ELECTRICAL ENGINEERING TEXTS PRINCIPLES OF ALTERNATING CURRENTS ELECTRICAL ENGINEERING TEXTS A series of textbooks outlined by a com- mittee of well-known electrical engineers of which Harry E. Clifford, Gordon Mc- Kay Professor of Electrical Engineering, Harvard University, is Chairman and Consulting Editor. Laws ELECTRICAL MEASUREMENTS Lawrence PRINCIPLES OF ALTERNATING-CUR- RENT MACHINERY Lawrence PRINCIPLES OF ALTERNATING CUR- RENTS Langsdorf PRINCIPLES OF DIRECT-CURRENT MACHINES Dawes COURSE IN ELECTRICAL ENGINEER- ING Vol. I. Direct Currents Vol. II. Alternating Currents Dawes INDUSTRIAL ELECTRICITY PART I Dawes INDUSTRIAL ELECTRICITY PART II ELECTRICAL ENGINEERING TEXTS PRINCIPLES OF ALTERNATING CURRENTS BY RALPH R. LAWRENCE ASSOCIATE PROFESSOR OF ELECTRICAL ENGINEERING OF THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY; MEMBER OF THE AMERICAN INSTITUTE OF ELECTRICAL ENGINEERS FIRST EDITION FOURTH IMPRESSION McGRAW-HILL BOOK COMPANY, INC, NEW YORK: 370 SEVENTH AVENUE LONDON: 6 & 8 BOUVERIE ST., E. C. 4 1922 TK / 14-1 COPYRIGHT, 1922, BY THE McGRAW-HiLL BOOK COMPANY, INC. PRINTED I N THE UNITED STATES OF AMERICA THE MAPLE PRESS - YORK PA PREFACE This book has been developed from notes on Alternating Cur- rents used for several years at the Massachusetts Institute of Technology with the junior students in Electrical Engineering. The portions of the notes dealing with single-phase currents were originally written by Professor H. E. Clifford, Gordon McKay Professor of Electrical Engineering at Harvard University, who was formerly Professor of Electrical Engineering at the Massa- chusetts Institute of Technology. The general arrangement and much of the material of these portions of the book are sub- stantially in the same form as originally written. No attempt has been made to include problems other than those used to illustrate the principles discussed, as two sets of problems on alternating currents, much more comprehensive than could have been incorporated in the book, were already published by Professor W. V. Lyon under the titles " Problems in Electrical Engineering" and "Problems in Alternating Cur- rent Machinery." The author wishes to express his indebtedness to Professor H. E. Clifford for the care with which he edited the manuscript and read the proof. The author also wishes to thank Professor W. V. Lyon for his many suggestions. RALPH R. LAWRENCE. MASSACHUSETTS INSTITUTE OF TECHNOLOGY, CAMBRIDGE, January, 1922. 640709 NOTATION In general the notation recommended by the American Institute of Elec- trical Engineers has been used. When the significance of the letters differs from that given in the following table, it is so stated in the text. A line over a letter indicates that it is a vector or complex quantity. b = Susceptance. C = Capacitance. D = Distance. d = Distance. E = Root-mean-square value of a voltage or an electromotive force. E m = Maximum value of a voltage or an electromotive force. e = Instantaneous value of a voltage or an electromotive force. / = Frequency in cycles per second. f = Function. g = Conductance. h = Height above the earth of the conductor of a transmission line. 3C = Field intensity. / = Root-mean-square value of a current. I m = Maximum value of a current. i = Instantaneous value of a current. j = Operator which produces a counter-clockwise rotation of ninety degrees. k = Constant. L = Coefficient of self-induction, or self-inductance, or a length. M = Coefficient of mutual-induction, or mutual inductance. m = Integer. N = Number of turns. n = Integer. O = Neutral point. P = Average power. p = Instantaneous power. p.f. = Power-factor. Q = Root-mean-square value of a charge or steady value of a charge. Q m = Maximum value of a charge. q = Instantaneous value of a charge. (R = Magnetic reluctance. r = Resistance. S = Coefficient of leakage induction, or leakage-inductance. T = j = Time in seconds of a complete cycle. vii viii NOTATION t = Time in seconds. V = Root-mean-square value of a voltage or an electromotive force. V m = Maximum value of a voltage or an electromotive force. v = Instantaneous value of a voltage or an electromotive force. W = Energy. x = Reactance. y = Admittance. Z = Number of inductors. z = Impedance, a = Phase angle. /3 = Phase angle. (B = Flux density. = 2.718 = Base of Napierian logarithms. = Phase angle. X = Wave length. M = Permeability, co = 2irf = Angular velocity. 2 = Summation. v = Flux. CONTENTS CHAPTER 1 PAGE ALGEBRA OF VECTORS AND OF COMPLEX QUANTITIES USED IN ELECTRI- CAL ENGINEERING 1 Quantities Involved in the Solution of Problems in Alternating Currents Types of Vectors met in Electrical Engineering Prob- lems Solution of Problems Involving Vectors by the Method of Trigonometry Vector Algebra Method of Complex Quantities Representation of Vectors by the Use of the Operator 3 Real and Imaginary Components of a Vector and Real and Imaginary Axes Operator (cos a j sin a) Operator which Rotates the Reference Axes through an Angle a Successive Application of Rotating Operators, Powers and Roots of Operators, Reciprocal of an Operator Reciprocal of the Operator (cos a. j sin a) Oper- ators which Produce Uniform Angular Rotation Solution of Vector Equations when the Vectors and Complex Quantities Involved are Expressed in the Complex Form, i.e., in the Form a + jb Expo- nential Operator e - j Exponential Operator which Produces Uniform Rotation Polar Form of Operator Square Root, Product and Ratio of Vectors or Complex Quantities by the Use of the Exponential Operator Complex Quantity The nth Root and the nth Power of Vectors and Complex Quantities Logarithm of a Complex Quantity or a Vector Representation of an Oscillating Vector, whose Magnitude varies Sinusoidally with Time, by the use of Two Exponential Rotating Operators. CHAPTER II ALTERNATING CURRENTS ' . 27 Direct Current or Voltage Pulsating Current or Voltage Con- tinuous Current or Voltage Alternating Current or Voltage Oscillating Current Instantaneous Value Cycle Periodic Time or Period Frequency Wave Shape or Wave Form Simple Har- monic Current or Voltage Alternating Current Calculations Based on Sine Waves Phase Generation of an Alternating Elec- tromotive Force Strength of Current Ampere Value of an Alter- nating Current Volt Value of an Alternating Voltage Advantages of Defining the Strength of Alternating Currents by the Square Root of their Average Square Value Relation between the Root- mean-square and Average Values for Simple Harmonic or Sinu- soidal Currents Form-factor Measurement of the Effective or Root-mean-square Value of a Current or Voltage Vector Repre- sentation of Simple Harmonic Currents and Voltages. ix X CONTENTS CHAPTER III PAGE POWER WHEN CURRENT AND VOLTAGE ARE SINUSOIDAL 48 Power Absorbed and Delivered by a Circuit Instantaneous Power Average Power Power when the Voltage and Current Waves are both Sinusoidal Volt-amperes, Apparent or Virtual Power Power-factor Reactive-factor Measurement of Average Power Measurement of Power-factor Energy and Wattless or Quadra- ture Components of Current Energy and Wattless or Quadrature Components of Voltage Measurement of Reactive Power, i.e., Reactive Volt-amperes Vectors Representing Voltage Rise and Voltage Fall Expression for Power when the Voltage and Current are in Complex. CHAPTER IV NON-SINUSOIDAL WAVES 69 Wave form of Alternators Representation of a Non-sinusoidal Current or Voltage Wave by a Fourier Series Effect of Even Harmonics on Wave Form Waves which have the Halves of the Positive and Negative Loop Symmetrical Changing the Reference Point from which Angles and Time are Measured in a Complex Wave Fourier Series for Rectangular and Triangular Waves Measurement of Current, Voltage and Power when the Wave Form is not Sinusoidal Determination of Wave Form Effective Value of a Non-sinusoidal Electromotive Force or Current Power when the Electromotive Force and Current are Non-sinusoidal Waves Power-factor when the Current and Voltage are not Sinu- soidal Effective Value of an Alternating Electromotive Force or Current from a Polar Plot Analysis of a Non-sinusoidal Wave and the Determination of Its Fundamental and Harmonics Fischer- Hinnen Method of Analysing a Periodic Wave into Com- ponents of Its Fouries Series Form Factor Amplitude, Crest or Peak Factor Deviation Factor Equivalent Sine Waves Equiva- lent Phase Difference Addition and Subtraction of Non- sinusoidal Waves CHAPTER V CIRCUITS CONTAINING RESISTANCE, INDUCTANCE AND CAPACITANCE. . 115 Coefficient of Self-induction or the Self-inductance of a Circut Coefficient of Mutual-induction or the Mutual-inductance of a Circuit Henry, Secohm or Quadrant Energy of the Field Effect of Self-inductance for a Circuit Carrying an Alternating Current Capacitance Effect of a Condenser in an Alternating- current Circuit Circuit Containing Constant Resistance and Con- CONTENTS xi PAGE slant Self-inductance in Series Solution of the Differential Equa- tion of a Circuit Containing Constant Resistance and Constant Self-inductance in Series Energy in the Electromagnetic Field Breaking an Inductive Circuit Impedance and Reactance Vector Method of Determining the Steady Current for a Circuit Having Constant Resistance and Constant Self-inductance in Series Polar Expression for the Impedance of a Circuit Containing Constant Resistance and Constant Self-inductance in Series Circuit Containing Constant Resistance and Constant Capaci- tance in Series Solution of the Differential Equation for a Circuit Containing Constant Resistance and Constant Capacitance in Series Energy of the Electrostatic Field Impedance and React- ance Vector Method of Determining the Steady Component of the Current in a Circuit Having Constant Resistance and Constant Capacitance in Series Polar Expression for the Impedance of a Circuit Containing Constant Resistance and Constant Capacitance in Series Solution of the Differential Equation for a Circuit Con- taining Constant Resistance, Constant Self-inductance and Con- stant Capacitance in Series Vector Method of Determining the Steady Component of the Current in a Circuit Containing Con- stant Resistance, Constant Self-inductance and Constant Capa- citance in Series Polar Expression for the Impedance of a Circuit Containing Constant Resistance, Constant Self-inductance and Constant Capacitance in Series Equation for the Velocity and Displacement of a Mechanical System Having Friction, Mass and Elasticity. CHAPTER VI MUTUAL- INDUCTION ^.-r. . . . 173 Mutual-induction Coefficient of Mutual-induction or Mutual- inductance Voltage Drop across Circuits Having Resistance and Self- and Mutual-inductance Coefficients of Mutual-induction, Mi 2 and M 2 i, of Two Circuits in a Medium of Constant Permea- bility are Equal Magnetic Leakage and Leakage Coefficients Relation among the Mutual- and Self-inductances of Two Circuits Containing No Magnetic Material, i.e., Having Constant Magnetic Reluctance Coefficient of Electromagnetic Coupling between Two Electric Circuits Having Mutual-inductance General Equations for the Voltage Drops Across Two Inductively Coupled Circuits Each Having Constant Resistance, Constant Self-induc- tance and Constant Capacitance in Series Leakage-inductance of Coupled Circuits Relations Among the Fluxes Corresponding to the Self-inductance, Leakage-inductance and Mutual-inductance Voltages Induced in the Windings of an Air-core Transformer Vector Diagrams of an Air-core Transformer. xii CONTENTS CHAPTER VII PAGE IMPEDANCES IN SERIES AND PARALLEL, EFFECTIVE RESISTANCE AND REACTANCE 196 Impedances in Series Complex Method Series Resonance Free Period of Oscillation of a Circuit Containing, Resistance Induc- tance and Capacitance in Series Impedances in Parallel Con- ductance, Susceptance and Admittance of a Circuit Vector Method Resistance and Reactance in Terms of Conductance and Susceptance Polar Expression for Admittance Parallel Reson- ance Impedances in Series-parallel Filter Circuits Effect of Change of Reactance and Resistance with Current Non-sinus- soidal Voltage Impressed on a Circuit Containing Constant Imped- ances in Series and in Parallel Effective Resistance Effective Reactance Equivalent Resistance, Reactance and Impedance of a Circuit and Also Equivalent Conductance, Susceptance and Admittance of a Circuit General Summary of the Conditions in Series and Parallel Circuits. CHAPTER VIII POLYPHASE CURRENTS 245 Generation of Polyphase Currents Double-subscript Notation for Polyphase Circuits Wye and Delta Connections for Three-phase Generators and for Three-phase Circuits Relative Magnitudes and Phase Relations of Line and Phase Currents and of Line and Phase Voltages for a Balanced Three-phase System Having Sinu- soidal Current and Voltage Waves Relative Magnitudes and Phase Relations of Line and Phase Currents and of Line and Phase Voltages for a Balanced Four-phase System Having Sinu- soidal Current and Voltage Waves Relative Magnitudes of Line and Phase Currents and Line and Phase Voltages for Balanced Star- and Mesh-connected N-Phase Systems Having Sinusoidal Current and Voltage Waves Balanced and Unbalanced Y-con- nected Loads. CHAPTER IX KIRCHHOFF'S LAWS AND EQUIVALENT Y- AND A-CONNECTED CIRCUITS. 271 Kirchhoff's Laws Balanced Y-connected Loads Connected across Three-phase Circuits Having Balanced Voltages Equivalent Unbalanced Y- and A -connected Three-phase Circuits Relations between the Constants of Unbalanced Equivalent A- and Y- connected Three-phase Circuits Equivalent Wye and Delta Impedances for Balanced Loads CHAPTER X HARMONICS IN POLYPHASE CIRCUITS 287 Relative Magnitudes of Line and Phase Currents and Line and Phase Voltages of Balanced Polyphase Circuits when the Currents CONTENTS xiii PAGE and Voltages are not Sinusoidal Wye Connection Delta Con- nection Equivalent Wye and Delta Voltages of Balanced Three- phase Systems Having Non-sinusoidal Waves Harmonics in Balanced Four-phase Circuits Harmonics in Balanced Six-phase Circuits. CHAPTER XI POWER AND POWER-FACTOR OF POLYPHASE CIRCUITS, RELATIVE AMOUNTS OF COPPER REQUIRED FOR POLYPHASE CIRCUITS, POWER MEASUREMENTS IN POLYPHASE CIRCUITS 314 Power and Power-factor of Balanced Polyphase Circuits Power- factor of an Unbalanced Polyphase Circuit Relative Amounts of Copper Required to Transmit a Given Amount of Power a Fixed Distance, with a Fixed Line Loss and Fixed "Voltage between Con- ductors, over a Three-phase Transmission Line under Balanced Conditions and over a Single-phase Line Relative Amounts of Copper Required to Transmit a Given Amount of Power a Fixed Distance, with a Fixed Line Loss and a Fixed Voltage between Conductors, over a Four-phase Transmission Line under Balanced Conditions and over a Single-phase Line Relative Amounts of Copper Required to Transmit a Given Amount of Power a Fixed Distance, with a Fixed Line Loss and a Fixed Voltage to Neutral, when the Loads are Balanced Power Measurements in Three- phase Circuits Proof of the Three-wattmeter Method for Measur- ing the Power in a Three-phase Circuit The N-wattmeter Method for Measuring the Power in an N-phase Circuit Two-wattmeter Method for Measuring Power in a Three-phase Circuit and the (N-l)-wattmeter Method for Measuring Power in an N-phase Circuit Relative Readings on Balanced Loads of Wattmeters Connected for the Two-wattmeter Method of Measuring Power in a Three-phase Circuit when the Current and Voltage Waves are Sinusoidal Determination of the Power-factor of a Balanced Three-phase Circuit, when the Current and Voltage Waves are Sinusoidal, from the Readings of Two Wattmeters Connected to Measure the Total Power by the Two-wattmeter Method Measurement of the Reactive Power of a Balanced Three-phase Circuit whose Current and Voltage Waves are Sinusoidal. CHAPTER XII UNBALANCED THREE-PHASE CIRCUITS. 337 Unbalanced Circuits Direct, Reverse and Uniphase Components of Three-phase Vectors Determination of the Direct, Reverse and Uniphase Components of a Three-phase System A Simple Graphical Construction for Finding the Direct- and Reverse-phase Components of a Three-phase Circuit whose Vectors are Sinusoidal xiv CONTENTS PAGE and Contain no Uniphase Components Mutual-induction between a Three-phase Transmission Line and a Neighboring Tele- phone Line Power in an Unbalanced Three-phase Circuit when the Currents and Voltages are Sinusoidal Phase Balancer Copper Loss in an Unbalanced Three-phase Circuit in Terms of the Direct, Reverse and Uniphase Components of the Currents Effect of Impressing an Unbalanced Voltage on a Three-phase Alternating-current Motor or Generator. CHAPTER XIII REACTANCE OF A TRANSMISSION LINE 355 Reactance of a Single-Phase Transmission Line Average Reac- tance per Conductor of a Completely Transposed, Ungrounded, Three-phase Transmission Line Transfer of Power among the Conductors of a Three-phase Transmission Line Mutual-induc- tion between Transmission Lines or between a Transmission Line and a Telephone Line Voltage Induced in a Telephone Circuit by a Three-phase Transmission Line which is Parallel to the Telephone Line Voltage Induced in a Telephone Line by an Adjacent Three- phase Transmission Line which Carries an Unbalanced Load Third-harmonic Voltage Induced in a Telephone Line which is Adjacent to a Transmission Line that is Fed from a Bank of Transformers which are Connected in Wye on Both Primary and Secondary Sides. CHAPTER XIV CAPACITANCE OF A TRANSMISSION LINE 376 Capacitance of Two Straight Parallel Conductors Charging Current of a Transmission Line Capacitance of a Balanced Three-phase Transmission Line with Conductors r,t the Corners of an Equilateral Triangle, Neglecting the Effect of the Earth Voltage Induced in a Telephone or Telegraph Line by the Electro- static Induction of the Charges on the Conductors of an Adjacent Transmission Line. CHAPTER XV SERIES-PARALLEL CIRCUITS CONTAINING UNIFORMLY DISTRIBUTED RESISTANCE, REACTANCE, CONDUCTANCE AND SUSCEPTANCE . . . 405 Series-parallel Circuits Equations for Voltage and Current of a Line whose Series Resistance and Reactance per Unit Length of Line and whose Parallel Conductance and Susceptance per Unit Length of Line are Constant Propagation Constant, Attenuation Constant, Wave-length Constant, Velocity of Propagation and Length of Line in Terms of Wave Length Evaluation of the Equa- tions for Current and Voltage when Tables of Hyperbolic Functions of Complex Quantities are not Available. INDEX. . 425 J J \ 'J " PRINCIPLES OF ALTERNATING CURRENTS CHAPTER I ALGEBRA OF VECTORS AND OF COMPLEX QUANTITIES USED IN ELECTRICAL ENGINEERING Quantities Involved in the Solution of Problems in Alternating Currents. All quantities involved in the solution of ordinary problems in direct currents are simple algebraic quantities. All equations are simple algebraic equations and may be handled by any of the ordinary algebraic methods.. These statements are equally true when applied to the currents, voltages, power, etc., - existing at any instant of time in an alternating-current circuit, i.e., when applied to the so-called instantaneous values of current, voltage and power. Except in special cases, instantaneous values are not important. What is desired is the average power and the effective voltage and the effective current. Effective voltage and effective current are not simple algebraic quantities and cannot be handled by ordinary algebraic methods. They are vectors and must be handled by methods which are applicable to vectors. For this reason a knowledge of vector algebra and the algebra of complex quantities is necessary to an electrical engineer. Types of Vectors met in Electrical Engineering Problems. There are two types of vectors, both of which occur in many alternating-current problems. There are vectors which are fixed in direction in space, i.e., space vectors, and vectors which are constant in magnitude and revolve with a constant angular velocity, i.e., revolving vectors. The latter are sometimes called time vectors. A constant force is a good example of a space vector. It acts in a fixed direction in space with a constant intensity and may be represented both in direction and magnitude 1 ALTERNATING CURRENTS l>v $ $t?aigktJi?iQ wjio&e tefcgth and direction represent, respective- ly} iKe ^ma^ttwJe^amT the" direction of the force. An alternating current, which varies sinusoidally with time, may be represented by a revolving vector of fixed magnitude which revolves with a constant angular velocity. If the length of the vector represents the maximum value of the current, its projection on a fixed reference axis is the value of the current at the instant of time considered. The number of revolutions per second made by the revolving line is equal to the number of cycles gone through by the current per second. The phase difference between two vector quantities is the angle between the two vectors which represent the quantities. Revolving vectors may be handled by any of the processes which are applicable to space vectors by merely considering them at some particular instant of time or they may be treated by methods which are applicable to them alone. Solution of Problems Involving Vectors by the Methods of Trigonometry. Vectors may be added or subtracted by the use of trigonometrical formulas for the solution of triangles but, when there are more than two vectors, this method of addition or subtraction becomes unnecessarily long and cumbersome and when applied to any but the simplest problems becomes hope- lessly involved. Vector Algebra. The vector algebra which is necessary for handling problems in alternating currents is comparatively simple. It makes easy the solution of alternating-current problems which would otherwise be difficult. A knowledge of vector algebra is therefore one of the most useful tools to the electrical engineer. There are two ways of treating vectors. They may be referred to rectangular coordinate axes and expressed in terms of their components along these axes or they may be expressed in terms of polar coordinates. Each of these methods has its advantages and both are useful. The former is better for addition and subtraction of vectors, but can be used for multiplication and division of vectors as well. The latter is better when only multi- plication and division are to be performed. It cannot be used for addition and subtraction. To change the expression of a vector from one form to the other is a simple matter. VECTORS AND COMPLEX QUANTITIES 3 Method of Complex Quantities. The method of handling vectors when they are referred to coordinate axes is known as the Method of Complex Quantities. In this method all vectors are resolved into two components, respectively along and at right angles to some conveniently chosen axis of reference. An operator j is attached to the component at right angles to the axis of reference to distinguish it from the component along that axis. The name "Complex" as applied to this method does not indicate complexity of method. The method, so far as its application is concerned, should be called the " Simplex Method. " The name " complex" comes from the fact that each vector involved is resolved into a so-called real and imaginary compo- nent, neither of which, however, is actually imaginary. The components to which j is attached are called the imaginary components or simply " imaginaries. " The other components are called the real components or simply "reals." The two rectangular axes along which these components lie are called the axis of imaginaries and the axis of reals. The axis of reals is the axis from which the angles are measured which show the phase relations of the vectors. Although j = \/ 1 is an imaginary quantity, neither the com- ponent of the vector to which j is attached nor the axis along which it lies is imaginary. Both are real, just as real as the other component and the other axis. Let OA, Fig. 1, be a vector A making any angle with the reference axis OX. Let the vector OA = A be resolved into two components, OB. = B and OC = C respectively along and at right angles to the axis OX. The vector has for its magnitude \/B 2 + C 2 and makes an angle tan" 1 ^ with OX. The expression for the vector A, in terms of its components, may be written A = B + C (1) where the addition must be considered in a vector sense. To make it possible to distinguish between vector and non-vector 4 PRINCIPLES OF ALTERNATING CURRENTS quantities in equations, it is customary to use a short line or a dot over letters or numbers representing vector quantities. For example A = B + C (2) A = B + C (3) Sometimes the dot is placed under the vector instead of over it. Since nearly all expressions met in alternating current problems are vector expressions, the use of dots or dashes is often unneces- sary. They are frequently omitted. The simple algebraic expressions which occur are easily distinguishable from the vector expressions without the use of any special symbols. Operator j. Some notation must be adopted which will make it possible to distinguish readily between the components along the two axes. The letter j is used for this purpose. The letter j is an operator which indicates that a vector to which it is attached has been rotated through ninety degrees in a positive direction. Counter-clockwise direction is always considered positive and the horizontal direction is usually taken for the axis of reference. Left to right along this axis is considered positive The operator j does not differ, except in the effect it produces, from other common operators, such as plus and minus signs, multiplication and division signs, exponents and radical signs, log, sin, cos, etc. For example, the exponent 3 with A 3 is merely an abbreviated way of writing A X A X A . The exponent 3 is an operator that indicates that a certain operation is to be per- formed on A, namely, that it is to be multiplied by itself twice. In a similar way the operator j indicates that the vector to which it is attached has been rotated through ninety degrees in a counter-clockwise direction. Using the operator j, equation (2) becomes A = B + JC (4) B + jC is one form of vector expression for the vector A . C, or in general the part of a vector to which j is attached, is the component at right angles to the reference axis, and B, or the component without j, is the component along the reference axis. C without the j attached would lie in a positive direction along the reference axis. The letter j indicates that it has been rotated through ninety degrees in a positive direction from the axis of VECTORS AND COMPLEX QUANTITIES 5 reference. Since the OX or horizontal axis was taken as the axis of reference, jC lies vertically Upward or along the OF axis. The expression given for the vector A in equation (4) is known as its complex expression. The reason for the term complex will be explained later. It does not indicate complexity of expression. It has other significance. If the operator j rotates a vector to which it is attached through + 90 degrees, applying j twice or f once should rotate it through + 180 degrees. Applying j twice or j 2 once reverses a vector and is equivalent to multiplying the vector by 1. Therefore j X j = f = - 1 and Applying j three times or j 3 once gives j* =j X j 2 =JX (-1) = -j The operator* j therefore rotates any vector to which it is attached through 90 degrees or through 90 degrees in a clock- wise or negative direction. Representation of Vectors by the Use of the Operator j. The four vectors A, jA, j 2 A = A Fig. 2. -A and j 3 A = jA are shown in FIG. 2. Four vectors of equal magnitude are given by the following equations. Ii = a + jb (5) A* = -a+jb (6) A* = -a - jb (7) I 4 = a - jb (8) 6 PRINCIPLES OF ALTERNATING CURRENTS These have magnitudes of A = \/a 2 + b 2 and lie respectively in the first, second, third and fourth quadrants. They are shown in Fig. 3. Equations (5), (6), (7) and (8) are the complex ex- pressions for the vectors. FIG. 3. The four vectors make angles, with the axis of reference, of 2 , tan a\ = - and 6 a where sin i = tan 2 = tan 0:3 = tan 4 = + b sin 2 = 7= a -b a -b sm sm a 4 = -& b 2 -b COS tt2 = COS <* 3 = COS 4 = 7= (11) Va 2 + b 2 +~b (12) When expressing the value of the tangent of the angle , which a vector makes with the reference axis OX, it should be left in the form of a ratio with the proper signs attached to both numerator and denominator. Unless this is done it is impossible to tell in which of two quadrants the angle lies. Neither the sine nor the cosine of the phase angle is alone sufficient to fix the position of a vector. For example: referring to equations (9), (10), (11) and (12), sin a = , = might refer to an angle in V a 2 + b 2 either the third or the fourth quadrant and cos a = ^= a 2 + b 2 might indicate an angle in either the second or the third quadrant. Real and Imaginary Components of a Vector and Real and Imaginary Axes. The two parts a and b of the vector expression VECTORS AND COMPLEX QUANTITIES 7 A = a + ft are called respectively the real part and the im- aginary part of the vector. The two axes along which they lie are called the axis of reals and the axis of imaginai ies. The expres- sion a + jb is called the complex expression of the vector A. Neither the imaginary part, 6, of the vector nor the axis of imaginaries, OF, is imaginary. Both are just as real as the com- ponent a, and just as real as the axis of reals. The term im- aginary comes from the mathematical significance of the operator j which is attached to the so-called imaginary component. Mathematically, the operator j = \/ 1 is an imaginary quan- tity. When used as a rotating operator it does not make the component of a vector to which it is attached any less real than the so-called real component. Operator (cos a j sin a). The four vectors AI, A z , A 3 and A 4 , given in equations (5), (6), (7) and (8), page 5 and shown in Fig. 3, may be written Ai = Ai(cos i + j sin i) (13) A.2 = A 2 (cos 2 + j sin 2 ) (14) Az = A 3 (cos 3 + j sin 03) (15) A 4 = A 4 (cos ou + j sin 4 ) (16) In all the equations the angles a are positive angles, i.e., they are measured in a positive direction from the axis of reference. If a, = 0, A i = AKcos + j sin 0) = A^l + JO) A i would lie therefore in a positive direction along the axis of reals. When i has any value other than zero, such as 0, the vector would make a positive angle with the axis of reals, (cos + j sin 0) is therefore an operator which will rotate the vector A i through a positive angle 0. In general (cos a -f- j sin a) (17) is an operator which rotates a vector to which it is applied through a positive angle a. It makes no difference whether the vector to which the operator is applied lies along the axis of reals or in any other direction, the operator will rotate it from its original position through a degrees in a positive or counter-clockwise direction. 8 PRINCIPLES OF ALTERNATING CURRENTS A little consideration will show that (cos a j sin a) (18) is an operator which rotates a vector to which it is applied through an angle a in a negative or clockwise direction from its original position. In both of the operators (cos a + j sin a) and (cos a j sin a) the numerical value of cosine and sine must be taken for the positive angle a. Whether positive or negative rotation is produced depends on the sign attached to j in the operator. Although the sine and cosine are for the positive angle a, it must not be forgotten that the sine and cosine of certain angles are negative. The general operator is (cos a j sin a). If a =0, cos a = 1 and sin a = 0, and (cos a. j sin a) = 1. If a = 90, cos a = and sin a = 1, and (cos a j sin a) = j. If a = 180, cos a .= 1 and sin a = 0, and (cos a j sin a) = 1. If a = 270, cos a = and sin a = -1, and (cos a j sin a) = +j. Operator which Rotates the Reference Axes through an Angle a. In certain cases it is necessary to refer a vector to a new axis of reference which is displaced from the original axis by some definite angle, such as a. Rotating the axes with respect to a vector is equivalent to rotating the vector in the opposite direction with respect to the axes. Therefore, if the operator (cos a + j sin a) rotates a vector through a positive angle a with respect to the axes, the same operator may be considered to rotate the axes through a negative angle a with respect to the vector. The operators (cos a j sin a) and (cos a + j sin a) are therefore two operators which applied to a vector will rotate the axes with respect to it through an angle a respectively in a positive and in a negative direction. In problems there will fre- quently occur vectors which are referred to different reference axes. Before these vectors can be added or subtracted, multi- plied or divided they must all be referred to the same reference VECTORS AND COMPLEX QUANTITIES 9 axes. This can be easily done, provided the angle between their reference axes is known, by applying the operator (cos a + j sin a). The operator with the negative sign produces a positive or counter clockwise rotation of the axes. With the positive sign the opera- tor produces a negative or clockwise rotation of the axes. Successive Application of Rotating Operators, Powers and Roots of Operators, Reciprocal of an Operator. Consider the two operators ki = cos 0i + j sin 0\ k 2 = cos 02 -h j sin 2 They rotate a vector to which they are applied through angles 0i and 2 respectively. Applied in succession they should rotate it successively through angles 0i and 2 or through a total angle (0i + 2 ). fci X k 2 = (cos 0i + j sin 0i)(cos 2 + j sin 2 ) = (cos 0i cos 02 sin 0i sin 2 ) -h ./(sin Si cos 2 + cos 0i sin 2 ) = cos (0i + 2 ) + j sin (0! + 2 ) (19) The product of two operators is thus a new operator which produces a rotation equal to the algebraic sum of the rotations produced by the operators individually. Similarly the product of any number of rotating operators is a new operator which produces a rotation equal to the algebraic sum of the rotations produced by the operators separately. kiXk 2 Xk 3 X . . X k n = (cos 0i + j sin 00 X (cos 2 + j sin 2 ) X (cos 3 + j sin 3 ) X . . X (cos n + j sin n ) = COS (01 + 0* + 03 + + 0n) + j Sin (01 + 02 + 03 + + O n ) = cos 20 + jsin 20 (20) If 0i, 2 , 3 , etc. are all equal ki X k 2 X k 3 X . . . X k n = k n = cos (nO) + j sin (n8) If n0 = 180 degrees or IT radians k n = k 9 = COSTT +j shiTT = 1 (21) 2 * = \/ - 1 = / = cos + j sin (22) 10 PRINCIPLES OF ALTERNATING CURRENTS 2 Therefore \/ I = j n is an operator which rotates a vector through - radians. Since the operator %/ 1 is equal to the nth root of minus one it must have n roots or values, each of which will produce a definite rotation. One of these roots is (cos h j sin -J , which produces the rotation of radians. Adding any number of 2ir radians to an angle does not alter the value of its sine or of its cosine. Equation (21) may therefore be written 1 = cos TT + j sin TT = cos (2q + !)TT + j sin (2q + I)TT where q is any positive or negative integer. f = i = cos V + jsin r (23) \ 9V I \ Tt / 2 It might appear from equation (23) that the operator j n has an infinite number of values, since q may have an infinite number of values. There are, however, only n different roots of minus one. After the nth root, the roots repeat. For example, let n = 3. In this case there are only three different roots. These are: For q = 0, j% = - radians or 60 degrees. o For q = 1, j** = TT radians or 180 degrees. For q = 2, j** = - TT radians or 300 degrees. o For any greater values of q the roots repeat. For example, for q = 3, the root is ~ TT, which is equivalent to ~. This is the o o same as the first root. If q is given negative values, the same three roots will result. For example, if q = 1, the root is radians or 60 degrees. This is the same as plus 300 degrees or is the same as the third root obtained using positive values of q. The operator j = -\/ l has two roots, which produce re- spectively -f-90 and 90 degrees rotation, When using j, VECTORS AND COMPLEX QUANTITIES 11 however, as an operator in the complex expression for a vector, the positive root is arbitrarily used. In such expressions j is used as an operator which produces a rotation of plus ninety degrees. Although the operator j is universally employed in work involving alternating currents to produce a rotation of +90 degrees, the roots of j which produce rotations of fractional parts of 90 degrees are not employed, as other more convenient forms of operator, which are single valued, are available for this purpose. Reciprocal of the Operator (cos a j sin a). Consider the reciprocal of the operator (cos a + j sin a), which produces a positive rotation of a degrees. 1 1 cos a j sin a cos a + j sin a cos a + j sin a cos a j sin a cos a j sin a . . /CM , = ^ . , = cos a j sin a (24) cos 2 a + sin 2 a Therefore, the reciprocal of an .operator which produces a rotation of a is an operator which produces a rotation of a. Dividing a vector by an operator which produces a rotation of any angle a gives the same result as multiplying it by an operator which produces a rotation of a in the opposite direction. In general (cos ai + j sin qQ X (cos a 2 + j sin a 2 ) X (cos |8i + j sin ft) X (cos 2 + j sin 2 ) X X (cos a n -f j sin ) _ X (cos /3 n +j sin ftO " cos (or 0o) + j sin (or 0o) where is the algebraic sum of the angles a, and is the algebraic sum of the angles |8. Operators which Produce Uniform Angular Rotation. Let A be any vector of constant magnitude which rotates at a uniform angular velocity of 2wf = co radians per second. The number of revolutions made by the vector per second is/. Let time, I, be reckoned in seconds and let it be considered zero when the vector A lies along the axis of reference in a positive direction. At any instant of time t seconds after t = 0, the vector will have rotated through ut radians and will make an angle of wt radians 12 PRINCIPLES OF ALTERNATING CURRENTS with the axis of reference, i.e., the axis of reals. The operator which will produce this rotation is (cos cot + j sin co). By making use of this operator the vector A may be represented by A = A (cos co* + j sin coZ) = Aj* l = Aj* ft ~ The operator (cos cot + j sin wt) produces a uniform rotation of 2irf = co radians per second. Solution of Vector Equations when the Vectors and Complex Quantities Involved are Expressed in the Complex Form, i.e., in the Form a + jb. In any vector equation the algebraic sum of the real terms on one side of the equation must be equal to the algebraic sum of the real terms on the other side of the equation, and similarly the algebraic sum of the imaginary terms on one side of the equation must be equal to the algebraic sum of the imaginary terms on the other side of the equation. Since the magnitude of a vector is equal to the square root of the sum of the squares of the real and the imaginary parts, the square of the sum of the real terms plus the square of the sum of the imaginary terms on one side of a vector equation must be equal to the square of the sum of the real terms plus the square of the sum of the imaginary terms on the other side of the equation. The operator j is omitted in taking the squares as it is not a part of the magnitude of any component to which it is attached. In a direct-current circuit, current is given by voltage divided by resistance. A similar expression holds for the current in an alternating-current circuit. In the alternating-current circuit the resistance must be replaced by the impedance. If the circuit is inductive the numerical value of the impedance is \/r 2 + x 2 , where r is the resistance of the circuit and x is the reactance and is equal to the inductance of the circuit multiplied by 2w times the frequency of the impressed voltage. The current multiplied by the resistance is the voltage drop in the circuit caused by the resistance. It is a vector which is in phase with the current. The current multiplied by the react- ance is a voltage drop in the circuit caused by the reactance. It is a vector which is in quadrature with the current and lead- ing it. Impedance is a complex quantity and its complex expression is r + jx. V = l(r + jx) VECTORS AND COMPLEX QUANTITIES 13 from which we have 7 = r + jx Let V = 100 + j50 and let the impedance be 4 + j3. Then the magnitude of I is V(TOO) 2 + (50) 7 = -== 111.8 V(4) 2 + (3) 2 = 22.36 amperes. The vector expression for the current is j 100 + J50 4 +J3 This must be rationalized, to get rid of the j in the denominator, by multiplying both numerator and denominator by the denomi- nator with the sign of its j term reversed. j 100 + J50 4 - J3 4 + J3 - 4 - J3 = 400 - J30Q + J200 + 150 16 - J12 +~J12 + 9 = 550 100 25 J 25 3 The component of the current J which is along the axis to which V is referred is 22 and the component at right angles to the axis is 4. The magnitude of 7 is V(22) 2 + (4) 2 = 22.36 amperes. 4 The vector 7 makes an angle tan ~ x -^o" 50 the reference axis. V = tan~ ] 100' e v = +26.6 degrees. The vector diagram for 7 and V is shown in Fig. 4. = 22-j4 FIG. 4. 0/ = tan- -4 "22 ' = - 10.3 degrees. = e v - e, = 26.6 - (-10.3) = 36.9 degrees. 14 PRINCIPLES OF ALTERNATING CURRENTS It frequently happens, in solving a vector equation, that the magnitude and not the phase angle of one of the terms is known. In such a case the equation cannot be solved as a vector equation. If, however, the term which cannot be expressed in vector form can be separated from the others, the equation may be solved by turning it into a simple algebraic equation by equating the square of the sum of the real terms plus the square of the sum of the imaginary terms on one side of the equation to the square of the term which cannot be expressed in vector form, which is on the other side. Consider the following example in which V is taken as the axis of reference, i.e., as the axis of reals. 6000(cos + j sin 9) = F(cosT) + j sin 0) + 50(cos 30 + j sin 30) (3 + j4) (26) The angle 6 is unknown. Substituting the values for the sines and cosines of the known angles in equation (26) gives 6000(cos S+j sin 6) = V(l + jO) + 50(0.866 + j0.500)(3 + j4) = (V + 29.9) + j(248.2 + 0) (27) (6000) 2 = (V + 29.9) 2 + (248.2) 2 V + 29.9 = -v/35,938,397 = 5995 V = + 5965 or -6025 Putting the positive value of V in equation (27) gives 5965 + 29.9 , . 248.2 = 0.9992 + jO.0414 cos = 0.9992 and sin = 0.0414 = +2.37 degrees. Using the negative value of V gives cos + j sin = -0.9992 + jO.0414 cos = -0.9991 and sin = 0.0414 S = +177.6 degrees. A vector diagram for equation (26) is shown in Fig. 5, in which ,50 (0.866 + j 0.600) V 6026(1+ JO) V-5965U+JO) . FIG. 5. the vectors corresponding to the negative solution are shown VECTORS AND COMPLEX QUANTITIES 15 dotted. The angles and vectors in Fig. 5 are not drawn exactly to scale. Their relative magnitudes are purposely altered to make the diagram clearer. Exponential Operator e J$ . When vectors are to be added or subtracted, the operator (cos a j sin a) should be used with each in order that the vectors may have the proper phase rela- tions. Vectors when expressed in the complex form, i.e., in the form a + jb, may be added or subtracted, multiplied or divided. When, however, the operations of multiplication and division only are to be performed, the exponential operator e ' e is much more convenient and should ordinarily be used. The exponential operator is particularly convenient in such cases since the pro- cesses of multiplication and division then involve only the addi- tion and subtraction of exponents. The exponential form of operator should not be used when vectors are to be added or subtracted, since these operations cannot be performed by the mere addition and subtraction of exponents. The expansion of e> e by Maclaurin's theorem* gives , - 2 , 4 6 s . - {2 + ~ j + etc - 3 5 f)~ + ^-| + ^-| + etc - ) (28) where, for example, |5 means 1X2X3X4X5 and e is the base of the Napierian logarithms, i.e., 2.718. Similarly, the expansion of the sine and cosine of gives sin = -| + ^ - ~ + etc. (29) f) 2 ft* # 6 cos = 1 - ^ + ^ - ^ + etc. (30) Therefore '* = cos + j sin (31) * See any standard book on calculus. 16 PRINCIPLES OF ALTERNATING CURRENTS The expansion of e~ ] ' e gives I + -f etc. / 2 n '13 r 4 e - + 2 4 6 4 |6 H - etc. j (*-?H 05 _ etc. etc. ) = cos j sin (32) e ie and e~'* are therefore two operators producing on a vector to which they are applied the same effect as the operators (cos 6 + j sin 0) and (cos 6 j sin 6) . e> e is an operator which rotates a vector to which it is applied through an angle 6 in a positive direction. e~ J ' is an operator which rotates a vector to which it is applied through an angle in a negative direction. The angle 6 in the two operators should be expressed in radians to be mathematically correct, but it is usually more convenient to express it in degrees. A (cos +j sin 6) = Ae* e (33) A (cos e - j sin 0) = Ae-" (34) represent two vectors which make angles of plus 6 and minus with the axis of reals. The operator (cos j sin 0) refers a vector to which it is applied to rectan- gular coordinates. The operator e 30 re- fers a vector to polar coordinates. The latter operator is called the exponential operator. The vectors given in equations (33) and (34) are shown in Fig. 6. Let x = (a + jb) and y = (c + jd) be two vectors making angles 6 X = tan" 1 - and 6 y = tan" 1 with some common reference d C axis. Then xy - (o + jb) (c + jd) = xy(cos 6 X + j sin O x ) (cos 6 = xy\cos(0 x + e y ) + j sin(0 x (35) FIG. 6. j sin y ) 8,)} VECTORS AND COMPLEX QUANTITIES 17 Similarly if q = q(cos q + j sin q ) and z = z (cos 0, j sin 2 ) are two other vectors qz = qz (cos g + j sin g ) (cos Z j sin Z ) = qz{cos(0 q - 2 ) + j sin(0, - 0,)) = qze j (e " ~ fc) (36) x = a+jb = /x\ /cos x +jsin0A y~c+jd \y) Vcos y + j sin y / _ a^ /cos X + j sin 0A /cos0 v jsin0 y \ y Vcos tf + j sin V 7 Vcos y j sin dj = x /cos(0 a - V ) + j sin(0 z - y )\ y \ cos 2 v + sin 2 v / = -{cos(0 z - e y ) + j sin(0 x - y )} y - *J(9 f ~ .) / (37) An Example of the Use of the Exponential Operator e J0 . Suppose that the product of three vectors, of magnitudes 10, 15 and 20 making angles of + 15, + 20 and 120 degrees respectively with a fixed reference axis, is to be divided by a fourth vector of magnitude 10 making an angle of + 10 degrees with the same reference axis. The four vectors are 10{cos 15 + j sin 15 } = 15 1 cos 20 +j sin 20 } = 15c> 200 20{cos(- 120) + jsin( - 120)} = 20e- J ' 120 10 {cos 10 +j sin 10 } = ICe' 10 " To be mathematically correct the angles in the exponents of the e's should be in radians. It is, however, more convenient to leave them in degrees. The result of the operation of multiplication and division, using the exponential operators, is ( 10 X 15 X 20] f' lfi X e> 20 X e-> 120 A ~ 10 6' 10 ' = 300 e^ 15 " + 20 ~ 120 " 100) = 300e-' 95 = 300 (cos 95 - j sin 95) = 300(- 0.0872 - J0.9962) = - 26.16 - J298.8 18 PRINCIPLES OF ALTERNATING CURRENTS If the result had been obtained by using the operator (cos 6 j -sin 6) for rectangular coordinates, it would first have been necessary to rationalize the expression. There would then have been four quantities of the form (a jb) to multiply to- gether in the numerator and two to multiply together in the denominator. To get the result in the final form of (a jb), the real terms in the numerator would then have to be added and divided by the denominator to get the real or a part of the resultant vector and the imaginary terms in the numerator would have to be added and divided by the denominator to get the imaginary or b part of the resultant vector. The saving of time by the use of the exponential operator is obvious in a case of this kind. Exponential Operator which Produces Uniform Rotation. If the angle 6 in the operator e <7 is replaced by an angle which is proportional to time, the operator will produce continuous rotation of any vector to which it is applied. Let = ut, where co is an angular velocity of 2irf radians per second and t is the time measured in seconds. Then e jut is an operator producing a continuous rotation of 2irf radians or / revolutions per second. A =Ae +jut (38) A' = A f e~ iut (39) are two rotating vectors which have magnitudes A and A' and rotate with a uniform angular velocity of 2wf = co radians per second. They make / revolutions per second. A revolves in a positive or counter-clockwise direction. . A' revolves in a nega- tive or clockwise direction. B = Be jut (40) C = Ce i(ut ~ e) (41) are two rotating vectors having magnitudes of B and C and revolving in a counter-clockwise direction with an angular velo- city of co = 27r/ radians per second. C lags behind B by 6 radians. These continuously rotating vectors may be expressed in terms of rectangular coordinates instead of in polar coordinates by the use of the operator (cos co j sin co). (See page 12.) B = B = AB{cos (a + ft + j sin ( + ftj (48) A = Ae 30i m A (cos a + j sina) Z {cos( + TT) + j sin FIG. 8. The product of two vectors is a new vector having a magnitude equal to the product of the magnitudes of the original vectors. It makes an angle with the reference axis equal to the sum of the phase angles of the original vectors. This was shown on page 9, but less simply, by the use of the operator (cos 6 + j sin 8) . A A fja A A = = ( " ~ ft} = icos (a ~ j sin (a ~ (49) The ratio of two vectors is a new vector having a magnitude equal to the ratio of the magnitudes of the original vectors. It is displaced from the reference axis by an angle equal to the difference of the phase angles of the original vectors. (50) The reciprocal of a vector is a new vector having a magnitude equal to the reciprocal of the magnitude of the original vector. VECTORS AND COMPLEX QUANTITIES 23 It is displaced from the reference axis by an angle equal in magni- tude but opposite in direction to the phase angle of the original vector. VAB = = VAB | cos (^-^) + i efa(^jfy (5i) i = v'I /( ^ ) The fundamental constants which appear in the exact solution of the transmission line for current and voltage are the square root of the product and the square root of the ratio of the series impedance per unit length of conductor and the parallel admit- tance to neutral, also per unit length of conductor. Impedance gj)H fulmitfl.nfift nrt* not vectors. They are complex, quantities. They are really special operators. If a current vector is operated on with impedance, the result is a voltage vector which differs both in phase and in magnitude from the current vector. Simi- larly if a voltage vector is operated on with admittance, the result is a current vector which differs both in magnitude and in phase from the voltage vector. So far as concerns the purely mechani- cal processes of multiplication, division, extraction of roots, etc. of complex quantities, they are carried out in the same manner for complex quantities as for vectors. For the solution of the equations_fpj^the_Jrajnsjnission line it__is necessary to be able to take the square root of the product and the square root of_ the ratio of complex quantities. Complex Quantity. A complex quantity is a quantity which multiplied by a vector gives a new vector of a different kind and magnitude from the original vector and displaced from it in phase. A complex quantity is a simple algebraic quantity combined with an operator. For example, the inductive impe- dance of an electric circuit is a complex quantity. It is written r +jx = zirect Current or Voltage. A direct current or voltage is unidirectional, but it may pulsate. However, the terms direct current and direct voltage, as ordinarily used, designate currents and voltages which are practically steady and non- pulsating. Pulsating Current or Voltage. A pulsating current or pulsat- ing voltage is one that pulsates regularly in magnitude. The term pulsating as ordinarily applied to a current or to a voltage rftfers to a unidirectional current or voltage. Continuous Current or Voltage. A continuous current or voltage is one which is practically non-pulsating. Alternating Current or Voltage. An alternating current or voltage is one that alternates regularly in direction. The term when applied to a current or voltage usually refers to a current or voltage which is periodic and whose successive half waves are of the same shape and area. Oscillating Current. An oscillating current is a periodic current whose frequency is determined by the constants of the circuit in which it is produced. Instantaneous Value. The instantaneous value of a current or voltage is its value at any given instant of time t. Cycle. A cycle of an alter- nating current or voltage is one complete set of positive and negative values of the current or voltage. These values re- peat themselves at regular intervals. Periodic Time or Period. The periodic time or period of an alternating current or voltage is the time required for it to pass through one complete cycle of values. It is expressed in seconds and is ordinarily denoted by T. A cycle is represented in Fig. 10 by the portion of the wave between a and b. 27 28 PRINCIPLES OF ALTERNATING CURRENTS Frequency. The number of cycles passed through by an alternating current or voltage in a second is its frequency. It is denoted by the letter /. For example, a 60-cycle circuit is one in which the current and voltage pass through 60 complete cycles per second. Its frequency and periodic time would be, respectively, / = 60 cycles and T = j = go = 0.01667 second. The frequencies commonly used in America are 60 and 25 cycles, although 50 cycles is also used. Abroad, 50 cycles is common and there are a few installations for railway work with frequencies lower than 25 cycles. In general, frequencies as low as 25 cycles are unsatisfactory for lighting, on account of the noticeable flicker in the lights produced in many cases by so low a frequency. Sixty cycles is almost universally used for lighting, although 50 cycles is perfectly satisfactory. Twenty-five cycles was formerly used for long-distance power transmission and for installations where power was the primary object. However, due to the better design of apparatus and also to a better understanding of the problems of long-distance power transmission, sixty cycles is now largely being used for long-distance power trans- mission as well as for ordinary power and lighting. In general, 60-cycle apparatus is lighter and cheaper than apparatus built for 25 cycles. The difference in weight and cost between 60-cycle and 25-cycle apparatus is essentially the same as between high- and low-speed direct-current motors and generators. The preceding definitions for the most part are taken with slight modifications from the revised (1921) Standardization Rules of the American Institute of Electrical Engineers. Wave Shape or Wave Form. The shape of the curve obtained when the instantaneous values of a voltage or current are plotted as ordinates against time as abscissas is its wave shape or wave form. The abscissa for one cycle is taken as 2ir radians or 360 degrees, and would correspond to one complete revolution of the armature of a two-pole alternator. Two alternating currents or voltages are said to have the same wave form, when their ordinates, at corresponding positions in degrees measured from the zero points of the waves, bear a constant ratio to each other. ALTERNATING CURRENTS 29 Simple Harmonic Current or Voltage. The form of periodic current or voltage most easily dealt with mathematically is one whose instantaneous values follow one another according to the law of sines. In general this wave form is most desirable from the standpoint of the generation, transmission and utili- zation of power. The sine-wave form of a periodic voltage, i.e., a sinusoidal voltage, is given by equation (1), where the time, t, is reckoned from the instant when the voltage is zero and in- creasing in a positive direction. v = V m sin 27r t= v m sin 2irft = V m sin at (1) V m is the maximum value of the wave or its amplitude. T is the duration of one complete cycle, i.e., the periodic time, and / is the number of complete cycles per second, i.e., the frequency. A simple harmonic or sinusoidal wave is positive during one half of each cycle and negative during the other half of the cycle. A simple harmonic current is plotted in Fig. 10 with angles as abscissas. Fig. 10 is its wave form. The abscissas might equally well have been time, 2ir radians corresponding to the time of one complete cycle, i.e., to T = j second. The abscissas are indicated in both these ways in the figure. Alternating Current Calculations Based on Sine Waves. Alternating-current calculations are commonly based on the assumption of sinusoidal waves of current and voltage. Where the waves are not sinusoidal, it is possible to resolve them into component sinusoidal waves, known as the fundamental and harmonics, having frequencies which are 1, 2, 3, 4, 5, 6, etc. times the frequency of the circuit. The components which have frequencies equal to 2, 4, 6, etc. times the frequency of the circuit, i.e., the so-called even harmonics, are not present as a rule in waves of current or voltage. (See Chapter IV.) The effects of the different component sinusoidal waves may be determined separately and then combined. Since any periodic alternating current can be analyzed into a series of simple sinusoidal waves of definite frequencies, it is desirable to discuss in considerable detail the conditions which hold for circuits carrying simple sinusoidal or simple harmonic currents. 30 PRINCIPLES OF ALTERNATING CURRENTS A periodic current or voltage need not follow the simple sine law, but its instantaneous values must be some function of time and it must go through a complete cycle in -j second. Its positive and negative loops need not be symmetrical, but on ac- count of the way in which periodic waves are generated, the positive and negative tm ^ *~ T loops are symmetrical with respect to the axis of time, except in very rare cases. A periodic current which does not follow the simple sine law is shown in Fig. 11. Phase. Time need not be reckoned from the instant when a current or voltage wave passes through zero and in many cases it is not so reckoned. If time is considered zero at a point a radians (equivalent to ^- X 7 second) after the wave has passed Zir j . through zero increasing in a positive direction, equation (1) becomes v = V m sin(co + a) (2) The angle a. is known as the phase angle of the wave. It indicates the number of radians between the point where the wave is zero and increasing in a positive direction and the point from which time is reckoned. The angle may be either positive or negative. When positive it is an angle of lead and the voltage goes through zero increasing in a positive direction before time, t, is zero. When the angle is negative it is an angle of lag and the voltage goes through zero increasing in a positive direction after time, t, is zero. Equations (3) and (4) represent a voltage and a current of the same frequency, both of which are sinusoidal but which do not go through their zero values at the same instant. v = V m sin(orf + ai ) (3) i = I m sin(orf + 2 ) (4) ALTERNATING CURRENTS 31 Although both waves have the same frequency, they are not in phase. Their difference in phase is the difference between their phase angles. When the phase angles are angles of lead as in equations (3) and (4), the angle of lead of v with respect to i is the phase angle of v minus the phase angle of i or (ai 2 ). The angle of lead of i with respect to v is the phase angle of i minus the phase angle of v or ( 2 i). The corresponding angles of lag are (a 2 i) for v with respect to i and (i 2 ) for i with respect to v. When 2 is less than i, (i 2 ) is positive and v actually leads i. When 2 is greater than a\, (i 2 ) is negative and ( 2 i) is positive. In this case v leads i by a negative angle and lags i by a positive angle. Since a negative angle of lead is equivalent to a positive angle of lag, when (a\ a*) is negative v actually lags i. A negative angle of lead is always equivalent to an actual angle of lag, and, similarly, a negative angle of lag is always equivalent to an actual angle of lead. Although the phase angles in equations (3) and (4) should be expressed in radians to be mathematically correct, it is usually more convenient to express them in degrees. When this is done, both wt and a must be reduced to the same units, i.e., radians or degrees, before they can be added. Generation of an Alternating Electromotive Force. The electromotive force generated or induced in a coil through which flux is varying is equal to the rate of change of flux through the coil with respect to time multiplied by the number of turns in the coil, that is, it is equal to the time rate of change of flux linkages for the coil. The flux linkages are equal to the flux through the coil multiplied by the number of turns with which this flux links. If e represents the instantaneous voltage induced in a coil by a change in the flux linking it = - 4: (5) The time rate of change of flux through the coil is -ri and N -r. is the time rate of change of flux linkages with the coil. The voltage e in equation (5) is a voltage rise. When the minus sign is omitted it is a voltage fall or drop. (For the significance of rise and fall applied to a voltage see page 64.) 32 PRINCIPLES OF ALTERNATING CURRENTS The electromotive forces induced or generated in the armature turns of a direct-current generator are alternating but they are rectified with respect to the external circuit by means of the commutator. Figure 12 shows a simple two-pole alternator with a revolving armature and a single armature coil. N and S are the poles. A is the armature which carries a winding placed in two diametrical slots a and 6. The terminals of this winding, in an actual machine, would be brought out to two insulated slip rings mounted on the armature shaft. The current would be taken from these slip rings by means of brushes. If the poles of the alternator are so shaped that the flux through the coil varies as the cosine of the angular displacement of the coil from the vertical position (see Fig. 12) a sinusoidal wave of electromotive force will be generated. In this case e = ~ N % =-**&*-<*<* = coN(p m sin ut (6) N is the number of turns in the armature winding and co is the angular velocity of the armature in radians per second. The maximum value of the flux through the coil is

