hypotenuse of the right-angled triangle and the side representing the ohmic
pressure drop.
From the vector diagram we can easily determine the equation expressing
the relation between current and pressure in an inductive circuit. It is
evident from Fig. 229 that
e² = (i. R)² + E².
By substituting for E, its value Lw.i and transposing we get
8
e
√ R² + (∞L)²
………….(121).
This may be called the Ohm's law for alternating current. It emphasises
the fact that self-induction causes an apparent increase in the resistance of a
circuit, since the denominator is increased from R to √R² + (@L)².
This increased resistance is represented graphically by the hypotenuse of
a right-angled triangle, the sides of which are equal to the resistance R and
the quantity oL respectively. Since this triangle (Fig. 230) is similar to the
corresponding pressure triangle in Fig. 229, it follows that the angle between
its hypotenuse and the side R is equal to the phase difference or angle of
lag 4, and we have therefore
tan &
=
wL
R
.(122).
The quantity R²+(L) is called the apparent resistance or, more
generally, the impedance, while the quantity oL is known as the reactance or
236
Electrical Engineering
inductive resistance. The quantity oL can only be looked upon and treated
as a resistance if its dimensions are those of resistance. It would otherwise
be quite impossible to add R2 and (wL). Now the dimension of the
R²
coefficient of self-induction is simply a length, and that of angular velocity
is the reciprocal of time, so that
the dimensions of reactance are L. T-1.
These are, however, the dimensions of velocity, which were seen in Section 40
to be also those of resistance. Moreover, since a henry is equal to 10º absolute
units and an ohm is also equal to 10° absolute units, the product oL must
give the inductive resistance directly in ohms.
With the help of equation (121) we are now in a position to calculate the
current for any given values of e, R and wL. We are no longer compelled to
follow the reverse course and find the terminal pressure for an assumed value
of the current. We shall assume the same data as already used on page 234,
viz.
e=118, R=10, L=6.28.

Im
√R²+(WL)

Reactance
Impedance.
R
Fig. 230.
From equation (121) we have
i
118
√10² + 6.282
Resistance.
Fig. 231.
10 amperes.
For the phase-lag of the current behind the terminal pressure we have
wL 6.28
tan &
=
0.628.
R 10
π
This corresponds to an angle of 32° 10′ or This is in evident agreement
6
with the length CD in Fig. 228 which was drawn for the same conditions.
75. Resistance and inductance in series.
We shall now consider the case in which two parts of a circuit are con-
nected in series, each part containing both ohmic and inductive resistance, as
represented in Fig. 232. The total pressure e is, in this case, the hypotenuse
of a right-angled triangle, one side of which is equal to the sum of the
pressure drops i. R₁ and i. R₂ due to the ohmic resistances, while the other
side is equal to the sum of the inductive pressure drops E,, and E,. At the
same time, however, the total pressure e is the geometrical sum of the
terminal pressures e, and e, of each half of the circuit. The triangle of
pressures can be drawn for each half of the circuit. The terminal pressure &
81
Resistance and inductance in series
237
of the left half forms the hypotenuse to the sides i. R₁ and E, and makes an
angle 4, with the current vector or side i R₁, the tangent of which is given
by the equation
Es
wL₁
tan 1 i. RTR
82
In a similar manner the terminal pressure e, of the right-hand half of
the circuit forms the hypotenuse to the two sides i. R₂ and E„, and makes
an angle, with the current vector, the tangent of which is given by the
equation
82
1
Es₂_wL₂
tan 2 i. R₂ R₂
The phase-lag of the current behind the terminal pressure e of the whole
circuit will lie somewhere between 4, and 2. Hence, although the current
in all parts of a series unbranched circuit must necessarily have the same
phase, the pressure differences between various parts of the circuit generally
differ in phase (Fig. 233).

e
iR
Es,
Ø
Fig. 232.
Ess
The circuit made up of an alternating current generator working on an
external circuit containing both resistance and inductance is similar to that
just considered. As an example we shall assume the following data:
Terminal pressure of generator
Current
Armature resistance
Self-induction of armature.....
Frequency
Phase-lag in external circuit
e = 2,000 volts,
i = 50 amperes,
Ra = 1 ohm,
L = 0·04 henry,
50,
cos + = 0·8.
We wish to determine the pressure drop in the generator and the electro-
motive force induced in the generator, both as regards its magnitude and its
phase. For the construction of the diagram shown in Fig. 234, we have the
following values:
i. Ra=50.150 volts,
Esinternal wL.i=0·04.2.3·14.50.50
628 volts.
From these two values we find the internal drop of pressure OF as the
hypotenuse of a right-angled triangle, one side of which is equal to the ohmic
238
Electrical Engineering
[
•
i
resistance drop, while the other side is equal to the E.M.F. due to the armature
self-induction. We thus get
OF= √50² +6282: = 630 volts.
We now find the ohmic resistance drop in the external circuit, thus:
If cos
i. R= e. cos & ⇒ 2,000.0·8 = 1,600 volts.
cosp⇒
0.8 we know that sin 06, so that the inductive drop in the
external circuit can be found in the following manner:
=
Esext. = e. sin $2,000. 0.6 1,200 volts.
The total ohmic resistance drop in the whole circuit is equal to
i. Ra+i.R= 50+ 1,600 = 1,650 volts,
and the total inductive drop is equal to
Egint. Egext=628+ 1,200 1,828 volts.
=
From these two values we find the E.M. F. of the machine E, as the
hypotenuse of a right-angled triangle, one side of which is equal to the sum


iR₁
Es
Fig. 233.
Es₁
iR2
~B
Ն
E,
F
Es int.
Es ext
Fig. 234.
iR
iRa
of the resistance pressure drops, while the other is equal to the sum of the
inductive pressure drops. We thus have
E₁ = √1,650² + 1,828² = 2,460 volts.
The angle 4, by which the current lags behind the E. M.F. of the generator
can be found from the equation
Σ(i. R) 1,650
cos 01:
= 0.67.
E₁ 2,460
Neglecting secondary effects, we can say that the terminal P.D. on open
circuit will be equal to the electromotive force E₁. On loading the generator
the pressure drops from 2,460 to 2,000 volts, i.e. 460 volts, whereas the
pressure required to drive the current through the armature was seen to be
OF=630 volts. This apparent contradiction is due to the fact that the
pressure drop OF in Fig. 234 is subtracted geometrically and not algebraically
from the E. M.F., to obtain the terminal pressure e.
The drop of 460 volts which we have found between no-load and full load
represents 18.6 per cent. of the no-load pressure. This is approximately the
usual figure found in practice and is, as we have seen, almost entirely due to
the internal self-induction of the machine. This brings out very clearly the
detrimental action of the self-induction in increasing the apparent internal
resistance of an alternating current generator
Resistance and inductance in parallel
239
76. Resistance and inductance in parallel.
In the case illustrated in Fig. 235 two branches are connected in parallel,
each containing both ohmic and inductive resistance. The terminal pressure
e is exactly the same both in magnitude and phase for each branch. This is
true even if one or both of the branches contain either ohmic or inductive
resistance alone, instead of each branch being partly non-inductive and partly
inductive as shown in Fig. 235. We have now to determine the magnitude
and phase of the total current, and the current in each branch, from a know-
ledge of e, R₁, wL₁, R₂ and wL2. The line OG in Fig. 236 is drawn to
represent the terminal pressure e and a semicircle is described upon it. OA
is drawn so as to make an angle 4, with OG. This angle is given by the
formula
tan 1
@L₁
R₁
The line OA represents the ohmic pressure drop in the upper branch of
Fig. 235. If we divide this by the resistance R₁ we get the current i̟, which

R₁
wL,
i
R₂
wL₂
D
B
F

Fig. 235.
e
Fig. 236.
may be represented by the length OC. In a similar manner the line OB is
drawn at an angle 2 with the diameter, where
tan 2
wL2
R₂
The line OB represents the ohmic pressure drop in the lower branch, and
by dividing it by the resistance R₂ we get the current 2, which may be
represented by the length OD. The resultant OJ of i, and i gives us the
total current i both in magnitude and phase. If the vector i be produced to
meet the semicircle at F, the length OF represents the ohmic drop, and the
length FG the inductive drop of a single coil or piece of apparatus which
could replace the branched circuit of Fig. 235. The total current lags behind
the terminal pressure by the angle FOG.
77. Effect of phase difference on A. C. power.
If the current is not in phase with the pressure, the instantaneous value
of the power is positive whenever the pressure and current are momentarily in
the same direction (Fig. 237). On the other hand, the power is momentarily
negative whenever the current and pressure are in opposite directions
(Fig. 238). By the direction of the pressure we mean the direction from the
240
Electrical Engineering
positive terminal or terminal of high potential to the negative or low potential
terminal. We shall not consider the internal electromotive forces, but only
external potential differences. When we speak above of power being positive, we
mean that the apparatus or part of the circuit under consideration is absorbing
electrical energy and transforming it into some other form such as thermal or
mechanical energy. When an apparatus has a negative power it is generating
electrical energy at the expense of energy in some other form. In a dynamo,
for example, the current flows from the negative terminal to the positive, that
is, against the terminal pressure, and the power is therefore negative, or, in
other words, electrical energy is being generated. In the motor the current
flows from positive to negative and the power is positive, that is, the motor is
absorbing electrical energy. We are thus led to the conclusion that a simple
alternating current generator in which the current is out of phase with the
E.M.F. does not act continuously as a generator, but, twice in each period,
takes electrical energy from the circuit and is driven for a moment as a
motor.


Fig. 237.
Fig. 238.
The same rule can be applied to any apparatus to which electrical energy
is supplied. If, at any moment, the current flows in the direction of the
terminal pressure, i.e. from + to —, electrical energy is being absorbed, but if
the current is flowing in the other direction, i.e. from to +, the apparatus
is acting momentarily as a generator and returning electrical energy to the
circuit.
Fig. 239 refers to a coil or other apparatus in which the current i lags by
an angle & behind the terminal pressure. This lag is due to the self-induction
of the coil. The time is plotted as abscissae and the current and P.D. as
ordinates. The angle by which the current lags behind the P.D. corresponds
to the time AB.
A curve is now plotted giving the momentary values of the power found
by multiplying together the corresponding values of current and P.D. These
values of the power are positive during the intervals OA and BC, but negative
during the shorter intervals AB and CD, as shown by the dotted curve P,
which represents the rapidly changing power to a certain scale. The area
enclosed between this curve and the horizontal axis represents the energy
supplied to the apparatus. This area lies partly above the axis and partly
below. The mean power is found by subtracting the area below the line
Effect of phase difference on A. C. power
241
from the area above the line, and dividing the difference by the time. It is
evident that the power supplied to the apparatus is less than we should
expect from multiplying together the pressure and the current as read on
suitable instruments.
In order to calculate the mean power, we shall imagine the curves to be
plotted to a base representing the angular movement of the vector, instead of
the time. If the strength of the current at a given moment be imax. sin a,
the terminal pressure at the same moment will be emax. sin (a + ), and the
momentary value of the power will be equal to their product, thus
Pemax. max. sin a. sin (a + ).
We may imagine this power to remain constant over the infinitely small
angle da, so that the area of an infinitely narrow vertical strip will be
P. da = emax • imax. sin a. sin (a + þ). da.

E
Fig. 239.
The mean value of the power is obtained by integrating this expression
between the limits 0 and π and dividing the resulting area by the base π.
In this way we get
Now,
1
π
P= emax. max. sin a. sin (a + p). da.
π 0
sin (a+) sin a. cos + cos a. sin o.
Substituting this value for sin (a +
:S
Π
), we have
emax · Umax · sin² α. cos p. da +
1
Р
Since
f
Π
0
•
1
訂
​emax. max. sin a. cos a. sin &. da.
0
sin a. cos a = 0, the second integral vanishes and we have
1
P =
π
We saw on page 223 that
T. E.
P =
emax · imax · COS $ ["s
ф
π
S
sin² a. da =
2
0
1
•
π
>
sin² a. da.
wherefore
emax. max. cos.
πT
2
16
242
Electrical Engineering
This may be written in the following form:
P
emax imax
√2√2
cos p.
Since the maximum values divided by √2 are equal to the R.M.S. or
effective values, we have
P=e.i.cos o
.(123).
The law that the power is equal to the product of current and pressure
is therefore only applicable to instantaneous values and not to effective
values. In any case where there is difference of phase, the true power is
found by multiplying the apparent power by the cosine of the angle of phase
difference. For this reason cos is generally called the power-factor. This
designation is the more suitable since the curves of current and pressure
depart in practice more or less from the sine form and rob cos o of the exact
meaning which we have attached to it in the case of pure sine-curves. The
power-factor is then generally defined as the ratio of the actual power, as
indicated on a wattmeter, to the apparent power, as calculated from the
current and pressure. Thus
cos
P
e.v
.(124).
In the example considered in Section 75, for example, the terminal pres-
sure e of the generator was 2,000 volts, while the current i was 50 amperes.
The apparent power in this case is 2,000. 50 = 100,000 watts. Since, how-
ever, the power-factor was given as cos = 0·8, the true power is only 80 per
cent. of this, or
P = e.i.cos &= 100,000. 0·8 = 80,000 watts.
The power transmitted to the generator from the steam engine corre-
sponds to the load of 80,000 watts only, so that no direct loss is entailed by
this reduction of output from 100,000 to 80,000 watts. On the other hand,
however, both the operation of the generator and the efficiency of the whole
system are adversely affected by the phase displacement or lag of the current.
The output of 80,000 watts is only maintained by a disproportionately great
E.M.F., which has to overcome the back E.M.F. of self-induction, in addition to
the resistance. The generator must therefore be capable of working at a
higher pressure, a part only of which is applicable to the external load.
This is perhaps more evident if we imagine the whole load to consist of
resistance and self-induction. We have then
P=.R,
which may be written in the form
Pi. R.i.
It is evident from Fig. 229 that i. R may be replaced by e. cos o,
giving us
P = e. cos.i.
This shows us clearly that, in calculating the actual power or output, we
must only consider the component e. cos & of the total pressure e, that is, the
component in phase with the current.
Effect of phase difference on A. C. power
243
There is still another point of view from which we may consider this
equation. We may look upon the power Pe.i. cos p as the product of
the pressure e and the component i. cos &
of the total current. This component is
equal to the projection OC (Fig. 240) of the
current vector i upon the pressure vector e.
Hence, the true power is found by multiply-
ing the pressure by the component of the
current in phase with it.

Es
i-cosy
Fig. 240.
VD
This component OC=i.cos & is called
the watt-component of the current, or the
energy-component, while the component OD
at right angles to it is called the wattless
current. Since a generating station works
at a constant terminal pressure, this last
method of imagining the current to be divided into two components is the
more convenient, and is the one generally adopted, although it would be
more in accordance with the real nature of the case to resolve the pressure.
We saw above that the lag of the current behind the pressure had the
effect of unduly increasing the necessary E.M.F. for a given load and current.

Lo
www
Fig. 241.
We might now express this in another way and say that the current must be
very large for a given load and pressure, since only a component of the
current enters into the calculation of the output. This increased current
necessitates a greater cross-section of copper in generators, cables and motors,
or else increased losses due to heating. Every effort is made therefore to
reduce the self-induction of generators and motors to a minimum.
The effect of phase difference on the power or output of a generator is
clearly illustrated by the following interesting experiment (Fig. 241). A
variable non-inductive resistance is connected in series with a highly induc-
tive coil across the terminals of an A.C. generator. The terminal P.D. of the
generator is maintained constant. The output of the generator is measured
by means of a wattmeter. When the rheostat handle is over to the right and
the resistance consequently large, the current and therefore also the power
will be small. As the rheostat handle is turned to the left, the resistance is
decreased and the power increases as one would expect. Beyond a certain
16-2
244
Electrical Engineering
1
point, however, the wattmeter reading decreases in spite of the increased
current caused by cutting out the resistance. The reason for this is evident,
if we turn to Fig. 229 on page 235, in which the area of the triangle is equal
to half the product of the base and height. The area is therefore equal to
te.coso.wL.i. Since, in our case, wL is constant, the area is proportional
to e.i.coso, and is therefore a measure of the output of the generator.
With a constant hypotenuse e, the triangle has a maximum area when the
other two sides are equal, that is, when the resistance R is equal to the
reactance wL, and the angle of lag is 45°. The output is then a maximum
for the given terminal pressure and inductance.
If the rheostat arm be put right over to the left, the current naturally
increases still further, while the wattmeter reading falls almost to zero. This
is explained by the fact that the resistance of the circuit is almost entirely
inductive, so that the angle of lag, as found from the equation
tan =
=
wL
R

e
i.n
B
A
Fig. 242.
is almost 90°. The power-factor cos & is therefore nearly 0, and the output
consequently very small in spite of the large current. The latter is almost
entirely wattless.
In order to show this in a graphical manner, we have drawn in Fig. 242
the sine curves for a circuit containing much self-induction, but little ohmic
resistance. The curve of ohmic pressure drop is so flat that it could almost
be neglected in finding the curve of terminal pressure. The latter has little
to do beyond counterbalancing the back E. M.F. of self-induction. The curve e
is consequently nearly 180° out of phase with the curve E, of self-induction,
and therefore almost 90° ahead of the current. By forming the products of
the momentary values of pressure and current, we find that the power is
positive from 0 to A, and negative from A to B. Since these two intervals.
are almost equal, the resultant mean power is very small. It would vanish
entirely if the winding of the coil had no resistance and the iron were free
from hysteresis.
Effect of phase difference on A. C. power
245
It follows from these considerations that a choking coil can be used to
cause a large drop of pressure without absorbing very much power. They
have been largely employed as steadying resistances in series with A.C. arc
lamps. Their employment is limited, however, owing to the disadvantage,
pointed out above, of loading the station and mains with wattless current.
If the pressure has to be reduced considerably, it is far preferable to use a
small transformer instead of a choking coil.
A further use to which choking coils have been put is in connection with
the series running of a number of glow lamps on high pressure mains. Each
lamp has a small choking coil connected in parallel with it, through which
the current passes when the lamp breaks, thus enabling the other lamps in
series with it to continue burning. If the broken lamp were short-circuited
the current through the whole circuit would be increased and the other
lamps endangered. This does not happen if choking coils are used instead of
short-circuiting devices. When the lamp is burning, the current divides
between the lamp and the coil, but on the lamp breaking, the whole current
passes through the coil, with the result that the total current is slightly
reduced. This reduction of current is hardly appreciable when the number
of lamps in series is very large, and it is further limited by making the
choking coils so as to have a variable coefficient of self-induction. When the
lamp breaks, the whole current is forced to pass through the coil and the
flux is thereby increased. This increase in the magnetic flux causes an
increase in the back E.M.F. and consequently a decrease in the current. If,
however, the iron core of the choking coil is already highly saturated, a large
increase in the current has but a small effect on the flux and therefore also
on the back E. M. F.
Choking coils are sometimes used in connection with the parallel running
of alternators. We saw in Section 35 that self-induction acts as a sort of
electrical inertia, opposing sudden variations of the current. There is often
a danger of heavy currents surging backwards and forwards between two or
more alternators when connected in parallel (see Sections 106 and 108).
These currents can be largely prevented by connecting choking coils between
the generators and the common bus-bars at the switch-board. The ad-
vantage gained in this way will more than compensate for the small pressure
drop in the coils. The power lost in such a coil is very small, as the current
is nearly 90° out of phase with the terminal pressure of the coil.
78. Effect of capacity.
A condenser consists essentially of two metal plates separated from each
other by a thin sheet of insulating material. If the two plates are connected
to the terminals of a battery or dynamo, the one plate becomes charged
positively and the other negatively. Although this way of looking at the
matter agrees with the general ideas of frictional electricity, it is hardly in
accordance with the supposition of Maxwell that the electric current is due
to the motion of positive electricity only. The two ideas can be brought into
agreement, however, by assuming that the negative plate is charged by the
246
Electrical Engineering
flowing away of positive electricity instead of the flowing into it of negative
electricity. We thus have a current round the circuit at the moment of
charging; this current flows from the positive terminal of the generator, to
the condenser, and from the negative terminal of the condenser back to the
generator. This current flows until the condenser is fully charged, that is,
until the back E. M.F. of the condenser is equal to the P.D. applied to it. A
certain quantity of electricity will have to flow into the positive plate (and
out from the negative plate) in order to charge the condenser to the given
back electromotive force E..
The effect of capacity in A.C. circuits is of great practical interest on
account of electric cables acting as condensers. To simplify the problem as
much as possible, we shall assume that the cable has neither ohmic resistance
nor self-induction. For the instantaneous value of the terminal pressure of
the machine we have
e = emax • sin a= ¤max ⋅ sin (wt).
If this pressure increases by an amount de in the succeeding interval dt,
we have
de
= Cmax • COS (wt). w .dt.
The quantity of electricity dQ, which passes into the positive plate of the
condenser during this interval is proportional to the increase of pressure de
and to the capacity of the condenser (see page 101). The capacity K must
be expressed in units corresponding to the ampere and volt, i.e. in farads.
We have then
dQe
dt
dQ.=K.de = K. emax. cos (wt). w.dt.
Now, is the rate at which electricity flows into the condenser in
coulombs per second, i.e. in amperes, during the short interval dt. It is the
instantaneous value i of the condenser current, and can be expressed as
follows:
dQe
dt
K.w.emax.cos a.
Hence, the current is a cosine function of the angle a which the terminal
pressure vector makes with its zero position. When a=0, cos a = 1, and the
current has its maximum value
imax = K. w. emax•
By dividing both sides by
which is therefore given by the
2 we obtain the effective value of the current,
equation
i = K.w.e
.(125).
Since the terminal pressure of the machine and the back E. M.F. of the
condenser are exactly equal and opposite to each other at every moment,
they must be numerically equal, so that we may write the above equation in
the following form:
2
Ec:
K.w
i
2π.~.K
.(126).
We see then that there is another pressure in the circuit in addition to
the generator pressure, viz. the back E. M. F. of the condenser, which has to be
Effect of capacity
247
overcome by the generator. We must now determine the phase of this new
electromotive force.
The thin full-line curve in Fig. 243 represents the terminal pressure of
the generator. The abscissae represent the angular displacement of the
vector e from its initial position. On the assumption that the circuit
contains neither resistance nor self-induction, we saw above that the current
is proportional to the cosine of the angle a. It will therefore have a positive
maximum when a = 0 and will vanish when a = 90°. In this way we obtain
the heavy line curve i for the current. Since we have seen that the back
E.M.F. of the condenser is directly opposed to the terminal pressure, the
dotted curve E, must represent the former. The current curve i lags 90°
behind the back-pressure curve Ec, but is 90° ahead of the terminal pressure
curve e.
It is necessary, however, to check this mathematical result by considering
the relation between current and pressure at several characteristic moments
during the period. The curve e cuts the axis at the point O and is steepest

e
1
A
B
Ec
Fig. 243.
at this point. This is the moment at which the terminal pressure is changing
most rapidly, and at which the quantity of electricity passing into the con-
denser per second has its maximum value. The current has therefore its
maximum value. So long as the terminal pressure continues to increase,
that is, during the interval OA, the current will have the same direction as
the pressure, and the ordinates of both are positive.
When the terminal pressure reaches its maximum value at the moment A,
it is constant for a moment and the current into the condenser ceases. When,
however, the terminal pressure decreases, it is overcome by the back E.M. F.
of the fully charged condenser, and a current is driven back through the
generator in opposition to its terminal pressure. This continues during the
quarter period AB, in which the ordinates of current and terminal pressure
are of opposite sign. Hence, the relative positions of the curves in Fig. 243
are quite correct, and we see clearly why the current curve is 90° ahead of
the curve of terminal pressure, or why the back E. M. F. of the condenser is 90°
ahead of the current.
248
Electrical Engineering
79. Resistance and capacity in series.
We assumed in the last section that no ohmic resistance was interposed
between the generator and the condenser, so that the terminal pressure of
the condenser was identical with that of the generator. If, however, a resist-
ance R lies between the generator and the condenser, the terminal pressure
e of the generator has to cover the drop of pressure i. R, in addition to over-
coming the back E.M.F. of the condenser. These two components are shown
in Fig. 244, in which the thick black line represents the pressure drop i. R,
while the dotted line represents the back E. M.F. of the condenser leading 90°
ahead of i. R.
At the moment O there is no back E. M.F., and the terminal pressure e has
only to supply the drop in the resistance. Its momentary value will there-
fore be OF.
F
G

A
in
B
K
A
Fig. 244.
At the moment A there is still a considerable ohmic drop and, in addition
thereto, a back E.M.F. to overcome. The momentary value of the terminal
pressure must therefore be GA.
At the moment B the current is passing through zero and there is
consequently no ohmic drop; the terminal pressure HB has only to counter-
balance the equal and opposite back E. M.F. of the condenser.
Finally, at the moment C the ohmic pressure drop is exactly covered by
the back E.M.F. of the condenser, which is now acting with the current. The
terminal pressure of the generator is therefore equal to 0.
When the curve e has been drawn in this manner, we find that it lags
behind the current curve. In Fig. 244 the current is ahead of the terminal
pressure by an angle represented by BC. The effect of capacity is thus the
opposite of that of self-induction, in that it causes the current to lead ahead
of the pressure, whereas self-induction causes the current to lag.
Circuit containing resistance, inductance and capacity 249
ᎥᎡ
- Es
Ec
80. Circuit containing resistance, inductance and capacity.
We must now consider the case in which ohmic resistance, self-induction
and capacity are all connected in series. At the moment represented in
Fig. 245 the current i is passing through its zero
value and the vector i. R is therefore horizontal.
The vector E, of the back E.M. F. due to self-induc-
tion lags 90° behind the current vector and is there-
fore vertically downwards. This vector must be
counterbalanced by an equal and opposite component
-E, of the terminal pressure of the generator.
From i. R and - E, we get the resultant OA. There B
is another E.M.F. acting in the circuit, however, viz.
the back E.M.F. E of the condenser, which is 90°
ahead of the current and is therefore represented
by a vector vertically upwards from 0. This must
be counterbalanced by an equal and opposite compo-
nent of the terminal pressure. The resultant OC
of the vectors OA and E, is the required vector
of the terminal pressure e.
Under certain condi-
tions the terminal pressure e may be much smaller than either the pressure
E, due to the self-induction, or the pressure E, across the condenser.

8
C
e
Fig. 245.
ож-
Es
-Ec
By means of the diagram in Fig. 245 we are now able to calculate the
current for any given values of the terminal pressure, ohmic resistance, self-
induction and capacity. From the figure, we have
BC= E¿- E.,
so that, for the right-angled triangle OBC, we have
e² = (i. R)² + (Ec — E。)².
From equation (126) on page 246, we have
and from page 229
i
Ec
@K'
8
E.=wL.i.
By substituting these values in the above equation, we get
i
2
e² = (i . R)² + („'/− wL. i)*,
(a
wK
which is equivalent to
e
1
R2 +
@K
wI)°
2
..(127).
This may be called the general expression of Ohm's law for an alternating
current circuit, since the denominator is the apparent resistance or impedance
of the circuit. It is represented by the hypotenuse of a right-angled triangle,
one side of which is equal to the ohmic resistance, while the other is equal to
the difference
1
ω Κ
wL. We must not forget that K is to be expressed in
250
Electrical Engineering
:
farads, so that the value of the capacity in microfarads must be multiplied
by 10-* before it is introduced into the above equation.
It is a point of great importance in the above equation that the angular
velocity w occurs in the numerator of one of the two terms within the
brackets, and in the denominator of the other. This indicates that there is
a value of the frequency for which the two terms will be equal and their
difference consequently vanish. To find this frequency we simply put
or
1
@K
@L=0,
1
W
.(128).
NK.L
1
The frequency at which the difference
@L vanishes is therefore
wK
ய
1
.(129).
2π 2π√K. L
For this value of the frequency, equation (127), which we have called the
general expression of Ohm's law, becomes simplified to the following form:
e
R
The current reaches its maximum value at this frequency, and is as great
as if both self-induction and capacity were absent, and the pressure e were
applied directly to the resistance R.
Let us consider as an example an alternating current generator with an
E.M.F. of 10,000 volts and a coefficient of self-induction L of 0.4 henry. We
shall assume that it is connected to a length of cable, the far end of which is
open, and which has a total resistance of 5 ohms for the double length.
The two cores of the cable act as the plates of a condenser, and have a
capacity of 2.5 microfarads. A condenser with this capacity may be assumed
to replace approximately the distributed capacity of the cable, if connected
between the cores half-way between the generator and the free end of the
cable. The resistance in series with this condenser will be half that of
the whole cable, viz. 2.5 ohms. On this assumption the capacity current
at a frequency of 50 can be found from equation (127):
10,000
i
2.52+
(6.28
106
6.28 .50 . 2·5
8.7.
2
6.28.50.0.4
.0·4)
This current flows in the cable in spite of the fact that it is open at
the further end. The current leads nearly 90° ahead of the pressure, so that
cos & is practically equal to 0 and the power is negligibly small.
It is interesting to determine the frequency at which the current in this
example would reach its maximum value. From equation (129) we find the
critical frequency, as follows:
1
1
2π√K.L 2π√2·5.10.04
160.
Circuit containing resistance, inductance and capacity 251
At this frequency the denominator in equation (127) becomes equal to
R and, although the further end of the cable is open, the current reaches
an enormous value, viz.
E 10,000
= 4,000 amperes,
R 2.5
The P.D. across the condenser is then equal to that across the self-
induction, and for both we find the following incredible value:
i
wK
=wL.i= 2π. 160. 04. 4,000 = 1.62. 106 volts.
The cross-section of the conductor is naturally unable to stand the large
current, and the insulation would be broken down by the enormous pressure.
This dangerous condition which arises at the critical frequency is known as
resonance. The above example is exaggerated on account of the value
chosen for the self-induction of the generator being much larger than the
actual value for a practical alternating current generator. For smaller values
of L the critical frequency is higher than 160. The two characteristic
peculiarities of resonance are large currents and enormous pressures, in
spite of open circuits and normally excited machines.
81. Self-induction and capacity in parallel.
In Fig. 246 is shown a circuit containing two parallel branches, one of
which consists of a choking coil in series with ohmic resistance, while the

K
R₂

2
L
e
R₁
P2
e
Fig. 246.
B
Fig. 247.
other contains a condenser, also in series with ohmic resistance. The terminal
pressure e is common to both branches. Its vector must consequently be
the hypotenuse of two right-angled triangles, one of which has sides equal
to i̟. R₁ and L., while the sides of the other are equal to 2. R₂ and ω Κ
(Fig. 247). The current & must lag behind the pressure e by an angle $1,
which is given by the equation
iz
wL₁
tan 41
R
252
Electrical Engineering
{
The current 2, on the other hand, leads ahead of the terminal pressure
by the angle 2, where
Ec
1
tan 2 i2. R₂ wK. R₂'
=
A circle is drawn on OG e as diameter, and a chord OA drawn so as
to make an angle with the diameter. OA represents the drop in the
lower branch due to ohmic resistance, while AG is equal to the inductive
drop. The chord OB is drawn so as to make an angle 2 with OG. OB
represents the ohmic pressure drop in the upper branch, and BG represents
the pressure lost in counterbalancing the back E. M. F. of the condenser. The
angles 41 and 42 must naturally lie on opposite sides of the diameter. By
dividing the chord OA by the resistance R₁, we get
OA
i
OC.
R₁
The current & is obtained in a similar manner by dividing the chord OB by
the resistance R₂, thus
Г2
OB _OD.
R₂
&₁
The resultant OJ is found from the components = OC and i₂ = OD by
the parallelogram of forces. The vector OJ represents the resultant current
both in magnitude and phase. It is noteworthy that its displacement from
the terminal pressure vector is very small. It is evident that by a suitable
choice of capacity the displacement between resultant current and terminal
pressure could be altogether prevented. Unfortunately, however, condensers
are not robust enough to permit of their employment on a large scale for
this very desirable purpose.
purpose. In Section 111, however, it will be shown that
an over-excited synchronous motor, the back E.M.F. of which exceeds the
terminal pressure, resembles a condenser, in that it takes a leading current
from the mains. Such motors have therefore been used to reduce the dis-
placement between the current and pressure vectors caused by self-induction.
CHAPTER XI.
82. The electromotive forces induced in a transformer.-83. The magnetising current. -84. The
hysteresis current.-85. Transformer on non-inductive load.-86. Transformer on induc-
tive load.-87. The effect of magnetic leakage.
82. The electromotive forces induced in a transformer.
The principle of the transformer has already been explained in Section 34.
The magnetic flux produced by the alternating current alternately grows and
dies away, first in the one direction and then in the other. In so doing,
it cuts through the turns of two separate windings on the core, and induces
in them electromotive forces proportional to the number of turns in each.
We shall consider, in the first place, the conditions existing in the primary
winding when the transformer is on open circuit, that is, when no current
is taken from the secondary winding. The primary winding is then nothing
more than a choking coil, in which the flux produced by the alternating
current induces a back E.M.F. of self-induction. We shall not, however, make
use of the idea of self-induction, but shall speak of the induced or back
electromotive force. If N be the highest value reached by the flux linking
the winding, ~ the frequency and S, the number of turns in the primary
winding, then the primary E. M. F. can be found from equation (120) on p. 230,
as follows:
E₁ = 4·44 . N . ~ . §₁ . 10–³
.(130).
The external P.D. applied to the primary terminals has, as we have already
seen in the fundamental diagram on page 235, a double function to perform,
viz. to overcome this back E. M. F. and to drive the current through the ohmic
resistance of the winding. The resistance of transformer windings is generally
so small that the latter component of the terminal pressure can be neglected.
The induced electromotive force E, is therefore almost exactly equal and
opposite to the terminal pressure e₁, and only allows that current to flow
through the winding, which is necessary to produce the magnetic flux. Since
the magnetic circuit of a modern transformer consists entirely of iron and
has a low reluctance, a very small current suffices to produce the necessary
flux, so that the no-load current is always very small.
Whether the secondary circuit be open or closed, the secondary winding
is cut by the alternating flux, which induces in it an E.M.F. E, in phase with
the primary E.M.F. If the number of turns in the secondary winding be S2:
we have, from equation (120) on page 230,
2
E₂ = 4·44 . N . ~ . S½. 10−ª .
……..(131).
254
Electrical Engineering
The electromotive forces are thus proportional to the number of turns
in the windings. At no-load there is no secondary current, and the electro-
motive force E, is equal to the secondary terminal pressure e2. Neglecting
the primary copper drop, we have
2
e2 S₂
e S
..(132).
Hence, at no-load the terminal pressures are proportional to the number
of turns in the windings.
83. The magnetising current.
Since the unloaded transformer is equivalent to a choking coil, the
magnetising current is given by equation (119) on page 230, viz.
Bmax. 1
iμ 1.78. S₁.μ
Bmax is the highest value reached by the number of lines per sq. cm.,
l is the length of the magnetic path, and µ is the permeability of the iron.
The value of µ is not quite constant during the period, but since, in order
to reduce hysteresis loss and magnetising current, transformers are always
worked at low flux densities, i.e. on the first part of the magnetisation curve,
the value of μ varies very little.
Instead of taking the value of μ from magnetisation curves found by the
ballistic method, we can determine it by actually measuring the magnetising
current taken by a transformer or annular ring built up of the same iron.
The maximum flux N can be found by equation (130) on p. 253, and the flux
density Bmax is obtained by dividing N by the cross-section of the iron. The
permeability can then be found from the following formula:
μ
Bmax. 1
1·78.in. S₁
μ
.(133).
The value of μ found in this way is, of course, the correct one to employ
in the transformer calculation. It is not, however, exactly the same as
the value of μ corresponding to Bmax in the magnetisation curve, but is
a sort of mean value which can hardly be defined, except by means of
equation (133).
The curve μ in Fig. 248 has been determined in this manner. It is
taken from Kapp's "Dynamo Construction," and gives what we may call the
mean permeability for the values of Bmax shown on the abscissa axis. As
one would expect from the shape of the magnetisation curve in Fig. 56
on p. 62, the permeability increases up to a certain value of Bmax beyond
which it decreases. The highest permeability has hardly been reached in
the figure, which only goes up to a flux density of 7,000. It is easy to see
from Fig. 56 that the permeability B/H would reach its maximum at the
point where the tangent to the curve would pass through the origin.
We shall now determine the phase of the various pressures and no-load
current by means of a vector diagram (Fig. 249). The induced electromotive
forces lag 90° behind the magnetising current i. If the vector i, is drawn
μ
The magnetising current
255
1
horizontally to the left, the vectors representing E, and E, must be drawn
vertically downwards. Since the electromotive forces are proportional to the
number of turns, we have
S₁
E₁ = E₂ · S₂
S₁
Hence, the vector of the primary electromotive force represents to a
certain scale the secondary electromotive force. This greatly simplifies the
diagram. All secondary electromotive forces are therefore multiplied by S₂
to reduce them to equivalent primary forces, which are then drawn on the
diagram.
Watt
Kilo


1,5
Q,5
રે
50
Watt pro Kilo
Mo
3000
2000
1000
ίμ
る
​جهود شد
1000
2000
3000
4000
5000
6000
7000
Fig. 248.
Fig. 249.
We have seen that the primary terminal pressure e, is almost exactly
equal and opposite to the primary electromotive force E. The e, vector will
therefore be vertically upwards. The angle between the terminal pressure e
and the primary current i is 90°, so that no power is taken by the open-
circuited transformer. We have, however, not only neglected the power lost
in heating the copper, which is very small, but we have also neglected the
power absorbed in continually reversing the magnetic field in the iron. This
latter loss is by no means negligible, and we must now investigate this effect
of the so-called hysteresis of the iron.
84. The hysteresis current.
If the power supplied to a transformer on open circuit be measured by
means of a wattmeter, it is found to be by no means negligible. This power
is principally absorbed by the hysteresis of the iron. If Ph is the power
measured on the wattmeter, the primary current and e, the primary
terminal pressure, we have
Ph
cos o
•
e1. 20
The primary no-load current lags therefore behind the terminal pressure
by an angle 4, while the primary back E.M.F. is, as before, exactly equal and
256
Electrical Engineering
t
opposite to e. Now the back E. M. F. must lag 90° behind the magnetic flux,
by the variation of which it is produced. When, for example, the E.M.F.
is a maximum, as shown in the figure, the magnetic flux
must be passing through its zero value, since this is the
moment when the steepness of the sine curve or rate of
change of the flux is at its maximum. The vector of the
flux density B must therefore be drawn horizontally to the
left. Hence, we are led to the conclusion that the magnetic
flux density B lags behind the no-load primary current i, by
an angle B.
In order to make this clearer, a hysteresis loop is drawn
in Fig. 251, in which the instantaneous values of the no-load
current are plotted as abscissae and the values of the flux
density B as ordinates. If the applied terminal pressure e
is a sine function of the time, the induced E. M. F. must also
Fig. 250.
be a sine function. The curve representing the flux density is drawn on
the right, and since it is merely displaced 90° from the induced E. M. F., it
is also a sine curve. Its maximum value is the same as that of the hysteresis
loop on the left of the figure.


E
AB C
FR-
R
0'
B
→t
Fig. 251.
A point is found in the sine curve having an ordinate equal to the
remanent magnetism OR. This point corresponds to the moment O' at
which the current is consequently zero.
At the moment A, on the other hand, the flux density has fallen to zero,
and we must measure off the corresponding current OA from the loop and
set it up as the ordinate at the point A.
At the moment B, the flux density is seen from the right-hand figure
to be EB, which corresponds on the loop to a current OB. The latter is
therefore set up as the ordinate at the point B.
The hysteresis current
257
Both the flux density and the current reach their maximum values at
the moment C. The point D at which the curve cuts the base line is
determined by finding the point for which the ordinate DG is equal to the
remanent flux density OG in the hysteresis loop.
It is evident from Fig. 251 that the current is out of phase with the
magnetic flux. The current curve, moreover, is distorted, so that, although
both curves reach their maximum values at the same moment, they pass
through their zero values at different times. We are met with the difficulty
that our rotating vectors represent true sine waves, so that we cannot exactly
represent the no-load current by a vector. For practical purposes, however,
the distorted current wave can be replaced by an equivalent sine wave, i.e.
one with the same effective value . The flux wave will lag behind this
sine wave of current by an angle 4, which, however, will be less than the
angle represented by O'A in Fig. 251.
In Fig. 252 the current i, is resolved into two components, one, ir, in
the direction of the terminal pressure e, the other, i, in
the direction of the induction. The former is the hyste-
resis current, while the latter is the wattless magnetising

current.
We see from the figure that
is not neglected, to E. i. cos p.
in
io
Now the power absorbed at no-load by hysteresis,
viz. Ph, is equal to e. i. cos o, or, if the copper drop
We have then
in = i, . cos p.
iμ
-ix
Fig. 252.
Ph
in
E₁
or
iµ = √i² — in
Ph=E₁.io.cos o,
We then find i from the equation
2
.(134).
(135).
The physical meaning of this is made plainer, perhaps, by drawing the
vector - in in Fig. 252 equal and opposite to in. Then the theoretical
magnetising current i is the resultant of the total current i, and the
current in, which may be called the demagnetising effect of hysteresis.
Hence, we may consider the current i, to be made up of two components,
viz. the true magnetising current i, and the component in, which counteracts
the demagnetising effect due to hysteresis. If hysteresis could be eliminated
the no-load current would be identical with the wattless magnetising current
, although the curve would still be distorted from the sine form owing
to varying permeability.
We shall now apply the above results to an actual example. The follow-
ing data apply to a 36-kilowatt transformer, described by Kapp:
Cross-section of iron
Length of path in iron
Number of primary turns
Frequency....
A = 900 sq. cms.
1 = 100 cms.
S₁ = 315,
= 50,
~=
T. E.
17
258
Electrical Engineering
Primary terminal pressure ...
Current at no-load
Power taken at no-load
Weight of iron....
e₁ = 2,000 volts,
¿。 = 0·36 ampere,
Pr=400 watts,
= 765 kilogrammes.
We are required to determine the permeability μ of the iron. From
equation (134) we have
Ph
400
0.2 ampere,
in
ራ
2,000
and from equation (135), for the magnetising current,
2
2
iµ = √i ² — in² √/0·36² — 0·2² = 0·3 ampere.
Before we can determine the permeability we must calculate the maxi-
mum flux density. This can be obtained by means of equation (130) on
page 253, thus
N
Bmax
e1. 108
A 4·44.~. S₁. A
= 3,180.
The permeability can now be found from equation (133) on page 254,
με
Bmax.l 3,180.100
1.78.i.S 1.78.03.315
=
= 1,900.
This is approximately the value given by Fig. 248 for a flux density of
3,180. The power absorbed per kilog. at a frequency of 50 works out to
400
765
= 0.52 watt. The curve of iron loss per kilogramme in Fig. 248 was
experimentally determined in this manner. This curve gives a somewhat
lower value than that just found, viz. 0·42 watt per kilogramme at 50 cycles
per second. When accurate curves are available, giving the iron loss and
permeability for the material to be used, the losses and no-load current of a
transformer can be accurately calculated. This is done by simply working
backwards through the calculation by which the curves were obtained from
the experiment. It should be borne in mind that the experimentally deter-
mined iron losses are not entirely due to hysteresis, but contain the eddy
current losses in the iron.
85. Transformer on non-inductive load.
To anyone who has clear ideas of self-induction and of the action of a
choking coil, the behaviour of a transformer on open circuit, as dealt with in
Section 82, will have presented no difficulties. A clear understanding of the
action of a loaded transformer is, however, much more difficult to obtain. It
is not easy to see that any current we choose can be taken from the secondary
winding, and that the primary current will automatically adjust itself to
a corresponding value. Whereas on open circuit the back E.M.F. prevented
all but a very small current from flowing through the primary winding, this
current grows immediately on current being taken from the secondary
winding. We naturally ask the reason for this sudden increase of primary
current, or apparent decrease of self-induction of the transformer. The
f
Transformer on non-inductive load
259
difficulty is due, to some extent, to the lack of visible connection between
the two windings which are so completely insulated from each other.
The relation between the primary and secondary currents can be easily
determined from a consideration of the conservation of energy. The supply of
energy to the primary must be equal to the sum of the energy given out from
the secondary and the losses within the transformer. If R, be the resistance
of the primary winding and R₂ that of the secondary, we can write down the
following equation for non-inductive loads, neglecting the small displacement
caused by the wattless magnetising current iμ:
2
2
e₁. i₁ = e₂. i₂ + i₂². R₁ + i₂². R₂+ Ph.
e1.i
Now, the losses are very small, even when compared with a fraction of the
full load. We shall therefore neglect them, and write
or
e₁. i₁ = e₂. 12,
ez Sz
Նշ
C1
S₁
.(136).
Hence, the currents are inversely proportional to the pressures or inversely
proportional to the number of turns in the respective windings.
Now, although we have obtained this result from a consideration of the
principle of the conservation of energy, which we know from experience can
be applied with perfect confidence, yet the difficulty mentioned above will
not have been removed. It will remain with us until we study the magnetic
effect of each winding and draw a diagram showing the magnetomotive
forces.
With a non-inductive load, the secondary current is in phase with the
induced E. M. F. The primary current, on the other hand, is practically in
phase with the primary terminal pressure, which is almost exactly opposed
to the induced E. M. F. It is therefore obvious that the primary and secondary
currents are nearly 180° out of phase, and consequently neutralise each other
in their magnetic effects. The apparent decrease in the self-induction as the
load is increased can therefore be explained by the neutralisation of the
larger part of the primary ampere-turns by the opposing ampere-turns of
the secondary.
We have already seen that, if we neglect the ohmic drop in the primary
winding, the electromotive force E, must be numerically equal to the con-
stant primary terminal pressure. For the maximum value of the total flux
we have, then,
N
e1.108
4·44 . ~ . S₁ °
Since everything on the right-hand side of the equation is constant, the
maximum value of the flux must also be constant, quite irrespective of the
load on the transformer. If an unloaded transformer be suddenly loaded,
the sudden rush of secondary current exerts a powerful demagnetising effect.
The magnetic flux and back E.M.F. are momentarily decreased, but the
primary current instantly rises to such a value that the demagnetising
effect is neutralised, and the magnetic flux maintained at its no-load value.
In other words, the combined magnetic effect of the primary ampere-turns
17-2
260
Electrical Engineering
S₁ and the secondary ampere-turns iS, is equal to the effect of the no-load
ampere-turns iS. The vector iS, in the diagram is the resultant of the
vectors S₁ and i2S2.
Before making the diagram, however, we divide all the ampere-turns by
the number of primary turns S. The vectors then represent currents
K
B

D
E'
А
in
A
in

C
ሃ
B
B
iu
' ἐμ
༥༠༥༩
E,
E,
1527=175
Fig. 253.
Fig. 254.
reduced to the primary winding. The diagram of currents in Fig. 253 is
obtained in this way, and the vector i。 is the resultant of the vectors & and
iz S2
S₁
(In several of the figures the Greek § is used instead of S.)
To construct the diagram we must first draw the triangle OAB of no-load
current from the three vectors io, in and ip. If the vector i is horizontal, the
S₁
vector of electromotive force E₁ = E2.
S.
will be vertically downwards, since
the induced E.M.F. lags 90° behind the magnetic flux. In Fig. 253
S₁
OE = E₁ = E₂ · S₂
Transformer on non-inductive load
261
Since the load is non-inductive, the secondary current vector must be in
the same direction as the induced E. M. F. As the diagram is derived from
the ampere-turns, the actual secondary current must be reduced to an equi-
valent current in the primary winding, so that
OC=
iz. Sz
S₁
Since the no-load current i, is the resultant of the reduced secondary
current and the primary current, the latter is easily determined by com-
pleting the parallelogram OCAD. We find then that
OD=i.
This primary current vector lags behind the primary terminal pressure
by an angle 1, which is exaggerated in Fig. 253 for the sake of clearness,
but which, in reality, is negligibly small. A transformer on a non-inductive
load is therefore equivalent to a non-inductive load in the primary circuit.
We must now investigate the effect of the primary pressure drop R₁,
which we have so far neglected. The vector OG is drawn along the primary
current vector, and made equal to i̟. R₁ (Fig. 254). OE' is made equal and
opposite to OE. The resultant OH of the vectors OG and OE' represents
the terminal pressure e, both in magnitude and phase.
The correctness of the diagram can be checked by means of the principle
of the conservation of energy. By projecting the vector of terminal pressure
upon the vector of primary current, we get
OK =
= e₁. cos $1,
GK = OK—OG=e̟₁. cos &₁—i₁. R₁,
GH = E₁,
OD = i₁,
S2
S₁
DB = i₂. ~²+ in.
It follows from the similarity of the triangles GKH and DBO that
or, substituting the above values,
DB
OD
GK
GH'
By multiplying across, we get
iz. S₂
S₁
+ ir
e₁. cos p₁ — i̟. R₁
21
E₁
E₁.
i₂. S₂+
i.
₁—¿‚³.
S₁
S₂
Putting E₁.
E₁, this becomes
S₁
2
+ E₁ . in = e̟₁ . i₂ . cos ₁ — ¿¹³. R₁.
1
e. i. cosp₁ = E½. ¿½ + ¿₂². R₂+ E₁ . in
2
==
2.
.(137).
The product E,. i, represents the total secondary power, both internal
and external. The product 3. R, is the power lost as heat in the primary
winding, while E. in represents the iron loss. The total supply of power is
thus equal to the sum of the secondary output and the transformer losses.
262
Electrical Engineering
It is important to notice that the induced back E. M. F. is affected by the
ohmic pressure drop in the primary winding. If the applied primary pressure
e is maintained constant, the magnetic flux decreases as the load comes on.
The no-load test should therefore be made at a pressure equal to the induced
E.M.F. when loaded. The flux will then be the same both on open circuit
and on load, and the no-load triangle OAB will be exactly correct at that
load, but not at any other load. The difference is so small in actual practice,
however, that this refinement may be safely neglected.
We turn now to the efficiency of the transformer on non-inductive load.
If the secondary winding have a resistance of R₂, the secondary terminal
pressure e, will be less than the induced E. M.F. E₂ by an amount i₂. R₂. The
secondary copper loss will be 22. R₂, and the real output e. 2. The efficiency
on non-inductive load is, therefore,
n
ez. in
2
eg. in
e₁. i. cos $1 е2 . ¿½ + i₁² . R₁ + ¿½². R₂+ Ph
..(138).
As this equation shows, the efficiency is very bad on small loads, on
account of the iron loss being relatively so large.
With increasing load the


a
I
· b
- b
D
ん
​I
αν
Fig. 255.
Fig. 256.
iron loss becomes of less relative importance, and as the copper losses are
small compared with the load when the latter is not too large, the efficiency
rises rapidly with the load. It reaches a maximum when the variable copper
losses are together equal to the constant iron loss. If the load be increased
beyond this point, the copper losses, which increase as the square of the
current, become so important that the efficiency decreases.
We pass now to a consideration of the most efficient distribution of losses
between the two windings for a given winding space.
S₁
Since i₂ = S
the total loss in the two windings is given by the following
formula:
P₁ = i² . R₁ + i₂² . R₂ = ¿² (R₁ + R₂ ·
Si
S₁2
S2
If l₁ and l, be the mean lengths of a turn in the two windings, we have
R₁ = p.
4. Sı
A₁
12. S₂
and R₂ = p.
A₂
By substituting these values in the above formula, we get
12
P₁ = "². S². P. (S². A, + S., A.)
•
p
S₂.
..(a),
Transformer on non-inductive load
263
where A, and A, are the cross-sections of the wire with which the transformer
is wound.
With a flat coil winding (Fig. 255) the turns of each winding have the
same mean length, so that
b₁ = l₂ = 1.
Ն
We may assume approximately in Fig. 255 that, with the omission of
a constant factor,
S₁. A₁ = a.h, S₂.A½=(b−a). h.
The combined copper losses are therefore given by the formula
l
Pi
P₁ = b² . B . p. 1. ( + b = a)
1
a).
h
..(b).
To find the minimum value of this expression, we must differentiate it
with regard to a and equate the result to 0. This gives us
1
1
+
a² Т (b − a)²
0,
b
a
or
2°
That is to say, the most efficient distribution of space is an equal distri-
bution. The cross-section of the wire will then be inversely proportional to
the number of turns, and directly proportional to the current. The current
density will be the same in each winding. If we put a =
we see that the losses are the same in each winding.
b.
2
in equation (b),
If the winding consists of cylindrical coils, as shown in Fig. 256, we have
l₂ = (d+α)π,
l=[d+2h-(ha)]. π = (d+h+α) π,
S₁. A₁ = b (h-a), S₂. A₂ = b.a.
Equation (a) will then take the following form,
P₁ =
2
i₂². S₁². p.π [d+h+a, d+a
b
+
(h− a) α
By differentiating with regard to a and equating the result to 0, we get
a² + a.d-
h.d
2
0.
This is the same equation as we obtain if we equate the two terms in the
bracket in the last equation for P. The efficiency for a given winding space
is therefore, as before, a maximum when the copper losses are equally distri-
buted. The current density will not be the same in each winding. These
calculations, however, are only rough approximations, since the relative loss
of space due to insulation will be much greater in the high pressure than in
the low pressure winding. A further inaccuracy is caused by the necessary
space between the two windings in Fig. 256.
As an example, we shall consider a transformer with a primary terminal
pressure of 2,000 volts, taking a current of 20 amperes, and having a ratio of
10 to 1, so that the secondary current is 200 amperes and the secondary
264
Electrical Engineering
terminal pressure at no-load 200 volts. If 3 per cent. of the power is lost in
heating the conductors, the pressure drop due to resistance will be 3 per
cent., and this drop must be equally divided between the primary and
secondary windings. We have then
1.5
i₁. R₁
•
2,000 = 30 volts,
100
1.5
iz. R₂
=
200 = 3 volts.
100
30
Hence
R₁:
= 1.5 ohms,
20
3
R₂
= 0.015 ohm.
200
The error introduced by subtracting the drop from the terminal pressure
algebraically instead of geometrically is negligibly small. We have therefore
E₁ = e̟₁ — i₁. R₁ = 1,970 volts,
E₁ 1,970
E:
= 197 volts,
and
10 10
e₂ = E₂ — i₂. R₂ = 194 volts.
The drop of pressure between no-load and full load is 200-1946 volts,
i.e. 3 per cent.
86. Transformer on inductive load.
We pass now to the consideration of the case where the secondary
terminals of the transformer are connected to a load consisting of both ohmic
and inductive resistance. If R' be the resistance and wL, the reactance of
the external secondary circuit, the phase difference between the secondary
terminal pressure and the secondary current is given by the equation
2
włą
tan 2=R'₂
The difference of phase between the induced E. M.F. E, and the current i
is given by the equation
tan o
=
2
@L₁
R'₂+ R₂
E2. S₁
S₂
which is the secondary
The vector OE in Fig. 257 is made equal to
induced E.M.F. reduced to its primary equivalent. A right-angled triangle is
constructed on the hypotenuse OE, so that the side OM lags behind OE by
the above angle 4. OM is then equal to the secondary pressure used in
overcoming ohmic resistance, the part OL being the internal drop and the
remainder LM the pressure used against the external ohmic resistance.
Since the whole diagram is reduced to the primary turns, we have
S₁
OM= i₂(R'₂+R₂).
S₂'
OL=i₂R₂.
S₁
S₁
Transformer on inductive load
265
Similarly EM represents the equivalent inductive drop in the external
circuit, or
S₁
EM = wL₂. i₂. S₂
The secondary terminal pressure is found by subtracting geometrically
the internal drop OL from the induced E. M. F. OE. We have therefore
S₁
D
H
i
A
LE=2• S.

eq

B
ре
Su
e2 51
23
inĘ z
E
Fig. 257.
G
M
42
ૐ
E
Fig. 258.
M
The figure shows that the terminal pressure is the hypotenuse of the
right-angled triangle formed by the ohmic and inductive pressure drops in
the external circuit.
We must now draw the primary and secondary current vectors. The
secondary current is in phase with the ohmic pressure drop produced by it.
When reduced to the equivalent primary current, we get
i2. S₂
OC=
S₁
The vector OD must now be drawn so that its resultant with OC is the
266
Electrical Engineering
I
no-load current OA. Since the no-load ampere-turns are the resultant of
the primary and secondary ampere-turns, the vector OD must represent the
primary current both in magnitude and phase.
The primary ohmic drop OG must be equal to . R and must be in
phase with the primary current. Equal and opposite to OE we draw OE' to
represent the pressure required to overcome the back E.M.F. The resultant
OH of OG and OE' gives us the applied primary terminal pressure e₁. From
an examination of the diagram we learn the following important facts :
1. The primary and secondary current vectors are now almost exactly
opposite to one another, and the no-load ampere-turns are almost equal to
the algebraic difference between the primary and secondary ampere-turns.
Since the no-load current is actually much smaller than shown in the figure,
S₂
the equation i₁ = i₂.
&
is still very nearly correct.
S₁
2. Since the ohmic drop in the windings is usually very small, the
terminal pressure vectors e, and e, come almost exactly vertical, that is, they
coincide in direction with the induced E.M.F.
3. The angle of lag ₂ in the external secondary circuit is consequently
nearly equal to the angle between the primary terminal pressure and
primary current. Hence, the secondary phase displacement reacts directly
on the primary circuit.
This last result enables us to look upon the transformer as a single piece
of apparatus, and to bridge over, in imagination, the separation of the two
windings. Thus, we speak of the pressure drop and copper loss of the trans-
former as a whole, as if it had but a single winding. The dotted line joining
the points G and E in Fig. 257 is parallel with, and equal to, the primary
terminal pressure e. Moreover, as the ohmic pressure drops OG and OL are
practically in the same straight line, they can be added together to give a
total drop, and we can write
GL=OG + OL=i. R₂+ig. Rg.
·
S₁
S₂
Putting & equal to
¿1. Sı
S₂
we have
S',2
GL = i₁. R₁ + ¿½. R₂ ·
S₁₂
= ₁₂ (R₂+ R₂ ·
·
S₁₂
S₂2
It would appear from this equation that the primary current passed
S₁² 2
through the two resistances R, and R. in series. The latter is the
S₁₂
secondary resistance reduced to its equivalent primary value. After over-
coming this total resistance the initial terminal pressure e, is reduced to the
secondary terminal pressure e. The transformer is somewhat similar to a
generator with an E. M.F. e, an internal resistance R= R₂ + R₂ .
S₁2
and a
>
To make this quite clear, the corresponding dia-
S₁
terminal pressure e·
S₂
gram for the generator is drawn in Fig. 258, in exactly the same position as
it has in the transformer diagram. The electromotive force e, is the hypote-
Transformer on inductive load
267
nuse, the side GM is the total ohmic drop, both internal and external, while
the other side EM is the external inductive drop. The generator must be
assumed to have no internal self-induction.
If, for example, we assume that the transformer considered in the last
section is connected to a secondary circuit with a power-factor of cos &₂ = 0·7,
and that the primary current & is 20 amperes, we can find the primary
terminal pressure which is necessary to maintain a secondary terminal
pressure of 200 volts. We have
S₁2
R₁ + R₂.
S₂2
= 1·5 + 0·015.100 = 3 ohms.
The total internal ohmic drop reduced to the primary winding is therefore
Si
GL = ₁₂ (R₂+ R₂ ·
S.
21
2
60 volts.
is
The drop of pressure in external ohmic resistance, when reduced to the
primary winding, is found as follows:
S₁
S₂
=
LM = e2. cos $2 • =200.07.10 1,400 volts.
The side GM representing total ohmic drop both internal and external is
therefore equal to 1,460 volts. Since cos ₂ = 0·7, sin &, must equal 0·714,
found as follows:
and the inductive drop EM is
or
S₁
EM= e. sin 2
•
$2
=200.0-714. 101,428
200. 0·714.10 = 1,428 volts.
For the hypotenuse EG we have √GM² + EM²,
e₁ = √1,460² + 1,428² = 2,040 volts.
This then is the primary pressure required to give a pressure of 200 volts
on the secondary terminals. This, however, is on the assumption that every
line of force produced by the primary winding passes through the secondary
winding and vice versa, so that the two windings have a common field and
no other. As a matter of fact, there is a partial leakage of lines, which
increases the drop of pressure due to increasing load. In the following
section it will be shown that, with these actual relations, we can still imagine
the two windings to be combined to form, as it were, a single winding.
87. The effect of magnetic leakage.
In Section 82 we compared the back E.M.F. induced in the primary
winding with the E. M. F. of self-induction of a choking coil. Since then,
however, we have purposely avoided the use of the term self-induction and
have always spoken of the induced E. M. F. This was necessary because, in
addition to this useful induced E.M. F., there is, in both windings, a useless
and harmful self-induction due to magnetic leakage.
In order to make this clear, we shall consider a very bad arrangement,
which would never be adopted in practice, viz. a core-type transformer with
the primary winding on one limb and the secondary on the other (Fig. 259).
268
Electrical Engineering
The lines of force produced on no-load pass, for the most part, completely
round the iron circuit and thus thread the secondary winding, but some of
the primary lines pass through the air and constitute the no-load leakage.
The number of lines threading the secondary winding is therefore smaller
than the total number threading the primary winding. The E. M. F. induced
in the secondary winding must consequently be less than that induced in the
primary winding. We may consider that a part of the primary terminal
pressure is used in overcoming the back E.M.F. due to this leakage self-
induction, leaving the remainder to overcome the back E.M.F. E, due to the
flux which links both windings. The latter E. M. F. will therefore be smaller
than the terminal P.D., and consequently the secondary terminal pressure
will be correspondingly less. The effect of magnetic leakage is, however,
quite negligible at no-load, as the reluctance of the iron path is so small,
compared with that of the air, that the leakage flux bears a very small ratio
to the useful flux.
I
BI
I


B₁
I
Fig. 259.
II
Fig. 260.
The effect of leakage increases immensely, however, directly the trans-
former is loaded. We have seen that the secondary current is almost exactly
opposed to the primary current. If, at the moment represented in Fig. 260,
the primary current is flowing inward on the side of the coil which is turned
towards us, then the secondary current will be flowing outward. By applying
the right-handed screw rule, it is easily seen that the two windings are in
opposition with regard to the main flux threading the two windings, but are
acting together with regard to the leakage flux. The primary and secondary
ampere-turns are in opposition for the iron path, but in parallel for the
leakage through the air. This causes a rapid increase in the ratio of the
leakage to the useful flux.
The increase of leakage with load is explained very clearly if we consider
the difference of magnetic potential between the points A and B. At no-load
(Fig. 259) this difference is only sufficient to drive the magnetic flux through
the lower iron core. When loaded (Fig. 260) the difference of magnetic
potential has also to overcome the counter magnetomotive force of the
secondary ampere-turns. This increased difference of magnetic potential
The effect of magnetic leakage
269
naturally drives a larger leakage flux through the air. It is analogous to
forcing water through a leaky pipe; the greater the back-pressure opposing
the passage of the water, the greater will be the quantity of water lost
through the leaks.
I
Fig. 260 cannot represent the actual distribution of flux, since, in the
lower half of the transformer, the leakage lines are shown passing through
the iron in the opposite direction to the useful lines. The actual flux
through any part of the magnetic circuit
must be the resultant of these opposing
fluxes, and the actual distribution will be as
shown in Fig. 261 at any moment when the
primary current exceeds the secondary current.
The actual distribution changes entirely in
character as well as in magnitude during a
period, and since the primary and secondary
currents are not exactly 180° out of phase,
there are moments when there is no primary
current, but yet a small secondary current
which produces a flux distribution the reverse
of that shown in Fig. 259. Hence, we see that the leakage at any moment
and at any point is the resultant effect of the instantaneous values of the
primary and secondary currents.

Fig. 261.
In order to reduce the magnetic leakage as much as possible, the two
windings are put as close together as possible (Figs. 255 and 256), or even
subdivided and alternate primary and secondary coils sandwiched together.
There is, however, always a small drop due to magnetic leakage.
8)
is
To show the effect of leakage on the vector diagram, we shall neglect the
small no-load current, and assume that the primary and secondary currents
are exactly opposed to each other*. The angle by which the secondary
current i, lags behind the induced electromotive force OE ( = E,
chosen arbitrarily in Fig. 262. It depends, as a matter of fact, upon the
relation between the total secondary self-induction and the total secondary
ohmic resistance. The pressure drops due to the ohmic resistance of the
windings are drawn in the direction of the currents, viz.
OG = i₁. R₁,
OL= i₂. R₂ .
S₁
S₂
We know from page 231 that the induced E. M.F. due to the self-induction
of leakage lags 90° behind the current. To overcome that caused by the
secondary current, an equal pressure ON, 90° ahead of the current i, will be
necessary. The total pressure drop in the secondary winding will be equal
to the resultant OR. The line RE will represent the secondary terminal
pressure both in magnitude and phase, so that
S₁
RE=e₂.
S₂
* The exact diagram will be given when considering the induction motor in Section 124.
270
Electrical Engineering
The effects of resistance and self-induction are greatly exaggerated in
Fig. 262. In reality they are much smaller, so that the terminal pressure
differs very little from the induced E.M.F.
i̟.
A component OP will be required in the primary circuit to counter-
balance the back E.M.F. due to the leakage of the primary winding. OP will
be 90° ahead of . The resultant OQ of the primary ohmic and inductive
drops must represent the total primary pressure drop. By compounding OQ
with the pressure OE', which counterbalances the back E.M.F. OE, we get the
applied primary terminal pressure OH=e̟.
We have thus determined the magnitude and phase of all the vectors,
and we can now determine the equivalent resistance and leakage self-induc-

H
&
E'

和
​iz
S
1
R
Fig. 262.
Fig. 263.
tion of a single apparatus to replace, in imagination, the two separate
windings. For this purpose we join QG and QE, and drop perpendiculars
from R and E on to QG produced. The figure thus obtained is given in
Fig. 263 and shows the various pressures with which we are concerned. The
hypotenuse QE is equal to OH, that is, to the primary terminal pressure e₁.
QU represents the sum of the pressure losses due to magnetic leakage, and
RU the sum of the ohmic pressure losses. The total drop of pressure in the
transformer is given by QR, leaving RE for the secondary terminal pressure.
We need hardly mention that the internal pressure drop is in reality much
smaller than indicated in Figs. 262 and 263.
It is of great importance that the triangle QRU can be found experi-
mentally. By short-circuiting the secondary terminals we make RE = 0, so
The effect of magnetic leakage
271
that the point E falls on the point R. The primary pressure must naturally
be chosen so small that the normal current flows in the short-circuited trans-
former. This primary pressure is then directly equal to QR. The resistances
are easily measured and the ohmic pressure drop RU calculated from them.
The triangle QRU can then be drawn and the effect of magnetic leakage
determined.
We can now construct Kapp's transformer diagram for constant current
but variable external power-factor (Fig. 264). The triangle QRU, which we
have just determined for the normal full load current, is drawn in the position
shown. With a radius equal to e, two circles are drawn, one with the centre
Q, and the other with the centre R. For any given power-factor 4, in the
external secondary circuit, RE gives the secondary terminal pressure, or

E
ይላ
12
R
?
V
S
Fig. 264.
rather, its primary equivalent, while QE or RV is equal to the constant e.
The difference between them is EV, which gives the fall of secondary pressure
between open circuit and full load. For any other value of the current, QR
will be different, so that a new diagram must be drawn.
The diagram shows that the fall of pressure between open circuit and
full load increases as the point E moves to the right, that is, as the power-
factor decreases.
If the current leads ahead of the terminal pressure, so that the point E
moves over to the left, the drop of pressure may not only disappear, but may
be turned into a rise, so that the terminal pressure increases with the load.
This may occur if the secondary circuit contains condensers or over-excited
synchronous motors. (See the corresponding diagram in Section 98, also
Sections 79 and 111.)
CHAPTER XII.
88. Types of A. C. generators.-89. The mean E. M. F. of a generator.-90. The effective E. M. F.
with sinusoidal field.-91. The E.M. F. of a single-slot winding.-92. The E. M.F. of a double-
slot winding.-93. The E. M. F. of a treble-slot winding.-94. The E.M.F. of a distributed
winding.-95. The E.M.F. of a closed D. C. winding.-96. The E. M. F. of a creeping bar or
wave winding.-97. The E. M. F. of a creeping coil winding.
88. Types of alternating current generators.
Many very different types of alternating current generators have been
proposed and constructed from time to time, but, as in so many things, there
has been a gradual elimination and standardization, until, to-day, nearly
every A. C. generator is constructed with an external stationary armature and
an internal revolving field magnet system. It is important, however, to study
the various types of machines, and this we shall do, not so much in their
historical order as in the order of their simplicity. We shall deal in a general
way with both single-phase and polyphase machines, but shall reserve the
consideration of the points peculiar to polyphase generators until we come to
the theory of polyphase currents. The single-phase machine has a single

N
Ls
N
L

S
Fig. 265.
N
Fig. 266.
winding on its armature. The side of each coil may be placed in a single
hole or slot, as shown in Fig. 265, or distributed between several slots as
shown in Fig. 266. In both of these cases there is a single coil-side under
each pole. In polyphase machines, on the other hand, the armature carries
two or three distinct windings, which may or may not be connected, but
which are displaced from each other around the armature. For the present,
we may consider these phases or windings to be entirely disconnected and
quite distinct, so that a two-phase generator is nothing more than a machine
with two single-phase windings. In Fig. 267, for example, we consider only
the winding 1 and neglect the winding 2. In this way we shall be able to
consider single-phase and polyphase machines together.
External pole machines were formerly made with ring and with drum
Types of alternating current generators
273
armatures.
In the ring armature shown in Fig. 268 all the coils are wound
in the same direction but are connected up alternately in opposite directions.
By drawing arrows on the ends of the coils to show the direction of the
induced E. M.F., and tracing the circuit from coil to coil, we find that all the
induced electromotive forces are added together.

N
وپے
N
L

९९
S
چچچ
Fig. 267.
N
Fig. 268.
Machines with external poles and drum armatures (Fig. 265) are still
built for small powers and high speeds. There is, as a rule, one coil per pair
of poles and this coil, as in the D. C. machine, spans an arc equal to the pole-
pitch. In Fig. 265, for example, the wire passes up slot 1 from back to front
and then down slot 1' from front to back, and so on. When this coil is com-
pletely wound, we pass to the coil 2, 2' by means of the dotted connection at
the back end of the armature.
The internal pole machine (Fig. 269) is strikingly simple both in principle
and construction, and is the type almost universally adopted. The external


-·R=3000·
---R=3625
625
Fig. 269.
Three-phase generator of the Société Electricité et Hydraulique, Charleroi.
and stationary armature is built up of sheet-iron stampings, secured in a
cast-iron frame. The winding is carried in slots or tunnels on the inner
surface of the annular ring. These slots are lined with insulating material,
micanite in machines for high pressures, and pressspahn, or some other such
material, in low pressure machines. The winding is similar to the ordinary
drum windings, or like that shown on the rotating armature in Fig. 265, but
here the ends are brought out to fixed terminals, so that the collection of
current at high pressures by means of slip rings is avoided. The poles, which
are alternately north and south, are fixed on the periphery of a flywheel,
T. E.
18
274
Electrical Engineering
which rotates within the armature. The magnetising winding on the poles.
consists either of insulated wire or of bare copper strip wound on edge, with
a thin layer of pressspahn between successive turns. The exciting current is
supplied by a direct-current dynamo or an accumulator battery, and is led
into the field coils through brushes and slip rings.
The machines which were used for the celebrated transmission of power
from Lauffen to Frankfort in 1891 had a single magnetising coil for all the
poles. The pole-wheel consisted of a cylinder, made in two parts, which were
bolted together, and surrounded by the magnetising coil (Fig. 270). The
cylinder carries claw-like projections which spring alternately from the right
and from the left-hand side of the coil, which they partially embrace. The
lines of force run, for example, from left to right through the cylinder, pass

Fig. 270.
(From Kapp's "Dynamomaschinen.”)
outward by the claws shown in section, which are therefore north poles, cross
the air-gap into the external armature and there divide into two fluxes, to
return by the adjacent claws to the left-hand end of the cylinder. A large
number of the lines of force pass, however, from claw to claw without going
through the armature. The magnetic leakage of this type of machine is con-
sequently extremely large, and the advantage of a single magnetising coil is
too dearly purchased.
We have, so far, considered machines with a single ring of poles. In
machines with disc armatures, however, there are two crowns of poles, so
arranged that unlike pole faces are opposite to one another. We shall con-
sider, in the first place, those machines in which the poles of each ring or
crown are alternately north and south. Such is the case in the old Ferranti
type of machine (Fig. 271). The armature consists of a wave-shaped copper
band, which rotates in the narrow gap between the two crowns of poles. The
Types of alternating current generators
275
poles shown in the figure lie behind the paper, while a second set of poles is
arranged in front of the paper so that unlike poles are exactly opposite. This
machine shows very clearly how the wire passes down, in front of a north
pole, and up, in front of the succeeding south pole. The winding has quite
lost its coil-like character and has become a pure wave winding.
The newer Ferranti machines have the same field system, but the
armature is built up of thin disc-like coils. The same construction was
adopted in the earliest Siemens and Halske alternators. To show the
essential similarity of disc and drum windings, a winding is shown in Fig. 272
which is equally applicable to either. For a Ferranti machine the poles have
simply to be bent round to form an arc of a circle, like Fig. 271, and another
set of poles arranged in front of them. The figure may equally well
represent the development of a drum winding, in which case the opposite
crown of poles is produced by induction in the armature iron.
In order that the electromotive forces may be added together, the coils
must be connected up alternately in opposite directions, as is shown in

funt

Fig. 271.
Fig. 272.
Fig. 272. We notice that we have here the sides of two adjacent coils under
one pole. The electromotive forces induced in all the wires under one pole
are in the same direction and the wires virtually form the side of a coil of
twice the number of turns.
There is another type of machine, in which there are also two crowns of
poles but in which all the poles on one side are north poles and all those on
the other side south poles. The Mordey alternator was of this type and is of
considerable historic interest. Its magnet system is very similar to that of
the Lauffen generators already mentioned, but the claws do not pass right
across the machine, alternately from left to right and from right to left; here,
they face one another with a small air-gap in between (Fig. 273). On one
side we have a row of north poles, and on the other a row of south poles. As
this pole wheel rotates, the lines of force cut the thin flat coils of the fixed
armature (Fig. 274). The number of these coils is double the number of
poles on one side. If the coils were as wide as the distance between the
centres of adjacent poles, the electromotive force of one side of the coil would
neutralise the electromotive force of the other side, but by winding the coils
as shown in Fig. 275, so that they span only half the pole-pitch, this is
prevented. When one side of a coil is in front of a pole and therefore having
18-2
276
Electrical Engineering
E.M.F. induced in it, the other side is in the neutral zone between the poles,
and consequently inactive. In the ordinary drum windings, on the other
hand, both sides of a coil are simultaneously active in producing E.M. F.
Hence, for the same number of wires on the armature, the same frequency


Fig. 273.
Fig. 275.

السنة
..
Fig. 274.
(From Kapp's "Dynamomaschinen.")
and the same flux per pole, the E.M.F. of a Mordey alternator is only half that
of a machine with alternate poles.
The difference can also be explained in the following manner: in the
alternate pole machine, during a complete period, the flux per pole is passed
through an armature coil, withdrawn, passed through in the opposite direction
and again withdrawn. In the Mordey or similar pole type, a period is made
up of a single introduction and withdrawal of the flux.
Types of alternating current generators
277
The armature coils must, of course, be so connected that their electro-
motive forces are added together. The adjacent sides of two successive coils
form a common group which is under a pole at the same time. If all the
coils are similarly wound, they must be connected up alternately in such a
manner that, in following round the winding, we pass clockwise round one
coil and anticlockwise round the next and so on (Fig. 275).
The so-called inductor alternators are similar to the machines just
described, in that there is no reversal of flux, but simply an introduction and
withdrawal of lines through the armature coils. The characteristic feature of
these machines is that neither the armature winding nor the field winding
rotates, the only moving part consisting of masses of iron. These masses of
iron conduct the magnetic flux across the gap, first at one point and the next
moment at another. The flux through any coil on the armature is con-
tinually changing and thus inducing alternating electromotive forces in the
coil. Machines of this type are specially suitable for running at high

A
-G-
F
B
Л
G
Fig. 276.
peripheral speeds. A portion of an inductor generator by the A. E. G. of
Berlin is shown in Fig. 276. There are two armatures A and B, built up
of sheet iron, and secured by the frame C which acts as the magnetic yoke.
At the bottom of the annular space, thus formed, the magnetising coil F is
wound. On the left is shown the flywheel which carries a number of iron
blocks G. These blocks offer a path of low reluctance to the lines of force,
which consequently pass, for the most part, through those parts of the
armature between which the blocks are momentarily situated. The magnetic
flux practically rotates with the iron blocks, which are therefore equivalent to
rotating electromagnets. The armature coils must naturally span a distance
equal to half the pitch of the blocks.
Were magnetic leakage entirely absent, the armature wires in the neutral
zone, i.e. between the blocks, would be entirely inactive. As the result of
leakage, however, each coil has opposing electromotive forces induced in its
two sides. If the flux passing across the block is N, while the leakage flux,
that is, the flux which crosses between the blocks, is N, then the E.M.F. is
proportional to N-N. When we take into account that the flux does not
278
Electrical Engineering
reverse, but simply varies between zero and the maximum, it is evident that,
for N in equation (143) on page 279, we must put
N₁—– Nı
N=
2
…………….(139).
in
The great leakage in this type of machine caused it to be given up
favour of the alternate pole type. The same was the case with the Lauffen
type. The chief disadvantage is not so much the additional ampere-turns
required to drive the larger flux through the yoke as the heavy drop of
pressure caused by the increased leakage when the machine is loaded (see
Section 103).
89. The mean E.M.F. of a generator.
As a general rule, the armature windings of D.C. and A. C. generators differ
from one another in that the wires of the D. C. winding are equally distributed
over the whole armature, while the wires of a coil-side of an A.C. winding are
either contained in a single slot or distributed between two or three slots in
close proximity. The width of a coil-side is therefore less than the width of
the pole, and, as a rule, even less than the width of the interpolar arc. The
instantaneous values of the induced electromotive forces in the various wires
are added together at every moment. To obtain the mean value of the E.M.F.
induced in the armature, we must multiply the number of wires connected in
series by the number of lines cut by each in a complete revolution, divide
the product by the time taken for the revolution, and by 10 to get the result
in volts. If, then,
N be the flux per pole,
p
the number of pairs of poles, and
60
taken by the revolution is
n
the number of conductors in series,
then the change of linkages per revolution will be 2p. N.z'. Since the time
of a second, we have for the mean or average
E.M.F.
Eav
60/n
2p.N.2.10-8-2N.p·
n
60
z′. 10-8 ...
.(140).
N
Putting
p.
60'
this becomes
E=2N~2.10-8 volts
.(141).
It is interesting to compare this result with that obtained for a series
wound D.C. armature. Since there are only two parallel paths through the
armature, a = 1 and equation (79) on page 120 may be written
n
=p.N.
•
2.10-8
60
z is here the total number of wires on the armature, of which only a half
are in series, so that '=. This gives for the D.C. armature
2
z.
E=p. N. 7. 22'. 10 = 2N ~ %′ . 10–º.
60
The mean E.M.F. of a generator
279
This is exactly the same equation as was obtained above for the mean
E.M.F. of the alternator.
If an alternator and a D.C. dynamo have the same total number of wires
on the armature, the E. M. F. of the former will be double that of the latter.
We assume that the D.C. machine has a series armature and that all the
armature conductors of the alternator are connected in series. The current
from the alternator will only be a half of that from the D.C. machine, other
things being equal, since the external current flows through all the wires on
the alternator armature, while in the D.C. armature the current divides
between two parallel paths.
90. The effective E.M.F. with sinusoidal field.
On the assumption that the flux density around the armature varies from
point to point in accordance with the sine law, and that the conductors of a
coil-side are concentrated in a single slot, the maximum E. M.F. is given by
equation (112) on page 220, viz.
Emax
=T.N.~.Z′.10-8.
As the effective or R.M.S. value is equal to the maximum divided by √2,
we have
.N.~.2.10-82.22. N~ z'. 10-8.........(142).
π
E
√/2
The ratio of the effective to the mean value is known as the form-factor.
It depends on the shape or form of the field distribution curve and on the
type of armature winding. In our case the form-factor is = 1·11.
2.22
2
We must now pass to the consideration of the effective E.M.F. for other
cases, differing from the ideal one just considered. We therefore write
formula (142) in a more general form, viz.
E=k.N.~.z. 10-8
(143).
By distributing the coil-side between two or more slots the E. M. F. is
reduced. If the field is sinusoidal it is a simple matter to calculate the
E.M.F. for each slot from equation (142). The electromotive forces for the
several slots can then be compounded by the parallelogram or polygon of
forces, taking into consideration the phase difference due to the displacement
of the slots around the armature.
Suppose, for example, that we have a sinusoidal field distribution and a
double slot winding, that is, one in which the coil-side is laid in two slots, and
that the distance between these two slots is equal to a third of the pole-pitch.
The pole-pitch corresponds to an angle of π radians, so that the electromotive
forces induced in the two slots differ by a phase angle of π/3 or 60°. The
whole winding may therefore be considered as made up of two parts, and the
E.M.F. in each part is given by equation (142) as
E' = 2·22. N~ 10-8 volts.
2
When we compound two electromotive forces E', having a phase difference
280
Electrical Engineering
{
of π/3, we find that the resultant is equal to 2. cos π/6. E' = 173E'. The
E.M.F. of the whole armature is therefore
E = 1·73E' = 1·92. N ~ z′.10–8 volts.
·
A case of special interest is that in which the coil-side is distributed
between so many slots that the winding is practically equivalent to a smooth
core winding, in which the wires lie side by side upon the surface. The pole-
pitch is represented in Fig. 277 by the angle π. The coil-side has a width
2y and its centre is displaced from the neutral axis by the angle a at
the moment represented in the figure. We shall consider a narrow element
of the coil-side of width dø, displaced from the neutral axis by an angle 4.
If the total number of wires connected in series on the armature is z', the
number in the strip do and in all equivalent narrow strips under other poles
will be do.
The maximum value of the E. M. F. induced in these wires by
2y
the sinusoidal field is given by equation (112) on page 220 as
T.N
do. z
27
10-8 volts.
•
π
नै
α
atz

Fax-j bodood
Fig. 277.
The instantaneous value at any moment will be proportional to the sine
of the angle & between the narrow strip and the neutral axis at that moment.
The instantaneous E.M.F. of these strips will therefore be
π
dE N ~ z'. sin &. dø.10–8 volts.
27
We see from Fig. 277 that the limiting values of ø are a − y and a+y,
and the instantaneous value of the E.M.F. for the whole armature is obtained
by integrating between these limits, thus
Now,
Pa+y
a-y
π
E
27
. N~ z'. 10-8
fo
•aty
(*** sin 4.dp.
a-y
sin 4. do = cos (a− y) — cos (a + y) = 2 sin a. sin y,
so that the previous equation becomes
E=T.
sin Y. N~ . 10–8. sin a.
Y
z.
The momentary value is thus proportional to the sine of the angle between
the centre of the coil-side and the neutral axis, and for the maximum value
we have
Emax
π.
sin y
γ
NZ. 10-8 volts.
The effective E.M.F. with sinusoidal field
281
sin y
This is exactly the same as the maximum value of the E. M. F. in a winding
with each coil-side concentrated in a single slot, except for the ratio
The effective E. M.F. will be reduced in the same proportion, so that we get,
for a distributed winding, the formula
sin y. N~ 2. 10-8 volts
E = 2·22 Y.
Y
Y
..(144).
As an example, we may consider a three-phase distributed winding, in
which each coil-side has a width equal to a third of the pole-pitch. We
have here
π
2y = 3
and from equation (144)
E = 2·22.
3
.
π
π
sin y = sin
0.5,
6
sin y-05-2
3
π
Y
7
=
N ~ z′ . 10—³ – 2·12 . N ~ z' . 10~º.....
..(145).
For a width of coil-side equal to two-thirds of the pole-pitch, such as
we have in a creeping three-phase winding, the relations are as follows:
2y = &π,
용​ㅠ​,
π
sin y = sin
0.866,
3
and from equation (144)
E = 2·22.
0.866
π/3
=
Nz'.10 1.84. N~ z'. 10-8......(146).
These equations (145) and (146) will prove very useful to us when we
come to the consideration of induction motors, but for single and polyphase
generators other equations will be required, which we now proceed to
establish.
91. The E.M.F. of a single-slot winding.
Instead of assuming, as we did in the previous section, that the field is
distributed according to a sine law, we shall now assume that the distribution
is rectangular, having a constant value under the pole, and falling to zero at
the pole-tip. The induced E. M.F. will then have a similar wave form. As a
matter of fact, such a distribution of the field is impossible, as there is
naturally a falling off near the edges of the pole and a fringing from the
pole into the interpolar space. This results in the E. M. F. curve approaching
more nearly to the sine form, and gives an effective E. M. F. about 10 per cent.
smaller than that theoretically calculated.
The theoretical curve of E.M.F. for a single-slot winding is shown in
Fig. 278. The E.M.F. attains its maximum value directly the slot comes
under the pole, and remains at this value until the slot emerges from under
282
Electrical Engineering
the pole. The time taken by a slot to move through the angle is
1
second, and to move through the angle ẞ the time taken is
2
1 B
2~π
2~
second.
This is the time taken by each wire to cut through the flux N from each
pole. The rate of cutting is therefore
N.2 π
B
and as there are z' wires in
series, the maximum E.M.F. will be
Emax 2
π
N~z. 10-8 volts
.(147).
B
In Fig. 279 we have plotted the square of the E.M.F. from moment to
moment and constructed a rectangle on the base 27 with an area equal
to the shaded areas. The height of this rectangle is.
π
E2 max. By taking
the square root of this height we find the effective value of the E.M. F., viz.
X
N
S
ß
九
​Fig. 278.
B
E=
Emax
π
Fig. 279.
..(148).


If, for example, the polar arc is of the pole-pitch, we get from
equation (147)
and from equation (148)
Emax = 3N~ 2.10-8,
2
E =
Emax
2·45N ~ z′ . 10-8 volts.
3
If we take into consideration the fact that the actual E. M.F. is roughly
B 2
π 3
10 per cent. less than this, we get, for the ratio
almost the same
value as we obtained for the single-slot winding and sinusoidal field.
92. The E.M.F. of a double-slot winding.
In a double-slot winding the width of the coil-side 2y is equal to the
pitch of the holes (Fig. 280). We need only consider the case in which this
distance is less than the interpolar arc, since this condition is always satisfied
in actual machines. If the pole-pitch is π and the width of pole ß, the
interpolar arc is π-B, and we assume that 2y <π- B.
The E.M.F. of a double-slot winding
283
The E.M.F. curve is made up of two rectangles, displaced relatively to
each other by an angle 2y. During the time that both slots are simul-
taneously under the pole, that is, over the angle - 2y, the E.M.F. is a
maximum, and, since it is quite immaterial whether the wires are in one or
two slots during this interval, the maximum E. M. F. is given by equation (147)
Emar=2 2. N~2.10.
When only one of the two slots is under the pole, i.e. over the angle 2y,
the E.M.F. has half this value.
X
X
N
27
B-23
27
Fig. 280.
S

2
max
J
ß
Fig. 281.
max

We set up, as before, the squares of the E.M.F. from moment to moment
(Fig. 281), and calculate the shaded area thus obtained. This is equal to
max
2.2y. (
E 2
2
2
+(B-2y). E max = (B − y). E²
max.
By dividing this area by the base π and taking the root of the quotient,
we get the effective value of the E. M.F., viz.
B Y
E
Y. Emax
π π
.(149).
This equation is equally applicable to the windings shown in Figs. 282
and 283. The portion of winding shown in Fig. 282 is equivalent to a single


To
Fig. 282.
sp
Fig. 283.
coil distributed between two slots per coil-side. The whole armature will
contain p such coils, whereas the winding shown in Fig. 283 will consist
of 2p single coils with a separate slot for each coil-side. For the purpose of
calculation, however, Fig. 283 is to be treated as a double-slot winding with
p coils on the armature. The winding must naturally be so carried out that
the current flows in the same direction through two adjacent slots.
An interesting winding, which is virtually a double-slot winding, is shown
284
Electrical Engineering
in Fig. 284.
N
I
II
III
S
N
Fig. 284.
This is known as a short-coil winding, since each coil spans an
arc equal to of the pole-pitch. This arrangement has the advantage of
allowing three separate windings or
phases to lie side by side, displaced from
each other by 120°, without any over-
lapping or crossing of the coils. The
fact that each slot is shared by wires
belonging to different phases is a dis-
advantage. That the winding of each
phase is electrically equivalent to those considered above, is to be seen from
coil I in the figure, since one side of it is under the middle of the pole and
the other side is displaced by the angle π/3 from the middle of the next pole.
The angle π corresponds, as before, to the distance from the centre of one
pole to the centre of the following pole. The electromotive forces induced
in the two sides of the coil are displaced by a phase angle of π/3, so that the
coil-breadth 2y is equal to π/3, and the E.M.F. is to be calculated from equa-
tions (147) and (149).

93. The E.M.F. of a treble-slot winding.
We assume, as before, that the breadth of the coil-side is less than the
interpolar gap, so that
2y <π-B.
The curve of E. M. F. is obtained by adding the ordinates of three rectangles,
displaced from each other by the pitch of the slots or half the breadth of the
L
poq
N
S


max
E³
1
max
j j p-23 g g
Fig. 285.
B
Fig. 286.
coil-side. For the sake of clearness the rectangles are shown in Fig. 285 as
having unequal heights. At first, when only one slot is under the pole, the
E.M.F. corresponds to a third of the armature wires. This lasts for a time
corresponding to y, where 2y is the breadth of the coil-side. Over an equal
arc y the E.M.F. corresponds to of the total wires, and over an arc B-2y
it has a constant value corresponding to the whole number of armature wires.
As the coil-side leaves the pole, the E.M.F. drops in the same way as it rose
when the coil-side came under the pole. The ordinates in Fig. 286 represent
The E.M.F. of a treble-slot winding
285
the squares of the momentary values of the E.M. F., and the shaded area is
found to be equal to
-
2y.
Emax
3
+ 2y (§. Emax)² + (ß − 2y) . E³max:
If we divide this area by the base π and extract the square root of the
resultant height, we get, for the effective value of the E. M. F.
E=
E =
B
8 y
Y. Emax
π 9 π
For Emax we have equation (147) on page 282, viz.
....(150).
2π
Emax
N-2.10-8
В
We may consider as an example a three-phase generator with treble-slot
winding, or, as it is more often expressed, with 3 slots per pole and phase.
Since there are 3.39 slots per pole, the pitch of the slots must be π/9.
Hence y = π/9, and, assuming that the ratio ẞ/π of polar arc to pole-pitch is
, we have
Emar=
2.2
1
=
N~ 2'. 10-8 — 4 . N ~ z′. 10-8.
Then, from equation (150), it follows that
E=
√
1 8 1
29.9
Emax=2:53. N~ z'. 10-8.
94. The E.M.F. of a distributed winding.
In the foregoing considerations we have always assumed that the breadth
of the coil-side was less than the polar arc ß. We shall now, however,
consider a case in which the coil-side is broader than either the polar arc or
the interpolar gap, so that
N
ос
Fig. 287.
2y > B.

N
Fig. 288.
N
S
poo
B.
Fig. 289:
At the moment of maximum E. M.F. (Fig. 289) the coil-side will project
beyond the pole, so that the value of Emax depends on ẞ and not on 2y. We
shall consider, in the first place, the moment when the coil-side is symmetrical
about the neutral axis (Fig. 287). The electromotive forces induced under
each pole will exactly neutralise each other. If the coil moves forward
through the arc a, the E.M.F. induced by the south pole will increase by
an amount proportional to a. The opposing E. M. F. induced under the north
pole will decrease simultaneously by the same amount. The total E.M.F.
286
Electrical Engineering
will therefore correspond to an arc 2a. Since the maximum value of the
E. M.F. depends on the polar arc ẞ, we have, for the E.M.F. at any moment,
E = Emax
2a
B'
This applies to the interval between the positions shown in Fig. 287 and
Fig. 288, i.e. until the coil-side is beyond the influence of the north pole.
If the breadth of the coil-side exceeds the interpolar arc by an amount 8,
it is evident from Fig. 288 that
-
8 = 2y (π-B) = 2y + Bπ
..(151).
The distance through which the coil moved during the interval considered
above is seen from Fig. 287 to have been 8/2. We draw ordinates equal to
the squares of the E. M.F. from moment to moment during this interval, and
calculate the area F, thus produced (Fig. 290)
r8/2
E2
F₁ = ["E?
max
(20) ²
).da
4. E2
B2
max
By substituting the limits, we get
E2 S3
αν
3
378/2
0
max
F₁ =
6. B²

§. p-s 23-p
Fig. 290.
The second interval to be considered is that between Fig. 288 and
Fig. 289, when the coil-side is partly under a single pole. If a represent
now the amount of the coil which is under the pole at any moment, then
the instantaneous value of the E.M.F. will be given by the equation
α
E = Emax.
B
At the beginning of the interval (Fig. 288) a is equal to 8, while at the
end (Fig. 289) a is equal to ß. We must square the values of the E.M.F.,
and integrate between the limits a 8 and a = B. In this way we get for
the area F, in Fig. 290
"B
=
FE
max
a²
da
B2.
E2
max
•
(B³ — S³)
382
During the third interval the whole pole face is covered with the coil-side
and the E.M.F. has a constant maximum value. This interval corresponds to
an arc 2y –ß (Fig. 289). By squaring the E. M.F. we find the area F in
Fig. 290, viz.
mar.
F₁ = E2 (2y — B).
The E.M.F. of a distributed winding
287
Although the various areas are shown separated in Fig. 290, they are,
of course, parts of a single area, divided into five parts by four vertical lines.
For the total area we have
2F₁+2F₂+ F₂ = Emax ·
E³ max. (2y
B
83
3
3 362
If this be divided by π and the root extracted, the effective E. M. F. is
obtained; thus
E = Emax
Emax N
27 B
π 3п
83
3ẞ². π
(2y> B).........(152).
This equation is based on the assumption that the coil-side is broader
than the pole face. The value of Emax will be smaller than the value given
B
by equation (147) on page 282 in the ratio since only this fraction of
27
the total number of wires on the armature can be under the poles at the
same moment. For the value of Emax to be put in equation (152), we have
therefore
Emax
π
Y
N~ z. 10-8 (2y> B) ............(153).
If, however, the coil-side is narrower than the pole face, Emax must be
found from equation (147). In calculating the effective E. M.F., moreover, we
must notice that the maximum E. M. F. depends on the coil-side 2y and not on
the polar arc B. The limits of integration for the second interval will be d
and 2y, and the third interval will spread over the arc B-2y. Proceeding
in a similar manner to the above, we find that
B 2y
Emax 1/B
E = Emax
π
83
3п 12². п
Emax is, in this case, given by equation (147).
(2y< ß) …………………….(154).
If the coil-side 2y is even narrower than the interpolar arc, equation (151)
will give a negative value for d. It must then be put equal to 0, that is, the
last term in equation (154) is to be omitted. The proof of this follows from
the consideration that there is no area F₁, and that the limits between which
the area F₂ is to be calculated are 0 and 2y. Both these conditions are
fulfilled by putting 8 = 0.
2
95. The alternating E.M.F. of a closed D. C. winding.
A very important example of distributed winding is found in the rotary
converter. This is practically an ordinary D.C. machine with taps taken from
two diametrically opposite points on the armature winding and connected
to two slip rings. If the machine is multipolar, and has a parallel wound
armature, the connections to the slip rings are joined to every equipotential
point on the winding, so that each ring has p connections.
These machines can be used for the simultaneous supply of both D.C.
and A.C. current, or for transforming either of these into the other. In the
vast majority of cases, however, it is driven as a motor from the A.C. side,
and supplies D.C. from the commutator side for the purposes of electric
288
Electrical Engineering
traction. When the coils to which the slip rings are directly connected are
in the neutral zone (Fig. 291), the brushes on the slip rings are equivalent
to those on the commutator. The P.D. between the slip rings must, at this
moment, be equal to the P.D. applied to the D.C. side of the converter. From
equation (79) on page 120, we have then
Emax
n
P. N.
α 60
z. 10—8,
where z is the total number of wires on the armature. Now, the number of
wires in one branch of the armature winding is
2
so that
2a'
2
z
2a®
S
N
Fig. 291.
S

x

y
x
N
Fig. 292.
Further,
n
~=
= P
60°
The above equation for the maximum E.M.F. on the A.C. side may there-
fore be written as follows:
Emax = 2N ~ 2′ . 10−8.
We could have found this value directly from equation (153).
When the coil to which the slip ring connection is made passes under
a pole (Fig. 292), each branch of the armature becomes the seat of two
opposing electromotive forces, proportional respectively to the arcs y and x.
The E. M.F. decreases until, when the slip ring connections come under the
mid-points of the poles, it is equal to 0.
T=
To find the effective E. M. F. of a single-phase converter, we must put
2y equal to π in equation (152) on page 287. If we assume that B/
which is about the usual value of this ratio, equation (151) on page 286
becomes
910
2y. B
+ 1.
T
π π
3*
210
The alternating E.M.F. of a closed D. C. winding 289
From equation (152) we get for the effective value
2y B
83
E = Emax
= 0-745 Emax,
П 3π 36²π
and, substituting the value of Emax found above,
E = 0.745.2N~ z'. 10-8 1.49. N~ 2. 10-8.
=
The effective alternating E.M.F. is therefore, in our example, 0·745 of the
P.D. between the D.C. brushes, or the D.C. pressure is 134 times the A.C.
pressure. This constant ratio of the pressures is very important. If we
wish to convert high-pressure alternating current into low-pressure direct
current, we must first transform the high-pressure alternating current to
alternating current of a pressure equal to 0·745 of the required D.C. pressure.
This is done in ordinary static transformers. Commutation difficulties limit
the D.C. pressure to about 1,000 volts, generally speaking, and high pressures
are not suitable for lighting purposes, except in exceptional cases. An
alternative to the use of static transformers and rotary converters is the
employment of motor-generators, in which the high-pressure alternating
current is supplied directly to an A.C. motor which is coupled to an ordinary
D.C. dynamo. Rotary converters have been used almost exclusively in America,
whereas, on the Continent they are quite the exception. In England, both
systems have been largely adopted, and their relative merits are continually
under discussion.
96. The E.M.F. of a creeping bar or wave winding.
A so-called creeping bar winding for a three-phase generator is shown
in Fig. 293, and offers a further example for the calculation of the E.M.F.
by means of equations (152) and (153). The winding-step is a little more
or less than the pole-pitch, as in a D. C. series winding, so that, after making
2p steps, the winding returns to a point adjacent to that from which it
started. Space is left between them, however, for the returning wire of
another winding. After making 2p steps each equal to y, we shall arrive
at a wire having a number differing by 2 from that from which we started.
If be the wires per phase, the total number of wires will be 32, and
we have
or
32=2p.y±2,
3z ± 2
2p
Y
y must, of course, be an odd number. If, for example, the total number
of wires be 30 (Fig. 293), z′ will be 10, and for a 4-pole machine, we have
Y
30+ 2
8 or 7.
2.2
8 being an even number is not permissible, but could be replaced by
Y₁ = 7 and y₂ = 9. The other alternative is y = y₁ = y₂ = 7, and this has been
Y2
adopted in the figure. Starting from the central point we pass down wire 1
yı Y2
T. E.
19
290
Electrical Engineering
and come up wire 8, then, passing across the front, we go down wire 15 and
so on. The winding schemes for the three phases will be as follows:
I
1— 8
15-22
29-6
13-20
27— 4
II
11-18
III
21-28
25- 2
5-12
-16
19-26
23-30
3-10
7—14
17-24
The six ends could be connected to six slip rings, or, as is more usual
in actual practice, the three starting points or beginnings of the three
windings can be connected together within the armature, and the three

N 28
27
29
30
2
나
​5
S
26
25+
I
24+
TII
I
23
22
21
20
12
S
19
13 N
18
17
15
16
7
10
8
Fig. 293.
remaining ends taken to three slip rings. This is shown in the figure, where,
however, only one winding is drawn, for the sake of clearness. If, now, we
consider the exact position of each wire with regard to the poles, we see
that wire 8 is equivalent to a wire between 1 and 30, but nearer to 1.
Similarly, the wires 8, 15, 22 and 29 may be imagined to lie, equally spaced,
between the actual positions of 1 and 29. All the wires of phase I lie
virtually between 1 and 27, or rather, between 1 and a point 4' equivalent to
the wire 4.
The coil-side is evidently equal to two-thirds of the pole-pitch, so that
2y/π=2/3. If we assume that ẞ/π = 1/2, the coil-side is broader than the
pole, and we have to use equation (153) on page 287, viz.
Emax
πT
Y
.N.10-83. N~.10-8.
The E.M.F. of a creeping bar or wave winding
291
From equation (151) on page 286, we have
8 27 B
+
π
π π
2 1
1
1
+
1
2
and, from equation (152) on page 287,
2 1 1
E= Emax
=0493 Emar•
3 6 162
Substituting the value found above for Emax, we get
E=211. N~2.10-8.
97. The E.M.F. of a creeping coil winding.
The three-phase creeping coil winding shown in Fig. 294 can also be
referred to an equivalent distributed winding. The winding consists of three
parts, belonging to the three phases. It is wound in such a way that each