t + 0). The phase of the current is so determined that at a time t = the vector OA = I m makes a positive angle 0, its phase angle, with the horizontal reference axis from which both angles and time are reckoned. The rotating vector exactly represents the current in both magnitude and phase, since its projection on the vertical axis OF at any instant represents the magnitude of the current at that instant and the angle (at + 6) which it makes with the horizontal axis represents the phase of the current at that instant. FIG. 15. When considering several simple harmonic currents and volt- ages of the same frequency, each may be represented in both magnitude and phase by a revolving vector of proper length and position, all vectors being drawn from the same point. Consider a circuit consisting of two branches in parallel carrying simple harmonic currents i\ and i? given by. the following equations : i! = I m i sin M + 0i) (16) iz = Imi sin ( = i\ + it Referring to Fig. 15, the resultant current xc at the instant of time marked tf is equal to the algebraic sum of the instanta- neous values xa and xb of the two component currents. The resultant current at the time t 1 is equal to the projection of 7 m0 , the resultant of the two component vectors I m \ and I m2 , on the vertical axis at time if. In the left-hand part of the figure the three revolving vectors are shown in the positions they occupy when t = 0. At the instant of time t io = Imo sin (at -f ) = I ml sin (ut + Oi) + /, sin (ut + 2 ) (18) The angle (0i 62) is the difference in phase between the two component currents whose maximum values are I m \ and 7 m2 . Im2 lags I m i by an angle (0i 2 ), i.e., it goes through its maxi- T mum positive value (0i 2 ) radians or (0i 2 ) ^- seconds later than I m \. The phase differences between the components I m \ and 7 m2 and the resultant, whose maximum value is 7 m0 , are respectively (0i ) and (0 2 ). The resultant goes through its maximum value (0i ) radians behind I m i and (0 62) radians ahead of 7 m2 . (0i ) is its angle of lag with respect to Imi and (0 2 ) is its angle of lead with respect to 7 m2 . Expanding the sine terms in equation (18) gives io = 7 m0 (sin o> cos + cos ut sin ) = I m i (sin wt cos 0i + cos ut sin 0i) + 7 m2 (sin co cos 2 + cos at sin 2 ) (19) Equation (19) must hold for all values of t. When t = 0, sin t = 1 and cos erf = 0. In this case i Q = I m0 cos = Imi cos 0i + 7 m2 cos 2 (21) 42 PRINCIPLES OF ALTERNATING CURRENTS Equations (20) and (21) represent two right-angle components of the resultant vector 7 m0 . Its magnitude must therefore be /mo = C/mi cos 0i + I m z cos 2 ) 2 + (/mi sin 0i + 7 m2 sin 2 ) 2 = V(Sf 7 m cos 0) 2 + (SJ 7 TO sin 0) 2 (22) m i sin 0i + 7 m2 sin 2 = ml cos ! m2 2 S} 7 TO cos Equations (22) and (23) are not limited to two currents or in general to two revolving vectors. They may be extended to apply to any number. If there are k currents of the same fre- quency to be added 7 MO = (S*!/ cos 0) 2 + (S{/ sin 0) 2 (24) 2\I m sin tan " = Equations (22) and (23) and also equations (24) and (25) may be obtained more easily by the use of the vector or complex expressions for the currents. Take the axis of time as the axis of reals (See page 6 Chap. I) and consider the vectors at the instant t = 0. At this instant 7 w0 , I m \ and 7 w2 make angles , 0i and 2 with the axis of reals. Their complex expressions at this instant are therefore ImO = ImO (COS + j SU1 ) = tt + jb I mi = I mi (cos 0! + j sin 0i) = ai + jbi 7m2 = Im2 (cos 2 -f- j sin 2 ) = a 2 + jb z 7 TO = Iml ~\~ Im2 a Q + jb = (ai + j 61) + (o 2 + jbz) + a 2 ) 2 + (61 tan = - ALTERNATING CURRENTS 43 If there are k vectors tan = a 2 . . + &2 + . . + + 2 + - + a* Jmo = 2* /, cos -h j 2* J TO sin /o = (Si/m cos 0) 2 + (S*/ w sin 0)* 2* I m sin tan t/o = ^irr --- ;r 2* I m cos Since the relations existing among the maximum values of rotating vectors and also among their phase angles are inde- pendent of the instant of time considered, it is not necessary to take t = when adding currents or voltages. The value of t which will give the simplest and easiest solution should be chosen. Usually this is t = 0, although it is better, in many cases, to give t such a value that one of the vectors will lie along the axis of reals. When the solution of a problem involves both currents and voltages and there is a common vector, such as current in a series circuit or voltage in a parallel circuit, this common vector may well be taken for the axis of reals. For example, in a circuit consisting of a number of branches in parallel the same voltage is impressed across each. It is common to the currents in the branches. In general it should be taken as the axis of reals. In the case of a series circuit the current is the same in all parts. In general, for a series circuit, the current should be taken for the axis of reals. In most problems arising in alternating-current work, interest centers on the effective values of currents and voltages, i.e., root- mean-square values. Maximum values are seldom used or desired. Since root-mean-square values and maximum values are proportional (J ro = \/2 /r.m..) it is customary to write the vector expressions for all alternating currents and voltages in trms of their root-mean-square or effective values. An Example of Addition of Currents. A load on a certain line consists of an induction motor in parallel with a synchronous motor. Since the motors are in parallel, the voltage across their terminals must be equal at every instant. The vectors repre- senting the voltages impressed across the terminals of the motors must therefore be in phase and must also be equal in magnitude. 44 PRINCIPLES OF ALTERNATING CURRENTS The induction motor takes a current of 100 amperes (r.m.s.) which lags the impressed voltage by 30 degrees. The current taken by the synchronous motor is 50 amperes. This leads the impressed voltage by 60 degrees. What is the resultant current which must be supplied to these motors in parallel and what is its phase with respect to the voltage impressed across their terminals? Take the voltage as the axis of reals. The vector expressions for the voltage and currents, using root-mean-square values, will then be V = 7(1 + JO) / = 100 (cos 30 - j sin 30) = 86.6 - j'50 /. = 50 (cos 6,0 + j sin 60) = 25.0 + J43.3 /O = Ii + Is = (86.6 - j'50),+ (25.0 + J43.3) = 111.6 -J6.7 7 = V(H1.6) 2 + (6.7) 2 = 111.9 amperes r.m.s. tan = -yfre = -- 0600 80 = 3.44 degrees The resultant current 7 is 111.9 amperes and it lags the voltage impressed across the motors by 3.44 degrees. In other words, it goes through its cycle 3.44 degrees or 3.44 X second ouu /\ J later than the voltage, / being the frequency of the circuit. Another Example. A two-pole, three-phase, sixty-cycle alter- nator has three exactly similar windings on its armature, spaced 120 degrees from one another and each containing 100 turns. Two of these windings are connected in series with each other in such a way that the voltage between their free terminals is equal to the difference between their induced voltages. If these two windings in series are connected to a constant non-inductive resistance of 10 ohms, what are the maximum and effective values of the current in the resistance? How much power is absorbed ALTERNATING CURRENTS 45 by the resistance? What is the phase relation of the current in the resistance to the voltage impressed across its terminals? What is the phase relation between the voltage across the two windings in series and the voltage induced in each winding? The field flux of the alternator per pole is 10 6 maxwells. This is distributed in such a way that the flux linking each armature winding varies sinusoidally with time. The voltage induced in any armature winding is where # is the flux linking the winding at the time t and N is the number of turns in the winding. According to the assumption that the flux linking each armature winding varies sinusoidally with time,

m COS tot The maximum value of this voltage is obviously The effective or root-mean-square value is \/2 = 4.44 Nfom abvolts = 4.44 NfamlQ-* volts The voltage induced in each winding of the alternator is therefore E= 4.44 X 100 X 60 X 10 6 X 10~ 8 = 266.4 volts effective. Let the armature windings be numbered 1, 2, 3 and assume that the direction of rotation of the armature is such that the phase 46 PRINCIPLES OF ALTERNATING CURRENTS order is also 1, 2, 3, i.e., the voltages induced in the armature winding go through their cycles in the order 1, 2, 3. The three voltages must be 120 degrees apart in time phase since the arma- ture windings are spaced 120 degrees apart on the armature. Let the windings 1 and 2 be the ones connected to the resistance. Take the voltage induced in phase 1 as along the axis of reals. The vector expressions for the three voltages will then be li = #i(cos - j sin 0) = 266.4(1 - JO) = 266.4 - JO E z = # 2 (cos 120 - j sin 120) = 266.4(-0.5 -j 0.866) = -133.2 - J230.6 E 3 = E 3 (cos 240 - j sin 240) = 266.4 ( -0.5 +j 0.866) = -133.2 +j 230.6 The voltage between the terminals of windings 1 and 2, connected in such a way that their voltages subtract, is E =Ei-E 2 = (266.4 - jQ) - (-133.2 -.7 230.6) = 399.6 + j 230.6 E = V(399.6) 2 + (230.6) 2 = 461.4 volts. tan = = - 578 = 30 degrees. The voltage E = E^ E 2 is therefore equal to 461.4 volts and leads the voltage EI by 30 degrees. It leads the voltage E% by 120 + 30 = 150 degrees. The expression for its instantane- ous value is eo = ^/2 X 461.4 sin (27r60< + 30) = 652.4 sin (377* + 30) Since the circuit contains nothing but resistance, the voltage impressed across it must be equal to the resistance drop at each instant. Since the resistance is constant, the current i through ALTERNATING CURRENTS 47 the resistance must be proportional to the voltage at each in- stant. Hence E sin (at + 30) i = J-r.m.s. r E r , m 461.4 r 10 = 46. 14 amperes. I m = \/2 X 46.14. = 65.24 amperes. Since the current is in phase with the voltage, the expression for the instantaneous current is i = V2 X 46.14 sin (377 t + 30 0) = 65.24 sin (377* + 30) The complex expression for the current in terms of its root- mean-square value is _ _ # _ 399.6 + J230.6 : T : 10 = 39.96 + J23.06 The vector expression for the current may also be found by making use of the operator which will rotate a vector through a given angle. Since the current is in phase with the voltage, it must be displaced from the axis of reals by the same angle as the voltage EQ producing it, i.e., by an angle of 30 degrees. The operator which will displace a vector to which it is applied by an angle is (cos + j sin 6). (See Chapter I, page 7.) There- fore, 7 = 7 (cos 6 + j sin B) = 46. 14 (cos 30 + jsin30) = 46.14 (0.866 + jO.500) = 39.96 + j 23.07 From the definition of the effective or root-mean-square value of a current it follows that the average power absorbed in the resistance is P = Pr = (46. 14) 2 X 10 = 21,290 watts. CHAPTER III POWER WHEN CURRENT AND VOLTAGE ARE SINUSOIDAL Power Absorbed and Delivered by a Circuit. The problems concerning power which arise in dealing with direct-current circuits are usually simple and there is seldom any doubt as to whether power is absorbed or delivered. The conditions are not always so simple with alternating currents. It is necessary therefore to determine definitely the conditions under which power is absorbed and delivered. The fundamental conditions are the same for direct- and alternating-current circuits. When a direct-current dynamo acts as a generator, the current flow is from the plus to the minus terminal through the external circuit and it is from the minus to the plus terminal through the armature of the dynamo. The machine is delivering power. The load is absorbing power. Through the load the current flow is from a higher to a lower potential or in the direction of decreas- ing potential, i.e., it is in the direction of the potential drop. Through the dynamo the current flow is from a lower to a higher potential or in the direction of increasing potential, i.e., it is in the direction of the potential rise. When the dynamo acts as a motor, the current flows through the armature from the plus to the minus terminal or in the direction of the potential drop. In this case power is being absorbed by the dynamo. In general, if there is a rise in potential in the direction of the current flow, power is being generated and power is delivered. If there is a drop or fall in potential in the direction of current flow, power is absorbed. The conditions under which power is absorbed or power delivered by an alternating-current circuit are exactly the same as those for a direct-current circuit. Due, however, to the differ- ence in phase of the current and voltage in most alternating-cur- rent circuits, the current is seldom in the direction of either the voltage rise or the voltage fall during all parts of a cycle. Con- sequently, power is absorbed during part of each cycle and de- 48 POWER 49 livered during the remaining part. If the average power absorbed during a cycle is greater than the average power delivered, the net effect is power absorbed. If the average power delivered during each cycle is greater than the average power absorbed, the net effect is power delivered. Instantaneous Power. The instantaneous power, p, in a circuit is given by the product of the instantaneous values of the current and voltage. p = ei If e is a voltage rise in the direction of current flow, p represents power delivered. When e is a fall in voltage in the direction of current flow, power is absorbed. Whether power delivered or power absorbed is considered positive or negative will depend upon the convention adopted for the positive direction of a voltage rise. If the voltage e is considered positive when it represents an actual rise in voltage in the direction assumed positive for current flow, the expression p = ei represents power delivered. When the product of e and i is positive, power is actually delivered, since e will then actually be increasing in the direction of current flow. When the product of e and i is negative, e will actually be decreasing in the direction of current flow and power will be absorbed. In the latter case there is a fall of potential, i.e. a negative rise in potential, in the direction of current flow and the power delivered is negative. Negative' power delivered is power absorbed. /According to this convention a voltage rise in the positive direction of current flow is positive. The negative of a voltage rise is a voltage fall.! If the positive direction for a voltage rise is taken opposite to the positive direction for current flow, the expression p = ei will represent power absorbed. Power will actually be absorbed when p = ei is positive and it will actually be delivered when p = ei is negative. According to this convention a voltage drop, or fall in potential in the positive direction of current flow, is considered positive. Curves of e, i and p = ei are plotted in Fig. 16. If e represents a voltage rise, considered positive in the positive direction of current flow, power will be delivered when e and i have the same sign, i.e., when both are positive or both are negative. There 4 50 PRINCIPLES OF ALTERNATING CURRENTS will then be an actual rise in voltage in the direction of current flow. When e and i have opposite signs power will be absorbed. In this case there will be an actual fall in voltage in the direction of current flow. Power delivered will be positive and power absorbed will be negative. Dashed Line Voltage Full Line Current Dotted Line Power FIG. 16. When only power delivered is to be considered, it is best to take the positive direction for a voltage rise in the direction assumed positive for current flow, since this convention makes power delivered positive and avoids the use of negative signs before the expressions for power delivered. This convention is best and also most logical when power delivered and power absorbed are involved in the same problem. When, however, only power absorbed is to be considered, it is best to adopt the convention that voltage drops are positive in the direction assumed positive for current flow, since then power absorbed is positive and the necessity of using minus signs before the expres- sions for power absorbed is avoided. Average Power. The power delivered or absorbed by an alternating-current circuit is the average power considered over a complete cycle, i It is seldom that the instantaneous power is of interest. The average power is given by 1 C T P = T eidt (1) POWER 51 This expression gives the mean ordinate of the power curve, i.e., the dotted curve in Fig. 16, and is equal to the net area enclosed by the power curve divided by its base. In order to evaluate the integral, it is necessary to know the form of the functions deter- mining the voltage, e } and the current, i. Power when the Voltage and Current Waves are both Sinu- soidal. Assume the voltage and current waves to be sinusoidal. CASE I. VOLTAGE AND CURRENT IN PHASE. Let e = E m sin co i = I m sin ut p = ei = E m l m sin 2 at =E m I m i -~i*^ (2) 1 C T E m l m 1 C T E I Average power = P = I m dt I n ' m cos 2ut dt = E m I m jEn_ 7^ 2 ~ Vl " V? = EI (3) where E and / without subscripts are the root-mean-square or effective values of the voltage and current. Therefore, when the voltage and current are in phase, the average power is equal to the product of the effective values of voltage and current. It is equal to the effective volt-amperes, i.e., the product of the effective voltage and effective current. The expression for the instantaneous power, p, (Equation 2), may be written p = p _ _^ cos 2wt = p - p C os 2wt (4) This equation and the equations for voltage and current are plotted in Fig. 17. It should be noted that the power curve is sinusoidal in form and has double frequency as compared with the voltage and current. Its axis of symmetry is at a distance P, equal to the average power, above the axis of the voltage and current curves. Its amplitude is equal to one-half the maximum volt-amperes or is equal to the effective volt-amperes. Effective volt-amperes are equal to average power when the voltage and current are in phase. Although the power is not constant, its 52 PRINCIPLES OF ALTERNATING CURRENTS flow is always in the same direction when the voltage and current are in phase. The corresponding instantaneous values of voltage and current are always of the same sign. The current flow is therefore always either in the direction of the voltage rise or in the direction of the voltage drop, according as e, in the equation for voltage, represents a voltage rise or a voltage drop. If e t=o FIG. 17. represents a voltage rise, the expression for average power gives the power delivered. If e represents a voltage drop, the expres- sion for average power gives the power absorbed. If e, represent- ing a voltage rise, and i are opposite in phase the expression for power will be negative and will represent power absorbed. If e, representing a voltage drop, and i are opposite in phase the expression for power represents power delivered. CASE II. VOLTAGE AND CURRENT IN QUADRATURE. Let e = E m sin cot i = I m sin (at - 90) p = ei = E n l m (sm 2 ut cos 90 - sin cat cos co sin 90) = E m l m (0 sin co cos wt) - sin 20 (5) 1 C T F 1 Average power = P = ~ I i22 (0 - sin 2wt)dt * t/o ^ = When the voltage and current are in quadrature the average power is zero. The curves for instantaneous power (equation (5)) and for the POWER 53 instantaneous voltage and current are plotted in Fig. 18. The power curve is again a simple harmonic curve of double frequency with respect to the voltage and current and it still has an ampli- tude equal to the effective volt-amperes, but its axis of sym- metry now coincides with the axis of symmetry of the voltage and current. This latter condition follows from the fact that the average power, P, is zero. t-o FIG. 18. Although the average power is zero, the power at any instant is not zero except at four points during each cycle. There is an oscillation of power between the source and the load, the average value of which is zero. The amplitude of this oscillation is equal to El, the effective volt-amperes of the circuit. If e represents a voltage rise, the positive power loops represent power delivered. The negative power loops represent power absorbed. During two quarters of each cycle, power is delivered by the circuit. During the other two quarters of each cycle an equal amount of power is absorbed. The power which is absorbed is stored as kinetic energy either in the rotating part of a motor or generator or in a magnetic field produced by the current, or it may be stored as potential energy in the electrostatic field of a condenser. This stored kinetic or potential energy is given back to the circuit as dynamic electrical energy. CASE III. GENERAL CASE. VOLTAGE AND CURRENT NEITHER IN PHASE NOR IN QUADRATURE. Let e = E m sin ut i = I m sin (at e) 54 PRINCIPLES OF ALTERNATING CURRENTS where 6 is neither nor 90 degrees. p = ei = E m l m sin ut sin (wt 6) (6) Since 2 sin a sin = cos (a |8) cos (a + 0), equation (6), may be written P = E m L cos ( + (9) - cos (2co - (7) i r r jr f Average power = P = _ I fl^fcos ( + 0) TJo 2 -C' TO-* cos i f\ - w* + 0= ^ cos (2ui L V2" V2 COS0 cos (8) The average power is equal to the product of the effective values of voltage and current multiplied by the cosine of the phase angle between the voltage and current. When the phase angle is zero, P = EI cos 6 becomes EI. When the phase angle is 90 degrees, P = EI cos 6 becomes zero. Equation (7) for the instantaneous power may be written p = EI cos 6 - EI cos (2co/ - 0) = P - EI cos (2ut - 6) (9) From equation (9) it may be seen that, in the general case as well as in the two special cases first considered, the power curve p t-o 1!). is a double frequency curve. As in the first two cases it has an amplitude equal to the effective volt-amperes and its axis is dis- placed from the axis of voltage and current by a distance equal to the average power. Curves of power, voltage and current are plotted in Fig. 19. POWER 55 If e is taken as a voltage rise, equation (8) represents power delivered. Power will actually be delivered when El cos 6 is positive. Power will actually be absorbed when El cos is negative. It will be positive when is less than 90 degrees, since the cosine of an angle which is less than 90 degrees is positive. It will be negative when 6 is greater than 90 degrees (but less than 270 degrees) , since the cosine of an angle which is greater than 90 degrees (but less than 270 degrees) is negative. Since an angle of lag of 6 degrees is the same as an angle of lead of 360 minus degrees and similarly an angle of lead of 6 degrees is the same as an angle of lag of 360 minus degrees, it is customary to use the smaller of the two angles when expressing the phase difference of a current and a voltage. Equal angles of lead and lag pro- duce the same effect so far as average power is concerned. When = the power curve, Fig. 19, is entirely above the axis of current and voltage. This corresponds to the conditions shown in Fig. 17 page 52 for Case I where the voltage and current are in phase. When is greater than zero and less than 90 de- grees, the axis of the power curve is displaced in a positive direc- tion from the axis of the voltage and current curves. In this case (assuming E represents a voltage rise) power is delivered. If is greater than 90 degrees, the axis of the power curve is dis- placed in a negative direction from the axis of the voltage and current curves. In this case the average power is negative and power is absorbed. If is equal to 90 degrees, power is neither delivered nor absorbed. This corresponds to Case II. Volt-amperes ; Apparent or Virtual Power. The volt-amperes, or apparent or virtual power of a circuit, is equal to the product of the effective or root-mean-square values of voltage and current. This product is not equal to the true power except, when the voltage and current are in phase. Although volt-amperes do not represent true power Except in the case mentioned, a know- ledge of the volt-amperes of a circuit is usually of considerable importance, since volt-amperes and not watts determine the limit of output of most electrical apparatus. The limit of output of all alternating-current apparatus, such as motors, generators, transformers, etc., is determined chiefly by the rise in temperature produced in the windings. The increase in temperature is caused principally by the core and copper losses. Core losses 56 PRINCIPLES OF ALTERNATING CURRENTS depend upon frequency and flux density and are fixed by the operating voltage and frequency. Copper losses are determined by the current. For fixed voltage and current these losses are practically independent of the phase relation between current and voltage. The limit of output of a generator, motor or trans- former, therefore, is determined by the volt-amperes which produce the limiting rise in temperature in the windings. All types of alternating-current apparatus, which can operate with different phase angles between current and voltage, i.e., at different power factors, are rated in volt-amperes. Full load is reached when they carry full rated current at rated voltage and frequency. Their output under this condition would be zero if the current and voltage were out of phase by 90 degrees. It will be a maximum when the current and voltage are in phase. This latter condition is most to be desired and is approached in prac- tice, although seldom exactly attained. The common unit for apparent power for large power apparatus is the kilovolt- ampere (abbreviated kv-a.). This bears the same relation to the volt-ampere as the kilowatt bears to the watt. Power-factor. For steady voltages and currents the power is always given by the product of volts and amperes, that is by the volt-amperes. For alternating currents this is true only when the voltage and current are in phase. In general, for sinusoidal waves, power is (equation (8)) El cos 6. It is equal to the volt- amperes multiplied by a factor, cos 0, which is determined by the phase relation of the voltage and current. This factor, cos 0, is called the power-factor of the circuit. When the current and voltage waves are not sinusoidal, the factor cos 6 has no significance except when considered with respect to the equivalent sine waves. These will be explained later. Theoretically, the power-factor may have any value from zero to unity, both inclusive. Although it is possible to obtain zero power-factor experimentally by the use of a synchronous motor which receives enough power mechanically to supply its losses, zero power-factor is not reached in practice. It would be a very undesirable condition in most cases. Unity power-factor may be obtained and is not uncommon in alternating-current work. The power-factor of a circuit is always the ratio of the true POWER 57 power to the apparent power, i.e., the ratio of the watts to the volt-amperes. It is merely a ratio and therefore independent of the units used except that the true power and the apparent power must be expressed in the same system of units. Power-factor is commonly expressed in per cent, although it is frequently given as a fraction. It must always be taken as a fraction when used in numerical expressions. Power-factor is the cosine of an angle only for sinusoidal waves of current and voltage. Power-factor = , T ,, (10) Volt-amperes This expression for power-factor holds regardless of wave form. Reactive-factor. The expression sin 6 = V 1 -- (cos 0) 2 = sin [cos" 1 (Power-factor)] (11) is called the reactive-factor. It is true, however, only for sinusoidal waves. When the waves are not sinusoidal, the re- active-factor is given by Reactive-factor = \/l (Power-factor) 2 (12) The reactive-factor is of considerable importance, since a knowledge of its magnitude is necessary when determining the size of the apparatus required to raise the power-factor of a circuit to unity, or by any desired amount. The reactive-factor is positive for lagging currents and negative for leading currents when 6 is the lag angle of the current with respect to the voltage. Measurement of Average Power. The average power in a circuit may be measured by an electrodynamometer. If the fixed coil of such an instrument is placed in series with one main of the circuit in which the power is to be measured and its movable coil is connected in series with a large non-inductive resistance and then shunted between the two mains, the average torque developed between the two coils will be proportional to the aver- age product of the instantaneous values of current and voltage. The torque exerted between two coils which are free to turn about a common diameter is proportional to the product of the currents they carry. In the electrodynamometer used as a wattmeter, the fixed coil (usually called the current coil) carries line current. The moving coil (usually called the potential coil), in series 58 PRINCIPLES OF ALTERNATING CURRENTS with non-inductive resistance, carries a current which is pro- portional to the voltage across its terminals, provided the non- inductive resistance is large compared with the inductance of the potential windings of the instrument. Average torque = i f ? -*J," - r Pi J Jo /t -.' where ^ and * c are, respectively, the instantaneous currents carried by the potential and current coils, ft is the non-inductive resistance in series with the potential coil plus the resistance of the potential coil and e p is the instantaneous voltage across the terminals of the potential coil circuit. P is the average power. The moving coil will deflect until the average torque developed in it is just balanced by the torsion of the control springs attached to its shaft. By proper calibration the instrument may be made to indicate the average power. When the power in high-voltage circuits is to be measured, it is necessary to use transformers with both the current and potential coils of the instrument. Measurement of Power-factor. Although there are special instruments which will indicate the power-factor of the circuit in which they are connected, they are not very accurate and are not in very general use except on switchboards. The power- factor of a circuit is generally determined by measuring the voltage, current and power by means of a voltmeter, an ammeter and a wattmeter and then taking the ratio of the watts to the volt-amperes. The reactive-factor is calculated from the power- factor by equation (12), page 57. Energy and Wattless or Quadrature Components of Current. When dealing with sinusoidal waves, it is often convenient to resolve the current into two right-angle components, one in phase Awith the voltage, the other in quadrature with it. These two 'components are I cos and I sin 0. The reason for selecting ; these two components will be made clear by what follows. Let e = E m sin cat .i = I m sin (cat - Q) POWER 59 Then since i = I m cos sin cot I m sin cos at p = E m sin cot (I m cos 0) sin cot E m sin cot (7 TO sin 0) cos cot (13) i r * P = ^ I E m sin cot (! cos 0) sin cotdt 7 Jo i r r I E m sin cot (7 ro sin 0) cos cotdt (14) i Jo The second integral vanishes and therefore contributes nothing to the average power or watts of the circuit. The power may be regarded as due to a voltage e = E m sin cot and a current i = (I m cos 0) sin cot, whose maximum values are E m and I m cos 6 respectively and whose phase difference is zero. Since the power is contributed wholly by the component of the current I m cos 0, which is in phase with E m , this component is known as the energy, power or active component of the current, or simply the energy, power or active current. The same terms are applied to the corresponding component of the effective current. In fact when energy, power or active current is mentioned, the energy, power or active component of the effective current is understood. The second integral may be regarded as representing the power due to a voltage e = E m sin cot and a current i = (I m sin 0} cos cot or i (I m sin 0) sin (cot 90), whose maximum values are respectively E m and I m sin and whose phase difference is 90 degrees. It has already been shown that when a voltage and a current are in quadrature the average power is zero. The component I m sin of the current contributes nothing to the average power of the circuit and is therefore known as the quad- rature, wattless or reactive component of the current or simply the quadrature, wattless or reactive current. The same terms are applied to the corresponding component of the effective current. Although the component I m sin contributes nothing to the average power of the circuit, its presence increases the resultant current and therefore increases the copper loss for a given amount of power. The resultant current is always equal to the square root of the sum of the squares of the active and the reactive com- ponents of the current. The resultant current causes the copper loss but only its active component contributes to the average power. 60 PRINCIPLES OF ALTERNATING CURRENTS If E and / represent the root-mean-square values of the voltage and current, the expression P a = El cos 6 = El X power factor represents the average power due to the active or energy com- ponent of the current. P a is the true average power of the circuit. It is sometimes called the active power or the active volt-amperes, i.e., the volt-amperes due to the active component of the current. The second term of the second member in equation- (14) is the average power due to the wattless or reactive component of the current. This average is zero. The expression for the instan- taneous power due to the reactive component of the current is p r = E m l m sin sin ut cos ut = -^^ sin sin 2co (15) This has double frequency and represents an oscillation of power between the generating source and the load, of maximum value Emlm sin = El sin (16) 2 Its average value over one complete loop, i.e., over one quarter of a period of the current wave, is - El sin 7T This expression is of no particular use in practice. In addition to representing the maximum value of the power oscillation caused by the reactive component of the current. Tji j expression (16), i.e., -^~ sin = El sin 0, also represents the root-mean-square volt-amperes due to the reactive component of the current. It is sometimes called the reactive power. A better name is reactive volt-amperes, i.e., the volt-amperes due to the reactive component of the current. Reactive power or reactive volt-amperes is the product of the total volt-amperes and the reactive-factor. The total volt-amperes of a circuit are equal to the square POWER 61 root of the sum of the squares of the active and reactive powers. Volt-amperes = V(E I cos 0) 2 + (E I sin 0) 2 = V(Active power) 2 -f (Reactive power) 2 = EI (17) It follows from equation (17) that with sinusoidal waves, power-factor may be defined as the ratio of the active power to the square root of the sum of the squares of the active and reactive powers. Although the average value of the oscillation of power caused by the reactive component of the current is zero, the oscillation cannot take place without producing a copper loss. The oscilla- tion cannot contribute to the available power, but it does increase the copper loss in a circuit. For this reason it is desirable to operate circuits at as nearly unity power-factor as possible. The active or energy and reactive or wattless components of current for a circuit carrying / amperes at a power-factor cos are Active current = I cos = / X Power-factor (18) Reactive current -= I sin 6 = 1 Vl cos 2 9 = I Vl - (Power-factor) 2 = / X Reactive-factor (19) For sinusoidal waves, the watts will always be given by the product of volts and energy amperes. t-o FIG. 20. Curves of voltage, active and reactive components of the current and the power curves corresponding to the two compo- nents of the current are shown in Fig. 20. The two power curves 62 PRINCIPLES OF ALTERNATING CURRENTS are obtained by plotting the two terms of the second member of equation (14) separately against time. The two component currents I m cos and I m sin 6 are assumed to be equal. This cor- responds to a power-factor of 0.707 and a phase angle of 45 degrees between voltage and current. The phase angle is assumed to be an angle of lag of the current with respect to the voltage. The power curves corresponding to both components of the current are double frequency curves and represent a periodic flow of power. The power due to the energy component of the current, I m cos 6, is always positive and represents the power given out by the circuit. The power due to the reactive compo- nent of the current, I m sin 6, alternates between plus and minus, and the average power is zero. The combined effect of the two components produces a power curve partly above and partly below the axis of time similar to that shown in Fig. 19, page 54. At certain instants the resultant power developed is positive and at others it is negative (see Fig. 19). The average power is found by subtracting the area enclosed between the axis of time and the negative loops from the area enclosed between this axis and the positive loops and dividing the remainder by T = j- The resultant area is the same as the area enclosed between this axis and the power curve corresponding to the energy component of the current, I m cos 6. The conditions shown in Figs. 17 and 18 represent limiting cases. In the first, the reactive component of the current is zero. In the second, the active component is zero. It is important to remark that the power in a single-phase circuit is always fluctuating, becoming zero or even negative at certain instants of time. This means that a circuit may supply power to the generator during part of a cycle. The conditions are different in a balanced polyphase circuit. Here the deficiency of power at any instant in one phase is made up by an excess of power in the other phases. The total power of such a circuit, i.e., the resultant power for all the phases, does not fluctuate. Energy and Wattless or Quadrature Components of the Volt- age. Instead of dividing the current into energy and wattless components with respect to the voltage, the voltage may be similarly divided into energy and wattless components with re- spect to the current. POWER 63 The two components of the voltage are e a = (E m cos 0) sin ut (20) e r = -(E m sin 0) cos at = (E m sin 0) sin (at - 90) (21) (See equation (13)) In this case the power may be considered to be due to a voltage e = (E m cos 0) sin ut and a current i = I m sin orf in phase with it. The component e = (E m sin 0) cos ut of the voltage is in quadrature with the current i = I m sin ut and therefore produces no average power. The component E m cos is known as the energy, power or active component of the volt- age. The component E m sin is known as the wattless, quadrature or reactive component of the voltage. Measurement of Reactive Power, i.e., Reactive Volt-amperes. The reactive power of a circuit may be measured by an elec- trodynamometer connected in the same way as for measuring true power, provided the current through its potential circuit can be made to lag the voltage across its terminals by 90 degrees. For measurement of true power, the current in the potential circuit -must be in phase with the voltage across its terminals. The necessary lag of 90 degrees for the current in the potential coil may be obtained by connecting it in series with a large inductance instead of a large non-inductive resistance as is done for power measurements. (See page 57.) Due to the fact that an inductance cannot be made without some resistance, the current in the potential coil will not lag by exactly 90 degrees if the series inductance alone is used. It may, however, be made to lag exactly 90 degrees by shunting the potential coil with a suitable non-inductive resistance. Since the effect of induc- tance depends upon frequency, a reactive-power meter will indi- cate El sin only when the frequency is that for which the instrument is adjusted. The effect of the inductance and shunted resistance will be understood after studying Chapter VII. The average torque developed between the coils of the electro- dynamometer will be Average torque = ~ I M T Average torque = k 7 sin 6 = kEI sin = k X (Reactive power) By suitable calibration the instrument may be made to indicate reactive power directly. Vectors Representing Voltage Rise and Voltage Fall. When plotting vector diagrams and in making computations involving currents and voltages, it is always necessary to distinguish between voltage rise and voltage fall or drop. If the direction ab in any circuit is taken positive for current flow, the current is actually positive and actually flows from a to b when the revolving vector representing the current 7^ lies in the first or second quadrants. It is actually negative and flows from b to a when the revolving vector representing it lies in the third and fourth quadrants. If ab is taken as the positive direc- tion for the current vector, V a b, representing a voltage associated . with the current 7 a &, is a voltage rise, if on the average it is actually increasing in the direction ab, i.e., there is an actual rise in the direction ab, when the current vector lies in the first and second quadrants. In other words, if the direction ab is assumed positive for the current flow 7&, the voltage V a b is a voltage rise and is considered positive when its active component, i.e., its component in phase with the current, is an actual rise in voltage in the direction ab when the current flow is positive. V a b is a voltage drop or fall and is negative when its active component, i.e., its component in phase with the current, is an actual drop in voltage in the direction ab when the current flow is from a to b. A voltage vector may be expressed either as a voltage rise or a voltage fall. If it represents a rise in voltage and power is absorbed, its component in phase with the current, i.e., its active component, will be negative with respect to the current vector. In this case the voltage, on the average, is actually a voltage fall in POWER 65 the direction of the current flow. On the average there is actually a voltage drop in the direction of current flow. However, the voltage may be considered to be actually rising in a negative direction with respect to the current. If a vector V a b is considered .to represent a voltage rise in the direction ab, there is an actual rise in voltage between a and b when V a b lies in the first and second quadrants. There is an actual fall in voltage in the direction ab when V a b lies in the third and fourth quadrants. If V^ represents a voltage rise in the direction ab, V a b is a negative rise in the direction ab or is a drop in the direction ab. There is an actual drop in voltage in the direction ab when the vector F a & lies in the first and second quadrants and an actual rise in the direction ab when it lies in the third and fourth quadrants. In general, when only power delivered or both power delivered and received are to be considered in the same problem, it is best to let a positive vector represent a voltage rise. A negative vec- tor will then represent a voltage fall or drop. According to this convention a vector V would represent a voltage rise in a circuit in the direction which is assumed positive for the current. A vector V would be the voltage drop in the same direction or a rise in the opposite direction. However, as has been stated, when dealing only with power absorbed it is convenient and customary to consider that a positive vector represents a voltage drop. When this latter convention is adopted, V would represent a voltage drop in the direction in which the current is considered positive and V would represent a voltage rise in the same direction. When a voltage rise is considered positive, power delivered is positive and power absorbed is negative. When a voltage drop is considered positive, power absorbed is positive and power delivered is negative. The mere reversal of an alternating current or voltage during a cycle does not change the sign of the vector representing it. If a voltage vector represents a voltage rise, i.e., its component in phase with the current is rising in a positive direction when the current flow is positive, it will still be a voltage rise when its position has reversed due to its rotation, since the position of the current vector has also reversed. Its component in phase with the current will still represent a voltage rise in the direction of cur- 66 PRINCIPLES OF ALTERNATING CURRENTS rent flow. It will be rising in a negative direction with respect to the direction assumed positive for current flow, but the cur- rent flow will also have reversed and will be negative. Expression for Power when the Voltage and Current are in Complex. Let E = e + je' FIG. 21. where E and I are the effective or root-mean-square values of the voltage and current. The small letters without primes are the real components; with primes they represent the imaginary com- ponents. The two vectors, E and ' x /, are shown in Fig. 21. From Fig. 21 P = El cos 6 = El cos (0i - 2 ) = EI(cos 0i cos 2 sin B\ sin ei + e'i' (22) The component e of the voltage cannot produce power with the component i' of the current, since e and z'-are in phase quadra- ture. Neither can the component e' of the voltage produce power with the component i of the current, since they are also in phase quadrature. The components in the products ei and e'i' are in phase. They are the only components that can contribute to the power. The expression e'i ei' is the reactive power or reactive volt-amperes. This may be shown as follows: Reactive power = P r = E I sin = EI sin (0! - 2 ) = EI(sm 0i cos 2 cos 0i sin 2 ) POWER 67 When is taken as the angle of lag of the current with respect to the voltage, the current will lag the voltage when the reactive power is positive. E I = V (active volt-amperes) 2 + (reactive volt-amperes) 2 = V (true power) 2 + (reactive power) 2 = volt-amperes. It should be noted that the power P = ei + di' is not given by the product of the complex values of current and voltage. This product is evidently (ei e'i') + j(ei' + e'i) and has no significance. Example of the Calculation of Power, Power -factor, etc. Let an alternating voltage E = 100 + J50 be impressed on a circuit whose constants are such that the current resulting is / = 20 ;30. E and / are both root-mean-square or effective values. E = V(100) 2 -I- (50) 2 = 111.8 volts 7 = V(20) 2 + (30) 2 = 36.06 amperes Let BE be the phase angle of E with respect to the reference axis and let 0/ be the corresponding angle for the current. tan B E = ~ = 0.5 B E = +26.57 degrees. QQ tan B, = -=JT- = -1.5 fl/ = -56.31 degrees. 20 The angle of lag of the current with respect to the voltage is then Oi = (+26.57) - (-56.31) = 82.88 degrees. The current / lags the voltage by 82.88 degrees. Volt-amperes = 111.8 X 36.06 = 4030 Power = 100 X 20 + 50 X (-30) = 500 watts. p Power-factor = -^j = cos B* E 500 0.1240 4030 B r E = cos- 1 (0.1240) = 82.88 degrees. Reactive-factor = sin 0* = 0.9923 68 PRINCIPLES OF ALTERNATING CURRENTS In finding the reactive-factor, 0J, is taken as positive since it represents a lag of the current behind the voltage. Reactive power = E I sin O r E = 4030 X 0.9923 = 3999 reactive watts or reactive volt- amperes. Active component of the current = I a = I cos Q' E = 36.05 X 0.1240 = 4.47 amperes. Reactive component of the current = I r = / sin B 1 E = 36.05 X 0.9923 = 35.78 amperes. CHAPTER IV NON-SINUSOIDAL WAVES Wave Form of Alternators. The wave form of commercial alternators is never exactly sinusoidal and under certain condi- tions it may differ considerably therefrom. The armature windings of multipolar alternators consist of as many identical elements per phase as there are pairs of poles. These elements are spaced 180 electrical degrees apart on the armature and are usually connected in series. Polyphase alternators have as many groups of windings as there are phases. These are dis- 360 placed from one another by - - electrical degrees, where n is the 71 number of phases. Since the groups of windings are always identical, the voltages induced in them will be equal in magnitude 360 1 but will be displaced in time phase by - - degrees or by - of the time of a complete cycle. The windings of a polyphase alternator are always interconnected in star or in mesh (See hapt. VIII) in order to diminish the number of terminals. In general, except for single phase, there are as many terminals as there are phases. Most commercial alternators are three-phase, although two-phase or four-phase alternators are occasionally used. If the sides of the coils of an armature winding are 180 electrical degrees apart and if one side of a coil occupies any given position with respect to a north pole, the other side of the coil will occupy an exactly similar position with respect to the adjacent south pole. If the poles are similar so that the flux distribution pro- duced by each is the same, the voltages induced in the two coil sides will be equal in magnitude but opposite in direction. That is, if one acts from the front to the back of the coil, the other acts Tom the back to the front. At every instant, the voltage gene- rated in the coil will be the difference of the voltages generated n its two sides. 70 PRINCIPLES OF ALTERNATING CURRENTS The voltage generated in a coil side is equal to e = &LsZ X 10- 8 (1) where L is the effective length of the coil side, i.e., the length of the part cutting the flux, and Z is the number of inductors in each coil. An inductor is one of the two active sides of each turn of an armature coil. Each turn has two inductors. The inductors are the parts of an armature coil which cut flux. (B is the flux density in gausses at the inductor and s is the velocity of the inductor in centimeters per second. The velocity s is fixed by the radius of the armature, the fre- quency and the number of poles. The only variable is the flux density (B over the pole face. This is determined by the total flux per pole and the shape of the pole face. Since the only variable is the flux density, the shape of the voltage wave must be the same as the shape of the curve of flux distribution in the air gap. If the pole faces are concentric with the armature surface, the flux distribution in the air gap will be nearly constant over the pole faces and will drop rapidly to zero at points midway between poles. If the edges of the poles are chamfered so that the air gap at the edges of the pole is longer than at the middle, the flux density will no longer be uniform over the pole face but will be greatest at the centre. By properly shaping the pole face, the! flux density may be made to vary approximately sinusoidally in passing from a point midway between any pair of poles to a point midway between the next pair. If it could be made exactly sinusoidal, the voltage generated in each coil side would also be sinusoidal, since it is proportional to the flux density at each instant. If the pole faces are nearly concentric with the armature, the total flux for a given permissible maximum flux density will be greater than when the pole faces are chamfered, but the voltage wave will be flat. If the pole faces are chamfered too much, the wave form will be peaked, i.e., more peaked than a sinusoid. Even if the poles of an alternator could be shaped to give an exactly sinusoidal flux distribution at no load, the wave form of the alternator under load might differ considerably from a sinusoid, due to the effect of armature reaction in distorting the flux distribution. In general, except when the current is in.; NON-SINUSOIDAL WAVES 71 quadrature with the induced voltage, armature reaction tends to crowd the flux towards one side of each pole and to give rise to a distorted voltage wave which may differ greatly from a sinusoid. Although the flux distribution may become badly distorted under load, it is possible to so arrange an armature winding that the voltage between its terminals will still be nearly sinusoidal. Nearly all alternators have distributed armature windings. That is, the coils for any one phase are distributed among several pairs of slots per pair of poles instead of being placed in a single pair of slots per pair of poles. In this case the voltage induced in the coils will not be in phase but will differ in phase by an angle equal to the angle in electrical degrees between the slots. Also, the coil pitch is often less than 180 electrical degrees, i.e., the coil sides are less than 180 electrical degrees apart. A very common pitch is % X 180 = 150 electrical degrees. One important effect of distributing a winding and also shortening its pitch is to smooth out the wave form and make it more nearly sinusoidal by reducing or eliminating the components in the voltage which cause it to differ from a sinusoid. As will be explained later, any distorted voltage wave may be resolved into a series of sinusoidal components having frequencies which are 1, 3, 5, 7, 9, etc. times the fundamental frequency, i.e., the frequency determined by the speed and the number of poles. If the angle between the slots is a electrical degrees, the voltage generated in coils which are in adjacent slots will be out of phase by a degrees for the component of fundamental frequency, 3a degrees for the component of third frequency, 5a for the component of fifth frequency, etc. The effect of this difference in phase is to reduce the distorting components in the voltage wave and thus to make the voltage more nearly sinusoidal. A similar effect is produced by shortening the pitch. By the use of a suitably shortened pitch, any one component in the voltage wave may be completely eliminated in the coil 4 voltage. For example, if the coil sides are > X 180 = 144 elec- o trical degrees apart, the components of fifth frequency in the two sides of any coil will be 144 X 5 = 720 degrees out of phase. They will be displaced two whole wave lengths and will there- 72 PRINCIPLES OF ALTERNATING CURRENTS fore be in phase. Since the coil voltage is the vector difference 4 between the voltages generated in the two coil sides, a ^ pitch will completely eliminate the voltage component of fifth fre- quency. Any component may also be eliminated by distributing the winding among a suitable number of properly spaced slots. The triple frequency component in the voltage is eliminated in a three-phase alternator by the inter-connection of the phases. Instead of trying to completely eliminate certain of the dis- torting components by the use of a suitable coil pitch coupled with distribution of the winding, a pitch and distribution are usually chosen which considerably reduce a number of the distorting components without completely eliminating any one of them. Better average wave form may be obtained in this way than when certain components are completely eliminated. Although it is possible to design an alternator, by the use of shortened pitch, a distributed winding and certain other devices, to give practically a sinusoidal wave of voltage under load as well as at no load, alternators are not commercially so designed except when they are to be used for special work, such as cable testing, which requries sinusoidal voltage. Under ordinary conditions, the wave form of commercial alternators does not differ greatly from a sinusoid. Even though the voltage wave of an alternator were exactly sinusoidal under all conditions, the current wave would not necessarily be sinusoidal. It is there- fore necessary to consider the conditions existing when the voltage and current waves are not simple harmonic functions of the time. Representation of a Non- sinusoidal Current or Voltage Wave by a Fourier Series. Any single-valued periodic function may be resolved into a Fourier series consisting of sine and cosine terms of different relative magnitudes and with frequencies which are in the ratios of 1, 2, 3, 4, 5, etc. to the frequency of the fundamental period of the function.* For example, any single- valued periodic quantity, like the voltage or current of an alter- nator, may be represented by the following series. For the alternator A Q is zero. e = AQ + AI sin ut + BI cos ut + A 2 sin 2cof + B 2 cos 2ut + A 3 sin 3orf + 5 3 cos 3otf + etc. (2) * See Fourier's Series and Spherical Harmonics, Byerly. NON-SINUSOIDAL WAVES 73 The sine and cosine terms are called harmonics. The first harmonics are called fundamentals. Theoretically an infinite number of terms is required to represent most non-sinusoidal voltages or currents, but for most practical purposes the first few terms are sufficient. It is seldom necessary to go beyond the term of eleventh frequency. For most wave forms the series converges rapidly. Under ordinary conditions, the first harmonic, or fundamental as it is called, (this includes both the sine and cosine term of fundamental frequency) is the most important term in the series and it has by far the largest amplitude. Any harmonic may be absent from a voltage or current wave. On the other hand, certain conditions may very much exaggerate a harmonic in a current wave. Also, certain harmonics may be present in a current wave which are not present in the voltage causing that wave. This always occurs when a voltage is impressed on a circuit whose resistance, inductance or capacitance is a function of the current. The sine and cosine terms of the Fourier series may be com- bined to give a series involving either sine or cosine terms alone. For example, equation (2) may be written e = A + Ci sin (co + Oi) + C 2 sin (2o>* + 2 ) + C 3 sin (3orf + 3 ) + etc. (3) or e = Ao + Ci cos (cot - 0/) + C 2 cos (2orf - 2 ') + C 3 cos (3wt - 3 ') + etc. (4) The angles in equation (3) are the angles of lead between the harmonics of the sine series and the corresponding sine com- ponents in equation (2). The angles 0' in equation (4) are the angles of lag between the harmonics of the cosine series and the corresponding cosine components in equation (2). All the angles and 6' in equations (3) and (4) are measured on the scales of angles for the harmonics. A phase displacement of degrees for /\ the nth harmonic corresponds to a displacement of - degrees for the fundamental. See Fig. 22. If the phase angles for the har- monics are measured on the scale of angles for the fundamental, equations (3) and (4) become 74 PRINCIPLES OF ALTERNATING CURRENTS i sin (ut + i) + C 2 sin 2(co + . 