N
N
8
S
Ꮽ
S
S
3
N
6
N
Fig. 294.
slot contains two separate coil-sides. The pitch of the slots is either a trifle
greater or smaller than the pole-pitch. Each phase occupies a third of the
armature periphery; phase I is wound in slots 1, 2, 3 and 4, phase II in
slots 4, 5, 6 and 7, and phase III in slots 7, 8, 9 and 1. The machine has
8 poles, but 9 coils. The number of coils must, in general, be one more or
less than the number of poles. Since in a three-phase machine the number
of coils must necessarily be divisible by 3, it follows that 2p + 1 must also be
divisible by 3.
If we consider phase I we see that the wires are virtually spread over
19-2
292
Electrical Engineering
1-
?
half the breadth of a pole, or a third of the pole-pitch, so that 2y/π=1/3. If
we assume that ẞ/π=2/3, we get from equation (151)
8_2y, B
+ - 1 =
TT π
1 2
+ -1=0.
3 3
As the breadth of the coil-side is less than the polar arc, we must use
equations (147) and (154) on pp. 282 and 287; thus
π
•
N~2.10-83. N~2.10-,
and
max=4a •
B
B 2y
E = Emax
π
3п
Sa
12². п
2 1
= Emax 3-9.
9°
Substituting for Emax, we have
E=2·24. N~ 2'. 10-8.
In actual practice the adjacent coil-sides of two neighbouring coils would
probably be wound in two separate slots. This type of winding has the dis-
advantage that an inequality in the strengths of the poles causes unequal
electromotive forces in the three phases.
If the various formulae established in this chapter be compared, it must
be remembered that we have made arbitrary and varying assumptions as to
the ratio B/π. As already pointed out, the theoretical results will differ
from those actually obtained, on account of our assumption of a rectangular
field distribution.
•
CHAPTER XIII.
98. The E. M. F. diagram for an A. C. generator.-99. The ampere-turns diagram for an A.C.
generator.-100. Calculation of armature reaction.-101. Experimental determination of
armature reaction and armature leakage.-102. Predetermination of exciting current and
pressure regulation.-103. Effect of polar leakage.
98. The E.M.F. diagram for an A. C. generator.
If the assumption be made that the coefficient of self-induction of the
armature of a machine has a constant value for all conditions of load and
excitation, the vector diagram is extremely simple. The electromotive force
E₁, induced in the armature by the rotating magnetic flux, forms the hypo-
tenuse of a right-angled triangle, one side of which is equal to the total
inductive pressure drop, both internal and external, while the other side
is equal to the sum of the internal and external ohmic pressure drops. The
direction of the current vector in Fig. 295 is vertically upwards. The self-
B


E,
iR
0 Es int.
F
D
Es ext.
Fig. 295.
A
iRa
f1
Fig. 296.
induction induces an E.M.F. lagging 90° behind the current, which must be
counterbalanced by a component OA of the E.M.F. OA consists of two
parts, one Esint. lost in the armature, and the other Esext. overcoming the back
E.M.F. in the external circuit. The component AB consists similarly of a
part AD lost in the machine, and a part DB which overcomes the external
ohmic resistance. To find the terminal pressure we must complete the
rectangle FDA and join OF and FB. The latter represents the terminal
pressure e, for it is the hypotenuse of a right-angled triangle with sides.
equal to the external inductive and ohmic pressure components. The line
OF represents the total drop of pressure in the generator. We have already
•
294
Electrical Engineering
1
seen in Section 75 that this drop cannot be subtracted algebraically from the
E.M.F., but must be subtracted geometrically, as shown in Fig. 295.
The angle FBD is the phase difference between the terminal pressure
and the current; it is the angle of lag in the external circuit and could
be determined by means of an ammeter, voltmeter and wattmeter, suitably
connected to the main generator leads. The angle ₁, on the other hand, is
the angle by which the current lags behind the induced E. M.F. To make
the meaning of this latter angle quite clear, a pole is drawn in Fig. 296
in the position it occupies at the moment of maximum armature current.
Were the angle ₁ equal to 0, the current would be a maximum at the
moment when the centre of the pole coincided with the armature conductor
or coil-side. In the present case the current lags behind the E.M.F. so that
the pole has got past the conductor by an angle o, before the current in the
conductor reaches its maximum value. It is, of course, only in a 2-pole
machine that the actual angle 1 (Fig. 296) is equal to the angle 4, in the
vector diagram (Fig. 295). If a machine has p pairs of poles, the actual
angle will only be equal to p₁/p, since one revolution of the generator will
correspond to p complete periods or revolutions of the vector diagram.
Before we can construct Fig. 295 we must determine both the open-
circuit characteristic and the self-induction of the generator armature. The
characteristic can easily be found experimentally by running the generator
at normal speed and observing the terminal pressure, while the exciting
current is varied, either by means of a rheostat or by altering the P.D. of a
separate exciting dynamo. A curve is drawn with abscissae equal to the
exciting current or to the ampere-turns per pair of poles, and ordinates
equal to the terminal P.D. which, on open circuit, is equal to the induced
E. M. F.
The curve obtained in this way is variously known as the magnetisation
curve, static characteristic, or open-circuit characteristic. Such a curve is
shown in Fig. 320. So long as the magnetic flux is small the curve is
practically a straight line, since the air-gap forms the principal part of the
reluctance, and the flux, and consequently the E.M.F., is proportional to the
exciting current. As the iron becomes saturated, its permeability decreases,
and the curve bends over more and more.
Having determined the open-circuit characteristic either experimentally
or by calculation, as described in Section 57, we have now to find the self-
induction. We shall assume that it is independent of the relative position
of poles and armature winding. That this assumption is hardly justified
can be seen from Figs. 297 and 298, where the paths of the lines of force are
indicated for two positions of the rotating field system. We are here con-
sidering the flux produced by the armature currents only, and it is evident
that the self-induction of the armature is not constant during a period.
Moreover, the phase relation of the current, that is, the angle ₁, must have
an effect on the self-induction, as it is evidently of some importance whether
the current in a coil-side has its maximum value when it is opposite a pole
or when in the neutral zone.
*
The E.M.F. diagram for an A. C. generator
295
If we neglect these effects, we can find the self-induction of the armature,
as we would that of any coil or piece of apparatus, by applying an external
alternating P.D. to the terminals of the stationary machine. The P.D. must
be of so low a value that the normal full-load current flows through the
armature*. If we neglect the small resistance of the armature, the applied
P.D. is equal to the back E.M.F. of self-induction, and the quotient obtained
by dividing this P.D. by the current is equal to the reactance 27~ L.
By means of the open-circuit characteristic and the self-induction of the
armature, we can now predetermine the terminal pressure for a given load
and a given exciting current. The power factor cos of the load must also
o
be known. From the open-circuit curve we find the value of E, for the
given exciting current, and, with the centre O, describe an arc with a radius
OB= E₁ (Fig. 299). The triangle OCF is made with the base OC=2π~L.i

x
Fig. 297.
N
S
B
&


Fig. 298.
Fig. 299.
and the side CF-i. Ra. To construct the diagram for a given power factor
cos in the external circuit, a vector is drawn from 0, making an angle &
with the vertical. This indicates the direction of the vector of terminal P.D.,
and a line is drawn parallel to it through the point F. This line cuts the
circle at the point B. By referring to Fig. 295 we see that FB represents
the required terminal pressure e, the letters OFB having the same signi-
ficance in both diagrams. The algebraic difference between OB and FB
represents the fall of terminal pressure between no-load and full load. In
order that this difference may be easily seen we have drawn another circle
with the same radius E₁, but with the point F as centre. The line FB is
produced to meet this second circle at G, and the intercept BG is evidently
equal to the pressure drop E₁-e. The figure shows plainly that this drop of
pressure increases as the point B moves round to the right, that is, as the
angle increases.
The pressure drop has a maximum value when the angle is equal to
90° (Fig. 300). The external circuit will then consist entirely of inductive
resistance, and the P.D. will be equal to the back E.M.F. of self-induction
* This is often a dangerous experiment to carry out on a large machine, since the alternating
flux produced by the armature induces very high electromotive forces in the field windings.
296
Electrical Engineering
in the external circuit. The internal pressure drop OF is subtracted almost
algebraically from the E.M.F. OB, and the terminal pressure FB reaches its
lowest value.
If the load consists entirely of glow-lamps, the power factor cos & will be
equal to 1, and the current will be in phase with the terminal pressure. The
diagram for this case is given in Fig. 301, from which it is seen that the fall
G


B
e-ik
F
B
G
C
Fig. 300.
C
Fig. 301.
of terminal pressure between open circuit and full load is very small. In
a modern generator the pressure will fall about 5 per cent. in such a case,
while on an external circuit, with a power-factor of 0.75, the drop will be
nearly 20 per cent.

B
E₁
e
F
C
Fig. 302.
A case of special interest is that in which the external circuit causes the
current vector i to lead ahead of the terminal pressure vector e by an angle &
(Fig. 302). This can occur when the circuit contains condensers or over-
excited synchronous motors.
In such a case the terminal pressure may
exceed the E.M.F. (Fig. 302). This striking result will be more fully dealt
with in Sections 107 and 111. We might mention here, however, that the
back E.M.F. of an A.C. motor may be greater than the applied terminal P.D.,
or even than the E. M. F. of the generator. If, for example, the E.M.F. of the
generator is 1,000 volts, while the back E. M. F. of the motor is 1,200 volts,
their common terminal pressure will automatically adjust itself to some inter-
mediate value. The state of affairs, so far as the generator is concerned, is
shown in Fig. 302.
The ampere-turns diagram
297
99. The ampere-turns diagram.
The diagram of electromotive forces described in the previous section
gives one a very clear insight into the relations determining the terminal
pressure under various conditions of load. If it be used, however, to deter-
mine the excitation necessary to maintain the P.D. between the terminals at
various loads, the calculated results will be found to differ from the results
obtained by actual experiment. This is due to the fact that the drop of
pressure is largely caused by a decrease in the magnetic flux entering the
armature. This decrease in the flux is caused by the demagnetising effect
of the armature current, which opposes, to a greater or less degree, the
ampere-turns on the field-magnets, and therefore leads to a decrease in the
E. M. F. induced in the armature.
We must now divide the loss of pressure, which we have attributed to
the self-induction of the armature, into two parts. The first part, which we
shall call E, is caused by the flux which follows the path indicated in
Fig. 303. It is assumed in the figure that the field-magnets are rotating



N
N
TUT TUT FUT
Fig. 303.
Fig. 304.
Fig. 305.
clockwise, and that the current lags behind the induced E. M. F. by such an
amount that the former has just reached its maximum value in the figure.
It is evident that the magnetomotive force of the armature winding is
opposed to that of the poles.
The second part of the loss of pressure we shall call Er and refer to
as the armature leakage drop. This is due to the flux which is produced
round the armature conductors, but which does not link the field winding.
A part of this flux follows the path shown in Fig. 304, that is, it encircles the
conductor which is embedded in the slot, some of it even crossing the air-gap
and running through the pole face. The remainder of this leakage flux
encircles the end connections, that is, those parts of the armature winding
which project beyond the iron core, as shown in Fig. 305.
1
By keeping these two sources of pressure loss separate, we obtain Fig. 306
in the place of Fig. 295. E, is the E.M. F. which would be induced by the
given field current, were there no armature reaction. The effect of the
armature reaction in weakening the flux is represented by the drop E,,
so that the E. M. F. corresponding to the flux actually entering the armature
is represented by E. From this we have to take the leakage drop E, and the
i. Ra drop, leaving the pressure e across the terminals.
It is evident that the electromotive forces E, and E, are purely fictitious,
and further, that the very lines of force by which they would be induced do
298
Electrical Engineering
not exist, except as imaginary components of the resultant flux which pro-
duces E. As a matter of fact, the E.M. F. E is also fictitious, as the effect of
the self-induction or armature slot leakage is to prevent the flux which enters
the armature from cutting the conductors by driving a part of it across the
slots. Hence, although the flux which crosses the gap is sufficient to produce
an E.M.F. E, the flux which actually cuts the conductors is only sufficient to
produce the E.M.F. BC in Fig. 306. Since the poles are excited by a steady
direct current, and the armature reaction is rapidly varying between 0 and
a maximum, we might be led to expect a pulsating flux. This is prevented,
however, to a great extent by the self-induction of the field windings and the
eddy currents in the poles, so that the flux passing through the poles has a
practically constant value N as the resultant effect of the constant ampere-
turns X, on the field and the mean value X, of the pulsating armature
reaction,
1
2
Hence, neither the electromotive forces E, and E, in Fig. 306, nor the
fluxes N₁ and N₂ in Fig. 307, have any real existence. The ampere-turns X,
1
2
B
B


E,
E
iRa
e
iR
G Er
Fig. 306.
Es
N₁
N
B
0
N₂ G
GO X₂ G
Fig. 307.
Fig. 308.
and X2, on the other hand, are really present, and act on a common magnetic
circuit, in which their resultant X produces a flux N. The E.M.F. E corre-
sponding to X ampere-turns is found from the open-circuit characteristic.
2
If the load be suddenly thrown off the machine, the ampere-turns X, of
armature reaction entirely disappear, and the points G and O in the figure
are brought together. The flux, and with it E, increases to a value corre-
sponding to X, ampere-turns. The flux and E.M.F. will not increase, however,
in the ratio of X, to X, on account of magnetic saturation.
1
100. Calculation of armature reaction.
The calculation of the armature reaction is more difficult in the case of
single-phase generators than in the case of two or three-phase machines. If
the armature current lags 90° behind the E. M.F., the current reaches its
maximum value at the moment that the coil-side is midway between the
poles (Fig. 310). The lines of force due to the armature current are shown
dotted and are evidently in direct opposition to the field-magnets. The
conditions both before and after Fig. 310 are shown in Figs. 309 and 311;
Calculation of armature reaction
299
the fact that the current is smaller is indicated by lighter crosses and dots
on the armature conductors.
If there are wires in series on the armature, then the wires per pole or
For the armature reaction at the
the turns per pair of poles will be
Z
2p'
moment represented in Fig. 310, we have
X2 max
imax amp.-turns per pair of poles.
2p
This is the maximum value, and, assuming a sine-wave current, the mean
armature reaction will be
or
z 2
z' 2
X2
Vmax
√2.i,
2p π
21
2
X₂ = 0·9 ? · 2′
per pair of poles,
2p
where i is the effective value of the current.

Fig. 309.
Fig. 310.
Fig. 311.
It is more to our
affecting armature
The calculation is,
We have not taken into account the fact that the pole is directly under
an armature coil during a part of the period only*.
purpose, however, to understand the various factors
reaction than to calculate them with great accuracy.
after all, more or less uncertain, as the flux pulsates appreciably, in spite of
the self-induction of the field coils and the eddy currents in the poles.
The calculation is much more trustworthy in the case of polyphase
machines, in which there are two or three separate windings on the armature.
Although we have not yet studied the subject of polyphase currents, we may
make use of the simple fact that, when the current in any coil-side of a
three-phase winding is a maximum, the current in each of the adjacent coil-
sides has half this maximum value. If the current is lagging 90°, the coil-
side midway between the poles carries the maximum current, while the
coil-side on either side of it carries a current in the same direction but of
half the amount (Fig. 312).
In order to determine the armature reaction per pair of poles, we must
first find the ampere-turns in each coil. If there are z' wires per phase, the
wires per coil-side will be , and the ampere-turns of coil 2-5 will be
Ź
2p
* By taking everything into consideration Kapp has found a coefficient 0-736 instead of 0·9
when ẞ/T=3, and 0.8 when ß/π=}.
300
Electrical Engineering
imax Coils 3-6 and 1-4 will each have §.
mar ampere-turns. The
2p
2p
flux which passes between holes 3 and 4 links all three of these coils, and has
22
2p
a M.M.F. due to imax ampere-turns. The flux which passes to the left of
3 and to right of 4 links only the coil 2-5, as shown by the dotted lines in
Z
the figure. It is produced by 2p
2max ampere-turns. If the polar arc is two-
thirds of the pole-pitch, the space between holes 3 and 4 covers half the pole

ら
​X
Fig. 312.
Z
face so that half the pole face is crossed by flux produced by 2
imax
2p
z
ampere-turns, and the other half by flux produced by imax ampere-turns.
2p
Taking the mean value for the whole pole, the armature reaction of a three-
phase generator is
X₂ = 1.52 imax
2p
.(155).
It might be objected that we have only considered the conditions existing
at the moment represented in the figure. As the pole wheel rotates, how-
ever, the current in coil 2-5 decreases and that in coil 3-6 increases, so



X
X
N
N
Fig. 313.
Fig. 314.
Fig. 315.
that the armature reaction is hardly altered. When the wires 3 and 6 are
midway between the poles the current in these wires will be at its maximum
value, so that the conditions represented in the figure are continually re-
curring. As already mentioned, these results make no claim to mathematical
exactness, but merely serve to make the effect of armature reaction as clear
as possible.
Calculation of armature reaction
301
We have assumed, so far, that the current lagged 90° behind the E.M.F.,
so that the armature reaction was directly opposed to the field-magnets. We
must now consider the effect of power-factor on the armature reaction. If
the current is in phase with the E. M. F., the current in any coil-side is a
maximum when that coil-side is approximately under the middle of the pole
(Fig. 314). At this moment the current can produce no flux round the field
magnetic circuit. A moment earlier the growing current produced a flux
which strengthened the field, while a moment later the decreasing current
weakens the field. The nett result is that the armature neither strengthens
nor weakens the field.
If the current lags behind the E. M. F. by an angle 41, the effect of armature
reaction is shown in Fig. 316. The vertical line represents the current i
lagging behind the electromotive force E. This triangle corresponds to
BAG in Fig. 306. BH is the excitation X which would produce E on open

VH
K
G
E
Fig. 316.
1
circuit, and it must therefore be the resultant of X, and X₂. The triangle
BKH corresponds to Fig. 308. It is evident that the armature reaction X,
becomes of increasing importance as p, approaches 90°. When ₁ = 90°, X₁
and X, will be co-linear and X will equal X-X,. This corresponds to
Fig. 310.
2
1
Another way of looking at it is as follows: Of the total armature reaction
only a component HLX,. sin p, acts in the direction of the resultant
excitation. If we represent this direct demagnetising component by Xa, we
have
Xa= X2. sin ò̟₁ = HL.
It can be seen from the figure that HB+ HL is very nearly equal to X1,
so that
or
X₁ = X + Xd,
X = X₁- Xa··
302
Electrical Engineering
For negative values of ₁, that is, for leading currents, the demagnetising
effect also becomes negative. This can be shown in Fig. 310 by reversing
the direction of the armature current. We can sum this up by saying that
lagging currents act demagnetisingly, while leading currents act magnetisingly.
This explains the rise of P.D. with increasing load in Fig. 302.
101. Experimental determination of armature reaction and
armature leakage.
Generally speaking, the results of experiments will be found to agree
with the calculations of the previous section. An important experiment in
this connection is the short-circuit test of an alternator. The armature is
short-circuited through an ammeter and the machine is run at its normal
speed with very weak field excitation, as the armature current would other-
wise be excessive. The field current is varied and simultaneous readings
are taken of field and armature current. It may be safely assumed that the
armature resistance is very small and that the self-induction due to slot-
leakage, etc., is comparatively small, especially in a modern machine with
open slots, large air-gap and several slots per pole and phase.


E
E2
E
Bi
Fig. 317.
iRa
X₁
X₂
Fig. 318.
X
The diagram for short circuit will then be like Fig. 317, in which
1
E, is the E.M.F. which would be given by the field excitation if the
permeability were constant,
E is the E.M.F. which would be produced by the flux caused by arma-
ture reaction under like conditions,
and E is the actual E. M. F. induced by the resultant flux, which has to drive
the current against the resistance and inductance of the armature.
The corresponding diagram of ampere-turns is given in Fig. 318. It is
evident that the armature reaction on short circuit is almost equal to the
field excitation X₁.
It is not necessary to know the number of turns on either the armature
or field, but a curve can be drawn showing the relation between the current
in the short-circuited armature and the field current (Fig. 319). This short-
circuit characteristic tells us the amount of field current neutralised by the
reaction of any given armature current. The curve is almost exactly a
straight line. The field current is not exactly neutralised, however, but gives
a resultant X as shown in Fig. 318.
In order to calculate the field current required to give a certain terminal
P.D. on a given load, it is necessary to know, in addition to the armature
reaction X2, the self-induction L of the armature winding, due to slot-
leakage, etc. This is a very difficult quantity to calculate and considerable
Determination of armature reaction, etc.
303
experience is necessary in applying the various formulae which enter into the
calculation.
The method which we shall now describe is due to Potier. To apply the
method to a generator, it is necessary to make an experiment which we have
not yet mentioned; this consists in loading the machine with a very induc-
Short circuit Current.
Field Current.
0 10 20 80 40 50

50
100
150
60
200
Fig. 319.
tive load, made up, say, of choking coils, so that the armature supplies a
heavy lagging current with a power-factor as near as possible to zero. We
shall illustrate the method by applying it to an actual example, viz. the
three-phase generator shown in Fig. 269. The figures for this machine were
Volts

2800
2600
2400
2200
2000
1800
1600
1400
1200
1000
800
600
400
anuit Characteristic
Open: ant
Armature abrera = 100 Amperes
Cos/9
A
200
B
20
40
60
80
100
120
140
160 180 200
Field Current
Fig. 320.
given by Heyland in the E. T. Z. for 1900. Its short-circuit characteristic is
given in Fig. 319 and the open-circuit characteristic in Fig. 320. When
loaded with choking coils, the field current was 150 amperes, the armature
current 100 amperes and the terminal pressure 2,140 volts. This point A is
plotted on Fig. 320, as is also the point on the base line representing a field
current of 18 amperes, since this is the field current which we see from
304
Electrical Engineering
..
Fig. 319 would give 100 amperes on short circuit, that is, when the terminal
P.D. is equal to 0. The lower curve is drawn through all such points repre-
senting the relation between terminal pressure and field current for a wattless
current of 100 amperes. The manner in which this curve has been drawn
will appear immediately.
A horizontal is drawn through the point A and made equal to the field
current on short circuit, viz. OB = 18 amperes. Through the point D, thus
obtained, a line is drawn parallel to the lower portion of the open-circuit
characteristic. A perpendicular is dropped from the point F where this line
cuts the characteristic. A right-angled triangle FGA is obtained in this
way,
which fits in between the two curves over the whole range. The side
FG is always vertical and represents, therefore, a certain difference of
pressure; while the side GA is always horizontal and represents a certain
field current or excitation.
If, when the machine is working on the inductive load at the point A, we
could, in some way, do away with the demagnetising effect of the armature
reaction, the field current could be reduced by some such amount as AG,
without lowering the terminal pressure. Further, if we could then do away
with the self-induction of the armature coils, the drop of pressure due to this
cause would disappear, and the terminal pressure would rise by some such
amount as FG. Since we have successively removed both armature reaction
and self-induction, the machine is now on open circuit and the point F is a
point on the open-circuit characteristic. Now, both armature reaction and
self-inductive drop are proportional to the armature current, and are constant
when the current is constant, as it is for all points on the lower curve in
Fig. 320. Further, as the current is wattless for every point on the curve,
the armature demagnetising effect must be subtracted algebraically from the
field excitation, and the drop must be subtracted algebraically from the
induced E.M.F. Hence, there must be some such right-angled triangle which
will fit in everywhere, and it is easily seen that FGA is the only possible one.
It follows, therefore, that the armature reaction due to a current of 100 am-
peres is represented by AG, and the self-inductive drop due to slot-leakage
etc., by FG. Since they are both proportional to the current, the corre-
sponding values for other currents can be readily found. In our case
X,= AG=10 amperes,
==
which is a tenth of the armature current, and
EFG = 170 volts,
whence
Er
wL
=
2π L
= 1.7 ohms.
How these quantities can be used to determine the terminal pressure
under any condition of load of any power-factor will be explained in the next
section.
Predetermination of pressure regulation
305
102. Predetermination of exciting current and pressure
regulation.
We now proceed to determine the excitation necessary to maintain a
terminal pressure of 2,200 volts on the machine experimented on in the
previous section, under the following conditions:
A load of 200 amperes and cos &= 1,
1.
2.
>>
3.
""
""
""
""
cos &= 0·8,
cos + = 0.
We shall also determine in each case the value reached by the terminal
pressure when the load is suddenly switched off, and we shall express the
increase of P.D. as a percentage of the normal P.D., i.e. 2,200 volts.
The armature resistance Rɑ is 0·22 ohm.
α
In the first case cos = 10, so that the external load is non-inductive, and
the whole terminal P.D. is used in overcoming ohmic resistance. The ohmic
pressure drop in the armature is
i. Ra=200.0.22 = 44 volts.
The total E.M.F. must have a component in phase with the current, equal
to the sum of these ohmic pressure drops. This is represented by AB in
Fig. 321, where
AB = 2,200 + 44 = 2,244 volts.
The pressure required to overcome the back E.M. F. due to the self-induc-
tion of the armature is represented by GA, where
GA=i.wL
==
200.1.7 = 340 volts.
This is at right angles to the ohmic drop, and we get for the total E. M.F.
GB= √2,2442 + 34022,270 volts.
This E. M. F. is seen from the open-circuit characteristic to require a
resultant excitation of 134 amperes, so that
X = BH = 134 amperes.
We have seen that the field current neutralised by armature reaction is a
tenth of the armature current, so that in our case
2
X,= 20 amperes.
2
Adding X and X, vectorially, as in Fig. 316, we get for the field excitation
X₁, the length BK in Fig. 321. By scaling this off on the diagram, we
find that
X₁ = 138 amperes.
This is the necessary field excitation on non-inductive load. If the load
is suddenly switched off so that the machine is on open circuit, it can be
seen from Fig. 320 that the terminal pressure will rise to 2,310 volts, since
this is the P.D. for an excitation of 138 amperes. The rise of terminal
pressure on switching off the load is therefore 2,310-2,200 110 volts,
2,200=110
which is 5 per cent. of the terminal pressure. This is a very good result.
We now proceed to the second case, in which the power-factor is 0·8.
T. E.
20
306
Electrical Engineering
138a
2270r
134 a
We have (see page 237)
i. Ra 200.0·22 ..
=
44 volts,
i. Rext = e. cos $2,200. 0·8 = 1,760
Esexte. sin = 2,200.0·6 1,320
· =
o
Esint = wL. i 17.200
=
""
""
340
""
The sum of the ohmic pressures is 1,804 volts, and the total inductive
drop is 1,660 volts. The total induced E. M. F. is represented by BG in
Fig. 322, where
BG= E = √1,804² + 1,660² = 2,455 volts.
The resultant excitation corresponding to this E. M. F. is seen from Fig.
320 to be 162 amperes, or
2244 v
B
X = 162 amperes.

176 a
162a
e=2200 r.
44r
1760v
1320 v
340
G
A
K
20a
H
340r
G
A
20a
H
Fig. 322.
Fig. 321.
The armature demagnetising effect KH=20 amperes is added geometrically,
giving a field excitation BH, where
BH = X₁ = 176 amperes.
Hence, to maintain the terminal P.D. at 2,200 volts with the lagging arma-
ture current, it is necessary to increase the field current to 176 amperes.
On throwing off the load the P.D. will naturally rise to a higher value than
before, and Fig. 320 shows this value to be 2,520 volts, corresponding to a
field current of 176 amperes. The rise of the terminal pressure is therefore
2,520 - 2,200=320 volts, which is 14.6 per cent. of the normal pressure.
The last case which we shall consider is one which is hardly likely to
occur under normal working conditions, viz.
cos += 0.
The whole 2,200 volts will be used in neutralising the back E.M. F. of the
external purely inductive load. By adding the internal inductive drop i.wL
B

Predetermination of pressure regulation
307
we get the total inductive pressure to be 2,200+340 = 2,540 volts. The
only ohmic pressure component is the 44 volts internal loss, which is practi-
cally negligible, so that the E.M.F. is equal to 2,540 volts. The no-load
characteristic gives a resultant excitation of 180 amperes, or
X = 180 amperes.
The armature demagnetising effect of 20 amperes can be added algebraically,
since the armature and field are in direct opposition. For the field excita-
tion we have then
X₁ = 180+20= 200 amperes.
On switching off the load the terminal pressure will now rise to 2,620
volts, an increase of 420 volts or 19.1 per cent. of the normal pressure.
The results thus obtained will be modified, however, if the machine has
considerable magnetic leakage from pole to pole. The effect of this leakage
is considered in detail in the next section. The machine considered will
have little polar leakage, since the poles are short and separated from each
other by a considerable distance.
103. Effect of polar leakage.
It might appear at first sight as if the effect of magnetic leakage between
the poles had been already taken into consideration in the open-circuit
characteristic. If, however, a machine be carefully designed for a given load,
and the diagrams be constructed as already explained, it will be found that
the actual relations when loaded differ considerably from those calculated
from the diagrams.
This difference is due to the fact that a greater resultant excitation is
necessary to produce an E. M.F. when the machine is loaded than to produce
the same E.M.F. on open circuit. To understand this, we must consider the
machine under the two conditions of open circuit and full load, with the same
flux entering the armature in each case and consequently the same E.M.F.
being induced. The number of ampere-turns necessary to drive this flux
through the air-gap teeth and armature is equal to the sum X,+Xt+Xa.
This same number of ampere-turns acts across the interpolar space on open
circuit, and drives the leakage flux from pole-shoe to pole-shoe.
If the component of the armature reaction which acts demagnetisingly
when the machine is on full load be Xa, the above ampere-turns must be
increased to Xg+Xt+Xa+Xa, in order to maintain the same armature
flux. Since this increased difference of magnetic potential acts between the
pole-shoes, the leakage flux will be increased in the same proportion. The
total flux passing through the pole will be increased by this increased leakage,
and the ampere-turns required to drive the flux through the pole will be
increased. Hence, to maintain the same E. M. F., the field excitation has not
only to be increased by the amount Xa when the machine is loaded, but also
by an amount equal to the increase in Xm. We see, therefore, that the
resultant excitation taken from the open-circuit characteristic for a given
E.M.F. is too small if the machine is loaded.
1
20—2
308
Electrical Engineering
Another way of looking at the matter is to consider the leakage as
increasing the armature reaction and thus causing a further drop in the
E. M.F. for a constant field current.
It is evident that the polar leakage will have little effect if the pole-cores
are unsaturated, and Xm is consequently small. In modern machines, how-
ever, the poles are purposely worked at a very high flux density, in order to
minimise the fluctuation of terminal pressure with changing load. In such
machines the polar leakage has a marked effect, and the constancy of terminal
pressure will depend largely on the reduction of this leakage to a minimum.
The exact calculation of the effect of the polar leakage is complicated by
the varying value of the permeability of the iron. We shall first put the
question in the following form: How many ampere-turns would be necessary
on open circuit to produce the same flux in the poles as actually passes
through them on full load?
We have therefore to consider the machine under two conditions, viz. on
full load and on no-load, with the assumption that the flux in the poles is the
same in each case. We make the further assumptions that the whole leakage
occurs between the pole-tips, and that the reluctance of the armature core
and teeth is negligible compared with that of the air-gap.
If
and
R be the reluctance of the air-gap,
Ri
""
then we can make out the following table.
Ampere-turns used
in overcoming
reluctance
of air- of leakage
leakage path,

Air-gap flux
Leakage flux
Total ampere-turns
on pole
gap
path
At no-load
X go
X go
No=
0·4πXg.
Ni.
R
0·4πXgo
Ri
X。=Xg+xm
0·4πX
g
At full load | X,* | Xg+Xã | N=
N₁ =
0·4π (Xg+Xa) X₁ =Xg+Xa+Xm
R
R
Since the flux in the pole is the same in each case, the decrease in the
useful flux must be equal to the increase of leakage flux, so that
N₁-N-N-N
10
If we substitute the values given in the above table, and solve for
Xg.-Xg, we obtain
R
Xg。−Xg=Xa• R+R₂
.(156).
Since, under these conditions, the air-gap flux is greater at no-load than
at normal load, the ampere-turns per pole at no-load must be greater than
d
* The total ampere-turns acting across air-gap and armature are X+Xa. The part Xa is
neutralised, however, by the armature ampere-turns, leaving X, to overcome the reluctance. It
is analogous to charging an accumulator.
Effect of polar leakage
309
the resultant ampere-turns at normal load. This is expressed by the
equation
X-Xg=X₂-(X₁-Xa).
0
1
d
If in the magnetisation curve represented in Fig. 323 OD=X-X, and
OB = X., then DB must be equal to X₁-Xa, the resultant ampere-turns on
load. Since OD represents the decrease in the ampere-turns acting across
the air-gap, and since the initial straight part of the characteristic is
approximately the magnetisation curve of the air-gap, it follows that O'D
represents the decrease in the flux crossing the air-gap, i.e. in the useful flux,
which is therefore reduced from AB to AF. Hence, the point O' is the
origin of a new characteristic, the abscissae O'F of which represent the
resultant excitation X₁ – Xa, while its ordinates AF represent the useful
flux.
It is obvious from the figure that this displacement of the origin from 0
to O' is immaterial so long as the machine is working below the knee of the
curve.
N
A
A


E
O D
B
0 D
B
Хо
I
Fig. 323.
Fig. 324.
No
N
The equation for X-Xg can now be simplified. From the law of
divided electric circuits, we have by analogy (see equation (12)),
This gives us for the leakage coefficient at no-load the value
Ri
R
0
λ
No + N₁₂ R+ R₂
.(157).
No
Ri
It follows from this that
λ
R+ R₂
λ - 1
R
λ-1
and
Xgo-Xg=
d
X₁ =
.(158).
λ
Hence, from a knowledge of the open-circuit leakage coefficient λ, the
value of OD=X-X, can be calculated for any given armature current and
power-factor.
It is obviously immaterial whether the above construction be applied to
the flux curve (Fig. 323), or to the open-circuit characteristic (Fig. 324).
CHAPTER XIV.
104. Alternator with constant excitation and constant terminal pressure.-105. Synchronising
power of armature.-106. The parallel connection of alternators.-107. The effect of field
regulation on alternators in parallel.-108. Phase-swinging of alternators.
104. Alternator with constant excitation and constant
terminal pressure.
If several alternators are connected in parallel to the same bus-bars, or
common switchboard terminals, the terminal pressure of each is no longer
dependent on itself alone, but is affected by the other machines in parallel
with it. If we assume that the other machines are very large compared with
the one under consideration, the terminal pressure will not be affected by
changes in the latter, so that it may be assumed to have a constant terminal
pressure e.
We shall assume, for the sake of simplicity, that the armature demagne-
tising effect can be included with the self-induction of the armature, due to
slot-leakage etc., by simply increasing the latter. The fundamental diagram
in Fig. 295 was drawn on this assumption and is therefore applicable to the
present case. It is drawn again in Fig. 325, in which E, represents the
induced E. M.F., assumed to be the same on load as on open circuit. So long
as the excitation remains unchanged, E, is a constant for all loads.
For our purpose we must construct Fig. 325 so that it represents the
given machine under certain conditions of load. With O as centre an arc is
struck with radius OB equal to the constant E.M. F. E. The line OM is
drawn at an angle a to the horizontal, so that
i. Ra Ra
tan a=
E
wL'
Both E, and L are used here in the same sense as in Section 98, that is, so
as to include the effect of armature reaction. Since E, is proportional to the
current, we can make OC to suit any given current i. A perpendicular
erected at C meets OM at F, and this point F is taken as the centre about
which to describe an arc with a radius equal to the constant terminal pressure
e. This arc cuts the former arc in the point B, which is joined to O and to F,
and from which a perpendicular is dropped on to OA.
The line BD should represent the ohmic resistance of the external circuit,
multiplied by the current i. Since, however, the alternator is one of several
connected in parallel we can no longer speak of its external circuit. We can
Alternators in parallel
311
say, however, that the machine is running under conditions equivalent to an
external load with an ohmic drop BD and an inductive drop FD.
Fig. 326 is a similar figure except that it is drawn for a larger current,
represented by a larger value of E¸, viz. OC' instead of OC. The values of
E₁, e and a are the same in both figures. There are two important points in
which the figures differ; the angle y between the E. M. F. and the terminal
pressure has changed, and the total power given out by the machine has
changed. Both the angle y and the power have been increased by the
increase of current.
We will not, at this point, inquire into the method by which this increase
of load can be effected; whether by some alteration on the generator, or by a
direct increase in the output of the engine. The point we must first consider
is the relation between the load on the generator and the angle y. The total
electric power generated is
N
α
E,
Du.?
P₁ = E₁.i.cos (☀ + y).

y
%
M
E,
B

FI
M
E,
C
A
Fig. 325.
Fig. 326.
C' A
If we assume that the ohmic resistance of the armature Ra is negligibly
small compared with its self-induction, the angle a vanishes and cos (+y)
becomes equal to sin d. At the same time OF becomes equal to i. wL, and
we have
sin d=
or
e
i. cl. sin y,
E₁.e
P₁ =
@L
sin y.
E₁
1
@L
Now, is the short-circuit current i,, so that P₁ = e. i. sin y.
The angle y is the angle by which the terminal pressure vector lags
behind the vector of induced E. M. F. We consider the terminal pressure as
being nearly in phase with the E.M.F. and not nearly 180° out of phase with
it. In a certain sense, however, they are in opposition.
The power generated is therefore proportional to the sine of the angle y.
If the vector e be set up vertically from the point 0 (Fig. 327) and a hori-
312
Electrical Engineering
zontal line equal to e. i, be taken as the diameter on which a semicircle is
drawn, then the chord OR, making an angle y with e, will represent the
power P₁.
If the resistance of the armature be
taken into account, the relations are as
follows:

From Fig. 325,
P₁ = E₁i. cos ($ + y) = E₁i. sin (a+8)
E₁i. (sin a. cos 8+ cos a . sin §).
R
BrA
Now,
cos &=
ON
OF
E₁- e .cos y
√ Ra²+(wL)²
Fig. 327.
FN
e.sin y
and
sin d
OF¯¯ i√ Ra²+(wL)²
Hence,
P₁ = E₁
sin a (E₁ — e. cos y) + e. sin y. cos a
√ Ra² + (wL)²
Now, the short-circuit current is given by the equation
En
20=
2
√ Ro² + (@L)²´
and we have
e.sin y. cos a e.cos y sin a = e sin (y — a),
therefore
P₁e.i.sin (y-a)+ E₁.. sin a ..........
F
R
α
Q
α
P₁
P₁
P
.(159).

+
50
+
100
150
200 Kilowatt
Fig. 328.
In Fig. 328 the vector e is drawn vertically, as before, and the line OP is
made equal to ei。, but is drawn so as to make an angle a with the horizontal,
where
tan α=
a
Ra
ωτ
Alternators in parallel
313
A semicircle is drawn on OP, and a line OE is set out at an angle y with
the vector e. If this line cuts the semicircle in a point Q, we have
OQ = OP. sin ẞ₁ = e. i。. sin (y − a),
because the angle ya between chord and tangent is equal to the internal
angle B₁.
If we add to OQ the constant amount
we have, from equation (159),
QR = E₁.i.sin a,
P₁ = OQ + QR.
The locus of all the points R obtained in this manner is represented by
the heavy black curve in Fig. 328. Since the angle a is always very small in
actual practice, the length QR which is equal to E₁.i.sin a, is negligibly
small, and the heavy curve is practically a circle. Fig. 328 is drawn for the
following values:
wL= 6,
Ra=1,
wL=6, e=1,000, E₁ =1,200.

T
e
X
J
50
α
P
100
150
200
Fig. 329.
We have, in the foregoing, considered only the total power generated by
the machine, and we now turn to the power given out by the machine to the
external circuit. For this we have
From Fig. 325 we see that
P = e.i.cos p.
cos += sin (y+a+8)=sin (y + a). cos d + cos (y+a). sin d.
By substituting the values found above for cos 8 and sin d, we get
(E₁- e. cos y). sin (y + a) + e. sin y. cos (y + a)
Pe.
√ Ra²+(wL)²
E₁
If we substitute i, for
and put
√ Ra² + (wL)²
we get
- e. cosy.sin (y+a)+e.sin y. cos (y + a) = — e. sin a
Pe. i. sin (y+a) — 2.sin a
e2
E₁
..(160).
In order to obtain this output graphically, the line OS in Fig. 329 is
314
Electrical Engineering
drawn so as to make an angle a with the horizontal, and is made equal to
e.io. On OS a semicircle is described. This semicircle has the same diameter
as that in Fig. 328, but the angle a is now above the horizontal. For the
length of any chord, making an angle y with the vector e, we have
OT= OS sin ẞ= e.i。.sin (a + y).
From OT we subtract a constant quantity XT, where
e². vo
XT=
.sin a,
E₁
and the remainder OX = OT — XT = P.

2
e
y
E,
오
​200
400
+
1 600
+
800
+ +
1000 1200 Volt
+ +
+
100
+
200
Amp
Fig. 330.
The length OX is therefore equal to the output P. The locus of all such
points as X is the heavy black curve, which, as before, differs but little from
a circle. The constants for Fig. 329 are as follows: Ra = 1, wL=6, e= 1,000
and E₁ = 1,200; a scale of kilowatts is added.
We proceed now to determine the current corresponding to any value of
the angle between E.M.F. and terminal pressure. The vertical OZ in Fig. 330
represents e and the vector OY the E.M.F. E. The line ZY corresponds to
the pressure OF in Fig. 325. Hence,
or
E
ZY=i.√Ra²+(wL)² = i.
20
го
i
ZY
E₁
…...(a).
Instead of measuring ZY in volts and multiplying it by i/E, we can
Alternators in parallel
315
make a new scale, as shown in the figure, on which the length ZY gives the
current directly. A point J is marked on the vector OY so that OJ is equal
to ZY. By repeating the construction for different values of y, and drawing
a curve through the points thus obtained, we get the locus i, which shows at
once the current corresponding to any value of y.
The three curves P1, P and i, from Figs. 328, 329 and 330, are shown
side by side in Fig. 331. We have already given the constants of the
machine, but repeat them here, together with other data necessary for the
construction of the diagrams.
Ra = 1, wL=6, e1,000, E₁ = 1,200,
tan α =
a
wł
Ra = 1 = 0·166,
Ra
0.166, sin a =
= 0·164,
6
io=
√ Ra²+(wL)²
2
√ Ra²+(@L)²
E₁
=
197 amperes.

20
E
P
P
50
100
150
200°
+
250 Kilowatt
+ +
+
100
200
Amp.
Fig. 331.
For the diameters OP and OS, we have
OPOS= e. i,
e.i。 =
197,000 watts,
QR E₁. i. sin a = 38,600 watts,
=
€2
XT:
i。. sin a =
27,000 watts.
E₁
The scales of pressure and power have been chosen arbitrarily. The
current scale is found by multiplying each figure on the pressure scale by
i。/E₁=0·164. Hence, the same length will represent either 200 amperes or
1,220 volts.
It is evident that each value of the displacement between E.M.F. and
terminal pressure corresponds to a definite current, total power and output.
316
Electrical Engineering
1
The part of the vector E, intercepted between the curves P and P₁ represents
the power lost in heating the armature winding. It increases with the angle
y, while the output increases slowly beyond a certain point and soon reaches
a maximum. To obtain a good efficiency it would be necessary to keep to
the left-hand part of the diagram.
It is interesting to notice that the power taken from the prime mover
and converted into electrical power increases as the angle y increases, until
y = 90° + a, when the power P₁ reaches its maximum value. This maximum
is seen from Fig. 328 to be equal to OP+QR, or
Pimaxe.io+ E₁io. sin a.
The increase of power with increasing displacement between E.M.F. and
terminal pressure is of the utmost importance for the parallel working of two
or more alternators, as we shall see in the following section.
105. Synchronising power of armature.
If several exactly similar alternators are working in parallel, and their
exciting currents are exactly equal, and, moreover, the power given to each
by its prime mover is the same, then Fig. 331 tells us that the displacement
between induced E. M.F. and terminal pressure must be exactly the same in each
machine. Now, as the terminal pressure is common to all the machines,
it follows that their electromotive forces must coincide in phase. This
means that an armature wire on one machine occupies, at any moment, the
same position relatively to the poles as the corresponding wire on any other
machine. The machines are then said to be running synchronously, by
which we mean that they are not only running at exactly the same speed,
but that they are also in phase at every moment, the E.M. F. curve of each
machine reaching its maximum at the same instant.
We are not concerned, for the present, with the way in which the
machines were brought into synchronism, but will first consider how this
agreement of speed and phase, when once obtained, is continually maintained.
The importance of this question is obvious. As in the parallel working
of direct-current dynamos, so here, the electromotive forces must be so con-
nected that terminals of like sign are connected together. If we consider
the individual electromotive forces, we see that each one is in opposition
to all the others. From this we see that the electromotive forces of all the
machines in parallel must have the same value and direction at every
moment, and must also reverse simultaneously. This can only be attained
by absolute equality of speed, or, if the machines have unequal number of
poles, by absolute equality of frequency.
Such equality of speed could never be obtained by external means. It is
fortunate, therefore, that the generators themselves exert a mutual influence
on one another, tending to maintain synchronism. The effect of this can be
seen from the fact that the connecting rods of two engines driving alter-
nators in parallel will remain perfectly in step hour after hour. The cause
of this can be seen by assuming that one of the generators, as a result of
Synchronising power of armature
317
some inequality in the turning moment of the engine, gets a little ahead.
This is equivalent to an increase in the angle y by which the E. M.F. vector of
that machine leads ahead of the vector of terminal pressure. The latter is
dependent on the other machines and will not be materially affected. This
increase of the angle y causes, as Fig. 331 shows, an increase of the output or
load on the machine. Since this load exceeds the mechanical power supplied
by the engine, the speed falls and brings the machine back to its normal
position, that is, the angle y is reduced to its normal value.
If, on the other hand, one of the generators lags slightly behind the other
machines, its load is decreased in accordance with Fig. 331. The mechanical
power supplied by the engine will then exceed the electrical load, and the
excess of power will accelerate the machine until its poles occupy the same
position, relatively to the armature, as the poles of the other machines. In
this way the alternators keep the engines in synchronism with mathematical
accuracy.
We have assumed that the machines are working on the left-hand part of
Fig. 331, where an increase of the angle y causes an increase of the electrical
power P₁. If, however, the machines are working in the neighbourhood of
their maximum power, an increase of y causes a very slight increase in the
load, so that the tendency to maintain synchronism is small. If we go still
further, an increase of y causes a decrease in the load, as is shown by the
dotted part of the curve P₁. As a consequence, the mechanical power
supplied to the alternator exceeds the electrical load by an ever-increasing
amount, which accelerates the machine more and more until it is 180° out of
phase, when it is evident that enormous currents will circulate between the
parallel machines and cause the whole station to shut down. It is necessary,
therefore, to keep to the left-hand side of the diagram, so that the inevitable
tendencies of the engines to get ahead or lag behind are counterbalanced by
the powerful synchronising forces of the alternators. The importance of this
in alternators with a large number of poles is evident when we consider that
a small mechanical displacement of the pole-wheel is equivalent to a large
displacement of phase.
In the foregoing we have looked upon the synchronising force as being
due to the increase or decrease of electrical output or load caused by the
leading or lagging of the pole-wheel. In practice it is more usual to speak
of the synchronising current, which circulates between the two machines,
adding itself to the normal current of the leading machine and subtracting
itself from that of the lagging machine. This current constitutes a load on
the leading machine, which is thereby retarded, while to the lagging machine
the current is a motoring current and causes it to accelerate.
It is evidently important that a small displacement should cause a large
change in the value of P₁, since this is the condition for a powerful syn-
chronising force. We might define the synchronising power of the machine
as the value of the differential quotient dP/dy. Now, in equation (159) on
page 312, we found that
P₁ = e. i. sin (y — a) + E₁. i。. sin a.
1.
318
Electrical Engineering
Hence,
dP,
1
= e. i cos (y — a).
dy
By means of this equation we can study the effect of armature self-
induction on the synchronising power. We shall consider the moment when
the machine is first switched into parallel, when, as we shall see in the next
section, the E.M. F. and terminal pressure are exactly in phase, so that y=0.
At this moment
io=
E₁
E₁
Ra
E₁
sin a,
√ Ra² + (wL)²¯¯ Ra˚ √ Ra² + (wL)²¯ Ra
so that, for y=0, we have
dP₁
dry
e. E₁
Ra
sin a. cos (− a) = E₁.
e. sin 2a
2. Ra
For given values of E₁, e and Ra this expression becomes a maximum
when a = 45°, i.e. when the reactance and resistance are equal. Although it
is not practicable to make the self-induction of the armature so small, it is
obvious that it should be made as small as possible. If, on the other hand,
@L=0, then a = 90° and sin 2a=0. The synchronising power of the arma-
ture disappears, therefore, with its self-induction, so that, although the latter
should be small, its presence is absolutely essential to the parallel running of
the machine.
106. The parallel connection of alternators.
The connection of an alternator to bus-bars on which other alternators
are already working is, in many respects, like the similar operation with a
direct-current generator. If a shunt dynamo has to be put in parallel with
other machines which are already running, it is run up to speed on open
circuit, and its field current adjusted until its E.M.F. is equal to the P.D.
between the bus-bars. The switch is then closed and the terminals of the
dynamo connected to the main terminals or bus-bars of the same sign.
With alternators the machine is started on open circuit and run up to
speed. The speed must, however, be carefully adjusted so that the frequency
is that of the bus-bars. The field current is then adjusted until the E.M.F. is
equal to the P.D. between the bus-bars. As in the case of direct current, the
switch must not be closed unless we are certain that the machine terminal
which is positive at the moment is switched on to the bus-bar which is
positive at that moment. The momentary value of the E.M.F. must also be
equal to the momentary value of the P.D. between the bus-bars. If these
conditions are fulfilled, the bus-bar P.D. will be exactly counterbalanced by
the equal and opposite E.M.F. of the machine, and no rush of current will
occur on closing the switch.
The principal difficulty in connecting an alternator in parallel with others
is to obtain this exact opposition of phase. It is almost impossible to main-
tain it for very long, and some indication is required as to the correct moment
for closing the switch. Glow-lamps are generally used for this purpose, as
shown in Figs. 332 and 333. In the first figure the lamps are connected
The parallel connection of alternators
319
across the terminals which will be connected by the switch. When the
speed of the machine is almost correct, the lamps will slowly alternate
between darkness and brightness. The switch should be closed during the
period of darkness, as the P.D. of the bus-bars and the E.M.F. of the machine
are then in opposition, as shown by the signs in Fig. 332. These signs refer,
of course, to one moment only.
In the arrangement shown in Fig. 333 the correct moment is indicated
by the lamps burning with their maximum brilliancy. The plus and minus
signs indicate the state of affairs at the correct moment for closing the
switch, and it is evident that, so far as the two lamps are concerned, the
bus-bars and the generator are connected in series, thus causing the lamps to
burn brightly.
Some experience is required to be able to close the switch at the correct
moment. It is evident that it is a far more difficult operation than switching
in a D.C. dynamo, since neither the frequency nor the phase can be determined
with mathematical accuracy. Such great accuracy, however, is not necessary,
+
+


+
Fig. 332.
+
Fig. 333.
as the synchronising power of the machine comes into play directly the switch
is closed, and almost immediately pulls the machine into synchronism.
So long as the machine remains unloaded the conditions are the same as
with direct current. The P.D. between the bus-bars and the E.M.F. of the
machine are equal in magnitude and in direct opposition, so that both the
current and the load on the machine are zero. The fundamental difference
between the parallel operation of D.C. and A.C. machines is seen, however,
when we consider how the newly connected machine can be made to take its
share of the load on the station.
We may consider a D.C. shunt dynamo running light, i.e. without delivering
current to the bus-bars, to which it, in common with several large dynamos,
is connected. To make the machine take a share of the load its field current
is strengthened, and thereby its E.M.F. increased. The armature current
increases in accordance with the equation
ia
E-e
Ra
320
Electrical Engineering
This increase of current causes an increase in the electrical power generated
by the dynamo. The former supply of mechanical power is not sufficient to
meet this new demand, and the machine slows down. As the engine slows
down, the balls of the governor fall and admit more steam, until the
mechanical power supplied by the engine is equal to the electrical power
or load E. ia*.
When this new state of equilibrium has been reached the machine runs
steadily at a speed somewhat lower than that at no-load. That the speed is
lower follows from the fact that the governor balls must occupy a lower
position in order to admit the increased steam supply. The fall of speed is
generally very small. We have seen, therefore, that the load on a shunt
dynamo is varied by means of the excitation. The dynamo reacts on the
prime mover and causes a small variation of speed, by means of which the
supply of power is varied between wide limits. It might be said that the
supply adjusts itself to suit the demand, i.e. the steam-engine supplies what
the dynamo demands from it.
In an alternating current system, however, the relations are quite
different. We shall assume that the machine under consideration is small
and that there are several large machines working on the same bus-bars, so
that the terminal pressure and speed of the large machines is not appreciably
affected by manipulations on the smaller one. We can therefore assume that
the terminal P.D. and the frequency are constant. The question is: How
can we load the small alternator which we have put in parallel with the
other machines? The first suggestion is naturally to increase the excitation
and with it the E.M.F. As we should expect, we obtain an increased current,
but, to our surprise, neither the wattmeter deflection nor the supply of steam
are increased. Since the power is unchanged in spite of increased current
and E.M.F., we conclude that the phase displacement between E, and i must
have increased.
On further examination, however, we see that it is impossible that the
load on the generator could be varied by means of the excitation. The
machine either runs at the synchronous speed or it falls out of step, so that
the increase of excitation has no effect on the speed of the engine or on the
position of the governor balls. The steam supply is the same as before, and
the power is consequently unchanged. The mechanical power transmitted
from engine to generator is the same as when running light before switching
into parallel, and is not affected by varying the excitation. To increase the
power we must shift the weight on the governor or adjust it in some way,
either by hand or by means of a small motor controlled from the switch-
board. By this means the steam supply is increased without any variation
from the constant speed corresponding to the frequency of the station.
Hence, the load on an alternator is varied by altering the mechanical power
supplied to it.
* As we have assumed the total load on the station to be constant, the load on any machine
can only increase by a simultaneous decrease occurring in the load on the other machines. The
only way this can occur, however, is by a variation in their terminal pressure. This change in e
is very small, and we have neglected it.
The parallel connection of alternators
321
The question naturally arises as to the manner in which this automatic
adjustment of the electrical output takes place with every change in the
supply of mechanical power. The first result of an increased steam supply is
an acceleration of the generator. This acceleration lasts only a moment, since
a slight forward movement of the E. M. F. vector in Fig. 328 increases the load
until it corresponds to the steam supply. By steadily increasing the output
of the steam engine, that of the generator can be increased until the machine
is working in the neighbourhood of its maximum power (Fig. 328), when
there is great danger of falling out of step.
There is a similar difference between the procedure to be followed when
disconnecting D.C. and A.C. machines from the bus-bars. Before disconnecting
a shunt dynamo its load must be removed by weakening its excitation. The
current from an alternator, on the other hand, is reduced to zero by manipu-
lating the governor or gradually reducing the steam supply.
From what we have just said it might appear that there is no need for an
automatic governor on the steam engine, since the speed of the engine is
fixed by the frequency of the station, that is, by the speed of the other
machines. This conclusion is quite right so far as one machine is concerned,
since its governor could be clamped without any change in the working of
the station. If, however, we desire the load on the station to be equally
shared between all the generators, even when the load is very variable, it is
evident that the governor on each machine must automatically regulate the
steam supply. A large load will then cause a general drop in the frequency
and a greater supply of steam to every machine. The frequency can be
brought up to its previous value by adjusting the steam supply to every engine.
•
Another interesting question deals with the effect of a change in the
excitation, since we have seen that it is without effect on the power. It is
evident that the terminal pressure of the station would be increased by
strengthening the field of every machine. What we wish to consider, how-
ever, is the effect on a machine of strengthening its field current without
changing the excitation of the other machines, and therefore without
materially altering the terminal pressure. This question will be considered
in the next section.
107. The effect of field regulation on alternators in parallel.
In this section we shall consider the changes which occur in an alternator,
working in parallel with a number of other alternators, when its field current
is varied. The terminal P.D. will be practically constant, and we shall assume
that the mechanical power supplied to the generator by the engine is also
constant.
If we take equation (159) on page 312, multiply each side by
e
E₁
20. sin a
and then add ), we get the following equation, in which we have put
2 sin α
cos (90°+y− a) instead of sin (y − a).
e
E₁₂² +
2
2 sin a)²
E₁.e
sin a
•
cos (90° + y − a) =
P₁. E₁
+
i. sin a
(2 sin a
e
~)* ... (161).
21
T. E.
322
Electrical Engineering
E,
Since is the internal impedance of the machine, and is therefore
20
1
constant, we see that the right-hand side of the equation is constant for all
excitations, for P, and e are both unaffected by the variable excitation. For
the sake of shortness we may represent the right-hand side of the equation
by R2. Now, an equation
R².
a² + b² – 2a. b cos 0 = c²
represents a triangle with sides a, b and c, of which the sides a and b enclose
an angle 0. Equation (161) is of this form, and can therefore be represented
by a triangle with sides E, and R, in which the two former sides
enclose an angle 90°+y-a.
e
2 sin a
This leads to a very simple diagram, from
which the value of the angle y between the E.M.F. and P.D. vectors can be

A
R
e
2 sin a
E
F<
J
E
400
800
+
1200
1600
2000
2400
2800
3200
3600 Volt
20
60 100
200
seen for any given value of E₁.
300 Amp.
Fig. 334.
From the point O in Fig. 334 a vertical line
OC is drawn equal to the terminal pressure e. Another line OA is drawn at
e
2 sin a
an angle a with the horizontal, equal to
The point A is half as high
as the point C. With A as centre, an arc is drawn with a radius equal to R,
i.e. equal to the root of the right-hand side of equation (161). A line OE is
drawn, making any given angle y with OC, and meeting the arc at the point
E. The triangle OAE must be the above-mentioned triangle, for its two
sides 'OA and OE enclose the angle 90°+y-a and its third side is R. The
side OE is consequently equal to the E.M.F. E, and it is ahead of the P.D.
- vector OC by the angle y. The point E moves on the circular arc and gives
the corresponding values of E, and y. The diagram is actually drawn for the
relations previously assumed, viz.
e = 1,000, Ra = 1,
wL=6,
and the constant power P₁, which, it will be remembered, includes the loss in
Field regulation of alternators in parallel
323
1
the armature winding, is taken as 100,000 watts. It is evident that, for the
same value of E₁, viz. 1,200 volts, Fig. 334 must give the same angle y as is
found in Fig. 328 for P₁ = 100,000 watts.
1
The effect of varying the excitation can be seen very plainly from the
diagram. By gradually decreasing the field current, the length of the vector E₁
is decreased, the point E moves down the arc towards B, and the angle y grows
larger and larger. This is exactly what we should expect from a mechanical
point of view. On suddenly decreasing the E.M. F. the electrical power is
momentarily reduced, with the result that the supply of mechanical power
is, for a moment, greater than that converted by the machine into electrical
power. This causes a slight acceleration and an increase of the angle y,
which brings P₁ up to its former value, corresponding to the unaltered supply
of mechanical power.
If the weakening of the excitation be carried far enough, the point E
arrives at B so that OB is in line with OA. This length OB represents the
smallest E.M. F. with which the machine can take an input of 100,000 watts,
or inversely, 100,000 watts is the largest power which the machine can take
for an E. M. F. equal to OB. If the excitation is decreased any further, or
the input from the engine increased, or even if the angle y is momentarily
increased owing to any slight irregularity, the machine will fall out of step.
As in Fig. 328, the angle y corresponding to this maximum power is equal to
90° + a.
For ordinary practical working it is evident that y must be small,
so that there is no danger of the machine falling out of step.
α
It is of great interest to know which value of the excitation would give
the most efficient working for the given input P₁. If we assume that the
¿ªRa loss in the armature is the only variable loss, the efficiency will be
a maximum when this loss is a minimum, that is, when the current is a
minimum. The drop of pressure in the armature, which is represented in
the figure by the line CE, is proportional to the armature current. This will
be clearer if we turn to Fig. 325, where the triangle BOF corresponds to the
triangle OCE in Fig. 334. As in equation (a) on page 314
in which
io
i
CE,
E
E₁
io
√ R²a + (wL)²
=
By substituting the values R = 1, wL= 6, and E, 1,200 we obtain the
scale of amperes in Fig. 334, viz.
i
197
1,200
•
CE = 0·164 CE.
The scale at the foot of the figure expresses the length CE directly in
amperes. For highest efficiency the vector CE must be a minimum, that
is, it must be in a straight line with AC. CD must therefore represent the
minimum current, which is found from the scale to be 916 amperes, the
corresponding E. M.F. OD being 1,220 volts. The loss in the armature winding
will be
22. Ra = 91.62.18,400 watts.
21-2
324
Electrical Engineering
This is a very excessive armature copper loss for a machine of 100 kilo-
watts input, but it has been so assumed for the sake of clearness in the
diagrams.
1
Since the input P, is constant, the minimum value of the losses will give
the maximum nett output P.
For any other value of the excitation, whether the point E lies above or
below D, the current will be larger, and the efficiency consequently lower.
It is evident, however, that the variation is very small in the neighbourhood
of the maximum, since the excitation can vary considerably on either side
of D without any great change in the current.
This latter circumstance is shown very clearly by plotting a curve with
the E.M.F. as abscissa and the corresponding current as ordinate. The curves
obtained in this way are known as the V-curves, because of their peculiar
shape. Two such curves are shown in Fig. 335, in which the abscissae are
made equal to the various values of OE in Fig. 334, while the ordinates
Amp.

200
100 KW
100
55 KW
400
800
1200
1600
2000 Volt
Fig. 335.
are equal to the corresponding values of CE. It is evident that there is
a certain excitation for each load, at which the current is a minimum. The
left-hand branch of the curve is that along which we are running the danger
of letting the machine fall out of step by decreasing its excitation beyond
the allowable limit.
It is evident from Fig. 334 that the E.M.F. could be varied over a wide
range without much effect on the current, if the circular arc were further to
the right, so that CE were longer. Hence, machines with a large internal
inductive drop i. wL are comparatively insensitive to changes of excitation.
This would be shown by flat V-curves.
If the inductive armature drop is small the V-curves will be more sharply
pointed. Although we cannot vary wL we can decrease the drop i. wĽ by
decreasing the current i, that is, by working on a smaller load. If Fig. 334
be drawn for a small input P₁, CE will be shorter and the arc will be nearer
to the vertical OC. The result is shown in Fig. 335 in the lower curve,
Field regulation of alternators in parallel
325
which has been drawn for a value of P₁ equal to 55 kilowatts. It is decidedly
more acute than the upper curve.
We must now consider the power-factor of the power supplied by the
generator to the bus-bars. The angle & between the terminal pressure and
the current can be found in Fig. 334 by drawing a line EF so as to make
an angle a with EC, and then dropping a perpendicular from O on to EF.
The triangle EOF is evidently identical with the triangle OBA in Fig. 325
and the angle COF is the required phase displacement . Moreover, we
have from Fig. 334
OA =
e
2 sin a
or
012
e
OA. sin a.
The triangle OAC is therefore isosceles, and AC makes an angle a with
the horizontal. The three following equations are evident from the geometry
of the figure
and
€ = 90° + a,
$ + y = 90° − (a + d),
180° = € + B + y +8.
By adding up each side we find that
$ = B.
Hence, the angle by which the current vector lags behind the vector of
terminal pressure is given by the angle between the lines CD and CE. If
the excitation be adjusted to give the E.M.F. OD, the angle ß vanishes and
the current is in phase with the terminal pressure. This agrees with the
previous calculation for this point, for which the input was 100,000 and
the loss 8,400, thus giving an output of 91,600 watts. Since the current
was 916 amperes and the P.D. between the terminals 1,000 volts, it follows
that the power-factor cos & must have been unity.
It might, at first sight, appear preferable to excite the machine so that
cos = 1 and the losses are a minimum. If, however, the external circuit
be at all inductive and this one machine be so excited that its current has no
wattless component, the wattless component of the external current must
be borne entirely by the other machines. It would evidently be a much
better arrangement to share the wattless current equally between the various
machines, or rather, so as to make the power-factor the same for every
machine.
If the excitation of our machine be greatly increased, the point E moves
further up the arc and the current lags far behind the terminal pressure
vector. The machine delivers a heavy lagging wattless current, which causes
the necessary armature drop E₁- e. On the other hand, by weakening the
field current, we cause the point E to move down towards B. The angle ß
or becomes negative, which means that the current vector leads ahead
of the terminal pressure. The electromotive force E₁, which is equal to OE,
becomes smaller than the terminal pressure e. This is due to the magnetising
effect of the leading armature current. This case can occur if the load
326
Electrical Engineering
consists of a synchronous motor, since by strongly exciting the motor, its back
E.M.F. can be made to exceed the E. M. F. of the generator. The terminal
pressure
of both machines will then have some value intermediate between
these electromotive forces, and consequently greater than the E.M. F. of the
generator.
The importance of the diagram in Fig. 334 is obvious when we consider
that it enables us to read off directly the electromotive forces, the currents
and the power-factors, and gives us, moreover, a clear insight into the
mechanical and electrical changes caused by varying the excitation.
108. Phase-swinging of alternators.
We shall consider an alternator connected to bus-bars, the P.D. and
frequency of which may be taken as constant. For a prime mover we shall
assume a turbine with a perfectly uniform turning-moment. If, owing to
any sudden variation in the load or in the excitation, the machine be
momentarily a trifle ahead of its normal position, the load P₁, which is
practically proportional to the angle y, will exceed the supply of mechanical
power, and the machine will be retarded. It will not stop when it arrives at
its normal position, but will swing past it like a pendulum. It will then be
behind its normal position, and the corresponding value of P₁ will be less
than the supply of mechanical power. The machine will therefore be ac-
celerated and will pass again through its normal position. It is evident that
the machine has an oscillation superposed upon the uniform rotation, and
this oscillation is gradually damped out by the eddy currents which are
caused by it.
1
Now the rate at which the power or load P, increases with the angle y
was seen on page 318 to be e. i. cos (y - a) per radian. In actual practice
the angle a is very small, and the angle y does not exceed 10° or 20°. If
the latter were larger it would either mean that the self-induction of the
armature was big or that the overload capacity of the machine was small.
Very little error will be introduced by putting cos (y-a)=1. The torque
is given by dividing the power by the mechanical angular velocity 2π.
Again, a mechanical displacement of 1 radian represents an electrical phase
displacement of p radians, and, finally, 1 metre-kg. per second is equal to
9.81 joules per second, i.e. 9.81 watts. With the aid of these relations we
find for the controlling torque per radian of mechanical displacement
Ρ
M₁ =
Mt p.e.i
p². e. i.
9.81.2π.
9.81.2π.
p
met.-kgs................(a).
If the moment of inertia of the revolving part of the machine be
I kg.-met.², then we know from mechanics that the time of a complete
oscillation will be given by the equation
t = 2π
I
second
9.81Mt
..(b).
Phase-swinging of alternators
327
t
=
If, for example, I = 53,300, p = 32, ~ = 48·5, e = 3,000, i, 670, the time
of a complete oscillation works out at t=0.55 second. Hence, the machine
would swing through its neutral position about four times per second, twice
going forward and twice moving backward.
We turn now from these natural oscillations of the alternator to the forced
oscillations caused by the inequality of turning-moment of any reciprocating
engine. To simplify the matter we may imagine
the alternator to be replaced by a D.C. dynamo.
We shall let t represent the time which elapses
between two successive maxima of the turning-
moment. The turning-moment diagram will be
irregular but may be represented approximately
by a sine curve, as shown in Fig. 336a, in which
the ordinates measured from EH represent the
total torque of the engine, while, by measuring
from OD, we get the amount by which the torque
differs from its mean value. If M₁ = AG, the
maximum excess turning-moment, then we have
for the maximum angular acceleration

Torque
G
D
10
A B
C
E
H
a
b
A
B
C
ω:
C
10
A
B
d
A
B
E
C
а₁ = 9.81
Ar
Mt
I
F₂
2
e
0
A B
C
==
Fig. 336.
This is the rate at which the angular velocity
of the dynamo or flywheel is changing at any
moment. It is a maximum when the turning
moment is a maximum, as shown in Fig. 3366. So long as the acceleration is
positive the angular velocity will increase, and will therefore attain its maxi-
mum value at the moment B (Fig. 336c). At this moment it will exceed the
normal angular velocity by an amount w₁. To find w, we must multiply
the average value of the acceleration by the time, since the velocity was
normal at A. The mean value of a sine function is equal to its maximum
value multiplied by 2/π, so that
t₁
2 t₁ 2.9·81 t₁. Mt,
•
π 4
4.77
I
……..(c).
So long as the velocity is above the normal, i.e. while w is positive, the
forward displacement of the flywheel increases. It will reach its maximum
value of e, radians at the moment C. To find the value of e, we must multiply
the average value of a by the time t₁/4, thus
2
€1
W1
П
4
t₁ 9.81 t2. Mt.
4π72 I
……….(d).
This is due entirely to the steam engine, and is true for any load,
electrical or otherwise. We must now consider the effect of the alternator
on the angular displacement. We shall assume that the engine forces the
alternator to oscillate with the frequency determined by the turning-moment
diagram. When the displacement is forward the alternator load is increased,
and when the displacement is backward the load is decreased, which is equi-
328
Electrical Engineering
valent to an accelerating force exerted on the machine from the bus-bars.
Now, by comparing Figs. 336 a and d, we find that the machine has its
extreme backward displacement at the moment when the turning-moment
of the engine is a maximum. At the same moment the accelerating torque
supplied from the bus-bars is a maximum, so that these two forces are exactly
in phase, having a forward maximum when the machine is in its most back-
ward position and a backward or retarding maximum when the machine is in
its extreme forward position. Every change of displacement from the normal
position corresponds to a change in the power supplied to the bus-bars, and
these variations may be so considerable as to render parallel operation im-
possible.
Since we have just seen that the variation of engine torque from the
normal is always such as to tend to restore the machine to its normal position,
and is exactly in phase with the electrical synchronising torque due to the
displacement, these two effects can be algebraically added. If the maximum
values of the two torques be M₁, and Mt,, they can be added together to give
a resultant torque with a maximum value Mt,+Mt. The displacement e,
found in equation (d) will be increased by the addition of Mt, to a value e, in
accordance with the equation
2
€
4.π²°
9·81 t₂². (Mt,+ Mt₂)
I
………….(e).
The value of the electrical synchronising torque Mt, increases with the
displacement. When the oscillations have become uniform, the displacement e
will give a value of Mt,, which, together with the mechanical synchronising
torque Mt,, is just sufficient to produce the displacement e. The value of
Mt, increases with the displacement, and causes the latter to increase gradually
from €₁, as found from equation (d), to e, as given by equation (e).
In equation (a) we found the value Mt of the synchronising torque per
radian displacement, and the value of Mt, in equation (e) will therefore be
equal to e. Mt.
From equations (b) and (e) we have
Mtz
Mt,+Mt₂ t2
t₁2
We shall now consider three different cases.
Case 1. When the natural period of oscillation of the alternator t is
greater than the period t, of the forced oscillations due to the engine. This
is the usual case in practice. We see from equation (f) that Mt, + Mt, will
then be greater than Mt. This agrees with Fig. 336 a and d, in which
we saw that the maximum torque of the engine coincided with a light load
on the generator, thus apparently magnifying the excess of torque.
The ratio in which the amplitude of the oscillations is increased by the
synchronising effects of parallel running will be
€1
€ Mt, + Mt.
Mt₁
t2
1
t² — t₁²
1-
t2
……..(9).
·(g).
The increase in the amplitude is evidently smaller when the natural
Phase-swinging of alternators
329
period is long compared with the forced period ħ. The engine in the
example on page 327 was a tandem compound engine with a single crank.
85.2
60
It ran at 85 r. p. m., so that there were
impulses per second,
or
tr
60
85.2
0.353 second.
We found the natural period t to be 0.55, which gives an increase of
amplitude of
Ε
1
1.7.
€1
1
(0·353)
Case 2. If the natural period t of the alternator is exactly equal to the
forced period t₁, equation (ƒ) shows that the initial displacement due to
the engine alone will be negligible compared with the increase of amplitude
due to the synchronising forces. Equation (g) leads to the same result, viz.
that e/e, will be infinitely big. This is a case of resonance, and the oscillation
rapidly becomes so large that the machine falls out of step. This danger
must be avoided by making the natural and forced frequencies decidedly
unequal.
Case 3. This case, in which the natural period t is smaller than the
forced period t₁, does not occur, as a rule, in generators, but is met with
in synchronous motors and rotary converters. It can only occur when the
rotating masses are very small. Although there is no engine to cause
oscillations in the converter, these may arise owing to a periodic variation
of the load. Turning to equation (ƒ)
Turning to equation (ƒ) we see that, for t to be smaller than
t₁, it is necessary for the denominator on the left-hand side to have a minus
sign. The resultant synchronising torque is therefore Mt, - Mt,, which means
that the engine has its smallest, and not its largest, torque when the gene-
rator has its extreme backward displacement. The forward acceleration which
exists at this moment is due to the synchronising power of the armature,
which is more than sufficient to counterbalance the deficiency of torque from
the engine.
The ratio in which the amplitude of the oscillations is increased by the
synchronising effects of parallel running will be in this case
€ Mt₂ - Mt,
Mt₁
€1
t²
2
t₁² - ť²
1
1
t2
t₂²
As in the first case, so here, we see the advantage of making the natural
period t differ as much as possible from the period t, of the forced oscillations.
In the present case this difference is increased by decreasing the moment of
inertia of the revolving system.
In the case of generators, however, the natural period t is generally
greater than the forced period t, of the engine impulses, so that the parallel
operation is only made possible in many cases by increasing the value of t.
There are two ways in which this can be done. If choking coils are con-
nected in the leads between the machine and the bus-bars, the short-circuit
330
Electrical Engineering
current i, is reduced. This reduces the synchronising torque per radian M,
(equation a) and consequently increases the period t (equation b). The
introduction of choking coils is often the only way in which rotary converters
can be prevented from hunting. The second way of increasing t is by
increasing the rotating masses. This is a very effective method, since it
reduces both the amplitude due to the engine alone and also the ratio in
which this amplitude is increased by the synchronising forces.
A third method of reducing the amplitude of the oscillations is due to
Leblanc, and consists in the utilisation of the damping action of the eddy
currents produced by the oscillation. To increase the eddy currents the pole-
shoes may be solid, or, if laminated, may have copper rods embedded in holes
running axially through the pole-shoes, and connected to copper plates
flanking the pole-shoes on either side. Each pole-shoe has usually about
half a dozen of these copper bars, and the whole forms a short-circuited
squirrel cage winding. When the poles swing, currents are induced in this
so-called amortisseur winding due to its cutting the field produced by the
We know from Lenz's law that these induced currents
will tend to stop the motion producing them, that is, they will tend to damp
out the oscillations. This device has the further advantage of keeping the
machine, say a rotary converter, in step on a sudden heavy overload of
momentary duration. The damping winding acts as an induction motor on
the sudden retardation of the machine, slowing it down and diminishing its
amplitude.
armature currents.
The comparison of the damping winding with an induction motor enables
us to see, however, why the device is not always effective. The braking
action is not due entirely to energy dissipated in the damping coils, but we
have an actual induction machine, sometimes running below synchronous
speed and taking power from the bus-bars to drive it as a motor, and at
other times running as a generator above synchronous speed and supplying
electrical power to the bus-bars. The maximum value Ma of the damping
moment will occur when the machine is swinging through its neutral
position, for it is then that the oscillating velocity is a maximum. It will
be proportional to this velocity, and, since it always acts against the direction
of motion, it will have a negative maximum at the moment B in Fig. 336 c.
It is always 90° ahead of the synchronising torque Mt. The resultant total
torque will be the resultant of the three torques Mt,, Mt, and Mɑ.
To determine the effect of the damping we shall assume that we have
found the maximum excess torque Mt, of the engine and the maximum
synchronising torque Mt, of the generator, neglecting the damping. If we
represent these by the lines AB and OA in Fig. 337, then the line
OB = 0A + AB = Mt₂+ Mt, will represent the total maximum torque for the
general case where the two can be added. When the effect of damping is
added, let us assume that the excess torque of the engine Mt, is altered so
that the resultant maximum torque is still OB, in spite of the damping.
This will cause an oscillation having a certain velocity as the machine swings
through the normal position, and, corresponding to this velocity, there will
be a damping torque Ma, which is represented by OD 90° ahead of Mt¸·
Phase-swinging of alternators
331
D
Since we have assumed that the engine torque Mt, is so adjusted that the
resultant torque, and therefore the resultant amplitude, is the same with the
damping as it was without, the line OA, which repre-
sents M₁₂, will remain unaltered. The excess torque
Mt, of the engine must, therefore, be so altered, that
OB is the resultant of OA, OD and Mt. Hence, to
maintain the oscillation at its previous amplitude in
spite of the damping, M₁, must be altered from AB to
OF. The excess engine torque OF calls into existence
the synchronising torque OA and the damping torque
OD, and the three together produce an oscillation cor-
responding to the resultant OB.

Fig. 337.
Now, the actual excess engine torque Mt, is only
equal to AB, so that the whole figure must be reduced
in the ratio of AB to OF. When the machine was undamped the line OA
represented to a certain scale the variation of power supplied to the bus-bars.
This will now be represented by OE, the resultant of OA and OD, since both
these torques are produced electrically. In this case, however, OE must be
reduced in the above ratio, so that the relation between the variation of
power with and without the damping will be
AB
OE.
OA
OF OE AB
EB' OA
It is evident from the figure that OE : EB is nearer unity than AB:0A,
so that the latter ratio determines whether the variation of power will be
decreased or increased by the damping. If AB is smaller than OA, then the
damping is beneficial. In this case the excess engine torque is smaller than
the maximum synchronising torque, i. e.
t1
Mt₁< Mt₂ or
Mt₂+ Mt₂ > 2.
Mti
Mt, +
Mt₂
We saw in equation (g) that
is equal to the ratio in which
€
Mtı
€1
the amplitude of the oscillation is increased by the effect of parallel running,
and that, for this to be larger than 2, the moment of inertia of the moving
parts must be small. The effect of damping will consequently be very
advantageous in such cases as rotary converters, where there is no flywheel,
apart from the armature of the machine itself, and generators driven by two-
or three-crank engines with four or six impulses per revolution, in which
case the flywheel can be made very small on account of the uniform turning-
moment.
CHAPTER XV.
109. The principle of the synchronous motor.-110. Synchronous motor with constant P.D. and
excitation.-111. Synchronous motor with constant load and variable excitation.
109. The principle of the synchronous motor.
There are three headings under which all alternating current motors can
be roughly classified. The synchronous motor, which we shall consider in
the present chapter, runs at exactly the same speed whether loaded or
unloaded; the asynchronous or induction motor slows down slightly with
increasing load; the commutator motor is similar to the D. C. motor and has
similar characteristics. Synchronous motors are generally polyphase, induc-
tion motors either single phase or polyphase, and commutator motors
generally single phase.
Any alternating current generator will run as a synchronous motor,
whether single phase or polyphase. In Fig. 338 is shown the north pole of


N
N
Generator
Fig. 338.
Motor
Fig. 339.
a generator, which is rotated clockwise, inducing thereby an E.M.F. in the
armature wire shown. If the external circuit is closed, the wire will carry a
current which, neglecting phase displacement, will flow away from the reader.
The armature current and magnetic field will produce a torque which must
be overcome by the mechanical torque exerted by the engine. It follows
from this that, if the field be excited as before and alternating current be
supplied to the armature, the machine will rotate as a motor in the opposite
direction.
A moment later the north pole in front of the wire will be replaced by a
south pole. In order that the torque may still act in the same direction as
before, the current must have reversed its direction during the interval. It
is therefore necessary that the speed of the pole-wheel be in exact agreement
The principle of the synchronous motor
333
with the frequency of the alternating current, even at the moment of switch-
ing on the motor. If the motor have p pairs of poles and rotate at
lutions per second, then
N
60
revo-
n
1
60
p
a.C.
where~, is the frequency of the A. C. supply. The synchronous motor must
be brought up to the synchronous speed by some external means before the
armature is connected to the mains. There are several ways of doing this.
If the field current is supplied by a direct coupled D.C. dynamo, this can be
used as a motor and driven by a battery or other D. C. source if available.
This method is also available if the synchronous motor is used for driving a
D.C. dynamo. Sometimes a small induction motor is supplied specially for
starting up the synchronous motor. Another method, largely adopted in the
United States, is to open the field circuit, and supply the armature with an
A.C. at half or a quarter of the normal pressure by means of a transformer.
The machine is brought up to speed by means of the eddy currents induced
in the pole-shoes by the rotating field, as we shall better understand after
studying the induction motor. This latter method is only applicable to
polyphase motors. The necessity of bringing the synchronous motor up to
speed in some way or other, before switching it on, prohibits its employment
in a great number of cases.
The operation of switching in the motor when it has reached synchronous
speed is exactly the same as connecting an alternator in parallel with others.
To make this clear, we may assume that it is quite undecided whether to use
the machine as a generator or as a motor. Such an assumption is quite practi-
cable if the motor drives a D.C. generator connected across the terminals of a
battery of accumulators. If the excitation of the dynamo be increased until its
E.M.F. exceed that of the battery, the battery will be charged, and the D.C.
machine will work as a generator. If, on the other hand, the excitation be
weakened until the E. M.F. falls below that of the battery, the current will be
reversed, and the battery will drive the D.C. machine as a motor.
To connect the A. C. machine to the A.C. supply, we follow the procedure
of Section 106, and bring the machine up to the correct speed by means of
the D.C. machine acting as a motor. The field current of the A.C. machine is
then adjusted until its E. M. F. is equal to the supply pressure. The switch
must be closed at the moment when the E.M. F. of the generator is exactly
opposed to the P.D. of the mains, as indicated by the phase-lamps. If the
moment is correctly chosen the main ammeter will not be affected, because
of the exact opposition of E.M.F. and supply pressure. The power required
to run the machine light will still be supplied by the D.C. machine.
To make the A.C. machine act as a generator its supply of mechanical
power must be increased. This is accomplished by weakening the field
current of the D.C. machine. As a general rule, the speed of a D.C. motor is
increased by weakening its field. In the present case, however, the speed is
fixed by the frequency of the A. C. supply, and a weakening of the field must
334
Electrical Engineering
e E
Ra
result in a lowering of the E.M.F. This will enable the battery to send a
larger current through the armature, in accordance with the equation
ia
This larger current produces an increased torque, since the
weakening of the field is very small compared with the resulting increase of
current. The acceleration produced by this increase of torque brings the
poles and armature winding of the A.C. generator into such a relative position
that the electrical load or output is gradually increased, as shown in Section
104, until the increased mechanical torque is exactly balanced. The A.C.
machine has thus been made to act as a generator by accelerating it, or
rather, attempting to accelerate it, by an external mechanical torque.
We shall now reverse the procedure, that is, we shall strengthen the field
of the D.C. machine, instead of weakening it. Since the speed is fixed, the
effect of strengthening the field is to increase the E.M. F. until, when it
exceeds the E.M.F. of the battery, the current reverses and the cells are
charged by the D.C. machine acting as a dynamo. The A.C. machine must
now be acting as a synchronous motor. The power supplied by the battery
to the motor-generator set was first positive, then nil, and, finally, negative.
This must cause a retardation or lagging of the machines, which brings the
poles and armature windings of the alternator into such a relative position
that the torque exerted by the alternator corresponds to the load on the D.C.
dynamo. The A.C. machine has, therefore, been caused to act as a motor,
simply by the retardation, or lagging behind, of the rotating part. In the
next section we shall endeavour to make the action clearer by means of
vector diagrams.
110. Synchronous motor with constant P.D. and excitation.
The vector diagram of the synchronous motor is very similar to that of
the A.C. generator. In drawing the diagram (Fig. 340) we must consider the
induced E. M.F. E as being a back E. M.F., acting more or less in the opposite
direction to the current and applied terminal pressure e. The vectors e=OK
and E = OB have a resultant OF, which must be the pressure used in driving
the current through the armature against its resistance and self-induction.
The resultant OF may therefore be resolved into two components, one OD to
overcome the resistance, and the other OC to counterbalance the effect of
self-induction. As before, we shall designate the angle by which OF lags
behind OC by the letter a. The current i must be in phase with the com-
ponent OD, and lags therefore behind the terminal pressure e by the angle 4.
From the fundamental principles of a motor we know that the current must
be more or less opposed to the back E.M. F., which we see is the case in
Fig. 340. It is evident, moreover, that at no-load, that is, on first closing the
switch, the vector E will lie along the dotted line, since there is no current
and the resultant of E and e must consequently vanish. As the motor is
mechanically loaded, it will lag more and more behind its neutral position,
and the vector E will lag behind the dotted vector. With the load corre-
Synchronous motor with constant P.D. and excitation 335
K
sponding to Fig. 340 this angle of lag is y. (Compare the closing sentences
of the previous section.)
The power supplied to the motor is given by the equation
P₁e.i. cos o,
where is the angle between the terminal pressure and the current. From
the figure we see that FH = e cos o, so that
P₁=i. FH watts.
1
A certain amount of this power will be lost in heating the armature
winding, viz.
H
i². Ra= i. iRa = i. FC watts.


e
α
(e)
У
F
"
i
D
G
C
A
i
Fig. 340.
Motor
Generator
Fig. 341.
1
The mechanical output of the motor, including, however, friction and iron
losses in the motor itself, will be given by the equation
P = P₁ — ¿² . Ra = i (FH – FC)=i.CH.
1
If ₁ be the angle between the vector of E.M.F. and the current vector
produced backwards, we have
CH= E. cos $1.
This gives the following equation for the mechanical output of the motor:
.(162).
PE.i.cos $1
We might almost have expected this result from our former consideration
of the D.C. motor.
On comparing the triangle FBG with the triangle OBA in Fig. 325 a
great similarity will be noticed. For the sake of comparison the vector
diagram of the generator is drawn again in Fig. 341. The only difference
336
Electrical Engineering
1
between Figs. 340 and 341 is the interchange of E and e. The total power
P₁ supplied to the motor in Fig. 340 is FB.i. cos p, while the total power P,
generated electrically by the alternator in Fig. 341 is OB.i. cos. These
are evidently equal. Now, on page 312 we established the equation (159) for
the value of P, for a generator, and it is evident that it can be made applic-
able to a motor by putting E for e and e for E, giving us
1
P₁ = E. i. sin (y − a)+e. i. sin a....
...(163),
where P, is the power given to the motor, and the short-circuit current i is
given by the equation
e
20
α
√ Ra² + (wL)²
2
In a similar manner we see that the output P of the motor in Fig. 340 is
equal to OB.i.cos p₁, which is exactly equal to the output P of the generator,
after deducting the i2. Ra loss in the armature, viz. FB. i. cos in Fig. 341.

E
P
∞
100
-e
Fig. 342.
Thus, the equation (160) found on page 313 for the generator will be true
for a motor if E and e be interchanged. We have therefore
E2
P = E. i. sin (y + a) – i.. sin a
e
..(164).
y is here the angle by which the E.M.F. lags behind the backward pro-
jection of the terminal pressure vector.
We must now draw curves to show graphically how the power varies with
the angle y. As in Figs. 328 and 329, so here in Figs. 342 and 343, circles
are drawn passing through O and having their diameters inclined to the
horizontal by the angle a. In Fig. 342 the diameter is set out below the
horizontal, and in Fig. 343, above it. Its length in each case is E. i. Every
chord drawn from the point 0 to any other point on the circle, is produced
beyond the circle by an amount e.i。. sin a. The heavy curve P₁, obtained
in this way, is the locus of the extremities of lines drawn from 0 to represent
the total power supplied to the motor for the corresponding value of y. The
vector of the terminal pressure e will be vertically downwards from O, and y
is the angle between E and -e, as shown in the figure.
Just as Fig. 342 represents equation (163) so Fig. 343 represents
Synchronous motor with constant P.D. and excitation 337
E².i。.sin a
equation (164). In this case, however, a portion equal to e
is subtracted from every chord, giving a length representing to a certain
scale the output of the motor (including friction etc.).
The curves are drawn for the following values: Ra=1, wL=6, e = 1,200
and E=1,000. The output is evidently a maximum when y = 90-a. If the
load be still further increased, the angle y becomes larger but the output of
the motor is thereby decreased, and the motor falls out of step. Since this
would be equivalent to a short circuit of the supply mains it must be care-
fully avoided by working on the right-hand side of the diagram in Fig. 343.
If the normal full load causes y to be about 30°, the motor will have about
100 per cent. overload capacity.
In order that a given value of P may correspond to a position on the
right-hand side of the figure, it is necessary for the circle, by means of which
E

x
J
+
50
100
150
Fig. 343.
+
200 Kilowatt
the figure is constructed, to have a large diameter. Hence E. must be
large, since this is the diameter of the circle. The short-circuit current i
can be made large by keeping the self-induction of the armature as small as
possible.
What was said with regard to the synchronising power and phase swinging
of alternators is equally applicable to synchronous motors. We now pass to
the consideration of a motor on constant load but the excitation of which is
varied.
111. Synchronous motor with constant load and variable
excitation.
We shall now consider the effect of varying the excitation of a synchronous
motor which is connected to mains with a constant P.D. The load on the
motor, i.e. its mechanical output, is constant. If we take equation (164) on
e
e
2
page 336, multiply each side by
>
20. sin a
and then add (
2. sin a),
we get
T. E.
22
338
Electrical Engineering
the following equation, in which we have put cos [90° — (a + y)] instead of
sin (a + y),
2
E2+
(2. sin a)
e
E.e
sin a
•
−
e
cos [90° − (a + y)] =
(z
2 sin a
P.e
20. sin a
...(165).
e
e
2 sin a
Since is the internal impedance of the machine, and is therefore
20
constant, we see that the right-hand side of the equation is constant for all
excitations, and can, for the sake of shortness, be represented by R² where R
is some constant. As we saw on page 322, this equation represents a triangle
with sides E,
and R, in which the two former sides enclose an angle
90° - (a + y).
This triangle is drawn in Fig. 344. From the point 0 a
perpendicular OC is drawn equal to the terminal pressure e. On this line as
base an isosceles triangle OAC is constructed with each side equal to 2. sin a
and the angle at ▲ equal to 2a. With A as centre and radius equal to R, an
200
Ꭱ
F<
T
e

C
e
2sin x
B
α
0
2000 2400 2800 3200
400
800
1200
1600
Fig. 344.
3600 Volt
arc DEB is struck. The triangle corresponding to equation (165) will have
OA for a base, while its apex moves along the arc DEB. For any point such
as E, the line OE represents the back E. M.F., making the angle y with the line
OC, which is the backward projection of the terminal pressure vector. The
figure is drawn for the values formerly assumed, viz. Ra=1, wL 6,
e=1,200 and E=1,000, while the constant output P is 50,000 watts. As
the excitation is weakened, OE gets smaller by the point E moving down the
arc towards the critical point B. This E. M. F. OB is the minimum, that is to
say, this is the weakest excitation with which the motor can give an output
of 50,000 watts. If we reconstructed Fig. 343 for an E.M.F. equal to OB
instead of 1000 volts we should find the maximum value of P, corresponding
to y = 90° — a, to be 50,000 watts.
In addition to the angle y we can find the angle between terminal
pressure and current from the diagram in Fig. 344. A line CF is drawn
making the angle a with CE, and a perpendicular OF is dropped upon it
Synchronous motor with variable excitation
339
from 0. It is seen at once that the triangle OFC corresponds exactly to the
triangle BGF in Fig. 340. The angle COF is therefore equal to the phase
displacement p.
Adding up the three angles of the isosceles triangle AOC, we get
180° = 2x+(B+8) + (90° − a),
while, in the right-angled triangle OFC,
$ = 90 − (a + 8).
By adding these two equations, we have
$ = B.
When the point E coincides with D the angle ẞ vanishes, and the current
is consequently in phase with the terminal pressure. Seeing that the load is
constant the current must have its minimum value at this point, which agrees
with the fact that the armature drop CE will then have its minimum
value CD.
If the point E is below D, the current vector OF lags behind the terminal
pressure, and the angle ẞ represents the angle of lag ø. This is the case
whenever the E.M.F. is less than the terminal pressure.
By increasing the excitation we can make the E.M.F. greater than the
terminal pressure, which is an absolute impossibility in a D.C. motor, since it
thereby becomes a generator. If OE exceeds OD, the angle ẞ becomes
negative, that is, the current vector leads ahead of the terminal pressure.
We have previously mentioned that synchronous motors, when over excited
in this way, can be advantageously employed to neutralise the lagging current
taken by other apparatus. In this way the load on the generators in the
power station can be made to have unit power-factor.
|
22-2
CHAPTER XVI.
112. The rotary field in a two-phase motor.-113. The rotary field in a three-phase motor.-
114. Mesh or delta connection.-115. Star connection.-116. The measurement of power
in polyphase circuits.-117. General principles of the rotor.
112. The rotary field in a two-phase motor.
The armatures of machines which generate two-phase alternating current
contain two separate windings, or sets of coils, displaced from each other by
90°, the pole-pitch being considered 180°. When the coil-sides of one
winding are situated under the middle of the poles, those of the other
winding are midway between the poles (Figs. 345 and 346). The two
windings differ merely in their instantaneous position, i.e. in their phase.
For this reason the windings are generally spoken of as the phases of the
generator, and we shall often use the expression in this sense.
N
N
A


S
Fig. 345.
B
S
Fig. 346.
The drum-winding in Fig. 345 is quite simple. The two-phase ring-
winding in Fig. 346 is wound in such a manner that the end connections,
after one coil is completed, pass right across the armature, as in the drum-
winding. The practical identity of the two windings is evident. The wires
on the outside of the ring at A and B are equivalent to the wires on the
periphery of the drum, and the electromotive forces in all the wires of one
phase can be added together.
The currents produced by the generator can be used to magnetise the
The rotary field in a two-phase motor
341
stationary iron ring of an induction motor. This ring, which is called the
stator, is built up of sheet-iron stampings, and is wound in exactly the same
manner as the armature of the generator. The ends of the stator-windings
are connected to the corresponding ends of the generator-windings (Fig. 347).
If the armature of the generator rotates, the connections must, of course, be
made by means of slip-rings.
If the armature be rotated in the direction indicated, i.e. clockwise, then
an E.M.F. towards the observer will be induced in the wires under the south
pole, and away from the observer in those wires which are under the north
pole. This is shown by the arrows in the figure. If we neglect any phase
displacement due to self-induction, the arrows will represent current at its
maximum value. At the moment chosen in the figure there will be no
current in the other phase-winding. By tracing out the currents in the
stator, we find that the field produced will be as indicated by the dotted lines.
محمد

N
Generator
4
-S-
2
3
Motor
Fig. 347.
The lines of force leave the stator at N, pass across the internal air space, or
through an iron cylinder if one be placed within the ring, and enter the stator
again at S. The internal surface of the ring will have a north pole at N and
a south pole at S.
After the armature of the generator has turned through 45° the conditions
will be as shown in Fig. 348. The stator has not moved, but the armature
coils of both phases of the generator are now equally induced. The field
near the pole-tips will not be so strong as at the centre of the pole, and,
moreover, the armature coils are only partly under the poles. The current
will therefore be smaller than that in Fig. 347. We may assume that the
current varies according to the sine law, so that its value at the moment
indicated in Fig. 348 will be
i = imax. sin 45° = 0.707 imax.
==
The smaller current is indicated in the figure by lighter arrows.
If we trace out the current in the windings of the stator, we see that
every coil is carrying current and that, although coils 1 and 2 belong to
different phases, they are virtually equivalent to a single coil of double the
342
Electrical Engineering
number of turns. The current has the same direction in both, viz. outwards
in the end connections facing us. Similarly coils 3 and 4 may be looked upon
as a single coil, more especially as in actual motors the whole periphery is
utilised and no gaps are left between the windings.
The magnetic field produced by these stator currents is shown by the
dotted lines, the inner surface of the ring has still a north and a south pole,
between which the flux crosses the internal space. The position of these
poles has rotated through 45° from the position they occupied in Fig. 347.
On turning the generator through another 45°, the stator coils 1 and 3 will
have no current, while the current in coils 2 and 4 will have reached its
maximum. There will then be a north pole at 1 and a south pole at 3.
We have thus arrived at the surprising result that the rotation of the
armature of the generator causes the field of the motor to rotate, or causes
the poles of the stator to rotate around the stationary iron ring. We can
therefore imagine the stator to be replaced by a rotating field-magnet system
S

1
N
Generator
Fig. 348.
4
N
3
Motor
S
2
with salient poles pointing inwards. The flux of such a field-magnet would
continually cut through an iron cylinder placed within the stator ring (compare
Fig. 367).
This iron cylinder is called the rotor. It, also, is built up of sheet-iron
stampings, carried on a central spindle or shaft. This shaft is supported at
either end in bearings, so that the rotor can turn within the stator, with a
small air gap between the two. A short-circuited winding is contained in
slots in the outer surface of the rotor. Currents are induced in this short-
circuited winding by the rotating field produced by the stator, and these
currents, as we know from Lenz's law, will tend to oppose the relative motion
between field and conductors. The result is that the rotor is dragged round
by the rotating field. The direction of rotation of the motor is therefore the
same as that of the rotating field, and this depends on the way the stator
phases are connected up to the generator. If the two connections to one of
the phases be interchanged, it can be easily seen that the direction of rotation
will be reversed.
The rotary field in a three-phase motor
343
113. The rotary field in a three-phase motor.
The armature of a three-phase generator contains three separate windings,
displaced from each other by 120° (Fig. 349). Care must be taken, when
drawing the winding, that the beginnings 1, 2, and 3, of the three phases are
120° apart, and that, starting from these points, each phase-winding passes
round the armature in exactly the same direction. In Fig. 349, for example,
each winding commences by passing inwardly across the front end of the ring.
These windings are connected by means of slip-rings with the corresponding
terminals of the stator of the motor. The fact that the generator has been
numbered clockwise and the motor in the opposite direction is immaterial
and has been adopted to avoid crossing the connections.
Phase 1, 1' is shown in Fig. 349 under the middle of the pole and, if
phase displacement due to self-induction be neglected, the current in this
phase will have its maximum value. The instantaneous value of the current
in the other phases will be
Ꮪ
i = imax. sin 30° 0'5 'max
=

3
N
Generator
2
2
3.
Motor
24
3
Fig. 349.
By tracing out the current in the various coils of the stator, it is seen that
the coil-sides 3', 1 and 2' carry current in the same direction, and may be
grouped together with respect to their magnetic effect. The same is true of
the coil-sides 2, 1' and 3, although they belong to three separate windings.
The resultant magnetic field will evidently be as indicated by the dotted lines
and the poles as shown by the letters N and S.
The state of affairs a twelfth of a period later is shown in Fig. 350. The
generator has turned through 30°, and the phase-winding 3, 3′ is now in the
neutral zone and therefore without current. In the other two phases the
momentary value of the current is
¿=¿max. sin 60° = 0.866 imax
The coil-sides 1 and 2′ of the motor may be now grouped together, and
similarly 2 and 1'. The lines of force will be as indicated by the dotted lines
and the north pole of the stator will be situated about coil 3'. It is evident
344
Electrical Engineering
that the poles have rotated through 30° from the position they had in Fig. 349,
and that the rotation of the generator causes a corresponding rotation of the
motor field. If the stator has a 2-pole winding, as in the figure, the field
makes a complete revolution in one period.

3
2
S
13'
N
Generator
Fig. 350.
Motor
3
114. Mesh or delta connection.
The six wires between the generator and the motor in Figs. 349 and 350
can be reduced to three by interconnecting the three phases. This can be
done in two ways; one, known as the mesh connection, we shall consider in
this section, while the other, known as star connection, will be considered in
the next section.
In the mesh connection the end of each phase-winding is connected to the
beginning of the next. The principle of this winding is seen very clearly in
machines which are wound so as to supply
both direct and alternating current. The
three-phase current is taken from three slip-
rings connected to points on the armature
separated by 120°. In other respects the
machine is identical with a D.C. dynamo. (See
the left-hand part of Fig. 354.) The width of
each coil-side is equal to of the pole-pitch.
If the coil-sides be made equal to of the
pole-pitch, as represented diagrammatically in
Figs. 349 and 350, the phases are connected as
shown in Fig. 351.