2 ) + C 3 sin 3(co/ + as) + etc. (5) + Ci cos (to* - a/) + C 2 cos 2(co - 2 ') + C 3 cos 3M - 03') + etc. (6) where = 0^3 and 3 r = 03 ; . In general a n = - B n and ' = - n ' where n is the O 7i 7i order of the harmonic considered. Since sine and cosine functions of time differ in phase by 90 degrees, the cosine function leading, the A and B terms in equation (2) are in quadrature. Therefore VAS + Bi 2 tan 0i = -j- R tan 2 ' = 1-r tan n ' = 4" >n In general AJ 2 Bo 2 ! tan 2 = -; A 2 5 , = VA n 2 + A n 7 ' \^ Resultant \ ,x Fundamental FIG. 22. Since, ordinarily, the current and voltage waves of alternators operating under steady conditions are symmetrical with respect to the axis of time, the constant A in the Fourier series for a current or a voltage may usually be omitted. NON-SINUSOIDAL WAVES 75 A wave containing a fundamental and third and seventh harmonics is shown in Fig. 22. The fundamental and the har- monics are shown by full lines. The resultant wave is shown dotted. Effect of Even Harmonics on Wave Form. Symmetrical waves, i.e., waves which have exactly similar positive and nega- tive loops, cannot contain even harmonics, since the phase of any even harmonic with respect to the fundamental will be opposite in the two halves of the wave. Let the following Fourier series represent a voltage which contains both odd and even harmonics: e = Ei sin (o>* + ft) + #2 sin (2o>< + ft) + # 3 sin (3o>* + ft) + #4 sin (4o>* + ft) + E b sin (5co* + 0s) + etc. (7) If this wave is symmetrical with respect to its positive and negative loops, its instantaneous values at two instants of time, / T^ / "1 \ such as t and (t + 2" ) = ( t + s> ) , which are separated by one half period, must be equal in magnitude but opposite in sign. Equation (7) gives the instantaneous value of e for a time t = t. For t = ( ^ + of ) the instantaneous value of e is e' = E, sin (+ + 0i) + E 2 sin fat + ~ + ft) + E 3 sin fat + j|? + 3 ) + E, sin fat + y + 4 ) + # 5 sin fat + I* + ft) + etc. (8) Remembering that o> = 2?r/ and also that a phase displacement of any whole number of wave lengths, i.e., any whole number of 2w radians, is equivalent to zero displacement so far as phase relations are concerned, equation (8) may be reduced to the following form: e' = #1 sin (wt + TT + ft) + # 2 sin (2orf + ft) + # 3 sin (3o>J + TT + ft) + # 4 -sin (4o>* + 4 ) + # 5 sin (5co2 + 7i + 0s) = -#i sin (w$ + ft) + # 2 sin (2orf + 2 ) -# 3 sin (3o>< + 0,) + # 4 sin (4orf + ft) -# 5 sin (5orf + ft) + etc. (9) 76 PRINCIPLES OF ALTERNATING CURRENTS It will be seen, by comparing equations (7) and (9), that while the fundamentals and the corresponding odd harmonics for points one half a period apart are opposite in phase, the even harmonics are in phase. Therefore the two halves of a wave containing even harmonics cannot be alike in shape. That the two halves of a wave which contains even harmonics are not alike, i.e., that a wave containing even harmonics is not symmetrical, is also shown by Fig. 23, which shows a wave containing a fundamental and a second harmonic. Since the voltage waves of commercial alternators are symmetrical, they cannot con- tain even harmonics. Even FlG 23 harmonics may, however, oc- cur in a current wave even though they are not present in the voltage producing it, but the conditions under which they occur seldom arise in practice. Even harmonics will occur in the current wave when an alternat- ing voltage is impressed on the winding of an inductive circuit having an iron core which is magnetized in a fixed direction by any means, such as a constant direct current in a second winding surrounding the core. In this case the self-induction of the winding carrying the alternating current is a function of the current and will be different for positive and negative values of the alternating current. This should be better understood after studying Chapter V. Waves which have the Halves of the Positive and of the Negative Loop Symmetrical. Waves which are symmetrical with respect to their positive and negative loops cannot contain even harmonics. If the halves of the positive and of the nega- tive loop are also to be symmetrical, the fundamental and all harmonics must pass through zero at the same instant. For this condition to be fulfilled, either the sine or the cosine terms in the Fourier series given in equation (2) must be zero. All even harmonics and the constant A Q must also be zero if the positive and negative loops are to be symmetrical. If the equation for the wave is written in sine or cosine terms NON-SINUSOIDAL WAVES 77 only (See equations (3) and (4), page 73), the phase angles of all harmonics must be either zero or 180 degrees when the phase angle for the fundamental is made zero by considering time, t, zero when the fundamental is zero. Changing the Reference Point from which Angles and Time are Measured in a Complex Wave. It is frequently convenient and often necessary, when considering complex waves, to change the position point from which angles and time are reckoned. One case in which this is necessary is when the wave forms of two waves, whose Fourier equations are known, are to be compared. For two waves to be similar, they must not only contain like harmonics, but the relative magnitudes of the harmonics and fundamental and their phase relations must be alike in the two waves. A glance at the Fourier equations of any two waves, which are to be compared for wave form, will tell whether they contain the same harmonics. The relative magnitudes of the harmonics and the fundamental may be easily determined from the coef- ficients of the Fourier series, but unless the points from which angles are measured occupy the same positions with respect to the fundamentals, the relative phase of the harmonics in the waves will not be obvious. If the points from which angles are reckoned do not occupy the same relative positions with respect to the fundamentals, it will be necessary to shift one of them until they do. Consider the following two waves: e l = 100 sin (orf + 30) + 40 sin (3ut + 75) + 20 sin (but - 60) (10) 6 2 = 150 sin M - 10) + 60 sin (3coZ - 45) - 30sin(5coJ - 80) (11) The waves contain like harmonics. The magnitudes of the fundamental and harmonics in the second wave are fifty per cent, greater than in the first wave. The relative magnitudes of the harmonics and fundamental are the same in both waves. To compare the phase relations of the harmonics and funda- mentals, one wave must be shifted in phase by an amount equal to the difference between the fundamental phase angles of the 78 PRINCIPLES OF ALTERNATING CURRENTS two waves. This can be accomplished either by shifting the first wave minus 40 degrees or the second plus 40 degrees. Shift the second wave. Since any phase displacement of a degrees for the fundamental corresponds to a phase displacement of no. de- grees for the nth harmonic, it follows that if 40 degrees be added to the fundamental phase angle, 3 X 40 = 120 degrees must be added to the phase angle of the third harmonic and 5 X 40 = 200 degrees must be added to the phase angle of the fifth harmonic. Shifting the fundamental of the second wave plus 40 degrees in phase gives e 2 ' = 150 sin (* + 30) + 60 sin (3co + 75) - 30 sin (5co + 120) = 150 sin (cot + 30) -f 60 sin (3urf + 75) + 30 sin (5ut - 60) (12) From equations (10) and (12) it is obvious that the differences in phase between the harmonics and fundamental are alike in the two waves. Fourier Series for Rectangular and Triangular Waves. The Fourier equations for a rectangular and for a triangular wave are interesting since such waves represent extreme cases of a flattened and a peaked sine wave. Neither of these wave forms could be exactly attained in practice nor would either be desirable even if it could be secured. The Fourier series for rectangular and for triangular voltage waves are given by equations (13) and (14) respectively. e = E l sin cot + -E sin 3u> + K E 1 sin fat o O + 7 E l sin 7 (at + etc. (13) e = EI sin cot E l sin 3co + ^ EI sin 5 - ^ #1 sin 7co + etc. (14) Both the rectangular and the triangular wave have similar positive and negative loops and therefore contain only odd harmonics. Also since the halves of each loop are similar, all the phase angles are zero. NON-SINUSOIDAL WAVES 79 The manner in which the wave form given by equation (13) approaches that of a rectangular wave, as successive terms are added, is illustrated in Fig. 24. -- /< >\ f ,' TT e - J57j sin w+i E v sin Swt 1 X FIG. 24. Although an infinite number of terms would be required to exactly represent a rectangular wave, a fairly good approximation to such a wave is obtained with a comparatively few terms. Even with the four terms plotted in Fig. 24, the resultant is rapidly flattening out and approaching the rectangular form. Ordinary waves met in practice can usually be represented with sufficient accuracy for most work by a comparatively few terms of their Fourier equations. Measurement of Current, Voltage and Power when the Wave Form is not Sinusoidal. Since the average torque producing the deflection in the electrodynamometer and iron-vane types of ammeter is proportional to the average square of the current in their coils, such instruments may be used to measure current when the wave form is not sinusoidal. This statement is not strictly correct for the iron-vane type of instrument, since the eddy currents and hysteresis in the iron vane may seriously affect the readings when pronounced high-frequency harmonics are present. However, when used on circuits of commercial frequencies and wave forms, the readings are approximately correct. Voltmeters are usually of the electrodynamometer 80 PRINCIPLES OF ALTERNATING CURRENTS type. Iron vanes are seldom used except in the cheaper instru- ments. For an electrodynamometer type of voltmeter to indi- cate correctly, the current through it at each instant must be strictly proportional to the instantaneous voltage across its terminals. The general effect of inductance in a circuit is to damp out the harmonics in the current to a greater and greater degree as their frequency increases. (See Chap. VII, page 208.) There- fore, since the coils of an instrument cannot be made without inductance, the current in a voltmeter cannot be strictly propor- tional, at each instant, to the voltage across its terminals, except when the voltage is sinusoidal. If the non-inductive resistance in series with a voltmeter is large compared with the inductance of its coils, the current through it will be substantially propor- tional, at each instant, to the voltage across its terminals, pro- vided the voltage does not contain very pronounced harmonics of high frequency. Both the electrodynamometer and iron-vane types of voltmeter, when used on circuits of commercial fre- quencies and wave forms, indicate root-mean-square or effective voltage to a high degree of precision. For very high frequencies the hot-wire type of ammeter and voltmeter must be used. These instruments depend on the ex- pansion of a fine wire due to the heating caused by the current in the wire. The effect of the change in length of the wire is multiplied and causes a pointer to move over a graduated scale. To measure current the wire is placed in series with the circuit carrying the current to be measured. To measure voltage the wire is shunted across the circuit in series with a non-inductive resistance. Since the strength of a current is defined in terms of its heating effect, the hot-wire type of instrument will indicate the average square value of the current in the hot wire. The inductance of such an instrument is extremely small, being due to the inductance of a straight, fine wire only a few inches in length. For this reason hot-wire instruments may be used on very high frequency circuits. They do not hold their calibration well and are therefore not satisfactory for general use on circuits of commercial frequency. The electrodynamometer type of wattmeter will indicate true average power provided the current in its current coil is equal NON-SINUSOIDAL WAVES 81 or proportional, at each instant, to the current in the circuit, and the current in its potential circuit is proportional, at each instant, to the voltage across its terminals. The instrument is subject to the same limitations as the voltmeter, but by proper design it may be made to indicate true average power when used in circuits of ordinary commercial frequency and wave form. Determination of Wave Form. The usual way of determining the wave form of a current or voltage is by means of an oscillo- graph. The oscillograph, in its simplest form, consists of a vibrating element made of a single loop of very fine wire which is stretched over two bridges. The straight, parallel sides of the loop between the bridges are from 0.2 to 0.25 millimeter apart, and lie in a strong, uniform field between the poles of an electro- magnet which is excited with direct current. The plane of the loop is parallel to the axis of the magnetic field. A very small, light mirror is attached to the parallel sides of the loop midway between the bridges. When a current is passed through the loop it will flow down one side and up the other, causing the two sides of the loop to deflect in opposite directions. The mirror will tilt through an angle which is proportional to the strength of the current. If a spot of light from a source which approximates a point is re- flected on a screen by the mirror, the spot will move, when the mirror is deflected, through a distance which is proportional to angular displacement of the mirror. This assumes that the displacement of the mirror is small. The loop with its mirror is immersed in oil to damp its vibrations and make it dead beat. When the wave form is to be photographed, the spot of light is reflected on a revolving drum which carries the film. The rotation of this drum gives the time element, i.e., the abscissa of the wave, and the displacement of the spot, which must be parallel to the axis of the drum, gives the other element, i.e., the ordinate of the wave. To prevent overlapping of the waves on the drum, a shutter is interposed between the light source and the drum and remains open only during one revolution of the drum. When the wave is to be observed or traced on a screen, the time element of the wave is obtained by placing a revolving mirror between the source of light and the screen. To keep the image of the wave fixed on the screen this mirror must revolve synchronously with the frequency of the circuit. 82 PRINCIPLES OF ALTERNATING CURRENTS The dimensions of the vibrating system must be such that its natural frequency of vibration shall be at least fifty times as great as the frequency of the highest harmonic to be detected. The minimum free period which it is practical to obtain is about 0.0001 second, but for most commercial work a period of 0.0002 second is sufficiently low. For current measurements the vibrating element is shunted across the terminals of a non-inductive shunt which is placed in series with the circuit carrying the current whose wave form is to be determined. Since the vibrating element is sensibly non- inductive, the current it carries at each instant will be directly proportional to the current in the circuit. For voltage measure- ment the vibrating element is connected in series with a suitable non-inductive resistance and is then shunted across the circuit whose voltage wave form is to be determined. Effective Value of a Non-sinusoidal Electromotive Force or Current. Let e = E nl sin (coZ + 00 + E m , sin (3co* + 3 ) + E mb sin (5w< + 5 ) + etc. (15) be an alternating electromotive force of periodic time T = CO containing a fundamental, E m \ sin (at -f 61), and odd harmonics, E m3 sin (3co + 63), E mb sin (5wt + 5 ), etc., whose periods are odd multiples of the fundamental period . The effective or root- co mean-square value of the electromotive force is given by T sin M + 00 + Ems sin (3J + 3 ) . } dt (16) This involves squared terms of the general form and product terms of the general form T 1 C T ^ [E mk sin (kwt + 0,) X E mQ sin (qoit + 0,) }dt 1 Jo NON-SINUSOIDAL WAVES 83 Each of the squared terms, on integration over a complete cycle, becomes equal to one-half the square of its maximum value, since the average product of two sine terms of like fre- quency is one-half the product of the maximum values. Each of the product terms of unlike frequency becomes zero on integra- tion over a complete cycle, since the average value of the product of two sine terms of unlike frequency is zero. Consider the product of two sine terms of like frequency. Let o> = a. f (a) = A sin (a + 0) X B sin (a + 0') = AB (sin a cos + cos a sin 0) (sin a cos B' + cos a sin B') = AB : sin 2 a cos cos 0' -f sin a cos a cos sin B' 4- cos a sin a sin cos B' + cos 2 a sin sin B' | But 1 cos 2a 14- cos 2a . sin 2a sin 2 a = - , cos 2 a = - , Sin a cos a = . Z Z Z Therefore f(a) = AJ5J- C S - cos cos 0' + ^^o^" cos ^ s * n ^ . sin 2a , 14 cos 2a . H sin cos r H = - sin sin Z Z and 1 r 2 *" f 1 1 1 f (o)do = AB \ cos cos 0' 4 4 + = sin sin 0' [ 27T Jo 1 2 Z = -~ cos (0-0') (17) Now consider the product of two sine terms of unlike frequency, f (a) = A sin (a + 0) X B sin (no + n ) = AB{ (sin a cos 4- cos a sin 0) (sin na cos n 4- cos na sin n )j = AB{sin a sin na cos cos n 4- sin a cos wa cos sin W 4- cos a sin na sin cos B n 4- cos a cos na sin sin n J where n is any integer which is greater than unity. Since and n are constants f(a) = AB{Ki sin a sin na 4 ^2 sin a cos na 4- -K 3 cos a sin na 4 K 4 cos a cos na} 84 PRINCIPLES OF ALTERNATING CURRENTS But sin x sin y = ~ j cos (x y) cos (x + y) cos x sin y L 'si sin (x + y) sin (x y) sin x cos y = ~ { si n C& + 2/) + sin (x ?/) 1 If cos x cos y = s j cos (a; + y) + cos (a; y) Therefore f (a) = AB -^ [cos (1 n)a cos (1 + n)a] 2 rr + 2 [sin (1 + ri)a + sin (1 w)a] + -~ [sin (1 + ri)a sin (1 n)a] + -?r[cos (1 + n ) a + cos (1 ri)a] and = (18) The root-mean-square or effective value of a non-sinusoidal voltage is therefore E = etc. + + Ei* + etc. (19) (20) where the E's with the subscript m are maximum values of the components of the voltage. Without the subscript m they are the effective values. Similarly, if i = Imi sin (co + 0i -f ai) + I m a sin (3coi + 3 + 0:3) -f- 7 W 5 sin (5oo -j- ^5 -f- #5) 4~ etc. is an alternating current of periodic time T = , its root-mean- co square or effective value is given by 7 = + /s 2 + / 5 2 + etc. (21) NON-SINUSOIDAL WAVES 85 where the /'s with the subscript m are maximum values of the components of the current. Without the subscript m they are the effective values. Power when the Electromotive Force and Current are Non- sinusoidal Waves. Let e = E ml sin (<*t + 00 + E m3 sin (3orf + 0) + E m s sin (5orf + 5 ) + etc. and i = I m i sin M + 0/) + /, sin (3orf + *.') + 7 m5 sin (5orf -f 6 ') + etc. represent an electromotive force and current respectively, whose periodic time is T = . The average power is P = ~ CO T sm (cat + 0i) + E mZ sin + E m5 sin (5co2 + 5 ) + etc.} X {/mi sin (cot + 0/) + /ma sin (3co + 3 ') + /mo sin (5co^ -f- 0s') + etc.}dt This involves product terms of like and unlike frequency of the forms and 1 C T ^ {E mk sin (kwl + 0*) X /m* sin (kwt + ^ Jo 1 f r ^ {E mk sin (fc + A ) X I mq sin (g + 0/ -/Jo On integration, the product terms of like frequency become (see equation (17), page 83) ^^ cos (0, - 0*') while the product terms of unlike frequency become zero on integration. (See equation (18), page 84) The average power in a circuit having harmonics in both cur- rent and voltage is therefore p = E cog $i _ , i/ + E cog ( ^ _ 3/) + mbmb cos (05 - 5 ') + etc. (22) P = E 1 I 1 cos (0i - 0/) + #3/3 cos (0 3 - 3 ') + #5/5 cos (0 5 - 5 ') + etc. (23) 86 PRINCIPLES OF ALTERNATING CURRENTS The letters E and / in equations (22) and (23) with the sub- script m represent maximum values. Without the subscript m they represent root-mean-square values. If a harmonic occurs in the current and is not present in the voltage, or vice versa if a harmonic occurs in the voltage and is not present in the current, it will contribute nothing to the average power developed. Power -factor when the Current and Voltage are not Sinu- soidal. The power-factor of a circuit is denned as the ratio of the true average power to the volt-amperes. This definition is independent of wave form. f True power P Power-factor = ^^-v - =TTT (24) Volt-amperes E I #!/! cos (0i - 0i') + #3/3 cos (0 3 - 3 ') + etc. Although a harmonic which occurs in the current of a circuit and does not occur in the voltage does not contribute to the average power it does increase the root-mean-square or effective value of the current required to produce the power. Therefore the power-factor of a circuit containing a harmonic in its current which is not present in its voltage cannot be unity. Similarly, the power-factor of a circuit containing a harmonic in its voltage which is not present in its current cannot be unity. The only way the power-factor can be unity is for cos (0i - 0i') = cos (0 S - e z f ) = etc. = 1 and #1 E 8 j-=-j- = etc. Ii is When the current and voltage waves are of different form, the maximum power-factor, for fixed effective values of current and voltage, will obviously occur when the phase displacement between the current and voltage is that which makes the power a maximum. From equation (25) it is obvious that unity power-factor can occur only when the current and voltage waves are exactly similar in form and have no phase displacement with respect to each other. In other words, both waves must contain like harmonics and these harmonics must have the same relative magnitudes and the same relative phase relations. If there is a NON-SINUSOIDAL WAVES 87 harmonic in the current which is not present in the voltage, the power-factor cannot be unity, since this harmonic will contribute to the root-mean-square value of the current without adding to the power developed. A similar statement is of course true regarding a harmonic in the voltage. There are many cases where harmonics are present in the current and are not present in the voltage causing it. Such a condition always occurs when a voltage is impressed on a circuit whose inductance is a function of the current. Since the induc- tance of a circuit is defined as the flux linkages per unit current, i.e., as L = N -JT, the inductance of all circuits containing iron must be a function of the current. Since the power-factor of a circuit is equal to the ratio of true power to volt-amperes, power-factor might be defined as the ratio of the actual power to the maximum power that could be obtained with the given current and voltage. Example of the Calculation of Effective Values of Current and Voltage, Average Power and Power-factor for a Circuit when the Fourier Equations of its Current and Voltage are Known. When the voltage v = 200 sin (377* + 10) + 75 sin (1131* + 30) + 50 sin (1885* + 50) is impressed on a certain circuit, the current is i = 8.51 sin (377* + 1118) + 8.04 sin (1131* + 7433) + 7.50 sin (1885* + 8432) v /(200) 2 + (75) 2 + (50) 2 \ 2 = 155.1 volts. (8.51) 2 + (8.04) 2 + (7.50)' = 9.84 amperes. P= 2 , 75 X 8.( J. A . XO ) M ros (30 74W 2 so ; * 7 ' 5 nn f*n $U29^ 851 + 216 + 155 = 1222 watts. 1222 Power-factor - ^ = 0.801 88 PRINCIPLES OF ALTERNATING CURRENTS Effective Value of an Alternating Electromotive Force or Current from a Polar Plot. If an electromotive force or current wave has been analysed and the components of its Fourier series have been determined in magnitude, its effective value may be readily calculated. Frequently, however, the time re- quired in such an analysis is excessive for the purpose in hand. In such a case the following graphical method gives satisfactory results. Let Fig. 25-a represent an alternating-current wave which has been determined by means of an oscillograph. Divide either half of the wave into any convenient number of equal parts, the number depending on the preci- sion desired, and erect ordinates at the points of division. These ordi- nates represent instan- taneous values, i, of the current. Make a polar plot of the wave as shown in Fig. 25-&, in which any radius such as that at e is equal in length to the ordinate at e in the left-hand half of the figure and is laid off at an angle from CLC the base line equal to 180 X degrees. An element of area of the polar plot enclosed between two radii differing in direction by an angle d0 is a triangle of base idQ, altitude i and area ^ d6. The area of the polar plot is therefore abed (a) e f g FIG. 25. where / is the effective or root-mean-square value of the current and T is the time of a complete period. Planimeter the polar diagram and determine its area. Then Area = (26) NON-SINUSOIDAL WAVES 89 The effective value of the current is then given by /Area _ - no /-r / = / - = 0.798 V Area \ I Analysis of a Non -sinusoidal Wave and Determination of Its Fundamental and Harmonics. It is often necessary to analyse a voltage or current wave, obtained by an oscillograph or other means, into the components of its Fourier series in order to determine the harmonics present and their magnitudes. The Fourier series may be written, f (a) = A + AI sin a + BI cos a + A z sin 2a + J5 2 cos 2a + . . . + A n sin na + B n cos na The expression for the constant A may be found by multi- plying the equation by da and then integrating it between the limits and 2ir. For example: r* 2* /2ir /2ir f (a) da = A I da + AI \ sin ada + BI I cos ada + Jo Jo Jo . . . + A n I sin nada + B n I cos nada Jo Jo All terms on the right-hand side of the equation, except the first, reduce to zero leaving i{a)da = 2wA and 1 T 2ir A - ~ f(a)da **JO The integral I f (a) da is the net area enclosed by the curve Jo represented by f(a) for a complete period or cycle. Since the positive and negative loops of current and voltage waves of alternators under steady conditions are symmetrical with respect to the axis of time, the net area represented by the integral and hence the constant A are zero. To obtain the second constant, AI, multiply both sides of the equation of the wave by sin a da and integrate the resulting equation between the limits and 2ir. 90 PRINCIPLES OF ALTERNATING CURRENTS rf*2ir f (a) sin ada = A I sin ada Jo pr > [2, + AI I sin 2 ada + B\ I cos a sin ada Jo Jo r pi- sin 2a sin ada + B 2 I cos 2a sin ada + Jo /27T /27T % I sin na sin ada + B n I i Jo Jo 27T /27T + A n sin na sin ada + n I cos no. sin ada All the terms on the right-hand side of the equation, except the second, reduce to zero leaving rf(a) sin ada = TT AI 1 C 2 " A] = - f(a) sin ada The constant BI may be found in a similar manner by multi- plying the equation of the wave by cos ada and integrating between the limits and 2?r. In general, any constant may be found by multiplying both sides of the equation by da and the trigonometrical function which appears in the term containing the desired constant, and then integrating the result- ing equation between the limits and 27r. For example: any constant such as A n may be found by multiplying both sides of the equation of the wave form by sin no. da and then integrating the resulting equation between the limits and 2ir. Doing this for the different terms gives i r 2 * AI = - I f(a) sin ada ^Jo f (a) cos ada 1 T 27r AZ = ~ f(o) sin 2ada T Jo 1 f 27r B 2 = - f (a) cos 2ada T JJ i A n = - f(a) sin nadfy f (a) cos nada NON-SINUSOIDAL WAVES 91 The values of the constants may be found by evaluating the integrals by means of a graphical method. Let the wave repre- sented by f (a) = y be plotted as shown in Fig. 26. The base of this figure represents a complete cycle and is equal to 2ir radians. FIG. 26. Divide the figure into m equal parts by equally spaced ordi- nates. The distance between any two adjacent ordinates is equal to radians. The integral AI = - f(a) sin ada ""t/O may then be written as a summation. 27T m where y k is the ordinate of the curve at the middle of the kth space. A = - 1 in- 4 ' ? % 4 5 ?? , in < 2 w 2m 2m . /2m - 1 + 2/m sm )S In general 2 ( /I 2r 2r + 2/3 sin n- + . . . /2m - 1\ 2ir \ ir )n- + y, cos >-) + . + V . cos (?=L- )?? / m 92 PRINCIPLES OF ALTERNATING CURRENTS If the equation of the wave is desired in sine or cosine terms alone, the constant C and the angles or 0' for the sine or cosine series may be obtained by the method given on page 74. For any term such as the kth, the constant C and the angles are ., 6k = tan l -p A-k S k ' = tan-' -Ofc The precision of the constants and the angles is increased by making m large. For most commercial work it is not neces- sary to carry the analysis beyond the seventh or the eleventh harmonic. Theoretically it is easy to calculate the constants by the method just outlined. The real difficulty lies in the amount of time involved. This may be reduced by arranging the work in a tabular form. The forms in use follow a method introduced by C. Runge,* in which the calculation of the constants is carried out by the aid of a systematically arranged schedule. Fischer-Hitmen Method of Analysing a Periodic Wave into the Components of Its Fourier Series. TheFischer-Hinnenmethod,f which is also known as the " Selected-ordinate Method," reduces to a minimum the work of determining the constants for any definite number of terms of the Fourier series of a wave. It is based on two mathematical laws which can easily be demonstrated. Let the distance between any two points a and b corresponding to a whole wave length, i.e., corresponding to 360 degrees or 2r radians, be divided into m equal parts. Measure the length of the ordinate at the beginning of each part for any harmonic such as the kth. Call the ordinates Y m i, Y m 2, F m3 , etc. where the first subscript, m, indicates the number of parts into which the wave is divided and the second subscript indicates the position of the ordinate. * Elektrotechnische Zeitschrift, 1905, p. 247. t Elektrotechnische Zeitschrift, Vol. 22, 1901, p. 296. NON-SINUSOIDAL WAVES If these ordinates are added (360 ) k- - -f ka > m j + Ak sin 93 m + A k sin ^ 3fc + sin ((m- l)k + fca 1 (27) I w J The angle a is the angular distance, measured on the funda- mental scale of angles, between the first ordinate and the point where the harmonic considered passes through zero increasing in a positive direction. Ak is the amplitude of the harmonic. (See Fig. 27.) FIG. 27. Expanding the angles in equation (27) gives 27 Y m = A k sin ka , 360 , 360 . , sin k -- cos lea + Ak cos tc - sin Ka m m 360 .k sin 2fc - cos ka -f- A* cos 2k - - sin fca m m . , 1W 360 7 sin (m I) k - - cos fca m 1M 360 7 H- A* cos (m l)k - - sin fca m 94 PRINCIPLES OF ALTERNATING CURRENTS / / Pi ; 360 o; 36 = Ak { sm Ka 1 + cos K r cos 2k [ L m m 1M 36 1 . . . + cos (m l)k / r 1 36 . .7 36 . -f- Ak { cos ka sin k - \- sm 2k - { L m m . . + sin (m l)k 1 I (28) m J } Since the sine of any whole number times 360 degrees is equal to zero and the cosine of any whole number times 360 degrees is unity, it is evident from inspection of equation (28) k that when is a whole number m 2? Y m = mA k sin ka (29) The sum of m equally spaced ordinates for any harmonic such as the kth is therefore equal to m times the first ordinate when the order, k, of the harmonic is any whole number times the number of ordinates, m. k If is not a whole number for the harmonic considered, the m series represented by equation (28) reduces to zero. This can be shown by the aid of the two following trigonometrical formulas. cos 6 + cos 26 + cos 30 + . sin 6 + sin 20 + sin 30 + ...... . /m - l\ . mB sin ( o j sm ~9 + sin (m -1)6= -- (31) sin - /b360 k If 6 is put equal to - - and is not a whole number, equation (30) reduces to 1 and equation (31) reduces to zero. There- is not a whole number Y m = A k {sin ka (1 - 1) + cos fca(O)} = (32) k fore when is not a whole number m NON-SINUSOIDAL WAVES 95 The relations given in equations (29) and (32) may be used to determine the components of the Fourier series for any periodic wave. Let Fig. 28 represent a fundamental and a third harmonic for a wave containing only a fundamental and third harmonic. FIG. 28. Erect the two ordinates a and b 360 degrees apart, taking any convenient point such as a for the origin. Let the fundamental and harmonic of the wave each be resolved into a sine and a cosine component referred to a as the origin from which to reckon time. Then the equation of the wave is y = Ci sin (ut + = A i sin w< + -f + C 3 sin 1 cos cot s sin 3w + 3 ) cos (33) (34) Where + tan 0i = tan 3 3 = Divide the whole wave length between a and b into three equal parts by three equally spaced ordinates shown in Fig. 28. Call the lengths of these ordinates Y S i, F 32 and Y BS . At the origin t = and each sine component of the wave (see equation (34)) is zero, but each cosine component has its maximum value. The length of the ordinate erected at the point t = will therefore be equal to the sum of the cosine terms of the fundamental and harmonics. F 3 i = B, + B, 96 PRINCIPLES OF ALTERNATING CURRENTS . In general, if there are k harmonics, the ordinate erected at the point t = will be equal to the sum of the maximum values of the cosine terms for the fundamental and all harmonics. Y ml = B l + 3 + 5 + . . . + B k (35) Referring again to equation (34) and Fig. 28, it is obvious from equations (29) and (32) that if F 3 i, F 32 and F 33 are the lengths of the ordinates at the points 1, 2 and 3, i(F 31 + F 32 + F 33 ) = B 3 or the maximum value of the cosine component for the third harmonic. If the wave contained harmonics which were multiples of the third, one third of the sum of the ordinates F 3 i, F 32 and F 33 would be the sum of the maximum values of the cosine terms for the third and all harmonics which were multiples of the third. That is, in general ^(F 31 + 7 32 + F 33 ) = (# 3 + B g + 15 + etc.) (36) o Similarly if a wave containing harmonics is divided into five equal parts by five equally spaced ordinates, F 51 , F 52 , F 53 , F 54 and F 55 , the first ordinate being erected at the point a, *(F 5 i + F 52 + F 53 + F 54 + F 55 ) = (B 6 + Bn + # 25 + etc.) o Dividing the wave into seven and into nine parts gives (F 71 + 7 72 + + F 77 ) = (B 7 + B 2l + 35 + etc.) 92 F 99 ) = (B 9 + 27 + 45 + etc.) In practice it is convenient to erect the first ordinate at the point where the curve to be analysed crosses the axis of time. In this case Y m \ is zero and from equation (35) B l + B z + B 5 + etc. = Bi = - 3 - B 5 - etc. For most current and voltage waves met in practice harmonics of higher order than the seventh are not important. Relatively high harmonics may sometimes be produced by the armature NON-SINUSOIDAL WAVES 97 teeth of motors or generators. If present in any appreciable magnitude they are as a rule easily detected by inspection of the wave and may then be calculated. If harmonics of higher order than the seventh are negligible #1 = -- 3 - B 5 - B 7 (37) 3 = i(F 31 + F 32 + F 33 ) (38) F 52 + . . . + F 55 ) (39) B 7 = i(F 71 + F 72 + . . . + F 77 ) (40) If the ninth harmonic were present as well as the third, fifth and seventh F 32 + F 33 ) (41) 9 = (F 91 + F 92 + . . . . . .+ F 99 ) (42) B s = |(F 31 + F 32 + F 33 ) - B 9 (43) When the approximate equations (equations (37) to (40) inclusive) are used to analyse a wave, the base line may be ap- propriately divided to detect higher harmonics. If they exist in appreciable magnitude they must be corrected for by the method indicated for correcting the third for the presence of the ninth. (See equation (43).) Each of the sine terms in the wave has its maximum value one quarter of a period (measured on the scale of angles for the harmonic considered) from the initial ordinate, Y m \, or in Fig. 28, page 95, from the ordinate F 31 . If the ordinates on Fig. 28 are shifted in the direction of lag (to the right) one quarter of a period for the third harmonic, one third of their sum will be equal to the maximum value of the sine term for the third harmonic. |(F 31 ' + F 32 ' + F 33 ') = A 3 (44) The primes on the F's indicate that they have been shifted in the direction of lag from their original position, i.e., from the position in which they were drawn for determining the coeffi- 98 PRINCIPLES OF ALTERNATING CURRENTS cients of the cosine terms, by one-quarter of a period for the harmonic considered. If the wave contains harmonics which are multiples of the third, (1V + F 32 ' + F 33 ') = (A, - A 9 H- A 15 - etc.) (45) o Similarly, if the F 5 group of ordinates for the fifth harmonic is shifted in the direction of lag by one-quarter of a period for the fifth harmonic, i.e., by -=- degrees or -= X ~ radians on the funda- o o z mental scale of angles, ^(F 51 '+F 52 ' + . + F 55 ') = (A5-A 15 -fA 2 5-etc.) (46) If the Y 7 group of ordinates are shifted in the direction of lag by one-quarter of a period for the seventh harmonic, i(F 71 ' + IV + . . . + F 77 ') = (A 7 -A 21 + A 35 -etc.) (47) If the harmonics above the seventh are negligible, A 3 - *(F 31 ' + F 32 ' + F 33 ') (48) A 5 = ^OV + F 52 '+. . .+ F 55 ') (49) A 1 =l(Y 71 ' + Yn' + . . .+ F 77 ') (50) If the ninth harmonic is also present, A 3 as found by equation (48) must be corrected for that harmonic. For example: A 9 =-g(F 9 i'+ F 92 '+. . . + IV) (51) A 3 = i(r 81 ; + F 32 r + F 33 + A 9 (52) If other high order harmonics are present, the other coefficients, A 5 and A 7 , must similarly be corrected. The method of correc- tion is the same as indicated for the coefficients A 3 and B s . If the first ordinate, Y m \, of the group of ordinates for the cosine series is shifted in the direction of lag, one-quarter of a period for the fundamental, its length will be Y ml ' = A! - A 3 + A 5 - A 7 + A 9 - etc. A! = Y ml ' - (- A 3 + A 5 - A 7 + A 9 - etc.) (53) NON-SINUSOIDAJ. WAVES The even harmonics do not contribute to the ordinate Y, 99 That equation (53) is true when either odd or even harmonics or both are present will be understood when the phase relations among the fundamental and harmonics of the sine -terms are considered at the ordinate Y m i'. A change of a degrees in the position of any ordinate measured on the scale of angles for the fundamental, is equivalent to a change in position of ka degrees with respect to the kih harmonic scale of angles. The following table is based on this relationship. This table gives the change in phase produced in each harmonic by moving the axis of reference in the direction of lag through one-quarter of a period for the fundamental. i tl I 11J.VS11.II> ^ lUlllJi/ 111 LJilCfcO^ 1 1 i.l 1 1 I 1 ' M 1 l< V MilJl^v All |MUl.^' Fundamental 90 8th 8 X 90 = 720 o 2nd 2 X 90 = 180 9th 9 X 90 = 810 C o 90 3rd 3 X 90 = 270 10th 10 X 90 = 900 C =c= 180 4th 4 X 90 = 360 llth 11 X 90 = 990 o o 270 5th 5 X 90 = 450 12th 12 X 90 = 1080 o 90 o 6th 6 X 90 = 540 13th 13 X 90 = 1170 C o 180 o 90 7th 7 X 90 = 630 14th 14 X 90 = 1260 o 270 o 180 The cosine terms for the fundamental and all harmonics are in phase at the ordinate Y m i and add directly to it. The sine terms for the fundamental and all harmonics are zero at the ordinate Y m i and contribute nothing to its magnitude. Accord- ing to the table, the fifth, ninth, thirteenth, etc. harmonics of the sine series will have their maximum values at a point 90 degrees from the ordinate F ml , measured in the direction of lag on the fundamental scale of angles, and will add directly to the maximum value of the sine term of the fundamental at an ordinate, F m /, erected 90 fundamental degrees in the direction of lag from the ordinate Y ml . A shift in the position of the axis of 100 PRINCIPLES OF ALTERNATING CURRENTS 90 fundamental degrees in the direction of lag changes the phase of the third, seventh, eleventh, etc. harmonics in the direction of lag by 270 degrees, measured on their own scales of angles, or by 360 -270 = 90 degrees in the direction of lead. They will therefore be opposite in phase to the sine terms of the fundamen- tal, the fifth, ninth, thirteenth, etc. harmonics at the ordinate Y m \ and will subtract from it. The even harmonics of the sine terms are all changed in phase by either 180 or degrees by a shift in the reference axis of 90 fundamental degrees in the direction of lag. They will conse- quently all have zero values at the ordinate Y m i, as well as at the ordinate Y m i, and will contribute nothing to it. The cosine terms for the fundamental and all harmonics will have zero values at the ordinate Y m i and will contribute nothing to it. FIG. 29. Most current and voltage waves met in practice are symmetri- cal, i.e., their positive and negative loops are identical except in sign. When the positive and negative loops of a wave are symmetrical, it is necessary to construct only the first loop, as the length of any given ordinate for the second loop may be determined from a suitably placed ordinate in the first loop. Consider a third harmonic. Refer to Fig. 29. Let the three ordinates marked 1/2 and 3 at their upper ends be the ordinates for determining one component, such as the cosine component-, of the third harmonic. If there are no harmonics present which are multiples of the third B, = Vi + Y 32 + F 33 ) NON-SINUSOIDAL WAVES MGl. Instead of dividing the whole wave into k parts (in this case three), where k is the order of the harmonic to be determined, let the half wave be divided into k parts, as shown in Fig. 29. The numbers for the k ordinates for the half wave or the 2k ordinates for the whole wave have circles around them and are placed at the bottom of the ordinates in Fig. 29. The second of the six ordinates, i.e., of the 2k ordinates, is obviously equal in magnitude but opposite in sign to the third of the three ordinates marked at their upper ends. Therefore 3 = |(F.i - F 62 + F 63 ) (54) If in addition to the third harmonic the wave contains har- monics which are multiples of the third B 3 + B 9 + 15 + etc. = * (F 61 - 7 62 + F 63 ) (55) To determine the fifth harmonic, divide the half wave into five parts. This corresponds to dividing the whole wave into ten parts. Then B b + 15 + #25 + etc. = i(Fio i - F 10 2 + F 10 3 - FIO 4 + F 10 5) (56) Similarly, by dividing the half wave into seven parts, the seventh harmonic plus the others of its group may be found. Dividing the half wave into seven parts corresponds to dividing the whole wave into fourteen parts. B 7 + 21 + B 35 + etc. = i(7 14 l - Y u 2 + Y lt , - Fi4 4 + F 14 5 - Fi4 e + Fu r) (57) When the loop of a symmetrical wave is divided into k parts by equally spaced ordinates, to determine the kth harmonic the k 1 sum of the ^ ordinates with even numbers is subtracted from k + 1 the sum of the ^ ordinates with odd numbers. Zi It is convenient, when analyzing a wave containing only odd harmonics, to divide the half wave into 2k equal parts by 2k equally spaced ordinates, erecting the first ordinate where the wave crosses the axis of time. When 2k ordinates are thus drawn in the half wave, those with odd numbers are used for determining 102 OF ALTERNATING CURRENTS the coefficients of the cosine terms. Those with even numbers will be one quarter period (for the fcth harmonic) from the others, and are used for determining the coefficients of the sine terms. For example, suppose a wave contains a fifth harmonic but no others, such as the fifteenth, which would be included with the fifth. Let the half wave be divided into ten parts and number the ordinates Y l} Y 2 , F s , . . . F 10 . Then \ - Y, + F 5 - F 7 + F 9 ) As = d - F 4 + F 6 - F 10 ) (58) (59) That is, the ordinates with odd numbers are used with alternate signs to determine the B coefficients, and the ordinates with even numbers are used with alternate signs to determine the A co- efficients. In each case the first ordinate is considered positive. An analysis of a wave containing only odd harmonics will make this clear. When calculating the B and A coefficients of a wave by the Fischer-Hinnen method, it must be remembered that the coeffi- cients are not obtained separately but in groups. For this reason, it is always necessary to determine, by an approximate division of the wave, the highest order harmonic that is present in appreciable magnitude, in order that the coefficients given by the approximate equations (37) to (40) and (48) to (50) inclusive may be corrected for the presence of high order harmonics. FIG. 30. Example of the Analysis of a Wave Containing Only Odd Harmonics by the Fischer-Hinnen Method. The wave of the magnetizing current of a 60-cycle transformer will be analyzed. Analysis shows that this wave contains a pronounced third NON-SINUSOIDAL WAVES 103 harmonic, a fairly large fifth harmonic and small seventh and ninth harmonics. The harmonics above the ninth are negligible. The half wave, with the proper ordinates for determining the coefficients for the third harmonic, is shown in Fig. 30. Since the ninth harmonic is not negligible, these coefficients must be corrected for its presence. Measurements made on the original curve from which Fig. 30 was reproduced gave the following lengths of the ordinates in inches: F! = 0.00 F 2 = 0.99 F 3 = 1.80 F 4 = 1.01 F 5 = 0.51 F 6 = 0.30 0.51 1.29 1.80 1.01 The scale for the ordinates of the original curve was one ampere per inch. 3 + 9 = KFi - F 3 + F 5 ) = K0.51 - 1.80) = 0.43 inches = -0.43 X 1 = - 0.43 ampere. A, + A 9 = KF 2 - F 4 + F 6 ) = (1-29 - 1.01) = 0.093 inches = 0.093 X 1 = 0.093 ampere. The half wave with the proper ordinates for finding the fifth harmonic is shown in Fig. 31. 23456789 10 FIG. 31. Measurements made on the curve from which Fig. 31 was reproduced gave the following lengths of the* ordinates in inches. Y l = 0.00 Y 2 = 0.51 F 3 = 1.32 F 4 = 1.82 F 5 = 1.52 F 6 = 1.01 FT = 0.65 F 8 = 0.44 F 9 = 0.33 F 10 = 0.22 1.85 1.74 1.97 2.26 104 PRINCIPLES OF ALTERNATING CURRENTS i - Y, + F 5 - Y 7 + Y 9 ) = i(1.85 - 1.97) = -0.024 inches = -0.024 X 1 = -0.024 ampere. A 5 = i(F 2 - F 4 + F 6 - F 8 + 7 10 ) = i(1.74 - 2.26) = -0.10 inches = -0.10 X 1 = -0.10 ampere. Further analysis shows that B 7 = 0.027 ampere. A 7 = 0.003 ampere. B Q = 0.004 ampere. A 9 = 0.014 ampere. 3 = (#3 + 9 ) ~ #9 = -0.43 - 0.004 = -0.43 ampere. A, = (A 8 -A 9 ) + A, = 0.093 + 0.014 = 0.107 ampere. From equation (37), page 97 B l = -(B 3 + 5 + # 7 + #9) = -{(-0.43) + (-0.024) + (0.027) + (0.004)] = 0.42 ampere. From equation (53), page 98 A l = F ml ' - (-^ 3 + A, - A 7 + A 9 ) By measurement on the original figure, the ordinate 4 (see Fig. 30), which is displaced ninety fundamental degrees in the direction of lag from the ordinate marked 1, is equal to 1.01 amperes. A l = 1.01 - { -(0.107) + (-0.10) - (0.003) + (0.014)) = 1.21 amperes. i A i sin ut + BI cos ut + A 3 sin 3ut + Bz cos 3o> + A 5 sin 5co -f- B 5 cos 5co + A 7 sin 7co + 7 cos 7co + A 9 sin 9o>Z + 9 cos 9co NON-SINUSOIDAL WAVES 105 1.21 sin 377* + 0.42 cos 377* + 0.107 sin 1131* - 0.43 cos 1131* - 0.10 sin 1885* - 0.024 cos 1885* -f 0.003 sin 2639* + 0.027 cos 2639*' + 0.014 sin 3393* + 0.004 cos 3393* i) 2 + (o.42) 2 = 1.28 amperes. "'=li = rf = - 347 0i = +19.1 degrees. ( -0.43)2 = 0.443 ampere. I9 3 = - 76.0 degrees. C 6 = VA 5 2 + ^ 5 2 = V(-0.10) 2 + (-0.024) = 0.103 ampere. 5 = 166.5 degrees. C 7 = V^ 7 2 + B 7 2 = V(0.003) 2 + (0.027) = 0.027 ampere. . AT 0.003 = +84 degrees. 9 2 = V (0.014) 2 +(0.004)" = 0.0146 ampere. 9 = +16 degrees. 1.28 sin (377* + 19?1) + 0.443 sin (1131* - 76?0) + 0.103 sin (1885* - 1665) + 0.027 sin (2639* + 84) + 0.015 sin (3393* + 16) 106 PRINCIPLES OF ALTERNATING CURRENTS Form Factor. The form factor for an alternating current or voltage is the ratio of its effective or root-mean-square value to its average value. Form factor = Effective value Average value (60) For a sinusoidal wave the form factor is E m ^ r = 1.11 2E, The form factor is usually less than 1.11 for a flat-topped wave, i.e., a wave that is flatter than a sinusoidal wave. It is usually greater than 1.11 for a peaked wave, i.e., one that is more peaked than a sinusoidal wave. The form factor has not much practical im- portance, as two waves having totally different wave forms may have equal form factors. For example, the following wave has the same form factor as a sinusoidal wave but it is of totally different wave shape. FIG. 32. mi sin u>t + - I m i sin 3wt This wave and a sine wave having the same root-mean-square or effective value are plotted in Fig. 32. Amplitude, Crest or Peak Factor. The amplitude, crest or peak factor of an alternating wave is the ratio of its maximum value to its root-mean-square or effective value. A knowledge of the amplitude or peak factor is of importance in testing insula- tion, since the stress to which an insulation is subjected by a given impressed voltage depends upon the maximum value of the NON-SINUSOIDAL WAVES 107 voltage and not upon its root-mean-square or effective value. A knowledge of the form factor of a voltage in connection with insulation testing, however, is of little if any value. The peak factor of a sinusoidal wave is -\/2 = 1.41. Deviation Factor. The deviation factor of a wave is the ratio of the maximum difference between the corresponding ordinates of the actual wave and an equivalent sine wave of equal length, to the maximum ordinate of the equivalent sine wave, when the waves are superposed in such a way as to make the maximum difference between corresponding ordinates as small as possible. (For definition of equivalent sine wave see the next paragraph.) Except in special cases, a deviation factor of 0.10 is permissible in the wave form of commercial electrical machinery. Equivalent Sine Waves. In most of the alternating-current phenomena which are met in practice, neither the voltage nor the current is sinusoidal, although both are periodic. In many cases where the wave forms do not differ greatly from a sinusoid, it is sufficiently exact to replace the non-sinusoidal waves of voltage and current by equivalent sine waves. These equivalent sine waves have the same root-mean-square or effective values as the actual waves they replace and their phase is so chosen as to make El cos 6 for the equivalent sine waves equal to the actual power. The sign of the phase angle between the equivalent sine waves is made the same as that of the phase angle between the fundamentals of the actual waves. When the fundamentals are in phase, but the power-factor is not unity, the sign of the equivalent phase angle is indeterminate. Complex waves, which differ much in wave form, cannot be replaced by their equivalent sine waves, when they are to be added or subtracted, without danger of introducing considerable error. (See page 110.) Equivalent Phase Difference. The angle 0, determined by p = cos" 1 j^j when voltage and current are not sinusoidal waves, is known as the equivalent phase angle. I and E are the effec- tive or root-mean-square values of the non-sinusoidal waves andP is the average power. The equivalent phase difference is the phase angle that must be used with the equivalent sine waves of current and voltage to produce the true power. The equivalent phase 108 PRINCIPLES OF ALTERNATING CURRENTS difference may often be misleading, since the presence of har- monics in one wave which are not present in the other will lower the power-factor even though there is no displacement between p the waves. In general 6, as determined by = cos" 1 -^ has no real physical significance except when sine waves are considered. Example of Equivalent Sine Waves. The analysis of the voltage and current waves for a certain circuit has shown them to be of the forms e = 100 sin ut + 20 sin (3co + 60) + 15 sin (5^-40) i = 40 sin (* - 30) + 5 sin (3co + 20) E . ^(ioo)V+ ())' W , 72>9 volts 1(40) 2 + (5) 2 / = A/ - = 28.5 amperes P = - cos 30 + - cos 40" 2t 2 = 1770 watts Power-factor = m ^ 5 = 0.852 Equivalent phase difference = 6 = cos" 1 0.852 = 31.6 degrees. If the equivalent sine voltage is taken zero when time, t, is zero, the equivalent sine waves of voltage and current are e = v"2 x 72.9 sin ut = 103.1 sin cot i = \/2 X 28.5 sin (co* - 316) = 40.3 sin (co* - 316) If the equivalent sine current is taken zero when time, t, is zero, the equivalent sine waves are e = 103.1 sin (ut + 31?6) i = 40.3 sin ut It should be noticed that the fifth harmonic in the voltage contributes nothing to the power, since there is no component in the current of the same frequency. Although it contributes nothing to power it does increase the voltage and therefore NON-SINUSOIDAL WAVES 109 lowers the power-factor. Because the fifth harmonic contributes nothing to power, it cannot be neglected in finding the root- mean-square or effective value of the voltage. As has already been stated, the power-factor of a circuit cannot be unity unless the current and voltage contain like harmonics, and then the relative magnitudes and the phase rela- tions of the harmonics must be identical in the two waves. The power-factor of a circuit containing nothing but pure resistance cannot be unity, even though in such a circuit there would be no displacement between the current and voltage waves, unless the temperature coefficient of the material of which the resistance is made is zero or the dimensions of the resistance unit are such that there is no appreciable change in its resistance during a cycle. If a resistance unit is made of fine wire of high tempera- ture coefficient its resistance will change appreciably during a cycle. If a sinusoidal voltage is impressed on such a resistance the current will be flatter than a sinusoid and will therefore con- tain harmonics, among which will be a marked third. The power-factor of a circuit whose resistance varies with current during a cycle cannot be unity, even though the circuit con- tains nothing but pure resistance. For commercial circuits, the change of the resistance with current during a cycle is usually too small to produce any noticeable effect on the shape of the current wave. Consider a case where there is no phase displacement between the current and voltage waves but the current wave contains a harmonic which is not present in the voltage. Let 6 = 100 sin 2x60* i = 10 sin 2ir6Qt + 5 sin 6rr60< The voltage is a pure sinusoidal wave, while the current contains a 50 per cent, third harmonic. This third harmonic contributes nothing to the power, since the voltage has no third harmonic. The root-mean-square or effective value of the voltage and the current are respectively = 70.7 volts 110 PRINCIPLES OF ALTERNATING CURRENTS P = 1Q * 10 cos (0 - 0) + 2 = 500 watts Power-factor > ^^ . 0.895 Equivalent phase difference = = cos~ 1 0.895= 26.5 degrees. The equivalent sine waves are e = 100 sin 2^60* i = T/2 X 7.91 sin (27r60* 26?5) = 11.18 sin (27r60* 26?5) Although the power in the example just given is due entirely to the sinusoidal voltage, e = 100 sin 2^60*, and the sinusoidal current, i = 10 sin 2?r60*, the latter is not the equivalent sine wave of current since it does not have the proper ampere value or phase relation. Where an exact analysis of any particular problem is essential, the substitution of the equivalent sine waves for the actual volt- age and current is not permissible. Addition and Subtraction of Non-sinusoidal Waves. When non-sinusoidal currents or voltages are to be added or subtracted, each must first be expressed in terms of its Fourier series, i.e., in terms of its fundamental and harmonics. The fundamentals and the harmonics of like frequency may then be added or subtracted vectorially to give the fundamental and the harmonics of the re- sultant wave. Equivalent sine waves cannot be added or sub- tracted vectorially except when the wave forms are identical and the phase displacement is zero. If the wave forms are very different or the phase displacement between the waves is great, the error produced by adding or subtracting the equivalent sine waves may be large. Example of Addition of Non-sinusoidal Waves. Let the following voltage be impressed on a circuit having two branches in parallel, one consisting of a resistance of 5 ohms in series with a condenser of 132.7 microfarads capacitance, the other consisting of 10 ohms resistance in series with an inductance of 0.0398 henry. e = 200 sin (377* + 10) + 75 sin (1131* + 30) + 50 sin (1885* + 50) NON-SINUSOIDAL WAVES 111 The currents in the two branches will be (See Chapter VII) i' = 9.71 sin (377* + 8596) + 9.00 sin (1131* + 83?15) + 7.81 sin (1885* + 88?66) i" =11.09 sin (377* - 46?31) + 1.625 sin (1131* - 47?47) 4- 0.661 sin (1885* - 3240) Capital letters with the subscripts 1, 3 and 5 will represent root-mean-square values of the fundamental and harmonics. The subscript used with the subscripts 1, 3 and 5 will indicate resultants. Consider the vectors representing the fundamentals of the currents at the instant when time, *, is zero. Their vector expressions are \/2 // = 9.71(cos 8596 + j sin 85?96) = 0.684 + J9.68 \/2/i"= 11.09(cos4631 - j sin 46?31) = 7.67 - ./8.03 \/2 loi = 8.35 -f jl.65 V2 7oi = V (8.35) 2 + (1.65) 2 = 8.51 amperes maximum. tan 0oi = ^^ = 0.1976 i = 11.18 degrees. o.oO Consider the vectors representing the third harmonics at the instant * is zero. \/2 7 3 ' = 9.00(cos 8315 +j sin 8315) = 1.074 +J8.94 V2 h" = 1.625(cos 47?47 - j sin 47?47) = 1.098 - j 1.198 V2 7as = 2.172+/7.74 ^ = \/(2.172) 2 + (7.74) 2 = 8.04 amperes maximum. tan 003 = ' = 3.563 03 = 74.32 degrees. 112 PRINCIPLES OF ALTERNATING CURRENTS Consider the vectors representing the fifth harmonics at the instant t is zero. \/2 I*' = 7.81 (cos 8866 + j sin 8866) = 0.183 +/7.81 \/2 7 5 " = 0.661 (cos 3240 - j sin 32?40) = 0.558 - jO.354 \/2 7 05 = 0.741 + J7.46 \/2 7 5 = V(0.741) 2 + (7.46) 2 = 7.50 amperes maximum. 7 46 tan 005 = ~ 7 ^j = 10.06 05 = 84.32 degrees. The resultant current wave is to = 8.51 sin (377* + 11?18) + 8.04 sin (1131* + 7432) +7.50 sin (1885* + 84?32) 7 /(8.51) 2 + (8.04) 2 +"(7T5Q)' 2 /o = \/- ^ = 9.84 amperes effective. The resultant power is Po = 20 ^ 8 ' 51 cos (10 - 11?18) + 75 X f ' 4 cos (30 - 74?32) 5 X7 ' 5 cos (50 - 84?32) = 851 + 216 + 155 = 1222 watts. The power in each branch of the parallel circuit is P ' = 200X9,71 C0g( _ 75 o 96) 235.4 + 202.3 + 152.4 = 590.1 watts. 200 X 11.09 cos(+56?31) = 615.2 + 13.2 + 2.2 = 630.6 watts P =P'+P" = 590 + 631 = 1221 watts NON-SINUSOIDAL WAVES 113 The maximum value of the equivalent sine wave of voltage is = V(200) 2 + (75) 2 + (50) 2 = 219.3 volts maximum. The maximum values of the equivalent sine waves of the cur- rents are = \/(9.71) 2 + (9.00) 2 + (7.81) 2 : 15.37 amperes maximum. V27" = Vai.09) 2 + (1.625) 2 + (0.661) 2 = 11.23 amperes maximum. (Power-factor)' = = 0.3501 V2 ' V2 Equivalent phase difference = 0' = cos" 1 0.3501 =69.50 degrees. (Power-f actor) " = oino 630 '^ 00 = 0.5121 Equivalent phase difference = 9" = cos" 1 0.5121 =59.20 degrees. If the equivalent sine voltage is taken zero when t is zero, the equivalent sine waves are e = 219.3 sin 377* i f = 15.37 sin (377* + 69?50) i" = 11. 23 sin (377* - 59?20) Add the equivalent sine currents as if they were actually sinusoidal waves. Consider the vectors representing them at the instant * is zero. \/27' = 15.37 (cos 69?50 + j sin 69?50) = 5.38 + J14.4 V27" = 11.23 (cos 5920 - j sin 59?20) = 5.73 - J9.64 V2/o = 11.11 + J4.76 \/27o = V(H.ll) 2 + (4.76) 2 = 12.09 amperes maximum. 12 09 7 = 'j=- = 8.56 amperes effective. 8 114 PRINCIPLES OF ALTERNATING CURRENTS Adding the equivalent sine waves vectorially, in the example just given, gives 8.56 amperes for the resultant current instead of 9.84 amperes, the correct value. The error is 13 per cent. If both branches of the divided circuit had contained similar constants, the error of adding the equivalent sine waves would probably have been less. It could not have been zero unless the wave forms were identical and the two waves were in phase. For example, suppose the wave forms of the two currents had been identical and each had contained a third harmonic. A phase displacement of 60 degrees between the waves, i.e., between their fundamentals, would have made the third harmonic in the resultant zero. The effect of the third harmonics would not have canceled had the equivalent sine waves been added. It is readily seen from what precedes that whenever the voltage, current and power in the component parts of a circuit with parallel branches are measured, and the equivalent sine waves of current determined from the instrument readings are added vectorially as if they were really sinusoidal waves, the resultant current thus determined may be in considerable error if the wave forms of the component currents differ greatly from sinusoids. A similar statement holds regarding the addition of the equivalent sine waves of voltage drop across the component parts of a series circuit. CHAPTER V CIRCUITS CONTAINING RESISTANCE, INDUCTANCE AND CAPACITANCE Coefficient of Self-induction or the Self-inductance of a Circuit. In the neighborhood of an electric circuit carrying a current, there exists a magnetic field whose intensity at any point is dependent upon the strength of the current, the con- figuration of the circuit and the distance of the point from the circuit. If the current alters its value, the field is also altered, increasing with increase of current and decreasing with decrease of current. This magnetic field is a definite seat of energy and for its production requires, therefore, a definite expenditure of energy, determined in amount by the flux and the conducting ampere-turns of the circuit with which this flux is linked. The linkages of flux with turns constitutes one of the most important factors of any circuit. The change in the number of linkages per unit current for an electric circuit is called the coefficient of self-induction or the self-inductance. The self- inductance of the circuit is denoted by the symbol L. If d(p is the change in flux linking a circuit of N turns produced by a change di in the current, the coefficient of self-induction or the self-inductance of the circuit is It is the rate of change of flux linkages of a circuit with respect to the current it carries. If the flux linkages per unit current are constant, the coefficient of self-induction may be written where

where L is the self- inductance and / the current. For the circuit, the kinetic energy is the energy in the magnetic field set up by the current. Capacitance. The capacitance of a condenser is measured by the charge required to raise its potential by unity. It is equal to the ratio of charge to voltage. c- Q V A condenser is said to have a capacitance of one farad when a charge of one coulomb raises it to a potential of one volt. This unit is too large for practical use. For this reason the capacitance of condensers is ordinarily expressed in microfarads. The resistance to direct current of a well made condenser is very high and for most practical purposes may be considered infinite. This does not mean, however, that a condenser connected across an alternating-current circuit will take no current. It will alternately charge and discharge at the frequency of the circuit in which it is connected and will, therefore, take an alternating current of perfectly definite effective or root-mean- square value. The resistance of a condenser to an alternating current is equal to the average power it takes, when placed across an alternating-current circuit, divided by the square of the effective or root-mean-square current taken from the mains. The power absorbed is the Pr loss in the condenser caused by the alternating charging current, plus the hysteresis loss in the dielectric. The latter loss is caused by the varying stresses produced in the dielectric by the alternating voltage impressed across the condenser terminals. The dielectric hysteresis loss per unit volume depends on the nature of the dielectric, the maxi- mum potential gradient to which it is subjected and the fre- quency. The dielectric hysteresis loss for air is zero. It is small in most dielectrics at commercial frequencies, i.e., at frequencies of 60 cycles or lower. At ordinary frequencies the losses in commercial condensers are small and for most purposes may be neglected. RESISTANCE, INDUCTANCE AND CAPACITANCE 119 Effect of a Condenser in an Alternating-current Circuit. The effect of a condenser in a circuit, so far as the phase relation between current and voltage is concerned, is just the opposite to that of inductance. Inductance causes the current in a circuit to lag the voltage drop across its terminals. A condenser causes the current in a circuit to lead the voltage drop across its terminals. The voltage drop across the terminals of a condenser is where C is the capacitance of the condenser and q its charge. When the charge on the condenser is a maximum, the voltage drop across its terminals is also a maximum. When the charge is zero, the voltage drop is also zero. The current taken by a condenser without leakance at any instant is equal to the dif- ference between the voltage impressed across its terminals and the opposing electromotive force due to the condenser charge, divided by the condenser resistance. The opposing electro- motive force is e = ^. Therefore, when the charge is a maxi- C mum the current must be a minimum and when the charge is a minimum the current must be a maximum. There can be no charge in the condenser until there has been current flow, since q = fidt. The current must consequently lead the charge and, since the voltage drop across the terminals of a condenser is e = ^, it must also lead the voltage drop. C The effect of a condenser in an electric circuit is similar to the effect of elasticity in a mechanical system. As a con- denser charges, its back electromotive force rises with the charge and offers an increasing opposition to further current flow, the opposition increasing in proportion to the charge. This back electromotive force is similar in its effect on current flow to the reaction of a spring, which is being stretched. The reaction of the spring offers increasing opposition to further stretching. Circuit Containing Constant Resistance and Constant Self- inductance in Series. In any energy relation whatsoever, the sum of the actions and the reactions must be zero. The action for the 120 PRINCIPLES OF ALTERNATING CURRENTS ordinary electric circuit is balanced by the impressed voltage drop. The reactions for a circuit containing resistance and self-inductance in series are the ohmic and reactive drops, the latter being due to the presence of self-induction. The condition of the circuit is completely determined by the equation In general this can be solved only when

* + a + 0) (26) where I m is the maximum value of the current after steady con- ditions have been attained and 6 is its phase angle with respect to the voltage E m . Substituting the value of the current given by equation (26) in equation (24) gives E m sin (wt + a) = rl m sin (cot + a + 9) = rl m sin (ut + a + 0) -f coL/ m cos M + a + B) (27) Since a cosine function of time leads a sine function by 90 degrees, equation (27) may be written in the following form E m sin (ut + a) rl m sin (co + a + 0) sin ( + a + + 90) (28) The two terms in the right-hand member of equation (28) are two quadrature components of the voltage drop, E m sin (ut + ), across the circuit. The maximum values of these components are rl m and coL/ m . The component wL7 m leads the component rl m by 90 degrees. These two components are respectively the reactive and the active components of the voltage drop. 130 PRINCIPLES OF ALTERNATING CURRENTS The vectors corresponding to the three terms of equation (28) a + are plotted in Fig. 36 for the instant of time t = . At CO Fia. 36. this instant the vector represent- ing the current lies along the axis of reals. The waves corresponding to the vectors are plotted in Fig. 37, be- ing assumed greater than a. Referring to Fig. 36 it is obvious that = /mVrM 7 ^ (29) and FIG. 37. The voltage E m leads the current I m by an angle or the current lags the voltage by the same angle. Since in equations (27) and (28) is the phase angle of the current with respect to the voltage, it should be considered negative as the current lags the voltage. The particular integral or steady term in the general equation for the current in a circuit having constant resistance and con- stant self-inductance in series is consequently RESISTANCE, INDUCTANCE AND CAPACITANCE 131 (32) The complete solution of equation (24) is therefore rt Jf / j\ i = A~L -j m = sin tut -f- a tan" 1 ) (33) V** 2 + co 2 L 2 \ r / When t is zero the current i is also zero. Putting both t and i equal to zero in equation (33) gives -E m in [ a tan l Therefore -B t \/r 2 +a> 2 L 2 sin ( - taar' I. + Vr^T^ Sin (" < + a ~ ta "'T) (34) UMJ The angle 6 = tan" 1 is the angle of lag of the current behind the voltage after steady conditions have been attained. It is determined by the constants of the circuit and theoretically at least may have any value between and 90 degrees. It is zero when the inductance is zero, i.e., for a non-inductive circuit. It would be 90 degrees for a circuit having inductance but no resistance, if such a condition were possible of attainment. In practice, 6 may be very nearly 90 degrees but it can never be exactly 90 degrees, since it is impossible to have a circuit without some resistance. There is no difficulty in making a circuit prac- tically non-inductive for ordinary frequencies. The magnitude of the transient term in the expression for the current (first term in the second member of equation (34)) is deter- mined by the particular point on the electromotive force wave at which the circuit is closed. The magnitude of the phase angle a, in the expression for the instantaneous voltage, equation (24), page 128, fixes the value of the voltage e when t is zero. The 7T transient will be a maximum when (a 6) = 9 or wnen the circuit is closed at that point on the electromotive force 132 PRINCIPLES OF ALTERNATING CURRENTS wave which corresponds to maximum current after steady con- ditions have been established. The transient will be zero when ( a _ 0) = o, i.e., when the circuit is closed at that point on the electromotive force wave which corresponds to zero current after steady conditions have been reached. Fig. 38 shows the resultant current and its transient and steady components when a circuit containing constant resistance and constant self -inductance in series is closed at the point on the electromotive force wave which makes the transient a maximum. The figure is for a 25-cycle circuit having a ratio of coL to r of 15. Impressed Voltage shown by line of round dots. Current shown by full line. Transient and steady components of current shown by lines of short dashes. 7" = 0.0955. Frequency ~ 25 cycles, -f- = 15, Angle of lag = 86.2 degrees. Curves show the conditions existing when the circuit is closed at the point on the voltage wave which makes the transient a maximum, a- 6 - -|- FlG. 38. It is obvious from Fig. 38 that the maximum value the current can attain, when a circuit having constant resistance and constant self -inductance in series is closed, can never be equal to twice the maximum value of the current under steady conditions. It may reach nearly twice that value in a high-frequency circuit having a large time constant, i.e., having a large ratio of self-inductance to resistance, for under such conditions the transient will have diminished but little at the end of the first half cycle and will then add to the steady component to give a maximum nearly RESISTANCE, INDUCTANCE AND CAPACITANCE 133 equal to twice the maximum value of the current under steady conditions. Under ordinary conditions the transient is of little importance when circuits having constant resistance and constant self- inductance in series are closed, since it practically disappears after a few cycles and in general produces no dangerous rise in current. In the case plotted in Fig. 38, the transient has relatively little effect after two and a half or three cycles, that is after about a tenth of a second, and this is for a circuit having a relatively large ratio of L to r. The effect of the transient is to displace the axis of the current wave so that it lies along a logarithmic curve, which is the transient, instead of lying along the axis of time. Although the transient is of relatively little importance in most cases, it is of importance in switching operations on high- voltage transmission lines. It is of especial importance in certain cases where the inductance is not constant as, for example, when an alternator is short-circuited or when a large transformer is switched on a line. When an alternator is short-circuited, the initial maximum value of the short-circuit current may reach ten or more times the maximum value of the sustained short-circuit current, which itself may be several times the rated full-load current. This large transient short-circuit current is caused, in the case of the alternator, by a very great decrease in the apparent inductance of the armature winding during the transient period of the short circuit. The steady current for a circuit having constant resistance and constant self-inductance in series is given by E m i = -==== sin A/r 2 + co 2 L 2 ( cot + a tan" 1 ] \ r / = I m sin ( + a - 6) (35) The maximum value of the current is found by dividing the maximum value of the voltage by \A 2 + co 2 L 2 . It lags the maximum value of the voltage by an angle 6 whose tangent is equal to the ratio of coL to r. Since the effective value of a sinusoidal current is equal to its maximum value divided by the square root of two, it is evident that the effective value of the current is given by the effective value of the voltage divided by 134 PRINCIPLES OF ALTERNATING CURRENTS \/r 2 + o> 2 L 2 . The quantity V r 2 + cu 2 L 2 plays a similar part in alternating-current circuits to resistance in direct-current cir- cuits. If it is denoted by the letter z, the expression for current may be written in a form similar to Ohm's law for direct cur- rents, i.e., T E E I = , = = Vr 2 + co 2 L 2 z Impedance and Reactance. The quantity z = r 2 + o> 2 L 2 is called the impedance and is measured in ohms. It is a constant only when the resistance, self-inductance and frequency are constant. The expression coL is called the inductive reactance and is also measured in ohms. Reactance is denoted by the letter x. Thus wL = x and z = r (37) Inductive reactance is constant only when frequency and self- inductance are constant. It is the inductive reactance of a cir- cuit which causes the current to lag in phase behind the voltage and thus to have a quadrature component. It is because this component is caused by reactance that it is called the reactive component. Vector Method of Determining the Steady Current for a Circuit Having Constant Resistance and Constant Self-inductance in Series. The current in amperes will be taken along the axis of reals. Refer to Fig. 36, page 130. For the present purpose read I m and E m on the figure as 7 and E respectively, i.e., as root-mean- square or effective values instead of maximum values. The active and reactive components of the impressed electromotive force are then given by rl and juLI = jxl respectively. E = rl -{-jxl = I(r +jz) = z In complex, therefore, the impedance of an inductive circuit is given by z = r+jx (38) The value of z in ohms is \A* 2 + # 2 - Calling the component of E along the axis of reals, i.e., the real or active component, RESISTANCE, INDUCTANCE AND CAPACITANCE 135 E a and the component along the axis of imaginaries, i.e., the imaginary or reactive component, E r E = E a + jE r E (volts) = VE a 2 + E r 2 Rationalizing equation (39), to get rid of j in the denominator, by multiplying the numerator and the denominator by the denominator with the sign of the term involving j reversed, gives j = E a + jE r r-jx r +jx r - jx E r r -E a x * * =/a + Jlr where E a r + E r x E r r - E a x and I r - -- are respectively the active and reactive components of the current. But / is along the axis of reals, therefore I r = 0. Hence E r r = E a x and ' ' where 6 is the tangent of the angle of lag of the current behind the voltage E. The power is E a la = El COS 0. (42) It should be noted that impedance is not a vector but a com- plex quantity. When it is multiplied by vector current vector voltage results. Complex quantities have no reference axis in the ordinary sense but when they are multiplied or divided by a vector, such as A, the result is a new vector referred to the same axis as the vector A , but usually not in phase with A . Although complex quantities, such as impedance, are not vectors, they have real and imaginary parts and when multiplied or divided, added or subtracted must be treated as vectors in so far as these operations are concerned, 136 PRINCIPLES OF ALTERNATING CURRENTS Polar Expression for the Impedance of a Circuit Containing Constant Resistance and Constant Self-inductance in Series. Multiplying and dividing the complex expression for impedance by \/r 2 H- x 2 will not alter its value. = + jx) Vr* + x" { 77===, + = z (cos + j sin 0) (43) where 2 is the magnitude of the impedance and 6 is its angle, x i.e., the angle which is determined by the relation = tan" 1 ' From equation (43) it is seen that impedance is a scalar quantity multiplied by the operator (cos + j sin 6) which rotates through the angle 0. Multiplying current expressed in complex by impedance, also expressed in complex, therefore gives the correct value of the impedance drop rotated into the correct phase position with respect to the current. From equation (43) it is obvious that the polar expression for impedance is z = z\0 (44) 1" where B = tan" 1 -. For inductive impedance 6 is positive. Circuit Containing Constant Resistance and Constant Capaci- tance in Series. The condition of a circuit containing constant resistance and constant capacitance in series is completely deter- mined by the equation e=ri + % (45) where e is the impressed voltage drop. The component of this to supply the ohmic drop is ri. ^ is the component to overcome the counter electromotive force of the condenser. If q is the charge on the condenser in coulombs and C is the capacitance of the condenser in farads, ^ is the voltage drop across the condenser in volts. RESISTANCE, INDUCTANCE AND CAPACITANCE 137 The energy relation is shown by eidt = ri 2 dt + ^ (46) Consider the total energy concerned in any given time T. C T C T C T o eidt = ri 2 dt + fat (47) Jo Jo Jo L The first term of the second member is the energy dissipated in joule heating, i.e., in the i 2 r loss. The second term represents the energy stored in the electrostatic field as potential energy due to the strain in the dielectric of the condenser. Solution of the Differential Equation for a Circuit Containing Constant Resistance and Constant Capacitance in Series. The solution of equation (45) giving the current and charge of the condenser in terms of e, r, C and t, will evidently depend for its form on the impressed electro- motive force. Case I. Growth of Charge and Current in a Capacitive Circuit on which a Constant Electromotive Force is Impressed. (Fig. 39, key on a.) UMAA/W FIG. 39. e = E = constant. E = ri + % Since E = r ^ -f * (48) This is a linear differential equation of the first order. It is of the same form as equation (5), page 121 and its solution is of the same type. q = Y + u Obviously the steady state is represented by the final charge on the condenser. The particular integral, u, is therefore equal to EC. The transient Y is found as before by putting E = 0. The solution of the equation is of the form q = At at . 138 PRINCIPLES OF ALTERNATING CURRENTS Substituting E = and q = Ae at in equation (48) gives = raAe at + A ^ (49) O This holds for all values of t. = ra + ^ (50) Equation (50) is an equation of the first degree and therefore has but a single root. 1 a= ~Cr The complete solution of equation (48) is therefore q = Ae~^ + EC (51) The constant of integration is found by putting t equal to zero. Let Q Q be the charge on the condenser when the circuit is closed, i.e., when t is zero. Then A = Qo - EC (52) Putting this value of the constant of integreation, A, in equa- tion (51) gives q = EC + (Qo-EC)e & (53) If the initial charge is zero, equation (53) becomes q = = o(l-e~^) (54) Where Q = EC is the final charge on the condenser. When the initial charge on the condenser is zero, the counter electromotive force of the condenser is zero when t = 0. The TjJ current in the circuit at this instant has its maximum value. . t r As t increases q also increases, producing a counter electromotive force g, which cuts down the current and hence the rate at which the charge increases, and therefore diminishes the rate of decrease of the current. The charge approaches the limit Q = EC, while the current approaches the limit zero. RESISTANCE, INDUCTANCE AND CAPACITANCE 139 The rate of increase of charge and therefore the rate of decrease in current is determined by Cr, the time constant of the circuit. Like the time constant for a circuit containing constant resistance and constant self-inductance in series, it is equal to the time re- quired for the charge to reach 0.632 of its final or maximum value. Since i = -, the equation for current may be found by differ- entiating equation (53) with respect to time. . . _ (feff !.- _ C/7* (55) (56) When the initial charge on the condenser is zero, equations (55) and (56) reduce to 6 r (57) When the initial charge on the condenser is zero, the charge on the condenser increases logarithmically to the limiting value EC (equation (54)). The current, on the other hand, reaches its maximum value Tjl instantly and then di- minishes logarithmically to zero. The growth of the charge and the decay of the cur- rent, when the initial charge is zero, are shown in Fig. 40. The curves are all for the same impressed voltage, E, and the same resistance, r. The capacitance is least for curve a and greatest for curve c. Energy of the Electrostatic Field. From the energy equation (equation (46), page 137), it may be seen that energy is constantly being dissipated in heat and also is being stored in the electro- static field at a rate, i, which rate constantly decreases. The Time FIG. 40. 140 PRINCIPLES OF ALTERNATING CURRENTS total energy thus stored in the electrostatic field is evidently given by (58) where Q and E are respectively the final charge and the final voltage of the condenser. The expression ^CE 2 represents the electropotential energy of the condenser due to its charge. It corresponds to the potential energy in mechanics of a stretched spring or other elastic body which is under stress. When a condenser is charged through a fixed resistance from a constant potential, one-half of the energy given to the circuit is absorbed as i 2 rt loss or joule heating in the resistance. The other half is stored as electropotential energy in the condenser. The efficiency of a system involving the charging of a condenser from a constant potential source through a fixed resistance therefore cannot be greater than fifty per cent. Putting T, in the energy equation (equation (47), page 137) equal to infinity gives E\ idt = \ ri*dt+^ I qidt Jo Jo C Jo r^F -JL r /F\ 2 *L 1 r El - *dt = r\ (-) e Cr dt + ~\ qdq Jo r Jo \r/ CJ * * E E _ 1= Cr\ ) 27 2CJ o E*C = ~ + ~ (59) The first term of equation (59) is the energy supplied to the circuit. The second and third terms are respectively the energy loss in the resistance and the energy stored in the condenser. The second and third terms are each equal to one-half the total energy supplied to the circuit. Case II. Decay of the Charge in a Capacitive Circuit when the Impressed Electromotive Force is Removed by Short Circuiting. (Fig. 39, page 137, key on 6.) Since electromotive force, e, is equal to zero 0-H + -r + (60) RESISTANCE, INDUCTANCE AND CAPACITANCE 141 In this case the steady charge is obviously zero since the elec- tromotive force is zero. The term which represents the steady state in the equation for the charge is therefore zero. q = Y + u = Y + (61) The complete solution of equation (60) consists of a transient term only. The solution is _ t q = Ae~c~ r When t is equal to zero, q is equal to Qo ', the initial charge on the condenser, i.e., the charge at the instant it begins to dis- charge. Therefore A =Q ' and _ J. q = Qo' e CT (62) If the charge has reached its final value EC, where E is the charging potential, before the electromotive force is removed, that is, before the discharge is started, equation (62) becomes __t_ q = EC e cr (63) In this case when t = 0, i.e., at the instant of short circuiting or removing the electromotive force, the charge q is equal to Qo' = EC. The charge decreases logarithmically to zero. Since i = -j., the equation for current may be obtained by differentiating the equation for charge, i.e., equation (62), with respect to time. i = ~^ r ~^ r ( 64 ) When Qo' = EC, i.e., when the steady state has been reached before the condenser begins to discharge, the expression for the current becomes E __L I = Cr r = -/o'e~^ (65) In this case the current i is equal to its Ohm's law value, /o' = , when t is zero. 142 PRINCIPLES OF ALTERNATING CURRENTS At the instant of removing the electromotive force, i.e., when t is zero, the charge on the condenser is a maximum, being equal to Q = EC, where E is the voltage of the condenser at the instant discharge is started. As t increases, the charge diminishes logarithmically to zero. The current has its maximum value, , for t = and also diminishes logarithmically, approaching zero as a limit with the charge. Observe, however, that the current is negative, which means that the current is flowing out of the condenser. During the discharge the energy of the electrostatic field is being gradually dis- sipated in heat in the re- sistance of the circuit. The decay of charge and current are shown graphically in Fig. 41. The capacitance is least for curve a and greatest for curve c. Case III. Simple Harmonic Electromotive Force Impressed on a Circuit Containing Con- stant Resistance and Constant Capacitance in Series. In this case the electromotive force is e E m sin (coZ + a) and FIG. 41. equation (45), page 136 becomes E / j dq . q , . & m sin (cot -|- a) = r , -f- ^ (bo) This is a linear differential equation of the first order with constant coefficients. The first term is a function of t. The solution is of the form q = Y + u The complementary function, Y, is found in the same manner as is Case I, by putting the left-hand member of the equation RESISTANCE, INDUCTANCE AND CAPACITANCE 143 equal to zero. This gives the transient component of the charge. From equation (51) this is Y = A 6~c~' (67) The particular integral, i.e., the term representing the steady state, may be found by a method similar to that used in Case III, page 128, for an inductive circuit with a sinusoidal electromotive force impressed. Under steady conditions, the voltage drop across the circuit da will be made up of two parts: one, ri = r-r., to supply the resistance drop; the other, ^, caused by the charge on the condenser. When the charge varies sinusoidally with time, the resistance drop will also be sinusoidal in form, provided the resistance is constant, for then the drop will be equal to a constant multiplied by the derivative of a sine function. The derivative of a sine function is a cosine function, which is equivalent to a sine function advanced 90 degrees in phase. If C is constant, the voltage drop, p across the condenser terminals will obviously be of the same wave form as the charge. Since the sum of any number of sinusoidal waves of like frequency is a sinusoidal wave of the same frequency, it follows that if r and C are both constant and the charge varies sinusoidally with time, the voltage across the circuit required to produce the charge must also be sinusoidal in wave form, since it is the sum of two sinusoidal drops related to the charge. Conversely, if the voltage impressed on the circuit is sinusoidal, the sum of the reactions due to the charge must be sinusoidal. The only way the sum of these reactions can be sinusoidal, when both r and C are constant, is for the charge to be sinusoidal. Let the charge be q = Q m sin (orf + a + 0') (68) where Q m is the maximum value of the charge after steady condi- tions have been attained and 0' is the phase angle of the charge with respect to the impressed electromotive force E m . 144 PRINCIPLES OF ALTERNATING CURRENTS Substituting the value of the charge given by equation (68) in equation (66) gives d ,v, E m sin (co + a) = r -r \Q m sin (co + a + 6 ) \ + ^Q m sin (ol + a + 0') = rcoQm cos (co + or + 0') + ~ Q m sin (< + a + 0') (69) Since a cosine function leads a sine function by 90 degrees, equation (69) may be written E m sin (coZ + a) = ruQ m sin (cot + a + 0' + 90) 1 sn + a + (70) The two components of the second member of equation (70) are quadrature components of the voltage drop, E m sin (to + a), across the circuit. The maximum E m (cosd'+jsmd') values of these components are ruQ m and - The vectors corresponding to the three terms of equation (70) are plotted in Fig. 42 for the instant of Qm time t = At this instant C* CO FIG. 42. the vector representing the charge lies along the axis of reals. Referring to Fig. 42, it is obvious that E m = V cox r 2 + = tan" 1 = tan -1 o>rC (71) (72) (73) RESISTANCE, INDUCTANCE AND CAPACITANCE 145 The voltage E m leads the charge Q m by an angle 6' or the charge lags the voltage by the same angle. Since 0' in equation (70) is the phase angle of the charge with respect to the voltage, it should be considered negative, as the charge lags the voltage. The particular integral or steady term for the charge in a circuit containing constant resistance and constant capacitance in series is consequently q = Em - = sin M + - tan" 1 rcoC) (74) The complete solution of equation (66) is therefore - F q = Ae Cr + - = sin M + - tan- 1 ro>C) (75) If the initial charge is zero, i.e., the condenser is uncharged when the circuit is closed, q will be zero when t is zero. . Putting both t and q zero in equation (75) gives ~== = sin (a - tan- 1 rcoC) (76) Therefore E -- m sin (a - tan- 1 ro>C)e Cr sin (ut + a tan" 1 rC) Cr cos (* + a - tan- 1 rcoC) (78) 10 146 PRINCIPLES OF ALTERNATING CURRENTS Replacing 6' by 90 (see Fig. 42) and also remembering that cos (|8 90) = sin and sin (0 90) = cos /3, gives -%_ , cos (a + tan' 1 l \ ~^ r 1 o> 2 C 2 E m (79) where tan" 1 ~ = is the angle of lead of the current with respect to the impressed voltage after steady conditions have been established. It should be observed that is positive and therefore, after steady conditions have been reached, the current leads the impressed electromotive force in a circuit containing constant resistance and constant capacitance in series by an angle whose tangent depends upon the resistance, capacitance and frequency of the circuit. For a circuit containing constant resistance and constant self-inductance in series the current lags the impressed electromotive force by an angle whose tangent depends upon the resistance, self-inductance and frequency of the circuit. (See page 131.) The limiting value a current can attain when a sinusoidal electromotive force is impressed on a circuit containing constant resistance and constant self-inductance in series is twice its steady state maximum value. (See page 132.) No such limita- tion exists when a sinusoidal electromotive force is impressed on a circuit containing constant resistance and constant capacitance in series. For fixed r and C, the transient will be a maximum when the circuit is closed at a point on the electromotive force wave which will make (a + tan" 1 ^j = 0. (See equation (79).) Under this condition the steady term is zero when t is zero, that is at the instant the circuit is closed. The transient has its greatest value when t = 0, that is, at the instant the circuit is closed and will then decrease logarithmically to zero at a rate which will depend upon the time constant, Cr. The ratio of the maximum value of the transient term to the RESISTANCE, INDUCTANCE AND CAPACITANCE 147 maximum value of the steady term in the current equation (equa- tion (79)) increases as the ratio of ^ = x to r increases and ap- proaches infinity as a limit. The maximum value of the transient J? term approaches the Ohm's law value of the current, i.e., -> as the ratio of ^ to r increases, since the coefficient of the T? transient term becomes approximately equal to - when r is small compared with ^* The ratio of r to -7, at sixty cycles for a well made commercial static condenser may be as low as 0.01 or 0.02. If a condenser having a ratio of r to -7, equal to 0.01 were connected to a con- coC/ stant potential, 60-cycle circuit at the point on the electromotive force wave which would make the transient a maximum, the current would immediately rise to approximately 100 times the value it would have after steady conditions were established. For circuits having constants likely to be met in practice, the transient, although it may be initially very large, practically disappears in a very small fraction of a cycle. For example : in the case just mentioned, o> would be 377. Let r be 0.1 ohm. Then -- = 0.01 For the transient to fall to 0.1 its initial value e"^ = 0.1 The duration of the transient is too short to be of importance. 148 PRINCIPLES OF ALTERNATING CURRENTS The maximum value of the steady component of the current is found by dividing the maximum value of the voltage by J r 2 _j * . it leads the maximum value of the voltage by an angle whose tangent is equal to the ratio of ^ to r, i.e., by an angle whose tangent is ^- Since the effective value of a sinusoidal current is equal to its maximum value divided by the square root of two, it is evident that the effective value of the current is given by the effective value of the voltage divided by r 2 _| _. The quantity <*/r 2 -\ ^ plays the same part in a capacitive circuit that the quantity \/r 2 + co 2 L 2 does in an inductive circuit. Impedance and Reactance. The quantity ' C0 2 C 2 is called the impedance and is measured in ohms. It is a constant only when resistance, capacitance and frequency are constant. The expression ^ is called the capacitive reactance and is coO also measured in ohms. Reactance is denoted by the letter x. Capacitive reactance is constant only when frequency and capacitance are constant. The capacitance of a circuit under ordinary conditions is constant. It should be noted that inductive reactance X L = coL = 2irfL is directly proportional to frequency. Capacitive reactance, - on the other hand, is inversely proportional to frequency. In- ductive reactance is positive. Capacitive reactance is negative. The significance of the negative sign with capacitive reactance will be understood from the "Vector Method" which follows. RESISTANCE, INDUCTANCE AND CAPACITANCE 149 Vector Method of Determining the Steady Component of the Current in a Circuit Having Constant Resistance and Constant Capacitance in Series. The current vector, 7 amperes effective, will be taken along the axis of reals. The effective values of the active and reactive components of the impressed electromotive force are given by rl and j~r ^ ~ ~~J X I respectively. The reactive component must be negative since the current in a capacitive circuit leads the impressed electromotive force. E = rl -jxl = I(r -jx) = Iz The vectors corresponding to equa- tion (81) are plotted in Fig. 43. The waves corresponding to the vectors in Fig. 43 are shown in Fig. 44. Time is taken zero when the current is zero. _j In complex, therefore, the impedance w of a capacitive circuit is given by z = r jx (81) E (cos 8-j sin 0) FIG. 43. (82) FIG. 44. The value of z in ohms is, of course, Calling the component of E along the axis of reals, i.e., the real or active component, E a , and the component along the axis of imaginaries, i.e., the imaginary or reactive component, E r , gives 150 PRINCIPLES OF ALTERNATING CURRENTS E = E a - jE r E (Volts) = V#a' + E r 2 I = Ea ~ jEr (83) r - jx Rationalizing equation (83), to get rid of j in the denominator, by multiplying the numerator and the denominator by the denominator with the sign of the term involving j reversed, gives 7 = E a - jE r r+jx r jx r + jx _ E a r + E r x __ E r r - E a x _ 2 2 3 2 2 where , _ E a r + E r x E r r - E a x -- ~~ are respectively the active and reactive components of the current. But 7 is along the axis of reals. Therefore I r = 0. Hence, E r r = EaX and |r = x = tan e (85) -H/a T where 6 is the angle of lead of the current with respect to the voltage impressed across the circuit. The power is E a l a = El cos 6. Polar Expression for the Impedance of a Circuit Containing Constant Resistance and Constant Capacitance in Series. The polar expression for the impedance of a circuit containing constant resistance and constant capacitance in series is (see page 136) z = z\-B = zH (86) where the angle 6 is determined by the relation 6 = tan- 1 * For capacitance x is negative. Therefore 0, in the polar expression for the impedance of a circuit containing resistance and capacitance in series, is negative. Circuit Containing Constant Resistance, Constant Self- inductance and Constant Capacitance in Series. The conditions in a circuit containing constant resistance, constant self-indue- RESISTANCE, INDUCTANCE AND CAPACITANCE 151 tance and constant capacitance in series are completely deter- mined by the equation e = ri + L + - (87) where e is the impressed electromotive force drop and ri, L^r. and ^ are, respectively, the components of the impressed electro- motive force drop to supply the ohmic drop, the drop due to self-induction and the drop due to the condenser. If the electro- motive force, 6, is in volts, r must be in ohms, L in henrys and C in farads. The current and charge will then be in amperes and coulombs respectively. The energy relation corresponding to equation (87) is f T eidt = ( T ri*dt + f (88) The first term of the second member of equation (88) is the energy dissipated in heat in the resistance of the circuit. The second term is the energy stored in the magnetic field of the inductance and the third term is the energy stored in the electro- static field of the condenser. Solution of the Differential Equation for a Circuit Containing Constant Resistance, Constant Self-inductance and Constant Capacitance in Series. The solu- tion of equation (87) for current in terms of e, r, L, C and t will evidently depend for its form upon e. Case L Charge of a Condenser through a Circuit Containing Constant Resistance and Constant Self-inductance in Series on which is Impressed a Constant Electromotive Force. (Fig. 45, key on a.) E = ri + L d 4. + I- (89) FIG. 45. (90) 152 PRINCIPLES OF ALTERNATING CURRENTS Equation (90) is a linear differential equation of the second order with constant coefficients. Its complete solution for charge, q, is of the form q = Y + u (91) where, as in all previous cases, Y and u are, respectively, the complementary function and particular integral and represent the transient and steady states. As in Case I, for a circuit having constant resistance and constant capacitance in series, the current will be zero when the steady state is reached. Under this condition the charge on the condenser will obviously be constant and equal to CE. The term u in equation (91), repre- senting the steady state, is therefore CE. The term Y, repre- senting the complementary function or transient, is found by putting E = and q = e at . (See Murray, Differential Equa- tions, page 63, second edition.) Making these substitutions gives = ra+La 2 + re a < (92) This holds for all values of t. = ra + La 2 + g (93) Equation (93) is of the second degree and therefore has two roots. These are Bi _ -re + (94) and * 7 ~ 2 (95) The complete solution of equation (90) is therefore q = A^ +A 2 e a ^ +EC (96) where A\ and A 2 are arbitrary constants of integration. . . dq Since i = ~ at i = Aiaie ait +Ata 2 e a2t (97) RESISTANCE, .INDUCTANCE AND CAPACITANCE 153 The form taken by the solutions of equations (96) and (97) depends on the relation of r 2 C 2 to 4Z/C. If r 2 C is greater than 4L, the roots a\ and a 2 are both real. If r 2 C is less than 4L, both roots are imaginary. If r 2 C is equal to 4L, the roots are equal. Cose /-a. r 2 C > 4L or r > 2-/g. In this case it is evident that the two roots ai and a* are essentially negative. If t = 0, q = and i = 0. Putting these values in equations (96) and (97) and solving for the constants AI and A 2 gives 01 - a 2 2 2 C 2 - 4LC Hence, from equations (96) and 97) (rC + Vr 2 C 2 -4LC -(" 2LC q = tiL .frC . n - ' C 2 - 4LC -4LC 7l (99) -4LC 2LC y-i (100) - 4LC I = i?r \ f rC ~ v>- t g a -4z>c Nj ^ L~( 2LC frC + Vr*C* - 4LC N t (101) V 2 C 2 - 4LC , ~\ 2LC The charge on the condenser, starting at the value zero, approaches the value Q = EC as its limit. The current, on the other hand, starts at the value zero, rises to a maximum and then decreases, approaching zero as its limit. do Since i = TT, it is evident that the slope of the curve repre- senting the charge is proportional to the current at any instant. This curve must, therefore, have a point of inflection at the time 154 PRINCIPLES OF ALTERNATING CURRENTS when the current has its maximum value. In Fig. 46 are given the curves of charge and current for the following values of the constants. E = 2000 volts. r = 100 ohms. L = 0.0016 henry. C = 1 microfarad. 0.0001 Second 0.0002 FIG. 46. 0.002 Case I-b. r 2 C < 4L or r < 2x/^. In this case it is evident that the roots ai and a 2 , equations (94) and (95), page 152, are imagi- nary, since the radical in the expressions for both a\ and a 2 is negative and may be written A/ 1 X \/4LC r 2 C 2 . Using the operator j to indicate the imaginary, -\/ 1, the expressions for a\ and a 2 become and . V4LC - r 2 C'< 1 2LC r . V4LC - r*C' ~2L~ J 2LC = a+jb = a jb (102) (103) r - r 2 C 2 _ Here a = anc * o = - ^ Both a and 6 are real. Equation (96), page 152, now becomes q = A,e (n+ib)t + A* (n - jb]t +EC (104) (105) RESISTANCE, INDUCTANCE AND CAPACITANCE 155 From the relations (cos 6 + j sin 0) = V4LC - r 2 C 2 and EC + e" Al + sin in (bt - + ta "- l (-a)! = EC - ECf'Jl + - sin bt + tan" 1 ( - j) 2\/LC -- 4L or r > 2*/^. In this case the roots ai and a z are essentially negative and have the same values as in Case I -a, page 153 -(- : -*nz- -)' - 4LC rC - V^^^LC ^ ~ ( ~ -^~ JC > (12 o) and -EC e - 4LC _ ( rC - Vr*C2 - 4LC\ V 2LC ) -4LC) t \ - AJ\^f e 2LC [ (121) In Fig. 49 are shown graphically the curves of charge and current during the discharge of a condenser through a circuit RESISTANCE, INDUCTANCE AND CAPACITANCE 159 having constant resistance and constant self-inductance in series for the following constants: E = 2000 volts. L = 0.0016 henry, r = 100 ohms. C = 1 microfarad. FIG. 49. ~ In this case as in Case 7-6, CaseII-b.r 2 C < 4Lorr < page 154, the roots ai and a% are imaginary and have the same values as in that case. Equations (118) and (119) become . f\/4LC- q = EC V4LC - 2L sin 2LC + tan- rC - r 2 C 2 1 - 2EC - Jl . V4LC - r 2 C 2 . I = e 2L SID - t A/4LC - r 2 C 2 2LC (122) (123) The charge and current have the same initial and the same final values as in Case Il-a. They oscillate about these final values with an amplitude which decreases logarithmically. The time of one complete vibration for charge or current is the same as in Case I-b and is found in the same way. T = 2-rVLC 4L 160 PRINCIPLES OF ALTERNATING CURRENTS If r 2 C is negligibly small as compared to 4L, T = If r 2 (7 = 4L, T, the time of a complete vibration, is infinite- The oscillatory discharge of a condenser is shown graphically in Fig. 50 for the following constants: E = 2000 volts. L = 0.0125 henry. r = 100 ohms. C = 1 microfarad. J).OQ2 0.001 FIG. 50. Case II-c. r 2 C = 4L or r = 2-J -^. Here the roots a\ and a 2 are equal as in Case I-c, page 156, and have the same values as in that case. The method of solving the differential equation is the same as in Case I-c, except that the term EC, which represents the steady condition, is zero. The solution involves nothing but the transient term since the final charge must be zero. From equations (112) and (113), page 157, i = Aiae at + A^ate at + A 2 e at The solution of equations (125) and (126) gives q = i = 2L -i 1 * rt 2L (125) (126) (127) (128) RESISTANCE, INDUCTANCE AND CAPACITANCE 161 Where EC =Q and E are, respectively the initial charge and initial voltage of the condenser. Equations (127) and (128) are shown graphically in Fig. 51 for the following constants: E = 2000 volts. r = 100 ohms. L = 2.5 henrys. C = 1000 microfarads. 0.10 Second 0.20 FIG. 51. Case III. A Simple Harmonic Electromotive Force Impressed on a Circuit Containing Constant Resistance, Constant Self-in- ductance and Constant Capacitance in Series. In this case the impressed electromotive force is of the form e = E m sin (ut -\- a) and E m sin (ut + a) = ri + l + or (129) (130) The complete solution of this linear differential equation of the second order, the first term of which is a function of t, is the sum of the transient and the steady value of the charge. As in all preceding cases the solution is of the form 9 = Y + u where Y is the transient term, i.e., the complementary function of the differential equation, and u is the particular integral and 11 162 PRINCIPLES OF ALTERNATING CURRENTS represents the steady state. The transient term or comple- mentary function is found, as in all other cases, by putting the first term, E m sin (ut + a), equal to zero. The value of the complementary function evidently depends on the relation of r 2 C to 4L. Its value is of the same form as in Cases Il-a, b and c. It is q = A^ + A 2 e a2 < (131) i = A^e 111 + A 2 a 2 e a2t (132) It is oscillatory when r 2 C is less than 4L or r is less than 2* / ^> \ C the frequency being given by (equation (124), page 159). f = = j4f (133) ~ T ~ 27TA/IC which is determined by the constants of the circuit. The fre- quency is entirely independent of the frequency of the impressed voltage. The constants for the transient terms for charge and current when the impressed voltage is sinusoidal are determined from the charge and the current when t = 0. When the tran- sient terms are oscillatory, the constants are determined from equations (106) and (107), page 155. The transient is of importance in all switching operations on circuits containing resistance, self-inductance and capacitance in series, since dangerous oscillations may be produced if the resistance of the circuit is low compared with 2*/ It is important in all radio work involving tuning of series circuits. Most radio circuits have mutual-inductance as well as self-in- ductance, and for such circuits the conditions are not so simple as those just discussed. The transients in transmission lines may be of great importance during switching operations, short-circuits, etc., but the equations just developed do not apply to a transmission line, since a trans- mission line is not a simple series circuit. It is a circuit contain- ing series resistance and series inductance, but it has parallel capacitance and leakance between the conductors and between the conductors and the earth. RESISTANCE, INDUCTANCE AND CAPACITANCE 163 The particular integral, i.e., the term representing the steady state for equation (130), page 161, may be found most easily by a method similar to that used in Case III for a circuit contain- ing constant resistance and constant self-inductance in series and also in Case III for a circuit containing constant resistance and constant capacitance in series. Under steady conditions, the voltage drop across the circuit da will be made up of three parts, namely : ri = r -^> the voltage _ di T d z q drop across the resistance, L -r = L~jT 2 ' the voltage drop across the self-inductance, and , = I i dt, the voltage drop across the = ^ I i capacitance. If r, L and C are constant, the reasoning that was used in Case III for a circuit containing constant resistance and constant self -inductance in series and also in Case III for a circuit containing constant resistance and constant capacitance in series will show that both the charge and current must vary sinusoi dally with time when the impressed electromotive force is sinusoidal. Assume that the charge is given by q = Q m sin (* + + 0') (134) where tf is the phase angle of the charge with respect to the impressed electromotive force. Then *- *><- + > -rg + Lg + g = wrQ m cos (ut + a + 0') - o> 2 LQ m sin (ut + a + 0') + ^Q m sin M + + 0') (135) Since the cosine of an angle is equal to the sine of ninety degrees plus the angle, equation (135) may be written in the following form : E m sin ( + a) = corQ TO sin (coZ + a + tf + 90) -u*LQ m sin (arf+ a + 0') + Q m sin (;* + + 00 (136) 164 PRINCIPLES OF ALTERNATING CURRENTS The vectors corresponding to the terms of equation (136) are OL + Q plotted in Fig. 52 for the instant of time t = CO This instant of time was chosen because it puts the vector representing the charge along the axis of reference. -o> 2 LQ FIG. 52. From Fig. 52 it is obvious that Q m E m = and Qm = E m (137) (138) tan tf ~Q m (139) The charge lags the voltage, E m sin (wt + a), impressed across the circuit, by the angle 0'. The angle 0', therefore, represents the lag of the charge behind the impressed voltage or the lead of the impressed voltage with respect to the charge. When ~ is greater than coL, tan 0' is positive and 0' is less than oou 90 degrees. When is less than o>L, tan 0' is negative and the RESISTANCE, INDUCTANCE AND CAPACITANCE 165 charge lags the impressed voltage by more than 90 degrees but less than 180 degrees. If r is zero, 9' is also zero. In this case the charge is in phase with the voltage impressed across the circuit, which, under this condition, is the same as the voltage across the condenser. The expression for the charge when steady conditions have been reached is therefore sin < ut + a tan" 1 (140) The angle tf = tan" 1 in equation (140) is a lag angle since equation (140) is an equation for the charge in terms of the impressed electromotive force. The charge lags the impressed electromotive force E m sin (wt + ) by the angle 6'. Since -T., the current under steady conditions is E cos sin at + a tan" 1 at +a tan" 90 (141) 3 1 (uL~) u>G ' FIG. 53. The current, therefore, leads the charge by 90 degrees. The vector representing the current is plotted in Fig. 53. 166 PRINCIPLES OF ALTERNATING CURRENTS Let 6 be the angle which the current makes with the impressed electromotive force, E m . Then e = e f - 90 i c m 6 = tai = tan- Equation (141) may therefore be written + E m sm cot -ha - tan- 1 1 \ (142) The signs of coL and ^ in the radical A/r 2 + (coL ~J , of equation (142), are reversed in the expression (coL ^j to make it correspond to the like term in the expression for the tan- gent of the angle 6. This can be done since the term (coL ^j is squared and changing its sign does not alter its squared value. The complete solution of equation (130), page 161, for charge is given by adding equation (140) to equation (131), page 162. q = Y + u + A* a * + - A', (i - <*) sin ut -j- a /J_ COL) (143) The complete solution of equation (130), page 161 for current, is given by adding equation (142) to equation (132), page 162. RESISTANCE, INDUCTANCE AND CAPACITANCE 167 * = Y + u + - -- = sin a-tan- (144) After a brief interval of time, the transient terms in equations (143) and (144) become sensibly zero for circuits with constants ordinarily met in practice and the charge and current then be- come simple harmonic functions of time. The maximum value of the current under steady conditions is found by dividing the maximum value of the impressed electro- motive force by -Jr 2 + ^coL ^j . The current lags the " L ~ coC 1 voltage by an angle whose tangent is -- - -- . When Q is greater than coL, the angle 6 becomes negative and is then equiva- lent to an angle of lead. In this case the current actually leads the electromotive force impressed on the circuit by an angle 0. When ~ is less than o>L, the angle 6 is positive and is actually an angle of lag. In this case the current actually lags the voltage impressed on the circuit. Since the effective value of a sinusoidal wave is equal to its maximum value divided by the square root of two, it is evident that the effective value of the current is found by dividing the effective value of the voltage by A/r 2 + (o>L - ~\ . This current leads or lag greater or less than current leads or lags the impressed voltage according as ~ is E E 6 = tan- 1 I- (146) 168 PRINCIPLES OF ALTERNATING CURRENTS Impedance and Reactance. The quantity (147) is called the impedance and is measured in ohms. It is constant only when resistance, self -inductance, capacitance and frequency are constant. It may increase or decrease with an increase in frequency depending on the relative values of coL and -~. The expression coL = X L is the inductive reactance and the expression pr = x c is the capacitive reactance. (coL ~\ = X L x c = x is the resultant reactance. It should be noted that in- ductive reactance is always positive while capacitive reactance is always negative. The resultant reactance may be either positive or negative depending upon the relative values of XL and x c . Inductive reactance always increases with increase of frequency. Capacitive reactance always decreases with increase of frequency. 1 77F When coL = ^, XQ = and the current is given by -- and U)O T is in phase with the impressed electromotive force. Under this condition the circuit is said to be in resonance. More will be said about resonance later. Vector Method of Determining the Steady Component of the Current in a Circuit Containing Constant Resistance, Constant Self-inductance and Constant Capacitance in Series. The current vector, I amperes effective, will be taken along the axis of reals. The effective values of the active and reactive compo- nents of the impressed electromotive force are rl and J!(XL x c ) - ( 1 \ = jl ( coL ~ ) respectively. juLI = j!x L is the reactive com- ponent due to the self-inductance. ~J~r< = ~Jl x c is the reactive component due to the capacitance. jtaLI leads the current while j ~ lags the current. COL- E = r I + j(x L - x c )I = I[r+j(x L - x c )} = Yz (148) When substituting numerical values it must be remembered that x c is negative. RESISTANCE, INDUCTANCE AND CAPACITANCE 169 The vectors corresponding to equation (148) are plotted in Fig. 53, page 165. The complex expression for the impedance is therefore z = r + j(x L x c ) The value of z in ohms is \/r 2 + (X L - x c ) 2 > Calling the component of E along the axis of reals, i.e., the real or active component, E a , and the component along the axis of imaginaries, i.e., the imaginary or reactive component, ft, gives E = E a +jE r E (volts) = \/E a z + E r 2 (149) = z* r 2 + fe + #. E, - ^c) s ' J r 2 + (X L - X C ) 2 (15C 1 a a #r H - ftfe - X C ) and I r = ftr- E a (XLXC) r 2 H -(%.- Xc) 2 r 2 + (X L -X C ) 2 . - x c ) Rationalizing equation (149), to get rid of j in the denominator, by multiplying the numerator and the denominator each by the denominator with the sign of the term involving j reversed, gives = E a + jE r r - j(x L - x c ) * r + j(x L - Xc) ' r - j(x L - x c ) E a r + E r (x L - x c ) .E r r - E a (x L - x c ) where are respectively the active and reactive components of the current. But / is along the axis of reals. Therefore I r = 0. Hence E r r = E O (XL x c ) and ft = x^-^c = xo = E a r r where is the angle of lag of the current with respect to the voltage impressed on the circuit. If x c is greater than XL, the angle 6 becomes negative. A negative angle of lag is equivalent to a positive angle of lead. In this case the current actually leads the impressed voltage. If x c is smaller than X L , the angle is positive and is actually an angle of lag. In this case the current actually lags the voltage impressed on the circuit. The power is E a l a = El cos 0. 170 PRINCIPLES OF ALTERNATING CURRENTS Polar Expression for the Impedance of a Circuit Containing Constant Resistance, Constant Self-inductance and Constant Capacitance in Series. The polar expression for the impedance of a circuit containing constant resistance, constant self-induc- tance and constant capacitance in series is (See pages 136 and 150) z = z\e_ (151) where the angle is determined by the relation 6 = tan" 1 tan" 1 . If the inductive reactance predominates, 6 is positive, while if the capacitive reactance predominates, 6 is negative. Impedance may be written in four different ways, namely: z = r + jxo = z (cos 6 + j sin 6) Equation for the Velocity and Displacement of a Mechanical System Having Friction, Mass and Elasticity. Before leaving the consideration of circuits containing constant resistance, constant self-inductance and constant capacitance in series, it will be of interest to compare the equations developed for such circuits with the equations for displacement and velocity of a mechanical system having constant friction, constant mass and constant elasti- city. It has already been pointed out that self-inductance and capacitance are respectively analogues of mass and elasticity in mechanics. Electrical resistance corresponds to mechanical fric- tion. Current corresponds to velocity and charge to displace- ment in a mechanical system. Consider a torsional pendulum. Let c7 be the moment of inertia of the pendulum about its axis of oscillation. Also let and G be respectively the length of the torsion wire and its coefficient of torsional rigidity. The radius of the torsion wire is r. Then if co and a are respectively the angular velocity and the angular displacement of the pendulum from its mean position due to an applied couple M Reaction due to friction = M/ = fc/co Reaction due to inertia = Mi ^y at 7r7*^ fy Reaction due to elasticity = M e = ff ~n~ -^ & L RESISTANCE, INDUCTANCE AND CAPACITANCE 171 Since the displacing couple M Q must balance to the sum of the couples due to the reactions caused by the motion of the pendulum Mo = Mf -f- Mi -f- M e - _ - 1 _ da Since co = -37 equation (152) may be written M< da . + (152) (153) Equation (153) corresponds exactly to equation (90), page 151, for an electrical circuit containing constant resistance, constant self -inductance and constant capacitance in series and its solution takes the same form as the solution for that equation. When MQ is zero, i.e., the pendulum oscillates freely without applied accelerating or retarding couples, equation (153) becomes = k ^t + ki w + t ' (154) Equation (154) corresponds to equation (117), page 158, for the discharge of an electrical circuit containing constant resistance, constant self -inductance and constant capacitance in series. The solution of equation (154) takes three forms corresponding to cases a, 6, and c for equation (117). Case Electrical circuit Mechanical system (a) Charge Q is a maximum when t = 0. It then decreases ap- proaching zero as a limit. Cur- rent starts at zero when t = 0, rises to a maximum then de- creases approaching zero as a limit. There is no oscillation. Displacement a is a maximum when t = 0. It then decreases approaching zero as a limit. An- gular velocity co starts at zero when t = 0, rises to a maximum then decreases approaching zero as a limit. There is no oscilla- tion. 172 PRINCIPLES OF ALTERNATING CURRENTS Case Electrical circuit Mechanical system Charge and current have the same initial and final values as in Case (a). They oscillate about these values with an amplitude which decreases logarithmically. Displacement and velocity have the same initial and final values as in Case (a). They oscillate about these values with an amplitude which decreases logarithmically. (c) r = ZA / 2-K* Charge and current have the same initial and final values as in Cases (a) and (6). There is no oscillation. Displacement and velocity have the same initial and final values as in Cases (a) and (6). There is no oscillation. fe When kf < 2 */^ there is oscillation, i.e., if the pendulum is displaced it will come to rest after a series of oscillations about its mean position. These oscillations will decrease logarithmically with time. The period of vibration is When kf is small, T = T = 1 - k e Although the resistance of an electrical circuit may frequently be too high for electrical oscillations to take place, the friction of a pendulum is seldom too high to prevent oscillation unless the pendulum is specially damped. The vibrating element of an oscillograph is an example of a highly damped torsional pendu- lum. If it is displaced, it will come to rest without oscillation. CHAPTER VI MUTUAL-INDUCTION Mutual-induction. Up to the present point, the only reactions which have been considered, when a current in a circuit is varied, are those existing or arising in the circuit itself. The equations which have been developed hold so long as the circuit considered is not in the neighborhood of other circuits or is so related to any other circuit in its vicinity that there can be no interaction between them. In general, however, when two or more circuits are in proximity, any change in the current in one of them will cause inductive effects in the others. Consider two circuits which are in proximity to each other. When one of two such circuits carries current a magnetic field will be established. This field will link not only the circuit by which it is produced but, in general, a certain portion of it will also link the other circuit. The relative amounts of flux linking each circuit will depend chiefly on the relative size and shape of the circuits, their relative position and the permeability of the sur- rounding medium. All of the flux produced by one circuit can never link the other, although the difference between the flux linking one and that linking the other may be made very small by closely interwinding them. On the other hand, the difference may be very great when the circuits are far removed from each other or are placed with the axis of one in the plane of the other. If each circuit is carrying current, any change in the current of either will be accompanied by a change in the flux linkages of each circuit. The change in the flux linkages of each circuit will induce in each a voltage which will be equal to the time rate of change of flux linkages for the circuit. The rate of change of flux linkages of one circuit with respect to change in current in the other forms one of the most important and fundamental constants in the electric theory of circuits. If both circuits are now closed and carrying current, any change in the current in one will induce in the other a voltage 173 174 PRINCIPLES OF ALTERNATING CURRENTS which will cause a current and, according to the law of Lenz, this second current must have such a direction as to oppose the change in the flux producing it. When the current in the first circuit increases, the current induced in the second circuit must be in a direction opposite to that in the first circuit. When the current in the first circuit decreases, the current induced in the second circuit must be in the same direction as that in the first circuit. No change can take place in the current of one without producing a corresponding change by induction in the current of the other. Any change in the current of one will result in a cur- rent in the other as there is mutual-induction between two. Mutual-induction plays an important part in the operation of most alternating-current apparatus. The operation of certain types of apparatus, such as the transformer and the induction mo- tor, depends entirely upon it. Without the transformer and the induction motor, the present development of alternating-current systems of power distribution and utilization would be impossible. Coefficient of Mutual-induction or Mutual-inductance. The self-inductance of a circuit or its coefficient of self-induction was defined as the rate of change in the flux linkages of the circuit with respect to its current. When the permeability of the sur- rounding medium is constant, the self-inductance of the circuit is constant. When the permeability varies with current strength, the self -inductance also varies with current strength. When the permeability is constant, self-inductance may be defined as the change in the flux linkages of the circuit per unit change in its current. These definitions of self-inductance assume that no change is produced in the currents in other circuits in the neigh- borhood. When the current is varied in a circuit having a self- inductance L, a voltage of self-induction is induced in it which is equal to the time rate of change of the flux linkages of the circuit. The voltage drop due to self-inductance (assumed constant) is fti = L^ (1) Similarly, the mutual-inductance or coefficient of mutual- induction of two circuits is defined as the rate of change in the flux linkages of one with respect to current in the other. When the permeability of the circuits is constant, their mutual- inductance is constant. When the permeability varies with the MUTUAL-INDUCTION 175 current strength in either circuit, the mutual-inductance also varies with current strength. When the permeability is constant, the mutual-inductance of the two circuits may be defined as the change of flux linkages for either per unit change of current in the other. When the current is varied in one of the two circuits having a mutual-inductance M , a voltage of mutual-induction is induced in the other and is equal to the time rate of change of the flux linkages for that circuit. The voltage drop in circuit 2 due to the mutual-inductance (assumed constant) of circuit 1 on circuit 2 is c 2 = M n ^ (2) Self-inductance and mutual-inductance are both measured in the same unit, the henry. The mutual-inductance of two circuits is one henry when the rate of change of flux linkages of either, with respect to the current in amperes in the other, is 10 8 flux linkages per ampere. When the permeability is constant, two circuits have a mutual-inductance of one henry when a change of 10 8 flux linkages is produced in either by a change of one am- pere in the current of the other. These definitions assume that the current in one circuit is constant while the current in the other circuit is changed. When the coefficient of self- or mutual- induction is expressed in henrys and the rate of change of current is in amperes per second, the induced voltage is in volts. If the permeability of a circuit is constant and there is no magnetic leakage between its turns, i.e., if all the flux produced by each turn links all the turns, the coefficient of self-induction is Ll = T^ 1 abhenrys (3) In where Ni is the number of turns and (R is the reluctance of the magnetic circuit. If N 2 is the number of turns in a second circuit so related to the first that all the flux produced by the first circuit links all of the turns of the second circuit, the coefficient of mutual- induction of the first circuit on the second circuit will be Mi2 = - N 2 abhenrys. (4) 01 All the flux produced by the first circuit can never link all the turns of the second circuit, and all the flux produced by the 176 PRINCIPLES OF ALTERNATING CURRENTS second circuit can never link all the turns of the first circuit. There will always be some leakage of magnetic flux between the two circuits. By closely interwinding the two circuits, the magnetic leakage may be made very small. It is, however, often very great. Although it is possible to calculate the self-inductance and mutual-inductance of certain very simple circuits with a fair degree of accuracy, accurate calculation of either self-inductance or mutual-inductance is usually impossible. Voltage Drop across Circuits Having Resistance and Self- and Mutual-inductance. From the definitions of self- and mutual- inductance, it is obvious that the voltage drops across two circuits having resistance and self- and mutual-inductance are given by the following differential equations. t>! = nt ! + L, 1 + M 21 ^ 2 (5) v, = r* + L 2 ^ 2 + M^l (6) When no magnetic material is present, the coefficients of self- and mutual-induction in equations (5) and (6) are constant. When the flux produced by the currents i\ and i% is in magnetic material, the coefficients of self- and mutual-induction will vary with the currents. They will be complicated functions of the currents. In general, when magnetic material is present, no exact solution of equations (5) and (6) can be obtained. It will be shown that the coefficient of mutual-induction of two circuits, in the absence of magnetic material, is the same whether it is taken for circuit 1 with respect to circuit 2 or for circuit 2 with respect to circuit 1. Under this condition the subscripts on the coefficient of mutual-induction, M, have no significance and may be omitted. The Coefficients of Mutual-induction, Mi 2 and M 2 i, of Two Circuits in a Medium of Constant Permeability are Equal. Let the permeability of the medium be constant. This assumes that there is no magnetic material present. The coefficients of mutual-induction M i 2 , of circuit 1 on circuit 2, and M 2 i, of cir- cuit 2 on circuit 1, are equal. This statement is true without regard to the size, shape or position of the two circuits. MUTUAL-INDUCTION 177 The total electromagnetic energy in an electrical system consisting of two circuits 1 and 2, which have constant self- and mutual-inductance and carry currents I\ and /2 respectively, is obviously independent of the order or manner in which the cur- rents are established. This follows from the law of conservation of energy. Let 1 1 and 1 2 be the final values of the currents in the circuits 1 and 2 respectively. Consider 7 2 to be zero and establish 1 1. The energy due to establishing /i will be due entirely to the self -inductance of circuit 1. While I\ is increasing, a voltage will be induced in circuit 2 by the mutual-induction of circuit 1 on circuit 2. No work can be done in circuit 2 by this voltage, since 1 2 is zero. The total electromagnetic energy due to estab- lishing the current 7i is (7) Now consider the current /i to be maintained constant while the current 1 2 is established. The change produced in the electromagnetic energy of the system by the establishment of 7 2 will be made up of two parts : one due to the effect on 1 2 of the voltage induced in circuit 2, the other, due to the effect on I\ of the voltage induced in circuit 1. The first part is due to the self-inductance of circuit 2, while the second part is due to the mutual-inductance of circuit 2 on circuit 1. The change pro- duced in the electromagnetic energy of the system by the estab- lishment of 7 2 is = L 2 /2 2 + Jlf will, (8) The total change produced in the electromagnetic energy of the system by the establishment of the currents /i and 7 2 is therefore W = W l +W 2 = 1 LJS + il, 2 / 2 2 + M 21 /i/ a (9) 178 PRINCIPLES OF ALTERNATING CURRENTS If 7 2 had been established first, the energy equation would have been W ' = W i = (10) Since, according to the law of the conservation of energy, W Q and Wd must be equal, M i2 and M 2 i must also be equal. In other words, the mutual-inductance of circuit 1 with respect to circuit 2 is equal to the mutual-inductance of circuit 2 with respect to circuit 1. In general, when there is no magnetic material present, the coefficient of mutual-induction of two circuits is the same whether it is considered with respect to circuit 1 on circuit 2 or with respect to circuit 2 on circuit 1. FIG. 54. Magnetic Leakage and Leakage Coefficients. Fig. 54 shows two circuits having self- and mutual-inductance. All the flux that links circuit 1 does not link circuit 2, and all the flux that links circuit 2 does not link circuit 1. The flux that is common to both circuits, i.e., the mutual flux, and the fluxes that link one circuit without linking the other, i.e., the leakage fluxes, are indicated. In Fig. 54,