Fig. 351.
The admissibility of interconnecting the three phases in this way can be
seen from the vector diagram in Fig. 352. The resultant electromotive force
R of phases 2 and 3 is equal and opposite to that of phase 1. If, then, we
connect two of the phase-windings in series (Fig. 353), their joint E. M. F. is
equal at every instant to the E.M.F. of the other phase, and the two voltmeters
in the figure give the same reading. The equal instantaneous electromotive
Mesh or delta connection
345
forces act in opposite directions around the ring, so that the whole ring can
be connected up without any current circulating around it. With regard to
the external circuit the two electromotive forces are, however, in parallel.
This is evidently analogous with the two halves of a D.C. armature, or with

2
İR
3
₹

Fig. 352.
L
Fig. 353.
two machines in parallel on the same bus-bars. It is quite clear that the
P.D. between any two of the three external leads is equal to the E.M.F. of one
phase-winding, neglecting of course, any drop of pressure due to resistance
when current is flowing.
The distribution of current in the mesh connection is apt to prove a
difficulty on first consideration. We know that electricity cannot be heaped
III

S
İ
2
2
وحن
2
3
N
Generator
I
Fig. 354.
3
Motor
up or stored at any point of the circuit, so that each wire in turn must act as
the return lead for the other two. To make the matter quite clear, we shall
consider the currents in generator, leads and motor for the two cases pre-
viously considered.
At the moment represented in Fig. 354 phase 1 of the generator is
directly under the pole, and its current will be a maximum, if we neglect any
effect of self-induction. The current in each of the other phases will have
half this value. The current in external lead I will be equal to 1-5 imax, if
346
Electrical Engineering
imax is the maximum current in each phase of the generator-winding. Of
this external current a part imax flows through phase 1, 1′ of the stator, while
a current of half this value flows through phases 3′, 3 and 2′, 2 in series.
Both in the generator and in the motor, the coil-side carrying the maximum
current has on either side of it coil-sides carrying current in the same
direction but of half the strength. The two stator currents combine between
coil-sides 1' and 2, and a current of 15 imax flows back to the generator through
the lead marked III, while lead II is momentarily without current.

S
Ⅲ
3
160
+60%
I
N
Generator
Fig. 355.
3
3
Motor
2
A twelfth of a period later, the generator will have turned through 30°, as
shown in Fig. 355. Phase 3 of the generator is now in the neutral zone and
therefore without current, while phases 1 and 2

carry a current of
imax • sin 60° = 0·866 imax⋅
•
These currents combine to produce a current of
1.73 max in the external lead I. The vector dia-
gram in Fig. 356 shows that this is the maximum
value reached by the external current. Half of this
current passes through the stator phase 3′, 3 and
returns to the generator by lead II, while the
other half passes through phase 1, 1' and back to
the generator through lead III. Hence, at the
moment that the current in one of the external
leads reaches its maximum value, one of the phases of both generator and
motor is without current.
Fig. 356.
We find, therefore, that the maximum external current is greater than
the maximum current in one phase of the mesh connection, and that these
maxima do not occur at the same time. This is very clear in the vector
diagram (Fig. 356). We must remember, however, that the starting points.
of two phase-windings have not been joined, but the starting point of one to
the end of the other. For this reason, when two current vectors are com-
pounded to find the external current, one of them must be reversed. The
external current OR is, for example, the resultant of OC and the dotted vector
OA. It is 1.73 times as great as OC and is 30° out of phase with it.
Star connection
347
115. Star connection.
Whereas in the mesh connection the end of each phase was joined to the
beginning of the next phase, in the star connection the three starting points
are connected together within the machine, and the three ends connected to
the three slip-rings or terminals (Fig. 357). That such an interconnection of
the three windings is permissible can be readily seen from Fig. 358. The
vectors OA and OB, representing the current in two of the phases, have, as
their resultant, the vector OR, which is exactly equal and opposite to the
vector OC of the current in the third phase-winding. The total current
flowing to the neutral point or common connection is therefore nil, or, in
other words, the current flowing towards the neutral point is exactly equal, at
every moment, to the current flowing away from this point. In Fig. 357 the
current of phase 1 divides into two parts and passes through phases 2 and 3
in parallel.
It is obvious that, with this arrangement, the current in each external
lead is identical both in magnitude and phase with the current in the

1
در
3
2
А
C

R
Fig. 357.
Fig. 358.
corresponding winding. The P.D. between two external leads, however, will
not be due to a single winding, but to two phase-windings connected in series,
or, more accurately, in opposition. The voltmeter in Fig. 357 is connected
across phases 1 and 3. At the moment indicated in the figure the induced
E. M.F. in coil 1 is a maximum, while in coil 3 it is equal to
Emax. sin 30° = 0·5Emax.
If the E.M.F. of coil 1 acts towards the neutral point, that of coil 3 will act
away from it, since they are 120° out of phase. We must therefore add the
two electromotive forces to find the voltmeter reading, which will therefore be
equal to 1.5Emax"
A twelfth of a period later the coil 2 will be in the neutral zone and have
no induced E. M.F. Coils 1 and 3 will have equal electromotive forces induced
in them, viz. Emax. sin 60° 0.866 Emax (Fig. 359). As before, we see that the
voltmeter will read the sum of these electromotive forces, that is, 173Emax
It can be shown very easily that this is the maximum momentary value
348
Electrical Engineering
attained by the P.D. between two external leads. If, for example, the vectors
OA and OB in Fig. 360 represent the electromotive forces induced in the coils
1 and 3, we must remember that these coils are connected in opposition, that
is, their starting points are connected together, so that in passing through the
armature from lead I to lead III (Fig. 359) we pass inwardly through coil 1
but outwardly through coil 3. To find the resultant E.M.F., which is the
R
2
S

I

120°
B
N
м
Fig. 359.
Fig. 360.
voltmeter reading, one of the vectors must therefore be reversed, giving OA
and OB' with the resultant OR. This resultant is 173 times as large as
either of the phase pressures and is displaced from them with regard to phase,
as shown in the figure.
It is quite allowable to have a star-connected generator driving a mesh-
connected motor, or vice versa. For long distance power transmission the
star connection is preferable both for motors and generators since the line
pressure is then 1.73 times the pressure per phase of the machines. If the

wwww
ww
*
*
Fig. 361.
load consists of resistances such as lamps, they can be arranged either in star
or mesh connection. If the former arrangement be adopted, care must be
taken that the number of lamps is the same in each branch.
This limitation can be removed and the branches loaded unequally without
affecting the constancy of pressure, so necessary for glow-lamps, by the use of
Star connection
349
a fourth lead connecting the neutral points of generator and load (Fig. 361).
This is analogous to the neutral or middle wire in the 3-wire system which is
now so largely used for direct-current lighting.
The generator-windings which we have described have been quite sym-
metrical, since the three windings were exactly similar and equally spaced on
the armature. Unsymmetrical windings have been proposed and used, the
most important being Steinmetz's monocyclic system. In this system the
main armature winding is an ordinary single-phase winding, in which the
coils span from the centre of one pole to the centre of the next pole. An
auxiliary winding of thinner wire is wound with a displacement of 90° from
the main winding. One end of this auxiliary winding is connected within
the machine to the mid-point of the main winding. The armature has three
terminals between two of which the single-phase lighting load is connected,
while the three terminals are used for a three-phase power load consisting
principally of induction motors.
A

A

E'
B
DC
B←
E
C
Fig. 362.
If the number of turns in the auxiliary winding be suitably chosen, the
P.D. between any two of the terminals will be equal. If the E.M.F. induced
in the main winding be E, and that in the auxiliary E', then the pressure
between the auxiliary terminal and either main terminal will be the resultant
of E/2 and E', which are separated by 90° as regards their phase. This re-
sultant will therefore be equal to
or
E\2
+ (E')².
If this is to be equal to the P.D. between the main terminals, we have
E
E\2
2
+(E')²,
3
E'
E = 0·86E.
4
The number of turns in the auxiliary winding must therefore be 0.86 of
the turns in the main winding.
+
!
350
Electrical Engineering
116. The measurement of power in polyphase circuits.
We shall consider three-phase power supply in this section. The measure-
ment of two-phase power is merely a duplication of the single-phase case and
need not be further considered. The following symbols will be used inde-
pendently of the method of connection employed (star or mesh).
e1
e₁ = the P.D. per phase (i.e. per limb in star connection),
i=the current per phase (i.e. per limb in mesh connection),
=
the angle between e, and i,
e = the P.D. between external lines,
i = the current in each external line,
P = the total power in watts.
The power per phase for either connection is equal to
e.i. coso,
and the total power is three times this amount.
For star connection we have
and
e = 1.73e₁,
i=ir,
e
P = 3е₁. i. cos
= 3.
•
1.73
i.cos &= √√3.e.i.cos p.
For mesh connection we have
i = 1.73₁₁,
e = er,
and
i
P = 3e, . i . cos & = 3e .
.cos√3.e.i. cos......(166).
1.73'
The formula obtained by using the line current and line pressure is thus
exactly the same, no matter what the connection employed.
This is a suitable point at which to compare the percentage losses of
power and pressure in the lines of three-phase and D. C. transmissions. We
assume that the same power P has to be transmitted the same distance
in each case.
We assume also that the effective P.D. between the lines
is the same in each case, and also the loss of power in the lines. We proceed
to determine the relative amounts of copper necessary in each case. Let
R₁ be the resistance of each single D.C. line,
R₂
3-phase
""
""
A₁
section
D.C.
""
>>
""
""
A₂
>>
3-phase
"
We have:
For D.C.
P = e. i,
i
P
e
Loss in the two wires.
P². R₁
2.2². R₁ = 2
e2
For 3-phase.
P3.e.i.cos o,
i =
P
√/3.e.coso
Loss in the three wires.
P². R₂
312. R₂ =
e². cos²
The measurement of power in polyphase circuits
351
Since these losses have to be the same in each case, we equate them, thus
or
R₂
2R₁
cos²
R₁
1
R₂ 2 cos²
Since the lengths are the same in both cases, the cross-sections must be
inversely proportional to the resistances. Hence
A, R₁
1
A₁ R₂ 2 cos² & *
Again, as the lengths are equal, the volume of copper in each case is pro-
portional to the sum of the cross-sections of the wires, so that
or
es
www
is
V₂
V₁
1
wwwww
iz
ww
wwwww
13
2
V₂ 3A₂
1
1
3 R₁
2 R₂
0.75
cos²

www
Fig. 363.
Hence, if cos = 1, i.e. if the current is in phase with the pressure, the
three-phase system requires only 75 per cent. of the copper necessary for the D.C.
transmission. If the load is at all inductive, it is evident that this advantage
rapidly disappears. These conclusions depend entirely on our assumption
that the effective pressure between the lines is the same in each case. The
same method is applicable, of course, whatever assumptions we make as to
the pressure.
We shall now consider the actual measurement of the power in a three-phase
circuit. With star connection and an accessible neutral point, the measure-
ment is very simple. The pressure coil of the wattmeter is connected
between the neutral point and the end of one of the phase-windings, while
the current of this phase is passed through the main coil of the wattmeter.
The wattmeter reading will give the power in one phase, and if the three
phases are equally loaded, the total power will be three times this amount.
352
Electrical Engineering
}
If the loads on the phases are unequal, the measurement must be carried out
in each phase and the three results added together. If three wattmeters
are available, the three readings can be taken simultaneously; if only one
wattmeter is used, it must be connected successively in the three phases.
If the neutral point is not available, another method, known as the two
wattmeter method, must be adopted. This method is equally applicable to
star and mesh connection, and is represented in Fig. 363. The current coils.
of the wattmeters are introduced into two of the leads, and their pressure
coils are connected between the same lead and the lead which has no watt-
meter coil in it. If e, e, and e, are the instantaneous values of the phase
pressures, and i, i, and is the corresponding values of the currents, then the
value of the power at the given moment will be
Since the sum of the
moment is zero, we have
or
P = e̟₁i₁ + €½¿½ + eziz.
currents flowing away from the generator at any
i₂ = − (¿½ + iz).
If this value be substituted for i in the above equation for P, we get
P = e̟₁i₁ − е½ (¿1 + is) + esis,
P = (e̱₁ − e₂) . ¿₁ + (e3 — €2) . i3.
e!
Now (e,e₂) is the momentary P.D. across the pressure coil of the upper
wattmeter, while (e-e) is the momentary P.D. across the pressure coil of
the lower wattmeter. The product (e, — e½). ¿½
gives the momentary value of the torque on
the upper wattmeter, while (es-e). is gives
that on the lower one. The reason for the
differences of the phase pressures appearing e"
in the brackets and not the sums, is that the
two phase-windings are connected in opposi-
tion across the pressure coils of the watt-
meters, so that to pass from one pressure
terminal to the other we go through two
phase-windings, one towards the neutral point
and the other away from the neutral point.

03
iz
in
iz
Lez
Fig. 364.
In consequence of the inertia of the moving
systems of the wattmeters, they take up sta-
tionary positions corresponding to the mean
value of the power. The total three-phase power will therefore be given by
the sum of the two wattmeter readings. One must be careful, however, to
observe the signs of the two terms in the last equation for P. As a rule,
they are both positive, and the wattmeters, if connected up as shown, will
deflect in the same direction. In this case their readings must be added.
If, however, the load be very inductive, one of the wattmeters will deflect
in the opposite direction. The direction of the deflection can be reversed by
reversing the pressure coil connections of this wattmeter. The total power
is then equal to the difference between the two wattmeter readings. The
reason for this reversal is evident from Fig. 364. The vector e' is the
The measurement of power in polyphase circuits
353
resultant of e, and e2, and represents the pressure across the upper watt-
meter. Similarly, e" is the pressure across the lower wattmeter. The
currents in the main wattmeter coils are represented by the vectors i, and
i3. The angle of lag in each phase is p. It is evident that an increase
of the lag & will increase the reading of the upper wattmeter, which is equal
to e'..cos (30), until = 30°, beyond which it will again decrease. In
the same way, the reading of the lower wattmeter is equal to
e". iz. cos (30+ $),
and will gradually decrease as the load becomes more inductive, until when
= 60° the wattmeter will read zero. If the lag increases still further, the
reading of the wattmeter will reverse.
$
The measurements can be carried out very conveniently with one watt-
meter, by means of a board containing six mercury cups, which can be con-
nected up by means of copper bridges as shown in Figs. 365 and 366. In
the first figure the current coil of the wattmeter is in series with phase 2,
while in the second figure phase 2 is unbroken and the wattmeter is placed

1

2
Fig. 365.
2
3
Fig. 366.
in phase 1. In either case, the pressure coil is connected between the current
coil and the unbroken line 3. The small mercury commutator is for the
purpose of reversing the pressure coil if necessary.
The duplicate readings can be avoided by the use of special wattmeters
consisting of two separate wattmeters with a common spindle, so that the
reading depends on the sum or difference of the two torques, as the case
may be.
117. General principles of the rotor.
Having considered the production of the rotating magnetic field and the
winding of the stator, we shall now study the effect of the rotating field
on the rotor. The rotor resembles a D. C. armature in that it is built up
of stampings of sheet iron, assembled on the shaft or on a spider or boss
carried on the shaft. The winding is carried in slots in the usual way, and
may consist of a permanently short-circuited arrangement of bars or coils, or
may be wound with a three-phase winding very similar to that on the stator.
In the latter case the three ends are brought to three slip-rings, between
which starting resistances can be introduced.
T. E.
23
354
Electrical Engineering
We have seen that the field of a stator, with a two-pole winding, makes
one complete revolution during every period. In order to follow this rotating
field more easily, we shall imagine, for the time being, that the stator is
replaced by a magnet ring with salient poles, excited with direct current.
This is shown in Fig. 367, where the poles are supposed to rotate around the
rotor in an anticlockwise direction. The currents induced in the rotor
windings will be exactly the same as if the poles were stationary and the
rotor turned in the clockwise direction. These currents are indicated in the
figure, and are away from the observer under the north pole, and towards
him under the south pole.
We assume, moreover, that the field magnet represents not merely the
field produced by the stator currents, but the resultant flux N, produced by
the combined action of stator and

rotor currents. In this case the in-
duced E.M.F. and the rotor current
will both have their maximum values
under the middle of the poles.
We know from Lenz's law that
the currents induced in the rotor
wires will tend to oppose the relative
motion between themselves and the
rotating field, that is, the rotor will
be dragged round in the same direc-
tion as the field. To make this clearer,
we may imagine ourselves to be swim-
ming down a rotor wire at A and
looking out towards the north pole,
N
S
Fig. 367.
then we know from Ampère's rule that the north pole would be driven in
the clockwise direction as shown by the dotted arrow. This arrow might
represent a hand put out from the rotor to prevent the approach of the pole;
the evident result is the anticlockwise rotation of the rotor.
If the motor is unloaded and the bearings are entirely frictionless, the
rotor revolves at the same speed as the primary field. There would then be
no cutting of lines, no induced E. M. F. and no current in the rotor. When
a load is applied to the motor spindle, a certain rotor current is essential
to the production of the necessary torque. This current can only be produced
by the rotor running at a lower speed than the field, so that the latter can
cut the rotor wires and thus induce the necessary current in them. This
decrease of rotor speed is known as the slip. Under ordinary conditions the
resistance of the rotor coils is so small that a very small E.M. F., and therefore
a very small slip, is sufficient to produce the necessary current. The slip,
that is, the difference between the speeds of field and rotor, is sometimes no
more than 1 or 2 per cent. on full load.
The fact that the rotor runs at approximately the same speed as the
primary field excludes the employment of two-pole windings with the high
frequencies commonly used. To ensure a steady illumination without any
General principles of the rotor
355
appreciable flicker, the frequency is generally about 50 per second. A two-
pole motor driven from these mains would have a speed of nearly 3,000
revolutions per minute. In systems where the load consists almost entirely
of motors, and little account need be taken of lighting, the frequency is
sometimes taken as low as 25. Even then, a two-pole motor would make
1,500 revolutions per minute, which is generally too high for motors above
about 10 H.P. The speed is reduced by using multipolar windings. In a four-
pole winding the simple two-pole winding, which we have already considered,
is compressed into a half of the total periphery, and connected in series
with a similar winding on the remaining half of the stator. In Fig. 368, for
example, the coils 1 and 1' of a three-phase stator are no longer diametrically
opposite, as they were in Fig. 349 on page 343, but are now separated by 90°.
The other coils 2, 2′ belonging to the same phase are connected in series
with them. For the sake of clearness, only the one phase has been drawn in


τη
N
N
ވ
N
N
2
Fig. 368.
Fig. 369.
Fig. 368, but the direction of the currents in the other phases is evident.
Whenever the current in any coil is a maximum, the coils on either side
of it carry currents in the same direction, but of half the magnitude. If the
current in the phase shown in the figure is a maximum at the given moment,
the dotted lines will represent the lines of force and show the positions of
the poles.
The rotating field will now make only half a revolution during each
period. If p be the number of pairs of poles, and the frequency of the
supply, then the field will make n, revolutions per minute, where
Ny
60
p
If the speed of the rotor ben, the difference will be n-n
revs. per
60
second. The effect on the rotor is the same as if it rotated backward at
this speed in a stationary field. The frequency of the currents induced in
the rotor will therefore be
N1 n
60
.p
…………….(167).
23-2
356
Electrical Engineering
If the rotor is wound, it must be arranged to have the same number
of poles as the stator. The winding, however, can be either two or three-
phase, quite irrespective of the number of phases of the stator-winding.
A four-pole drum-winding is shown in Fig. 369, where one phase only is
shown connected up, the currents in the other phases being represented
by crosses and dots. It is assumed that the current has its maximum value
in the phase shown connected up, while the other currents have half this
value. In actual practice drum-windings are always used; we have only
used the ring-windings to make the rotating field as simple as possible.
The direction of rotation of the rotor is determined by that of the primary
field. It can be reversed by changing two of the supply leads on the stator
terminals.
The polyphase motor differs in this respect from the single-phase induc-
tion motor. Such a motor can be obtained by disconnecting one of the
supply leads to a three-phase star wound stator. The motor will not start
from rest, but, if the rotor be started in either direction, the speed will
increase in that direction until it reaches the speed corresponding approxi-
mately to the primary frequency. Such motors are generally started as two-
phase motors by means of an auxiliary winding on the stator. The action of
the single-phase induction motor is very complicated, and we shall postpone
its further consideration until Section 132.
CHAPTER XVII.
118. Distributed windings and the electromotive forces induced in them.-119. The magnetic
flux of an induction motor.-120. The effect of the iron in the magnetic path.-121. Torque
of an induction motor.-122. Calculation of slip.
118. Distributed windings and the electromotive forces
induced in them.
For the sake of simplicity we have always assumed that all the wires.
constituting a coil-side are contained in a single slot, or that the width of
the coil-side is so small that there are considerable gaps between the wires
of different phases. As a matter of fact, however, the wires of each coil-side
are distributed between several slots, so that the coil-side has considerable
width. In this way the field strength changes gradually from point to point

Lilio
=
(X
ca
어​!
Fig. 370.
round the stator, and sudden jumps are avoided. Moreover, the number of
ampere-wires in each slot is kept small in order to decrease the self-induction
and leakage, as will be seen in Section 131. As a general rule each stator
coil-side is distributed between from 2 to 5 slots, while each coil-side of the
rotor occupies from 3 to 7 slots. The rotor and stator must have a different
number of slots, as otherwise there is a tendency for the motor to remain
stationary and act as an ordinary transformer, instead of starting.
358
Electrical Engineering
A four-pole distributed coil-winding is represented in Fig. 370. The three
phases are distinguished by full, dotted, and chain-dotted lines, of which the
latter is assumed to be carrying its maximum current, while the others have
half this current. The strength of the current is indicated approximately by
drawing the crosses and dots heavily or lightly. It is easily seen that the
mid-points of the magnetic poles are situated at the junction of opposite
currents, so that the neutral axes lie always at the centres of groups of wires
carrying current in the same direction.
To calculate the E. M.F. induced in the winding we make the assumptions
that the field is distributed around the periphery according to the sine law,
and that the coil-side is distributed between so many slots that it is equi-
valent to a smooth winding.
and
Let
N, be the stator flux per pole*,
z, the number of stator wires per phase,
~ the frequency of the stator current.
On the above assumption the breadth of the coil-side is equal to } of the
pole-pitch, and we can employ equation (145) on page 281, viz.
E₁ = 2·12N~2.10-8.
Now, on account of magnetic leakage, the flux N cutting the rotor-
winding is less than that cutting the stator-winding. Moreover, the rotor
wires are not cut with the primary frequency ~ but with the frequency of
slip The E.M.F. induced in the distributed rotor-winding is therefore
E₂ = 2·12N ~ 22′ . 10–³,
of the
where the rotor-winding is also three-phase with z wires per phase.
In these coil-windings the breadth of each coil-side is equal to
pole-pitch. Each coil is completely wound with the requisite number of
turns before proceeding to the corresponding coil under the next pair of poles.
Another type of winding which can be used for both stator and rotor
is the creeping bar-winding (wave-winding). In this winding (compare
Section 96) the step is always in one direction, that is, it is a simple wave-
winding, and, after going completely round the armature, arrives at a point
near that from which it started. On proceeding with the winding, those
wires which belong virtually to the same coil-side will lie together. If the
total number of wires be z₁, the winding-step will be given by the formula
21 ± 2
y
2p
y must be an odd number, and z, must be divisible by 3, since the whole
winding is to be divided into three phases. The principle of the winding is
much clearer when a large number of wires is chosen, on account of a small
unavoidable dissymmetry which is exaggerated when the total number of
wires is small. The values chosen for Fig. 371 are 2, 54, i.e. 18 wires per
phase, and p = 2, which gives a four-pole winding. For the step, we have
=
54 +2
y =
4
=
14 or 13.
* No is the total flux in the stator. It is made up of the common flux linking both stator
and rotor, together with the leakage flux.
Polyphase wave-windings
359
For a singly reentrant winding, 14 is impossible, so that y = 13. The
winding is connected up as a star-winding in Fig. 371, the external leads
coming to wires 6, 24 and 42. The current is assumed to have its maximum
value in the first phase, which, starting from the neutral point, passes along a
to wire 1, where it goes down to the further end of the stator or rotor, crosses
54
2
52
X
X
4
50Ⓡ
X
X
6
53
51
N
48
49
46
X
47
45
44
X
/43 9
42
41
40
38
37
36
a
Ъ
11
18
13
014
15
X
Ø 17
19
21
33
23
N
25
X
341
27
x
X
X
24
32
X
(x)
X
30
26
28
Fig. 371.
to 1+13=14, up which it comes to the front, and so on, in accordance with
the following table:
1-14
27-40
53-12
25-38
51-10
10
10
23-36
49 8
21-34
47— 6
If we continued in this way we should obtain an ordinary direct-current
winding. It would be closed on itself and by tapping three points, separated
from each other by of the pole-pitch, we should obtain a mesh-connected
armature. By reference to Fig. 354 it is evident that the breadth of the
coil-side would be equal to of the pole-pitch. In Fig. 371, however, we have
chosen star connection; the end of wire 6 is led to a terminal or slip-ring,
and wire 19 is taken as the beginning of a new phase. Whether star- or
1

360
Electrical Engineering
mesh-connected, the breadth of the coil-side is the same, viz. of the pole-
pitch.
To make the diagram as clear as possible, the slots are shown in two tiers.
A thick line is drawn around each coil-side of the first phase, a thin line
around each coil-side of the second phase, and a dotted line for the third
phase. Since the sum of any two of the currents is at any moment equal
to the third, the currents in phases b and c must flow towards the neutral
point, so that, in following the winding of these phases from the neutral
point, we shall continually oppose the direction of the current. The winding
of phase b is given by the following table:
19-32
45- 4
17-30
43- 2
15-28
Similarly, the winding of phase c is as follows:
37-50
کو
/
9-22
35-48
28
41-54
13-26
39-52
11-24
46
5-18
31-44
3-
16
29-42
7-20
33-46
The lines which have been drawn so as to enclose the various coil-sides
make it very evident that the phases overlap. It is for this reason that the
breadth of each coil is of the pole-pitch; without overlapping the breadth
could not exceed of the pole-pitch. If the number of slots per coil-side
be very large, and the field distribution sinusoidal, the E.M.F. induced in
each phase of the stator is given by equation (146) on page 281:
E₁ = 1.84N~, 21. 10-8.
1·84N。
For a given terminal pressure, i.e. for a given back E.M. F. of the motor,
the magnetic flux will therefore be greater in the ratio 2·12: 1.84 than
it would be for a coil-winding. This increase of flux requires a greater
magnetising current, more especially as the creeping bar-winding is less
efficient in this respect than the coil-winding, as we shall see in the next
section. The overlapping or consequent breadth of the coil-side is therefore
avoided in actual practice by making the winding slightly different from
that shown in Fig. 371. Instead of winding the whole of phase a straight
forward from 1 to 6 as shown in the table on page 359, we stop when a half
is completed, viz. 1, 14, 27, 40, 53, 12, 25, 38. In the ordinary way we
should have gone from 38 to 51, but we now pass from 38 to 2, at least,
the actual connection is equivalent to this. From 2 we proceed in the
usual way to 15, 28 etc., and pass through 54 and 52. The wires constituting
Polyphase wave-windings
361
the coil-side are now 53, 1, 52, 54, 2, and they cover about of the pole-
pitch.
The method actually adopted is not exactly that explained above, for,
after winding half the phase, the direction of the winding is reversed. This
is merely to get symmetrical end-connections, and makes no difference in
the principle of the winding. Another method by which the want of
symmetry in Fig. 371 can be avoided consists in making the wires per
phase a multiple of the number of poles, and neglecting entirely the con-
dition 2py = z ±2. In Fig. 372, for example, there are 4 wires, or 2 slots,
per pole per phase. For the four poles this gives a total of 24 slots with
48 wires. The slot-step is exactly equal to the pole-pitch, viz. 6 slots.
If we start from the neutral point O and follow round phase I, we pass
from 0 to 1, down 1, across the back to 7, up 7, across the front to 13, and

9
22
o
x
i
118
17
A
X
I
×
16
X
15
14
13
X
X
O
Der
10
Fig. 372.
so on. The regular step would take us from the bottom of slot 19 to the
top of slot 1, which is already filled. We therefore make the connection
a little longer so as to pass to the top of the next slot 2. This longer
connection is drawn slightly heavier. The winding then proceeds in the
regular manner through 2 (top), 8 (bottom), etc. until we come to the bottom
of 20, from whence we pass by a special connection to the bottom of 2, and
reverse the direction round the winding. After passing through 2, 20, 14, 8
a special long connection, shown by the heavy line, connects across to the
bottom of 1, whence by 19, 13, 7 we come to the end of the winding. This
winding allows the end-connections to be arranged very conveniently; all
those at the further end of the armature are exactly alike, and span the
362
Electrical Engineering
pole-pitch. This is also true of the front end, except, in the present case,
for two slightly longer connections, which are adjacent to the beginning and
the end of the winding. In general, if there are q' wires per coil-side, the
winding will go round the stator or rotor times in each direction, and
2
the number of long end-connections will be q'- 2.
The beginning of
of the pole-pitch,
The other phases are wound in exactly the same way.
phase II should be separated from that of phase I by
but we can naturally start from the diametrically opposite but corresponding
wire 17 instead of from 5. This is only true, of course, because this happens
to be a four-pole winding.
The E.M.F. of the above winding is exactly the same as for a coil-winding.
Another type of winding, if we may so call it, is the squirrel cage, or
the similar winding consisting of a number of separately short-circuited loops.
Both of these are used for rotors. The E. M. F. induced in each wire is given
E.M.F.
by equation (142) on page 279, as
2.22. N.~.1.10-8.
The electromotive forces in the various wires are out of phase, and there-
fore cannot be added. If, however, we are merely concerned with the heating
effect, we can imagine that a third of the wires are connected in series, if, at
the same time, we take as the resistance the sum of the resistances of the
third of the wires. We can then speak of the E. M. F. "per phase," which will
be given by the equation
In general, we have
where
E = 2·22. N~'. 10-8.
E=k. N~z'. 10-8,
k = 2·22 for squirrel cage windings,
2.12 ordinary coil-windings,
""
1.84
""
creeping bar-windings.
119. The magnetic flux of an induction motor.
In the present section we shall seek to prove that the magnetic field of an
induction motor has an approximately sinusoidal distribution. Moreover, we
must calculate the magnetic flux produced by the stator and rotor currents
with the different types of windings. Fig. 373 represents diagrammatically
a coil-winding with a large number of slots per coil-side. For the sake
of clearness, the stator periphery is shown straight and the coil-sides are
slightly separated from each other. The current in one phase is just at
its maximum value, while in the other two it has half that value, as indicated
by the crosses and dots.
We see now that D is the centre of a north pole, while at the points A
and & the north pole ceases and south poles commence. The strength of the
field falls off on both sides of D, and corresponds at every point to the number
of ampere-turns which are effective at that point. This is represented by the
length of the arrows in the figure.
The magnetic flux of an induction motor
363
In order to calculate the excitation at various points on the periphery we
must observe that any line of force is produced by the ampere-wires which it
encircles. If there are q' wires in each coil-side, i.e. q' wires per pole per
phase, then
at A the excitation is 0,
""
B „
>>
""
D"
""
q.max ampere-wires,
q'. imax
2
+ q' • i max +
=2q'. imax
D
F G
q'. imax
2
....." 0.000 0....
A
B

A F B T D T
퓽
​Fig. 373.
π
F A G
If we plot the excitation at every point as ordinates on a base representing
the pole-pitch π, we obtain the lower part of Fig. 373. Between A and B,
and between B and D, the increase is uniform and is represented by straight
A
B
C
D E F
G
H
ဝဝဝဝ ဝဝရီ ၁ဝဝဝဝဝ


3
I-
Fig. 374.
I
3
lines. The mean excitation is obtained by finding the area of the shaded
figure and dividing it by the base π. In this way we get
1
X mean ( ·
π
q'. imax +3.. q'. imax) = 1·166. q'. imax
Since the strength of the field at the various points of the periphery is
proportional to the excitation at those points, it follows that the field curve
has the same shape as the excitation curve, and is roughly a sine curve.
364
Electrical Engineering
We turn now to the other extreme case, in which one phase is without
current, while the current in the other two phases is equal to
imax. sin 60° = 0.866. imax (Fig. 374).
The wires between D and H form a coil-side of 2.q'. 0.866. imax = 1.73. q'. imax
ampere-wires. This is the effective excitation between points B and D, and
is represented by the equal arrows between these points. Outside these
points the excitation falls off uniformly, as shown by the arrows and by the
shaded figure. By calculating the area of this figure and dividing by the
base, we get
I mean
1
π
.2..173.q.imax = 1·155. q'. imax •
This differs but little from the value found above for the other extreme
moment; hence, the mean excitation and the mean flux density in the
A
B C
D

* X
X
X
x
[X]
X
X
X X x X

-7 ÷ 7 ÷ 7.
朵朵
​Fig. 375.
air-gap is approximately constant. Taking the mean of the two limiting
values, we have
X mean
1.166 + 1.155
2
•
· q'. imax = 1·16 . qʻ. imax
.(168).
The flux produced is proportional to the excitation. The rotating field
has therefore an approximately constant number of lines of force, but a
maximum flux density which varies, according to Figs. 373 and 374, in the
ratio of 2 to 1.73. The two-phase motor is inferior to the three-phase in this
respect, since the fluctuations in its maximum flux density are much greater,
as can be seen on reference to Section 112.
The mean value of the excitation can be determined in a similar manner
for a wave-winding of the type shown in Fig. 371 (Fig. 372 is analogous to
the coil-winding just considered). In Fig. 375 the current in one phase has
its maximum value, while in the other phases it is equal to .max The
conductors between A and B neutralise each other. The field strength
between these points is therefore constant and corresponds to the ampere-
wires between B and D. The number of these ampere-wires is
q'. imax +2.2
q
imax
2
= 1·5 q'. i'max⋅
If we make the same assumption as before as to the large number of
The magnetic flux of an induction motor
365
slots per coil-side, the flux density will decrease uniformly from B to a point
between B and D, where the north pole ends and the south pole begins.
If the excitation is plotted on a base representing the pole-pitch, the shaded
figure in Fig. 375 is obtained.
x
(x
x
A
B
C
000
Q
π
π
D

O O O O O
(x)
X
Fig. 376.
The same is repeated for the other extreme case, when the current in one
phase is passing through zero (Fig. 376). The wires between A and D may
now be looked upon as belonging to one large coil-side, in the middle of
which the field strength is zero. From this point the field increases uniformly
on either side, until point B is reached, at which the excitation will be
2. imax • sin 60° = 0.866. q'. imax. From B to A the field still increases
uniformly but at a smaller rate. The excitation at A is due to all the
ampere-wires between A and D, viz. 2q. max. sin 60° 1.73.q.imax.
2
=

Field cutting the
Rotor Bars
B +y 무
​D
-Y
B

Field produced by
Rotor Currents
B
Fig. 377.
On comparing Figs 375 and 376 with Figs. 373 and 374, we find that
the shaded curves have exactly the same shape, but that they have been
exchanged, the flat topped curve being now obtained for the conditions of
phase which previously gave the pointed curve. The ordinates of the present
366
Electrical Engineering
curves are only 0.866 of the previous ordinates, so that equation (168)
becomes
mean = 0·866. 1·16. q'. imax = 1'005. q'. imax... .(169).
We shall now consider the excitation produced by the currents in a
squirrel cage rotor. The upper curve in Fig. 377 represents the field which
cuts the rotor bars. It varies from point to point around the rotor according
to the sine law. The E. M. F. induced in the various rotor bars, and the rotor
current i, will also be a sine function of the position on the periphery. This
is indicated in the figure by the crosses and dots being correspondingly
thickened. The magnetising effect of these rotor currents will be zero at B,
and will increase on either side up to the points A and C, where it will be a
maximum.
It hardly requires to be proved that the curve of the field produced by
the rotor currents is also a sine curve, displaced 90° from the field cutting
the rotor bars*. If the total number of conductors on the rotor be z₂, the
Z2
and the mean current taken around the rotor
conductors per pole will be
2p
is imax. The excitation at C is therefore
2
π
2
22
X max
imax
•
2p
πT
Xme
Since we have just seen that the field has a sine curve distribution,
mean
2
π
X In order to compare the squirrel cage with the other
•
types of winding, we may imagine it to be divided into three parts or phases,
and put
We have then
22=3.2p.q'.
2 2
X
mean
imax
π π
3.2p.q 12
2p π-2
. q'. imax = 1·22 q'. imax …………..(170).
X,
.(171),
In general, whatever the type of winding, we may write
·X mean = c. q'. imax
in which the coefficient c has the following values:
c = 1005 for creeping bar-windings,
1.16
coil-windings,
1.22 squirrel cage rotors.
""
If the area of the pole-face be A,, the flux per pole can be found from
the formula
N = Bmean · Ag
•
=
0·4π. X mean · Ag 04π.c.q.imax · Ag
lg
_
* The average current in the band of conductors DE is
sin y
γ
lg
......(172).
1
27
に
​+y
•
imax ⋅ cos a. da=imax
Y
and the number of conductors in this band is
Z2 27
2р T
2P
The ampere-wires producing the field at
22
P.π
·
D and E are therefore imax. siny. Hence the field is proportional to siny where y is
measured from B.
The magnetic flux of an induction motor
367
We have here neglected the reluctance of the iron and considered merely
that of the double air-gap of length l„.
If the above values of the coefficient c are compared with those of k on
page 362, the corresponding values will be found to be almost proportional.
120. The effect of the iron in the magnetic path.
We shall now make use of the formulae which we have just established,
to calculate the magnetising current of an actual 150 H.P. three-phase motor

0000
Stator
00000000000000
[000000000000000000000000(
Rotor
2 4
6
8 10 12 14 16 18 20 cm
Fig. 378.
made by the Oerlikon Company of Zürich. The following particulars of this
motor, together with Fig. 378, are taken from Arnold's "Konstruktionstafeln.”
Terminal pressure...
Primary frequency
Number of pairs of poles …….
Axial length
Diameter of rotor
The double air-gap
Total number of stator conductors
Combined length of a stator and rotor tooth...
Ratio of tooth pitch to tooth width
From these data we get the following:
Pressure per phase..
e = 3,300 volts.
=
50.
p = 6.
L= 32.5 cms.
D= 90 cms.
lg = 0·15 cm.
≈₁ = 2,016.
= 6 cms.
= 2·5.
e1
G₁ =
√3
3,300
1,910 volts.
368
Electrical Engineering
Conductors per phase
zí = 3/2 = 672.
21
Conductors per pole per phase ...
q' =
56.
3.2p
TD.L
Polar area
Ag
763.
2p
Let N. be the total flux per pole produced by the stator at no-load.
Since the pressure per phase e, is almost exactly equal and opposite to the
induced electromotive force E₁, we can apply the formula found on p. 362
for coil-windings, viz.
e₁ = E₁ = 2.12 No.~.z. 10-8.
Hence
N₁ =
N.
e₁.108
2.12.1.21
1,910.108
2.12.50.672
2.69.10º.
If we assume that 2 per cent. of this flux is lost in leakage and that
98 per cent. of it passes through the rotor, the flux N in the latter will be
N=0·98N₁ = 2·64. 10º.
From equation (172) on page 366, we have
N
=
0.4π. c. q'. imax
lg
•
Ag,
in which c = 1.16 for coil-windings. Introducing the effective value i
we get
io
N.lg
√2.0·4π.c.q'. Ag
4.5 amperes.
2max
√2›
This would be the magnetising current if the iron had no reluctance.
The actual current taken by this motor on open circuit was 6 amperes, which
is 33 per cent. more than the above value.
The main part of the reluctance of the iron lies in the teeth, on account
of the flux density in them being very high, and the idea will suggest itself
at once of adding something to the length of the air-gap to allow for this.
Since the ratio of the tooth pitch to the thickness of the tooth is 2:5, and
about 15 per cent. of the cross-section of the iron is lost in insulation, the
flux density in the teeth will be 2.5/0.85 = 2.95 times that in the air-gap.
The length of air-gap equivalent to the teeth will be found by multiplying
the length of the teeth by 2.95 and dividing by the permeability μ. The
mean flux density in the air-gap is
N 2.64.106
Bmean
Ag 763
3,450.
The mean flux density in the teeth is 2.95 times as high, and is therefore
3,450 . 2·95 = 10,200. The magnetisation curve for the iron is given in Fig.
379, and from this we find that a value of B equal to 10,200 requires 2·6
ampere-turns per cm., i.e.
Hence
X
X
2-6, and H = 0·4π .
3.25.
Τ
B 10,200
μ = H
= 3,140.
3.25
The effect of the iron in the magnetic path
369
The combined length of a stator and rotor tooth is 6 cms., so that the
path of a line of force has a length of 12 cms. in the teeth.
12.2.95
3,140
= 0·011 cm.
This means an
This will increase
equivalent addition to the air-gap of
the value of l, from 0·15 to 0·16 and the value of & will increase in the same
proportion, viz. from 45 to 48 amperes. This is still very much smaller
than the actual value, and proves that the above method of allowing for the
reluctance of the iron is incorrect. The reason for this difference between
the calculated and actual current is to be found, mainly, in the fact that the
iron has a particularly high permeability at the mean value of the flux
density, viz. 10,200 lines per cm. We have assumed that the permeability
was equally good at every point, which is far from being the case.
A more
reasonable method would be to take the mean permeability at the various
flux densities, instead of the permeability at the mean flux density. Even
then, the result is little better.
Bt
5000
Bg


6000
5000
13000
4.000
11000
9000
3000
7000
2000
5000
1000
2 4 6 8 10 12 14 16 18 20
100 200 300 400 500 600 700 800 900
Xg+Xt
Fig. 379.
Fig. 380.
(+)-Ampere-turns per Cm.
The only reliable method of accurately determining the magnetising
current is to draw a curve giving the relation between the flux density at
any point of the air-gap and the ampere-turns necessary to produce this flux
density*. Then, knowing the excitation acting at each point of the periphery
we can find from this curve the corresponding flux density. To plot the
curve we choose several values of B, and calculate the ampere-terms neces-
sary for the gap alone from the formula
Xg=0·8. Bg . lg,
or, since in our case l, = 0·15 cm.,
Xg=0·8. Bg. 0·15 = 0·12B,.
We then calculate the corresponding flux density in the teeth by multi-
plying the density in the gap by the ratio of the cross-sections. In our case
then we have
T. E.
B₁ = 2·95Bg.
* See Kapp's "Dynamo Construction."
24
370
Electrical Engineering
I
The ampere-turns per cm. of path for the flux density B in the teeth is
found from the magnetisation curve for the stampings (Fig. 379). We have
then
Xt
X₁ = (+)...
祭
​.lt.
t
For lt we must put the total length of path in the teeth, that is, two
stator teeth and two rotor teeth, so that
X₁ = (†).,.2.6-12 (†).
Xt
t
t
Adding X, and Xt, we have the total excitation X necessary to produce
the assumed flux density. This calculation has been carried out for values
of B, from 2,000 to 5,200, and the results are given in the following table :
B, X=0·12B, B=2-95B, (4), 1,=12 (†), X,+X,

2,000
240
5,900 1·1
13
253
3,000 360
8,850 1.9
23
383
4,000 480
11,800 3.6
43
523
4,500 540
5,000 600 14,800 10.7
5,200 624 15,300 18.5
13,300
5'4
65
605
128
728
222
850
g
In Fig. 380 we have plotted the values of X,+ Xt as abscissae, and the
corresponding values of B, as ordinates.
We must now assume a value of the no-load current and find the flux
produced by it. If the result comes out far from the actual flux, we shall
have to repeat the calculation with another value of the current. Seeing
that our approximate calculation gave 48 amperes, we shall estimate the
current at 5.5 amperes, so that
and
i=5·5 amperes,
20 max = √/2.107.78 amperes.
Half the pole-pitch is taken as a base line and the excitation is plotted at
every point, giving the curve X in Fig. 381, similar to the curve obtained in
Fig. 373 for the same conditions. As before, it can be shown that the maxi-
mum excitation is
2q'iomax = 2.56.7·78 = 870.
For an abscissa of π/6 the excitation has half this value. The scale of
ampere-turns is plotted on the left-hand vertical axis. Points have been
marked on this axis corresponding to the values of X + Xt in the above
table, viz.
X = Xg+Xt=253, 383, 523, 605, 728, 850,
and horizontal lines have been drawn through these points to meet the curve
X. Ordinates have been drawn through the points of intersection, and made
equal to the corresponding values of B, in the table, viz.
B₁ = 2,000, 3,000, 4,000, 4,500, 5,000, 5,200.
The effect of the iron in the magnetic path
371
The scale for B, has been arbitrarily chosen and is plotted on the right of
the figure. The curve B, is drawn through the points found in this way, and
represents the field distribution for a current of 5.5 amperes. The average
flux density is determined by measuring the area enclosed between this curve
and the base, and dividing it by the base.
The result obtained with the aid of a planimeter was
Bg mean = 3,635.
The same construction has been carried out in Fig. 382 for the other
extreme condition, corresponding to Fig. 374. As found in connection with
the latter figure, the maximum excitation is here
1.73 q'iomax=173.56.778-750.
The result obtained from Fig. 382 was
X
1000
850
Bg
728
605
523
383
253
FLO
Bg mean = 3,583.

Ba
5000
Ba
4000
誓
​吾
​3000
2000
1000
Fig. 381.
Fig. 382.
The mean of these two values of the flux density in the gap is
Bgmean
3,635 + 3,583
2
= 3,609.
The total flux entering the rotor per pole is therefore
N = Bg mean · Ag = 3,609. 763 = 2·75 . 10º.
This value is so near the required flux of 2.64.10 lines that we may
safely assume proportionality between current and flux. The magnetising
current required for a flux of 2.64 million lines will therefore be
cores.
го
5.5.2.64.106
2.75.106
5.3 amperes.
Now, we have neglected throughout the reluctance of the stator and rotor
In addition to this, we have taken the flux density in the air-gap as
uniform, whereas the slots will cause it to vary, thus increasing the maximum
24-2
372
Electrical Engineering
density and the necessary excitation. Finally, we must remember that the
observed no-load current of 6 amperes contains an energy component to cover
the friction and iron losses. We cannot be surprised, therefore, that the
calculated current of 5-3 amperes is about 10 per cent. smaller than the
measured current of 6 amperes. Our last result is certainly more reliable
than that obtained by taking the permeability for an average value of the
flux density.
121. The torque of an induction motor.
To calculate the torque we shall assume that the rotating field is exactly
sinusoidal, with the same flux per pole as the actual field. We shall confine
our attention in the first place to a coil-side of the rotor-winding, with a
breadth 2y and with its centre displaced by the angle a, from the point
where the field is zero (Fig. 383). We saw in Section 90 that the E.M.F.
induced in such a coil-side, and therefore the current in it, is proportional to
the sine of the angle a₁. The instantaneous value of the rotor current is
therefore
i2i2max. sin a₁.