i X L 2 . Coefficient of coupling = M i X L (18) - fci) X V(i - fci) x (i - (19) The coupling between two circuits is said to be close when the coefficient of coupling is large. In radio work, circuits which have a coupling greater than 0.5 are said to be close coupled. The degree of coupling that is desirable depends on the purpose for which the circuits are to be used. For commer- cial transformers with iron cores the coupling between the primary and secondary windings is very close and may be as high as 0.98 or 0.99. Close coupling is generally desired in a commercial transformer, since the voltage regulation of a trans- former depends very largely on the closeness of coupling between its primary and secondary windings. Good voltage regulation requires small magnetic leakage between primary and secondary windings and therefore close coupling. Such close coupling as is used in commercial power transformers is undesirable in transformers for radio work. General Equations for the Voltage Drops Across Two In- ductively Coupled Circuits Each Having Constant Resistance, Constant Self -inductance and Constant Capacitance in Series. Let r, L and C with subscripts 1 and 2 be the constants of the two coupled circuits and let M be their mutual-inductance. The diagram of connections is given in Fig. 55. If vi and vz are the instantaneous voltage drops across the two circuits when their currents are i\ and t" 2 , the following differ- ential equations hold FIG. 55. 182 PRINCIPLES OF ALTERNATING CURRENTS vi = riii + L^ + i I i^t + M^ (20) Cit C i,/ Clt (21) Equations (20) and (21) are the fundamental differential equations for inductively coupled circuits such are used for radio-telegraphy. A simple and important case of the solution of equations (20) and (21) occurs when the circuits have only resistance and self- and mutual-inductance and the second circuit is short-cir- cuited. This is equivalent to saying that the series capacitances Ci and Cz are infinite. In this case equation (21) becomes diz _ ( M dii Trti ~dt ~~ \Lz~di ~Lz Substituting the value of -rr from equation (22) in equation (20) and remembering that Ci is infinite v = r i + I L - \^lri (23) 1 L 2 / dt Lz 2 ' If TZ is small, equation (23) may be written PI = riii + s LI - -=- > -7,-, approximately. (24) [ LZ ) dt Equation (24) is of the same form as equation (1), page 120, for a simple series circuit containing resistance and self-induc- f M 2 } tance in series except that \ LI -j } replaces LI. I Liz ) The effect of the short-circuited coupled circuit is to diminish the apparent self -inductance of the primary circuit. If the coeffi- cient of coupling could be made unity, y- would be equal to LI. f M 2 } In this case the apparent inductance, \ LI -y- M of the primary circuit would be zero and it would act like a circuit containing only pure resistance. Although it is not possible to make the MUTUAL-INDUCTION 183 apparent inductance of the primary circuit exactly zero, it may be made very low by interwinding the two circuits so as to make the magnetic leakage between them very small. A closely coupled, low resistance, short-circuited secondary winding may be used to suppress the arc when a relay circuit, which carries direct current, is opened. For this purpose the main relay winding would probably be wound on a copper cylin- der which would serve as the low resistance, closely coupled, secondary winding. This cylinder would have no effect on the operation of the relay so long as the exciting current was steady. When the exciting current varied, however, the current induced in the low resistance, closely coupled winding would decrease the apparent inductance of the exciting winding. The presence of the coupled winding would cause the exciting current to reach its final steady value more quickly after the circuit was closed. By decreasing the apparent inductance of the exciting winding, it would also practically suppress the arc when the exciting circuit was broken. The effect of the coupled winding is to de- crease the apparent time constant of the primary winding. Leakage -inductance of Coupled Circuits. When dealing with inductively coupled circuits, it is often desirable to split the self-inductance of each winding into two parts. One part, called the leakage inductance, is produced by the leakage flux. The other part is due to that portion of the total flux of self- induction which links both windings. The separation of self- inductance into these two parts often much simplifies problems involving inductively coupled circuits. The treatment of the alternating-current transformer and also of the induction motor is very much simplified by the use of this device. Self-inductance, L, has been defined as the change in the flux linkages ef a circuit per unit change in its current. The flux concerned includes all of the flux produced by the current when it acts alone. Similarly, the mutual-inductance, M, of two circuits has been defined as the change produced in the flux linkages of one circuit by unit change of current in the other. In this case the flux concerned includes only that portion of the total flux which links both windings when one alone carries current. It does not include any leakage flux, since the leakage flux, by its definition, cannot link the second circuit. Leakage-indue- 184 PRINCIPLES OF ALTERNATING CURRENTS tance is defined in a similar manner. The leakage-inductance, Sj of a circuit is the rate of change of the leakage-flux linkages per unit current. The flux concerned in leakage-inductance includes only that portion of the total flux which does not link the second circuit. Leakage-inductance, as well as self- and mutual-inductance, is measured in henrys. The leakage-in- ductance of a circuit is one henry when a change of one ampere in its current causes a change of 10 8 leakage-flux linkages. The leakage-inductance of a circuit in henrys multiplied by 27r/, where / is the frequency, is its leakage reactance in ohms. Just as eiL = LI-JJ is the voltage drop produced in circuit 1 by self- inductance, and CZM = M-TJ is the voltage drop produced in circuit 2 by mutual-inductance, so eis = $1777 is the voltage drop produced in circuit 1 by leakage-inductance. The corresponding root-mean-square or effective values of the voltages are Eu, = 2n/Li/i = coLJi, E*M = 27T/MI, = uMIi, EIS = 2firfSlIi = C0$i/i. If the inductances are expressed in henrys, the currents in amperes and the frequency in cycles per second, the voltages will be in volts. Let (pis and (p 2S be the leakage fluxes for circuits 1 and 2. Assuming that all of the leakage flux links all of the turns of the circuit in which it is produced, the leakage inductances are S, = Ni ^ 10- 8 henrys (25) ai\ S 2 =N 2 d ^ 10- 8 henrys (26) 6^2 where the currents are expressed in amperes. If the leakage fluxes per ampere are constant S, = N^~ 10~ 8 henrys (27) ^i Si = N 2 ^ 10- 8 henrys (28) MUTUAL-INDUCTION 185 Relations Among the Fluxes Corresponding to Self -inductance, Leakage -inductance and Mutual-inductance. Consider two coupled circuits having N\ and Nz turns. Let the inductances be expressed in henrys. Assume the inductances to be constant. This is equivalent to saying that there is no magnetic material in the vicinity of the circuits. For 7\ amperes in circuit 1, circuit 2 being open, ~ 1 1 X 10 8 = (p\ maxwells = total flux linking circuit 1 due to its ^ 1 self -inductance. Cf -j^- 1 1 X 10 8 = 2 maxwells = total flux linking circuit 2 due to its self -inductance. o j~ J 2 X 10 8 = (pzs maxwells = leakage flux linking circuit 2 due ^ 2 to its leakage-inductance. -=;- I z X 10 8 = (p M \ maxwells = flux linking circuit 1 due to the mutual-inductance of circuit 2 on circuit 1. Obviously, i Mi maxwells = part of MR -f" 2S = lL + 190 PRINCIPLES OF ALTERNATING CURRENTS Equation (45) is an equation of induced voltage drops. Equa- tion (46) is an equation of the fluxes producing the voltages given in equation (45). The corresponding voltages and fluxes are placed directly under each other in these two equations. The vector diagram corresponding to equation (44) is shown in Fig. 57. The primary and secondary currents are assumed to be 145 degrees out of phase, with the primary current leading. FIG. 57. The numbers of turns in the primary and secondary windings are taken equal. If the primary selfrinduced voltage drop, JuLJi, is split up into two parts, one, juSJi, due to leakage flux, the other, - N l rr i -fir, due to the component,

In an air-core transformer, I,, and Ei are in quadrature, since there is no core loss. The only losses are the copper losses. These are taken care of toy the Jtfi and 7 2 r 2 drops in the primary and secondary windings. In Fig. 58, // = 7 2 , since the ratio of the turns in the primary and secondary windings was assumed to be unity, i.e., Ni was assumed to be equal to JV 2 - Since 1^ produces the resultant mutual flux,

S 2 by the leakage reactance drops jliXi and jHzXz. Ei is the voltage drop produced in the primary winding by the resultant mutual flux V M R. The corresponding voltage drop in the secondary winding is E 2 . E z , that is the voltage rise, is used on Fig. 59 in place of the voltage drop to keep the diagram more open. Also, since power is absorbed on the primary side of a transformer and delivered on the secondary side, it is better to draw E\ as a voltage drop and Ez as a volt- MUTUAL-INDUCTION 193 age rise. Since E\ and Ez are produced by the same flux, i.e., by the resultant mutual flux

m = 35.71 X 10 8 4.44 X 2000 X 60 = 6.70 X 10 3 maxwells. CHAPTER VII IMPEDANCES IN SERIES AND PARALLEL; EFFECTIVE RESISTANCE AND REACTANCE Impedances in Series. The current in a series circuit is the same in all parts and the resultant voltage drop across the entire circuit is equal to the vector sum of the voltage drops in its component parts. This may be expressed analytically by the following equations. 7 = / 1 = / 2 = /, = etc. (1) 7o = # i + E 2 + # 3 + etc. (2) = 7ozi + 7oz 2 + 7oz 3 + etc. (3) where 7o and Vo are the resultant current and the resultant voltage drop across the entire circuit. 7, E and z with the subscripts 1, 2, 3, etc., are the currents, voltage drops and impedances for the component parts of the circuit. The sum- mations in equations (2) and (3) must be made in a vector sense. It is common practice to use a dot or a short line over a letter, when the latter is used to represent a vector or a complex quan- tity, in all cases where confusion or misunderstanding would result without such designation. In all alternating-current work currents and voltages must be added or subtracted vectorially. No special designation therefore is necessary in most cases to indicate this. However, for the sake of greater clearness, a short line will be used over all letters which represent either vectors or complex quantities. Power is not a vector, neither is resistance nor reactance, but resistance and reactance drops are vectors. Impedance is a complex quantity. Impedances, therefore, must be handled like vectors so far as operations of addition, subtrac- tion, multiplication and division are concerned. Consider a circuit consisting of an inductive impedance, Z L , in series with a capacitive impedance, z c . The diagram of connec- tions is shown in Fig. 60a. The inductive part of the circuit has a 196 IMPEDANCES IN SERIES AND PARALLEL 197 resistance r L and an inductive reactance X L = 2irfL = coL. The capacitive part of the circuit has a resistance r c and a capacitive reactance x c = o~fC = ~~C' The vector diagram of the circuit is shown in Fig. 606. (a) Referring to Fig. 606, E L = IoVr L 2 + x L 2 is the voltage drop across the inductive impedance Z L . This drop is made up of two parts: one, 7 r L in phase with the current, the other, IQX L in quadrature with the current and leading it by 90 degrees. The voltage drop E L leads the current by an angle whose tangent is ~ = ~- EC = ^c 2 -f- x c * is the voltage drop across the capacitive impedance z c . This drop, like the voltage drop E L across the inductive impedance Z L , is made up of two parts: one, I r c in phase with the current, the other, IQX C = ^ in quadrature with the current but lagging it by 90 degrees. The voltage drop E c lags the current by an angle B c whose tangent is x-c _ -1 r c wrcC The resultant voltage drop Vo across the two impedances in series is equal to the vector sum of the voltage drops EL and 198 PRINCIPLES OF ALTERNATING CURRENTS EC- This resultant voltage drop is made up of two parts, /oro = IO(TL H- r c ) and 7 Zo = ?O(ZL #c), respectively in phase and in quadrature with the current. Whether the resultant quad- rature component leads or lags the current depends on the rela- tive magnitudes of the inductive and capacitive reactances, X L and x c . If XL is larger than x c it will lead. On the other hand, if x c is larger than XL it will lag. From Fig. 606 it is obvious that EQ COS 00 = lofQ = Io(r L + r c ) (4) EQ sin 60 = IQXQ = Io(x L - X C ) (5) In general, if there are k impedances in series, it is clear from equations (1) and (2) that Eo cos 8 = I r = 7 (ri + r 2 + . . >+ r k ) (6) E Q sin 0o = IOXQ = IQ(XI + x z + . . . -f x k ) (7) and r = ri + r 2 + . . . -f r* = 2*r (8) TO = Xi + x 2 + . . . + x k = Z\x (9) For a series circuit, therefore, the resultant resistance is equal to the sum of the resistances of its separate parts. Similarly, the resultant reactance of a series circuit is equal to the sum of the reactances of its separate parts. A series circuit, therefore, acts like a simple circuit having a single resistance TQ in series with a single reactance XQ. Resistances are always positive. Reactances, on the other hand, may be either positive or negative according as they are due to inductance or to capaci- tance. It is obvious, therefore, that reactances must be added algebraically, i.e., with regard to their signs. Since EQ cos = / O ro and E sin = loXo are two quadrature components of the voltage Eo, it follows that EQ = 7o\/ro 2 + XQ Z = / Zo (10) and 7 _ The current in amperes is always given by the voltage in volts impressed on the circuit divided by the resultant impedance in IMPEDANCES IN SERIES AND PARALLEL 199 ohms. The resultant impedance z is not the sum of the sepa- rate impedances except in complex. - (xi + x 2 -t- . . . -h r = V(SW 2 + (S*x) 2 (12) Referring to Fig. 606, it will be seen that tan 0o = - (13) cos = ~ (14) z /IC N sin 0o = {*&) ZQ Complex Method. Since the current is the same in all parts of a series circuit, i.e., it is common to all parts of the circuit, it will be taken as the axis of reference, that is, as the axis of reals, In complex the voltage drops E L and E c across the two parts of the series circuit shown in Fig. 60 are "E L = 7oZ L = 7p(r L + jx L ) (16) >E C = IQZC = IO(TC jx c ) (17) V = E L + EC = [Q{ (TL + r c ) + j(x L - x c ) } = 7o(r + jxo) = /o2o (18) zo = r + jx (19) XL Xc XQ /nr\\ tan = i r = ~ ^ U ^ ri + T 4= , = T ^ (21) V(r L + r c ) + (x, - *cY * XL - X C X Q , . In general, if there are k impedances in series Ei = JoZi = Jo(ri + jxi) E 2 = lozz = 7o(r 2 + 3x2) E k = loZk = Io(rk = J { (ri + r z + . . . + r t ) + j(xi +X2+ . . . + (23) 200 PRINCIPLES OF ALTERNATING CURRENTS The two terms, 7o?*o and jloXo, in equation (24), are respectively the active and the reactive components of the resultant voltage drop across the circuit tan * = X f = | (25) TO ^]/ cos 0o = -7^= = = - = -r- 2 * r (26) Vro 2 + zo 2 zo V (Z*r) 2 + sin = j= = = = - ^- (27) Vr 2 + *o 2 zo V (2W 2 + (S?z) 2 Although the resistances and the reactances of a series circuit may be added directly to give respectively the resultant re- sistance and reactance, impedances may not be so added. Impedances must always be added in complex. For example, Z = Zi + Z 2 + . . . + Zjfc can only mean 20 = (ri + r 2 + . . . + r k ) +j( Xl + z 2 + . . . + x k ) (28) It can never mean anything else. The resultant impedance in ohms is (29) + z 2 (30) where the r's and x's are expressed in ohms. When only the ampere value of the current in a series circuit is desired, it is found by dividing the magnitude of the impressed voltage in volts by the magnitude of the resultant impedance in ohms. EQ (volts) T , -^r 4 = / (amperes). (31) z (ohms) The phase of the current with respect to the voltage is fixed by equations (25), (26) and (27), in which is the angle of lead of the voltage with respect to the current. The angle of lead of the current with respect to the voltage would be do. Since a negative angle of lead is equivalent to an angle of lag, the angle Bo in equations (25), (26) and (27) may equally well be considered as the angle of lag of the current with respect to the voltage. Whether actually represents an angle of lead or an angle of lag in any particular case will depend upon its sign as fixed by the IMPEDANCES IN SERIES AND PARALLEL 201 resultant reactance of the circuit. When , as determined by the resultant reactance XQ, is positive, i.e., when the sum of the inductive reactances, 2x L = 22-jrfL, is greater than the sum of the capacitive reactances, S ( x c ) = 2 ' ^ * S actuan< y an an gle of lead of the voltage with respect to the current or an angle of lag of the current with respect to the voltage. In this case the volt- age leads the current or the current lags the voltage. The circuit as a whole is inductive. When is negative, it is actually an angle of lag of the voltage with respect to the current or an angle of lead of the current with respect to the voltage. In this case the voltage lags the current or the current leads the voltage. The circuit as a whole is capacitive. Example of Impedances in Series. A circuit, which consists of two impedances and a non-inductive resistance in series, is connected across a 230-volt, 60-cycle circuit. One of the im- pedances has a self -inductance LI = 0.1 henry and a resistance 7*1 = 5 ohms. The other impedance has a capacitance C = 100 microfarads and a negligible resistance. The non-inductive resistance r& = 10 ohms. What is the current and what is its phase with respect to the impressed voltage? How much power does the circuit absorb and what is the power-factor? What are the potential drops across each of the impedances and the non- inductive resistance, and what are the phase angles of these drops with respect to the current in the circuit? What are their phase angles with respect to the impressed voltage? 7*1 = 5 ohms xi = 27r60 X 0.1 = 377 X 0.1 = 37.7 ohms. r 2 = 10 6 r 3 = 10 ohms. x z = zo = (n + r 2 + r 3 ) + j(xi + x z + x 3 ) = (5 + + 10) + j(37.7 - 26.53 + 0) = 15+J11.17 zo = V(15) 2 + (11. 17) 2 = 18.70 ohms. 230 /o = - = 12.30 amperes. 202 PRINCIPLES OF ALTERNATING CURRENTS Power-factor = cos = - = -^|n = 0.8021 ZQ 1O.7U 0o = 36.67 degrees. Total power = P = 230 X 12.30 X 0.8021 = 2264 watts. The voltage impressed on the circuit leads the current by 36.67 degrees or the current lags the voltage by the same angle. The circuit as a whole is inductive, since XQ is positive. EI (drop across Zj) = = 12.30 X V(5) 2 + (37.7) 2 = 12.30 X 38.03 = 467.8 volts. 07 7 tan 0! = ^r- = 7.54 o 0i = 82.45 degrees. The voltage drop Ei, across the inductive impedance, leads the current by 82.45 degrees. Ez (drop across zz) = IoZ 2 = 12.30 X A/(0) 2 + (26.53) 2 = 12.30 X 26.53 = 326.3 volts. -26.53 tan 2 = ~ -- = oo 2 = -90 degrees. The voltage drop Ez, across the capacitive impedance, lags the current by 90 degrees. (drop across 2 3 ) = = 12.30 X V(lW 2 ~+W 2 = 12.30 X 10 = 123.0 volts, tan 03 = j Q = 3 = degrees, IMPEDANCES IN SERIES AND PARALLEL 203 The voltage drop E?., across the non-inductive resistance, is in phase with the current. Since the voltage impressed across the circuit leads the current by 36.67 degrees, to get the phase of the voltage drops across the impedances and the non-inductive resistance with respect to the impressed voltage Fo, 36.67 degrees must be subtracted from the phase angles of the drops with respect to the current. Therefore, the phase angles of the drops with respect to the voltage are 0i' = 82.45 - 36.67 = 45.78 degrees. 2 ' = -90.0 - 36.67 = -126.7 degrees. 3 ' = o.O - 36.67 = - 36.67 degrees. Ei leads the voltage drop Fo by 45.78 degrees. E% leads the voltage drop Fo by 126.7 degrees or lags it by_126.7 degrees. Es leads the voltage drop Fo by 36.67 degrees or lags it by 36.67 degrees. It should be noted that the voltage drop across each impe- dance is much greater than the voltage drop across the entire circuit. This is always possible when a circuit contains both inductance and capacitance in series with a low resistance. In this particular problem the drops across the two impedances are 82.45 + 90.0 = 172.45 degrees apart in phase and therefore contribute little to the resultant voltage across the circuit. More will be said about this condition under " Resonance." The preceding problem could have been solved by the complex method, but in this particular problem the ordinary algebraic method gives a somewhat shorter solution. Another Example of Impedances in Series. A certain circuit consisting of two impedances, one of which has a resistance of 10 ohms and an inductive reactance of 10 ohms, takes a leading current of 15 amperes at 0.8 power-factor when connected across a 200- volt, 60-cycle circuit. What are the resistance and reactance of the second impedance? The use of the complex method will give the most direct solu- tion in this case. Call the known impedance Zi and the unknown impedance z 2 . Take the current as the- axis of reference. Then 204 PRINCIPLES OF ALTERNATING CURRENTS To = 15(1 + JO) = 15+ JO Fo = 200(cos 0o - j sin ) = 200(0.8 - jO.6) = 160 - J120 V 160 - J120 - Jo = 10.67 - J8.0- r = 10.67 ohms. XQ = 8.0 ohms. r 2 = 10.67 - n = 10.67 - 10.00 = 0.67 ohm. Xz = 8.0 x\ = -8.0 - 10.0 = -18.0 ohms. The second impedance is therefore capacitive, since Xz is negative. Its complex expression is z 2 = 0.67 - J18.0 It has a resistance of 0.67 ohm and a capacitive reactance of 18.0 ohms. If Xz is expressed in ohms and Cz in microfarads -10 6 * 2 27T/C 2 -10 6 27r60(-18.0) = 147.4 microfarads. Series Resonance. A series circuit containing inductance and capacitance is said to be in resonance when the sum of the inductive reactances is equal to the sum of the capacitive reac- tances. For resonance therefore . X = zi + x 2 + z 3 + . . . + x k = (32) = Zx L - 2xc = (33) Consider a circuit containing a resistance r, an inductance L and a capacitance C, in series. E I = IMPEDANCES IN SERIES AND PARALLEL 205 For resonance,

    or co 2 LC = 1 (34) coC and (35) With constant frequency, resonance may be produced by adjusting either L or C or both, equation (34), or with L and C constant, it may be produced by adjusting the frequency, equa- tion (35). Since the resultant reactance is zero, the current at resonance is in phase with the voltage impressed across the entire circuit and follows Ohm's law. It is given by ' H '-7 Since the resultant reactance at resonance is zero, the resultant impedance is a minimum and is equal to the resultant resistance of the circuit. The current is therefore a maximum and is limited only by the resultant resistance of the circuit. It is entirely independent of the magnitudes of the inductive and capacitive reactances. The drop in potential due to the inductance is 7coL. Due to the capacitance it is These two potential drops at resonance are without effect on the resultant potential drop across the circuit as a whole. They may be much greater than the voltage impressed on the entire circuit and may easily reach excessive values when the resultant resistance of the circuit is small in comparison with its inductive and capacitive reactances. Series resonance, for constant potential circuits, is usually a very undesirable condi- tion, on account of the excessive current and high voltages that may be produced when the resistance is small compared with the inductive and capacitive reactances. If a low resistance series circuit is tuned for resonance by varying the capacitance, and curves of current and of voltage drops across the inductance and capacitance are plotted against capacitance, they will all show very marked peaks at resonance. The maximum current and maximum voltage drop across the inductance will occur at resonance, but the maximum voltage 206 PRINCIPLES OF ALTERNATING CURRENTS drop across the capacitance will not occur at resonance. If the capacitance is constant and the inductance is varied, the maxi- mum drop across the capacitance will then occur at resonance, but now the maximum voltage drop across the inductance will not occur at resonance. The steepness with which the curves rise as resonance is approached depends upon the magnitude of the resistance of the circuit compared with the magnitudes of the inductive and capacitive reactances. The point of resonance is most pronounced when the resistance is low in comparison with the reactances. V= 100 volts / = 60 cycles L = 0.1 henry r = 1, 4 and 8 ohms for curves tt b and c respectively 50 60 70 80 90 100 110 120 Capacitance in Microfarads FIG. 61. When a series circuit is tuned for resonance by varying either L, C or /, low resistance permits very sharp tuning. With high resistance, the resonance curves become flattened and sharp tuning is then impossible. With high resistance compared with the reactances, the point of resonance will not be well defined. This is illustrated in Fig. 61. Three current curves, showing series resonance, are plotted in Fig. 61 for a 60-cycle, 100-volt circuit having an inductance of 0.1 henry and a variable capacitance. Curve a is for a resistance of 1 ohm. Curves b and c are for resistances of 4 and 8 ohms respectively. The lower parts of the three curves so nearly IMPEDANCES IN SERIES AND PARALLEL 207 coincide, that for b and c only the portions of the curves near resonance are plotted. For curve a the voltage drop across "the inductance is 3770 volts at resonance. It is one-quarter and one-eighth this amount at resonance for curves b and c respec- tively. 1800 1600 1400 Juoo 8*1000 c 5 a BOO 400 200 -200 -400 -600 !- l-iooo g-1200 -1400 -1600 -1800 0.4 0.6 0.8 1.0 1.2 14^^ 1.8 2.0 2.2 24 Frequency, Cycles ! per sec. xjlO' 6 Voltage = 1 volt Capacitance - 0.00235 microfarad Inductance = 377 microhenrya Resistance 25 ohms 30 40 10 FIG. 62. Fig. 62 shows the effect of varying the frequency to produce resonance in a circuit containing constant resistance, constant inductance and constant capacitance in series. Curves of inductive reactance, X L = uL, capacitive reactance, x c = 208 PRINCIPLES OF ALTERNATING CURRENTS resultant reactance, X L x c = uL > and current are Ceo plotted for a circuit having r = 25 ohms, L = 377 microhenrys and C = 0.00235 microfarads. Resonance occurs at a fre- quency of 1.69 X 10 5 cycles. Free Period of Oscillation of a Resonant Circuit Containing Constant Resistance, Constant Inductance and Constant Ca- pacitance in Series. The natural or free oscillation frequency of a circuit containing constant resistance, constant inductance and constant capacitance in series is , = _, JL (36) (See equation 124, page 159.) When r 2 is small compared with 4L equation (36) reduces to C Therefore, the resonant frequency of a series circuit, which has low resistance compared with the ratio of its inductance and capacitance, is the same as its free oscillation frequency. This makes it possible to tune a series circuit containing inductance and capacitance to have a sharp current maximum for a definite frequency. A circuit on which a non-sinusoidal electromotive force is- impressed may be in resonance for a certain harmonic and yet be very far from resonance for the fundamental or other har- monics (see equation (35), page 205). A circuit can be in resonance for only one frequency at the same time. By proper tuning, as by varying either the inductance or the capacitance or both, it is possible to exaggerate any harmonic in the current for which there is a corresponding harmonic in the impressed voltage. Tuning is made use of in radio-telegraphy in order to make a receiving circuit respond to electromagnetic waves of a definite frequency. On account of the very high frequency used in radio-communication, the resistance of radio circuits is usually very small in comparison with their reactances, making very sharp tuning possible, IMPEDANCES IN SERIES AND PARALLEL 209 A Problem in Resonance. A series circuit, which consists of a non-inductive resistance of 5 ohms, an impedance of 1.5 ohms resistance and 0. 1 henry inductance and a variable capacitance, is connected across a constant potential 220-volt, 60-cycle cir- cuit. If the capacitance, starting with a value too small to produce resonance, is gradually increased, for what value of the capacitance will the voltage drop across it have the greatest value? For what value of the capacitance will the voltage drop across the inductance have the greatest value? As the capacit- ance is increased, which of the two voltage drops will have its maximum value first? XL = 377 X 0.1 = 37.7 ohms V L = IwL = Ix L where V c and V L are the potential drops across the capacitance and inductance respectively. Vro 2 + (XL - Xc) 2 Vc Ixc / V -t ' lVr.'+( t - *c) 2 J VrV+ (X L - x c Y} X (XL ~ X C ) 2 dx c ro 2 + (X L - x c ) 2 = r 2 + (X L - x c ) 2 + X C (X L - xc) = r 2 + x L 2 2x L x c + x c 2 + X L X C x c 2 = X c = ro 2 + x L 2 + 37.7 = 38.8 ohms. 10 6 10 377 X 38.8 = 68.4 microfarads. The capacitance for maximum voltage drop across the capacitance is therefore 68.4 microfarads. 14 210 PRINCIPLES OF ALTERNATING CURRENTS The maximum voltage drop across the capacitance is 220 V c (maximum) = , = = X 38.8 V(6.5) 2 + (37.7 - 38.8) 2 220 X 38.8 6.593 = 1293 volts. Since L is constant, the voltage drop across the inductance must be a maximum when the current is a maximum. The current is a maximum at resonance. Therefore \7 V L (maximum) = - X L TO - 22 X377 6.5 X = 1276 volts. At resonance V c and V L are equal in magnitude. V c at resonance is therefore 1276 volts, which is less than its maximum value. X C = X L C (at resonance) = = 70.3 microfarads. The voltage drop across the capacitance reaches its maximum before the voltage drop across the inductance reaches its maxi- mum. As resonance is approached (C increasing, x c decreasing) the current rises very rapidly for a small decrease in x c . The increase in the current will more than balance the decrease in x c until the point of resonance is nearly reached. As resonance is approached, the current curve flattens out preparatory to de- creasing after resonance is passed. At some point just before resonance is reached, the decrease in x c will just balance the in- crease in 7. The maximum voltage drop across the capacitance will occur at this point. When the resistance is small in com- parison with X L , the maximum voltage drop across the capacitance will occur very near the point of resonance. When the resistance is large, the maximum voltage drop across the capacitance may occur quite a bit before resonance is reached. In the problem just solved, what additional resistance will it be necessary to place in series with the circuit in order that the maximum voltage drop across the capacitance shall be limited to 500 volts. IMPEDANCES IN SERIES AND PARALLEL 211 The capacitive reactance for maximum voltage drop across the capacitance is (equation (38)) ro 2 + x# Xc " ~ The resultant reactance is XQ = XL - Xc -.= XL - ro 2 + X L * r 2 The resultant impedance is The resultant impedance drop across the condenser is V c = Ix c = - - X x c ZQ 220 , r 2 -f x L 2 _ nn ., = X = 500 volts Vr 2 + x L 2 = 500 r 220 X L 37.7 /500\ 2 / /500\ 2 \220/ \ \220/ = 18.5 ohms total resistance. Added resistance = 18.5 - 1.5 - 5.0 = 12.0 ohms. Impedances in Parallel. The voltages impressed across the branches of a circuit consisting of a number of impedances in parallel are equal. The resultant current taken by the circuit is equal to the vector sum of the component currents taken by the branches. This may be expressed analytically as follows: y = EI = E 2 = E* = etc. (39) = IiZi IzZz = /s2s = etc. (40) 7o =7i + /2 + 73_+etc. (41) ' '0 ' / j f\\ 21 22 23 f i + i+ i + etc.} (43) L 3l 22 23 212 PRINCIPLES OF ALTERNATING CURRENTS Consider a circuit consisting of two impedances, z\ and 2 2 , in parallel as indicated in Fig. 63a. The impedance, z\, of branch 1 has a resistance ri and an inductive reactance Xi = coZ/i. The impedance, 22, of branch 2 -1 has a resistance r 2 and a capacitive reactance x 2 = g ' Since the impedances are in parallel the same potential, F , is impressed across their terminals. ?! (a) FIG. 63. The graphical construction for this circuit is shown in Fig. 636. Fo is the voltage drop impressed on the circuit and is also the potential drop across each of the branches. (Equation (39).) Since these drops are equal and in phase, F must be the common side of the right-angled voltage triangles. The intersection of the vectors representing the resistance and reactance drops in each of the branches of the circuit must therefore lie on a circle drawn with the vector Fo as diameter. All voltages are drawn to a common scale. Likewise all currents are drawn to a common scale, but the scales for voltage and current may be, and indeed usually will be, different. The sides OD and DF of the resultant triangle ODV Q are the resultant resistance^ and reactance drops, r 7o and z 7o, respec- tively. Knowing 7 , the resultant resistance, r , and the re- sultant reactance, XQ, may be found. From the diagram (Fig. 636) /o cos = Ii cos 0i + Ii cos 02 /o sin = /i sin 0] + 7 2 sin 2 (44) (45) IMPEDANCES IN SERIES AND PARALLEL 213 These equations may be written V \X l _l_ * \S '* ~ /\ /\ ~ A. ~ 2o ZQ Z\ Z\ Z% Z% V V r V 1 Tl 4- Tz 1 (46) (47) (48) (49) (50) CK11 'OA 0, >l)j 21^.2' 21 2/ TQ -p XQ 1 7"i -f- Xi r 2 -f- X 2 I 2g 2o 2i Zi Zi 2 2 T7 X X - V 1 Xl I ^ 1 ^OA 2 . 2 ' 1 21 2' 21 2l TO 'h ZQ I** T Zi r 2 ~r #2 J Hence r 2 + z 2 ri 2 + ii 2 ' r 2 2 + x 2 2 In general, if there are k branches in parallel P --^FTP (53) It must be remembered that x is positive for inductance and negative for capacitance. Therefore the summations in equa- tions (45), (48), (49), (51) and (53) must be made in an algebraic sense. Conductance, Susceptance and Admittance of a Circuit. The expression r -4^ = 9 ' ' ' (54) is known as the conductance of a circuit and is denoted by. the letter g. When there are a number of circuits in parallel, their resultant conductance, from equation (52), is given by the sum of their separate conductances. Therefore go = 0i + 2 + . . . + Qk = X*g (55) The expression - 6 (56) 214 PRINCIPLES OF ALTERNATING CURRENTS is known as the susceptance of a circuit and is denoted by the letter 6. When there are a number of circuits in parallel, their resultant susceptance, from equation (53), is given by the sum of their separate susceptances. Since reactance, x, may be either positive or negative, according to whether it is inductive or capacitive reactance, it is obvious from the expression for sus- ceptance, equation (56), that susceptance is likewise either positive or negative. Inductive susceptance is positive. Ca- pacitive susceptance is negative. Since susceptance may be either positive or negative, susceptances must always be added in an algebraic sense, i.e., with regard to their signs. bo = bi + &2 + . + bk = ^b (57) Since 7 cos = V go and 7 sin = Vob are two quadrature components of the resultant current 7 (see Fig. 636), it is obvious that /o = V cos = > sin = (61) go . 2/o 2/o Vector Method. From equation (43), page 211 fl 1 1 Fo I 7 + ~ 4- r + etc. 7o = Fo \ ~ 4- - 4- ~ + etc. (62) [ Z\ Z 2 Zz Putting the z's in their complex form, equation (62) becomes / = yl--L + _l_ + __l +et(j 1 (63) Ifj 4- JXi r 2 + jx 2 r 3 + jx s Rationalizing, it = Fo _ x \r\ 4- jxi r 3 + jx 3 r 3 - 4- X! r . _ +( r -7f^- J '77f^') + etc -! (64) == Vo { (gi + 02 + ft + etc.) - j(6x + 6 2 + 6 3 4- etc.) | (65) = Fo (0o - jbo) go = 0i + 02 4- #3 4- etc. (66) 60 = 61 4- b 2 4- 63 4- etc. (67) It should be noted that while impedance is z = r 4- jz, admit- tance is y = g jb. Since inductive reactance is positive and capacitive reactance is negative, the impedance of an inductive circuit is actually z = r -}- jx and the impedance of a capacitive circuit is actually z = r jx. Since inductive susceptance is positive and capacitive susceptance is negative, the admittance of an inductive circuit is actually y = g jb and the admittance 216 PRINCIPLES OF ALTERNATING CURRENTS of a capacitive circuit is actually y = g + jb. The signs of the second terms in the complex expressions for impedance and admittance are opposite. For impedances in parallel, the resultant conductance is the sum of the component conductances and the resultant suscep- tance is the algebraic sum of the component susceptances. Con- ductances, susceptances and admittances, as has already been stated, are measured in reciprocal ohms, i.e., in mhos. Admittances cannot be added directly. Admittance is a complex quantity. Admittances, therefore, like impedances, can only be added in complex. For example: 2/0 = y\ + 2/2 + 2/3 + . . + 2/Jfc can only mean 2/0 = 2/1 + 2/2 H- 2/3 + . . . + 27* = (01 + #2 + 03 + . - - + fiffc) -j(bi + 62 + 6 3 + . . . +6*) (68) It can never mean anything else. Resistance and Reactance in Terms of Conductance and Susceptance. It is often necessary, when dealing with a parallel circuit, to determine its resultant resistance and reactance, i.e., the resistance and reactance of the simple series circuit which would replace it so far as total power, resultant current and power- factor are concerned. The expressions for r and x in terms of g and b may be found from the relation between impedance and admittance z = r + jx y g - 1 g + ft -jrJFp + JjTfp (68) Evidently IMPEDANCES IN SERIES AND PARALLEL 217 Polar Expression for Admittance. It is often convenient, when dealing with the more complicated problems such as those associated with the transmission line, to use the polar expression for admittance. The polar expression for impedance is given in Chapter V, pages 136, 150 and 170. J -'-' + 6 l j (72) (73) The expression y\~~0 or y\B is known as the polar expression for admittance. The angle is the angle of the admittance and is determined by the relation = tan- 1 - (74) g Admittance is a scalar quantity, y, multiplied by an operator = (cos j sin 0), which rotates through an angle 0. When vector voltage is multiplied by admittance either in complex or polar form, it gives vector current in its proper phase relation with respect to the voltage, regardless of the axis to which the voltage is referred. An Example of the Solution of a Parallel Circuit by the Use of Conductance, Susceptance and Admittance. An impedance coil of 10 ohms resistance and 0.1 henry inductance, a con- denser of negligible resistance and 100 microfarads capacitance and a non-inductive resistance of 20 ohms are connected in paral- lel across a 200-volt, 60-cycle circuit. What are the resultant current and power taken by the impedance coil, condenser and resistance in parallel? What is the resultant power-factor of the circuit? What current and power does each branch of the circuit take? What is the power-factor of each branch? 218 PRINCIPLES OF ALTERNATING CURRENTS u = 2 impressed on the circuit by 12.95 degrees. Ii = V Q yi = 200(1 + jO) (0.006575 - jO.02479) = 1.315 - J4.958 /i = \/(1.315) 2 + (4.95S) 2 = 5.13 amperes. Pi = TV0i = (200) 2 X (0.006575) = 263 watts. cos 0, = 1 315 = -^ ^ - = 0.2563 = power-factor for branch 1. O. lo 6 1 = 75.15 degrees. Since tan 0i = is positive, 0i is positive and the current 7i J7i lags the voltage Fo impressed on the circuit by 75. 15 degrees. J 2 = = 200(1 + jO) (0 + jO.03770) = + J7.540 I 2 = V(0) 2 + (7.540) 2 = 7.54 amperes. P 2 = TV<72 = (200) 2 X (0) = watts. cos 62 = = =-^ = 0.0 = power-factor for branch 2. = 90 degrees. 220 PRINCIPLES OF ALTERNATING CURRENTS Since tan 62 = is negative, 2 is negative and the current 7 2 02 leads the voltage Vo impressed on the circuit by 90 degrees. / 3 = To 3 = 200(1 + jO) (0.0500 - JO) = 10.00 - jO / 3 = V(io.oo) 2 + (o) 2 = 10.00 amperes. = (200) 2 X (0.0500) = 2000 watts cos 03 = 10.00 - , . TTTT = 1.0 = power factor for branch 3. 6 3 = degrees. The current 7 3 is in phase with the voltage Fo impressed on the circuit. Po = Pi + P 2 +P 3 = 263 + + 2000 = 2263 watts. Parallel Resonance. A parallel circuit is said to be in reson- ance when its resultant susceptance is zero. Io = V y Q = F (<7o - J6 ) (75) cos (9o = 7 =M= (76) At resonance 6 = and /o = Fofifo (77) cos 6 Q = 1 (78) For fixed resultant conductance, the resultant admittance of a parallel circuit is a minimum at resonance, being equal to the resultant conductance. The resultant current is also a minimum. Consequently the resultant impedance is a maximum and would be infinite if it were possible to have the resultant conductance zero. For the resultant conductance to be zero the resistances of IMPEDANCES IN SERIES AND PARALLEL 221 all parallel branches of the circuit would have to be zero. For a series circuit with fixed resistance, the resultant impedance is a minimum at resonance and is equal to the resultant resistance. For both series and parallel circuits, the power-factor is unity at resonance and the resultant current is in phase with the voltage. For a series circuit in resonance, the reactive components of the voltage drops across the component parts of the circuit add up to zero. For a parallel circuit in resonance, the reactive components of the currents in the parallel branches of the circuit add up to zero. For a parallel circuit with constant impressed voltage, there can be no rise in voltage across any branch of the circuit at resonance. Although the resultant current is a minimum, the currents in the inductive and capacitive branches may be large if their impedances are small. Since the impedance is a maximum for parallel resonance, parallel resonance for a constant current circuit, i.e., for a circuit in which the current is maintained con- stant, would be undesirable in most cases since the potential drop, VQ = , across it at resonance would in general be excessive. The currents in the inductive and capacitive branches probably would also be excessive. In general, parallel resonance is a highly desirable condition for a constant potential power circuit, since it gives a minimum resultant line current for a given amount of power transmitted. All commercial power circuits, except those for street lighting with series arc or series incandescent lamps, are operated at constant potential and the loads are in parallel. A few series power circuits, other than those for street lighting, are used abroad, but these are all direct-current circuits. Parallel reson- ance not only permits the power to be transmitted with a mini- mum line loss, i.e., with a minimum 7 2 r loss in the line, but what is more important it also permits the use of the minimum gen- erator capacity for a fixed amount of power transmitted. The output of alternating-current generators is determined by the product of current and voltage, i.e., by volt-amperes, and not by watts. Their output is therefore a maximum at unity power- factor, which corresponds to resonance. Special devices are used in many cases to raise the power-factor of commercial 222 PRINCIPLES OF ALTERNATING CURRENTS power circuits by producing resonance or partial resonance by placing capacitive loads in parallel with the existing commercial inductive loads. The saving in generator and transformer capacity, line copper and line transmission losses will in many FIG. 64. cases much more than balance the cost of the power-factor correcting apparatus. The saving in space occupied by the generating and distributing apparatus is an important factor. Series resonance would be a highly desirable condition for a constant-current power circuit, since it would make the resultant IMPEDANCES IN SERIES AND PARALLEL 223 voltage impressed across the circuit a minimum for a fixed amount of power transmitted. With the current maintained constant, there would be little danger of excessive voltage drops across the component parts of the circuit. Fig. 64 shows the effect of varying the frequency to produce parallel resonance in a circuit consisting of two branches in parallel, one containing an inductance, L = 377 microhenrys, in series with a resistance, r L = 10 ohms, the other containing a capacitance, C = 0.00235 microfarads, in series with a resistance, r c = 10 ohms. The impressed voltage is 1 volt. The admit- tance, which due to the low resistances is substantially equal to the resultant susceptance, and also the current are a minimum at resonance, which occurs at about 1.7 X 10 5 cycles. The con- stants of the circuit for which Fig. 64 is plotted are of magnitudes such as might occur in radio work. Networks consisting of series-parallel branches tuned to be in series resonance for one frequency and in parallel resonance for another are of great importance in radio work and in "carrier systems" of wire telephony, where they are used to damp out waves of one frequency and to let through those of another frequency. More will be said of this under "Impedances in Series-parallel. " The curves in Fig. 64 should be compared with those for series resonance in Fig 62, page 207. Although minimum resultant current occurs in a constant- potential parallel circuit at resonance when the conductances of the parallel branches are constant, minimum resultant current does not occur at resonance if the resistances instead of the conductances of the parallel branches are constant. Since conductance is a function of reactance as well as of resistance, when the resistances of the parallel branches are constant the conductances of these branches will not remain constant when their reactances are varied to produce resonance. Consider a circuit with two parallel branches on which a constant voltage at constant frequency is impressed. Let one branch consist of a constant resistance in series with a constant inductance. Let the other branch contain a constant resistance in series with a variable capacitance. The diagram of connec- tions of the circuit is shown in Fig. 64a. 223a PRINCIPLES OF ALTERNATING CURRENTS Refer to Fig. 646. The current I L in the inductive branch is fixed in direction and magnitude with respect to the voltage V. V= constant f= constant A X L - constant ?*/,= constant X = variable = constant FIG. 64a. When the capacitance of the condenser is infinite, x c is zero and the current I c in the condensive branch is in phase with the im- pressed voltage V. Under this condition, V = I c \/r c 2 +x c 2 = I c r c . when C=.A<.oc / Q when C = FIG. 646. If the capacitance is now decreased, I c will decrease and will lead the voltage V which is still equal to I c \/r c 2 + x c 2 , but in this case x c is not zero. As the capacitance is decreased, the end of the vector representing the I c r c drop will trace a semicircle with the impressed voltage V as a diameter. This semicircle is shown dotted in Fig. 646. Since r is constant, I c must vary directly as I c r c . The end of 7 C , therefore, will also trace a semicircle. The diameter of this semicircle is the vector 7 C for the condition C = c . This circle is shown by a dot-and- dash line. Since 7 is found by adding to I c a vector of fixed magnitude and fixed direction, if l c travels on a semicircle, 7 must also travel on a semicircle. This latter semicircle is shown by a full line. Its diameter lies between the end of the vector 7 for the condition C = and the end of the vector I L . Resonance will occur when the vector 7 is in phase with V. As the capacitance of the condenser circuit is decreased from infinity, resonance will first occur when the end of 7 lies at the point 6. If the capacitance is further decreased, 7 will decrease IMPEDANCES IN SERIES AND PARALLEL 2236 and resonance will again occur when the end of 7 lies at the point a. Neither of these resonant conditions corresponds to minimum IQ. Minimum 7 will not occur until the capacitance is further decreased to the value which makes the vector 7 normal to the semicircle c a 6, on which its end lies. When 7 has its minimum value, the extension of the vector which represents it will pass through the center of the semicircle cab. A line drawn between the end of the vector I L and the point c is equal in magnitude and direction to the vector represent- ing the current taken by the capacity branch when the resultant current 7 is a minimum . The conditions which produce minimum resultant current can easily be calculated from Fig. 646. Since the diameter of the circle on which 7 swings is equal to V V , when I L sin 6 is equal to s~ this circle will be tangent to the line a6, Fig. 646. Under this condition, there will be only one value of the condenser capacitance which will produce resonance. V When L L sin B L is greater than , resonance cannot be produced by any value of the condenser capacitance. When resonance is produced in power circuits by use of an over-excited synchronous motor which takes a leading current, resonance corresponds to the condition of minimum resultant current since the equivalent conductances of the loads are con- stant for any fixed loads. In radio circuits resonance produces practically minimum current since in high-frequency radio circuits the resistances are small compared with the reactances. Example of Parallel Resonance. An impedance of 0.1 henry inductance and 10 ohms resistance and a capacitance of 70 micro- farads and negligible resistance, are connected in parallel. For what frequency will the circuit act like a non-inductive resistance? What will be the value of this resistance? X L = 27T/L = 6.284 X / X 0.1 = 0.6284 / ohms. -1 -10 6 _ 1 For the circuit to act like a non-inductive resistance, the result- ant susceptance of the circuit must be zero. For the susceptance to be zero, the circuit must be in resonance. For resonance 224 PRINCIPLES OF ALTERNATING CURRENTS 26 = X L -X C 0.6284/ 1430 = 100 + 0.3949/ 2 1330 X3949 = 58.03 cycles X L = at resonance = 2u- X 58.03 X 0.1 = 36.46 ohms -L r c 10 -I- _ - n !\2 I ff\\Z I xv. 2 / (10) 2 + (36.46) 2 (0) = 0.006994 -JO _!_ 1_ ZQ ~ y ~ 0.006994 - jO = 143.0 + JO r = 143.0 ohms. Impedances in Series-parallel. When a circuit consists of a number of series elements, some of which are made up of two or more impedances in parallel, the elements containing the impedances in parallel must be replaced by their equivalent simple impedances by means of equations (70) and (71), page 216. The circuit may then be solved like a simple series circuit. An example will make this clear. Two impedances z\ and z z in parallel are connected in series with a third impedance z 3 . How much power and current will the entire circuit take when it is connected across 200 volts? What will be the currents in the impedances z\ and 2? IMPEDANCES IN SERIES AND PARALLEL 225 z\ = 5 + j2 ohms. 2 2 = 6 j4 ohms. 23 = 1 + J3 1 n x, 9 I M 9 v*. 91 M 9 2] rj" T #l TI T 3Ji 5 2 (5) 2 + (2) 2 J (V) + (2) 2 = 0.1723 - jO.0690 mho. + 3 (6) 2 + (4) 2 = 0.1154 + jO.0769 mho. The resultant admittance of z\ and z 2 in parallel is $12 = yi + 2 = 0.2877 + J0.0079 = 0.2877 - j(- 0.0079) mho. The general form for admittance is y = g jb. Therefore 612 is negative, as indicated. The resultant impedance of the impedances z\ and 22 in parallel is 912 &i2 12 (<7i 2 ) 2 + (6i 2 ) 2 ''(ait) 1 + (6i 2 ) 2 0.2877 -0.0079 + r. (0.2877) 2 + (0.0079) 2 ' J (0.2877) 2 + (0.0079) 2 3.475 - jO.0954 ohms 212 = V (3.475) 2 + (0.0954) 2 = 3.476 ohms. The resultant impedance of the entire circuit is 20 = 23 + 2i2 = (1 +j3) + (3.475 - jO.0954) = 4.475 -f J 2.905 ohms. 15 226 PRINCIPLES OF ALTERNATING CURRENTS Take the impressed voltage drop F as the axis of reals. Then Fo =200+ JO j Fo = 200 + JO z 4.475 + J2.905 200 + jO 4.475 - J2.905 4.475 + J2.905 4.475 - j2 905 = 31.45 - J20.42 amperes. Jo = \/(31.45) 2 + (20.42) 2 = 37.5 amperes. Po = total power = 200 X 31.45 + X (-20.42) = 6290 watts. = JoVo = (37.5) 2 X 4.475 = 6290 watts. The voltage drop across the impedances z\ and z 2 in parallel is l2 = o = (200 + JO) - (31.45 - J20.42)(l + J3) = 107.3 - J73.93 volts. F 12 = V(107.3) 2 + (73.93) 2 = 130.4 volts. = /ozi2 = 37.5 X 3.476 = 130.4 volts. 130.4 130.4 " V(5) 2 + (2p ~ 5.39 = 24.19 amperes. Fi2 = 130.4 = 130.4 22 " VW+~W ~ 7.21 = 18.08 amperes. The fact that series resonance gives minimum impedance for fixed total resistance and parallel resonance gives maximum impedance for fixed total conductance has already been men- tioned. This difference between the impedance of a series circuit and of a parallel circuit at resonance is of great importance in radio-telegraphy as, by its use, it is possible to so tune a receiv- ing circuit that it will respond to a definite frequency and at the same time will suppress an undesirable frequency. This is IMPEDANCES IN SERIES AND PARALLEL 227 accomplished by the use of what is known in wireless work as coupled circuits. Such networks consist of series and parallel parts which permit of independent tuning for series resonance for the desired frequency and for parallel resonance for the frequency to be suppressed. A simple coupled circuit is shown in Fig. 65. L 2 is an inductance. Ci and C 3 are capacitances. D is a detector. The electric oscillations received by the antenna are impressed on the circuit at E. With CB fixed, L 2 and Ci are first adjusted for parallel reson- ance for the frequency to be suppressed. Their resultant par- allel impedance for this frequency will, therefore, be a maximum. Since the detector circuit is in series with L 2 and Ci in parallel, FIQ little current of the frequency to be suppressed will flow through the detector. 3 is now ad- justed to produce series resonance in the series circuit con- sisting of the detector and Ca in series withL 2 and C\ in parallel. This makes the impedance of the circuit as a whole a minimum for the frequency to be detected. By this double series-parallel tuning, minimum impedance is produced in the detector circuit for the frequency to be detected and maximum impedance is produced for the frequency to be suppressed. The above method of tuning assumes that the frequency to be suppressed is higher than the one to be detected. If the fre- quency to be detected were the higher, C 3 would have to be replaced by an inductance. An Example of Series-parallel Tuning in a Radio Circuit to Prevent Response to an Interfering Frequency. The wireless receiving circuit shown in Fig. 65 is to be adjusted to respond to a frequency of 50,000 cycles and to damp out an interfering frequency of 100,000 cycles. L 2 is an impedance of 10,000 micro- henrys self-inductance and 80 ohms resistance. D is a ticker detector having negligible inductance and 15 ohms resistance. The circuit consisting of L 2 and C\ in parallel will first be adjusted for parallel resonance for 100,000 cycles. Then, with- out changing L 2 or Ci, the capacitance Cs will be adjusted to 228 PRINCIPLES OF ALTERNATING CURRENTS produce series resonance in the circuit consisting of the detector D and the capacitance Ca in series with the capacitance C\ and the inductance L 2 in parallel. The resistances of the capacitances will be assumed to be negligible. At 100,000 cycles 22 = 80 + J10- 2 X 27T X 10 5 = 80 + j6280 ohms. At 50,000 cycles z 2 = 80 + jlO- 2 X 27! X 5 X 10 4 = 80 +J3140 ohms. For parallel resonance of L 2 and Ci at 100,000 cycles, the resultant susceptance of the circuit consisting of L 2 and Ci in parallel must be zero. Therefore, since the resistances of the capacitances are assumed to be negligible 2 9T co 2 L io- 2 farads , -,, farad. (80) 2 + (6280) : Ci = 0.0002537 microfarad. At 50,000 cycles 1 80 3140 22 (80) 2 + (3140) 2 ''(SO) 2 + (3140) 2 = 8.1 X IO- 6 - J318 X IO- 6 mho. yi = \ = + j2ir X 5 X IO 4 X 0.2537 X 10~ 9 = +J79.7 X IO- 6 mho. 2/12 = y\ + #2 = admittance of 1 and 2 in parallel = 8.1 X IO- 6 - J238.3 X IO- 6 mho. 2 12 = _L= 142 + J4197 ohms. ^12 For series resonance at 50,000 cycles for the entire circuit, the resultant reactance of the entire circuit must be zero. Therefore TT- = 4197 ohms 4197 X 27T X 50,000 = 0.0007585 microfarad. IMPEDANCES IN SERIES AND PARALLEL 229 Total impedance of the entire circuit at 50,000 cycles is 2 = (15 + JO) + (142 + #197) + (0 - #197) = 157 + jO ohms. z = 157 ohms. At 100,000 cycles. J. _ 80 6280 : 2z ~ (80) 2 > (6280) 2 J (80) 2 + (6280) 2 = 0.00000203 - jO.0001594 mho. yj = J27T X 10 5 X 0.0002537 X 10~ 6 = jO.0001594 mho. 2/12 = 2/i + fa = 2.03 X 10- 6 - jO mho. _ . y l2 2.03 X 10- 6 - JO = 492,000 + jO ohms. 3 = ~~ 2* X 10 5 X 0.7585 X 10~ 9 = - J2100 ohms. zo = (15 + JO) + (492,000 + JO) + (0 - J2100) = 492,000 - J2100 z = 492,000 ohms. The ratio of the impedances of the entire circuit at the two frequencies is go (at 100,000 cycles) = 492,000 z (at 50,000 cycles) " 157 = 3130 If the interfering waves and the waves for which response is desired were of the same intensity, the current in the receiving circuit due to the interfering frequency would be only ^- of that caused by the waves of the frequency to be detected. Filter Circuits. The principle of the so-called filter circuits, which are used in wireless telephony and also in the carrier-wave system of wire telephony to filter out waves of certain frequency, are nothing more than a combination of series and parallel circuits that can be tuned for both series and parallel resonance to suppress waves of a certain frequency and to allow waves of another frequency to pass. 230 PRINCIPLES OF ALTERNATING CURRENTS Jn the carrier system of wire telephony, speech is transmitted by very high frequency currents, which are above audible frequency. The intensity of these high frequency waves is modu- lated, i.e., varied to correspond to the variation in intensity produced in the air waves by the voice. Several of these high frequency carrier waves, each of a different frequency, are superposed on a single circuit. At the receiving end of the line each wave is filtered out and delivered to the proper receiving circuit. At the present time (1921) four two-way carrier- wave conversations, i.e., four each way, may be carried on over a single circuit, in addition to the ordinary telephonic and tele- graphic messages. Effect of Change of Reactance and Resistance with Current. Reactance and Resistance Functions of Current. In all the equations involving resistance or reactance or both, which have been considered thus far, both resistance and reactance have been assumed constant. The equations are true only under these conditions. When a sinusoidal voltage is impressed on a circuit of constant impedance, the resultant current under steady conditions is also sinusoidal. When, however, either the resistance or the reac- tance of the circuit is not constant but varies with the current, i.e., is a function of the current, a sinusoidal voltage impressed on the circuit will give rise to a non-sinusoidal current. The inductance of a circuit containing iron cannot be constant, since the flux is not proportional to the current. The flux per ampere is a variable. The reactance of such a circuit at any fixed frequency is a function of the current, except at very low flux densities when it may be approximately constant. The resistance of a circuit, which has so small a heat capacity that the variation in the 7 2 r loss during a cycle produces an appreciable change in temperature, must vary with the current during each cycle, unless the coefficient of resistance variation with temperature is zero. If the temperature coefficient is posi- tive like that of a metal, a sinusoidal voltage will produce a current wave which is flatter than a sine wave. The current will contain harmonics. In such a case the power-factor of the circuit cannot be unity, even though the circuit contain no reactance, IMPEDANCES IN SERIES AND PARALLEL 231 since the product of current and voltage cannot be equal to the true power absorbed. The average power absorbed in heating by a circuit containing a resistance which is a function of the current is T i*rdt .f Tjo -If Tjo (79) This integral can be equal to (7 r .m..) 2r only when r is inde- pendendent of the current strength. The apparent resistance of a circuit is always r = p (80) where P is the average power absorbed due to the resistance and I is the root-mean-square value of the current. (See Effective Resistance, page 234.) The expressions p v r = TO and r = - i 2 i where r, p and i are instantaneous values of resistance, power and current and v is the instantaneous voltage drop due to resistance, are always equal. In practice the change in temperature of a conductor with current during a cycle is usually too small to produce appreciable effect. Low candle-power, high-voltage, incandescent lamps, however, especially those like the tungsten lamp with extremely fine filaments, do show an appreciable variation in resistance with current during a cycle even at frequencies as high as sixty cycles. If a circuit contains iron, there may be a very appreciable change in its inductance with current during each cycle, especially when the iron is worked at high saturation. Due to this varia- tion, marked harmonics may be present in the current which are not present in the impressed voltage. With a sinusoidal voltage impressed on the circuit, a third harmonic with a maximum value as great as 25 to 40 per cent, of the maximum value of the fundamental would not be excessive. A third harmonic of this magnitude often occurs in the no-load current of a transformer with sinusoidal impressed voltage, 232 PRINCIPLES OF ALTERNATING CURRENTS The voltage drop across a circuit containing constant resis- tance and constant inductance is given by e = ri + L A i (81) If L = N-J-. decreases with increase of current, as it must when iron is present, then for a given impressed voltage, -r must di increase more rapidly than when L is constant in order that L-r plus ri shall be equal to the impressed voltage at each instant. A sinusoidal voltage impressed on a circuit containing iron, therefore, will give rise to a current which will be more peaked than a sinusoid. Non-sinusoidal Voltage Impressed on a Circuit Containing Constant Impedances in Series and in Parallel. When a non- sinusoidal voltage is impressed on a circuit consisting of constant resistances, constant inductances and constant capacitances in series or in parallel, the fundamental and each harmonic must be considered separately. They cannot be combined until the final solution is reached when the root-mean-square value of the current and voltage drops may be found in the usual manner by taking the square root of the sum of the squares of the compon- ents caused by the fundamental and each harmonic acting separately. It must be remembered that reactance, susceptance and conductance are functions of frequency. For constant resis- tance, inductance and capacitance, inductive reactance is propor- tional to frequency and capacitive reactance is inversely pro- portional to frequency. Conductance decreases with increase in frequency for a circuit containing inductance and resistance only and increases with increase in frequency for a circuit containing capacitance and resistance only. Conductance must be found for each frequency from the resistance, inductance and capaci- tance. Susceptance may either increase or decrease with fre- quency, depending on the relative magnitudes of resistance and reactance and the way the reactance changes with frequency. Susceptance, like conductance, must be found for each frequency from the resistance, inductance and capacitance. IMPEDANCES IN SERIES AND PARALLEL 233 Example of a Simple Series Circuit on which a Non-sinusoidal Voltage is Impressed. A voltage whose equation is e = 300 sin 314* + 75 sin 942* is impressed on a circuit consisting of a condenser of negligible resistance and 25 microfarads capacitance in series with an impedance of 40 ohms resistance and 0.1 henry inductance. What is the equation of the current and what is the root-mean- square value? The impressed voltage contains a fundamental of and a harmonic of / = -^ = 50 cycles ZTT 942 / = - = 150 cycles. It therefore contains a fundamental and a third harmonic. For the fundamental E ml 300 = _ 300 __ = 300 " \/(40) 2 + (31.4 - 127.4) 2 ~ 104.0 = 2.884 amperes. 