гд
000000
Fig. 383.
This current is the same for every wire in the coil-side, but the field in
which each wire is situated is different. To find the value of the torque at
this moment, we must determine the average strength of the field over the
arc 2y. The flux density at any point is equal to Bmax. sin a, where a is the
angle between this point and the point of zero field strength. The average
over the arc 2y is found as follows:
1
raity
B
Bmax. sin a. da = Bmax
•
sin y sin dı.
27
Y
To calculate the torque we shall require the product of the mean field
strength and the momentary value of the current. We have
i2. B=Bmax. i2max
sin y
Υ
sin² α1.
In this equation i, is the momentary value of the current, and B the
mean value of the field strength over the coil-side at that moment.
Now the coil-side moves steadily through the field, i.e. the angle ₁ varies
from 0° to 360°, and, since the value of the product i. B is proportional to
the square of the angle a₁, its mean value can be found in exactly the same
The torque of an induction motor
373
way as the mean power was determined in Section 71. Analogous to
equation (115) on page 223, we
(B. ¿½)mean
have
Bmax. 2max sin y
2
Y
The force on the coil-side in dynes is found by multiplying together the
strength of the field, the absolute current in the wire, and the total length of
effective wire in the coil-side. Since, however, the same mean force is given
by every coil-side during a period, the total tangential force on the rotor is
equal to f, where
f:
=
Bmax. 2max sin y
2.10
. 22. L dynes
Y
.(173).
This force acts at an arm equal to the rotor radius r. Since the
cylindrical surface of the rotor is equal to the product of the polar surface
A, and the number of poles 2p, we have
or
2πг. L=2p. Аg,
p. Ag
T. L'
To convert the torque or turning-moment from dyne-cms. into kg.-metres,
we must divide by 981,000. 100. Thus we get
Mt
=
Since
f.r
9.81.107
2
π
2p. Bmax. Ag. 22. 2max sin y
π.40.9.81.107
. Bmax = Bmean
Υ
..(174).
2
Bmax · Ag= N.
π
Substituting, further, the effective current i₂ for i2max/√/2, we get
sin y
Mt = 3·6.p. N. z2 . ¿½ . 10−10 .
γ
.(175).
The breadth y becomes 0 for a squirrel cage, is equal to π/3 for a coil-
winding, and 2π/3 for a creeping bar-winding.
The ratio
sin y
Υ
entered into the coefficient k in the equations (144), (145)
and (146) for the electromotive force. We come, therefore, to the conclusion
that, with various rotor-windings, the torque is proportional to the coefficient
k for the E. M. F. induced in the rotor.
The same result, together with the formula for the torque, can be obtained
directly by imagining that the rotor is fixed, and that the whole power
supplied to it is dissipated in heating the rotor or a resistance connected in
series with it. Let N be the actual field cutting the rotor bars, as the
resultant of the stator and rotor currents. This field cuts the stationary
rotor-winding with the primary frequency=p.
tions made by the field per minute. The current in the rotor will be in
phase with the E. M. F. induced by this field. If this E.M. F. in each phase be
•
N1
60
where m
is the revolu-
374
Electrical Engineering
E₂, and the total number of rotor conductors z,, then the total power supplied
to the rotor will be
2
~1 · 22 · 10˜³. ¿½. … … …..(176).
P₂ =3E,. i₂ = 3.k. N. ~₁.z½.10~8 . ¿½ = k . N . ~1. 22. 10−8. ¿½..
This power is supplied to the rotor by a field which rotates with an
angular velocity
w= 2π.
NĮ
60
2π.
Ρ
Now, the power in met.-kgs. per second is equal to the product of the
torque and the angular velocity. Remembering that equation (176) gives
the value of P₂ in watts, and that a met.-kg. per second is equal to 9.81 watts,
we have
2
P₂ = Mt. w. 9·81.
Substituting for P, and w, we get
2
Mt=
k.p.N.z.i2.10-8
2π.9.81
(177).
We have assumed here that the rotor is fixed, but this is quite immaterial
so long as the values of N and i₂ remain the same. If we substitute for k the
values found for it on page 362, we get the same values for the torque as
those found above. This is therefore a check on the above result.
It is interesting to compare the torque of a three-phase induction motor
with that of a D. C. motor. For the latter we found, in equation (98) on
page 191, that
ia
2α
M
p.N.z.ia.10-8
а.2π.9.81
ia
= 3·25. p. N. z.
10-10
2a
From equation (175) for
is the current in each armature conductor.
a squirrel cage rotor we get a coefficient of 36 instead of 3·25, but the
equation is otherwise the same. Hence, other things being equal, the
squirrel cage three-phase motor gives a 10 per cent. greater torque than the
direct current motor.
122. Calculation of slip.
We saw in Section 117 that the completely unloaded rotor runs synchro-
nously with the field, i.e. at the speed
NĮ
~1.60
Ρ
We saw, moreover, that, as the motor is loaded, the speed decreases to a
value n such that the rate at which the lines of force are cut is sufficient to
induce the necessary E. M. F. E, in the rotor. In this way a current is produced
in the rotor, and this current produces the necessary torque, in accordance
with the equations established in the previous section.
* In this case we must use the mechanical angular velocity and not the phase or vectorial
angular velocity 2π~1=2π · P · 60
•
N
Calculation of slip
375
The slip, or the frequency ~, with which the field cuts the rotor-winding,
corresponds to the difference between the speed of the rotating field and that
of the rotor, and
Nr
N
60
.p.
The small E.M.F. which is thus induced in the rotor-winding is given by
the equation
E₂-k. N~z. 10-8.
2
By N we understand the field which actually cuts the rotor conductors, so
that there is no difference of phase between E, and the rotor current 2. If
the resistance of each phase of the rotor-winding be R₂, we have
E₂ = i2. R₂.
2
The power wasted in heating the three-phase windings of the rotor is
therefore
3. i. R₂ = 3. E2. ik. N~ 2. 10-8. iz............(178).
z, has been substituted for 3z. By dividing this equation by equation (176)
we get
3122. R₂
P₂
.(179).
Hence, the slip, expressed as a percentage of the synchronous speed, is
equal to the percentage of the power supplied to the rotor which is lost in
heating the rotor-winding. This explains why motors are always made with
a small slip. It is evident that the slip can be made to have almost any value
by suitably altering the rotor resistance.
It is interesting to compare the behaviour of a three-phase motor with
that of a direct-current shunt motor. For the latter we have the equation
e.ia = E. ia + iàª. Ra·
2
α
In this equation e. ia is the power supplied to the armature, while E. ia is
the power transformed into mechanical output. These two powers are in the
ratio of e to E, that is, the ratio of the speed no at perfect no-load to the
speed n when loaded. Hence
and
по
По
E.ia
e.va
n
По
n__e.ia - E. ia ia².Ra
e.va
e.va
The percentage drop of speed from no-load to full load is therefore equal
to the percentage loss of power in the armature winding.
Let us now consider the case in which the slip is 100 per cent. This
occurs at the moment of starting, before the motor begins to rotate. The rotating
field cuts the rotor-winding at the high primary frequency, and induces a
very large E.M.F. in it. The current in both rotor and stator might then be
so great as to damage the motor and cause a complete breakdown. It is for
this reason that squirrel cage rotors are generally confined to small motors.
They can be used, however, for large motors, if they are brought up to the
normal speed before switching the stator on the supply mains, or if they are
376
Electrical Engineering
started up together with the generator. Large squirrel cage motors are
largely used in the United States; they are generally started at a reduced
terminal pressure, a special transformer being supplied for the purpose.
These
The majority of motors above about 5 H.P., and many smaller ones, have
rotor-windings into which resistances are introduced at starting.
windings can be connected either in star or mesh. Although a mesh-connected
rotor-winding is closed on itself, it is electrically open until the three corners
of the mesh are connected together, either through the external starting
resistance or by short circuiting the slip-rings. The starting resistances are
most conveniently arranged in star connection.
Liquid resistances are often used for starting induction motors. The
resistance is gradually reduced by dipping the electrodes further into the
liquid. The starting resistance does not merely reduce the current taken by
the motor but materially increases the starting torque (compare Section 126).
It is evident from equation (179) that the slip is increased and the speed
consequently decreased by putting resistance in the rotor circuit. This is
analogous to putting resistance in the armature circuit of a shunt motor and
leads to the same large waste of power. The speed varies, moreover, with
every change of load.
It has been tried to vary the speed by altering the number of poles of the
stator, but the necessary connections are very complicated. If two motors
are coupled, as is generally the case in traction, they can be made to run at
half speed by supplying the primary of one from the secondary of the other.
This so-called cascade arrangement has been adopted on an Italian railway.
It cannot be denied, however, that the constant speed of the induction motor
and the lack of an efficient speed regulation make it very unsuitable for a
great many applications.
The consideration of a motor with its rotor fixed and considerable resist-
ance in series with the rotor-winding is of special interest. The whole power
supplied to the rotor is dissipated as heat in the rotor and starting resistance,
and the motor is nothing more than a transformer. If the current is the
same, the total power supplied to the rotor is the same as when the motor
was running. The heat produced in the starting resistance is equal to the
mechanical output of the motor when working normally at the same current.
The study of the induction motor is much simplified by looking upon it as a
transformer, to which the laws and equations which we have established for
transformers can be applied. The fact that the field is here a constant
rotating one, and not a simple alternating one, introduces no material
difficulty.
CHAPTER XVIII.
123. Rotor current, torque, and output in relation to the slip, neglecting leakage.—124. The
circle-diagram, neglecting primary losses.-125. Output, torque and slip from the circle-
diagram.-126. Normal load, starting torque and maximum torque.-127. The circle-
diagram, corrected for primary copper loss.—128. The corrected values of output, rotor
current and slip.-129. The most convenient form of circle-diagram.-130. Practical
example.-131. The leakage factor.
123. Rotor current, torque, and output in relation to the
slip, neglecting leakage.
We shall assume, for the sake of simplicity, that there is no magnetic
leakage, so that no lines of force pass across the slots in stator or rotor.
There is then only one flux which crosses the air-gap and links both stator
and rotor conductors. This flux is produced by the combined action of the

X
x
+
X
X
A
X
X
Χ
B
D\
节
​Fig. 384.
stator and rotor currents. We shall assume that this flux N has a positive
maximum at ▲ and a negative maximum at B, so that A is the mid-point of
a north pole and B the mid-point of a south pole. The lines of force pass
vertically downwards, and the vector OX may represent the ampere-turns or
resultant excitation producing this flux.
If the field be rotating in the clockwise direction, it will induce a current
in the rotor with maximum values at A and B.
378
Electrical Engineering
By means of the right-hand rule it is seen that the current at A is
towards the observer, while at B it is away from him. Each coil-side of the
three-phase rotor-winding is distributed between 4 slots. The points and
crosses in the slots at A and B are drawn heavy to indicate the maximum
current. In the other two phases the current has half this value.
2
The vector OX, of the excitation due to the rotor currents alone must be
drawn horizontally from left to right, since this is the direction of the flux
which the rotor currents alone would produce.
1
In order to give the resultant excitation OX, it is necessary for the stator
excitation to be equal to OX, both in magnitude and phase. The stator
currents have their maximum values at C and D, with crosses at C and dots
at D.
The stator has 3 slots per pole per phase. The current will have its
maximum value in the phase CD and half that value in the other phases.
From Fig. 384 we see
(1) that the primary and secondary currents are in approximate
opposition, as in the ordinary transformer;
(2) that the current induced in the rotor is strongest at the point
where the field cutting the rotor is strongest; and
(3) that the rotor-current vector is at right angles to the rotor field
vector.
When, now, we remember that the rotating field cuts the stator as well
as the rotor, we see that the E.M.F. induced in the stator-winding will be a
maximum at points A and B, whereas the maximum stator current is at
C and D.
We see then
(4) that the stator current lags behind the stator pressure by the
angle in the figure.
This comparatively large angle of lag is due to the fact that the induction
motor has to produce its own field, and that, since the magnetic path neces-
sarily includes the air-gap, there must be a large wattless or magnetising
current. Although the air-gap is sometimes reduced to a quarter of a
millimetre or less, it is only in large motors that cos & can be got up to 0·9,
and only in extremely large ones that this value of the power-factor can be
exceeded. The wattless current, as we have already seen, causes a needless
loss in the wire in generator, mains, and motor, besides having a bad effect on
the pressure regulation of the former. For this reason, induction motors
must be constructed relatively larger than direct-current and synchronous
motors. Finally, the large magnetising current is especially unfavourable
when the motor is running light.
We shall now calculate the most important values for an ideal motor free
from magnetic leakage. If the ohmic pressure drop in the stator is small,
the induced E. M.F. in the stator will be almost equal to the constant pressure
e applied to each phase. It also depends on the magnetic flux N which, in
the absence of leakage, links both stator and rotor. The induced E.M.F. in
the stator is given by the equation
E₁k. N~. 10-8.
Rotor current, torque, output, etc.
379
If the value of E, is equal to that of e, it must be constant, and, as is
evident from the above equation, the flux N must also be constant.
If this field, which we have seen to be common to both stator and rotor,
cuts the rotor conductors with a frequency ~ corresponding to the slip, an
E.M.F. will be induced in the rotor-winding, in accordance with the equation.
E₂ = k . N~ z2 . 10−8.
2
We shall assume, for the sake of simplicity, that the windings of stator
and rotor are similar, so that both have the same value of k.
2
The rotor current i will be equal to E₂/R₂, and is therefore proportional
to the slip. We can express this by the equation
where c is a constant.
i₂ = c₁.~,
The torque is proportional to the product of the rotor current 2 and the
flux N (equation 177). In our case the latter is constant, while the current
is proportional to the slip. By introducing a constant c, we can write
Mt
= C₂. ~.
C2 •

Ma
P
iz
P2
B
A
-72·
-no
Fig. 385.
The power transmitted to the rotor is also proportional to the product
2. N (equation 176). Since the flux is constant, this power is therefore
proportional to the slip, and we can put
where c, is another constant.
P₂ = Cs.~,
C3
The mechanical power P, including the power wasted in friction, can be
found by subtracting the copper loss in the rotor from the total power trans-
mitted to the rotor. Now, the copper loss is proportional to the square of the
current, and therefore to the square of the slip. If c be another constant
and R, the resistance of each phase of the rotor, we have
P=P₂-3.12. R₂ = C3. ~ C₁. ~².
•
The rotor current, torque, transmitted power and mechanical power are
plotted in Fig. 385 on a base line representing speed. OA is equal to the
speed n。 at no-load and OB is equal to the speed at a certain load. The
difference OA - OB= AB is equal to the revolutions per minute lost by the
380
Electrical Engineering
slip. The first three of the above quantities are proportional to the slip and
will therefore be represented by straight lines through A. The mechanical
power, on the other hand, is given by the ordinates of the parabola in the
figure. It must be zero, both at O where the rotor is at rest, and at A where
the motor is running light. It will reach its maximum value for a speed n
equal to
No
2
, i.e. for a slip of 50 per cent. The normal working position is
much further to the right where the slip is small and the efficiency con-
sequently high.
An actual motor differs from this ideal motor in almost every respect, but
especially with regard to the starting torque and overload capacity.
At starting, the speed is nil and the torque, according to Fig. 385, is very
large. As a matter of fact, the effect of leakage in an actual motor is to
make the starting torque extremely small. Resistance has to be inserted in
the rotor circuit to enable the motor to start against the normal load.
This can be explained by the large leakage flux caused by the excessive
currents in stator and rotor when the slip is 100 per cent. The flux which
cuts the rotor conductors is thereby so reduced that there is very little torque
in spite of the large rotor currents. Hence, the starting resistance not only
reduces the otherwise excessive current taken by the motor, but also enables
the motor to exert a large starting torque. This will be more clearly under-
stood after studying the circle-diagram of Heyland, to which we now proceed.
124. The circle-diagram, neglecting primary losses*.
We shall consider, in the first place, the general effect of the leakage,
before passing on to the vector diagram and an accurate determination of the
various quantities involved.
When there is no load on the motor, the total stator field is made
up of
two fields, one, the flux which crosses the air-gap and enters the rotor, the
other, the primary leakage flux N. Both fluxes are produced by the no-load
stator current, and are therefore in phase with each other, and inversely pro-
portional to the magnetic reluctances of the air-gap path and the primary
leakage path.
When the motor is loaded, the flux N,, which crosses the air-gap, is pro-
duced by the combined action of the stator and rotor currents.
It was formerly the custom to consider this air-gap flux as linking the
rotor-winding (Fig. 386) and inducing therein an E. M. F. sufficient to maintain
* The historical development of the circle-diagram is very interesting. Heyland published
the diagram in the E. T. Z. on the 11th Oct. 1894, and gave further developments on pages 649
and 823 for the year 1895. In the E. T. Z. 1896, page 63, Behrend developed the diagram
analytically, but made a small error in the determination of the rotor current. The convenient
determination of the slip and losses was given by Heyland in the E. T. Z. for 1896, p. 138. (See
also Heyland's "Eine Methode zur experimentellen Untersuchung an Induktionsmotoren,"
published in Voit's "Sammlung," Vol. п, 1900.) Emde corrected Behrend's error in a letter to
the E. T. Z. 1900, p. 781, which opened an interesting discussion. In the "Z. für E." Vienna,
for 1899, Ossanna gave the diagram, corrected for stator loss. (See also an article by Ossanna in
the E. T. Z. 1900, p. 712, and also by Thomälen in the E. T. Z. 1903, p. 972.) It is interesting
to note, however, that Ossanna's circle was really included in Heyland's first publication.
The circle-diagram, neglecting primary losses
381
g
the rotor currents against the resistance and self-induction of the rotor-
winding. The ohmic pressure drop in the rotor would then be equal to the
resultant of two electromotive forces, produced by the fluxes Ñ, and N₁₂
respectively. This second flux is the rotor leakage flux and it is not in phase
with the air-gap flux. As a matter of fact, however, these two fluxes do not
exist separately in the rotor, but combine to form the resultant rotor flux N.
It is therefore preferable to consider that only a part N of the air-gap flux Ng
passes round the rotor-windings, while a part N₁, is forced along the leakage
paths by the back ampere-turns of the rotor current (Fig. 387).
In Fig. 388, let the vector LD represent the flux N which passes through
the rotor-winding. The E.M.F. induced in the rotor-winding by this flux will
lag 90° behind it, and will be in phase with the rotor current, since, in
accordance with Fig. 387, this E. M.F. has to cover merely the ohmic pressure
drop and not the self-induction. Hence LB = 2, is perpendicular to LD.


Ni
Ng
N2
N
Fig. 386.
Fig. 387.
The secondary leakage flux N₁₂ = LE = DA is drawn in the opposite
direction to the rotor current, because, from the point of view of Fig. 387, this
flux is produced by a current which exactly balances the back magnetomotive
force of the rotor current.
The flux N, in the air-gap is the sum of the rotor flux N and the
secondary leakage flux N. We have, therefore, N=LA.
The magnetomotive forces of the stator and rotor currents are nearly
opposed to one another. Their resultant, however, must be such as to
produce the flux N, in the air-gap. If c be a constant and R the reluctance
of the air-gap and if, moreover, both stator and rotor have the same number
of turns, Ohm's law for the magnetic circuit gives us the equation
c. ig
N, -
R
In this equation i, is the magnetising current which is necessary to maintain
the flux N, across the gap. It will be the resultant of the primary and
secondary currents. If the scale to which the current is plotted be suitably
chosen, the length LA can be made to represent the current ig as well as the
flux Ng. If LB represents i, to this same scale, the primary current & must
be represented by the line AB. For the sake of compactness, we have drawn
the triangle of forces instead of the parallelogram.
382
Electrical Engineering
The primary leakage flux N₁, which links the stator-winding without
crossing the air-gap, is in phase with the primary current, and is represented
by the line 40 in continuation of the line BA. The total flux N, in the
stator is represented by LO. With a constant terminal pressure, and
consequently, on our present assumptions, a constant induced E.M.F.
in the stator, the primary flux N₁ = LO is constant at all loads.
Let R₁, and R₂ be the reluctances of the primary and secondary leakage
paths, and let
Τι
=
Ꭱ
Ru
R
T2
R₁₂

B
i z
î,
D
Ni-Triz
NETi
Tie
B
G
Frio
io
No
L
B
N
Ng
E
i,
Fig. 388.
Then, from the magnetic Ohm's law, we have
c. i
Ru
C. 21
C
AO = N₁
c. vz
C
R
T2. 12 = T₂. 12.
DA = Ni₂ = R₂
It is to be noted that we have so chosen the scale of current in Fig. 388,
that the same length, viz. AL, represents both i, and Ng, which, since
No
=
R
c., is equivalent to putting
C
Ꭱ
=1. Now, the point A divides the
.ig
line OB in the ratio of T₁ to 1, and as DAF is parallel to BL, it follows that
Τι
FA=
12.12.
1+ T₁
It is also evident that the point F divides the primary flux LO in the
ratio 7₁: 1, that is, in the ratio of the reluctances of the air-gap to that of the
primary leakage path. Since the flux will divide between two parallel paths
1
1
The circle-diagram, neglecting primary losses 383
in inverse proportion to their reluctances, OF must be the primary leakage
flux at no-load, and FL the corresponding flux in the air-gap. If i be the
no-load stator current, it follows from the magnetic Ohm's law that
whence
FL
c.io
R
= io,
OF=T₁. FL = T1 • 10.
If we draw AG parallel to DL, i.e. perpendicular to FA, we get
Τι
i₂
FG = FL.
FA
FA+AD
1+ T1
Τι
1+ Ti
· i½ + T2. iz
T= T1+T₂+ T1 T2,
FG - 20 - T₂
By simplifying this and putting
we get
T
Hence, both F and G are fixed points on the constant line LO, and the
point A moves on a semicircle on GF as diameter. If, now, the whole figure
be enlarged in the ratio 7:1, and allowance be made for the fact that the
rotor conductors are not equal in number to those on the stator, but are only
of the latter, we have the following values:
Z2
21
OF = io,
FG = ²
T
1
i222
V 2 Z2
OA = i₁,
FA
1+7₁ 21
.(180),
..(181).
The end of the primary current vector moves on a semicircle on the
diameter 20 (Fig. 389).
T
It is evident that the conditions at various loads are largely dependent
on the leakage coefficients 71 and 7, and the resulting coefficient 7. The
values of T₁ and T₂ lie between 0·07 and 0·02. If 7, and 7₂ are assumed to be
equal, the value of 7 can be found from the table on page 387.
T1
T2
125. Output, torque and slip from the circle-diagram.
From equation (176) on page 374 we know that the power supplied to
the rotor is
P₂ = k. N~₁₂. 10-8.12.
2
Since the point F in Fig. 388 divides the line OL in the ratio T₁: 1, it
follows that FL
From the same figure, we have
No
1+71
No
=
N=LD FL.cos B
cos B
1+ 1
and, from equation (181),
i2. Z₂ = (1 + T₁) . FA . ≈₁ = (1 + 7₁) FA . 3 . zí.
(182),
384
Electrical Engineering
AH
If, now, we put FA = (Fig. 389), and substitute these values of
cos B
i2.22 and N in the above equation for the power supplied to the rotor,
we get
P₂ = 3. k. N, ~₁ ~, . AH.
2
1
On the assumption that there are no losses in the stator, the terminal
pressure e, will be represented by a vector perpendicular to N, and also to io,
as shown in the figure. The ordinate AH is therefore the energy component
cos o of the primary current. The product k. N~z, is equal to the
electromotive force E₁, and therefore also to the primary terminal pressure
e. On substituting these values, we get
P₂ = 3. e₁. AH=3. e. i. cos
2
..(183).
This proves that the power supplied to the rotor is equal to that supplied
to the motor itself, which is, of course, a necessity, since we have neglected
stator losses.
K

A
M
Bo
༤
G
N
H
F
ίο
1
io
Z
Fig. 389.
2
=
The power P, is transmitted to the rotor by means of a field which
rotates with an angular velocity w 2π/p, and it is immaterial whether
the rotor is fixed or rotating. The torque is obtained by dividing the power
by the angular velocity, hence,
Mt
P₂
@.9.81
P₁.p
2π. ~1 9.81
met.-kgs.
•
If we substitute for P₂ its value from equation (183), we get
2
Mt
2π.
3.e.p
~1.9.81
AH:
e.p
20.5.1
AH met.-kgs. ......(184).
Not only does the Heyland diagram, however, enable us to read off the
power transmitted to the rotor and the torque exerted, but it also gives us a
direct reading of the slip corresponding to any value of the stator current.
Equation (179) on page 375 expresses the important fact that the percentage
slip is equal to the percentage power lost in the rotor-windings, or
ΙΣ
3.i. R₂
P₂
Output, torque and slip from the circle-diagram
385
By squaring each side of equation (181) on page 383, we get
i3.2,2
(1 + 7₁)². %₁²
and from the figure it is evident that
=
FA2,
FA
GF
or FA²
=2.HF.
HF FA
T
It follows from these two equations that
(1 + T₁)² io 21²
12²
2.HF.
T
Z2
By substituting this for 22 in the above formula for~
and putting
P₂=3е₁. AH, and
HF
AH
tan B, we have
༤=
= C. tan B
where
C:
(1 + T₁)² 2₁² i。. R₂*
T
2
22 e1
.(185),
..(186).
The percentage slip is therefore proportional to the tangent of the angle
B. To determine it graphically, the line GK in Fig. 389 is drawn so as
to make an angle B. with the horizontal, where
1
tan B. = -
.(187).
If this line cuts the semicircle at J, when the load is such that the primary
current becomes OJ, the slip will be equal to ~1, because
~
C. tan B. 1.
The slip will then be 100 per cent., and the motor is therefore at rest.
OJ is the so-called short-circuit current, i.e. the current taken by the stator
when the rotor is fixed, or at the moment of starting.
A point K is taken on the line GJ so that the perpendicular KN on to
the base is exactly 100 mm. long. This perpendicular cuts the line GA in
the point M. When the primary current is OA, we have
C. tan B
=
tan B
tan B.
MN
KN'
and since KN is 100 mm. the length of MN in mm. is equal to the per-
centage slip.
We could determine the mechanical power, or output of the rotor, by
decreasing the power supplied to the rotor by an amount corresponding to
the slip. A simpler way of proceeding is to join FJ, cutting the ordinate AH
in the point P. The angle FPH is equal to ẞ。, and we have
* The resistance
212. R₂
222
tan B HF HE PH
tan B AH PH˜AH'
is the secondary resistance reduced to its equivalent primary value,
as we have already seen in connection with transformers. If the windings on stator and rotor
are dissimilar, k², and kż½ must be substituted for z, and z₂ in equations (181) and (186).
(Compare page 366.)
T. E.
25
386
Electrical Engineering
Since the percentage slip is equal to the percentage loss in the rotor,
3. i. R₂
P₂
2
PH
AH'
We have already seen that AH is a measure of the power transmitted to
the rotor, whence PH must represent to the same scale the loss in the rotor,
and the difference AP, the mechanical power or output. It is evident from
equation (183) on page 384 that the mechanical output P may be expressed
as follows:
P = 3. e₁. AP watts.
This output consists of the frictional loss and the useful output.
126. Normal load, starting torque and maximum torque.
The induced E.M.F. lags, roughly speaking, 90° behind the magnetising
current, and the E.M.F. vector should be drawn vertically downwards in
Fig. 390. If the stator losses be neglected, the terminal pressure vector will
be exactly opposite to the E.M.F. vector, and will make an angle ø with the
primary current, as shown in Fig. 390. The angle of lag will be evidently a

でっ
​ei
ዋ
Fig. 390.
io
Lo + io
o+ió
27
minimum when the current vector is a tangent to the semicircle, and the
motor should be so designed that the normal working conditions correspond
approximately with this position. The maximum value of the power-factor
is given by the following equation :
го
2T
1
(cos +)max
го
+ 20
2T
2T+1
..(188).
Hence, if the primary losses be neglected, the maximum value of the
power-factor depends only on the leakage coefficient 7.
The table given below shows the maximum power-factors attainable with
different values of the leakage coefficients, on the assumption that 7₁=72.
We see that a value of cos p=0·9 can only be obtained by making 7, and T₂
as small as 0·03, which is about as small as it is possible to reach *. It is
* Heyland makes cos =1 by fitting a commutator with a small number of segments to the
rotor and supplying the magnetising current to the rotor instead of the stator. Since the
Normal load, starting torque and maximum torque 387
further evident that, for a given value of 7, the value of the primary current
i, corresponding to the maximum power-factor, has a definite relation to the
no-load current i. This is expressed in the equation
in
i. 2T
=
(cot )max.
From this equation we can determine the most advantageous value of the
ratio i/. The values have been calculated and are given in the table below. We
see that under the best conditions the no-load current lies between and of
1348
the normal full-load current. This relatively large no-load current is caused
by the air-gap between stator and rotor, and, although a wattless current, is
one of the greatest disadvantages of an induction motor. Even when the
leakage of stator and rotor is reduced to 2 per cent. each, the no-load current
is still 0·197 of the normal current. For example, a 500 H.P. motor of
the Alioth Company*, which, on account of its size, should give very good
results, had a no-load current of 9.3 amperes, and a full-load current of
50 amperes. This gives a value of the ratio i, i, equal to 93: 50 or 0.186.
It is safe to conclude that this motor had leakage coefficients 71 and 7, of
2 per cent., since the normal current vector of 50 amperes would almost
certainly be tangential to the Heyland circle.
:
Τι
T2
We turn now to the overload capacity of the motor, still keeping to the
assumption that the motor runs at its normal full load with the smallest
possible value of 4. The primary current i, is the tangent to the semicircle,
and (cos $)max is a measure of the normal torque (equation (184)). The
maximum possible torque is given to the same scale by the radius The
overload capacity, or the ratio of the greatest possible to the normal torque,
is therefore
maximum torque
i0/2T
normal torque ¿½. (cos p)max
•
io/i
27. (cos )max
го
2T
The last column of the following table has been calculated from this
formula.
饔
​
1
T1=T₂ T=T₂+ T2+T1T2 (COS )max
=2T. (cot $)max
2r+1
Overload ratio
io/i1
2T . (COS P)max
0.07
0.145
0.776
0.355
1.58
0.06
0.124
0.802
0.33
1.65
0.05
0.103
0.83
0.305
1.78
0.04
0.082
0.86
0.275
1.95
0.03
0.06.1
0.892
0.24
2.2
0.02
0.040
0.925
0.197
2.63
frequency of the E. M. F. induced in the rotor is very small, a small pressure is sufficient to drive
the magnetising current through the rotor-windings. This small pressure is usually tapped off
from the stator coils and applied to the brushes. The wattless magnetising current is practically
negligible compared with the whole current supplied to the stator, so that the motor has a
power-factor of unity. (See the E. T. Z. 1901, p. 633; 1902, pp. 28 and 533; 1903, pp. 51, 72,
95 and 213.)
* See E. T. Z. 1901, page 547.
25-2
388
Electrical Engineering
*
It is evidently very difficult to obtain an overload capacity of 26 times
that obtained with the maximum value of cos p. It is possible, of course, to
design the motor so that the normal primary current vector falls below the
tangent. The value of cos is not greatly decreased, and the overload
capacity, with regard to the normal torque, is increased.
A point of special interest and importance is the torque at starting. If
the resistance of the rotor is negligibly small, equations (186) and (187) give
us the following result:
1
tan Bo
C
T. 22. e1
2
(1 + 7₁)² . z₁² . R₂. ¿。
=∞
The angle B, is then equal to 90°, and the points J and G in Fig. 389
fall together. OG is therefore the theoretical stator current at starting, on
the assumption that the rotor resistance is negligibly small. Under these
conditions the starting torque is evidently zero.
Even when the actual resistance of the rotor is put into the equation
for B., the line OJ is still very nearly perpendicular, and the point J, the
ordinate of which gives the starting torque, is still very low. It is evident,
however, that the resistance of the rotor circuit can be increased to any
desired extent by putting starting resistance across the slip-rings. As can
be seen from the above equation, the angle ẞ, will be thereby decreased, and
the ordinate of the point J, which gives the starting torque, increased. To
make the motor start with its maximum torque, the angle B, must equal 45°,
so that tanß₁ = 1. If the combined resistance of rotor and starting resistance
per phase be R₂, the conditions for this maximum starting torque are as
follows:
010
tan Bo
R₂
T. 22². C₂
(1 + 7₁)² . ¿¡ . 21² . R₂
T. 222. e₁
(1 + T₁)² · 20 · 21 2
•
•
=
1,
With this resistance per phase the motor will exert the maximum torque,
as given by the diagram, at the moment of starting.
127. The circle-diagram, corrected for primary copper loss.
i。.
Up to this point we have assumed that the resistance of the stator-
winding and the consequent losses could be neglected. On this assumption
the terminal pressure e, can be represented by a vector equal and opposite to
that of the induced E. M. F., and making an angle of 90° with the no-load
current . As a matter of fact, however, the terminal pressure e, has not
only to overcome the back E.M.F. E₁, but has also to drive the current i, through
the resistance R₁. In Fig. 391 the terminal pressure vector e, is drawn
vertically, as before, and the primary current vector OP lags behind e
by the angle . Along OP is drawn the primary copper drop OP, i₁ . R₁.
The vector E, which counterbalances the back E.M.F. is the side of a parallel-
ogram, of which OP, is the other side and e, the resultant diagonal. E, is
evidently smaller than e₁, and the magnetic flux and magnetising current
The circle-diagram, corrected for primary copper loss 389
1
must be correspondingly reduced. We may assume as an approximation
that the reduction of magnetising current is proportional to the reduction of
flux. The reduced magnetising current OF, is at right angles to E₁, and
OF₁: F₁G₁ = T. The end P of the primary current vector lies on the circle
on the diameter F₁G₁ with the mid-point P₁. The position and size of this
circle varies with every change in the load, but we shall use it to find the
locus of the point P. To this end we must find the equation to the circle
on FG₁, that is, we must express the co-ordinates of the centre P₁, and the
radius of the circle, as functions of the co-ordinates of P.
Let OFi, be drawn at right angles to e, to represent the no-load current
for a stator without resistance, that is, the current which would produce a
FM =r, OM = m,
flux to give an E.M.F. equal to e. Further, let FG = 1 T
and P₁F₁ = r₁, and let
Original circle.
G
T
x and y be the co-ordinates of the point P,
X1,, Yı
X2
""
Y2
""
""
Reduced circle.
""
""
P
12
""
""
P₂.
Auxiliary circle.

Pry R₁
བ
P
G₁
P
F
aj
F
M
m
Fig. 391.
Q
a
E,
On the assumption that the no-load currents are proportional to the
electromotive forces, we have
OP₁: 0M = E₁: e₁
1
and, since the angles OQP, and MOP, are equal, these triangles must be
similar. It follows from the figure that
X₁
e1 - Y₂
Ya
2
Moreover,
x₂ = x. R₁,
X2
m
Yı
m r₁ OF E
X2
e1
r.
OF &
Y₂ = y. R₁,
Since i. cos = y and i₂² = x² + y², we have
2
E₁² = e̟¸² + ¿‚². R² – 2é̟₁₁R₁ cos p.
X1
X₁ =
m. e₁ — myR₁
e1
mx R₁
Yı
Y₁ =
e1
p2
r₁²
2
² = 1 ½ {e¸² + (x² + y²) R₂² — 2é̟R₁y}.
2
-
+
390
Electrical Engineering
Now, the point P lies on the circle with the centre P₁, and its co-
ordinates must therefore fulfil the equation
(x − x₁)³ + (y — y₁)² = r².
We can substitute the values already found for a₁, y₁ and r₁, and put
m² - p² = 82 and 2e3+8². R₁₂².
We thus get the equation
2m.e²
28.e. R₁
82.612
x² + y² -
y
22
22
22
m.e
Putting
p
.(189),
22
82.e₁. R₁
22
.(190),

b
R
i
७
"}
Fig. 392.
adding p² + q² to each side of the above equation, and putting
2
we get
p² + qª _ sº, e,² = R²
22
(x − p)² + (y − q)² = R².
.(191),
Hence, the locus of point P is a circle (Fig. 392), the centre of which
has co-ordinates p and q, and the radius of which is equal to R.
This circle is known as the Ossanna circle.
If the values of p and q be substituted in equation (191), and the last
term be multiplied top and bottom by z² = e² + s². R₁², we get
2
•
2
§¹.e³. R₁₂ s².e². (e²² + s² . R₂²)
m².e¹ s².e
R2
m². e₁
24
+
24
whence
R2
24
or, since m² — s² = p²,
R = ".e₁²
r.
22
24
.(192).
From the circle with the radius R, the angle corresponding to any
primary current & can be directly determined.
The circle-diagram, corrected for primary copper loss 391
If the line OA be drawn in Fig. 391 so as to make the angle a with OP,
then OA will make the same angle a+ with e, in the original diagram
as OP makes with E, in the reduced diagram. Since both diagrams are
based on the same value of 7, we have
ОР
OF₁
E₁
OA OF e₁
Hence, for many purposes, it is sufficient to take the point A in the
original diagram, and to imagine the diagram reduced in the ratio OP: 0A
and, at the same time, turned up by the angle a. The end of the primary
current vector will thus be brought to the point P on the Ossanna circle.
Further, since the triangles OPA, QP₂O are similar, the angle OAP must
be equal to 4. It follows from this that the points A and P lie on a circle
to which OQ is a tangent at O, and which has its centre on the line OG.
Hence, the point P on the Ossanna circle corresponding to the point A
on the original circle must lie on the circle through the points O and A,
which has its centre on the base line OG. This is the auxiliary circle in
Fig. 391.
It must be carefully noted that neither of the three circles in Fig. 391
is the Ossanna circle. We proceed now to give a graphical construction,
whereby the Ossanna circle may be determined. The points F and G in
Fig. 393 are the ends of the original vectors of no-load current and short-
circuit current respectively. F" and G' are the corresponding points on the
Ossanna circle. We know, from what we have just seen, that F" must lie on
a semicircle on OF, and G' on a semicircle on OG, the centre of which is M₁.
The triangles OFF and OG'G are therefore right-angled triangles. Since
the circle on OG touches the original circle on FG at the point G, the
Ossanna circle must be tangential to the circle on OG at the point G'.
From this it follows that the centre M。 of the Ossanna circle lies on the
line G'M₁.
0
If the line joining the points M and M, be produced to meet the vertical
through O in the point D, we have
m
p
q
m
OD
With the aid of equations (189) and (190) this gives us the simple result,
that
1
OD=
ex
R₁
Now, both at no-load and on short circuit with a rotor of negligible
resistance, the motor is merely a choking coil, so that the primary terminal
pressure e is the hypotenuse of a right-angled triangle, one side of which
represents the ohmic drop i, R. If each side of the triangle be divided by
R₁, the hypotenuse becomes which is OD. Since the side representing
the ohmic drop would become the no-load current OF" and, in the other
case, the short-circuit current OG', the angles OF'D and OG'D must be
right angles. As we have already found the angles OFF and OG'G_to
e1
R₁
392
Electrical Engineering
be right angles, it follows that the points GG'D and the points FF'D are
collinear. The points MMD were made in a straight line. Hence, to draw
the Ossanna circle we have simply to join the point G, where the line GD
cuts the circle on OG, with the point M₁. This join G'M, will cut the
line DM in a point M。, which is the centre of the Ossanna circle. Since
we have the points F and G' through which it must pass, the circle can
be drawn. The line OG' will cut the circle at a point E, while OF" produced

Mo
M
M
G
-m-
io
Fig. 393.
cuts it in C. It can be proved that the points C, M。, E lie on a horizontal
line, while, at the same time, C is on GD and E on FD.
This construction can only be conveniently carried out in the case of
small motors, in which the primary resistance plays an important part. In
large motors with a good efficiency, the length OD
would be enormous,
e1
R₁
compared with the remainder of the diagram. In such a case the only effect
of the primary resistance is to raise the centre of the semicircle almost
The circle-diagram, corrected for primary copper loss 393
vertically through a distance q, while the diameter is hardly changed
from
го
T
In a 600 H.P. Oerlikon motor*, for example,
e₁ = 1,900,
20 = 36,
R₁ = 0·4,
T=0·117;
from these data we find
го
= 154,
190,
2T
m = io +
io
2T
=
s² = m² — p² = 12,400,
2² = e²+s². R 3,600.10 +2.10³.
=
If we neglect the second term in the expression for 22, the error is only
2 in 3,600, and we can then put 22=e. This makes pm in equation (189),
and M, vertically above M in Fig. 393. It also makes the new radius R
equal to the original radius r in equation (192). The distance through which
the centre is raised is
Ꭱ
82.e₁.R₁
s. R₁
q=
2.6 amperes.
22
er
The smallness of this correction may be seen from the fact that the
length OG represents 344 amperes.
128. The corrected values of output, rotor current and slip.
In Fig. 394 we have drawn both the original circle, neglecting the stator
resistance, and the Ossanna circle, the former dotted and the latter as a full

P
A
D
es
Bo
G
Ն
AN
ai
iR
C
B
LOM.
MH
Bo
K
F
E
Fig. 394.
line. The auxiliary circle through O is also drawn as a full line, but lighter
than the Ossanna circle. We saw in the previous section that the triangles
OPA, QP₂O were similar. If we drop the ordinate PBK, the angle OPK
is equal to 4. The triangles OBP, OPA and QP,O are therefore similar, and
we have
PB i
i Re
* See E. T. Z. 1900, No. 52.
:
394
Electrical Engineering
}
The power lost in stator resistance is
312. R₁ = 3e₁. PB.
The power supplied to the stator is
P₁ = 3₁₁.i. cos = 3e,. PK.
&
By subtracting one from the other we get the power transmitted to the
rotor, viz.
P₂ = 36₁.BK.
2
The length FP in Fig. 391 is a measure of the rotor current, in ac-
cordance with equation (181). To find FP in Fig. 394 we must reduce FA
E, OP
in the ratio
or
ę
OA'
vz. Z z
(1 + 71) %1
=F,PFA.
OP
ΟΑ
To make this reduction a horizontal is drawn through B to meet FA in C.
The ordinate through C cuts the original circle in D. Then we have
Also
FD2
FV FC OB
FA²=FH-FA¯OA
OB ОВ ОР OP2
OA
OP'OA¯ ÕA²
From these two equations, it follows that
OP
FD-FA.
ig. Zg
OA¯¯ (1+7₁). %1
The line GJ is drawn at an angle B, with the horizontal, where ß。 is the
same as in equations (186) and (187) on page 385. Then we have
and
UV =
FV
tan B.'
FD2
FV =
FD2
i2.22
2r
io/T ¯¯¯ (1 + T₁)² . z‚² T
From this equation, together with equation (186) on page 385, we get
UV
i2². R₂
er
The loss in the rotor is therefore
312². R₂ = 3é̟₁. UV.
Since CV = BK, the mechanical power given out by the rotor must be
P = 3e₁. CU.
Further, since the relative slip is equal to the relative loss in the rotor,
UV LH
CV AH
For the convenient determination of the slip the method of Fig. 389
can be employed, using the point A in the original circle.
The most convenient form of the circle-diagram 395
129. The most convenient form of the circle-diagram.
Ossanna's method of correcting the circle-diagram, as described in the
foregoing section, has the advantage of strict accuracy, but the decided dis-
advantage of being extremely cumbersome. For this reason another method,
due to Heyland, is generally employed, in which the losses in the stator and
rotor are subtracted from the ordinates of the original circle-diagram. The
simplicity of the method more than compensates for the small error thus
introduced.
It was shown in Section 127 that the radius of the corrected circle should
be almost exactly and that its centre should be nearly vertically above
io
2T
the centre of the original circle by a small distance q. We shall now neglect

M
G
ау
M₁
A
P
C
g
Fig. 395.
ዋ
F
K
this small distance so that the corrected circle and the original circle will be
identical. The vector of the primary terminal pressure will be vertical, and
the power supplied to the motor will be (Fig. 395)
P₁ = 3é̟₁. i. cos & = 3e̟, AH.
From this supplied power we must now subtract the primary copper loss.
The line GK is drawn at an angle a, below the horizontal, so that
io. R₁
tan a₁ =
e1
(2
2 +
T
(193).
A circle is drawn with its centre at M, where the vertical through M
meets GK, and with a radius M,G. This circle meets the line GM, produced
at the point K vertically below F; it also cuts the line AG in the point P.
A horizontal is drawn through P to meet the ordinate AH at the point C.
It is evident from the geometry of the figure that the angles a₁, a, and a, are
396
Electrical Engineering
all equal, and that the triangles AHF, PCA are similar.
follows that
AC HF
AP FA
AP
and
tan as.
FA
Multiplying these equations together, we have
AC = HF. tan a₂ = HF. tan α₁
In the triangle OAF, we have
i² = FA² + i² + 2i。. HF,
го
and since
HF,
T
FAGF.HF =
i?
20
= ²/2 . HF + i¿ª + 2i, . HF.
T
Solving this equation for HF, we get
is² — i²
HF=
From this it
..(a).
1
6. (2 + 2)²
is. T
Substituting this in the above equation for AC, and using equation (193),
we have
AC=
i2 - i²
is (2 +
2)
T
2
(¿‚² — ¿²). R₁
tan ag
e1
We shall now neglect the term 22. R. which is always very small, more
especially as we can allow for it later when correcting for the no-load losses.
For the primary copper loss, we have therefore
2
3. i₁². R₁ = 3е. AC.
By simple subtraction we find the power transmitted from the stator to
the rotor to be
P₁ = 3. e₁. CH3. e. PQ.
To determine the effect of the rotor copper loss we draw the line GL
in Fig. 396, so that
tan y₁ — tan αı
(1 + T₁)² . i . 21² . R₂
T. 22. e₁
_ C ..
=
.(194).
This is the same expression as we used in Section 125 (equation (186))
for determining the slip. The vertical through M cuts this line in the
point M₂. With the centre M, and radius M₂G we draw the circle GRFL.
As before, the angles y₁, Y, and y, are equal. A horizontal through R meets
AH in S. Then
RA
FA
tan and
Ys
AS HF
ᎡᎪ FA
Multiplying these equations together, we have
We saw above
AS HF. tan y₁ = HF tan Yı
that AC HF tan a₁, whence
=
CS= AS-AC-HF. tany, - HF tan a₁ = HF (tan y₁ — tan a₁).
..(b).
The most convenient form of the circle-diagram
397
If we substitute for HF its equivalent
value from equation (194), we get
FA2
'
io/T
and for tan y₁-tana, its
CS
2
FA² (1+T₂)². i . z₁². R₂_ (1 + 7₁)². FA². z² R
io/T T. 22². G
From equation (181) on page 383, we know that
(1 + 7₁)² . FA². 212
2
22
so that
2,22
i2. R₂2
CS:
e₁
We get, therefore, for the copper loss in the rotor
3122. R₂ = 3е₁. CS.

P
Ꭱ
T'
H
T
M
G
81
M
M₂
1
Fig. 396.
Subtracting this from the power transmitted to the rotor, we have
P3.e₁.SH = 3. e₁. RT.
This is the power which is transformed into mechanical power in the
rotor. It includes the no-load losses, which, however, can be allowed for by
drawing a horizontal line at a suitable distance above the base line. This
line, which is shown dotted, is then taken as the foot of the output ordinates.
For the sake of simplicity we add the stator hysteresis loss to the friction
loss, and remember that we have still a part of the stator copper loss, viz.
i. R₁ to allow for. The sum of these losses is practically equal to the power
taken to run the motor light at the normal voltage. Calling this P., we
have
P₁ = 3.&.TT.
398
Electrical Engineering
The ordinate RT' is then a measure of the nett mechanical output of the
motor.
We must now construct a scale of slip in our diagram. For this purpose
the line GJ is drawn at right angles to GM, (Fig. 397). This line cuts the
original circle in the point J. Since GJ is a tangent to the lowest circle, it
does not intersect it; in other words, the point R in Fig. 396 coincides with
the point G and the mechanical output is nil. The motor is therefore
stationary and the point J corresponds to a slip of 100 per cent. The vector
OJ represents the primary starting current.
To find the slip for any primary current i, we join GA, and drop a per-
pendicular from any point D in GJ produced, on to GM₁. Since the sides of
any triangle are proportional to the sines of the opposite angles, we have
sin B
XY
GY sin [90° — (α₁ + B)]
sin B
cos (α₁+B) *
Ꭰ

A
i
G
BIY
Oly
M
H
F
M
M₂
Fig. 397,
Similarly, since the angle at D is equal to y,
GY
sin (yı — α₁)
sin
a₁, we have
(Y₁ — α₁)
COS Y1
DY¯¯ sin (90° — Y1)
By multiplying together the two left-hand sides and the two right-hand
sides of these equations we get
XY
DY
tan Yı
Y₁- tan a₁
cotan B - tan a₁
From equations (a) and (b) of this section and from Fig. 397, we have
AH
cot B
HF
tan α₁ =
AC
HF'
AS
tan y₁ =
HF'
Hence
XY AS - AC CS
DY˜AH-AC¯CH'
The most convenient form of the circle-diagram
399
CS
CH
Now, this ratio represents the fraction of the power supplied to the
rotor which is lost in heating the rotor-winding. It must, therefore, represent
the ratio of the slip to the synchronous speed. Hence
XY
DY
If, now, the perpendicular DZ is so drawn that the length DY is equal to
100 mm., the length XY in mm. will give the percentage slip.
130. Practical Example.
We shall now construct the Heyland diagram for an actual motor, viz. a
600 horse-power three-phase motor by the Oerlikon Company of Zürich*.
The motor was designed for the abnormally low speed of 75 revolutions per
minute. For this reason the number of poles must be relatively large and
the leakage consequently above the average (see Section 131). For these
reasons a frequency of 50 would have been too high, so that a frequency of
22.5 per second was employed. For the number of pairs of poles we have
~
1
p =
n/60
= 18.
The diameter of the rotor was about 300 cms. and the air-gap 0.2 cm.
The terminal pressure e, of the delta wound stator was 1,900 volts. The
resistance R₁ per phase of the stator was 0.4 ohm, while that R₂ of the rotor
was 0.016 ohm. The stator had 324 slots each containing 8 wires, so that the
total number of wires z₁ in the primary winding was 324.8=2,592. The
rotor had a winding of bare copper bar, each of the 432 slots containing
a single bar, so that z,= 432. Since both stator and rotor had ordinary coil-
windings, the winding coefficients k, and k₂ are equal. The no-load current
is given as 62 amperes, and the normal full-load current as 170 amperes.
The corresponding currents in the stator-winding can be found by dividing
by √3. Collecting these data, we have
e₁ = 1,900,
R₁ = 0·4,
R2
R₂ = 0·016,
21
2₂ = 2,592,
Z₂ = 432,
го
11
62
36,
√3
170
21
= 100 (normal).
√3
From experiments which were carried out on the motor, and which will
be considered in the next section, the leakage factor under normal conditions
was found to be
0.117 or 7₁ = 7₂ = 0·057.
For the construction of the diagram, we have
OF = i。= 36,
io
FG
-
36
= 306.
0.117
T
* See E. T. Z. 1900, No. 52.
400
Electrical Engineering
From equations (193) and (194) on pp. 395 and 396, we have
Rr
tan α₁ =
i. R. (2 + 2) =
36.0.4
and
ራ
tan y₁ - tan α₂ =
R₂
(1 + 71)². io. 21². R2 – 0·103.
By subtraction therefore we have
T. 22². G
1,900
(2 + 0·117) = 0·08,
D
tan y₁ =0·103 – 0·08 = 0·183.