31.4 - 127.4 -96.0 - - -- - = -- = -2.40 0i = -67.38 degrees. For the third harmonic 75 75 = 75 V(40) 2 + (94.2 -^4&5p 65.4 1.146 amperes. 234 PRINCIPLES OF ALTERNATING CURRENTS KI 7 tan B 3 = j^- = 1.293 6 3 = 52.3 degrees. to = 2.88 sin (314* + 67?4) + 1.15 sin (942* - 52?3) j^ s = /^88)~ 2 = 2.19 amperes. Effective Resistance. When a direct current flows in a circuit, there is a loss in power due to the resistance of the circuit. This loss is the so-called copper loss and is equal to the current squared multiplied by the ohmic resistance of the circuit. The resistance is equal to the copper loss divided by the current squared. Similarly, when an equal alternating current flows in the same circuit there is also a loss in power, but, in general, the loss is greater with alternating current than with direct current. On account of this increase in loss, the apparent resistance of the circuit is greater with alternating current than with direct current, and under certain conditions it may be many times greater. The apparent resistance of a circuit with alternating current is always equal to the loss in power caused by the current divided by the current squared. The resistance of a circuit with direct current is not only equal to the loss caused by the current divided by the current squared, but it is also equal to the drop in potential across the circuit caused by the current divided by the current. The apparent resistance of a circuit with alter- nating current does not equal the drop in potential across it divided by the current, except in very special cases. The increase in loss with alternating current is due to two causes: first, to certain local losses produced by the varying magnetic field in the surrounding material and in the conductor itself, and second, to the non-uniform current density over the cross section of the conductor. With direct current, the current density is uniform over the cross section of ordinary conductors, i.e., conductors of a single material of uniform specific resistance. The local losses produced by an alternating current are: eddy- current and hysteresis losses in adjacent magnetic material, eddy-current losses in adjacent conducting material, and eddy- current loss and also hysteresis loss, if the conductor is magnetic, in the conductor itself. IMPEDANCES IN SERIES AND PARALLEL 235 When an alternating current flows in a circuit, the power absorbed is equal to the current multiplied by the active compo- nent of the voltage drop produced by the current. Any increase in the active component of the voltage drop, due to local losses produced by the current, is equivalent to an increase in the apparent resistance of the circuit. The active component of the voltage drop is increased by all local eddy-current and hysteresis losses caused by the current and also by any distortion of the current distribution over the cross section of the conductor. The distortion of the current distribution has much the same effect, so far as the resistance of the conductor is concerned, as decreas- ing the cross section of the conductor. It thus increases its apparent resistance. The apparent resistance of a circuit to an alternating current is known as its effective resistance. Effec- tive resistance is not a constant. It varies with frequency, and also with current strength if the circuit is adjacent to magnetic material. In general, effective resistance increases rapidly with increase of frequency. In all work dealing with alternating currents, effective resistance must be used in finding the I 2 r loss and the potential drop in the circuit due to resistance. The non-uniform distribution of the current over the cross-section of a conductor is due to the difference in the reactances of ele- ments of the conductor taken parallel to rIG. DO. its axis. Let Fig. 66 represent the cross- section of a conductor. Consider any circular element in the conductor of radius x and width dx. The flux lines caused by a current in a straight cylindrical conductor, which lies in a medium of uniform permeability, are concentric circles with their centers on the axis of the conductor. These flux lines not only surround the conductor but exist in the conductor itself. The field intensity at a point outside any cylindrical element, such as the one shown in Fig. 66, due to the 27 current in the element is equal to -^ /*, where / is the current in the element, R is the perpendicular distance between the point and the axis of the conductor and /* is the permeability of the medium at the point considered. The field intensity inside 236 PRINCIPLES OF ALTERNATING CURRENTS the element due to the current it carries is zero. Since the only flux which can link any element, such as the one of radius x shown in the figure, is that which lies without it, it is evident that the flux linking an element must increase as its radius decreases. All the flux produced by the entire conductor links the element at its center. This includes the flux within as well as without the conductor. The only flux that can link an element at the surface of the conductor is the flux which lies without the conductor. The self -inductance per unit axial length of any element is equal to the flux linking it per unit length per unit current. This is equal to all that flux outside the element which is contained between two planes drawn perpendicular to the axi's of the con- ductor at unit distance apart. It is obvious, therefore, that the self-inductance of the elements will increase with decrease in their radii, will be least for the element at the surface of the con- ductor and greatest for the element at the center. The reac- tance will therefore be least for the element at the surface and will be greatest for the element at the center. Since reactance (27T/L) is proportional to frequency, the difference between the reactance of an element at the center and at the surface of any conductor of fixed radius will increase with frequency. It will also increase with the radius of the conductor and its permea- bility. It will be much greater for iron conductors than for those of copper. The magnitude of the current in any element of a conductor will be inversely as the impedance of the element. The current will therefore be greatest in elements at the surface and least in elements at the center. At very high frequencies very little current will flow through the central portion of a large conductor. Nearly all of the current will be carried by the portion near its surface. This crowding of an alternating current towards the surface of a conductor is known as skin effect. It is small for conductors of small cross section carrying low frequency currents, but for large conductors or high frequencies it may become very great. At very high frequencies, such as are used for radio- communication, the resistance of a cylindrical tube may be very nearly as low as the resistance of a solid conductor of equal radius and the same material. The skin effect can be calculated for non-magnetic conductors IMPEDANCES IN SERIES AND PARALLEL 237 of circular section which are not adjacent to other conductors. For conductors of non-circular section it must be determined experimentally. In certain simple cases, when conductors are embedded in iron, as are the armature conductors of an alternator, it is possible to calculate the skin effect with a fair degree of approximation. The skin effect at 60-cycles for straight, stranded, copper wires of circular cross section is shown in Fig. 67. 150 1-10 \m 100 / Skin Effect at 60 Cycles for Stranded Copper Cable / / ' / / / / / "0 500,000 1,000,000 1,500.0002,000,0002,500,0003,000,000 Circular Mils FIG. 67. In general, the current density in a conductor is least at the points where the flux linkages per ampere due to the current in the conductor are greatest. The apparent or effective resistance -of a circuit not containing a source of electromotive force may be found by measuring the power absorbed when a known current of definite frequency is passed through it. If P is the power absorbed due to the current 7 at a frequency /, the effective resistance of the circuit at the frequency / is r, = p (82) Effective Reactance. The effective reactance of a circuit to alternating current, especially if there be iron present, is always less than the true reactance. The true reactance is equal to 2wfL where L is the self-inductance of the circuit and / is the frequency. The self -inductance, L, is equal to the flux linkages 238 PRINCIPLES OF ALTERNATING CURRENTS of the circuit per unit current producing the flux. The entire current in a circuit is not effective in producing flux except when no eddy-current or hysteresis losses are caused by it. The power absorbed in any circuit, exclusive of that due to copper loss, is always equal to El cos 0, where E and I are the back electromotive force and the current respectively. The angle 0f is the phase angle between E and 7. If there is no power absorbed by the circuit other than the true ohmic cop- per loss, the angle 0f between E and I must be 90 degrees and E and I will be in quadrature. Under this condition, El cos0f = and VI cos f = Pr ?, . is the true copper loss, where V is the voltage drop impressed across the entire circuit and r h. is the ohmic resistance of the circuit. If there be magnetic or conducting material near the circuit, in which hysteresis or eddy-current losses occur, the current must have a component in phase opposition to the voltage rise induced by the flux, to supply these losses. This component does not contribute to the flux. Its effect, so far as flux is con- cerned, is canceled by the reaction due to the eddy-current and hysteresis losses. The only component of the current which is effective in producing flux is that in phase with the flux and, therefore, in quadrature with the voltage induced by the flux. This component is equal to the current that would be required to produce the flux were there no eddy-current or hysteresis losses. It is called the magnetizing component of the current or simply the magnetizing current. Consider a coil of wire wound on an iron core. If an alternat- ing current is passed through the coil, a flux will be set up which will link the turns of the coil. The current, as has been stated, will not be in phase with this flux if there are eddy-current or hysteresis losses. Let L be the flux linkages with the circuit per unit current producing the flux, i.e., per unit of the component of current in phase with the flux. Call this component 1^. Then the induced voltage drop through the coil, caused by its true self-inductance L, is 2-jrfLI^. Let p, Fig. 68, be the flux produced by the current 7 and let E = 27r/L7^ be the voltage drop caused by this flux, where 7^ is the component of the current 7 in phase with the flux when the vector Eab represents a voltage rise. If the circuit ab contains inductive reactance in addition to resistance, lab will lag the voltage drop by an angle whose cosine is the ratio of the resistance to the impedance, that is by the angle = cos- 1 -T= (2) V r 2 + x 2 where r and x are the resistance and reactance respectively of the circuit ab. The current 7o& must always be used with the voltage Eab. The power in the circuit is Eablab cos and never E ab l ba cos 0. If E ab is a voltage rise, Eablab cos is power generated. If 250 PRINCIPLES OF ALTERNATING CURRENTS is a voltage drop, EaJ.ab cos is power absorbed. If Eablab cos 6, representing power generated, is negative, it actually represents power absorbed, since negative power generated is power absorbed. HEablab cos '0, representing power absorbed, is negative, it actually represents power delivered or generated. The voltage drop Eab due to a current in an impedance is Eab = lab V^ 2 + X 2 It is never E ab = Ib a \/r 2 -+- x 2 . It is often necessary to use double subscripts with resistance, reactance and impedance and also with conductance, susceptance and admittance, but in such cases the subscripts have no other significance than to indicate between what two points of a circuit the quantities mentioned are measured. The order of the sub- scripts can mean nothing, since resistance, reactance, impedance, etc. are not vectors. When in specific problems exact numerical results are required, an analytical solution by the method of complex quantities is necessary. However, experience has shown con- clusively that, even though a particular problem is to be solved analytically, an approximate vector diagram should always be drawn as it facilitates the correct interpretation of the work and serves as a check against errors. It is usually advisable to draw all polyphase vectors radially from a common center. When vectors are drawn radially from a common center, there can be no question as to their phase relations, and if each vector is lettered with two subscripts, ihere is no excuse for mistaking the angle between any two vectors, such as E a b and 7 C d, for the angle between Eab and Idc- Neither is there any excuse for mistaking the angle between Eab and 7b a for the angle between Eab and 7 a &. The one angle is the supplement of the other. Example Illustrating the Use of Double -subscript Notation. A certain three-phase generator has one end of each of its three armature windings connected together in such a manner that the voltages generated in the windings, when considered in a direction from common junction to free terminal, are 120 degrees apart in time-phase. The three voltages will ""Be as- sumed to be equal in magnitude. The diagram of connections is shown in Fig. 7 la. POLYPHASE CURRENTS 251 Let inductive loads, n + jxi and r 2 -f jx 2 , be apph'ed between the terminals ab and ac respectively, as shown in Fig. 7 la. Call V, with proper subscripts, the voltage rise across the termi- nals of the generator windings. Let the internal impedance of each winding be r + jx. The internal or induced voltage rises in the three phases are desired. (a) Ic'a'-I Ij.r + ix FIG. 71. Referring to the vector diagram shown in Fig. 716, Voa, V b and V oc are the voltage rises across the terminals of phases oa, ob and oc of the generator. For simplicity, these voltages are assumed to be equal in magnitude and 120 degrees apart in time-phase. Actually, the voltage rises E^, E b and E oc induced in the windings would be equal in magnitude and 120 degrees apart in time-phase and the terminal voltages would differ slightly in magnitude and from the 120 degree phase relationship, due to the impedance drops in the armature windings of the alternator. However, for the illustration of the use of double- subscript notation, the assumption of equal terminal voltages, 120 degrees apart in time-phase, is made, since it permits a defi- nite solution of the problem without the use of Kirchhoff s laws and therefore better suits the present purpose. The voltage V a b between any pair of terminals such as a and b is equal to the vector sum of the voltages generated in the two phases connected between those terminals, considered in the direction ab. The voltage rise V a b is therefore equal to V ao + 252 PRINCIPLES OF ALTERNATING CURRENTS Vob = V oa + Vob- It is equal to the vector difference between the voltage rises V b and V oa - The voltage rises between the terminals ab, be and ca are Vab = V_ a 'b' = -Vo a + Vob (3) V bc = Vb'c' = -Vob + Vo C (4) V ca = F C V = -Foe + V oa (5) The corresponding voltage drops are -V_ab = -Va'b> = Vo a ~Vob (6) -Vbc = -Vb'c' = 'Vob -Voc (7) ~V ca = -Vc'a' = Voc -Voa (8) The voltage drops V a b and V ca are shown in Fig. 716. The current J a > v in the impedance Zi, connected between the terminals 'a' and b', is equal to the voltage drop V a b = V a >b' across a'b' divided by the impedance z\ = r\ + jx\. This current lags the voltage drop V a > v by the angle 0i = tan" 1 Simi- larly, the current 7 c v, in the impedance z 2 connected between the terminals c' and a' , is equal to the voltage drop V ca = V c v across c'a' divided by the impedance 2 2 = r 2 + jx^. This cur- rent lags the voltage drop V c > a > by the angle 2 = tan" 1 The current in the phase oa of the generator must be equal to the current in the line connected to that phase. This line current must be equal to the vector sum of the currents in the two impedances connected to this line. Therefore loa = laa' = la'b' I la'c' = la'b' ~ /c'a' (9) The currents carried by the other two phases of the generator, i.e., phases ob and oc, are Lb = hb' = /6V (10) loc = I cc' = I c'a' (11) The voltage rise induced in any armature winding, i.e., in any armature phase, is equal to the vector sum of the voltage rise across its terminals and the impedance drop produced in it POLYPHASE CURRENTS 253 by the current. If E^ E b and E oc are respectively the voltage rises induced in the phases oa, ob and oc, then Eob = V b + Iob(r + jx) (13) Eoc = Voc+Ioc(r+jx) (14) In order to illustrate the analytical solution ^f this problem, assume that the three terminal voltages Voa, V b and V oc are each equal to 200 volts. They are also assumed to be 120 degrees apart in time-phase. Let the two impedances which serve as loads be 21 = n + jxi = 20 + j5 ohms (15) 22 = 7*2 + jx 2 = 15 + J15 ohms (16) Let the impedance of each phase of the generator be z = r + jx = 0.1 + jl.O ohm (17) Take the terminal voltage rise Voa along the axis of imagi- naries, to correspond to the vector diagram of Fig. 716. Then Voa = 200(cos 90 + j sin 90) = 200(0 + jl) = -f J200 Vob= 200(cos 30 - j sin 30) = 200(0.866 - jO.500) = 173.2 - jlOO Voc= 200(cos 150 - j sin 150) = 200( -0.866 -jO.500) = -173.2 -jlOO From equations (6) and (8) , the voltage drops, V a b and F^,, across the impedances z\ and 32 are -Vab = -Va'b' = Voa ~ V ' ob = (0 + J200) - (173.2 - J100) = -173.2 + J300 volts -V^ = -V C 'a'= V c -Voa = (-173.2 - jlOO) - (0 = -173.2 -J300 volts la'b' ~ - 173.2 -f J300 20 + J5 254 PRINCIPLES OF ALTERNATING CURRENTS - 173.2 -f J300 20 - j5 20 + j5 20 - j5 = 4.62 + J16.14 amperes. -173.2 - J300 15 + J15 -173.2 - J300 15 - j!5 15 + j!5 15 - j!5 = 15.78 J4.23 amperes. J- oa = J-a'b' 'c'a' = (-4.62 + J16.14) - (-15.78 - J4.23) = 11.16 + J20.37 amperes. lob = Ib'a' ~Ia'b' = 4.62 J16.14 amperes. J-oc J- c'a' = -15.78 - J4.23 amperes. Eoa = Voa + Ioa(r + jx) = (0 + J200) + (11.16 + j20.37)(0.1 + jl.O) = -19.25 + J213.2 E oa = (-19-25) 2 = 214.1 volts. E ob = V ob + I ob (r + jx) = (173.2 - J100) + (4.62 - J16.14)(0.1 +J1.0) = 189.8 - J97.0 E ob == \/(189.8) 2 + (-97^ 2 = 213.2 volts. E oc = V oc + Ioc(r + JX) = (-173.2 - jlOO) + (-15.78 -j4.23)(0.1 + jl.O) = -175.9 - J115.2 E oc = V(-175.9) 2 + (- = 210.2 volts. Call the angles made by the vectors E oa , E b and E oc with the axis of reals B oaj O ob and d oc . Then POLYPHASE CURRENTS 255 Since the real part of the voltage Eoa is negative and the imaginary part is positive, the angle , must lie in the second quadrant. Therefore Boa = 95.2 degrees. -97.0 tan B b = 189.8 -0.511 Since the real part of the voltage E ob is positive and the im- aginary part is negative, the angle B ob must lie in the fourth quadrant. Therefore B b = 27.2 degrees. -115.2 tan d oc = -175.9 = 0.655 Since the real part of the voltage E oc is negative and the imaginary part is also negative, the angle B oc must lie in the third quadrant. Therefore BOC = 146.8 degrees. i The polar expressions for the three voltages Eoc are E M = 214.1 952 E b and 213.227?2 Eoc = 210.2 1468 Wye and Delta Connections for Three-phase Generators and for Three-phase Circuits. The windings of a three-phase alter- nator may be represented dia- grammatically by three windings displaced 120 degrees from one another, as shown in Fig. 72. If the windings are assumed to rotate in a counter-clockwise di- rection, the voltages generated in them will be equal and will differ 120 degrees in time-phase. The voltage generated in the phase marked 01 will lead the voltage generated in phase 02 by 120 degrees. It will lead the voltage generated in phase 03 by 240 degrees. Three-phase voltages will be generated in the FIG. 72. 256 PRINCIPLES OF ALTERNATING CURRENTS windings. These may be represented by three vectors, # i, #02 and Eos, which are equal in magnitude and 120 degrees apart in time-phase. These are shown in Fig. 73. The order of the subscripts used with the vectors indi- cates the direction in which the voltages are considered with respect to the windings. All voltages must be con- sidered in the same direction around the armature if they are to differ by 120 degrees in time-phase. The voltage vector Eoi is taken along the axis of reals. The symbolic expres- sions for the three equal voltages are #01 = (1 - J0)# (18) FIG. 73. -<-$-*> J^ rn ( ~ (19) (20) where E is the magnitude of the voltage generated in any one phase. From equations (18), (19) and (20) + #02 +#03 =0 (21) Since the vector sum of the voltages generated in the three windings is equal to zero when considered in the same direction around the armature, the terminals of windings 01, 02 and 03 may be connected to form a closed circuit, as shown in the left- hand half of Fig. 74, and no current will flow since the resultant voltage acting in the closed circuit formed by the armature windings is zero. The terminals of the alternator will be taken from the junction points between the phases, i.e., from the points 1, 2 and 3. The windings connected in this manner form a closed delta. This connection is known as the delta connection. Instead of connecting the phases in delta, they may be con- nected to form a wye by joining the corresponding terminals of the windings of the three phases. Either the terminals marked or those marked 1, 2 and 3 may be joined for wye connection. Delta and wye connections, usually written A and Y, are illus- trated in Fig. 74. The field poles are omitted in the figure. POLYPHASE CURRENTS 257 A A-connected alternator can have only three terminals. A Y-connected alternator need have only three, but in some cases a fourth is brought out from the common junction between the windings or neutral point, as this junction is called, to permit grounding the alternator. When loads are connected in wye the neutral connection is usually employed. When the neutral point is available, the load may be applied between the three pairs of line terminals, i.e., between 1 and 2, between 2 and 3 and between 3 and 1, or it may be applied between the terminals 1, 2 and 3 and the neutral point. In the first case the load is A -Connection 3f-Connection FlG. 74. A-connected. In the second case it is Y-connected. When the windings of an alternator are connected in delta, the load is usually connected in delta, although it may be connected in wye; but when the load is connected in wye there can be no connection between the neutral of the load and the generator since no neutral connection for the alternator exists. Although both Y- and A-connected loads are used, A-connection is the more common. Y-connection is the more common for alternators. Y-connected alternators are somewhat better than those con- nected in delta for several reasons, the most important of which are: there can be no short-circuit current in their armatures due to harmonics; for a fixed terminal voltage and output, the ratio of the amount of insulation to copper required in their armature windings is somewhat less than for A-connection; they permit grounding the neutral point of the system. If an alternator has more than two phases, there are always two ways in which its armature windings may be connected. These are known as the mesh and the star connections and 17 258 PRINCIPLES OF ALTERNATING CURRENTS correspond to the delta and wye connections for a three-phase alternator. Alternators with more than four phases are not used. They would possess no advantage over three-phase or four-phase alternators and would have marked disadvantages from the standpoint of power transmission. They would also be more complicated. Very few four-phase alternators are built, except for special purposes such as for use in existing four- phase plants. From the standpoint of power transmission, three-phase is superior to all others. It requires only three line conductors, as against four for four-phase and six for six-phase, and for a fixed amount of power transmitted a fixed distance with a fixed line loss and fixed voltage between any two con- ductors, (not necessarily adjacent), it requires only three-quarters as much copper as single-phase, four-phase or six-phase. From the standpoint of power transmission alone, four-phase trans- mission possesses no advantage over single-phase transmission. All power transmission lines are three-phase. Relative Magnitudes and Phase Relations of Line and Phase Currents and of Line and Phase Voltages for a Balanced Three- phase System Having Sinusoidal Current and Voltage Waves. By a balanced system is meant one in which the voltages in all phases are equal in magnitude and differ in phase by equal angles. The currents must also be equal in magnitude and they must also differ in phase by equal angles. For a balanced system the phase angles between the voltages and also between the currents 360 are equal to degrees, where n is the number of phases. A n balanced load is one in which the loads connected across all phases are identical. For Y-connection, there can obviously be no difference between the current in any phase and the current in the line to which the phase is connected. Line and phase currents are the same for Y- connection. Line and phase voltages, however, are not the same. The voltage between any two terminals, i.e.j the line voltage, is the vector difference of the voltages in the phases connected between the two terminals considered. The line voltage is neither equal in magnitude to the phase voltage nor is the line voltage in phase with the phase voltage. Refer to POLYPHASE CURRENTS 259 Line current In Line current 7 22 Line current / 3 3 FIG. 75. Fig. 75. For Y-connection, the same diagram may be used for either a diagram of connections or a vector diagram. Fig. 75 will serve for both. Obviously = phase current loi. = phase current I 2. = phase current /os. The line voltage Viz is not equal in magnitude to either Voi or Voz or in 3 phase with either. Viz = Vio + Fo2_ = Voi + Voz Viz = 2V cos 30 = \/37 in magnitude (22) where V is the magnitude of the phase voltage. Viz is equal in magnitude to the phase voltage multiplied by the square root of three. It differs in phase from the phase voltage Voz by 30 degrees and from the phase voltage Voi by 150 degrees. For a balanced, three-phase, Y-connected circuit, line voltage is equal in magnitude to phase voltage multiplied by the square root of three. Line voltage differs in phase from the voltages in the phases connected between the lines considered by 30 or 150 degrees, according to which of the two phase voltages is considered. Viz lags Voz by 30 degrees and lags Voi by 150 degrees. (23) Vzs lags Vos by 30 degrees and lags Voz by 150 degrees. (24) Vzi lags Voi by 30 degrees and lags Voz by 150 degrees. (25) If the phase rotation is opposite to that shown in Fig. 74, i.e., if Voz leads Voi by 120 degrees and Vos leads Voi by 240 de- grees, Viz would lead Voz by 30 degrees instead of lagging it by 30 degrees, Vza would lead Vw by 30 degrees instead of lagging it by 30 degrees and Vai would lead Voi by 30 degrees instead of lagging it by 30 degrees. The relations existing between phase and line currents of a balanced A-connected system are similar to those existing 260 PRINCIPLES OF ALTERNATING CURRENTS between phase and line voltages of a . balanced Y-connected system. Let the left- and right-hand diagrams, Fig. 76, be, respectively, a diagram of connections and a vector diagram of currents in the branches, /u/ i.e., in phases, 12, 23 and 31 of a A-connected arma- ture or A-connected load. The line current 7ir in line 11', which is con- nected to the common junction point of phases 12 and 31, is equal to the vector sum of the currents in these phases, both con- 6 sidered in a direction to- wards the junction point, 1. This follows from the fact that the vector sum of the currents at any junction point must be equal to zero. For a balanced load /!!' = /21 + /31 = -7l2 + hi Iu>. = 21 cos 30 = \/37 in magnitude, (20) where / is the magnitude of the phase current. The line current 7ir is equal in magnitude to the phase current multiplied by the square root of three. It differs in phase from the phase current /si by 30 degrees and from the phase cur- rent 7i2 by 150 degrees. For a balanced three-phase A-connected circuit, line current is equal to phase current multiplied by the square root of three. Line current differs in phase from the currents in the phases to which the line considered is connected, by either 30 or 150 de- grees according to which of the two phases is considered. In' leads 7si by 30 degrees and leads In by 150 degrees. (27) /22' leads 7 12 by 30 degrees and leads J zs by 150 degrees. (28) 7 3 3' leads 7 2 a by 30 degrees and leads 7 3 i by 150 degrees. (29) POLYPHASE CURRENTS 261 If the phase rotation were opposite to that shown in Fig. 76, Iu> would lag 7 3 i by 30 degrees, 7 22 ' would lag 7i 2 by 30 degrees and /sa' would lag 7 23 by 30 degrees. In a A-connected system, the voltage between any pair of lines, such as 1 and 2, is obviously equal to and in phase with the voltage in the phase connected between the lines considered. It is also equal to the vector sum of the voltages in the other two phases when these are considered in the proper direction. Fi2 = Fi3 + F 32 (30) F23 = 721 + Fl3 (31) Fn = F 32 + Fi (32) For a A-connected system, line and phase voltages are equal. If a single-phase load is applied between any pair of terminals of a A-connected alternator, the current will divide between the two parallel paths formed by the armature windings inversely as their impedances. The impedance of the windings is the same for each phase of an alternator. Since one path consists of a single phase and the other path consists of two phases in series, the impedances of the two paths between which the current divides will be in the ratio of one to two. Therefore, the phase across which the load is applied will carry two-thirds of the current. The other two phases in series will carry the remaining third. The currents in the two branches will be in phase, since the ratio of the resistance to the reactance is the same in each branch. When a balanced three-phase load is connected across the terminals of an alternator which has balanced voltages, the three line currents will be equal in magnitude and will differ by 120 degrees in phase. The three phase-currents in the windings of the alternator will also be equal in magnitude and will differ by 120 degrees in phase. The phase currents will be the same as the line currents for a Y-connected alternator and will be equal to the line currents divided by the square root of three for a A-connected alternator. For the A-connected alternator, there will be a shift in phase between line and phase currents, as is shown by equations (27), (2'8) and (29). The phase angle between the phase currents and phase voltages will be the same for the alternator as for the load. These statements are entirely independent of whether the load and generator are connected alike, i.e., both either in wye or 262 PRINCIPLES OF ALTERNATING CURRENTS in delta, or whether they are connected differently, i.e., one in wye the other in delta. Relative Magnitudes and Phase Relations of Line and Phase Currents and of Line and Phase Voltages of a Balanced Four- phase System Having Sinusoidal Current and Voltage Waves. |p A four-phase alternator has four identical armature windings which are displaced ninety electrical degrees from one another. These windings may be connected either ~^ in star or in mesh, corresponding to the wye and delta connections for a three-phase alternator. The voltages induced in the ^ windings of a four-phase alternator are ninety degrees apart in time-phase and may therefore be represented by four equal vectors displaced ninety degrees from one another as shown in Fig. 77. The vector expressions for the four vectors are 7oi = 7(0 +jl) 7o2 = v(i + jo) 7 03 = 7(0 - jl) 704 = 7(-l +JO) where 7 is the magnitude of the phase voltage. Since 7oi + 7 02 + 7 03 + 7 04 = 0, (33) no current will flow if the four armature windings are connected in order around the armature to form a closed circuit, i.e., are connected in mesh, in the same way as the armature windings of a three-phase generator are connected in delta. Instead of connecting the armature windings in mesh, they may be con- nected in star by joining their corresponding ends in the same way that the windings of a three-phase alternator are joined for wye connection. Mesh and star four-phase connections are shown in Fig. 78. For simplicity, gramme-ring windings are shown, instead of windings like those actually used. The windings actually used would be similar to those illustrated in Fig. 70, page 247, for a three-phase alternator. A four-phase alternator must have four terminals. When POLYPHASE CURRENTS 263 connected in star, a four-phase alternator may have a fifth terminal brought out from the common junction between the windings, i.e., from the neutral point. In general, for star connection it is obvious that there can be no difference between the current in any line and the current in the phase to which the line is connected. Line and phase cur- rents are the same for star connection. Line and phase voltages are not the same for star connection. The voltage between any pair of line terminals, such as 1 and 2, of a star-connected alter- nator, i.e., the line voltage, is equal to the vector sum of the voltages in the phases connected between the line terminals 1 and 2, both voltages being considered in the direction from 1 to 2 through the windings. Mesh Connection Star Connection FIG. 78. In general, for mesh connection the voltage between any pair of adjacent line terminals, i.e., the line voltage, is equal to the volt- age in the phase to which the lines considered are connected. Line and phase voltages are equal for mesh connection. Line and phase currents, however, are not equal for mesh connection. The current per terminal, i.e., the line current of a mesh-con- nected alternator, is equal to the vector sum of the currents in the two phases to which the line considered is connected, both currents being taken in the same direction with respect to the junction point between the phases and line, i.e., both being taken either towards or away from the junction point. Refer to Figs. 77 and 78. For a four-phase, star-connected alternator, Line current 7n> = phase current 7 i. Line current 1W Line current 7 3 3' = phase current 7o2- = phase current Jos. Line current 7 44 ' = phase current 7 04- (34) (35) (36) (37) 264 PRINCIPLES OF ALTERNATING CURRENTS The line voltage Viz is neither equal to nor in phase with phase voltage 7oi or 7 2, but Fi 2 = 27 cos 45 = \/2V in magnitude, where V is the magnitude of the phase voltage. The line voltage of a four-phase, star-connected alternator is equal in magnitude to the phase voltage multiplied by the square root of two. It differs in phase from the voltage in one of the phases between the lines considered by 45 degrees and from that in the other by 135 degrees. The line and phase currents are equal. For a mesh-connected, four-phase alternator, the current in any line, such as line 1' (see Fig. 78), is equal to the vector sum of the currents in phases 1 and 2 both taken towards the junction point between the line and the phases. Ill' = loi ~\~ 720 = 7oi 7()2 where 7oi and 7o2 are the currents in phases 01 and 02 respectively. For a balanced load Iiv = 27 cos 45 = -\/2I in magnitude, (38) where 7 is the magnitude of the phase currents. The line current of a four-phase, mesh-connected alternator, carrying a balanced load, is equal in magnitude to the phase current multiplied by the square root of two. There is a phase difference between the current in any line and in the phases to which it connects of 45 or 135 degrees according to which of the two phases is considered. The line and phase voltages of a mesh-connected, four-phase alternator are equal. Relative Magnitudes of Line and Phase Currents and Line and Phase Voltages for Balanced Star- and Mesh-connected N -phase Systems Having Sinusoidal Current and Voltage Waves. The line voltages of any balanced mesh-connected system are always equal to the phase voltages. The line and phase currents of any balanced star-connected system are always equal. These POLYPHASE CURRENTS 265 relations, which have already been stated, are obvious from an inspection of Figs. 74 and 78. For any balanced n-phase system, the phase voltages and also the phase currents differ in 360 time-phase by degrees, where n is the number of phases. The complex expressions for the phase voltages of an n-phase system are Foi = V jcosO - jsinO ) (39) T7 T7 f 360 . . 360 \ y 02 = V | cos - j sin _ j (40) 360 360 \ F 03 = V < cos 2 j sin 2 - - > (41) ^ H H } Tr v\ ( ^360 . . t 1N F 0n = F < cos (n - 1) - - - j sin (n - 1) L /i 71 where F is the magnitude of the phase voltages. If the system is star-connected, the voltage Viz between line terminals 1 and 2 is Fi2 = Fi^+ T r 01 = -Foi -f F 02 360 The voltages Foi and Fo2 are - - degrees apart in time-phase. Therefore Foi and Fo2 are ( 180 - Vdegrees apart in phase. Hence for a balanced system, Viz is equal to the vector sum of two equal voltages which differ in phase by ( 180 j degrees. i / ^firt\ Fi 2 = 2F cos (180 - - - ) in magnitude. 2i \ 71 I 1 / ^fiO\ Vune = 2F 8tor cos 5 (180 - - -) in magnitude. (42) 2 V fi I A similar relation can be shown to exist between the magni- tudes of the line and phase currents of a balanced mesh-con- nected system. For a mesh-connected system 1 / ^fiO\ I line = 21 mesh cos ( 180 - - -)ia magnitude. (43) z \ n i The following relations between the magnitudes of the line and phase currents and between the line and phase voltages for balanced systems may be obtained from equations (42) and (43) and the general statements that precede. 266 PRINCIPLES OF ALTERNATING CURRENTS Number of phases Star connection Mesh connection Line current equals Line voltage equals Line current equals Line voltage equals 3 Iphase V3 V phase V 3 Iphase ' phase 4 Iphase V2 V phase V2 Iphase Vphase 6 Iphase ' phase Iphase Vphase 12 Iphase 0.518 Vphase 0.518 Iphase Vphase Example. A certain experimental, 60-cycle, alternating- current generator has six identical armature windings which are displaced 60 electrical degrees from one another. When the alternator is driven at rated speed and has normal excitation, the voltage generated in each winding is 50 volts. What are the line and phase voltages of this alternator when the armature windings are connected for six-phase, for three-phase and for single-phase? The complex expressions for the voltages in the armature windings may be written Foi = 50 (cos - j sin 0) = 50.0 - JO F 02 = 50 (cos 60 - j sin 60) = 25.0 - J43.3 Fos = 50 (cos 120 - j sin 120) = -25.0 - J43.3 F 4 = 50 (cos 180 - j sin 180) = -50.0 - JO F 6 = 50 (cos 240 - j sin 240) = -25.0 + j'43.3 Foe = 50 (cos 300 - j sin 300) = 25.0 + J43.3 SIX-PHASE One winding would be used for each phase. Star Connection Mesh Connection Viz = Fio 4- Fo2 Line and phase voltages = Foi 4- Fo2 are equal. = -(50 - JO) 4- (25.0 - J43.3) V line = 50 volts. = -25.0 -J43.3 Fi2 = V(-25.0p + (-43.3) 2 = 50 volts Vline = 50 VOltS POLYPHASE CURRENTS 267 THREE-PHASE Two windings would be used in series for each phase. The phase voltages would be Voa = Foi + F 02 = (50.0 - JO) + (25.0 - J43.3) = 75.0 - J43.3 Vob = Fos + Fo4 = (-25.0 -J43.3) + (-50.0 -JO) = -75.0 -J43.3 Voc = Fo5 + Foe = (-25.0 + J43.3) + (25.0 + j'43.3) = 0+./86.6 Star Connection Mesh Connection b = Vao_ + Vob_ Line and phase vc = -Voa + Vob are equal. = -(75.0 - J43.3) V liM = V(75.0) 2 + (-43.3) 2 + (-75.0 -J43.3) = -150.0 - jO = 86.6 volts V ab = \/(-150) 2 + (0) 2 = 150 volts. V line = 150 VOltS SINGLE-PHASE If all six windings were connected in series, their resultant voltage would be zero. For single-phase, the windings may be divided into two groups of three each. The windings in each group would be connected in series and then the two groups paralleled. First group V' = Voi + Fo 2 + Foa = (50.0 - JO) + (25.0 - J43.3) + (-25.0- J43.3) = 50.0 - J86.6 Second group V" = F 4 + Fos + Foe = (-50.0 -JO) + (-25.0 + J43.3) + (25.0 +J43.3) = -50.0+J86.6 The voltages V' and V" are equal in magnitude but opposite in phase. By reversing the connections of one group of windings, the voltages in the two groups of windings could be brought into 268 PRINCIPLES OF ALTERNATING CURRENTS phase. The two groups of windings could then be paralleled, giving a single-phase alternator with a voltage of - 50.0) 2 + (86.6) 2 = 100 volts and a current capacity of twice the current capacity of each winding. Example of a Balanced A-connected Load. Three equal impedances, each having a resistance of 10 ohms and an inductive reactance of 15 ohms at 60 cycles, are connected in delta across a balanced, three-phase, 230-volt circuit. What is the line current and what is the total power absorbed by the load? The magnitude of the phase currents is j * phase " 230 230 V(10) 2 H- (15) 2 ~ 18.03 = 12.75 amperes I line = A/3 X 12.75 = 22.08 amperes. * phase \-l phase) s\, 'phase = (12.75) 2 X 10 = 1626 watts. Ptotal = 3 X Pphase = 3 X 1626 = 4878 watts. Example of an Unbalanced A-connected Load. Three im- pedances, z 12, 223 and 2 3 i, are connected in delta across the lines 1-2, 2-3 and 3-1 respectively of a 220- volt, three-phase, 60-cycle circuit having balanced voltages. Each of the impedances has a resistance of 5 ohms. Impedances 212 and z 23 have inductive reactances at 60 cycles of 5 and 10 ohms respectively. Im- pedance 2 3 i has a capacitive reactance at 60 cycles of 10 ohms. If the phase order of the voltages, Fi 2 , F 23 and F 3 i, between the terminals 1-2, 2-3 and 3-1 of the circuit is clockwise, i.e., if 7i 2 leads F 23 and 7 23 leads 7 3 i, what will be the three line cur- rents? What will be the total power consumed by the delta- connected load? POLYPHASE CURRENTS 269 A diagram of connections is shown in Fig. 79. Take the voltage Viz between lines 1 and 2 as the axis of reference. The complex expressions for the three line voltages will then be V u = 220 (1 + JO) = 220 + JO F 23 = 220 ~ -j = -110 - J190.5 ?ai = 220 - +j = -110 + J190.5 -. FIG. 79. The phase currents, i.e., the currents in the branches of the A-connected load, may be found in complex by dividing the phase voltages in complex by the impedances also expressed in complex. The phase currents are 220 + JO /12= = = 22 - J22 -110 - J190.5 5+jlO - 19.64 +jl. 18 -110+ J190.5 5 - J10 = -19.64 - jl.18 From Fig. 79, it is obvious that the line currents are In = Iiz + In = /i2 /si = (22 - J22) - (-19.64 -jl.18) = 41.64 - J20.82 In = V(41.64) 2 + (-20.82)' = 46.55 amperes. 270 PRINCIPLES OF ALTERNATING CURRENTS /2'2 = 1 23 + 1 21 = 1 23 1 12 = (-19.64 +J1.18) - (22 - J22) = -41.64 +J23.18 / 2 '2 = V(-41.64) 2 +(23.18) 2 = 47.65 amperes. /3'3 = /31 + /32= /31 ~ /23 = (-19.64 -jl.18) - (-19.64 + jl.18) = - J2.36 + (-2.36) 2 = 2.36 amperes. The total power is equal to the sum of the copper losses in the three phases. Piotai = (/i 2 ) 2 X r 12 + (7 23 ) 2 X r 23 + (7 3 i) 2 X r 3i = !(22) 2 + (22) 2 ) X 5 + {(19.64) 2 + (1.18) 2 ) X 5 + {(19.64) 2 + (1.18) 2 )} X 5 = 4840 + 1936 + 1936 = 8712 watts. If the opposite phase order, i.e., counter-clockwise, had been assumed, the total power would have been the same as for clock- wise phase order but the three line currents would have been different. Balanced and Unbalanced Y-connected Loads. Balanced and unbalanced Y-connected loads can best be taken up after the application of Kirchhoff's laws to alternating-current circuits has been considered. They will be considered in the next chapter. CHAPTER IX KIRCHHOFF'S LAWS AND EQUIVALENT Y- AND A-CONNECTED CIRCUITS Kirchhoff's Laws. KirchhofFs laws may be applied to single- phase or to polyphase alternating-current circuits as well as to direct-current circuits. When applied to alternating-current circuits, either instantaneous values of currents and voltages must be used in the equations, or all currents and voltages must be considered in a vector sense and must be referred to some conveniently chosen axis of reference. When instantaneous values are used, the resulting equations are algebraic equations. When vectors are used, the resulting equations are vector equa- tions, and all currents, voltages and impedances must be ex- pressed in their complex form and all referred to the same axis of reference. The solution of the resulting equations will give the unknown quantities in their complex form. Kirchhoff's laws applied to an alternating-current circuit may be stated as follows: For instantaneous values : (a) The algebraic sum of the instantaneous values of all cur- rents flowing towards any junction point in a circuit is zero at every instant. ii + 1*2 + t' 8 + . . . + in = SI * = (1) (b) The total rise or fall of potential at any instant in going around any closed circuit is zero. 61 + e 2 + 3 + . . . + e n = s; e H se = 2" e/ii = (2) For vectors: (a) The vector sum of all currents considered towards any junc- tion point in a circuit is zero. 1 1 + 7 2 + h + . . . + 7n = rj = o (3) 271 272 PRINCIPLES OF ALTERNATING CURRENTS (b) The vector sum of all the potential rises or potential falls taken in the same direction around any closed circuit is zero. Ei + #2 + EB + . . + Er. = S* Erise = S; Efall = (4) KirchholFs laws hold under all conditions for instantaneous values, but when applied to vectors, sinusoidal currents and voltages are presupposed. The current and voltage of a circuit can both be simple harmonic functions of time only when the resistances, inductances and capacitances of the circuit are con- stant and independent of current strength. When the re- sistances, inductances and capacitances of a circuit are constant, KirchhofFs laws may be applied to it when the currents and voltages are not simple harmonic functions of time by resolving the currents and voltages into their Fourier series and then considering the fundamentals and harmonics separately. The resultant current or voltage in any branch may then be found in the usual way by taking the square root of the sum of the squares of the root-mean-square values of the fundamental and harmonics. When vectors are used to represent equivalent sine waves, the application of KirchhofFs laws to a circuit may and usually will give very inaccurate and generally useless results, especially when there is capacitance as well as inductance in the circuit. Before attempting to solve any problem involving KirchhofFs laws, a diagram of connections should be made. Each corner or junction point of this diagram should be numbered or lettered. Double subscript notation should be used. (See page 248.) Example of the Application of Kirchhoff's Laws to a Simple Three- phase Circuit. Three impedances z oa = 10 + JO, z cb = 1 + J10 and FlG 80 z oc = jlO are connected in wye to the lines a, b and c respectively of a three-phase, 230-volt circuit with balanced voltages, Fig. 80. If the_voltage drop Vdb between lines a and b leads the voltage drop Vbc, what will be the three line currents and the voltage KIRCHHOFF'S LAWS 273 drops between the lines and common junction point of the im- pedances? If Vab were taken lagging V bc instead of leading it, a different solution would result. If the voltage V ab is taken as the axis of reference, the vector expressions for the three line voltages will be V<* = 230 + JO ^ (4) Vbc = -115 -J199.2 (5) Vca = -115 + J199.2 u (6) From KirchhofFs laws V ab = 230 + JO = 7o(10 + JO) + 7o&(l + ./ 10) (7) Vbc = -115 - J199.2 = 7U1 + J10) + 7 OC (0 - J10) (8) V ca = -115 + J199.2 = 7 CO (0 -J10) + 7 oa (10 + JO) (9) 7ao + ho + Jco = (10) From equation (7) j 230 - 7 Q fe(l + J10) 10 = 23.0- O.llob-jToo (11) From equation (8) j -115 -J199.2 -7^(1 +J10) -jio = -jll.5 + 19.92 - jO.l/bo + Ibo (12) Substituting the values of 7 ao and I co from equations (11) and (12) in equation (10), remembering that 7 OC = I co , gives 23.0 - O.l7o6 - jlob + ho + jll.5 - 19.92 + J0.l7 bo - ho = 7o b (0.1 + jl.l) = 3.08 + jll.5 (13) - _ 3.08+ jll.5 0.1 -jl.l 0.1 +J1.1 X 0.1 -jl.l = 10.62 - jl.834 (14) Substituting the value of Job from equation (14) in equation (7) gives - _ 230 - (10.62 - jl.834)(l + jlO) 10 = 20.104 - jlO.437 (15) 18 274 PRINCIPLES OF ALTERNATING CURRENTS Substituting the value of J bo from equation (14) in equation (8) gives -115 - J199.2 - (-10.62 + jl.834)(l + J10) Ioc ~ -jlO = 9.483 - J8.604 (16) Then lao = 20.10 - jlO.44 (17) ho = -10.62 +J1.834 (18) I co = -9.483 + J8.604 (19) The sum of the three currents J ao , ho and 7 CO is zero as it should be. lao + ho + ho = +0+JO y oa _ j o( ^ 2 ta a = (-20.10 + J10.44)(10 + jQ) = -201.0 + J104.4 (20) Vob = lob Zob = (10.62 -jl.83)(l +J10) = 28.92 + J104.4 (21) V oc == J- oc Zoc = (9.48 - j8.60)(0 - jlO) = -86.0 - J94.8 (22) As a check, the voltages V oa , V b and V oc may be combined to give the voltages between lines. Vob = Vao + Vob = 201.0 - J104.4 + 28.92 + J104.4 = 229.9 + jO Vbc = Vbo + Voc = -28.92 - J104.4 - 86.0 - J94.8 = -114.9 - J199.2 Vca = Vco + V. a = 86.0 + J94.8 - 201.0 + J104.4 = -115.0 -hj'199.2 Balanced Y-connected Loads Connected across Three-phase Circuits Having Balanced Voltages. When three equal imped- ances are connected in wye across a three-phase .system whose voltages are balanced, the common junction of the impedances is the true neutral point of the three-phase system. The cur- rents may therefore be found both in phase and in magnitude by dividing the wye voltages of the system by the impedances, both voltages and impedances being expressed in complex. KIRCHHOFF'S LAWS 275 Refer to Fig. 81. Let Zi = z, z 2 = z and 2 3 = z be three equal impedances connected in wye across the terminals of a three-phase system having balanced voltages. Let the line voltages be V 12, V& and Fsi. The diagram of connections is shown in the left-hand half of Fig. 81. The vector diagram of voltages is shown in the right-hand half. Foi, Fo2 and Fos are the phase voltages of a balanced Y-connected system having line voltages equal to Fi2, F 2 3 and F 3 i. The voltage Fi2, between lines 1 and 2, is taken as the axis of reference. The vector expressions for the three line voltages are Fi 2 - F (1 .+ JO) (23) (25) where F is the magnitude of the voltages between the lines 1-2, 2-3 and 3-1. Applying KirchhofTs laws to the circuit gives (26) 02 Fl2 = /102 + / F 2 3 = /202 + /03 2 F 3 1 = /302 + /Ol Z Joi + /02+ /03 = (27) (28) (29) 276 PRINCIPLES OF ALTERNATING CURRENTS From equations (26) and (27) Vi* ~ 7o2 = ~ r 7 T^23 /20 g T^23 -7 /0 ., x 103 = - - - = "~ 120 (61) Substituting the values of 7i and 7 3 from equations (30) and (31) in equation (29) gives ^12 , 7 , 7 , F23 7 - -- r ^02 + 1 02 + -=- - 120 = g Replacing Viz and F 23 by their vector expressions from equations (23) and (24) gives 1 = (cos 30 + j sin 30) X _ (33) ^ V(cos 2 30 -f sin 2 30 I line = V = -T= in magnitude, (34) where V is the magnitude of the voltages between lines, i.e., of the line voltages. The delta voltage or line voltage of a balanced three-phase system is equal in magnitude to the wye voltage multiplied by the square root of three. (See equation (22), page 259.) It is dis- placed from the wye voltage by 30 degrees. Conversely, the wye voltage is equal in magnitude to the delta voltage divided by the square root of three, and is displaced from the delta voltage by 30 degrees. By referring to the vector diagram in Fig. 81, it will be seen V that ~/ ^cos 30 + j sin 30) is the wye voltage of the system KIRCHHOFF'S LAWS 277 for phase 2. The current in phase 2 is therefore equal in magnitude to and in phase with the current obtained by dividing the wye voltage for phase 2, expressed in complex, by the impedance of the phase, also expressed in complex. Expressions similar to equations (33) and (34) may be found for the other line currents. It follows from equations (33) and (34) that the phase or line currents for a balanced Y-connected load, which is connected to a three-phase system whose voltages are balanced, is equal in magnitude to the line voltage divided by the square root of three and by the phase impedance. It is equal to the wye voltage of the system divided by the phase impedance. The vector expres- sion for the current in any phase, such as phase 1, is equal to the wye voltage for phase 1, expressed in complex, divided by the phase impedance, also expressed in complex. It must not be forgotten that the preceding statements are true only for a bal- anced system. If either the voltages or the impedances are unbalanced, the statements are not correct. Example of a Balanced Y-connected Load. Three equal impedances, each having a resistance of 10 ohms and an inductive reactance of 15 ohms, are connected in wye to a 230-volt three- phase system whose voltages are balanced What are the line currents and the total power taken by the load? The wye voltage of the system is V3 Vy I line = Iy= 230 \/3 132.8 + (15)* 18.03 = 7.36 amperes. Total power = P = 3 X (I phaS e) 2 X r phase = 3 X (7.36) 2 X 10 = 1625 watts. An Example Involving Balanced Y-and A -connected Loads in Parallel Across a Three-phase System Whose Voltages are Balanced. Three equal impedances, each having 5 ohms 278 PRINCIPLES OF ALTERNATING CURRENTS resistance and 5 ohms inductive reactance, are connected in wye across a 230-volt, three-phase circuit whose voltages are balanced. Three other equal impedances, each having 10 ohms resistance and 5 ohms capacitive reactance, are connected in delta across the same circuit. What is the resultant line current? What is the total power taken by the two loads in parallel? The complex expressions for the impedances are ZY = 5 + J5 g A = 10 - J5 The diagram of connections and a vector diagram of the wye and delta voltages are shown in Fig. 82. The voltage F i between neutral and line 1 is taken along the axis of imaginaries. FIG. 82. Since the load is balanced, it is necessary to consider the current in one line only. According to Kirchhoff's laws, line current In/ = I i + /2i + 7ai. The vector expressions for the voltages Foi, Viz and Vn are = + J132.8 ' Vl2= FlO + 702= -Foi+702 .1\ J2/ 230 ^-/ 2 115 - J199.2 KIRCHHOFF'S LAWS 279 = 115+J199.2 /21= ~ = I -115 + J199.2 10 - jb -115+ J199.2 10 - j5 N 10 + jb = - 17.17 +J11.34 Li = 115+ J199.2 10 - jb 115 + J199.2 10 + jb 10 - j5 10 -f j5 = 1.232 + J20.54 j Foi -/Ol = ^^ = + J132.8 5 + J5 + J132.8 5 - j5 5 + jb X 5 - J5 = 13.28 + J13.28 = (13.28 + J13.28) + (-17.17 + jll.34) + (1.232 + J20.54) = - 2.66 + J45.15 Jir = V(-2.66) 2 f (45.15) 2 = 45.23 amperes. Total power = P = 37 A 2 rA + 3/ K 2 r> = 3{(-17.17) 2 + (11.34)'} X 10 + 3j(13.28) 2 + (13.28) 2 } X5 = 17,990 watts. 280 PRINCIPLES OF ALTERNATING CURRENTS This problem could have been solved somewhat more easily by replacing the A-connected load by its equivalent Y-connected load. Equivalent Y- and A-connected loads will now be con- sidered. Equivalent Unbalanced Y- and A-connected Three-phase Circuits. If an equivalent three-phase system is defined as one which takes the same line currents at the same line voltages with the same phase relations between the line currents and line voltages as the three-phase system it replaces, then any un- balanced delta system of currents may be replaced by just one equi- valent wye system of currents ; but any unbalanced delta system of voltages may be replaced by an infinite number of wye systems of voltages. Fixing either the voltage or the constants of one of the equivalent wye branches fixes both the voltages and constants of the other branches. Conversely, any unbalanced wye system of voltages may be replaced by just one equivalent delta system of voltages, but any unbalanced wye system of currents may be re- placed by an infinite number of delta systems of currents. Fixing either the current or the constants of one of the equivalent delta branches fixes both the currents and constants of the other branches. If the above definition of an equivalent three-phase system is further qualified by the condition that the equivalent system must not only be equivalent as a whole but also equivalent between each pair of mains, i.e., if any main is opened the system must take the same current from the remaining two mains with the same phase relation between this cur- rent and line voltage as the three-phase system it replaces would have under like conditions, then there is just one delta system which will exactly replace a wye system and conversely there is just one wye system which will exactly replace a ^ delta system. That any A-connected system may be replaced, so far as voltages alone are concerned, by an infinite number of Y-connected systems will be evident by referring to Fig. 83. Let V a b, Vbc and V ca be the voltages of the A-connected system and V oa Vob and Voc the voltages of a Y-connected system which has KIRCHHOFF'S LAWS 281 the same line voltages. Obviously any three wye vectors having their ends at a, b and c will replace the three delta voltages. The common point, o, from which the three wye vectors are drawn may be anywhere either within or without the delta. Relations between the Constants of Unbalanced Equivalent A- and Y-connected Three-phase Circuits. Let z's with primes be the impedances of the branches of a Y-connected circuit and let z's without primes be the impedances of the equivalent A-con- nected circuit. Refer to Fig. 84. Let line c be open. For the A-connected system, z\ will then be in parallel with z 2 and z 3 in series. For the Y-connected system z\ and 2 2 ' will be in series. For equi- valence between lines a and b the following relation must hold 2i(2 2 + Z 3 ) /''./ 2a6 = ^ r7= i r. = 2i + Z 2 2i 2.3 With line a open 2 6c = 22(23 + 22 2 3 With line b open 2 2 ) 2l 2 2 ' 2 3 ' Subtracting equation (37) from equation (35) gives . - - 2 3 2 2 = 2-! +ir+7 3 Adding equations (36) and (38) gives 22 == ^ j - j ~ (35) (36) (37) (38) (39) 282 PRINCIPLES OF ALTERNATING CURRENTS The impedances z 3 ' and z\ may be found in a similar manner. *' - mri, ' All impedances must be expressed in their complex form. The neutral of an unbalanced three-phase A-connected circuit may be considered to be the neutral point of the equivalent Y-connected circuit as determined by the impedances z\ , zj and 2 3 ', given by the preceding equations. Although it is occasionally convenient in certain problems to be able to replace an unbalanced A-connected circuit by its equivalent unbalanced Y-connected circuit, little is saved, as a rule, in the amount of time and labor involved in obtaining a complete solution. The work of changing the given constants to the constants of the equivalent circuit is usually as great as the labor saved in solving the new equivalent unbalanced circuit over what would have been required to solve the original circuit. The conditions are very different when the loads are balanced, as the. transfer from a balanced delta connection to the equiva- lent balanced wye connection or vice versa may be made quickly and easily merely by the use of the factor 3. Equivalent Wye and Delta Impedances for Balanced Loads. It is often desirable, when solving certain types of problems aris- ing in engineering, to replace the impedances of a balanced A-con- nected load by impedances connected in wye which will take the same power at the same power-factor from the three-phase mains. It may also occasionally be desirable to replace a Y-connected load by its equivalent A-connected load. In either case, in order to retain the same power-factor, the ratio of the resistance to the reactance for the equivalent impedances must be the same as for the impedances they replace. If this ratio is maintained, it is merely necessary, when substituting a A-connected load for a Y-connected load or vice versa, to find three impedances which will take the same line current at the same line voltages as the original load. Suppose a balanced A-connected load is to be replaced by a KIRCHHOFF'S LAWS 283 balanced Y-connected load. The ratio between the equivalent wye (line) and delta currents of any balanced three-phase system is Each of the Y-connected impedances must, therefore, take the square root of three times as much current as each of the equivalent A-connected impedances, but the voltages impressed across the Y-connected impedances are only one over the square root of three times as great as the voltage impressed across the A-connected impedances. Since the current varies inversely as the impedance of a circuit and directly as the voltage, each of the Y-connected impedances must be V3 V3 3 times as great as each of the A-connected impedances, if they are to take the same line current. Conversely, the impedance of each branch of a balanced A-connected load must be three times as great as the impedance of each branch of the balanced Y- connected load it replaces. This relation between the imped- ances of equivalent Y- and A-connected balanced loads may be obtained from the general equations (39), (40) and (41), pages 281 and 282, by putting z l = z z = z 3 . Example of the Substitution of a Balanced Y-connected Load for a Balanced A-connected Load. The problem of balanced Y- and A-connected loads in parallel, which was solved on page 277, will be solved by replacing the A-connected load by its equivalent Y-connected load. The constants of the actual loads were Z Y = 5 + J5 Z A = 10 - J5 The impedance of the Y-connected load which will replace the A-connected load is _ , 10 .5 * = y-^3 The actual wye impedance and the equivalent wye impedance for each phase may be treated as two impedances in parallel across a voltage which is equal to the wye voltage of the system. 284 PRINCIPLES OF ALTERNATING CURRENTS The resultant admittance of the two impedances in parallel is 2/o = + , = (QY + QY') j(by + by') Zv Zv 10 5 3 + (5 , + (5 ,+ ( :y +(|) . = {0.1 +0.24} - j{0.1 - 0.12} = 0.34 + J0.02 e == * == ' to neutral X 2/0 230 . ,_ = -/^X V(0.34) 2 - = 132.8 X 0.3406 = 45.23 amperes. Total power = P = 3 X (V toneutral ) 2 X g = 17,990 watts. That the substitution of an equivalent Y-connected load for the actual A-connected load simplifies the solution of this problem is evident. Another Example of the Substitution of an Equivalent Y- connected Load for a Balanced A-connected Load. A balanced load, consisting of three equal inductive impedances, each having a resistance of 60 ohms and a reactance of 30 ohms, is connected in delta at the end of a three-phase transmission line which has a resistance of 1 ohm and an inductive reactance of 2 ohms per conductor. If the voltage of the line at the generating station is balanced and is maintained at 2300 volts between conductors, what will be the voltage between conductors at the load, i.e. } at the receiving end of the line? What is the efficiency of transmission? This is really a case of a Y-connected load in series with a A- connected load, the line impedances being the Y-connected load. KIRCHHOFF'S LAWS 285 The diagram of connections is shown in Fig. 85. The equivalent Y-connected load which replaces the A-connected load is shown dotted. FIG. 85. Let FSF and F fl y be the voltages to neutral, i.e., the wye voltages, at the station end of the line and at the receiving or load end respectively. Then 2300 SY 1328 volts. The equivalent Y-connected impedances which will replace the actual A-connected impedances are each = 20 + jlO Since the load is balanced, the neutral point at the load and at the station will be at the same potential and no current will flow between them even if they are connected. Therefore the voltage to neutral at the generating end of the. line may be as- sumed to be used up in the potential drop between the station and the neutral point of the equivalent Y-connected load. This drop will be equal to the line current multiplied by the resultant impedance of a single conductor and of one phase of the equiva- lent Y-connected load. Therefore /fine = SY 286 PRINCIPLES OF ALTERNATING CURRENTS where Z L = 1 + J2 is the impedance of the transmission line per conductor. 1328 I line = V(l + 20) 2 + (2 + 10) 2 1328 KA Qn ^. - = 54.89 amperes. At the load the voltage between line and neutral is VRY = IlineZRY _ _____ = 54.89 X V(20) 2 + (10) 2 = 1227 volts. At the load the voltage between lines is FA = A/3 X V RY = 1.732 X 1227 = 2125 volts. l TR * Efficiency of transmission = TRY _ = 0.952 TRY 20 20 + 1 = 0.952 X 100 = 95.2 per cent. CHAPTER X HARMONICS IN POLYPHASE CIRCUITS Relative Magnitudes of Line and Phase Currents and of Line and Phase Voltages of Balanced Polyphase Circuits when the Currents and Voltages are not Sinusoidal. Let 001, vo z and vo s be the instantaneous induced voltages of a balanced three-phase alternator, where (1) voi = V m \ sin (0 + V m3 sin 3(o>0 + 7 m5 sin 5() + etc. ^02 = V m i sin (tat - 120) + Vms sin 3(co - 120) + V m , sin 5( 03 = V m i sin (at - 240) + 7^ sin 3(co* - 240) + F m5 sin 5(co - 240) + etc. (3) The fundamental and the third and the fifth harmonics in equations (1), (2) and (3) are plotted in Fig. 86. The angular displacement between any harmonic of any phase and the corresponding harmonic of phase one is given in the following table. Phase Phase displacement in electrical degrees 1st 3rd 5th 7th 9th llth 1 o- 2 120 3 X 120 - 360 =0=0 5 X 120 = 600 O 240 7 X 120 = 840= 120 9 X 120 = 1080 =0= 11 X 120 = 1320 =0=240 3 240 I 3 X 240 = 720 5 X 240 - 1200 =0=120 7 X 240 = 1680 =0= 240 9 X 240 = 2160=G=0 11 X 240 = 2640 =0= 120 A phase displacement between any two current or voltage vectors of like frequency, of any whole number of wave lengths, i.e., of any integral number of times 360 degrees, does not alter their relative phase. 287 288 PRINCIPLES OF ALTERNATING CURRENTS It will be seen from the table that the harmonics of triple frequency are in phase, as are all harmonics which are multiples Phase FIG. 86. of this frequency, such as the ninth, fifteenth, etc. If the third harmonics are omitted, as well as all multiples of the third, the HARMONICS IN POLYPHASE CIRCUITS 289 phase order of the remaining harmonics, starting with the first, or fundamental, alternates from the order 1, 2, 3 to the order 1, 3, 2. If the fundamental of phase one leads the fundamental of phase two, the fifth harmonic of phase one will lag the fifth harmonic of phase two. The phase order of the fifth, eleventh, seventeenth, etc., harmonics is opposite to that of the funda- mental. The phase order of the seventh, thirteenth, nineteenth etc. harmonics is the same as that of the fundamental. The vectors for the fundamentals and harmonics are shown in Fig. 87. FIG. 87. Diagrams of wye and delta connections are shown in Fig. 88. FIG. 88. Wye Connection. If the alternator whose phase voltages are given by equations (1). (2) and (3) is Y-connected, its terminal voltages will be (see Figs. 87 and 88). 012 = -0oi + 002 = \/3F w i sin (co< -J.50 ) + + \/3F m6 sin (5ut + 150) sin (7wt - 150) etc. (4) 290 PRINCIPLES OF ALTERNATING CURRENTS V23 = ~v 02 + VK = V3Vmi sin (ut - 120 - 150) + sin (5* - 6 + 150) + \/3/m7 sin (lut - 6 7 - 150) + etc. (9) 294 PRINCIPLES OF ALTERNATING CURRENTS i2>2 = -i<>2 + ioi = V3I ml sin (ut - 120 - 6 1 - 150) + + \/3/m5_sin (5co + 120 - 5 + 150) + \/3/ w7 sin (7co* - 120 - 6 7 - 150) + etc. (10) i*'i = -im + i<>2 = VSImi sin (ut - 240 - 6 l - 150) + + A/3/^sin (5o>* + 240 - 5 + 150) + -v/37.,7 sin (7co* - 240 - 7 - 150) + etc., (11) where the 7 TO 's with subscripts 1, 3, 5, etc. represent the maximum values of the fundamental and the harmonics. The angles are the angles of lag between the component currents and the corre- sponding component voltages. The angles 6 will be different for the fundamental and the harmonics. For constant inductance, constant capacitance and constant resistance, the angle of lag for any harmonic, such as the nth, is x c nx ^~~ 6 n = tan~ 1 - r where X L and x c are the inductive and capacitive reactances for the fundamental frequency. Equivalent Wye and Delta Voltages of Balanced Three-phase Systems having Non-sinusoidal Waves. In general, there can be no third-harmonic voltage between the lines of any balanced three-phase circuit. If there can be no third harmonic in the voltage between the lines of a balanced three-phase system, obviously there can be no third harmonic in the equivalent wye voltage of such a system. The equivalent wye voltage is there- fore equal to the line voltage divided by the square root of three. When there is no harmonic of triple frequency or any multiple of this frequency, it is evident from equations (7) and (8), page 290, that the ratio of line to phase voltage is equal -to the square root of three for wye connection. This relation between the line voltage and the equivalent wye voltage of a balanced three-phase circuit is made use of in determining the power-factor of a balanced three-phase system which may contain harmonics. An Example Illustrating the Relations between Line and Phase Voltages of a Three-phase Alternator which may be Connected either in Wye or in Delta. Oscillograph records show that the HARMONICS IN POLYPHASE CIRCUITS 295 phase voltage of a certain sixty-cycle, three-phase, Y-connected alternator contains third, fifth and seventh harmonics, but no harmonics of higher order of appreciable magnitude. The expression for the phase voltage is e = 1880 sin 377* + 175 sin (1131* - 25) + 75 sin (1885* - 30) + 30 sin (2639* + 40) If time is reckoned from the instant when the fundamental of phase one is passing through zero increasing in a positive direction, what are the expressions for the voltages of the three phases? What are the expressions for the three line voltages? If the alternator is reconnected in delta, what will be the three line voltages? What are the root-mean-square values of the line and phase voltages for wye connection? What is their ratio? What is the ratio of the line voltages for wye and for delta connection? The phase voltages are (see table on page 287). e 01 = 1880 sin 377* + 175 sin (1131* - 25) + 75 sin (1885* - 30) + 30 sin (2639* + 40) e Q2 = 1880 sin (377* - 120) + 175 sin (1131* - 25 0) + 75 sin (1885* - 30 + 120) + 30 sin (2639* + 40 - 120) = 1880 sin (377* - 120) + 175 sin (1131* - 25 0) + 75 sin (1885* + 90) + 30 sin (2639* - 80) eoa = 1880 sin (377* - 240) + 175 sin (1131* - 25 0) + 75 sin (1885* - 30 + 240) + 30 sin (2639* + 40 - 240) = 1880 sin (377* - 240) + 175 sin (1131* - 25) + 75 sin (1885* + 210) + 30 sin (2639* - 200) For wye connection (see Figs. 87 and 88, page 289). 612 = e i + e 2 = V3 X 1880 sin (377* - 150) + + V3 X 75 sin (1885* - 30 + 150) + \/3 X 30 sin (2639* -f 40 - 150) = 3256 sin (377* - 150) + 130 sin (1885* + 120) + 52 sin (2639* - 110) 296 PRINCIPLES OF ALTERNATING CURRENTS 023 = 002 + 00? = V3 X 1880 sin (377* - 120 - 150) + + A/3 X 75 sin (1885* - 30 + 120 + 150) + V3 X 30 sin (2639* + 40- 120 - 150) = 3256 sin (377* + 90) + 130 sin (1885* - 120) + 52 sin (2639* + 130) 031 = 003 + 001 = V3 X 1880 sin (377* - 240 - 150) + + \/3 X 75 sin (1885* - 30 + 240 + 150) + A/3 X 30 sin (2639* + 40 - 240 - 150) = 3256 sin (377* - 30) + 130 sin (1885* + 0) + 52 sin (2639* + 10) The root-mean-square phase voltage is T7 /(I880) 2 + (175) 2 + (75) 2 -K30P y phase vwy w A/ 2 = 1336 volts. The root-mean-square line voltage for wye connection is = 2305 volts. The ratio of the line and phase voltages for wye connection is Vune (wye) = 2305 = V phasc (wye) 1336 This ratio would be exactly equal to the square root of three if there were no third harmonic in the phase voltage. For delta connection the third harmonic phase voltage is short-circuited in the closed circuit formed by the A-connected armature winding, and therefore cannot appear between the line terminals. If the delta connection is made as shown in Fig. 88, page 289, the line voltages will be 021 = 1880 sin 377* + 75 sin (1885* - 30) + 30 sin (2639* + 40) e u = 1880 sin (377* - 240) + 75 sin (1885* + 210) + 30 sin (2639* - 200) 032 = 1880 sin (377* - 120) + 75 sin (1885* + 90) + 30 sin (2639* - 80) HARMONICS IN POLYPHASE CIRCUITS 297 The root-mean-square line voltage for delta connection is V*. (delta) = = 1331. The ratio of the root-mean-square line voltages for wye and delta connections is Vune (wye connection) = 2305 = = / - Vn ne (delta connection) 1331 " V c This ratio is equal to the square root of three as it must be since neither voltage contains the third harmonic. An Example Involving Harmonics in a Balanced Three-phase Circuit. Three identical non-inductive resistances, when con- nected in delta across the terminals of a balanced Y-connected alternator, consume 12,000 watts. When these same resistances are connected in wye across the same alternator, they consume 4,750 watts when their common connection, i.e., their neutral point, is connected to the neutral of the alternator. Under this condition the current in the neutral is 15 amperes. The voltage of the alternator is assumed to be the same in both cases. What are the root-mean-square values of the voltages between the line terminals of the alternator and between its line terminals and neutral point? If there were no harmonics of triple frequency or multiples of this frequency present in the voltage of the alternator, the power consumed by the resistances when connected in wye would be equal to one third of the power they consume when connected in delta across the same voltage. This follows from the fact that power in a circuit is When there is only pure resistance, i.e., when the inductance and capacitance are each zero, the expression for power reduces to the voltage squared divided by the resistance. The expression P = V*g applies to one particular harmonic, that for which the conduc- tance, g, is computed. The conductance will have a different value for the fundamental and each harmonic, except when the 298 PRINCIPLES OF ALTERNATING CURRENTS inductance and capacitance are each zero and the resistance is constant. If the inductance and capacitance are each zero and the resistance is assumed to be constant, the expression for the power consumed by a circuit which has no reactance holds for all harmonics and fundamental, provided there are no harmonics in the voltage which are not present in the current. Power in delta _ (Delta voltage) 2 Power in wye (Wye voltage) 2 If there were no third harmonic or multiples of the third present in the voltage, the ratio of delta voltage squared to wye voltage squared would be (\/3) 2 = 3. Therefore, if no harmonics of triple frequency or multiples of this frequency were present in the current when the resistances were connected in wye, they would consume = 4) ooO watts. The difference between the actual power consumed by the three resistances when Y-connected, with their neutral point and the neutral point of the alternator interconnected, and 4,000 watts must be due to harmonic currents of triple frequency or multiples of this frequency which return on the neutral. The harmonic currents of triple frequency and multiples of this frequency in each of the three resistances when Y-connected must be equal to one-third the neutral current. 15 -^- = 5 amperes. o The copper loss, i.e., the Pr loss, due to harmonic currents of triple frequency and multiples of this frequency in each of the resistances when Y-connected, is 4,750 - 4,000 750 ^ - = -5- = 250 watts. o o Therefore (5) 2 X r = 250 r = 10 ohms per resistance unit. HARMONICS IN POLYPHASE CIRCUITS 299 When the resistances are A-connected the power consumed, which is 12,000 watts, or 4,000 watts per resistance unit, must be due to a current that does not contain any harmonics of triple frequency or multiples of this frequency. This current must be I A (\f\f\ I = J~p = 20 amperes. Therefore V. = 20 X 10 = 200 volts. Fr vV(v^* +( * = 125.8 volts. Harmonics in Balanced Four -phase Circuits. There can be no harmonic of triple frequency or any multiple of this frequency in the line voltage of a balanced three-phase alternator or circuit. With a four-phase alternator or circuit, the third harmonic and other harmonics whose frequency is a multiple of triple frequency are not cut out from the line voltages. Odd harmonics of any order may appear in both the line and the phase voltage for four-phase connection. None is suppressed by this connection. The ratio of the magnitudes of the fundamentals in the line and phase voltages of a four-phase, star-connected alternator is equal to the square root of two. The ratio of the line and phase voltages for each harmonic that may be present is also equal to the square root of two. Let the instantaneous values of the phase voltages of a four- phase alternator be given by the following equations : VQI = V m i sin &t -\- Vms sin 3co + V m s sin 5co -f V m7 sin Tut (12) H-etc. "02 = V m i sin (co* - 90) + Vms sin 3(* - 90) + V m s sin 5(cof - 90) + V ml sin 7(co - 90) + etc. (13) "os = Vmi sin (ut - 180)+ V m3 sin 3(o>* - 180) + V m s sin 5(co* - 180) + V ml sin 7(ut - 180) + etc. (14) "04 = Vmi sin (ut - 270) + V m3 sin 3 (* - 270) + etc. (15) 300 PRINCIPLES OF ALTERNATING CURRENTS Phase FIG. 89. HARMONICS IN POLYPHASE CIRCUITS 301 The V m 's with the subscripts 1, 3, 5 and 7 are the maximum values of the phase voltages. The fundamentals and third harmonics in equations (12), (13), (14) and (15) are plotted in Fig. 89. The angular displacement between any harmonic in any phase and the corresponding harmonic in phase one is given in the following table. Phase displacement in electrical degrees 1st 3rd 5th 7th 9th llth 1 2 90 3 X 90 = 270 5 X 90 = 450 90 7 X 90 = 630 270 9 X 90 = 810 C 090 11 X 90 = 990 270 3 180 3 X 180 = 540 01 80 5 X 180 = 900 0180 7 X 180 = 1260 0180 9 X 180 = 1620 0180 11 X 180 = 1980 180 4 270 3 X 270 = 8 10 090 5 X 270 = 1350 0270 7 X 270 = 1890 090 9 X 270 = 2430 270 11 X 270 = 2970 90 It will be seen from the preceding table that the harmonics of like frequency in the four phases of a balanced four-phase alter- nator or balanced four-phase load are ninety degrees apart in time-phase. It should also be noticed that the phase order of the fundamentals and harmonics of different frequencies alternates from the phase order 1-2-3-4 to the phase order 1-4-3-2. One phase order is opposite to the other. See Fig. 90. The vectors F i, F 02 , F 03 and F 04 , for the fundamentals and 302 PRINCIPLES OF ALTERNATING CURRENTS harmonics for a balanced four-phase alternator or load, are shown in Fig. 90. Star and mesh four-phase connections are shown diagrammati- cally in Fig. 91. FIG. 91. The line voltages for the fundamental or any harmonic, for star connection, are Fi2 = - Foi + F 02 (16) F 23 = -7o2 + Fo3 (17) F 3 4 = -F 03 + 704 (18) F 4 i = -Fo4 + Foi (19) Since the component voltages of like frequency in the four phases differ in time-phase by 90 degrees, the line voltage for the fundamental or the line voltage for any harmonic is Vure = 2V phase COS 45 = \/2 V phase (20) where Vu ne is the magnitude of the voltage between any pair of adjacent lines for the fundamental or any harmonic and Vphase is the magnitude of the corresponding voltage in the phases. The instantaneous voltages between lines, i.e., line voltages, for star connection, corresponding to the phase voltages given in equations (12), (13), (14) and (15) are (see Figs. 90 and 91). Vi2 = -VQI + v 02 = \/2V ml sin (otf - 135) sin (3* + 135) sin (5coZ - 135) \/2V m7 sin (7at + 135) + etc. (21) HARMONICS IN POLYPHASE CIRCUITS 303 sin (at - 90 - 135) sin (3co* + 90 + 135) sin (5co* - 90 - 135) sin (7wt + 90 + 135) + etc. (22) V34 = _ V03 + Vo4 = ^/2V ml sin (at - 180 - 135) sin (3a>t + 180 + 135) sin (5co< - 180 - 135) + A/2 F m7 sin (7co^ + 180 + 135) + etc. (23) sin ( w ' - 270 - 135) sin (3^ + 270 + 135) sin (5co* - 270 - 135) + V~2V m 7 sin (7co^ + 270 + 135) + etc. (24) The root-mean-square values of the line and phase voltages for star connection are etc. 17 ,2 _|_ V 2 I V 2117 21 A f ' ml I r m3 ' w5 HT ' w7 JT "to. ~T~ ^ = V2 (27) ' phase The ratio of line to phase root-mean-square voltage for a four- phase, balanced, star-connected alternator or circuit is always equal to the square root of two. This is entirely independent of wave form. Although the line and phase voltages will contain like harmonics in the same relative magnitudes, the wave forms of the two voltages will not be alike. The relative phase dis- placements of the harmonics in the two voltages will be different, as will be seen by comparing equations (21), (22), (23) and (24) with equations (12), (13), (14) and (15) on page 299. The line and phase voltages of a mesh-connected, four-phase 304 PRINCIPLES OF ALTERNATING CURRENTS alternator or circuit are the same both in magnitude and wave form, since no harmonic is short-circuited in a mesh-connected four-phase circuit and thus eliminated from the terminal voltage. For balanced conditions, the same relations hold between line and phase currents for four-phase mesh connection as hold between line and phase voltages for a balanced, four-phase, star-connected circuit. If the neutral of a balanced, four-phase, star-connected load is connected to the neutral of the source of power supplying the load, no current will flow in the neutral, since the fundamentals of the phase currents and all harmonics of any order that may exist in the phase currents are ninety degrees apart in time-phase and therefore add up to zero at the neutral point. The vector sum of any four equal vectors which differ in phase by ninety degrees is equal to zero. The only current carried by the neutral connection of a four-phase, star-connected system is that due to an unbalanced load. Since the harmonics as well as the fundamentals in the phase voltages of a balanced four-phase system are ninety degrees apart in time-phase, there can be no circulatory current in the armature of a mesh-connected four-phase alternator, whose voltages are balanced, since the vector sum of the component voltages of any given frequency will be zero. This is different from the conditions which may exist in the armature of a three- phase, mesh connected, i.e., delta-connected alternator. In this case the third harmonics and all harmonics whose frequency is a multiple of triple frequency are short circuited in the closed delta formed by the armature windings. Harmonics in Balanced Six-phase Circuits. Although no six-phase alternators are built, there is a demand for a consider- able amount of six-phase power for the operation of rotary con- verters, or synchronous converters, as they are sometimes called. This is always obtained from three-phase systems by means of ordinary static transformers connected for three-phase to six- phase transformation. In most of the larger cities, alternating- current generation and transmission of power are used, but direct-current distribution is usually employed in the business and thickly settled districts. The rotary converter is also used in connection with most street railways operating on direct HARMONICS IN POLYPHASE CIRCUITS 305 current. The rotary converter is the chief connecting link between the alternating-current and the direct-current systems. The rotary converter is essentially a direct- current generator which has taps brought out to slip-rings from equidistant points on its armature. It receives polyphase alternating-current power through the slip-rings and delivers direct-current power from the direct-current commutator. On the alternating-current side it acts as a synchronous motor and on the direct-current side it acts as a direct-current generator. A single armature with a single armature winding serves for both motor and generator action. The armature winding may be considered to carry a current which is the difference of the alternating-current supplied and the direct-current delivered. On account of the armature current being equal to the differ- ence between the two currents, a greater direct-current out- put may be obtained for a fixed average armature copper loss than could be obtained were the machine driven mechanically and loaded as a direct-current generator. For a fixed amount of power converted, the efficiency of a rotary converter is higher and the cost is less than for a motor-generator which would do the same work. The gain in output and efficiency increases with the number of phases used. For three-phase operation at unity power-factor, the output of a rotary converter is about thirty per cent greater than could be obtained from the same machine operated as a direct-current generator. For six-phase operation it is nearly double. There is not a proportionate gain in going to twelve-phase operation, although it may pay to build very large rotary converters for this number of phases. On account of the importance of the rotary converter as a connecting link between alternating-current transmission and direct-current distribution, it is worth while to give some con- sideration to the relations among the harmonics in six-phase circuits. Six-phase systems are always derived from polyphase systems, usually three-phase, by means of transformers. A six-phase system may be considered to be two three-phase systems which are superposed, with the voltages of one reversed with respect to those of the other. This reversal may be obtained by merely reversing the connections of the secondary windings of the trans- 20 306 PRINCIPLES OF ALTERNATING CURRENTS formers supplying one of the superposed systems. Consider Fig. 92. This figure shows the vectors representing the funda- mental voltages of a six-phase system. FIG. 92. The two superposed three-phase systems, to which the six- phase system is equivalent, are distinguished in Fig. 92 by using full lines for one and dotted lines for the other. Figure 93 shows the six voltage vectors for the fundamentals of a six-phase system. Star Connection Mesh Connection FIG. 94 Either star or mesh connection may be used for six-phase circuits, but for the rotary converter, mesh connection must be used, since the armature winding must serve to generate direct current and must therefore form a closed circuit. All direct- current armatures must have closed-circuit windings. Star and mesh connection are illustrated diagrammatically in Fig. 94. The following table gives the phase relations among the fundamentals and harmonics in a balanced six-phase system. HARMONICS IN POLYPHASE CIRCUITS 307 The table gives the angular displacement between any harmonic in any phase with respect to the corresponding harmonic in phase one. Phase Phase displacement in electrical degrees 1st 3rd oth 7th 9th llth 1 2 60 3 X 60 - 180 5 X 60 = 300 7 X 60 = 420 =0= 60 9 X 60 = 540 =0= 180 11 X 60 = 660 =0= 300 3 120 3 X 120 = 360 =C=0<> 5 X 120 = 600 =0= 240 7 X 120 = 840 120 9 X 120 = 1080 =0= 11 X 120 = 1320 =0= 240 4 180 3 X 180 = 540 =0= 180 5 X 180 - 900 01 80 7 X 180 = 1260 =0= 180 9 X 180 = 1620 =0=1 80 11 X 180 = 1980 180 5 240 3 X 240 - 720 00 5 X 240 = 1200 =0=120 7 X 240 = 1680 =0=240 9 X 240 = 2160 =0= 11 X 240 = 2640 =0= 120 6 300 3 X 300 - 900 =C= 180 5 X 300 - 1500 =0= 60 7 X 300 = 2100 =0=300 9 X 300 = 2700 =0= 180 11 X 300 = 3300 =0= 60 The vectors for the fundamental and for the harmonics of a balanced six-phase system are shown in Fig. 95. From inspection of the table it is obvious that for a balanced six-phase system, all harmonics which are present, except those of For 1st, 7th, 13th etc. 020406 For 3rd, 9th, 15th etc. FIG. 95. For 5th, llth, 17th etc. triple frequency and multiples of this frequency, differ by sixty degrees in time-phase among themselves. The harmonics of triple frequency and multiples of this fre- quency are either in phase or in phase opposition among them- 308 PRINCIPLES OF ALTERNATING CURRENTS selves, those in adjacent phases always being in phase opposition. Omitting the harmonics of triple frequency and multiples of this frequency, the phase order of the others, beginning with the first harmonic, i.e., the fundamental, alternates from the direct order, i.e., 1-2-3-4-5-6, to the reverse order, i.e., 1-6-5-4-3-2. These phase relations among the fundamentals and harmonics of the different orders are what would be expected and what would necessarily follow from the phase relations existing among the fundamentals and harmonics of a balanced three- phase system. For a balanced, star-connected, six-phase system, the line voltage for the fundamental and for any harmonic, except those of triple frequency and multiples of this frequency, is equal to the phase voltage in magnitude. Refer to Figs. 94 and 95. 7i2 = -Foi + Fo2 Voi = Vio (= ^04) and F 2 are two equal voltages which are 120 degrees apart in phase. Therefore Vi2 = Vline = 2V phase COS 60 = 2 X ~V phase = V phase (28) For the harmonics of triple frequency or multiples of this frequency in a balanced three-phase system, F i and Vo 2 are in phase. Therefore for a balanced, star-connected, six-phase system Fi2 = -Foi + Fo2 Vline = 2V phas6 (29) There can be no voltages of triple frequency or multiples of this frequency in the voltages between alternate line terminals of a balanced, star-connected, six-phase system. Consider terminals 1 and 3. The voltage between these terminals for star connection is (Fig. 94, page 306.) Fl3 = -Foi + F 3 By referring to the table on page 307 it will be seen that the voltage Fis is zero for the harmonics of triple frequency and multiples of this frequency. For mesh connection, the voltage between terminals 1 and 3 is, (Fig. 94), page 306,_ Vis = -(Foi + Fo 2 ) (30) HARMONICS IN POLYPHASE CIRCUITS 309 By referring to the table, it will be seen that this is zero for harmonics of triple frequency or multiples of this frequency. Although there can be no harmonics of triple frequency or multiples of this frequency in the voltage between alternate terminals, i.e., between the three-phase terminals, of a balanced six-phase system, there may be harmonics of triple frequency or multiples of this frequency in the voltage between any diametrical terminals, such as 1 and 4, of a balanced six-phase system. Similar relations exist among the line and phase currents of a balanced, mesh-connected, six-phase system to those holding between the line and phase voltages of a balanced, star-connected, six-phase system. When mesh connection is used for an armature, there can be no short-circuit current in the armature, as there may be in the case of three-phase delta connection, since the vector sum of all the component voltages acting around a mesh-connected, balanced, six-phase circuit is zero. By referring to the table on page 307 it will be seen that the vector sum of the six-phase voltages acting around a mesh-connected, balanced, six-phase circuit is zero for the fundamental and for each harmonic. The diametrical voltage of a six-phase, mesh-connected circuit between any two diametrical terminals, such as terminals 1 and 4, is Fl4 = Foi + F 02 + F 03 (31) By referring to the table on page 307 it will be seen that for harmonics of triple frequency or multiples of this frequency, this is equal to Fos or Voi. For star connection, the diametrical voltage between any dia- metrical terminals, such as 1 and 4, for balanced conditions is FH = -Foi + 7o4 (32) By referring to the table on page 307 it will be seen that, for the third harmonic or any multiple of the third harmonic, Vu = Therefore, for mesh connection, the third-harmonic, six-phase. line voltage is equal to the third-harmonic phase voltage. For star connection it is double the phase voltage The same state- ments hold for any multiple of the third. Balanced conditions are assumed. 310 PRINCIPLES OF ALTERNATING CURRENTS In general for a balanced six-phase system, no harmonics are suppressed in the voltage between adjacent terminals. Any harmonic voltage that exists in the phase voltage will be present in the voltage between adjacent terminals The harmonics of triple frequency and multiples of this frequency that exist in the phase voltage are suppressed between alternate terminals, i.e., between the three-phase terminals. For mesh connection, the phase and line voltages are identical. For star connection, phase and line voltages are the same only when the phase voltages contain no harmonics of triple frequency or multiples of this frequency. For star connection, the fundamental line and phase voltages are equal. The line and phase voltages for any given harmonic, except the third or its multiples, are also equal. For the third-harmonic or any harmonic whose frequency is a multiple of triple frequency, the line voltage for star connection is double the phase voltage. It is equal to the phase voltage for mesh connection. An Example Illustrating the Relation Among the Voltages of a Six -phase Mesh-connected System Having a Badly Dis- torted Wave Form. One type of rotary converter is purposely built in such a way that its phase voltage may be badly distorted by changing the flux distribution in the air gap in order to alter the root-mean-square alternating-current voltage obtained with a given pole flux. In this way it is possible to change the ratio of the alternating- and direct-current voltages. Under certain conditions, the phase voltage, i.e., the voltage between adjacent armature taps, was found to be e = 325 sin 377* + 110 sin (1131* + 90) + 50 sin (1885J + 50) If time is reckoned from the instant when the fundamental in the voltage of the phase between armature taps 1 and 2 is zero and increasing in a positive direction, what are : (a) The expressions for the six instantaneous six-phase voltages? (6) The expression for the instantaneous three-phase voltage between armature taps 1 and 3? (c) The expression for the instantaneous diametrical voltage between armature taps 1 and 4? (d) What are the root-mean-square values of the six-phase HARMONICS IN POLYPHASE CIRCUITS 311 voltages, i.e., the voltages between adjacent armature taps; the three-phase voltages, i.e., the voltages between alternate arma- ture taps; and the diametrical voltages, i.e., the voltages between any two diametrical armature taps such as 1 and 4? The voltage between armature taps 1 and 2 will be assumed to lead the voltage between taps 2 and 3. The armature of a rotary converter is mesh-connected. The expressions for the six instantaneous six-phase voltages are (see Figs. 94 and 95, pages 306 and 307 respectively and also the table on page 307) e n = e lo = 325 sin 377* + 110 sin (1131* + 90) + 50 sin (1885* + 50) e 23 = e 20 = 325 sin (377* - 60) + 110 sin (1131* - 180 + 90) + 50 sin (1885* - 300 + 50) e 34 = 630 = 325 sin (377* - 120) +110 sin (1131* - + 90) + 50 sin -(1885* - 240 + 50) e 45 = 640 = 325 sin (377* - 180) + 110 sin (1131* - 180 +90) + 50 sin (1885* - 180 + 50) 6o6 = 6b0 = 325 sin (377* - 240) + 110 sin (1131* - + 90) + 50 sin (1885* - 120 + 50) eei = eeo = 325 sin (377* - 300) + 110 sin (1131* - 180 + 90) + 50 sin (1885* - 60 + 50) The root-mean-square value of the six-phase voltage is * six-phase A/ (325) 2 + (110) 2 + (50) 2 = 245.2 volts. The expression for the instantaneous voltage between armature taps 1 and 3 is (see Fig. 94, page 306.) = A/3 X 325 sin (377* - 30) + + V3 X 50 sin (1885* + 50 + 30) = 562.9 sin (377* - 30) + 86.6 sin (1885* + 80) The root-mean-square value of the voltage between armature taps 1 and 3 is v v /(562.9) 2 + (86.6)' r 13 three-phase 2 402.7 volts 312 PRINCIPLES OF ALTERNATING CURRENTS The ratio of the three-phase and six-phase voltages is Vthfee-phase _ 402.7 _ ., _ .~ V six . phase 245.2 " When there are no third harmonics, or multiples of the third, present in the six-phase voltage, the ratio of the three-phase and six-phase voltages is equal to 1.732, i.e., the square root of three. The voltage between any two diametrical armature taps is equal to the vector sum of three equal vectors which are displaced by equal angles. The easiest way to find the resultant of these three vectors is to add the first and third vectors and then add their resultant to the second vector. For the fundamental and all harmonics, except those of triple frequency or a multiple of this frequency, the first and third vectors are 120 degrees apart in time-phase. The vector sum of these is equal both in phase and in magnitude to the second vector. The vector sum of the three vectors therefore is in phase with the second vector but has twice its magnitude. That this statement is true will be seen by refer- ring to Fig. 95, page 307. For the harmonic of triple frequency or for harmonics of any multiple of triple frequency, the first and second vectors are opposite in phase. (See Fig. 95, page 307.) Their vector sum is zero. The resultant of the three vectors for the harmonic of triple frequency or for harmonics of any multiple of triple frequency is equal in direction and magnitude to the third vector. It follows from the preceding statements that the expression for the instantaneous voltage between diametrical armature taps 1 and 4 is 014 = 010 + 020 + 030 = 2 X 325 sin (377* - 60) + 110 sin (1131* + 90 - 180) + 2 X 50 sin (1885* + 50 + 60) = 650 sin (377* - 60) + 110 sin (1131* - 90) + 100 sin (1885* + 110) The root-mean-square value of the voltage between diametrical armature taps 1 and 4 is v _ y /(650) 2 + (IIP) 2 + (100) 2 * 14 ~ V diametrical A/ ~ ^ 471.5 volts, HARMONICS IN POLYPHASE CIRCUITS 313 The ratio of the diametrical and six-phase voltages is Vdiametrical _ 471.5 _ V tit . p ^ 245.2 = When there are no harmonics of triple frequency or multiples of this frequency present in the voltages, the ratio of the diam- etrical and six-phase voltages is equal to 2. It should be noted that the wave forms of the six-phase, three- phase and diametrical voltages are all different. CHAPTER XI POWER AND POWER-FACTOR OF POLYPHASE CIRCUITS, RELATIVE AMOUNTS OF COPPER REQUIRED FOR POLYPHASE CIRCUITS, POWER MEASUREMENTS IN POLYPHASE CIRCUITS Power and Power -factor of Balanced Polyphase Circuits. Since a polyphase alternator has as many independent windings on its armature as it has phases, it is evident that the total output of such an alternator must be equal to the sum of the outputs of all its phases, no matter how they may be interconnected. In general, the power in any polyphase circuit whatsoever is equal to the sum of the powers developed in each phase. Total power = P = PI + P 2 + PS + etc. (1) = ZP where the P's with subscripts 1, 2, 3, etc. are the powers developed in phases 1, 2, 3, etc. respectively. For a three-phase circuit Po = Pi + P 2 + P 3 = TVi X (p./.)i + VJ* X (p./.)2 + TVs X (p./.)s (2) Vi, V z and V 3 are the phase voltages. 7i, 7 2 and J 3 are the corresponding phase currents, and (p./.)i, (p./.) 2 and (p./.) 3 are the corresponding phase power-factors. For sinusoidal waves, the power-factor is equal to the cosine of the angle between the phase current and the phase voltage. The total volt-amperes of a circuit is equal to the sum of the volt-amperes in its phases. Total volt-amperes = V l l l + V 2 I 2 + F 3 / 3 + etc. (3) = SF7 In a balanced polyphase circuit, all the phase currents are 360 equal in magnitude and differ in phase by - - degrees, where n is 314 POWER AND POWER-FACTOR 315 the number of phases. All phase voltages are also equal in mag- Oflfl nitude and differ in phase by degrees. It follows for a bal- anced polyphase circuit that the phase power-factors must all be equal. For a balanced circuit Total power = P = nP p = nVJp X (p.f.) P (4) where n is the number of phases and V p , I p and (p.f.) P are the phase voltage, the phase current and the phase power-factor respectively. For a balanced three-phase circuit, P = 3 P P = 3TVp X (p./.)* (5) Total volt-amperes = 3V P I P (6) For a A-connected circuit, the line and phase voltages are the same. For a Y-connected circuit, the line and phase currents are the same. For a balanced circuit having sinusoidal waves of current and voltage, the line current is equal to the phase current multiplied by the square root of three for delta connection. The line voltage is equal to the phase voltage multiplied by the square root of three for wye connection. Indicating line current and line voltage by I L and V L respec- tively, equations (5) and (6) for sinusoidal waves and a balanced A-connected circuit become Total power = P = 3V P L (p./.), = V3 VJ L cos P (7) Total volt-amperes = 2V P I P = 3V p = = V3VJ L (8) where 6 P is the phase power-factor angle, i.e., the phase angle between any phase current and the corresponding phase voltage. For sinusoidal waves and a balanced Y-connected circuit, equations (5) and (6) become Total power = P = 3 ^~ /p(p.f -) P y 3 = V3VJ L cos P (9) 316 PRINCIPLES OF ALTERNATING CURRENTS Total volt-amperes = SF P 7 P = 3 JL I p = V*VJ L (10) From equations (7), (8), (9) and (10), it is evident that the expressions for the total power and the total volt-amperes of a balanced three-phase circuit, having sinusoidal waves of current and voltage, are the same whether the circuit is A- or Y-connected. Equations (7), (8), (9) and (10) also hold for balanced condi- tions when the current and voltage waves are not sinusoidal, provided there are no harmonics present of triple frequency or any multiple of that frequency. This follows from the fact that when a balanced three-phase circuit contains no harmonics of triple frequency or multiples of this frequency the ratio of line to phase voltage for wye connection and the ratio of line to phase current for delta connection are equal to the square root of three. (See pages 290 and 294.) When equations (7), (8), (9) and (10) are applied to a balanced circuit having non-sinusoidal waves, but no harmonics of triple frequency or any multiple of this frequency, V L and IL must be understood to be the equiva- lent sinusoidal voltage and current respectively. The angle B p is the equivalent phase angle for the equivalent sinusoidal waves in each phase. Since there can be no third harmonics in the line voltages or line currents of a balanced three-phase system, there can be no third harmonics in the equivalent wye voltages or equivalent delta currents of such a system. There are only a few cases where there will be third harmonics in the actual wye voltages of a circuit or in the actual delta currents of a circuit, when balanced line voltages are impressed. From equations (9) and (10) it is obvious that the power-factor of a balanced three-phase circuit, which does not contain har- monics of triple frequency or multiples of triple frequency in either voltage or current in any of its phases, is given by (Pj.)p = /~ P v r (ID V3 VL IL For sinusoidal waves of voltage and current, this is equal to the cosine of the phase angle between the phase current and phase voltage. Power-factor is never the cosine of the phase angle between the line current and line voltage. POWER AND POWER-FACTOR 317 The power-factor of a balanced four-phase circuit is given by the following expression x (p ' f ' )p - Po For star connection and balanced conditions, the line voltage is equal to the phase voltage multiplied by the square root of two, and the line and phase currents are equal. This statement is independent of the wave form of the current or voltage. Therefore for a balanced, four-phase, star-connected circuit 4 VL_ T 2-s/2FJz, (13) \/2 The same expression holds for a balanced, mesh-connected, four-phase circuit. For a balanced, mesh-connected, four-phase circuit, the line current is equal to the phase current multiplied by the square root of two, and the line and phase voltages are the same. This statement is independent of the wave form of the current or voltage. Therefore, for a balanced, four-phase, mesh-connected circuit ( = P = - ^- AV IL ZVZVJL (14) ^ ' L 7^. V2 Equations (13) and (14) are true for any balanced four-phase circuit, whether star- or mesh-connected, and are independent of wave form. For sinusoidal waves of current and voltage, equa- tions (13) and (14) reduce to cos 6 P , where B p is the phase angle between the phase current and the phase voltage. Power-factor of an Unbalanced Polyphase Circuit. The power-factor of a balanced polyphase circuit is a perfectly definite thing which may be determined by simple measurements. For a balanced polyphase circuit the power-factors of all phases are equal. A satisfactory workable definition for the power- factor of an unbalanced polyphase circuit is still to be determined. The question of the definition of the power-factor of an un- balanced polyphase circuit is under discussion at the present time (1921) by the Standards Committee of the American Institute of Electrical Engineers. The power-factor of an unbalanced polyphase circuit might be 318 PRINCIPLES OF ALTERNATING CURRENTS defined as the average phase power-factor, but if so defined it would be impossible of determination, except by measuring the current, voltage and power of each phase separately. This is impracticable in the majority of cases and often would be impossible. For example, if the average power-factor of an unbalanced A-connected circuit were to be determined, it would be necessary to open up the delta and insert an ammeter and a wattmeter in each phase in order to measure the current and power of each phase separately. Line and phase voltage of a A-connected circuit are the same. Although in experimental work it might be possible to open a A-connected circuit in order to insert instruments, in commercial practice it would seldom be possible to do this. To determine the average power-factor of a Y-connected circuit, it would be necessary to have the common junction between branches of the load available, in order to measure the voltage and power of each phase. In many cases the neutral connections of Y-connected circuits are not brought out or are not available at the point at which the measurements have to be made. If the average power-factor of a load on a power station were to be determined, the only measurements that could be made would be the line currents, the voltages between the three pairs of lines or conductors and the total power. These measurements alone would not be sufficient to determine the average power-factor of the load. In many cases equation (11), page 316, is used to determine the power-factor of an unbalanced three-phase circuit by sub- stituting for VL and I L the average voltage between lines and the average line current respectively. The result obtained in this way has no particular significance. It is not the average phase power-factor or the total power divided by the total volt-amperes, or anything else in particular. If P, I and V with subscripts 1, 2 and 3 represent phase power, phase current and phase voltage of an unbalanced three-phase circuit, the average phase power-factor of such a circuit is given by the following expression 3 ( V I VI VI- Pl(F 2 /2)(F 8 /3) +Pl(FlJl)(1 (15) POWER AND POWER-FACTOR 319 The expression derived from equation (11) by substituting average line current for I L and average voltage between lines for V L is ( n = Pi + P* + P* (16) where Viz, V& and V 3i are the voltages between the pairs of lines 1-2, 2-3 and 3-1 respectively, and I L i, ILZ and ILS are the three line currents. It is obvious, by comparing equations (15) and (16), that equation (16) can give the average power-factor only when the load is balanced, i.e., when Viz = Vzs = FSI and I L \ = ILZ = /. For small amounts of unbalancing, the average power-factor and the value of the power-factor calculated from equation (16) do not differ greatly in most cases, but when equation (16) is used, the fact should not be overlooked that it does not give the average phase power-factor. Commercial circuits as a rule are not much out of balance. It has been suggested (1924), by a special committee appointed for the purpose of denning the power-factor of a polyphase circuit, that the power-factor of such a circuit be defined as the ratio of the total active power to the square root of the sum of the squares of the total active power and the total reactive power, i.e., , -, Total active power V (Total active power) 2 + (Total reactive power) 2 This definition does not give the ratio of watts to volt-amperes except when the circuit is free from harmonics. The total active power of any balanced or unbalanced three- phase circuit, that does not have a neutral connection which carries current, may be measured by the two-wattmeter method for measuring power or by a polyphase wattmeter (page 326). The total reactive power of a balanced or unbalanced three-phase circuit, that does not have a neutral connection which carries current, may be measured by two reactive-power meters (pages 63 and 327a) or by a polyphase reactive-power meter provided the current and voltage are sinusoidal. If power-factor is defined in the way suggested by the com- mittee, it is possible to determine the power-factor of a balanced 320 PRINCIPLES OF ALTERNATING CURRENTS or an unbalanced three-phase circuit as easily as the power-factor of a single-phase circuit. Since the maximum output of all alternating-current generators, motors and transformers and the regulation and capacity of transmission and distribution lines depend not only upon the amount of power developed but also upon the power-factor at which they operate, it is extremely important to have a definition of the power-factor of a polyphase circuit which makes it possible to determine the power-factor of such a circuit easily. Power-factor may then be taken into consideration in determining equitable power rates for customers who use power at relatively low power-factor. Relative Amounts of Copper Required to Transmit a Given Amount of Power a Fixed Distance, with a Fixed Line Loss and Fixed Voltage between Conductors, over a Three-phase Trans- mission Line under Balanced Conditions and over a Single-phase Line. Let P and V be respectively the power transmitted and the limiting voltage between conductors. Let I\ and 7 3 be the currents per conductor for the single-phase and three-phase transmission respectively. Then P(single-phase) = VI \ cos 6 P (17) P(three-phase) = V3F/ 3 cos 6 P (18) where B p is the power-factor angle, i.e., the angle between the line current and line voltage for single-phase transmission and the angle between the line current and the equivalent wye voltage for the three-phase transmission. Balanced three-phase transmission is assumed. From equations (17) and (18) Let ri and r 3 be the resistance per conductor of the single-phase and three-phase lines respectively. Then for equal copper losses for the single-phase and three-phase transmission 27, V, = 37 3 V 3 'J _ 3 7 3 * _ 3 1 __ 1 r 3 ~2 X 7?~2 X 3~2 Since the amount of copper required for a given length of transmission line is directly proportional to the number of POWER AND POWER-FACTOR 321 conductors employed and inversely proportional to the resistance per conductor Copper for three-phase transmission _ 3 1 _ 3 Copper for single-phase transmission 2 2 ~~ 4 From equation (21) it is evident that twenty-five per cent less copper is required to transmit a given amount of power a given distance, with a fixed transmission loss (copper loss) and a fixed voltage between conductors, over a three-phase line under balanced conditions than over a single-phase line, or thirty-three and a third per cent more copper is required for the single-phase transmission. Relative Amounts of Copper Required to Transmit a Given Amount of Power a Fixed Distance, with a Fixed Line Loss and a Fixed Voltage between Conductors, over a Four -phase Trans- mission Line under Balanced Conditions and over a Single-phase Line. The maximum voltage between conductors of a four-phase line exists between alternate conductors. If V represents this voltage, the voltage between adjacent conductors is A/2 -~ X V = 0.7077 If the limiting voltage, V, is taken as the maximum voltage between any two of the conductors of the four-phase line, the four-phase system will require the same amount of copper as a single-phase line for the transmission of a given amount of power, under balanced conditions, a fixed distance with a fixed line loss. The four-phase system is equivalent to two single- phase lines each transmitting half the total power at a voltage equal to the diametrical voltage of the system. Twice as many conductors are required as for single-phase transmission, but each conductor need be only half as large, as it carries only half as much current as each conductor of the single-phase line. If the voltage between adjacent conductors of the four-phase line is made equal to the voltage, 7, between conductors of the single-phase line, the maximum voltage between any two con- ductors of the four-phase line will be \/2 X V and will occur between alternate conductors. If V is taken as the voltage between adjacent conductors of the four-phase line and 7 4 is used as the current per conductor 322 PRINCIPLES OF ALTERNATING CURRENTS y P (four-phase) = 4 7= I* cos O p V 2 = 2A/27/4 cos P (22) P (single-phase) = 71 1 cos B p (23) Since the power is the same in the two cases i 4 = ^7= (24) Ii 2V2 If r 4 is the resistance per conductor of the four-phase line, for equal losses for single-phase and four-phase transmission 2/i 2 ri = 4 / 4 V 4 n = 2 x ^! = 2 x / 1 N 2 r 4 /i 2 Copper for four-phase transmission __ 4 1. = 1. . , Copper for single-phase transmission ~ 2 4 2 Relative Amounts of Copper Required to Transmit a Given Amount of Power a Fixed Distance, with a Fixed Line Loss and a Fixed Voltage to Neutral, when the Loads are Balanced. When the voltage to neutral is fixed, there is no difference in the amounts of copper required to transmit a given amount of power a fixed distance with a fixed line loss, whether the transmission be single- phase, three-phase or four-phase. Let V n be the voltage to neutral. Then P(single-phase) = 2V n I 1 cos 6 P (27) P(three-phase) = 3V a I 3 cos 6 P (28) P (four-phase) = 47 n / 4 cos P (29) Is _ 2 /4 = 2 Ii ~ 3 /i 4 2Ii*ri = 3I 3 2 r 3 ri_3l3!_34_2 r 3 ~ 2 X /! 2 ~ 2 X 9 " 3 2/iV! = 47 4 2 r 4 r 1 _47 4 2 = 44_l r 4 2 A 7i 2 2 A 16 2 Copper for three-phase transmission __ 3 2 __ ^ Copper for single-phase transmission 2 3 Copper for four-phase transmission _ 4 1 _ ^ Copper for single-phase transmission 2 2 POWER AND POWER-FACTOR 323 Power Measurements in Three-phase Circuits. The total power in any polyphase circuit is equal to the algebraic sum of the powers in each phase. If a wattmeter is placed in each phase, the sum of the wattmeter readings is the true power in the circuit. Consider the three-phase A-connected and Y- connected circuits shown in Fig. 96. The circles marked W in this figure represent the current coils of the wattmeters. The rectangles in the middle of the circles are the potential coils. The current coil of each wattmeter carries the current in one phase. The voltage of this phase is impressed across the poten- tial coil of the wattmeter. Obviously each wattmeter, connected as shown in Fig. 96, measures the power in a single phase. The FIG. 96. sum of the readings of the three wattmeters, for either the delta connection or the wye connection, must give the total power in the circuit. Under ordinary conditions, it is impossible to break into a A-connected circuit and thus place a wattmeter in each phase as shown in the left-hand half of Fig. 96. Neither is it always pos- sible, in the case of a Y-connected circuit, to get at the neutral point which is required for the connections shown in the right- hand half of Fig. 96. If the connection shown in the right-hand diagram, between the common junction of the wattmeter poten- tial coils and the neutral point of the circuit, is omitted, the sum of the readings of the three wattmeters really the algebraic sum will give the true power in the circuit for any degree of unbalancing or for any wave form, whether the circuit be A- connected or Y-connected, provided the load, if Y-connected, does not have its neutral point connected to the neutral of the 324 PRINCIPLES OF ALTERNATING CURRENTS FIG. 97. source of power. If the neutrals are interconnected, but the neutral connection does not carry current, the conditions are the same as if it did not exist. Even with a balanced load, the neutral connection, if it exists, will carry current if the phase voltage of the circuit contains a harmonic of triple frequency or any multiple of triple frequency. When there is no harmonic of triple frequency or multiple of this frequency in the phase voltage of the source of power, there will still be a pronounced third-harmonic current in the neutral connection under bal- anced conditions, when the load is inductive and contains iron, especially if the iron is worked at high saturation. Proof of the Three -wattmeter Method for Measuring the Power in a Three-phase Circuit. Let the instantaneous values of the phase cur- rents, phase voltages and phase powers be denoted by i, e and p, respectively, with subscripts 1, 2 and 3 to indicate the particular phase. Let the three wattmeters Be connected as shown in Fig. 97. The instantaneous power in any single-phase circuit is equal to the product of the instantaneous values of current and voltage. Therefore the total instantaneous power, po, in the three-phase circuit is Po = ioieoi + ^02^02 + iotfoa (32) According to Kirchhoff's laws ioi + ioz + ioa = (33) Let e a be the instantaneous potential difference between the points and a, Fig. 97, i.e., the difference in potential between the neutral point of the system and the common junction of the potential coils of the wattmeters. Then the following relation is obviously true. eoa(ioi -f io2 + ios) = eoaioi + e 0a ioz + e 0a ioa = (34) Subtracting equation (34) from equation (32) gives POWER AND POWER-FACTOR 325 oa) + ^02(^02 e oa) H- ias(eoz e 0a ) = PQ (35) But Therefore, the instantaneous power in the circuit is PQ = ioi a i + iotCaZ + ^03^3 (36) = Pi + P2 + P3 (37) where pi, p 2 and p 3 are the instantaneous powers corresponding to ioiBai, iotfaz and tosias respectively. The average power in the circuit is I f Tjo (38) which is the sum of the actual wattmeter readings. This method of measuring power in a three-phase circuit is known as the three- wattmeter method. Since any A-connected circuit may always be replaced by an equivalent Y-connected circuit, it is evident that the three-wattmeter method of measur- ing power may be used to measure the total power in a A-con- nected circuit as well as to measure the total power in a Y-connected circuit. The proof of the three-wattmeter method is based on the assumption that the sum of the instantaneous currents in the phases of the load is zero. If there is a neutral connection between the load and the source of power, and it carries current, the sum of the three instantaneous phase currents is no longer equal to zero. When the neutrals of the load and source of power are connected and the neutral connection carries a current, io, equation (33) becomes ion + 1*01 + ioz + ios = The sum of the currents ioi, i z and io 3 is no longer zero, except when the neutral current is zero. The proof of the three-watt- meter method for measuring power holds only when there is no neutral connection between the source of power and the load or when the neutral connection, if used, carries no current. If 326 PRINCIPLES OF ALTERNATING CURRENTS the neutral point, a, of the wattmeters is connected to the neu- tral point, 0, of the load (Fig. 97, page 324), the three- wattmeter method of measuring power will give the true power under all conditions, since there is a wattmeter in each phase of the load. The N-wattmeter Method for Measuring the Power in an N -phase Circuit. The proof just given is not limited to three- phase circuits. It may be extended to apply to a circuit with any number of phases. The power in an n-phase circuit may be measured by n wattmeters, each with its current coil in one line and with its potential coil connected between the line in which its current coil is placed and a common junction point, to which one terminal of each wattmeter is connected. The n-wattmeter method does not hold when there is a neutral connection, which carries current, between the load and the source of power, except when the common junction of the potential coils of the n watt- meters is connected to the neutral of the load. The three-watt- meter method for measuring three-phase power or, in general, the n-wattmeter method for measuring n-phase power is seldom used. Three-phase power is usually measured by the two- watt- meter method. In general the (n l)-wattmeter method is used for measuring the power in an n-phase circuit. Two -wattmeter Method for Measuring Power in a Three- phase Circuit and the (N 1) -wattmeter Method for Measuring Power in an N-phase Circuit. No assumption was made in the proof of the three-wattmeter method regarding the position of the common junction point, a, of the "wattmeter potential coils, which may be any point whatsoever. It may be on one of the lines. In this case, one wattmeter will read zero and may be omitted. The algebraic sum of the readings of the two remaining watt- meters will then give the true power. If there are more than three phases, the common junction point of the potential coils of the n wattmeters may still be on one of the lines making the reading of one of the n wattmeters zero. The algebraic sum of the readings of the other (n 1) wattmeters will be equal to the true power in the circuit. POWER AND POWER-FACTOR 327 In general, the power in an n-phase circuit may be measured by (n 1) wattmeters, each with its current coil in one line and its potential coil bridged between the line containing its current coil and the line which does not contain the current coil of a watt- meter. Applied to a three-phase circuit the (n 1) -wattmeter method becomes the two- wattmeter method. The connections for the two-wattmeter method for measuring power in a three- phase circuit are shown in Fig. 98, page 326. 