R
الخ
Χ
G
Y
M
2
Ma
Fig. 398.
TH F
Fig. 398 is drawn from these values of i, 2, 4, and y. The total no-load
==
losses were found to be P, 16,000 watts, and the dotted horizontal line is
drawn at a corresponding height above the base line, viz.
T'T=
Po 16,000
3.e 3.1,900
2.8 amperes.
We now proceed to determine the current, efficiency, power-factor and
speed for different values of the load, from no-load up to the maximum load,
at which the motor stops. By load we mean the useful turning-moment M,
as measured on the motor pulley. This is not the turning-moment exerted
on the rotor, but that obtained by dividing the useful output Pu by the
angular velocity; thus
Pab
Mu
น
met.-kgs.
n
2π.
9.81
60'
Practical Example
401
Since the useful output is equal to the product 3. e₁. RT', where RT'' is
measured to a scale of amperes, we have
3.1,900. RT'
RT'
Mu
= 5,550.
met.-kgs.
n
ஊ.
9.81
•
60
n
The corresponding value of the primary current i, is given by the length
OA on the scale of amperes.
The watt-component of the primary current is AH, so that the efficiency
is given by the ratio
n =
RT'
AH'
In order that the speed may be read off directly, the perpendicular from
D on to GM, is taken so that the length DY down to the point at which it
meets GO, is equal to 75 mm., 75 revolutions per minute being the synchronous
speed. It follows from the previous section that the length DX in mms. is
equal to the speed of the motor, i.e.
n = DX in millimetres.
In this way the values given in the following table for seven different
values of ¿ can be found very quickly. The third horizontal row represents
the normal full load conditions as shown in Fig. 398. The last vertical
column gives the values of the line current i, which, for the delta connection,
is equal to √3.4.

cos o
RT'
n=DX
AH
Mu=
ᎡᎢ '
i₁=OA
RT'
AH
η
5,550.
AH in mms.
OA
i =√3₁₂
n
45
21.2
24.2
0.87
74-4
0.548
1,565
78
60
38.7
42.7
0.91
74
0.718
2,900 104
100
73.5
80.5
0.916
73
0.805
5,600 173
196.3 121.2
142.7 0.85
70.5
0.73
9,520
340
220 123.9
150
0.83 68.3
0.682
10,000
380
243 121.2
152.5 0.793 66.6 0.63
(284 113.5 150.2 0.753
10,100
421
64.8 0.57
9,700
490)
The curves in Fig. 399 have been plotted from this table. The abscissae
represent the useful torque. This reaches a maximum value of 10,000 metre-
kilogrammes, which is nearly double the normal load of 5,600 met.-kgs. The
motor has, therefore, nearly 100 per cent. overload capacity. The corre-
sponding value of the current is 380 amperes, which is more than double the
normal current of 173 amperes. If the load be still further increased the
motor comes to rest. It is therefore impossible for the motor to run with the
point A on the left-hand side of the diagram in Fig. 398; when starting
without external resistance in the rotor circuit, the point A starts from J and
moves over the semicircle to the right-hand side. By putting suitable
starting resistance in the rotor circuit, the point A may be made to start on
the right-hand side of the diagram; in this way the starting current is greatly
reduced.
T. E.
26
402
Electrical Engineering
L
The speed of the motor under full load is 73, which represents a slip of 2
in 75, or 2.7 per cent. The speed drops continually as the load is increased
and reaches 68.8 revs. per minute at the maximum possible load.
The curves of efficiency and power-factor increase very rapidly at first,
reach their maximum values at the normal load, and then fall very gradually
as the load is still further increased. Both curves are practically horizontal
over a remarkably wide range in the neighbourhood of the normal load. An
efficiency of 92 per cent. and a power-factor of 0.81 might appear, at first
sight, to be very low values for a motor of 600 horse-power. We shall see in
the next section that the low speed results in a relatively great leakage, and
affects both the power-factor and the efficiency disadvantageously.
n i
An

70 350
60 300
50 250
7
40 200
30 150
i
20 100
10 50
n
77
cos y
1
COS
08
1000 2000 3000 4000 5000
6000 7000
8000 9000 10000
metre kilogrammes
0,6
04
02
Fig. 399.
Oerlikon standard motor of the same output for a speed of 370 revolutions
per minute, with a frequency of 50 ~ per second, had a power-factor of 0·92,
an efficiency of 95 per cent., and 1.5 per cent. slip.
We shall now consider the experimental determination of the leakage
factor 7, which we have seen to be of fundamental importance in the principle
and working of induction motors. We shall also consider the various factors
which affect the leakage, since an accurate knowledge of these factors is
essential to the design of an induction motor.
The leakage factor
403
131. The leakage factor.
If the motor has a two- or three-phase wound rotor, the leakage factor 7
can be determined experimentally in the following manner. A terminal
pressure & is applied to the stator, and the pressure across the slip-rings of
the open rotor is measured. The motor is simply a transformer on open
circuit and the terminal pressures should be proportional to the numbers of
turns z, and 22. Due allowance must be made, of course, for any differences
between the connections and windings of the stator and rotor.
Z2
21
We should expect the pressure across the rotor terminals to be e̟.
If, however, some of the lines which link the primary do not link the
secondary, the pressure e, will be smaller than e̟ . The ratio of the actual
Za
21
pressure e to the expected pressure è̟. is equal to the ratio of the flux
22
21
linking the rotor to the total flux produced by the stator.
ratio of the leakage flux to the rotor flux at no-load, or
Now, T₁ is the
+
Hence,
N 1
N
1
e2
N+N₁₂ 1+1
22
e1.
21
In the same way a three-phase pressure e₂ can be applied to the rotor, and
the resulting pressure e, across the disconnected stator terminals measured.
It should theoretically be equal to e., but leakage will cause it to be
smaller, and we have
21
22
1
&
1+ T₂
C2 •
22
·
As the difference between the theoretical and the actual value is very
small, great care is necessary if reliable results are required. Having found
7₁ and т, we can easily calculate the leakage factor 7 = 71+72 + T1T20
Τι
This method of finding the leakage factor has the advantage of simplicity
and depends directly on the definition of 7, and T₂. Unfortunately, however,
it leads to incorrect results in the majority of cases, because the leakage factor
is not a constant, as we have assumed, but decreases with large currents
owing to the saturation of the tips of the teeth. The current used in the
above method is very small, and the normal current is several times as large.
The power-factor of the motor is therefore better than one would expect from
the above experiments.
T1
For the same reason it is obvious that the Heyland diagram is not strictly
correct, except for one special load, at which the assumed values of T, and 72
hold true. It may be necessary, in extreme cases, to draw separate diagrams
for different values of the load. It is, however, of the greatest importance
26-2
404
Electrical Engineering
that the assumed values of 7 and 7½ should be correct at the normal load.
To do this, the rotor is short-circuited and fixed, and the primary current I
determined. If the rotor resistance were negligible, the short-circuit current
OJ = I, in Fig. 389 would coincide with OG. In any case, it will not differ
much from it as regards magnitude. Having measured also the no-load
current io, we have
or
T
Io = io+
го
I. — i.
T
T
.(195).
If the short-circuit test were made with the normal primary voltage, the
current would be far too large. This current would highly saturate the tips
of the teeth across which the leakage mainly occurs. The value of 7 obtained
by this method would be too small for the normal working condition, and
would only be true for the extreme left-hand side of the diagram*. The
stator pressure should be adjusted so as to give a short-circuit current
approximately equal to the normal working current. The reluctance of the
leakage paths will then be the same as under normal working conditions.
The theoretical value of I, at the full pressure can be found by multiplying
the measured current by the ratio of the full pressure to the reduced pressure
at which the test was made.
To make this clearer, we shall take, as an example, the motor considered
in the previous section. The experimental data are taken from the E. T. Z.
1900, No. 52. The short-circuit current was measured with values of the
primary P.D. of 600 and 390 volts respectively instead of the normal P.D. of
1,900 volts. The values found for the short-circuit current were 200 and 110
amperes in the external leads, which correspond to 200/√3 = 116 and
110/√/3=637 amperes respectively.
Considering in the first place the value found with a P.D. of 600 volts, viz.
116 amperes, we see that this would give a short-circuit current I, at the full
P.D. of 116.
368 amperes. In the second case a current of 637 am-
peres is measured at a P.D. of 390 volts, which corresponds to a current of
63.7.
1,900
390
1,900
600
= 310 amperes at the normal primary pressure.
These results are collected in the following table:
e1
io
observed
current
1,900
36
600
116
390
63.7
>
Io (calc.)
for 1,900 v.
io
I。 - io
368
0.109
310
0.131
1
Ít is evident that the leakage factor 7 decreases as the current increases,
so that the values of 7, and 72 in the table on page 387 decrease with increasing
current. The reason for this, as we have already pointed out, is that the tips
of the teeth become saturated and so prevent an increase of current from
* See "Electrician," 1904, page 87, for an interesting example.
The leakage factor
405
causing a corresponding increase of leakage flux. The percentage loss of flux
by leakage is therefore continually decreased as the current in stator and rotor
is increased.
T
The value of ▾ corresponding to the normal full load of the motor can be
found by interpolation from the foregoing figures. We have
I₁ = 116,
63.7,
T= 0·109,
0.131.
Hence, for the normal current of 100 amperes,
T = 0·117.
1
Now, from page 383, we know that T=T₁+T₂+T₁₂, and if 7₁ = 7₂ this
becomes 7=27₁ + 7₁², which for a value of of 0.117, gives
T1
= T2
T
0.057.
Although a large value for a motor of this size, it is by no means excessive
when the number of poles is taken into account.

Fig. 400.
The third method of constructing the diagram is perhaps the most
obvious one, viz. by making an actual brake test on the motor at different
loads, or by fixing the rotor and varying the regulating resistance in the
rotor circuit. We measure the line current i, the terminal voltage e, and, by
means of one or more wattmeters, the power P₁ supplied to the stator. We
have then
1
P₁
cos o
√3.e.i'
The primary P.D. is set up vertically, and the angle set out for each
value of the primary current, which is then marked off along its corresponding
line. Were invariable the points so obtained should lie on a circle. A
circle can generally be drawn through the point in the working part of the
figure, and then used for the calculation of T. The points shown in Fig. 400
were obtained by the author on a 14 H.P. motor.
406
Electrical Engineering
Since the power-factor and the overload capacity are both directly
dependent upon the leakage factor 7, the designer of an induction motor
must strive to obtain as small a value of 7 as possible, that is, he must reduce
the leakage to the smallest possible limit. It is obviously disadvantageous
to place the winding in closed tunnels or holes, on account of the small
reluctance offered to the leakage flux. This is especially the case if a con-
siderable thickness of iron is left between the hole and the air-gap, but if this
bridge of iron be reduced to a minimum by stamping the holes very near the
air-gap, there is little objection to the totally enclosed hole. The best
construction, from the point of view of reduced leakage and of ease of manu-
facture, is the rectangular open slot, into which former-wound coils can be
placed. This leads, however, to an unequal distribution of flux in the air-gap,
and to a consequently increased magnetising current.
It is necessary in all cases to divide the coil-side between several slots, in
order to keep down the leakage flux produced by the current in one slot. A
limit is set to the number of slots by the proportionately greater amount of
space taken up by the insulation. Generally speaking, the stator has from
3 to 5 slots per coil-side, while the rotor has from 3 to 7. It is obvious
why induction motors are often made with relatively large diameters and
small axial lengths, since the large pole-pitch enables the coil-side to be
distributed among a large number of slots. Unfortunately, however, in-
creasing the diameter at the expense of axial length adds considerably to the
cost of the motor. This is obvious from the consideration that the output
could be considerably increased by making the motor axially longer, without
adding very materially to the cost of manufacture. Another disadvantage of
the short axial length is the relative importance of the flank leakage.
A second way in which the value of r can be reduced is by decreasing the
air-gap. Since the product 717, is negligibly small, we may put TT₁+T₂,
=
and, as we can see from the definition of 7, and 7, on page 382, the value of τ
is directly proportional to the reluctance of the air-gap
1
It is interesting to consider the effect on the operation of the motor of
varying the air-gap. This cannot be done by using one motor and succes-
sively turning a little off the stator or rotor, since this would not only increase
the air-gap but would also increase the reluctance of the tips of the teeth.
Two motors should be compared which have the same slot dimensions, etc.,
but different air-gaps. The smaller the air-gap, the smaller is the value of 7,
and the larger is the maximum power-factor, in accordance with the equation
1
2T+1'
(cos +)max=
A decrease in the air-gap will naturally lower the no-load current, since a
smaller magnetising current is now required to drive the flux across the
reduced air-gap. The maximum torque is dependent on the radius of the
Heyland circle, or on 2. This has remained the same for both motors,
T
showing that the maximum torque is but little affected by the air-gap if the
motors are the same in other respects.
* See Behrend, "The Induction Motor," also Behrend, E. T. Z. 1904, p. 59.
The leakage factor
407
The value of τ is very greatly affected by the choice of frequency. In
order to make this clear we may consider two motors with stator and rotor
cores of the same dimensions and with the same number of slots. The
primary P.D., the output and the speed are the same in each case, but one
motor is designed for a frequency of 50 cycles per second, the other for 30.
We shall assume that the proportion of slot space occupied by copper is the
same in each case, as also the current density in the wires, and the number
of ampere-turns per cm. of periphery.
Since the output and the P.D. are the same, it follows that the current
will also be the same in each case, if we neglect any difference in the value
of cos 4. Moreover, since the number of ampere-turns per cm. of periphery
has been taken as the same in each case, the equality of current leads to the
equality of the number of wires on the stator and also of the number of wires
per slot.

}
Phase I
Phase II
se
Phase II
10000000 0 0 0 0
Phase I
Pole-pitch.
Fig. 401.

Phase III
Phase II
10 0 0 0 0 0 0 0 0 0 0 0 0 0 0
<
Pole-pitch
Fig. 402.
The motors will differ, however, in the number of pairs of poles, in accord-
ance with the equation
p =
n₁/60
The motor for a frequency of 50 must have a greater number of poles
than that for a frequency of 30, in the ratio of 5: 3. Since both motors
have equal peripheries, their pole-pitches will be in the proportion of 3 : 5.
If Fig. 401 represents the first motor with 3 slots per coil-side or 9 per pole-
pitch, then the second motor will be represented by Fig. 402, in which there
are 5 slots per coil-side or 15 per pole. The number of wires in each slot is
the same in each case, so that the wires per coil-side will be as 3: 5.
The E. M.F. per phase is given by the equation
ę = k. N。 . ~1. 21. 10-8 volts.
408
Electrical Engineering
I
We can put N, approximately equal to the flux Bmean. Ag which crosses the
n
air-gap, and ~, equal to p. 60 neglecting the slip; we thus get
n
e₁ =k. Bmean. Ag.p. z₁. 10–8 volts.
60
In this equation A,.p is equal to half the cylindrical surface, and is
therefore the same in each case. Since, moreover, e̟, k, n, and zí are the
same in each case, it follows that both motors must work with the same
average flux density Bmean.
With equal flux density the hysteresis loss is proportional to the fre-
quency, and will therefore be § times as great in the 50~ motor as in the
30~ motor.
To produce the same flux density in each case will require the same
number of ampere-turns per coil-side, and as the coil-side of the 50 ~ motor
contains only of the wires in the coil-side of the other motor, its no-load or
magnetising current will have to be greater in the ratio of 5:3, that is, the
no-load current will be directly proportional to the frequency*.
The question is not nearly so simple when we come to consider the
relative values of the leakage factor 7. If we consider the leakage flux due
to one coil-side in Fig. 401 we see that it has to cross 3 slots, whereas in
Fig. 402 it has to cross 5 slots, or, in other words, the reluctances of the
leakage paths are as 3:5. If we assume, temporarily, that the short-circuit
current is the same in each case, the ampere-wires per coil-side are in the
same ratio, viz. 3:5. Since both the ampere-turns and the reluctances have
changed in the same proportion, the resulting flux will be the same in each
case. In the first case, this flux links 3 slots with a frequency of 50, whereas
in the second case it links 5 slots with a frequency of 30; the E. M. F. induced
in the coil-side would be the same in each case, but as the first motor has
§ times as many coil-sides as the second, the total E. M.F. of the first motor
would be times that of the second. Hence, to produce the same short-
circuit current in each case, the primary pressures would have to be in the
ratio of 5:3, or, on the other hand, with the same P.D. the short-circuit
currents of the two motors are as 3:5.
We have seen above that the no-load currents are in the ratio of 5:3.
The alterations in both no-load current and short-circuit current are there-
fore in such a direction as to reduce T in the second motor, i.e. the one for
the lower frequency. If our assumptions were strictly correct, the value of τ
would vary as the square of the frequency. An important part of the leakage,
which we have not considered, is that around the ends of the coils, where
they project beyond the stator and rotor. Since the 30 motor in Fig. 402
would have much longer coil-ends, this part of the leakage would be increased
as the frequency was decreased. How far this increased leakage would
counteract the decreased slot leakage is not at all certain. If it exactly
counterbalanced it, the short-circuit current would be the same in both cases
~
* When considering any given motor, the primary P.D. of which is maintained constant, the
no-load current must evidently vary approximately inversely as the frequency.
The leakage factor
409
T
and would vary directly as the frequency. The general correctness of these
results has been proved experimentally by Behrend.
It is interesting to note that the two motors will have about equal over-
load capacity. The motor with the greater number of poles has the greater
magnetising current, but also the greater leakage factor 7. The radius of
the Heyland circle, and therefore the maximum torque, will be approximately
the same as in the other motor.
The important points of difference between the two motors lie in the
magnetising current and in the power-factor.
•
CHAPTER XIX.
132. Resolution of the primary M.M.F. of the single-phase motor into two constant rotating
components.-133. The E. M. F. induced in the actual stator.-134. The circle-diagram of
the single-phase motor.-135. Single-phase commutator motors.
132. Resolution of the primary M.M.F. of the single-phase
motor into two constant rotating components.
If one of the leads supplying current to a three-phase induction motor be
broken, it will continue to run as a single-phase motor in the same direction.
Coils 1 and 2 of the three-phase motor in Fig. 403 now form a single coil
with a breadth equal to two-thirds of the pole-pitch, carrying the same
current throughout. If the rotor circuit is open, the primary current pro-
duces an alternating flux, the axis of which has a fixed direction. In the
figure this direction is vertical. The centres of the poles remain at A and C
and the neutral points at B and D. It is difficult to see how a torque can be



B
2
O O
Fig. 404.
Fig. 403.
Fig. 405.
produced by means of this alternating field, and how the direction of this
torque can be determined by the initial direction of rotation which is arbi-
trarily given to the rotor.
Ferraris resolved the alternating field into two oppositely rotating fields.
We shall go a step further, and resolve, not only the flux, but also the
primary ampere-turns into two oppositely rotating sets of ampere-turns. We
must assume, however, that the conductors are distributed around the stator
according to the sine law, as is shown in Fig. 404 by circles of various sizes.
Theoretically this would lead to an infinite number of infinitely small wires.
We can represent this sinusoidal distribution of the ampere-turns in the
manner indicated in Fig. 405.
On this assumption, the effect of the alternating ampere-turns at any
moment and at any point is exactly equal to the combined effect of two
Magnetomotive force of single-phase stator
411
1
sinusoidal windings, carrying a constant current and rotating in opposite
directions. The constant current is such that the number of ampere-turns
in each rotating winding is a half of the total ampere-turns on the actual
stator, when the alternating current reaches its maximum value. In Fig. 406
the actual stator current is a maximum. At this moment, those sides of the
imaginary rotating coils (shown black) which carry current in the same
direction are superposed, and their individual effects can be added. They
are therefore equivalent to the stationary alternating ampere-turns, repre-
sented by the outer ring of the figure, which have their maximum value at
the given moment.
An eighth of a period later the stator current will be
Cmax. sin 45° = 0.707 Cmax:
We see that at this moment the rotating ampere-turns at A and C neutralise
each other (Fig. 407). The addition of the two sine curves in the lower part
of the figure gives a sine curve with a maximum at B, as before, but with the



x
A
OD
B
x
A
A
D
B
B

A
B
Fig. 406.
A
B
Fig. 407.
A
Fig. 408.
ordinates reduced in the ratio of 0.707: 1 as compared with Fig. 406. The
magnetic effect of the two rotating windings is still equal to that of the
actual stator-winding. The same agreement is found after another period,
when the actual stator current is zero (Fig. 408). The rotating ampere-turns
now neutralise each other at all points.
We shall now imagine the motor to be frictionless and running light
with a squirrel cage rotor of negligible resistance. We shall assume the
direction of rotation to be clockwise, and we shall designate the imaginary
stator-winding which rotates in this direction as the forward ampere-turns,
and that in the opposite direction as the backward ampere-turns. The
magnetic flux which the forward ampere-turns drive through the rotor
rotates synchronously with it and does not cut the rotor bars. This flux is
shown in Figs. 409, 410 and 411, for the three moments previously con-
sidered. In all that follows, the flux, electromotive force, and rotor currents
will naturally be distributed around the periphery according to the sine law.
With respect to the backward ampere-turns, the rotor has a slip of
200 per cent. In consequence of this, a large short-circuit current flows in
412
Electrical Engineering
the rotor, directly opposing the backward ampere-turns, and preventing the
flux from entering the rotor. The flux is driven, however, along the secondary
leakage path (Figs. 412, 413 and 414). The outer black circle in this figure
represents the backward ampere-turns, and the inner one, the rotor ampere-
turns. On the assumption that the rotor has no resistance, an infinitely
small number of lines is sufficient to induce the rotor currents necessary to
counterbalance the primary backward ampere-turns. When running perfectly
light with a rotor of zero resistance, we have therefore the forward rotor field,
the backward rotor leakage field, and the backward rotor ampere-turns.
Since the rotor itself runs synchronously, the rotor bars must carry an alter-
nating current of double frequency.
Turning now to the actual conditions in the motor, we see that the flux
in Fig. 409 is produced by the difference between the actual ampere-turns



X
X
Fig. 409.
Fig. 410.
Fig. 411.



+
Fig. 413.
Fig. 412.
Fig. 414.
on the stator and those on the rotor. Figs. 409 and 412 represent ampere-
turns and fluxes existing simultaneously in the motor; similarly for Figs. 410,
413 and 411, 414. The flux represented in Fig. 411 is 90° out of phase with
the actual stator ampere-turns both as regards time and space. Since the
actual stator current is zero at this moment, the flux in Fig. 411 must be
produced by the rotor currents in Fig. 414. The application of the corkscrew
rule shows that the rotor current is in the right direction to produce this flux.
The single-phase motor is thus reduced to a two-phase motor, one phase of
which is obtained from the difference between the stator and rotor currents,
while the other is obtained from the rotor currents alone. These relations
are somewhat altered, even on no-load, if the rotor resistance is taken into
account. They are considerably altered when the motor is heavily loaded.
The single-phase motor does not possess the simplicity and uniformity of the
polyphase induction motor. We shall proceed further with the theory of the
rotating ampere-turns, and see to what results it leads us.
Magnetomotive force of single-phase stator
413
When considering the polyphase induction motor, we saw that a rotating
field could be produced by a stationary polyphase winding. We can there-
fore imagine each of the primary rotating windings to be replaced by a
stationary winding of a great number of phases, each phase consisting of
a single turn. The wires are distributed uniformly round the stator, and
carry sine-wave currents of appropriate phase. The result is a uniformly
rotating sine-wave distribution of current similar to that in Fig. 384. The
maximum current in each wire is half the maximum current in the actual
stator (Fig. 404), in which we assumed a sine distribution of conductors.
The number of turns or phases is so chosen that the total number of ampere-
turns on either of the polyphase windings is equal to a half of the actual
ampere-turns on the stator, at the moment of maximum current. Thus, if
there are S₁ turns on the actual stator, and the primary current be I amperes,
the maximum ampere-turns on the stator will be √2.I.S₁. Each of the
two imaginary polyphase windings must have 2 IS, ampere-turns.
2
the effective current per phase be I/2, as we have assumed above,
average current in all the phases will be
If now there are
2/2 I
π
2'
phases or turns on the polyphase winding, the ampere-turns will be
2√2 I S₁π √2
π
2° 2 2
I. S₁,
which is the number we have seen to be necessary.
133. The E.M. F. induced in the actual stator.
If
the
S₁. π
2
Let Fig. 415 represent the diagram for one phase of the forward rotating
excitation, under certain conditions of load, slip, etc., while Fig. 416 represents
P


G₁
E
Fig. 415.
'N
-E,
P
屋
​Fig. 416.
the diagram for one phase of the backward rotating excitation under the
same conditions. In these diagrams the following relations must hold :
I
OF OF
i=OP, 2' F₁G₁ F₂P.
where i is the R. M.S. current in each phase of the imaginary polyphase
windings, and I is the actual stator current. We shall now show that, in
spite of their rotation in opposite directions, the geometrical sum of the two
electromotive forces E, and E, induced in each phase of the imaginary
414
Electrical Engineering
יז
D
A
α
a
B
polyphase windings, represents, to some scale, the E.M.F. induced in the actual
stator. Let the centre of the forward rotating excitation be at F, i.e. an
angle a ahead of its initial position in which the actual stator current was
a maximum (Fig. 417). The electromotive
force – E₁ (equal and opposite to the induced
E.M. F. E) will have its maximum at the point
G, the arc FG corresponding to the angle 1
in Fig. 415. As each phase consists of a
single winding, the R.M.S. electromotive force
E is a measure of the flux density. The flux
density is therefore greatest at G, while at E
it is proportional to E, cos e. Now the number
of wires per cm. of periphery of the actual stator
varies as the cosine of the angle measured from
B. The number of wires subtending the angle
de at E is proportional to cos BÔE. de, that is, to cos (a+1-e).de. The
momentary value of the E.M.F. induced in the wires of the actual stator,
covering the arc de at E, by the forward rotating flux is proportional to
E₁.cos e and to cos (a + 1 − e) de. To find the momentary value of the
induced E.M.F. E' in the actual stator, due to the forward rotating flux,

C
Fig. 417.
G
E
we must integrate between the limits e = GỐC and e = GÔA. Introducing
e
a coefficient c, we obtain the following expression for the momentary value
of E':
E'
(a+1+π/2
a+61-π/2
π
c. E₁.cos e . cos (a + 41 − e) . de = C E₁. cos (a + 1).
•
2
.
The momentary E. M.F. E" induced by the backward rotating flux is found
in the same way:
π™
E" = c
E₂. cos (a + 2).
2
The sum of the two electromotive forces
In our case ₂ = 90° (Fig. 416).
is therefore
E' + E" = c
π
2
[E₁ cos (a + 1) + E₂ cos (a + $2)].
2
This momentary value of the actual induced E. M. F. must be equal and
opposite to the stator terminal pressure. If now, in Fig. 418, the vector OP
represents the actual stator current I, the momentary value of which is
Imax. cos a, then the vector - E, must, according to the equation for E+E",
be drawn ₁ ahead of OP, and the vector - E₁, an angle 42 ahead of OP.
The resultant of — E, and — E, is constant, and, neglecting the scale, equal
to the terminal pressure. The angle between – E₁ and - E₂ is equal to
€ + y = $2 − $1 = 90° — ₁, and the actual angle of lag between e and I is
€+₁. The angle e has here quite a different meaning to that which it
had in Fig. 417. To make Fig. 418 as clear as possible, it has been drawn
for exactly the same conditions as to load, etc., and to the same scale as
Figs. 415 and 416.
1
The circle-diagram of the single-phase motor
415
134. The circle-diagram of the single-phase motor.
2
In Fig. 418, the magnetising current OF, for the forward rotating flux is
set out at right angles to E₁, and the magnetising current OF, for the
backward rotating flux, at right angles to E, that is, in the direction
of IOP. Since the magnetising currents OF, and OF, are proportional
to the corresponding electromotive forces - E, and — E, and are also at
right angles to them, the resultant of OF, and OF, must be proportional
to and at right angles to e, that is to say, OF is constant both in direction
and magnitude, whatever the conditions as to load, etc. To find the magni-
tude of OF, we make use of the fact that it is the resultant of OF, and OF,
for all conditions, and therefore also for an open rotor. With an open rotor,
however, the two rotating fields are exactly equal, OF, is equal to OF, and
both lie along OF, since ₁ = 2 = 90°. OF = i, is therefore twice as large

OP-i QM-m MP-r
OP-J OG-a PF-r,
P
Pa
M
€
у
B
F
مة
ε
→
Fig. 418.
as either individual rotating no-load current. Since the actual stator current
is twice as large as either rotating current, it follows that OF represents
the actual stator current with an open rotor.
io
FG is made equal to T
2'
OP=I is now halved, making OP.=i=1, the current in each phase
of the rotating excitation. Since OP, is the short-circuit current for the
backward rotating flux, it follows that OF: F2P, T. Moreover, the angle
between I = OP and – E₁ is equal to 1, whence the point P, lies on a circle
on the diameter F G₁, and we have the relation
2
- E,
OF₁
F₁G₁¯¯¯ F½Ð¸¯¯ FG
OF OF
= T.
=
0
1
The line
Then FF₂ is parallel to GP, and the angle OGP, is equal to e.
joining the centres M and P₁ is likewise parallel to FF₁, and the triangles
1
t
416
Electrical Engineering
·
OP₁M, OFF and GP.O are similar. Calling the abscissae positive when
measured to the left of O, putting OM = m and OG= a, and calling the
coordinates of the points P. and P₁, xo, yo and x, y, respectively, we have
the following relations:
m
X1 (α − xo), Yı
Xr
m
x₁ =
α Xo α
a
m
Yo•
α
If now r = MF, the radius of the large circle, then the radius r₁ = P₁F₁
is given by the equation
or, since GP2 = (a − x)² + y²,
r²
हाह
1
OF, GP.
OF
a
= ~ ~ [(a − x.)² + yo³].
p2
a²
Now, since the point P, lies on the circle with the radius r₁, we have the
equation
(x。 − x₁)² + (Y。 − Y₁)² = r₁².
We substitute in this equation the values found above for x, y, and r²,
and replace a, and y, by
end point of the vector
ac
and, where x and y are the coordinates of P, the
2
of actual stator current. By a little manipulation
the equation can then be put into a form in which the coefficients of ☛² and
y² are equal, while those of xy and y vanish.
The end point P of the actual stator current vector lies therefore on
a circle, the centre of which is on the x axis. The equation is much simplified
by putting a r for m. We can obtain the circle in a simpler manner by
determining the position of the point B. Let OB = I。, the actual no-load
stator current when the rotor is closed, but running synchronously. The
I。
no-load current in each of the imaginary polyphase windings will be 2
This will be the magnetising current of the forward rotating excitation, and
the short-circuit current of the backward rotating component. Hence OP,
and OF, will be identical, and lie along OF. Now OF= is the sum of
OF, and OF2, that is, the sum of OP, and OF₂, but OF₁ = OP。
I。
and
2
T I
T
OF₂ = OP..
therefore
1+T
2
1+ T
I。
I. I T
+
2
2'1 + T'
230
1+T
or
OB = I。
2i0.
T
1 + 2T
1+
1+T
.(196).
On the other hand, the primary current locus must pass through G, for
OG is the short-circuit current. The diameter of the primary current circle
is therefore
BG = OG – OB=i+
2i.
T
1+T
1 + 2T
20 1+T
......(197).
T1+2T
The circle-diagram of the single-phase motor
417
The following characteristic properties of the single-phase motor follow
from these equations.
As T is of the order of 0.1, the no-load current OB is nearly twice as
large as the current with open circuited rotor.
2. The maximum power-factor is smaller than that of polyphase motors.
3. Assuming the rotor to have no resistance, the slip is zero, and the
speed of the motor is therefore constant, depending simply on the frequency.
The turning-moment is therefore proportional to the power, and is represented
by the ordinates of the circle GB. The overload capacity is evidently much
smaller than that of the polyphase motor.
4. A single-phase motor will not start as such, even though resistance
be inserted in the rotor circuits, for, when the rotor is stationary, the two
rotating excitations are exactly similar, and there is no reason why the motor
should revolve in one direction in preference to the other.
Further investigation* shows that the end point of the primary current
vector lies on a circle, even when the rotor resistance is taken into account.
The centre of this new circle lies vertically above that of the circle obtained
above by neglecting the rotor resistance. The proof of this point is, however,
beyond the scope of this work. It can be shown, moreover, that the power
supplied, the mechanical output, the torque, and the rotor losses, for the
actual motor, are equal to the sum of the corresponding values obtained from
the diagrams for the two rotating excitations. The sum of the two slips
must be always 200 per cent., and the torque and mechanical output due to
the backward rotating excitation must be reckoned negative.
These results are very remarkable, when we consider the complicated
nature of the relations involved. For example, if the rotor resistance be
taken into account, the rotor currents are found to have a sine-wave dis-
tribution, the position of the maximum rotating in the opposite direction
to the motor, while its value fluctuates from moment to moment. It is
found also that the actual field which cuts the stator-winding has a sine-
wave distribution, and rotates in the same direction as the motor, while its
maximum value is subjected to periodic fluctuations (elliptic rotating field).
135. Single-phase commutator motorst.
The characteristics of the single-phase induction motor, mentioned in the
last section, render it unsuitable for most purposes. For traction purposes
especially, a large starting torque and a large overload capacity are required.
These can be obtained by supplying the rotor or armature with a commutator.
Great advances have been made during the last few years in this branch
of the subject, and motors of this type are meeting with a large amount of
practical success.
All single-phase commutator motors have a stator-winding of the type
shown in Fig. 419. This can be either a closed winding, similar to that
* See the article by Dr Thomälen in E. T. Z. 1905.
+ See articles by Osnos, Eichberg and Pichelmayer in the E. T. Z. 1904, also by Fynn in
"Elec. Review," 1906.
T.E.
27
418
Electrical Engineering
of direct-current machines, in which the current supplied divides between
two parallel paths, or a single-circuit open winding. Over one half of the
periphery of the stator, the current flows from front to back, and in the other
half from back to front. Such a winding can be represented diagrammatically
as shown on the right in Fig. 419. The axis of the winding, that is, the
direction of the magnetic flux produced, is horizontal in the figure. All
commutator motors possess also an armature or rotor, wound exactly like
a direct-current armature, and a commutator on which bear two or more
brushes (Fig. 420). The current is either supplied to the rotor from an
external source or induced in the armature, which, in that case, is short-
circuited by an external connection between the brushes. The rotor currents


h

Fig. 419.
Fig. 420.

N
Fig. 421.
produce, in either case, a field which passes through the rotor in the direction
of the brush axis, if we imagine the brushes to rub on the bare surface of
the winding itself. Such an armature is represented diagrammatically on the
right in Fig. 420.
For the sake of clearness, the distributed stator-winding of Fig. 419
has been replaced in Fig. 421 by a winding on salient poles. The stator
carries the field current, while the rotor carries the torque-producing motor
current. Both are connected in series (series motor). It is to be specially
noted, that in this and all the following figures, including Fig. 428, the
axis of the motor field is shown horizontal, and the axis of the torque-
producing winding vertical. With the current in the direction indicated,
the torque will be clockwise. Seeing that the direction of both field and
armature current is reversed at the same moment, the problem of the single-
phase motor would appear to be solved. Fig. 422 represents diagrammatically
such a simple series motor, with resistances R to regulate the speed.
Single-phase commutator motors
419
The series motor produces its own magnetic field and takes therefore, in
common with induction motors, a lagging current. Were this effect confined
to the useful horizontal flux, the phase difference would be allowable, but we
have, in addition, the vertical flux produced by the armature current. This
causes a harmful self-induction, and a very considerable phase displacement.
In addition to this, sparkless commutation is rendered very difficult.


R
www
Field
R
K
WW
WW
Field
Fig. 422.
Fig. 423.
It is therefore a great improvement if this cross-magnetisation be sup-
pressed by a compensating winding K on the stator. The plan which
naturally suggests itself is to connect the field winding, compensating
winding, and armature all in series (Fig. 423).
It is, however, better to short-circuit the compensating winding (Fig. 424).
The armature is then the primary, and the compensating winding the
secondary, of a transformer. Since the induced secondary current is directly
opposed to the primary current, the cross-magnetisation will be suppressed.


Fig. 424.
Fig. 425.
It is now only a step further to exchange the primary and secondary
of the transformer, that is, to short-circuit the brushes, and connect the
compensating winding in series with the field coils. Fig. 425 shows this
arrangement with salient poles, while in Fig. 426 we have the same arrange-
ment with the windings distributed around the stator periphery.
A further step is the substitution of a single resultant winding for the
two stator-windings. We have thus arrived at Elihu Thomson's so-called
repulsion motor (Fig. 427). In this motor the axis of the stator-winding
is at an angle with the axis of the rotor ampere-turns, that is, the brush axis.
27-2
420
Electrical Engineering
To understand the working of the motor, however, it is better to think of
the two rectangular components of the stator ampere-turns, one, at right
angles to the brush axis, producing the true motor field, and the other, along
the brush axis, inducing the armature current by its transformer effect. This
current in the motor field produces the torque, enabling the motor to start
up under heavy loads. An E.M.F. is induced by the rotation of the armature,
which, for a given motor field, is directly proportional to the speed, just


Az
Fig. 426.
A,
Fig. 427.
as in a D.C. motor. The speed of the motor is therefore largely dependent
upon the magnitude of the component of the stator excitation which produces
the motor field. The speed varies, therefore, with the position of the brushes.
A reversal of the motor can obviously only be obtained by shifting over the
brush axis. This can be obviated by having either two separate pairs of
brushes or two separate stator-windings, only one of which is used at a time.
E,
K
Ez
K₂
A
The compensation or suppression of the armature cross-magnetisation, as
carried out in the Winter-Eichberg motor, is of special importance (Fig. 428).
The horizontal true motor field is produced by
the armature, the brushes E, and E, being in
series with the stator-winding A. As a matter
of fact, the excitation current is supplied to
the armature from the secondary terminals of
a transformer, the primary of which is in series
with the stator-winding A. As the primary
and secondary currents are nearly 180° out of
phase, the principle of the motor is not altered
by this arrangement. It has the great advan-
tage of allowing the pressure across the excitation brushes E₁, E₂ to be
varied by altering the ratio of the transformer. In this way, the speed
may be regulated and the power-factor improved. The stator-winding A
carries the torque-producing current. The functions of stator and rotor have
thus been reversed. The armature, which is now the stator, is fixed, while
the field-winding rotates. The stator current produces a vertical cross flux,
which can be neutralised by putting a special winding on the armature,
and short-circuiting it along a vertical axis by means of the brushes K₁, K,.
We should then have exactly the same arrangement as in Fig. 424, except

Fig. 428.
2
Single-phase commutator motors
421
that rotor and stator would be reversed. As a matter of fact, however, the
second armature-winding is unnecessary, and in practice both pairs of brushes
rub on the same commutator. According to the law of the superposition of
currents, the effect of the actual armature current is equal to the combined
effects of the two windings, viz. the field-winding E₁E, and the compensating
winding K₁K₂, which we considered above.
2,
The action of commutator motors can be best understood by the con-
sideration of vector diagrams introducing the electromotive forces of self
and mutual induction*. The self-induction does not include merely the
leakage flux but also the flux crossing the air-gap. We shall use the fictitious
fluxes produced by the stator or rotor alone, although, as a matter of fact,
these fluxes combine to form a resultant flux. For the sake of simplicity we

B
A—
Ꮄ
800
a
E
он
Fig. 429.
shall neglect the resistances of windings and brushes, and also the magnetic
reluctance of the iron. For the same reason we shall not go into the
question of the short-circuiting of individual armature coils under the
brushes.
To calculate the mutual induction let us consider the arrangement shown
in Fig. 429, in which the rotor-winding is short-circuited by means of brushes
along an axis making an angle a with the axis of the stator-winding. As
in equation (68 a) on page 85, the coefficient of mutual induction M
is 10-8 times the flux which links each individual turn of the stator-
winding when a current of 1 ampere flows in the rotor. By linking
each individual turn, we mean that if the flux links two turns it must
be doubled, or, in other words, we must take the number of linkages. Only
those stator coils between B and D, i.e. subtending an angle π- 2a = 28,
* The following development, to the end of the present chapter, has been specially worked
out by the author for the English edition.
422
Electrical Engineering
need be considered, as the coils between D and F are not linked with the
flux produced by the rotor. Let
S₁ and S, be the number of turns in series in the stator and rotor
respectively,
and i, the R. M.S. currents of the stator and rotor (in the supply
wires and short-circuit connection respectively),
B₂, the highest value of the induction in the gap for a direct current of
1 ampere in the short-circuiting connection,
l, the double air-gap in cms.,
2a, the number of parallel paths for the stator current,
2a,, the number of parallel paths for the rotor current.
If now the current in the short-circuiting connection is 1 ampere, the
current in each rotor wire is
poles is
1
S₂. 2α2
Ρ
S2.2a2 S₂
2α2*
p
p
1
amperes. The number of turns per pair of
2α2
The number of ampere-turns per pair of poles is therefore
from which it follows that
0·4πS₂
B₂ = 1.p

ooooog
*
1
dx
K
Fig. 430.
In Figs. 429 and 430 this maximum value occurs at E. From this point
it falls off uniformly in both directions, reaching zero value at C and G. Now
dx S₁. 2a1
π p
the number of stator turns in a width dx is
The flux N linked
with these turns is partly positive and partly negative. N is obtained by
multiplying the shaded area in Fig. 430 by the armature length Lɑ and by
TD
the ratio of the pole-pitch
2p
to the angle π. We thus find that
N = .B. (1-4).
La.D.π
4.p
a
To find the mutual induction this must be multiplied by the above
Single-phase commutator motors
423
number of turns, viz.
between the limits
dx
π
201
S₁. and by 10-8; this must then be integrated
p
8 and +8. To find the coefficient M for the whole
stator, the result must be multiplied by the number of pairs of poles p, and
divided by the number of parallel paths 2α1. In this way we obtain
M
La. D
4p
-. B₂. S₁. 10-8.
10−
10. ft
•+8
(1-4) dz
dx.
If we perform the integration, put
k = 3
S₂
8
π
4
77-3
and
S₁
and substitute the value found above for B₂, we get
Let z represent
2
0.2π² S₁² La. D
M=c.k.
.10-8
3 p²
10-. w, where w is the angular velocity
2
0.2m² S₁² La.D
3 p² 1
of the alternating current vector, that is, 2π~. We have then
w.M=c.k.. 2.
The E.M.F. induced in the rotor, due to the current in the stator, is
E₁₂ = Mwi₁ = c.k.z.i.
The E.M.F. induced in the stator, due to the current in the rotor, is
=
E21 Mwi,c.k.z.i.
To find the self-induction of the stator we have to put a = 0, which makes
k=1. We must, moreover, replace S₂ by S₁, and multiply the result by
1+T, to allow for the stator leakage.
The E. M. F. induced in the stator due
to its self-induction is therefore

E₁ = 2 (1 + 7₁) ¿½.
The E.M.F. induced in the rotor
due to its self-induction can be found
in the same way. Owing to the
square of the number of rotor turns
entering into the equation, we get
E₂ = c. z. (1+T₂). 12.
26------
B
G
Π
D
Fig. 431.
We turn now to the E.M.F. in-
duced in the rotor by its rotation. No E.M. F. will be induced by the
rotation in the field produced by the rotor itself, nor in that produced by
the stator coils BD (Fig. 429). An E.M.F. will be induced, however, by the
rotor-windings cutting the field produced by the stator coils DF. The
breadth of these coils is 2a=π- 28, and the field is distributed as shown
in Fig. 431. At the moment of maximum stator current the maximum
value of B will be
B₁ ==
0·4π S₁ 2α
Ρ π
i max.
424
Electrical Engineering
#
By multiplying the shaded area in Fig. 431 by the length of the armature
La, and by the ratio of the pole-pitch
the maximum value of the flux:
πD
2p
to the angle π, we obtain, as before,
π+28
N=B₁.
La. D.
4.p
The total number of conductors on the rotor is
Z₂ = 2S2. 2α2.
If v is the ratio of the actual speed to the synchronous speed then
n
V
60
υ.ω
p P. 2π
>
p
The E.M.F. at the moment of maximum stator current is then found
exactly as in D.C. machines.
Er max
P.N
Az
n
22. 10-8 volts.
60
Substituting in this equation the values just found for N,
taking R.M.S. values on both sides, we get
where
E, c.k.v.z.i,
= c. k'. v. z. i,,
3
k'
-(1-4).
n
and z₂, and
60
This E.M.F. is in phase with the stator current.
12
We will now construct the vector diagram for the repulsion motor. As
in any transformer, the secondary current i, is more or less in opposition
to the primary current & (compare Fig. 429). If the primary current
vector i, be drawn vertically upwards (Fig. 432), the vector representing the
rotor current will not be far removed from the vertical in a downward
direction. The vectors E, and E2 will lag 90° behind the current &, while
the vectors E, and E will lag 90° behind the current 2. Since the total
E. M.F. in the short-circuited armature must be zero, the resultant of E₂ and
E₁ must be equal and opposite to Er, that is, it must be vertically down-
wards in the vector diagram. On the other hand, the resultant of E, and E2
gives the induced back E. M. F. in the stator, which must be equal and opposite
to the terminal pressure.
We have then
12
2
21
OA = E, and OE = e.
From the figure and our equations for the electromotive forces, we have
Ea kk'v
e cos &₁ = ОA .
z.ij.
E2
1+ T2
Similarly, putting 1+7=(1+7₁) (1+T₂), we find that
T=
e sin ₁ = E₁— E12 ·
Ea 1+7-k²
E2 1+T₂
tan $1
1+T-k²
kk'v
We have from these two equations
z.i.
Single-phase commutator motors
425
When the rotor circuit is open, the stator current io is given by the
equation
e=iz (1 + T₁).
Putting this value for e in the above equation for e sin o₁, we have
sin di
1+T-k²
(1 + 7). ∞
•
21.
The locus of the end point of the primary current vector is therefore
a circle (Fig. 433), the diameter of which is the short-circuit current I,
where
I。
(1+T). i
1+T
1 + T-k²

Ej2
i
AET
Ep
E2
ε
E21
Fig. 432.
The turning-moment can be found from the equation
Mt
e. i cos $1
e. i₁ cos $1
kg.-met.,
N
9.81.2π.
9.81.2π.
V
60
Ρ
or, using the above equations for i, and e cos 1,
*
M
p.k.k
=
e
9·81.2π.~'i (1+7)
i' kg.-metres.
The right-hand side of this equation is constant except for the primary
current.
-
426
Electrical Engineering
In Fig. 432, the angles 2, ʼn and e are equal, whence we obtain the
following relations for the phase of the rotor current:
OE. cos $1
e cos 01
sin o
Ea
21
E12
ki,
cos p₂
E₂
c. i₂ (1 + T₂) ´
With the help of the formulae for the E.M.F. we find that
tan 2
e. cos $1 (1 + T2)
k.z.i
or, substituting for e cos 1,
k
tan 2
V.
k
The rotor current can now be found from the formula for cos ₂, thus
k
i
c (1+T₂) cos 2

G
A
Q7
Fig. 433.
We have thus expressed the mechanical and electrical variables as func-
tions of the speed. If, in Fig. 433, OF be made equal to io, and FC be
drawn parallel to OA, it can be proved that
k.c
FA
iz.
1+T₁
In a similar manner, we can construct the vector diagram for the Winter-
Eichberg motor. We can imagine the rotor to have two similar windings,
displaced 90° from each other. The short-circuiting brush axis corresponds
with the axis of the stator-winding, whereas the axis of the magnetising
winding E₁E, (Fig. 428) is displaced 90° from it. The electromotive forces
can be found from the equations already established. It is to be noted,
however, that the rotating short-circuited winding cuts the flux produced
by the magnetising winding, including also the leakage flux. As a matter
of fact, both these windings are represented by the single armature-winding.
Let a represent the ratio of primary to secondary turns on the trans-
former in Fig. 428, and let the indices 1, 2 and 3 refer respectively to the
Single-phase commutator motors
427
stator, magnetising winding and short-circuited winding. Putting &₂ = a. i'
and using our previous formulae for the electromotive forces, we find that
E₁
E2
= 2 (1+7₁) ǹ, due to self-induction of stator,
= a . c² . z (1 + T₂) ¿, due to self-induction of magnetising winding,
c². z (1+™½) is, due to self-induction of short-circuited winding,
E₁ = c.z., due to the mutual induction from stator to short-circuited
E18
Es
winding,
E = c.z.i,, due to the mutual induction from short-circuited winding
Ers
118
3
ㅠ
​to stator,
c.v.z.i, induced in the magnetising winding by rotation in
field due to stator,

E2
Ej3
E
Er 32
i, L2
ET23
A
Eg
E31
E+ 20
Ersa
3
Tπ
3
π
fi3
a2
EV
Fig. 434.
a.c².v.z (1+T₂). ₁, induced in the short-circuited winding by
rotation in field due to magnetising winding,
c².v.z(1+T₂) is, induced in magnetising winding by rotation
in field due to short-circuited winding.
It is evidently impossible for any other E.M.F. to be induced. From the
above equations it follows that
Er
Ern-Er₂-
Eras
E13 E3 E2
3
V.
For the sake of simplicity the vector i, in Fig. 434 is drawn in the same
direction as the vector i, seeing that the introduction of the transformer
does not affect the principle of the motor. As in the repulsion motor, the
resultant of the electromotive forces E, and E₁ must be equal and opposite
13
428
Electrical Engineering
to Er, while the resultant of E and E, must be equal and opposite to the
applied stator pressure. In the magnetising winding, we have E,,, in phase
with and E,,, in phase with is, and these two electromotive forces have
a resultant OC. That this resultant must lie to the right is evident from
the fact that the power-factor of the Winter-Eichberg motor is nearly unity,
so that the vectors of the electromotive forces OB and E,, induced in the
stator and magnetising windings by self and mutual induction, must be com-
pensated, i.e. brought into the vertical, by means of OC. From the equality
of the angles a₁ and a,, and from the above ratios, it follows that the triangles
OCE and AOE₁ are similar. OC is therefore horizontal. The pressure
across the magnetising winding is E- OC and is perfectly wattless. The
pressure across the primary side of the transformer is a (E-OC). Since
OC is proportional to the speed, it depends only on the speed as to whether
the pressure across the transformer is zero, 90° ahead of the current, or 90°
behind it.
Таг
13
The wattless component of the stator pressure is OD, while the watt-
component is BD. The power-factor can therefore be found from the
equation
tan &=
OD+a (E₂- OC)
BD
From the figure and the formulae for the electromotive forces, it can be
proved that
tan & = a. c (1+73) {1.05
V
}
T.1.05
+
1.05 a.c(1+T₂) v
This equation shows that the power-factor is unity at a certain speed,
which depends to a small extent on the ratio of the transformer. For a
motor without leakage (r=0), the speed at which the power-factor is unity
lies in the neighbourhood of the synchronous speed.
It is worthy of note that, at starting, the stator-winding and short-circuited
rotor constitute a short-circuited induction motor, in which the flux is confined
to the leakage paths, as shown in Fig. 412. The starting current is, however,
at the same time, the magnetising current in the magnetising winding, in
which the flux passes across the air-gap in the usual way. Hence, at starting,
the pressure across the stator is very small compared with that across the
magnetising winding or across the transformer. As the speed increases,
the E.M.F. in the magnetising circuit due to rotation comes more and more
into play, until at a certain speed the pressure across the transformer vanishes
and the stator receives the full supply pressure
* The student who wishes to follow this subject still further, should refer to the E. T. Z.
1904-5-6, where articles will be found by Eichberg, Richter, Latour and others. Other articles
by Sumec will be found in the Vienna "Zeitschrift für Elektrotechnik.”
CHAPTER XX.
136. Relation between direct and alternating currents in rotary converters.-137. Armature
copper loss in rotary converters.-138. Comparison between rotary converters and direct-
current generators with regard to armature copper loss.-139. La Cour's Cascade converter.
136. Relation between direct and alternating currents
in rotary converters.
The principle of the rotary converter has been already considered in
Section 95. If suitable points on the armature-winding of a direct-current
motor are connected to slip-rings, an alternating current may be taken from
these rings when the motor is running in the usual manner. If this arrange-
ment is inverted, that is, if alternating current is supplied to the motor by
means of the slip-rings, it runs as a synchronous motor, and direct current can
be collected from the commutator. The peculiarity of this type of machine is
the simultaneous existence of both direct and alternating currents in the
same armature conductors, so that the alternating current may sometimes be
added to the direct, while, at other times, it is flowing in the opposite
direction and must therefore be subtracted. The heating of the armature-
winding of a rotary converter is consequently entirely different from that of
the same machine when mechanically driven as a direct-current dynamo on
the same load.
In order to compare the heat developed in both cases, we must know the
relation between the maximum value imax of the alternating current and the
value c of the direct current. These are the currents in the armature con-
ductors and not those in the external circuit. The relation between them
may be expressed by the equation
:
η
imax
C
..(198).
Now, neglecting losses, the alternating current power supplied must be
equal to the direct-current output. Let
v = the number of armature sections; thus, v = 2 for single-phase, 3 for
three-phase, and so on,
z = the number of wires per armature section,
i= = the alternating current per armature section,
imax
√/2
E the E. M. F. per armature section,
φ the phase difference between E and i.
430
Electrical Engineering
If P be the total alternating current power, we have
P=v.E.i.cos o.
By putting E-k. N~ 2'10-8 and i=
Imax nc
√2 √2
and substituting z, the
total number of wires on the armature, for vz, we get
P=k.N~ z cos $.10-8.
nc
√2
On the other hand, the direct-current power is given by the equation
n
P=2. N. .z.10-8.ia.
α
60


Fig. 435.
Fig. 436.


Fig. 437.
By putting ia = 2a.c and
we obtain the relation
n
p
60
Fig. 438.
=~, and equating these two values of P,
η
2/2
k cos o
(199).
In this equation the value of k is largely dependent upon the breadth of
the coil-sides or armature sections. In the single-phase converter (Fig. 435)
the breadth 2y of the section is equal to the pole-pitch, and 2y=π. In the
two- or four-phase converter (Fig. 436) the breadth 2y is equal to
π
In
2'
ல்
Current relations in rotary converters
431
three-phase converters (Fig. 437) 2y
2π
3
while in the six-phase converter it
π
is only (Fig. 438).
;
3
The coefficient k depends also upon the distribution of the magnetic field.
As before, we shall consider three cases, firstly, a sinusoidal field, secondly, a
pole breadth equal to two-thirds of the pitch, and lastly, a pole breadth of
half the pitch. For a sinusoidal field we have equation (144) on page 281,
sin y π sin y
k = 2.22.
Y
√2 Y
For pole breadths of or the pitch, equations (152) and (153) on
page 287 can be employed for single- or three-phase converters, since the
breadth of the coil 2y is greater than the polar arc B. In four- and six-phase
converters the breadth of the coil is less than the polar arc, and equations
(147) and (154) on pages 282 and 287 must be employed. The values thus
obtained are given in the following table:

Values of the coefficient k
= 150
Sinusoidal
β
field
===
π
single-phase 1
√2
three-phase
√6
four-phase
2
// √5 // √6
√15/10
√10 √3
six-phase
√√2
√5
√14
From these values of k, the value of ŋ =
2√2
k ços o
can easily be found. We
shall make use of these values in the next section in order to determine the
copper loss in the armatures of rotary converters.
137. Armature copper loss in rotary converters.
As a general rule alternating current is supplied to the rotary converter,
while direct current is taken from its commutator. It runs, therefore, as a
combination of A. c. synchronous motor and D. C. generator. The E.M.F. induced
in the armature opposes the alternating current, whereas the direct current,
being produced by the E. M. F. is in the same direction. It follows from this
that the direct and alternating currents within the armature must neutralise
each other to a large degree, thus modifying the amount of heat developed in
the armature-winding.
It does not follow that the heat developed will be necessarily less than in
the ordinary dynamo. This question will depend very largely on the phase
difference between the E. M.F. and the alternating current. Moreover, each
coil-side or section is at times under two unlike poles, and it is evident that,
at such moments, the alternating and direct currents in some wires must be
added together, while in other wires they must be subtracted. The ideal
432
Electrical Engineering
arrangement would have an infinite number of commutator segments and of
phases, so that each coil-side or section would be infinitely narrow.
To make the problem as clear and as simple as possible, we shall assume
that the number of commutator segments is very large, and we shall trace
the variations of current in a single armature conductor during a complete
period*. We shall choose, in the first place, a conductor in the middle of a
Fig. 439.
ӨӨ
Fig. 440.



Fig. 441.
section, i.e. midway between two adjacent slip-ring connections, and we shall
assume that the phase difference between E.M. F. and current is zero, which is
a condition easily obtained by suitable excitation.
At the moment when the conductor under consideration passes the
neutral zone (Fig. 439) both the direct and alternating currents in it are
reversed. If we assume that the direct current taken from the machine is
constant, the curve c in Fig. 440 will represent the direct current in the



C
៩
Fig. 442.
Fig. 443.
Fig. 444.
conductor, plotted to a time base. Similarly, curve i will represent the
alternating current. The resultant current curve is given in Fig. 441. To
obtain the power lost in the given conductor at any moment, the corresponding
ordinate in Fig. 441 must be squared, and multiplied by the resistance R of
the conductor.
The conditions are not so favourable for a conductor separated by an
angle e from the centre of the coil-side. As this conductor passes the neutral
zone at A (Fig. 442) the direct current in it is
reversed. The alternating current, however, being
in phase with the E.M. F., does not reverse until the
middle point of the section passes through the
neutral zone, i.e. an angle e later. The alternating +
current curve in Fig. 443 is displaced, therefore, to
the right, and by combining the ordinates, Fig. 444
is obtained as the curve of resultant current in the
given conductor.

Fig. 445.
If, now, the current lags behind the E.M.F. by the angle 4, the current
curve will be displaced still further to the right (Fig. 445). For the sake of
* See "Elements of Electrical Engineering" by C. P. Steinmetz, 1902.
Armature copper loss in rotary converters
433
simplicity the angle a has been plotted as abscissae instead of the time t.
The instantaneous value of the current in the given conductor is
c — imax · sin (a — e − p) = c {1 − n sin (a — e $)}.
We square the instantaneous value of the current and take the mean of
all such values between 0 and . If R be the resistance of the given con-
ductor, the mean power lost in it will be
c²R
π
[(1 − n. (α-
π.
0
π.
n² 4in
{1 [ — ŋ . sin (a — e − p)}². da = c²R {1+
c²R { 1 + 2/2
π
cos (e + $)}.
In carrying out the above integration we must square out the bracket,
and integrate separately the three terms thus obtained, remembering that
(€ + p) is constant. On examining the right-hand side of this equation, we
see that the mean loss of power in any conductor contains two constant terms,
independent of the position of the conductor, and a third term which varies
with the position of the conductor in the coil-side. The average value of this
third term, taken over the whole coil-side, is
1
4η
cos (e + 4). de
2y π
π
4n sin y
༡
cos o.
-Y
The mean loss of power per conductor, taken over the whole armature, is
therefore
n² 4n y
P₁ = c'R (1+2 in. cos).
c²R
π Y
2√2
If we substitute for n its value
we have
k cos o
4
3.6 sin y
P₁ = c²R (1+
k² cos²
k Y
sin y)
.(200).
ļ
We shall use this equation in the next section, in order to compare the
converter with the direct-current generator, in respect of the copper loss in
the armature.
138. Comparison between rotary converters and direct-current
generators with regard to armature copper loss.
The product c'R in equation (200) is the loss of power per armature con-
ductor, when the machine is working as a direct-current generator with a
current c in the armature-winding. The expression in the brackets,
T=1+
4
3.6 sin y
Υ
k² cos² + k
gives us the relation between the armature copper loss in the converter, and
that in the same machine when driven as a D.C. dynamo with the same
output. The greater the phase difference 4, that is, the smaller the value of
cos ø, the larger will be the second term in the above expression for T, and
the greater the loss. The influence of the width of coil-side or section is also
plainly seen in the above equation. If, for example, the field be sinusoidal,
the coefficient k is proportional to
sin y, and the last term in the equation
,
Y
28
T. E.
434
Electrical Engineering
becomes a constant. The width of coil-side can then affect the second
term only. The coefficient k increases as the width of coil-side is decreased.
With narrow sections the second term is therefore small, and consequently the
loss is small. The superiority of the six-phase converter, with its narrow
sections, is plainly evident. Any practical deviation of the field from the sine
form will not materially alter these relations.
The following table contains the values of г, i.e. the ratio of the losses in
the converter to those in the equally loaded dynamo, under various conditions.

3.6 sin y
T=1+
4
k2.cos2 p
k
Y
сов ф
Sinusoidal
β
field
=}
caps
-
π
single-phase {
1
1.38
1.26
1.10
0.8
2.50
2.28
1.94
three-phase {
1
0.56
0.53 0.49
0.8
1.23
1.13
0.99
1
0.38
0.36
0.35
four-phase
0.8
0.94
0.87 0.77
1
0.27
0.26
0.26
six-phase
0.8 0.77 0.71 0.63
The table shows how considerably the losses can be reduced by so adjusting
the field current that the power-factor is unity. It shows, further, the enor-
mous superiority of the polyphase converter, as compared with the single-phase
type. The influence of pole-form is seen to be very small and is, in reality,
still smaller, owing to the rounding off of the pole-shoes.
A more important question from the practical point of view is: How
great a load can be put on the converter without overheating its armature?
Assuming equal losses in both cases, let
c = the direct current in each conductor of the converter armature, and
c₁ = the current in each armature conductor when driven as a dynamo.
The loss per armature conductor in the dynamo is cºR, whereas in the
converter it is Tc²R (equation 200). The loss being the same in each case,
we have
c₁²R = гc²R,
C
1
or
On the assumption of equal losses in the armature, the relation between
the D.C. load on the converter and that on the dynamo is therefore equal to
1//T. The following table gives the values of 1///г for the various cases
considered.
Permissible load on rotary converter
435

1
cos o
Sinusoidal
100
field
cake
π
T
single-phase {
1
0.85
0.89
0.95
0.8
0.63
0.66
0.72
three-phase 8
1
1.33
1.37
1.43
0.90
0.94
1.00
four-phase {0.8
1
1.62
1.66
1.70
1.03
1.07
1.14
1
1.93
1.95
1.94
six-phase
0.8
1.14 1.18
1.26
Hence, with unity power-factor the output of a single-phase converter is
85 per cent. of the output of the same machine, used as a dynamo, and allowing
the same armature heating. Under the same conditions, however, the six-
phase converter has nearly double the dynamo output.
139. La Cour's cascade converter.
The disadvantages of the rotary converter with regard both to starting
and regulation of the terminal pressure on the D. C. side are obvious. Further
disadvantages are the necessity of static transformers for reducing the alter-
nating supply pressure, and the difficulty of commutation for any but low
frequencies. These difficulties are overcome to a certain extent by using
motor-generators, in which the two machines are quite distinct but coupled
mechanically together. We then have a free hand in the design of each
machine, and the number of poles in the dynamo is not affected by the
frequency of the A. C. supply. The motor can be designed to take the full
supply pressure, thus dispensing with transformers, and the D.C. terminal
pressure can be regulated in the usual way. On the other hand, we have two
machines each of the same output, which is also the full output of the set.
This increases the capital cost, requires more floor space, and decreases the
efficiency. The motor can be synchronous or asynchronous. With the
induction motor the question of starting is quite simple, but we then have
the disadvantage of a low and non-adjustable power-factor, whereas the
synchronous motor has all the advantages and disadvantages of the converter
with regard to starting and power-factor.
A type of converter which has come to the front during the last two or
three years is the so-called cascade converter of La Cour. It combines more
or less all the advantages of the rotary converter, the synchronous motor-
generator, and the induction motor-generator. It consists essentially of an
induction motor and D.C. dynamo side by side on the same shaft. In addition
to being keyed on the same shaft, the rotor and armature are connected
+
28-2
436
Electrical Engineering
electrically, and the power is transmitted partly mechanically along the shaft,
as in a motor-generator, and partly electrically from rotor to armature, so that
they are together equivalent to the armature of a rotary converter. The
three-phase current is supplied to the stator at any pressure up to about
10,000 volts, and the rotor revolves at a speed much below synchronism.
Currents are induced in the rotor with a frequency equal to the slip, which, if
both machines have the same number of poles, will be 50 per cent.
These
rotor currents pass across into the armature of the dynamo and are of just the
correct frequency to drive it as a synchronous motor at the speed corresponding
to 50 per cent. slip in the induction motor. Hence, half the total power
passes mechanically along the shaft, while the other half passes electrically
across from the rotor to the armature. The power-factor can be adjusted by
varying the field current of the dynamo.

3 phase supply
Luntumi
wwwwww
Lutunu
Starting
resistance
Stator
ми
D.C Field
www.
D.C.
Bus-bars
Rotor
D.C.Armature
Fig. 446.
The machine is started from the A.C. side as an induction motor. To
enable this to be done the rotor phases are disconnected at the neutral point,
and a three-phase starting resistance introduced into three of the phases.
When the correct speed is attained the rotor phases are all joined at the
neutral point. The rotor and armature can be nine or twelve-phase since
the connections are short and direct without the intervention of slip-rings.
This large number of phases improves the synchronising power and the
efficiency. This latter is slightly lower than that of the ordinary converter,
but much better than that of a motor-generator. Since only a part of the
power is converted into mechanical energy each machine is smaller than the
corresponding machine of a motor-generator of the same output. A cascade
converter with nine-phase connection between rotor and armature is repre-
sented diagrammatically in Fig. 446.
If the motor has pm pairs of poles and the dynamo pa, the field of the
motor will make revs. per second. The rotor will, in this case, run at
1
Pm+Pa
Pm
revs. per second. The currents induced in the rotor, and supplied by
La Cour's cascade converter
437
Pa
Since
Pm+Pa
revs. per second, its
it to the armature of the dynamo, will have a frequency ~
the dynamo has pa pairs of poles, and rotates at
frequency is also ~
Pa
1 Pm + Pa
the correct frequency.
Pm+Pa
and the current supplied to it from the rotor has
Equation (179) on p. 375, which is the equation for the slip of an induction
motor, may be put into the following form:
- slip __ energy converted into mechanical form
total energy supplied to rotor
It is immaterial whether the difference between the total supply of energy
and that part of it which is converted into mechanical energy, is dissipated as
heat, as in the induction motor, or used to drive a synchronous motor or
rotary converter, as in the present case. If we substitute for the frequency
of slip its value ~
-
1
Pa
Pm+Pa
Pa
Pm+Pa
>
we get
energy supplied mechanically
total energy
Pm
Pm + Pa
Hence, of the total output, a fraction
Pm
Pm+Pa
is transmitted mechanically
along the shaft, while the remainder
Pa
is transmitted electrically.
Pm+Pa
APPENDIX.
THE SYMBOLIC METHOD OF SOLVING ALTERNATING-CURRENT
PROBLEMS.
(See Steinmetz, " Alternating-Current Phenomena.”)
1. The polar diagram.
In considering alternating currents we have used rotating vectors, the
lengths of which were equal to the maximum values of the current or
potential difference. The direction of the vector corresponded to the plane
of an imaginary coil which rotated in a uniform vertical field, and the projec-
tion of the vector on a vertical line represented the momentary value of the
E.M.F. induced in the rotating coil.

S
E
X
B
α
A
Fig. 447.
Fig. 448.
We shall now look at the matter from another point of view. We shall
imagine the armature coil to be at rest and the poles to rotate, as shown in
Fig. 447. As the initial position of the polar axis we arbitrarily choose the
horizontal. In the polar diagram (Fig. 448) we shall also take the horizontal
OA as the initial line or line of reference. The vector OE Emax is set out
at the angle & which the fixed armature coil makes with the horizontal. A
circle is drawn on OE as diameter. Contrary to our previous method, we
keep everything stationary except the line OX which rotates in the anti-
clockwise direction. At the moment when this rotating line makes an angle
=
The polar diagram
439.
a with the initial line OA, it will cut the circle at the point B. The value
of the E. M. F. at this moment will be
Emax ⋅ cos ($− a) = OB.
•
2
OB is the projection of Emax on the rotating line OX. Emax is called the
amplitude of the sine wave and is called its phase. In polar coordinates
the instantaneous value is expressed as a function of the variable angle a.
It reaches its maximum value at the moment when OX coincides with OE.
In Fig. 449, for example, the electromotive forces E, and E, are of different
phase, because their maximum values OE, and OE, make different angles &,
and with the horizontal. The electromotive force E, is behind E, because
the rotating line OX passes through OE, before passing through OE. It is
evident that rotating OX in an anti-clockwise direction round the stationary
vector diagram gives exactly the same result as our former procedure of
giving the whole vector diagram a clockwise rotation, and projecting on to a
fixed vertical line.
2. Geometrical addition.
1
We have seen in Fig. 448 that the momentary value is represented by
the length OB, that is, by the projection of the diameter OE on the rotating
line OX. As in Section 72 (page 227), it can be shown that the two electro-
motive forces QE, and OE, in Fig. 449 can be replaced by a resultant with
an amplitude OR and with a phase angle 4.
1
E2
R

E,
X
42
Pi
α
Fig. 449.
In the following considerations we shall not trouble about momentary
values, but shall consider simply the maximum value, or amplitude, and the
phase of the wave. We can, naturally, use effective values instead of maxi-
mum values, since this simply alters the scale. The wave is completely
determined by its amplitude and phase. Instead of giving the length OE
and the angle 4, we can give the rectangular coordinates of the point E. In
Fig. 450, for example, the wave represented by the diameter OE, is com-
pletely determined by the abscissa a = 3 and the ordinate b = 4. We have
further
and
OE = √ a² + b²,
b
tan &
a
440
Electrical Engineering
We shall follow the usual plan of calling those abscissae positive which
lie to the right of O, while those to the left we shall call negative. Ordinates
which are above O we shall call positive, and those below, negative. We
shall multiply all ordinates (not abscissae) by the imaginary quantity √ — 1,
which, for the sake of shortness we shall represent by j. On this assumption
+3 represents a right-handed abscissa, while +j.4 represents an upward
ordinate, and the wave represented by Fig. 450, which has an amplitude of
E, is represented completely by the equation
E=3+j.4.
We shall follow the plan adopted by Steinmetz and use capital letters to
denote amplitudes, and small letters to denote the projections on the coordi-
nate axes.
We saw on page 227 that the projection of the resultant was equal to
the sum of the projections of the components. If we are given two waves of
different phase, we are now in a position to carry out the geometrical
addition to find their resultant, by simple calculation, instead of the graphical
+
E2
E +

H M

a
+
Fig. 450.
Fig. 451.
$1.
+
E₁
method previously employed. The two electromotive forces E, and E2, the
phases of which are 4, and 2 (Fig. 451) have a resultant E of phase .
The trigonometrical (calculation of this resultant would be very laborious,
and the graphical solution is not very easy when we have to set out the
angles, and 2. The symbolic method enables us to solve the problem in
the following manner:
E₂+3+j.2
E₁ = -4+j.3
2
E=E₁+ E₂ =-1+j. 5.
2
1
We see that geometrical addition is no longer a mere name but an actual
operation of addition. The sign between E, and E, signifies their
geometrical addition, and a glance at Fig. 451 shows us that the geometrical
resultant has the coordinates -1 and +j. 5. Passing now from imaginary
or complex quantities to the calculation of real quantities, we have
and
E = √(− 1)² + 5² = √/26,
+5
tan 4 ======= 5.
1
Rotation of a vector
441
3. Rotation of a vector.
Fig. 452 represents two electromotive forces of equal amplitude, one of
which, viz. E,, lags 90° behind the other. These are evidently represented
by the equations

and
E₁ = + 4+j. 3,
E2
E₂ = −3+j. 4.
2
Since (j) is equal to -1, it
follows that the wave E, can be
obtained from E₁, by multiplying
by j, thus
E₂=j. E₁=j.4-3.
2
Hence, multiplication by j
E2
E₁
0
Fig. 452.
simply rotates the diameter of the circle in Fig. 448 in the direction
in which the line OX is rotating, and gives a wave lagging 90° behind
the original wave.
In a similar manner we can obtain the wave E, by multiplying the wave
E, by -j:
E₁=-j. E₂ =−j(−3+j. 4),
or, since (j) (+j) = +1,
E₁ = 4+j. 3.
Hence, multiplication by -j denotes a rotation of the diameter in
the direction opposite to that of OX, and gives a wave 90° ahead of the
original wave.
If two waves have a phase difference of 180°, it is evident that the
coordinates of the one are the same as those of the other, except that they
have the opposite sign. Multiplication by 1 will therefore rotate a
wave through 180°.
-
If we are given the amplitude and phase of the current, we are now in a
position to express the ohmic pressure and the E. M.F. of self-induction in the
symbolic method. Let the current be expressed by the equation
I=3-j. 2,
M
let the resistance r be 2 ohms and the reactance 27 L, which for the sake
of simplicity we shall denote by x, 1.5 ohms. The pressure required to over-
come the simple ohmic resistance is found by multiplying I by r, so that
I.r=2.(3−j. 2) = 6 − j . 4.
Since the E.M.F. induced owing to the self-induction lags 90° behind the
current, its equation is found by multiplying the current equation, first by x
to get the magnitude, and then by j to put it back 90°. Hence
or, in the present example,
E¸=j.x. I,
8
E=j.15.(3-j. 2) =+3+j. 45.
This is confirmed by Fig. 453. If, instead of the self-induced E.M.F., we
442
Electrical Engineering
require the component of the terminal' pressure which has to counterbalance
the induced E.M.F., we must multiply the equation for E, by -1. This
gives us
1
-E=-3-j. 4·5.

-Es
Es
Fig. 453.
Ir
4. Inductance and Resistance in series.
The component of the applied terminal pressure which is equal and
opposite to the self-induced E. M.F., which it counterbalances, leads 90° ahead
of the current. It is obtained by multiplying the current I by the react-
ance x, and also by -j. This gives-j.x. I. The applied terminal pressure
is the resultant of this and the component which overcomes the resistance.
We obtain by geometrical addition
E=I.r+(−j.x.I)=I.(r−j.∞).
The apparent total resistance R-j.x is called the impedance and is
denoted by Z, so that
Z=r-j.x.
The impedance is therefore made up of the two sides of a right-angled
triangle, the hypotenuse of which gives its real value, thus
z = √ r² + (− x)².
Ohm's law for the alternating current circuit may be expressed as follows:
In the usual method.
In the symbolic method.
E
I=
E
Z r-j.x
e
e
2
√x² + 20²
We shall now determine the terminal pressure of a generator, the resist-
ance of which is r, and the reactance x. The external circuit has a resistance
r and a reactance x. The internal impedance Z, is equal to r。-j. ∞。, and
the external Z to r-j.x (compare Fig. 234 on page 238). If we call the
Inductance and Resistance in series
443
constant E. M. F. induced in the generator E, and the terminal pressure Et,
both expressed in the symbolic method, we have
and
E
I=
E.
Z+Zr₂+r − j . (x。 + x)'
E₁ = I.Z=
E..Z
r。+r−j. (x。 + x)
If we proceed now to find the real values, we have
Et
Now,
and
Hence,
E..z
E..z
√(ro +r)² +(x+x)² √r₂²+2ror+p²+x²+ 2x6x+x
r² + x² = 2²,
r² + x² = 2².
E。z
Et=
√20² + 2² + 2 (ror + xox)
The same result could have been derived directly from Fig. 234.
5. Capacity and Resistance in series.
We saw in Section 78 that the back E.M.F. of a condenser led 90° ahead
of the current. From equation (126) on page 246 it is evident that this
back pressure E, is to be obtained from the current I by dividing by 2π~ K
and multiplying by -j. Hence,
I
E c = − j · Z π ~ K ·
•
The component of the terminal pressure which exactly counterbalances
this back pressure is therefore +2 ~ K'
+j
I
2π~
If capacity be connected in
series with resistance, the terminal pressure will be given by the equation
I
E=I.r+j.
E = I. r + j • 2π ~ K
1
(r+irk)
I(r+j.
For the impedance of the circuit we have therefore
1
Z = r + j · 2π ~ K'
For the actual values we have
K
1
2
2=
p² +
2π K
Z=r-j. 2π~ I
1
Z=r+j.
2π~ K'
and
If we compare the two equations
1
we see that
2πK may be looked upon as a negative reactance. If we let
~ ~
the reactance a represent the component of the impedance which is at right
444
Electrical Engineering
angles to the current, whether due to self-induction or to capacity, the
formula
will be quite general. For inductance we have x=2π~ L, and for capacity
1
X
2π~ K
Z=r-j.x
1
Steinmetz calls
the condensance.
2π~ K
6. Resistance, Inductance and Capacity in series.
The equation
I=
E
r−j.x
can now be used when the circuit contains both inductance and capacity, if
≈ be taken to represent 2π ~ L–
I=
1
2π~ K
•
The equation then becomes
E
r−j. (2π ~ L
1
2π~ K
Passing from complex to real quantities, we have
I
E

1
√r + (2x ~ L - 2 = ~R)
K
The equation which we established for the terminal pressure of a
generator was
Et
2
E。.z
√zo² + z² + 2 (ror+x。x)
It can be seen from this equation that there is a certain value of the
external impedance z, for which the terminal pressure is equal to the E. M.F.
The exact value of this impedance depends on the internal impedance of the
generator. For Et to equal E, in the above equation, it is necessary that the
denominator becomes z, that is, that
zo²+2 (ror+x。x) = 0.
If we neglect the small internal resistance r, we can put 。。, and
we get
2
or
x² + 2xx = 0,
X
Xo
2°
The negative sign shows that the condensance of the external circuit
must exceed its reactance by an amount equal to half the internal reactance.
When this condition is fulfilled the terminal pressure of the generator will
be the same as its E. M.F. If the capacity of the external circuit is still
further increased, the terminal pressure will exceed the E. M.F. induced in the
armature. (Compare Fig. 302.)
Inductance and Capacity in parallel
445
7. Inductance and Capacity in parallel.
Fig. 454 represents an inductive resistance shunted by means of a con-
denser. For the upper branch we have
E
I₁:
r-j.x
The current I, in the condenser branch leads 90° ahead of the terminal
pressure E, and is therefore given by the equation
I,——j. 2π ~ K . E.
For the total current I we have, therefore,
I

- E
I = I₁+1₂ =
E
r−j.x
j. 2π ~ K. E.
By putting E outside, and multiplying top and bottom
by r+j.x, in order to remove the complex quantities
from the denominator, we get
I = E.
(r+j.x
p² + x²
-j.wK}.
r WI
I,
Fig. 454.
K
We have put o for 27~, the angular velocity of the vector. The equation
may be altered so as to separate horizontal from vertical components, thus
r
I = E.
зов + х2
+j. (m² 400
+ x²
wK
=)}
.(a).
To obtain real values we must add the squares of the two components
and extract the root. We thus obtain
r
2
XC
I = E .
+
r² + x²
r² + 20²
wk).
Κ
By suitably choosing the capacity it is possible to bring the external or
total current I into phase with the terminal pressure E, so that the whole
arrangement appears non-inductive. The condition for equality of phase
between I and E is evidently the disappearance of the imaginary component
in equation (a). This can only occur if
2π ~ K:
X
p² + x²
We notice that the necessary capacity depends on the frequency, on the
inductance, and also on the resistance. Wattless current in the external
circuit can therefore be eliminated by the use of suitable capacity, but a
change in either the frequency or the resistance upsets the compensation
and causes a phase difference between current and pressure.
8. Admittance, Conductance and Susceptance.
As we pointed out on page 10, it is often more convenient to consider
conductance than its reciprocal, resistance. In calculating the combined
resistance of several parallel paths, we obtained the equation
1 1 1 1
+ +
Ꭱ Ꭱ Ꭲ Ꭱ Ꭲ Ꭱ,
446
Electrical Engineering
This equation expresses the result directly and simply in terms of the
conductances of the various paths. Steinmetz introduced this and other
reciprocals into the symbolic system. The reciprocal of impedance he termed
admittance and represented it by the letter Y. We have then
and
1
1
Y
y
1
1
Y
r-1.x
r+j.x
p² + 20²
j.x
+
r² + x²
We see then that the admittance consists of two components, the one
real, the other imaginary. The real component is called the conductance
and is denoted by g, while the imaginary component is called the susceptance
and is denoted by b. Hence
r
g
11
p² + x²
8/9/2
r
b =
p² + x²
2/8
z2
or
Y = g+j.b.
Ohm's law may now be written thus:
E
I =
=Z=E.Y,
I = E.(g+j.b).
The actual value of the admittance is given by the equation
y = √ g² + b²
x²
+
1
1
or, since r² + x² = 2²,
y
22
√x² + 20²
……….(a),
.(b),
To illustrate the use of these
equations, we shall again consider the
problem of making the external current in Fig. 454 in phase with the
terminal pressure by suitably choosing the capacity of the condenser. To do
this we must make the imaginary component in the equation
I = E.(g+j.b)
vanish. That is, the susceptance b must vanish.
...(c)
If the upper branch in Fig. 454 has a reactance, its susceptance is
given by equation (b) as
b₁ ===
X1
2
X1
x²² + r₁²°
2
1
In the lower branch, x=—
wK
and r=0, so that equation (b) gives
b₂ = - wk.
For the combined susceptance we have
b = b₁+ b₂ =
X1
ⓇK.
x₁² + r₁²
Admittance, Conductance and Susceptance
447
If there is to be no phase displacement between E and I, the second
term in equation (c) must vanish, since it is this component which causes
the displacement. Hence
b:
X1
x² + r₁₂²
2
@K = 0,
or
It is noteworthy-
@K
2
X1
x₁² + r₁²
(1) that the conductance and susceptance of the individual
branches are not the reciprocals of their resistance and reactance
respectively, but are given by the equations
= and b=
b === ;
g=
22
(2) that neither the resistance nor the reactance of the combined
circuit is equal to the sum of the resistances or reactances of the
separate parallel branches.
In order to find b in the above example, it was necessary to find b₁ and b,
for each branch separately, and then to add them together.
7
LIST OF SYMBOLS USED.
CAPITAL LETTERS.
A
area of cross-section in sq. cms.
B
flux density per sq. cm.
C
D
E
E12
F
a constant, see p. 385
diameter of armature in cms.
electromotive force in volts (occasionally in c. G. 8. units)
E. M. F. induced in winding 2 by current in winding 1
force in kilogrammes
strength of magnetic field
H
I
current; moment of inertia
J
energy in joules or ergs
K
capacity in farads
Ꮮ
M
Mt
Mu
N
No
P
P₁
P₂
Q
2
axial length of armature; self-induction in henries
mass in grammes; coefficient of mutual induction
turning moment or torque
useful or nett turning moment
magnetic flux; flux in rotor
magnetic flux in stator
power in watts; power output from rotor
total power supplied to stator
power transmitted to rotor
quantity of electricity in coulombs
T. E.
Qr
quantity of heat in calories
R
resistance in ohms; reluctance
S
speed of car in kms. per hour
number of slots in armature
number of turns in coil
temperature Centigrade; duration of short circuit
T
V
volume in cubic cms.
W
weight in kilogrammes
X
ampere-turns
Y
winding pitch of slots; admittance
Ꮓ
impedance
α
b
C
SMALL LETTERS.
acceleration; atomic weight; pairs of parallel paths
breadth of slot; susceptance
current density; direct current in conductors of rotary;
various coefficients
29
#
450
d
e
e
f
g
h
i
depth of slot
List of Symbols used
terminal pressure in volts
pressure per phase
force in dynes; tractive force in kgs. per 1000 kgs.
acceleration due to gravity; conductance
horizontal component of earth's magnetism
current in amperes
ir current per phase
j
k
m
n
p
q'
8
t
√1 in symbolic treatment
valency; various coefficients
length of conductor
weight in milligrammes; strength of pole
revolutions per minute
pairs of poles
wires per coil-side, i.e. per pole per phase
resistance in ohms; radius in metres; distance in cms.
slope in parts per thousand
time in seconds; natural period of oscillating alternator
t₁ forced period of oscillating alternator
8_8_d_2
velocity in cms. per second; ratio of speed to synchronous speed
reactance
admittance; winding step
impedance; total number of wires on armature
wires in series or wires per phase
frequency in cycles per second; frequency of slip
frequency of supply in induction motors
GREEK LETTERS.
a
twice the angular brush displacement; variable angle
ẞ polar arc
γ
F9A
Δ
half width of coil-side; phase angle between pole and terminal
pressure of alternator
output of converter output of dynamo
2y + B-π
temperature coefficient (see p. 9)
€
phase displacement due to engine and alternator
€1 phase displacement due to engine only (see p. 328)
n
efficiency
nn hysteresis coefficient
!
λ
leakage coefficient of dynamo (see p. 140)
μ
permeability
ν
ρ
number of tappings per pole pair in rotary converter
specific resistance (see p. 8)
Τι
T2
T
φ
÷ ÷ 3
air-gap reluctance÷primary leakage reluctance
air-gap reluctance÷secondary leakage reluctance
T1 + T₂+ T1 T2
angular phase displacement
$1 angle between current and E.M.F. of alternator
angular velocity
1
1
Absolute system of units, 91
unit of current, 53, 96
electromotive force, 96
resistance, 97
self-induction, 100'
capacity, 101
Acceleration, 92
Accumulators, capacity, 36
charging and discharging, 35, 37
chemical action, 35
construction, 34
efficiency, 36, 37
in parallel with dynamo, 173.
terminal P.D.,
Acids, 27
37
Active material in accumulators, 36
Admittance, 446
Alternating current, effective value, 224
average value, 221
instantaneous value, 221
Alternators, 272
Ammeter, calibration, 7
for alternating currents, 225
principle and connections, 2, 7
Amortisseur, 330
Ampere (unit of current), 2, 53, 96
Ampere-hours, 2, 96
Ampere-turns, see also Excitation, 64, 68
Ampère's swimming rule, 52, 79
Angle of lag, 231, 233, 236, 240, 440
Angular velocity, 192, 219
Anions, 26
Anode, 26
INDEX.
Armature current in rotary converters, 432
leakage of dynamo, 145
alternator, 297, 302
reaction of dynamo, 120, 144
alternators, 298, 302
resistance of bipolar machine, 107
with parallel winding, 113
with series winding, 120
Asynchronous motor, see Induction motor
Atomic weight, 29
Auxiliary poles, 147
Average current, 221
Avogadro's law, 31
Back ampere-turns, 145, 302
Ballistic galvanometer, 62
Bar winding, 289
Bases, 26
Battery, see Accumulator
switch, 176
Bichromate cell, 39
Bifilar winding, 87
Biot-Savart's law, 53
Booster, 180
Brake, electrical, 189
Brush position, 106, 143
Buffer battery, 174
Building up of field of dynamo, 135, 137
Bunsen cell, 38
Cadmium cell, 4, 39
Calibration of ammeters and voltmeters, 7
A.C. instruments, 225
Calorie, 19, 99
Capacity of battery, 36
condensers, 100, 245
and resistance in series, 248, 443
and inductance in parallel, 251, 445
resistance and inductance in series, 249, 444
its influence on power factor, 252, 445, 446
Cascade arrangement of induction motors, 376
Cascade converter, 436
Cast steel for dynamos, 62
Cathode, 26
Cations, 26
Cells, 34, 38
connections for maximum current, 15
Centimetre, 71, 91
dyne, 98
Characteristic of separately excited dynamo on
no load, 160
on load, 166
alternator, 294, 303
induction motor, 369
series dynamo, 168
series motor, 207
shunt dynamo on no load, 169
on load, 169
Charging curve, 37
with single battery-switch, 177
double battery-switch, 178
booster, 179
Chemical action in accumulators, 34
in primary cells, 38
energy, 32
Choking coil, 229, 245
Circle diagram for polyphase motor, 382-397
single-phase motor, 415, 426
Coil-side, width and E.M.F., 279, 280, 362, 431
and excitation of induction motor, 366
and torque of induction motor, 374
Commutation, 143, 149
Commutating field, 155, 156
poles, 148
Commutator, 105
motor, single phase, 417
polyphase, 386 (footnote)
29-2
452
Index
Compensating for lag, 252, 445
Compensation of cross-magnetisation, 148
single-phase motors, 419, 428
polyphase motors, 386 (footnote)
Compound dynamo, 136, 172
Complex quantities, 440
Condensance, 444
Condenser, 100, 245
pressure, 246, 443
current, 246
Conductance, 10, 445
Conductivity, electric, 10
magnetic, 49, 61
Connection of cells, 15
Conservation of energy, 32, 216, 259, 261
Converter, see Rotary converter
Copper, specific resistance, 8
temperature coefficient, 9
voltameter, 30, 221
Coulomb (unit of quantity), 2, 96
Coulomb's law for electricity, 96
magnetism, 42, 95
Creeping bar winding, 289, 359, 362, 366
coil winding, 291
Critical frequency, 250
Cross ampere-turns, 145
Cross magnetisation in alternators, 297
commutating motors, 419
dynamos, 145, 146
Current, definition and unit, 2, 53, 96
effective, 224
mean, 221, 225
measurement, 2, 55
Current density under brush, 151, 153
in dynamos, 113
Damping of galvanometers, 89
by eddy currents, 89, 330
in alternators, 330
Daniell cell, 4, 38
Declination, 50
Delta connection, 344
Diagram of currents in induction motors, 382
-397
transformers, 260
ampere-turns in alternators, 298
pressures in alternators, 293
in synchronous motors, 335
in transformers, 271
Dielectric, 100
Dimensions, 105, 236
Direction of current, 80, 104
induced E.M.F., 79, 82
lines of force, 45
rotation of D. c. motor, 187
polyphase motor, 343, 356
single-phase motor, 356, 410
Discharge curve, 37
Double battery switch, 178
Double bridge (Kelvin's), 17
Drop, see Pressure drop
Drum, two pole, 107
parallel wound, 115
series wound, 123
series parallel wound, 131
Dynamometer, 77, 225
Dyne, 43, 94
Eddy currents, 88, 180, 330
brake, 89
instruments, 90
Effective E.M.F. and current, see Root-mean-
square
Efficiency of accumulators, 36
A.C. generators, 316, 324
D.C. generators, 180
D.C. motors, 195
induction motors, 401
traction motors, 211
transformers, 262
Electrical degrees, 220
Electro-chemical equivalent, 31
Electro-dynamometer, 77, 225
Electrodes, 26
Electrolysis, 26
Electrolyte, 26
Electrolytic cell, 26
Electrolytic mean value, 221, 225
Electromotive force, 2, 77, 95
its measurement, 18
of alternators, 279
of A.C. motors, 358
cells, 3, 34, 39
choking coil, 230
converter, 288, 430, 431
dynamo, 107, 114, 120
mutual induction, 82
polarisation, 31
self-induction, 84, 228
transformer, 253
End cells, 176
End cell switches, 176
Energy, definition and units, 20, 78, 98
conservation of, 32, 216, 259, 261
stored in magnetic field, 71, 87
lost in hysteresis, 50, 73
Equalising connections, 115, 133
Equivalent weights, electro-chemical, 31
Erg, 75, 98
Excitation of A. c. generators, 305
D.C. generators, 161
induction motors, 363, 367
synchronous motors, 334
Farad, 100, 250
Faraday's law of electrolysis, 28
swimming rule, 59
Faure's process of pasting plates, 34
Ferranti machine, 275
Field due to straight conductor, 51
coil, 55
solenoid, 59
magnets of dynamo, 138
alternator, 272
strength and its unit, 43, 61, 95
Flank leakage in alternators, 297
induction motors, 408
Flux density, 61, 95
Force, definition and unit, 43, 93
between two poles, 43
conductor and pole, 53
field and conductor, 53
pole, 44
Form factor, 279
Forming accumulator plates, 34
Foucault currents, see Eddy currents
Frequency, 75, 219, 221
its effect on leakage, 407
Frictional electricity, 1
Friction losses in machines, 180
Galvanometer, 2
Index
453
Galvanometer shunts, 14
Geometrical addition, 227, 440
German silver, resistance, 9
temperature coefficient, 10
Gradient of track, 209, 211
Gramme-calorie, 19, 99
Gramme ring, 105
H-armature, 102
Heat, unit, 19, 98
Heating due to hysteresis, 50
Hectowatt, 21
Henry, 85, 100
Heyland diagram, 382
Hoisting motor, 204
Hopkinson's yoke for testing iron, 61
method of testing dynamos, 185
Horizontal component of Earth's field, 50
Horse-power, 21, 99
Hot-wire instruments, 225
Hunting of A.c. generators, 326
Hysteresis, 50, 72, 225
coefficient, 75
current, 256
loss, 75, 257
Ilgner system, 203
Imaginary quantities, 440
Impedance, 235, 442
Inclination, 50
?
Inductance, see also Self-induction, 236, 444
and resistance in series, 236, 442
in parallel, 239
and capacity in parallel, 251, 445
and resistance in series, 249, 444
Induction coil, 83
of E.M.F., 77, 81, 103, 217
magnetic, see Flux density
motor, diagrams, 382-398
E.M.F., 358
elementary principles, 353
field, 362
output, 397
slip, 355, 374, 384
torque, 354, 372, 379, 384, 387, 400
windings, 359
Inductive load on alternator, 295, 303, 305
resistance, see Inductance
Inductor alternators, 277
Internal resistance of batteries, 15, 173
Interpoles, see Auxiliary poles
Ions, 26
Iron in magnetic field, 49
loss coefficient, 76
Joule, unit of energy, 20, 98
Joule's equivalent, 20
law of heating, 20, 99
Kapp's transformer diagram, 271
Kilogramme, 91, 94
Kilowatt, 21
Kilowatt-hour, 21
Kirchhoff's rules, 10
Lag of synchronous motor, 334
Lap winding, 115
Lauffen type of alternator, 274
Lead (phase advance), 248, 325, 339
Leads, loss in, 23, 350
Leakage, its calculation, 141
Leakage between poles of alternator, 307
in induction motor, 381, 403
I
in inductor alternator, 278
in transformers, 267
coefficient of dynamo, 70, 140
of induction motor, 382, 403
experimental determination,
1
403
variation with air gap, 406
Leblanc's amortisseur, 330
Leclanché cell, 4, 39
Left-hand rule, 187
Length, unit, 91
Lenz's law, 80, 81, 82, 187, 330, 342, 354
Lifting power of a magnet, 71
Lines of force, definition and unit, 45
Liquid starter, 376
Load characteristic of separately excited dy-
namo, 167
series dynamo, 168
shunt dynamọ, 170
normal of induction motor, 386
Loading alternators, 320, 333
synchronous motors, 333
Loss coefficient of iron, 76
Losses in D.c. machines, 180
leads, 23, 350
converters, 431
transformers, 255, 262
Magnet, 42
Magnetic axis, 42
circuit, 65, 162
conductivity, 49, 61
induction, see Flux density
line of force, 45
moment, 45
pole, 42
potential difference, 47, 69
resistance, 65
Magnetisation curves, 61, 63
of dynamos, 165, 206
of alternators, 303
of induction motors, 369
Magnetising current of induction motors, 367
transformers, 230, 254
force, 61
Magneto machine, 134
Magneto-motive force, 65
Manchester dynamo, 139, 140
Manganin, 10
Mass, 91
Maximum power-factor of induction motor, 386
single-phase motor, 417
Maximum value of alternating current, 219
Mean current, 221
Mesh connection, 344
Metallic oxides, 27
Metals, 26
Metre, 91
Metre-kilogramme, 20, 98
Micro coulomb, 96
Microfarad, 101, 250
Mirror galvanometer, 56
Molecular magnetism, 50
Moment of inertia (alternators in parallel), 326
Monocyclic system, 349
Mordey alternator, 276
connections, 114, 133
Motor, 187
Motor-converter, see Cascade converter
454
Index
Multiple circuit ring winding, 112
drum winding, 115
Mutual induction, 82, 422
Natural frequency of alternator, 326
Neutral wire, 24
zone, 106, 142
No load, see Open circuit
current of induction motor, 366, 371
single phase motor, 415
transformer, 254, 257
losses, 180
of induction motors, 397, 400
separation, 183
Non-inductive load on alternator, 296
transformer, 258
winding, 87
Non-metals, 27
Normal load of induction motor, 386
North pole, 42
Ohm, unit, 5, 97
Ohm's law, 5
for alternating currents, 232, 249
for magnetism, 64
Ohmic drop and its phase, 233, 441
Open-circuit characteristic of alternator, 303
dynamo, 161
Ossanna's circle diagram, 390
Output of D.C. motor, shunt, 193
series, 208
induction motor, 379, 394, 397
synchronous motor, 336
Over-excitation, 252, 339
Overload capacity of alternators, 316
D.C. motors, 147
induction motors, 387, 409
single-phase motors, 417
synchronous motors, 337
Pacinotti armature, 104
Parallel conductors, 76
connection of cells, 15
of D.C. generators, 318
of alternators, 245, 318, 333
of inductance and capacity, 251, 445
of resistance and inductance, 239
Parallel running of alternators, 310, 318
working with battery, 173
winding, ring, 112
drum, 115
Periodicity, see Frequency
Permanent magnet, 50
Phase, 439
of condenser P.D., 247, 443
ohmic drop, 233, 441
rotor current, 354, 365, 372, 378
self-induced E.M.F., 231, 441
regulation, 339
Planté accumulator plates, 34
Polar diagram, 438
Polarisation, 31, 38
Pole, 42
strength, 42, 94
width, its effect on E.M.F., 283, 285, 431
number, its effect on leakage, 407
Polyphase machines, 272
Potier, 303
Power, 21, 99
in A.C. circuits, 223
with phase displacement, 239
Power-factor, 241
325
effect on terminal P.D., 296–306
armature reaction, 300, 306
maximum, of induction motor, 386
variation with excitation of generator,
synchronous motor, 339
rotary converter, 434
Predetermination of excitation, see Excitation
Pressure drop of alternator, 238, 296
due to polar leakage, 307
dynamo, 166, 171
transformer, 271
Quadrant, 100
Quantity of electricity, 2, 96, 246
Reactance, 235, 442
voltage (in commutation), 154
Regulation of alternators, 238, 296
P.D. of shunt dynamo, 170
204
motor speed by varying excitation, 199,
-
by series parallel control, 215
by Sprague's method, 213
by series resistance, 197, 211
speed of induction motor, 376
commutator motor, 419, 420
Reluctance, 65
of iron in induction motor, 367
Remanent magnetism, 50, 73, 135
Repulsion motor, 419
Resistance, electric, 5, 97
its measurement, 8, 16
magnetic, 65
and inductance in parallel, 239
in series, 236, 442
and capacity in series, 249, 444
capacity in series, 248, 443
Resonance, 250, 329
Resultant current and pressure vectors, 227
excitation of alternators, 297
polyphase motors, 354, 377, 383
transformers, 261
Retardation method, 183
Reversal of magnetisation, 50, 73
Reversal of D.C. motors, 190
induction motors, 356
repulsion motors, 420
Right-hand rule, 81
Ring armature, 105
parallel wound, 112
series wound, 120
series-parallel wound, 127
Root-mean-square current, 224
pressure, 224
Rotary field in single-phase motor, 417
two-phase motor, 341
three-phase motor, 343
Rotary converter, 288, 429
armature currents, 430, 432
E.M.F., 431
losses, 431
Potential, 47
difference, 21
magnetic, 54, 69
Potentiometer, 18
output, 435
Rotor, 353
current, 354, 365, 372, 378
losses, 375, 379, 384, 394, 398
Index
455
Saturation, 63, 171
its effect on leakage, 141, 307, 403
Second, 91
Self-excitation of dynamos, 135, 137
Self-induction, 85, 228
of armature coil, 154
of alternator, 237, 294, 302
of transformer, 267
effect on commutation, 144, 151
parallel running, 329
V-curve, 324
and resistance in parallel, 239
and resistance in series, 236, 442
and capacity in parallel, 251, 445
and resistance in series, 249, 444
Separately excited dynamo, 134, 160
Separation of losses in D.C. machines, 182
Series dynamo, 135, 168
characteristic, 168, 206, 209
for arc lighting, 169
used as a motor, 188
motor, 205, 418
torque, 208
speed, 209
parallel control of motors, 215
winding, 120, 123
parallel winding, 127, 131
Short-chord winding, 118, 125
Short-circuit characteristic, 303
test of alternator, 303
induction motor, 404
transformer, 270
Short-circuited rotor, 353
Shunt dynamo, 136, 169
as motor, 189
motor, 194, 199
Shunt for galvanometer, 14
Shuttle armature, 102
Silver voltameter, 2, 40, 96
Single-phase commutator motors, 417
generators, 272
induction motor, 410
E.M.F., 413
field, 412, 417
circle diagram, 415
no load current, 415, 417
resolution of excitation into
rotating components, 410
Slip, 354, 375, 379, 385, 394, 399
Slope, see Gradient
Slot pitch, 126, 132
shape and leakage, 406
windings, 119
Slotted armature, 127, 133
Smooth core winding, 105, 108, 280
Spark coil, 83
on breaking circuit, 86, 201
Sparkless commutation, 144, 149, 153
Specific resistance, 8
Speed, 92
of D.c. motor, 192, 194, 198, 200, 205, 209
induction motor, 355, 376, 401
synchronous motor, 333
Sprague's method of series motor control, 213
Squirrel cage rotor, 353
Star connection, 347
Starter, 199, 376
non-sparking, 201
regulating, 197, 212
Starting booster, 204
current of induction motor, 375, 388
Starting D.C. motors, 199
induction motors, 376
resistance, 201
single-phase motors, 356, 417
synchronous motors, 332, 334
torque of induction motor, 387
Static electricity, 1, 97
characteristic of dynamo, 169
alternator, 294, 304
Stator of two-phase motor, 340
three-phase motor, 343
losses, 388, 396
Steinmetz coefficient, 75
Susceptance, 445
Swimming rule, 52, 79
Swinburne's short-cord winding, 118
Symbolic method, 438
Synchronising alternators, 316, 318, 333
force and moment, 317, 326
lamps, 319
power, 317
Synchronous motor, 332
with constant excitation, 334
variable excitation, 337
Tangent galvanometer, 56
Teeth, saturation and leakage in induction
motor, 404
leakage in alternators, 297
Temperature coefficient, 9
rise of field coils, 10
Terminal pressure of an alternator, 295, 305
a battery, 37
a compound dynamo, 172
a separately excited dynamo, 166
a resistance, 6
a series dynamo, 169
a shunt dynamo, 170
a transformer, 254, 259, 271
Thomson's double bridge (Kelvin), 17
Elihu, repulsion motor, 419, 421, 426
Three-phase transmission, 350
Three slots per pole per phase, 284
Three-wire dynamo, 158
system, 24
Time, unit, 91
Tooth, see Teeth
Torque of D.c. motor, 54, 187, 190
400
series motor, 205
shunt motor, 194
single-phase motor, 417
three-phase motor, 354, 372, 384, 387,
Traction problems, 208
Transformer, 83, 253
efficiency, 262
losses, 256, 262
on open circuit, 254, 257
on non-inductive load, 259
on inductive load, 264
ratio, 254, 259
Turning moment, see Torque
Two-part commutator, 104
Two-phase motor, 341
Two slots per pole per phase, 272, 279, 282
Units, 91
V-curve, 324
Valency, 28, 30
Vector diagram, 226
肃
​456
Index
Velocity, 92
Volt, 2, 78, 96
Voltameter, 40, 221, 225
Voltmeter, calibration, 7
construction and connections, 7
Volumetric relations in electrolysis, 31
Ward-Leonard motor control, 203
Watt, 21, 99
component, 243
Watt-hour, 21
Wattless current, 243
Wattmeter, 75
three-phase, 353
Wave winding, 124, 129
Weber, unit of pole strength, 94
Weston cell, 4, 39
Wheatstone bridge, 16
with A.C., 87
Wilde, Dr, 134
Winding step, 108–133
Winter-Eichberg motor, 420, 426
Work, 20, 75, 98
Wound rotor, 353, 376
12
CAMBRIDGE: PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS.

UNIVERSITY OF MICHIGAN
3 9015 06708 7331

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