1 C T I C T ^ = T i i01 e * ldt ~^ T I * The two-wattmeter method, or in general, the (n 1)- wattmeter method, gives the true power in a circuit without regard to balance or wave form, provided the neutral of the load, if star-connected, does not have its neutral connected to the neu- tral of the source of power. If the neutral connection carries no current, the conditions are the same as if the neutral connection did not exist. About the only conditions likely to occur in practice, when a neutral connection for a three-phase circuit will not carry current, are when the load is exactly balanced and there are no harmonics present of triple frequency or any multiple of that frequency. The connections of the wattmeters for the three- wattmeter method are symmetrical. The readings of the three wattmeters, for the three-wattmeter method of measuring the power in a three-phase circuit therefore, will either be all positive or all nega- tive, barring very exceptional conditions, and will add directly to give the true power in the circuit. The only case where the read- ings of the wattmeters will not either be all positive or all negative is when the load is very badly unbalanced. Under all conditions, the algebraic sum of the wattmeter readings is the true power. With the two-wattmeter method, the wattmeters are not connected symmetrically. By referring to Fig. 98, it will be seen that wattmeter No. 1 has its potential coil connected from line 1 to line 3 or in a left-hand direction with respect to the sequence of phases as numbered. Wattmeter No. 2 has its potential coil connected from line 2 to line 3 or in a right-hand direction. The algebraic sum, not the numerical sum, of the 327 'a PRINCIPLES OF ALTERNATING CURRENT readings of the two wattmeters always gives the true power. In order to read a wattmeter it must be connected so that its pointer will deflect up scale. It is necessary, therefore, to so connect the two wattmeters that their pointers will deflect up scale and then to determine, from the connections used, whether the readings of the wattmeters are alike in sign or opposite. For balanced loads, the readings will be opposite in sign whenever the power-factor of the circuit is less than five-tenths. To determine whether the readings of the wattmeters used in the two-wattmeter method for measuring power in a three-phase circuit are of like or unlike sign, it is merely necessary to see whether the wattmeters are connected alike or differently, i.e., to see whether the current is lead to corresponding current terminals and whether corresponding ends of the potential coils are con- nected to the common line. If the wattmeters are of different make or type they may both be placed in the same circuit to determine which terminals correspond. If the load is approxi- mately balanced, the sign of the readings of the wattmeters may be determined very easily by merely disconnecting from the common line (line 3 in Fig. 98) the potential coil of the watt- meter which has the smaller deflection and connecting it to the line not containing its current coil. If the reading of the watt- meter reverses when this is done, the signs of the wattmeter readings are opposite and the readings must be subtracted to give the true power. Changing the connections of the wattmeter with the smaller deflection, as indicated, connects the potential coils of both wattmeters alike with respect to the cyclic order of the three-phase circuit and serves the same purpose as placing both wattmeters in the same circuit. The Total Reactive Power of a Three-phase Circuit, whose Wave Forms are Sinusoidal and which has no Neutral Connec- tion that Carries Current, may be Measured by the Use of Two Reactive-power Meters Connected like the Wattmeters for Measuring Power by the Two-wattmeter Method for Measuring Power in a Three-phase Circuit. Let the current and voltage waves of a three-phase circuit be sinusoidal. They may then be represented by vectors each of which may be resolved in to a real and an imaginary component. Let small v's with and with- PO WER AND POWER-FACTOR 3276 out primes represent respectively imaginary and real components of the voltages. Let small i's with and without primes represent respectively corresponding components of the currents. Assume that there is no neutral current. Refer to Fig. 98, page 326. Let 7oi = ii + jii 7 02 = it + jit' 7os = is + jis Also let Voi = Vi + jVi 702 = Then, referring to page 66, the total reactive power is P r = Wii - wiii') + Wit - vtit') + Wit ~ fW) (39a) Since there is assumed to be no neutral current, 1 01 + /02 + /03 = The following relations must also be true. ii + 12 + *s = (396) ii' + U + i* = (39c) Substituting the values of i 3 and i s f from equations (396) and (39c) in equation (39a) gives P r = vi'i\ viii -h v*i* v 2 i2 v 3 '(ii + it) + v a (ii + it) = iiW - vs) + ii(-Vi + v 3 ) -f 12^2' - Vs') H- i2(-v z +v 3 ) - iiv^i) + (^2^32' - ^2^32) (39d) where P\ r and P 2r are respectively the readings of reactive-power meters connected as shown in Fig. 98, page 326. 328 PRINCIPLES OF ALTERNATING CURRENTS Problem Illustrating the Use of the Two-wattmeter Method for Measuring Power in a Balanced Three-phase Circuit. Two wattmeters are connected to measure the power taken by a certain balanced, three-phase inductive load. Wattmeter No. 1 has its current coil in line 1 and its potental coil between lines 1 and 3. Wattmeter No. 2 has its current coil in line 2 and its potential coil between lines 2 and 3. The readings of the watt- meters are respectively 40.96 and 4.36 kilowatts. When that end of the potential coil of wattmeter No. 2 which is connected to line 3 is connected to line 1, the reading of wattmeter No. 2 reverses and the pointer goes up against the stop. If the con- nection of the current coil of this wattmeter is now reversed, so that the pointer will deflect up scale, the wattmeter reads 40.96 kilowatts. (a) What is the power taken by the load? (6) If the line currents are each 100 amperes and the voltages between lines are each 500 volts, what is the power-factor of the load? (c) By what angle does the equivalent sine current in each phase lag the equivalent sine voltage impressed across each phase? (d) If the load is Y-connected without a neutral, what is the current in each phase and what is the voltage across the terminals of each phase? (e) If the load had been A-connected, what would have been the current in each phase and what would have been the voltage across the terminals of each phase? Since there is no neutral connection, when the load is con- nected in wye there can be no harmonic of triple frequency or multiple of this frequency in the current. There can be no harmonic of triple frequency or multiple of this frequency in the line voltage, since the load is balanced. There would be a third harmonic in the phase voltage, i.e., the wye voltage, if the load involved iron. It will be assumed that no iron is present. In this case there can be no harmonic of triple frequency or multiple of this frequency in any part of the circuit. Under this condi- tion, the ratio of the line and phase (wye) voltages will be equal to the square root of three. If no iron is present, there will be no circulatory current of triple frequency or any multiple of this POWER AND POWER-FACTOR 329 frequency in the closed delta, when delta connection is assumed. Under this condition, the ratio of line and phase (delta) currents will be equal to the square root of three. (a) Since wattmeter No. 2 reverses when the connections of its potential coil are changed, its reading must have been negative as originally connected. Therefore, Total power = F = 40.96 - 4.36 = 36.60 kilowatts. Po 36.60 = V3 /. V line - V3X100X500 = 0.4226 (c) Angle of lag of the equivalent sine phase current behind the equi- valent sine phase voltage for either wye or delta connection is 6 P = cos- 1 0.4226 = 65 degrees. (d) = 100 amperes. Wye connection 500 = 288.7 volts. V'6 w 7. Delta connection 100 phase = ~= = 57.7 amperes. V ph ase = 500 VOltS. Another Example of the Use of the Two-wattmeter Method. It is frequently convenient, when solving problems in three-phase circuits, to apply the principle of the two-wattmeter method to determine the total power taken by a circuit. When the solution of a problem gives the line currents and line voltages expressed in their complex form, the easiest way to determine the total power consumed is to assume that wattmeters have been inserted in the circuit to measure the power by the two-wattmeter method. 330 PRINCIPLES OF ALTERNATING CURRENTS The solution of the problem illustrating KirchhofTs laws, which was given on page 272, gives for the three line currents lao = 20.10 - J10.44 ho = -10.62 + jl.83 I co = -9.48 + J8.60 The line voltages are F6 = 230 + jO V bc = -H5 - J199.2 y ca = -115 + 199.2 Assume that one wattmeter is placed in line a (see Fig. 80, page 272), with its potential coil connected between lines a and b, and that the other wattmeter is placed in line c, with its potential coil connected between lines c and 6. The algebraic sum of the readings of the wattmeters, if actually connected as indicated, would be the total power in the circuit. Wattmeter in line a would carry a current I ao = 20.10 jlO.44 and would have a voltage Vab = 230 + JO impressed across its potential coil. This wattmeter would read P a = (20.10)(230) +'(-10.44X0) = 4623 + = 4623 watts. The other wattmeter would carry a current I co = 9.48 + J8.60 and would have a voltage V cb = -V bc = 115 + J199.2 impressed across its potential coil. This wattmeter would read p c = (_9.48)(115) + (8.60)(199.2) = -1090 + 1713 = 623 watts. Total power P = P a + P c = 4623 + 623 = 5246 watts. Relative Readings on Balanced Loads of Wattmeters Con- nected for the Two-wattmeter Method of Measuring Power in a Three-phase Circuit when the Current and Voltage Waves are Sinusoidal. Figure 99 shows the vector diagram of a balanced, three-phase, inductive load having a power-factor of cos 6 P . For an inductive load having a power-factor of cos 8 P , the phase currents lag the phase voltages by an angle B p . 7 i lags Foi by B p degrees, J 2 lags Voz by Q p degrees and I m lags ^03 by Op degrees. POWER AND POWER-FACTOR 331 Let the wattmeters be connected as shown in Fig. 98, page 326. The wattmeter in line 1 carries the current 7 i and has the voltage Vsi impressed across the terminals of its potential coil. The voltage and current are considered in the same direction in phase 01. The wattmeter in line 2 carries the current 7 2 and has the voltage V 32 impressed across the terminals of its poten- tial coil. The current and voltage are considered in the same FIG. 99. direction in phase 02. / O i lags F 3 i by (0 P - 30) degrees. 7 02 lags F 32 by (0 P + 30) degrees. The readings of the two watt- meters are therefore PiCwattmeter in line 1) = I 01 V 31 cos (0 P - 30) = I L V L cos (0 P - 30) P 2 (wattmeter in line 2) = cos (0 P + 30) = I L V L cos (0 P + 30) (40) (41) where the subscript L indicates line values. When P = 0, i.e., when the power-factor of the circuit is unity, both wattmeters read alike, namely IL\ T L cos 30. When P = 60 degrees, i.e., when the power-factor is 0.5, the current and voltage for the wattmeter in line 1 will be out of 332 PRINCIPLES OF ALTERNATING CURRENTS phase by 30 degrees, the current lagging the voltage, while the current and voltage for the wattmeter in line 2 will be out of phase by 90 degrees, the current also lagging. The reading of the wattmeter in line 2 will therefore be zero. The entire power developed in the circuit will be indicated by the wattmeter in line 1. For angles of lag greater than sixty degrees, i.e., for power-factors less than 0.5, the current and voltage for the watt- meter in line 2 will be out of phase by more than ninety degrees. Under this condition, the reading of the wattmeter in line 2 will 1.0 0.8 0.6 0.4 0.2 0.0 -0.2 -0.4 -0.6 -0.8 -1.0 / z / 7 8 1 2 3 4 5 o/ x e D 9 1( Powe / r-fae6 rinPe r cent / / / / / / Ord nates . cos (0P+30 ) coB(0-n FIG. 100. reverse, i.e., it will become negative, and must be subtracted from the reading of the wattmeter in line 1 to get the true power. The true power is always equal to the algebraic sum of the readings of the two wattmeters, but the sign of the reading of one of the wattmeters reverses and becomes negative when the power-factor of the circuit becomes less than 0.5. When the power-factor is less than 0.5, it is necessary to reverse the connections of the current coil of one of the wattmeters in POWER AND POWER-FACTOR 333 order to make it read up-scale. All readings taken after the reversal of the current coil must be considered negative. The ratio of the readings of the two wattmeters, connected for the two- wattmeter method, is the same for equal power-factors with leading and lagging currents, but the actual readings are in- terchanged. For example, the wattmeter in line 2 will read zero for a power-factor of cos (+60), i.e., a lag of the phase current behind the phase voltage of 60 degrees. For an angle of lead of 60 degrees, i.e., a power-factor of cos ( 60), the wattmeter in line 1 will read zero.* Both wattmeters will read alike for unity power-factor. They will read alike in magnitude but opposite in sign for zero power-factor. It must be remembered that the above statements are true only when the load is balanced and the current and voltage waves are sinusoidal. Clockwise phase order is assumed in the preceding discus- sion. Changing the phase order will interchange the readings of the wattmeters. The ratios of the readings of two wattmeters, connected to measure the power taken by a balanced, three-phase load with sinusoidal current and voltage waves, are plotted against power- factors in Fig. 100, page 332. Determination of the Power-factor of a Balanced Three- phase Circuit, when the Current and Voltage Waves are Sinu- soidal, from the Readings of Two Wattmeters Connected to Measure the Total Power by the Two-wattmeter Method. The power-factor of a balanced three-phase circuit may be determined by measuring the total power, the line current and the line voltage, and then applying equation (11), page 316. If a circuit is balanced and its current and voltage waves are sinusoidal, the power-factor may be found from the readings of two watt- meters which are connected to measure the total power by the two-wattmeter method. Refer to Fig. 99, page 331. From equations (40) and (41), page 331, the readings of the two wattmeters are Pi (wattmeter in line 1) = I L V L cos (0 P - 30) P 2 (wattmeter in line 2) = I L V L cos (O p + 30) * Angles of lag are taken positive in equations (40) (41). 334 PRINCIPLES OF ALTERNATING CURRENTS Expanding the cosine terms, Pi = lLV L ^cos0 p + sin0 p I 2 z } P 2 = from which ~Y cos P - I sin 6 P Pi - P 2 = sin 8 P Pi +P 2 ~ V3 cos 0/ tan0 p = Va^-qrl 2 (42) fi ~T *2 Since the tangent of the angle of lag between the phase current and phase voltage of a circuit is always equal to the ratio of the reactive phase power to the true phase power, it is obvious from equation (42) that \/3(Pi PZ) is reactive power. For bal- anced conditions, the square root of three times the difference of the readings of wattmeters connected to measure the power of a three-phase circuit by the two-wattmeter method is the reactive power of the circuit. An Example Involving the Use of the Two-wattmeter Method for Measuring the Power in a Balanced, Three-phase, Y-con- nected Circuit Containing Third Harmonics. Each branch of a Y-connected load consists of a condenser, an air-core induc- tance and a non-inductive resistance, in series. When this circuit is connected to a source of power with balanced voltages which contain no harmonics of higher order than the third, each of two wattmeters, connected to measure power by the two- wattmeter method, reads 15 kilowatts. Each voltage between line terminals of the load is 230 volts. When the neutral of the Y-connected load is connected to the neutral of the source of power, the wattmeter readings and the voltages between the line terminals of the load remain unchanged, but there is a current in the neutral connection of 60 amperes. Under this condition, each of the voltages between the line terminals of the load and the neutral point is 150 volts. What are the resistance, induct- ance and capacitance of each branch of the load? The funda- mental frequency of the impressed voltage is 60 cycles. Let the subscripts 1 and 3 attached to the letters P, I and V indicate, respectively, fundamental and third-harmonic power, current and voltage per phase. POWER AND POWER-FACTOR 335 Since the line voltages of a balanced three-phase circuit cannot contain third harmonics, the 230 volts impressed across the terminals of the load must contain the fundamental only. Therefore the fundamental voltage to neutral must be oorj V l = - = = 132.8 volts. V3 Since the wattmeters have their potential coils connected across the line terminals of the load, the voltages impressed across the potential coils must contain the fundamental only. The wattmeters can record only fundamental power, even though their current coils carry third-harmonic current, since their potential coils have only fundamental voltage impressed across them. There can be no power developed by a harmonic in the current if the corresponding harmonic in the voltage is absent. Since the two wattmeters read alike, the power-factor of the circuit for the fundamental alone must be unity. The circuit must therefore be in resonance for the fundamental. Pi = TVgn = 7,2 ^1 1 n 2 + Z! 2 where g\, r\ and x\ are the conductance, resistance and reactance per phase for the fundamental. For resonance, x\ is zero and _ Fi 2 _ (132.8) Pi 15,000 + 15,000 3 = 1.76 ohms. The third-harmonic voltage to neutral is y 3 = V050) 2 - (13278)' = 69.8 volts. Third-harmonic current per phase is equal to one-third of the third-harmonic current in the neutral. 73=-^=-;- =20 amperes. o o The third-harmonic impedance per phase is 2 3 = -OFT = 3.49 ohms. x, = V(3.49) 2 - (1.76) 2 = 3.01 ohms. 336 PRINCIPLES OF ALTERNATING CURRENTS Let XLI and x c \ represent the inductive and capacitive reac- tances for the fundamental. Let XLS and x c z represent the same quantities for the third harmonic. Then XLI Xci = XLS X C 3 = 3x L i ^Xci = 3.01 3Li(9 - 1) = 9.03 XLI = 1.13 ohms. XGI = 1.13 ohms. L = - = = 0.00300 henry. 10 6 10 6 = = 377XTI3 = 234? microfarads ' Measurement of the Reactive Power of a Balanced Three- phase Circuit, whose Current and Voltage Waves are Sinusoidal. If the potential coils of two wattmeters, which are arranged to measure the power of a balanced three-phase circuit whose current and voltage waves are sinusoidal, are interchanged, the reading of either wattmeter multiplied by the square root of three is equal to the reactive power in the circuit. For example, if the potential coil of the wattmeter with its current coil in line 1, Fig. 98, page 326, is connected between lines 2 and 3 instead of between lines 1 and 3, and the potential coil of the wattmeter with its current coil in line 2 is connected between lines 1 and 3 instead of between lines 2 and 3, the reading of each wattmeter will be I L V L sin Op If the potential coils of the two wattmeters are reconnected as stated above, it will be seen, by referring to Fig. 99, page 331, that the wattmeters will read . Pi = /oJ r 32 cos (90 - 6 P ) = I L V L sin 6 P Pi = /o2V,i cos (90 + P ) = I L V L sm (-0 P ) Reactive power = 37 pha seV phase sin e phas e = SIpVp sin B p L sin d p CHAPTER XII UNBALANCED THREE-PHASE CIRCUITS Unbalanced Circuits. The problems involved in the operation of balanced three-phase circuits or, in general, of balanced poly- phase circuits are comparatively simple and easy of solution. Except when there is interaction between phases, such problems may be treated like any single-phase problem by considering a single phase. Although in practice most polyphase circuits are nearly balanced, many cases of bad unbalancing exist. Moreover, certain types of polyphase apparatus are particularly sensitive to any unbalancing. This is especially true of the rotary con- verter, which was mentioned on page 304. This machine will develop bad commutation on its direct-current side and certain other bad operating features when operated from a polyphase circuit whose voltages are even moderately out of balance. For a fixed total output, the copper loss in all types of polyphase apparatus increases with the amount of unbalancing. For these reasons and many others, any method which will simplify the handling of problems which arise when dealing with unbalanced circuits is of great importance. A method of great power for handling many such problems, which although not new has come into prominence during the last few years through its application to problems arising in connection with unbalanced three-phase circuits, depends upon the fact that any unbalanced three-phase system of sinusoidal vectors may be resolved into three component systems of vectors. * The components of the first of these systems are identical. They are zero when the vector sum of the three original vectors is zero. These components may be called the Uniphase or residual components. The components of the second system together * R. E. Oilman and C. LeG. Fortescue, Trans. Am. Inst. of Elect. Eng., vol. XXXV, 1916, p. 1329. C. L. Fortescue, Trans. Am. Inst, of Elect. Eng., vol. XXXVII, 1918, p. 1027. W. V. Lyon, Electrical World, June 5, 1920. 22 337 338 PRINCIPLES OF ALTERNATING CURRENTS give rise to a balanced three-phase system whose phase order is the same as the phase order of the original vectors. The com- ponents of the third system together give rise to a balanced three-phase system, but the phase order of this system is oppo- site to that of the original vectors. In other words, any unbalanced three-phase system of sinusoidal voltages may be replaced by two balanced systems of sinusoidal three-phase volt- ages having opposite phase orders and three sinusoidal uniphase voltage vectors. Any unbalanced system of three-phase cur- rents, whose wave forms are sinusoidal, may be similarly replaced. Direct, Reverse and Uniphase Components of -Three-phase Vectors. Let Vi, V% and F 3 be the three voltages of an un- balanced three-phase system whose yoltages are sinusoidal. Each of these voltages may be resolved into any number of sinusoidal components of the same frequency, but since there are only three known quantities, namely, Vi, Vz and F 3 , there can be only three independent simultaneous equations to deter- mine the components. Let each voltage be resolved into three components. Then Vi = x + y + ~z (1) F 2 = ax + by + cz (2) F 3 = dx + ey + fz (3) where x, y and z are the three components of Vi, and a, b, c, d, e and / are complex coefficients. These coefficients must be so related that they may be determined from equations (1), (2) and (3), which are the only possible simultaneous equations. Since the operation of motors and generators is well understood under balanced conditions and can be easily handled, two of the coefficients, b and e, will be so chosen that the components y, by and ey form a balanced three-phase system of voltages, whose phase order is the same as that of the original three voltages. The coefficients c and / will be so fixed that the components z, cz and fz also form a balanced three-phase system of voltages, but the phase order of this system of voltages will be opposite to that of the original vectors. The remaining coefficients, viz., a and d, will be fixed so that the components x, ax arid dx are identical. The coefficients a and d are therefore each unity. UNBALANCED THREE-PHASE CIRCUITS 339 Then Fi + F 2 + F 3 = 3x + (1 + 6 + e)y + (1 + c + /) (4) and y + 5y + ~ey = (5) 2 + CZ + /* = (6) These conditions together with the known values of Fi, F 2 and V 3 are sufficient to definitely fix the components. Other systems might have been chosen, but this would complicate the conditions produced by unbalancing rather than simplify them. The particular advantage of the systems chosen is that two are balanced systems, for which the conditions are well understood. When the conditions are such that the vector sum of the three original vectors is zero, i.e., when Vi + v 2 + v 3 = o 3x must also be zero, since y + by+ey and z + cz + Jz are each always equal to zero. The vector sum of the three-line voltages of any three-phase load is always zero. Hence there cannot be any uniphase com- ponents in the line voltages of a three-phase load. There cannot be any uniphase components in the line currents of either a balanced or unbalanced A-connected load or in the line or phase currents of a balanced or unbalanced Y-connected load without a neutral connection, since the vector sum of the currents in each case must be zero. There probably will be uniphase com- ponents in the phase voltages of an unbalanced Y-connected load either with or without a neutral connection, since the vector sum of these voltages, in general, is not zero. It is possible, however, to have an unbalanced Y-connected load in which the uniphase components are zero. Since, in general, the vector sum of the phase currents in an unbalanced A-connected load is not zero, these currents usually will contain uniphase com- ponents. In this case also, it is possible to have a load in which the uniphase components are zero. There will be uniphase components in the line and phase currents of an unbalanced Y- connected load with a neutral connection which carries current, since the vector sum of the phase currents and therefore the vector sum of the line currents is not zero. 340 PRINCIPLES OF ALTERNATING CURRENTS Since (1 + b + e)y and ( 1 + c + /)z are each equal to zero it follows that Vi + F 2 + ? 3 = 3z i.Zt + F.+ V. (?) That is, the uniphase component in each phase is equal to one- third of the vector sum of the three original vectors. Vi-x = y_ + 2 (8) V*-x = by + cz (9) Vs - x = ey + fz (10) The vector sum of the drrecl_ajidj:everse components for each phase may be found by subtracting the uniphase component vectorially from each of the original voltage vectors. Instead of using x, y and z as the components of Vi, and the coefficients b, c, e and /, it will be simpler and more convenient in what follows to use the subscripts u, d and r with the letter V to indicate the uniphase, the direct-phase and the reverse-phase components respectively. Using this notation y x = Vu+ 7di + V_n (11) F 2 = Vu + Vd2 + Vr2 (12) F 3 = Vu + Fd3 + Fr-3 (13) Vi + F 2 + V, = 3Vu (14) A similar relation exists among the currents in any un- balanced three-phase system. For example, /!=/ + l*i + In (15) /2 = In + Jrf2 + Irt (16) /3 = In + Id + Jrt (17) /1 + /2 + /3 = 3/ (18) Figure 101 shows the direct- and reverse-phase components and also the uniphase components of the currents in an un- balanced Y-connected three-phase load with a neutral connection. The direct-phase components are shown by heavy solid lines. Heavy dotted lines are used for the reverse-phase com- ponents. The uniphase components are shown by solid lines. Dot-and-dash lines are used for the original vectors. UNBALANCED THREE-PHASE CIRCUITS 341 The three components, i.e., the direct-phase, reverse-phase and uniphase components, of each of the original vectors of an unbalanced three-phase system of voltages or currents may be determined either analytically or graphically from the known magnitudes and phase relations of the original three-phase vectors. FIG. 101. Determination of the Direct, Reverse and Uniphase Compo- nents of a Three-phase System. Consider the three voltage vectors given by equations (11), (12) and (13). It has already been shown that the uniphase components are Vu T 3 (19) Also V*i + V d z + Vu = (20) Vrl + VrZ + V* = (21) Let V's with primes indicate the vectors obtained by subtract- ing the uniphase components from each of the given three-phase vectors. Then if the phase order of the direct-phase components is clockwise "Vz* Vz Vu = TV = F 3 - Vu = + V,, = YJI - 24 + Fr, +240 (24) 342 PRINCIPLES OF ALTERNATING CURRENT'S If the system is balanced, the uniphase components, V v , and each of the reverse-phase components are zero. Let each of the vectors in equation (23) be rotated in a clock- wise direction through 120 degrees by applying the operator 1-120. Then TV ~ 120 = Vdi -240 + Vn J5! (25) Subtracting equation (25) from equation (22) gives TV _ -120 - Fj-240 -240 (26) The voltages { TV - TV "120 |, FjO and V dl -240 are shown in Fig. 102. FIG. 102. By referring to Fig. 102 it will be seen that TV - TV |- 120 1 = {TV + TVI+ 6Q 1 = A/3V d il-3Q(27) Hence (28) (29) Since Fi' =Fdi+F r , (Equation (22)) Vn = Fi' - Vdi (30) to UNBALANCED THREE-PHASE CIRCUITS 343 Referring to Fig. 103 it will be seen that equation (30) reduces 1 --tvi'l-ao" + TV -9Q Having determined the direct- and reverse-phase components, Vdi and Fri, for phase 1, the direct- and reverse-phase systems of components may be found by applying the proper operators to Vdi and Vn. Vdi = F dl |-0 (33) V dz = 7^1-120 (34) V* = V* 1-240 (35) Vn = Vn 1+0 .(36) y r2 = Vn 1 + 120 7 rl l+240 (37) (38) 344 PRINCIPLES OF ALTERNATING CURRENTS From equation (28) it is obvious that to find the magnitude of the direct-phase component, Vdi, for phase 1, F 2 ' must be rotated in a positive or counter-clockwise direction through 60 degrees, added to Vi and the result divided by the square root of three. The correct phase position of this direct-phase com- ponent is found by rotating it, as just found, through 30 degrees in a positive or counter-clock-wise direction. FIG. 104. FIG. 105. From equation (31) it is obviousjhat to find the magnitude of the reverse-phase component, Vri, for phase 1, F 2 ' must be rotated in a negative or clockwise direction through 60 degrees, added to V\ and the result divided by the square root of three. The correct phase position of this reverse-phase component is found by rotating it, as just found, through 30 degrees in a negative or clockwise direction. The method of finding the direct- and reverse-phase compo- nents for phase 1 is illustrated in Figs. 104 and 105. A graphical determination of the direct- and reverse-phase components is sufficiently accurate for many purposes. When the vector UNBALANCED THREE-PHASE CIRCUITS 345 expressions for the three vectors which are to be resolved into direct- and reverse-phase components are known, an analytical solution is simple and will require little if any more time than the graphical solution. A Simple Graphical Construction for Finding the Direct- and Reverse -phase Components of a Three-phase Circuit whose Vectors are Sinusoidal and Contain no Uniphase Components. When there are no uniphase components in the currents or voltages of a three-phase circuit, whose currents and voltages are sinusoidal, and only the magnitudes of the currents or voltages FIG. 107. are known, the direct- and reverse-phase components may be found by a very simple graphical construction. * The vector sum of the line voltages of a three-phase circuit is zero. They there- fore can contain no uniphase components. Also the vector sum of the line currents of a three-phase A-connected circuit, or Y-con- nected circuit without a neutral, must be zero. They can contain no uniphase components. Suppose the magnitudes of the line voltages of a three-phase *See Unbalanced Three-phase Circuits, W. V. Lyon, Electrical World, June 5, 1920. 346 PRINCIPLES OF ALTERNATING CURRENTS circuit, whose voltages are sinusoidal, are known. Since their vector sum must be zero, they must form the sides of a closed triangle, as shown in Fig. 106, where Viz', VM', and Vsi are the voltages. Construct an equilateral triangle 2-4-3 on Vzs' as a base. The side 2-4 of this triangle is Vw rotated through 60 degrees in a positive direction. From equation (28), page 342 the diagonal 1-4 = V\+ (-89.15 +J36.36)| = -8.28 +J2.27 Then 7ui = 19.00 - J9.33 7 M2 = 19.00 - J9.33 7 U3 = 19.00 - J9.33 348 PRINCIPLES OF ALTERNATING CURRENTS I dl = 89.3 + J7.05 I d2 = (89.3 + J7.05)(cos 120 - j sin 120) = -38.55 - J80.86 7 d8 = (89.3 + J7.05) (cos 240 - j sin 240) = -50.76 + J73.81 7 r i = -8.28 + J2.27 7 r2 = (-8.28 +j2.27)(cos 120 + j sin 120) = 2.174 - J8.305 7 r3 = (-8.28 + J2.27)(cos 240 + j sin 240) = 6.106 + J6.035. Mutual-induction between a Three-phase Transmission Line and a Neighboring Telephone Line. The vector sum of the direct-phase components and the vector sum of the reverse- phase components of the currents of a transmission line are zero. Therefore, if the conductors of a three-phase transmission line could be equidistant from the conductors of a telephone line, which runs parallel to the transmission line, these component currents could produce no inductive effects on the telephone line. Although it is impossible to have the conductors of a transmis- sion line equidistant from the conductors of a telephone line, they can, on the average, be made equidistant by properly trans- posing them. By proper transposition, the inductive effects of the direct- and reverse-phase components may be made zero. This is not true of the uniphase components. These are all in phase and no amount of transposition of the conductors of the transmission line will alter their inductive effects on the telephone line. To get rid of the inductive effects of the uniphase com- ponents, the telephone line must be transposed. The uniphase components or residuals, as they are called by telephone en- gineers, play a very important part in the interference effects produced on telephone lines by unbalanced transmission lines which are operated with grounded neutrals. The whole analysis of the interference between transmission lines and neighboring telephone lines is much simplified by resolv- ing the currents in the transmission line into direct-phase, reverse- phase and uniphase components. UNBALANCED THREE-PHASE CIRCUITS 349 Power in an Unbalanced Three-phase Circuit when the Cur- rents and Voltages are Sinusoidal. The power in a three-phase circuit is Po = Pi + P 2 + Pa (39) where PI, P 2 and P 3 are the powers in phases 1, 2 and 3 respec- tively, P! = Fi/i cos 0i (40) P 2 = F 2 / 2 cos 02 (41) PS = F 3 / 3 cos 6 3 where the F's, I's and 0's represent phase values of voltage, current and power-factor angle. If Vi and 1 1 are each resolved into direct-phase, reverse-phase and Uniphase components, the expression for PI becomes Pi = V d l d COS 07,; + V r lr COS B^ r \ + VJ U COS O + V d l r COS 07?; + F d / tt COS + v r i d cos 0/;; + v r iu cos 07: + F u / d cos 07 u d ; + VJ r cos where Fd, /<*, V r , I r , V u and 7 U represent the numerical values of the direct-phase, reverse-phase and Uniphase components of the voltages and currents. The limits on the phase angles, 0, indicate to which phase they refer. Expressions which are similar to equation (42) may be written for P 2 and P 3 . When the expressions for PI, P 2 and P 3 are added to give the total power, P , it will be found that the sums of the terms involv- ing unlike components are zero. For example: the sum of the three following terms is zero. V d l r COS 0^; + V d l r COS 07?! + VJ r COS 0^ = F d / r (cos 07:; + cos 07:; + cos 07::) = o (43) The components in equation (43) are plotted in Fig. 108. By referring to Fig. 108, it will be seen that equation (43) may be written in the following form, TV r {cos a + cos (120 + a) + cos (120 - )} = F d / r (cos a + cos 120 cos a sin 120 sin a + cos 120 cos a + sin 120 sin a] = (42) T7 r J 1 A/3 . 1 , A/3 - v */-{ cos a -cos a sma -cosa + -^ sin eel = Z 2 2 Z 350 PRINCIPLES OF ALTERNATING CURRENTS The sum of the other terms involving components which are unlike may similarly be shown to be zero. Therefore Po = 3V d l d cos Ojl + 3V T Ir cos + 3V U I U cos e v T : (44) The total power in an unbalanced three-phase circuit, which has no uniphase components, is equal to the sum of the powers developed by the direct- and reverse-phase components. When reverse-phase components are present in the currents and voltages of a three-phase circuit, i.e., when the circuit is unbalanced, the power developed in at least one phase is greater, and in at least one other phase is less than the average power per phase. If there were no reverse-phase components, the power developed by all phases would be the same. The effect of the reverse- phase components is to transfer 10o. power from one phase to another. The principle of the phase balancer depends on this trans- fer of power from one phase to another by reverse-phase components. Phase Balancer. It is necessary, in certain cases, for central stations to supply large amounts of single-phase power for special purposes, such as for the operation of electric railways using single-phase, alternating-current, series motors. This single- phase load not only badly unbalances the voltage of the system supplying it, but also very much decreases the permissible output from the generating equipment. The operation of certain types of polyphase apparatus, notably the rotary converter, is difficult on circuits whose voltage is out of balance. The impedance of a polyphase rotating machine, such as a synchronous motor provided with a damping winding or an induction motor, is much less for the reverse-phase components of an unbalanced system of voltages than for the direct-phase components. Such a machine, when connected to a circuit having unbalanced voltages, will take reverse-phase currents which are large compared with the direct-phase currents. The reverse-phase currents will be substantially opposite in phase to UNBALANCED THREE-PHASE CIRCUITS 351 the reverse-phase currents in the load which causes the unbal- anced voltages. For this reason, a synchronous motor with damping winding or an induction motor will partially restore the condition of balance of currents in an' unbalanced circuit to which it is connected, but such a machine alone cannot restore the condition of complete balance of currents. A type of apparatus has been developed within the last few years by means of which complete balance may be restored to an unbalanced three-phase circuit. This is accomplished by taking from the unbalanced circuit reverse-phase currents equal G= Source of Power S= Synchronous Motor Coupled to B B== Pha=e Balancing Alternator FIG. 109. in magnitude but opposite in phase to those caused by the exist- ing unbalanced load. Since unbalanced voltages are usually due to unbalanced line drops caused by an unbalanced load, balancing the currents of a circuit will usually balance the voltages. The Phase Balancer, as the machine is called, consists of a synchronous motor, with a low impedance damper, driving a three-phase synchronous alternator. One phase of the alter- nator is connected in series with each phase of the synchronous motor, but the phase orders of the alternator and motor are opposite. The alternator impresses reverse-phase voltages of such magnitude and phase that the motor takes reverse-phase currents from the line which are equal and opposite to those in the unbalanced load. An automatic device is used to adjust the magnitude and phase of reverse-phase voltages of the phase- balancing alternator. The magnitude of the voltages is ad- justed by means of the field excitation of the alternator. Their phase is varied by rotating the magnetic axis of the field with respect to the field structure by the use of two independently excited field windings displaced ninety degrees from each other on the field structure. 352 PRINCIPLES OF ALTERNATING CURRENTS The development of the phase-balancer would have been improbable without the knowledge that an unbalanced three- phase load, with sinusoidal currents and voltages and without a neutral connection, could be resolved into two balanced three- phase loads having opposite phase orders. The schematic diagram of connections for a phase balancer to balance currents is shown in Fig. 109. Copper Loss in an Unbalanced Three-phase Circuit in Terms of the Direct, Reverse and Uniphase Components of the Currents. The copper loss in phase 1 is -Pi copper = (7i phase) 2 X T\ The direct, reverse and uniphase components of the currents are shown in Fig. 110. If Ir\ and I u are each resolved into two quadrature components with respect to hi, it will be seen, by referring to Fig. 110, that Pi copper = I [Idl + Irl COS + I u COS ft] 2 + [I r i sin a + I u sin ft} 2 }r l = [Idi 2 + In 2 + 7 w 2 ]n + 2[/di/ r i cos a + I d j u cos ft + I r Ju cos a cos ft + I rl l u sin a sin ft]ri (45) Similarly, by resolving I r2 and I u each into two quadrature components with respect to 7dx = -~ dx maxwells. (11) This links the current i A in the conductor A. The total flux linkages with the current i A in conductor A are therefore 2i A 2 oc dx = 2t A 2 log e - (12) per centimeter length of conductor A. PART c. The flux due to the current i a , in conductor B, which links the current i Aj in conductor A, includes all the flux pro- duced by the current i B which lies between a distance D from the axis of conductor B and infinity. This assumes that the current IA in conductor A is concentrated in its axis. The flux linkages with the current i A due to this flux are (Fig. Ill) f JD log, (13) per centimeter length of conductor A. The flux due to the current i B which lies at a distance less than D from the axis of conductor B does not link the current in conductor A and therefore cannot produce mutual-induction on conductor A, REACTANCE OF A TRANSMISSION LINE 359 The total resultant flux linkages with the current i A in conductor A per centimeter length of conductor A are 2i A 2 log, y + ^i A 2 + 2i A i B log, -^ (14) Since a single-phase line is considered, i A must be equal and opposite to i B at every instant, no leakage being assumed. There- fore i A = IB = i and the expression for the resultant linkages becomes 2i 2 log, H &*" ( 15 ) per centimeter length of conductor A. Putting the current i equal to unity and multiplying by 10~ 9 gives for the resultant inductance in henrys per conductor per centimeter length of line L = (2 log, y + Q 10~ 9 henrys. (16) In terms of common logarithms, the inductance per conductor per centimeter length of line is L = (4.605 logio y + 0.5/*') 10~ 9 henrys. (17) For conductors of non-magnetic material such as are generally used for transmission lines, // is unity. For a single-phase transmission line, with copper or aluminum conductors or with conductors of any other non-magnetic ma- terial, the reactance per conductor per 1000 feet is ZIOGO ft. = 27r/ (140.4 logio y + 15.2) 10~ 6 ohms. (18) The reactance per mile per conductor is -r^. x per mile = 2*f (741 logio - r + 80) 10- 6 ohms. (19) It is important to observe that for a single-phase transmission line the reactance per conductor results from two parts: the self-inductance of the conductor considered and the mutual inductance due to the other conductor. For a polyphase line, the reactance per conductor also results from two factors : the self- inductance of the conductor considered and the mutual-induc- tance due to the other conductors. 360 PRINCIPLES OF ALTERNATING CURRENTS For a single-phase line, the flux due to the current i A which links conductor B is equal to the flux due to the current i B which links conductor A. Since the two currents, i A and i B , are not only equal but opposite in direction, the combined flux linkages produced by them with the current in each conductor are zero. The resultant inductance per conductor of a single- phase line is therefore due to the flux the current in the conductor considered would alone produce between its axis and the axis of the other conductor, assuming that the current is concentrated in the axes of the conductors. Average Reactance per Conductor of a Completely Transposed, Ungrounded, Three-phase Transmission Line. The lengths of the transposed sections will be assumed equal. The average reactance per conductor per unit length of line is equal to the average flux linkages per ampere per conductor per unit length of line, due to both self- and mutual-induction, multiplied by 27r/, where / is the frequency of the line. ft r r \ (a) (6), <*) FIG. 112. The transpositions required for complete transposition of the line are shown Fig. 112. Parts a, b and c of this figure show the three positions of the conductors. The D's with subscripts 1, 2 and 3 are the distances between the conductors, measured between centers. A, B and C are the conductors. The inductance of any conductor such as A will consist of three parts : (a) That due to the flux linkages with A resulting from the flux produced by the current in conductor A, i.e., by the current i A . These linkages per unit current are its self- inductance. (6) That due to the flux linkages with A resulting from the flux produced by the current in conductor B, i.e., by the current REACTANCE OF A TRANSMISSION LINE 361 i B . These linkages per unit current are the mutual-inductance of B on A. (c) That due to the flux linkages with A resulting from the flux produced by the current in conductor C, i.e., by the current i c . These linkages per unit current are the mutual-inductance of C on A. For the arrangement of the conductors shown in (a) of Fig. 112, the flux linkages with the current i A in conductor A are per centimeter length of line : (See equations (9), (12) and (13).) /./. = 2i A * log, -^ + I *V + 2i A i B log, - + 2i A i c log, ^- (20) But log, y = log, oc + log, - 1 log, - = log, oc -f log, jy 1 log, ^- = log, oc -f log, yr- i_J 3 '-'3 Therefore f.l. a = 2i A 2 log, - -}- 2* A 2 + 2z' A fc* log, p + 2t A ic log, -^ + 2i A (t A + f a + tc) log, (21) Since (i A + i s -f- i c ) = when there is no ground connection which carries current 2i A (i A + i B + i c ) log, oc = (22) The expression given in equation (22) is not indeterminate as might appear at first glance, but is actually zero. The linkages might equally well have been taken up to some distance such as x from the conductor A, instead of up to infinity. If this had been done, the term in question would have been 2i A (0) log, x, which is obviously zero. Increasing x to infinity would not change the value of the expression which would still be zero. Since the last term in equation (21) is equal to zero f.l. a = 2i A * log, - + ^ 2 + 2 *** lo & 5" + 2i * ic lo & D~ (23) 362 PRINCIPLES OF ALTERNATING CURRENTS For the arrangement of the conductors shown in (b) of Fig. 112, the flux linkages with current i A in conductor A per centi- meter length of line are f.l b = 2i A 2 log, - + ^ A 2 + 2i A i B log, 77 + 2i A i c log, w (24) r ^ x>/2 -L/i For the arrangement of the conductors shown in (c) of Fig. 112, the flux linkages with current i A in conductor A per centi- meter length of line are f.l e . = 2i A * log, - r + \ iA 2 + 2i A i B log, -^ + 2^'c log, ~ (25) The average flux linkages with current ^ in conductor A per centimeter length of line are f.Lav. = I (f*La + f.l+ + Ac) = 2t^ log, * + ^V + |W. log, STxlb^ ge (26) Since it is assumed that the line is ungrounded and has no neutral connection carrying current (27) IB H~ ic = i A and -ii lo 1 1 2 ' ' 1 1 3 Al & Di X D 2 X D 3 3^ c ge Di X D 2 X A = I**A 2 log, (D! X Z) 2 X Z) 3 (28) Combining equations (26) and (28) gives for the average flux linkages with the current i A in conductor A per centimeter length of line 2 log, ^ + log, VZ), 2 X D 2 2 X D, 2 ! (29) (30) Similar expressions hold for conductors B and C. It should be noticed that for a completely transposed three- REACTANCE OF A TRANSMISSION LINE 363 phase line, the average flux linkages per conductor per unit length of line depend only on the current in the conductor considered. The resultant reactance drop per conductor per unit length of a completely transposed three-phase line depends only on the dis- tances between the conductors and the current in the conductor considered. These statements assume that there is no neutral connection which carries current. If there is a neutral connec- tion which carries current, equation (30) does not hold, since under these conditions equation (27) is not true. The average reactance per conductor per centimeter length of a completely transposed three-phase line which has no ground or neutral connection carrying current is *. = ar/i log, + , io- (31) ohms per conductor per centimeter length of line. Per mile of line in terms of common logarithms this becomes x n . = 2irf {370.6 logic v ^i" X D ^ x D * + 80} 10~ 6 (32) ohms per conductor per mile of line. It is immaterial in what units the D's and r are expressed, provided they are expressed in the same unit. If the conductors are at the corners of an equilateral triangle, D l = D 2 = AJ = D and x av . = 27r/(741 log, ^ + 80} 10- 6 (33) ohms per conductor per mile of line. By comparing equations (33) and (19), page 359, it will be seen that the average reactance per conductor of a completely transposed three-phase line, with conductors at the corners of an equilateral triangle, is equal to the reactance per conductor of a single-phase line of equal length and with equal spacing between conductors. If the conductors of a completely transposed three-phase line are at the corners of an isosceles triangle, two of the distances DI, D 2 and D 3 will be equal. Two of the distances may also be 364 PRINCIPLES OF ALTERNATING CURRENTS equal with the conductors arranged in a plane. Let DI = D D and let Z> 3 = D'. Then x av . = 27T/1 370.6 log + 80 1 10 " 6 = 2^370.6 log,, x + 80,10- = 27r/{ 370.6 (2 logio y + | logic 5-') + 80}10- 6 = 27r/{ 741 logio y + 80 + 247 lo glo ^ } 10~ 6 (34) ohms per conductor per mile of line. When the conductors all lie in the same plane, with equal distances between the middle and each outside conductor, D' D = -x- and equation (34) becomes x av . = 27T/J741 logio ~ + 80 + 247 logio 2} lO' 6 = 27T/I741 Iog 10 ^ 4- 154} 10- 6 ohms per conductor per mile of line. (35) When DI = D 2 = D 3 = D, i.e., when the conductors are at the corners of an equilateral triangle, equations (23), (24) and (25), pages 361 and 362, all reduce to f.l. = 2i A * log e * + ~ i A * + 2i A (i B + i c ) log, ~ (36) Since it is assumed that there is no neutral connection IA + IB + ic = or IA = is + ic (37) Combining equations (36) and (37) f.l. = 2zV log, j + \ i A * (38) When the conductors are at the corners of an equilateral triangle, the flux linkages per conductor depend only on the cur- rent in the conductor considered and the distance D between conductors. The flux linkages per conductor and therefore the reactance per conductor of a three-phase line, with conductors at the corners of an equilateral triangle, are independent of the REACTANCE OF A TRANSMISSION LINE 365 currents in the conductors other than the one considered. This is true whether the load carried by the line is balanced or un- balanced, provided there is no neutral connection which carries current. Transposition is not necessary to maintain balanced line drops with balanced load. Transposition is necessary, however, to prevent mutual induction with other transmission lines or adjacent telephone lines. Problem Illustrating the Calculation of the Reactance per Conductor of a Completely Transposed Three-phase Trans- mission Line. A certain 110,000-volt, three-phase, 60-cycle, transmission line has its conductors arranged in a horizontal plane with 10 feet between the middle conductor and each outside conductor. The conductors have a diameter of 0.46 inch. What is the reactance per conductor per mile of line? From equation (35), page 364 10 V 1 2 x av . = 2 X 3.142 X 60(741 Iog 10 + 154) 10~ 6 = 377 (741 X 2.72 + 154} lO" 6 = 0.82 ohm per conductor per mile of line. Transfer of Power among the Conductors of a Three-phase Transmission Line. There is a transfer of power among the conductors of a three-phase transmission line except when the conductors are at the corners of an equilateral triangle. For simplicity, consider a three-phase line with conductors at the corners of an isosceles triangle. Refer to equations (23), (24) and (25), pages 361 and 362. Let D l = D 2 = D and let D 3 = D'. Under these conditions, if the instantaneous values of the currents are replaced by their vector values, the equations become f.l. a = 21 A * log, - r + \IS + 2~I A (I B log, ^ + Ic log, ^-,) = 21 A * log, y + ^L 2 + 2 I A Ic log, ^> (39) /.Z. t = 27 A 2 log, I + ^IA* -\~2I A (I B + I c ) log, ^ = 2/ A 2 log, y +\I A * (40) f.l. c = 21 A * log, * + \I A * + 2~I A (I B log, ^> + /clog, -^) = 27 A * log, j + \I A 2 + 2 I A I B log, ^7 (41) 366 PRINCIPLES OF ALTERNATING CURRENTS Let the vectors I A , //* and I c , Fig. 113, represent the currents in the conductors A, B and C respectively. Assume clock- wise rotation and balanced conditions for the currents. Refer to equations (39) and (41). Let D r be greater than D. Under this condition, log, jy is negative. For the relative position of the conductors shown in Fig. 112(a), page 360, there is a reactance drop, I A x = 2irfI A (2 log e h o) > see equation (39) , in conductor A due to the current J A . This reactance drop is 90 degrees ahead of the current J A and therefore represents no power with respect to /A. Since the logarithm of the ratio jy * s negative for the assumed relative magnitudes of D and D', the drop, I c x' = '2irflc (2 loge rv) j see equation (39), in conductor A due to the current 7c in conductor C is 90 degrees behind I c as shown. It has components in phase and in quadrature with the current I A , Fig. 113. The effect of the quadra- ture component is to increase the ap- parent reactance drop in conductor A. The effect of the in-phase component is equivalent to an apparent increase in the resistance drop in conductor A. This apparent increase in the resistance drop in conductor A does not represent FIG. 113. ic i a loss of power. It merely represents a transfer of power from conductor A to conductor C by mutual- induction. For the relative position of the conductors shown in Fig. 112(6), page 360, there is no mutual-induction between conductors B and A or between conductors C and A. See equation (40). There is therefore no transfer of power between conductors B and A or between conductors C and A. For the relative position of the conductors shown in Fig. 112(c), page 360, there is a reactance drop, J A x = 27rfI A (2 log, ~ + o) see equation (41), in conductor A due to the current I A . This reactance drop is 90 degrees ahead of the current 7 A and REACTANCE OF A TRANSMISSION LINE 367 therefore represents no power with respect to J A . Since the logarithm of the ratio jy is negative for the assumed relative magnitudes of D and D', the drop, J B x' = 2*fI B (2 loge^/j , see equation (41), in conductor A due to the current I B in conductor B is 90 degrees behind J B as shown. It has components both opposite in phase to and in quadrature with the current 1 A , Fig. 113. The effect of the quadrature component is to increase the apparent reactance drop in conductor A. The effect of the component which is opposite in phase to the current I A is to produce an apparent decrease in the resistance drop in con- ductor A. This apparent decrease in resistance drop in the conductor A represents a transfer of power from conductor B to conductor A by mutual-induction. In a properly transposed line, the resultant transfer of power among conductors is zero for the line as a whole. Mutual-induction between Transmission Lines or between a Transmission Line and a Telephone Line. The mutual-induction between a transmission line and a telephone line, or another transmission line which is parallel to the first line, may be found in exactly the same way as the mutual-induction between the conductors of a single transmission line. Consider the mutual-induction between the conductors of a three-phase transmission line and one conductor of a telephone FIG. 114. line which is parallel to it. Let the currents in the transmission line be i A , i B and i c and let i be the current in one conductor of the telephone line. Let D A , D B and D c be the distances in centimeters between the conductor of the telephone line and the conductors of the transmission line which carry the currents z A , IB and i c respectively. Refer to Fig. 114. The total flux linkages produced by the currents i A , i B and i c with the current i in one conductor of the telephone line per 368 PRINCIPLES OF ALTERNATING CURRENTS centimeter length of the telephone line are, from equation (13), page 358 f.l. = 2i A i log, -- + 2i B i log, - + 2i c i log, DA UB = 2i { (IA + IB + i c ) C + i A log, -=r- + IB log, jr- + i c log, -=- (42) -L/A ^B .L/c- J But (i A + fc's + ic) = since there is no neutral connection which carries current. Therefore f.l. = 2i { i A log, ^ + i B log, ^- + i c log, ^-} (43) .L/A L)B DC } The voltage induced in abvolts per centimeter length of the telephone line is Vabvoli, = 47T/ JA log, ~- + 7 B log,yr- + 7c log, yp (44) DA UB DC J where I A , IB and J c are the vector expressions for the currents in the conductors of the transmission line. If the portion of the transmission line which is parallel to the telephone line be completely transposed, the resultant flux linkages with the conductor of the telephone line will be zero, provided the transmission line has no neutral connection which carries current. If there be a neutral connection which carries current, the relation (i A + i B + i c ) = does not hold. With complete transposition, each conductor of the three- phase transmission line must occupy successively the positions 1, 2 and 3, Fig. 114. Each position must be maintained for one- third of the distance over which the complete transposition is made. The conductors must be given a complete rotation in position. The average flux linkages, produced by the transposed trans- mission line are, per centimeter length of the telephone line, 2f /. , 1 1 1 f.l. av . = ~ 3 [^A log, _ + ^ B log, -g- + ^ c lo ge ^- + (ic log, ^ + i A log, A- + i B log, ^- + (IB log, -^- + i c log, -p- + i A log, g-j } (45) REACTANCE OF A TRANSMISSION LINE 369 Since for an ungrounded transmission line with no neutral connection, (i A + i B + ic) = O -X log. + iog. + iog.-O (46) Therefore, with complete transposition, the average mutual- induction between an ungrounded, three-phase transmission line and each conductor of a parallel telephone line is zero. The resultant voltage induced in each conductor of the telephone line therefore must be zero. The above statement is not limited to a telephone line but holds equally for the mutual-induction between a completely transposed three-phase transmission line with no neutral connection and any other parallel conductor. The parallel conductor may be one of the conductors of another transmission line. The average mutual-induction between a three-phase power line and a telephone line may likewise be neutralized by trans- posing the conductors of the telephone line. If the conductors of the telephone line are transposed, the transmission line not being transposed, the voltage induced in adjacent transposed sections of the telephone line, by the mutual-induction of the transmission line, will be opposite in phase. If the two lines are parallel and the transposed sections of the telephone line are equal in length, these voltages will be equal in magnitude, and in phase opposition, and their resultant, therefore, will be zero. This is true without regard to the relative magnitudes of the currents, i A , IB and i c , carried by the transmission line. Their sum may or may not be zero. There may or may not be a neutral connection which carries current. Therefore, transpos- ing the telephone line will eliminate the effect of mutual-induction, due to a three-phase transmission line, even when the trans- mission line has a ground connection which carries current. Transposing the telephone line gives each of its conductors equal exposure to each of the conductors of the transmission line as well as to any neutral connection which may exist and must, therefore, make the resultant voltage induced in each conductor of the telephone line zero over any completely transposed length of telephone line. 24 370 PRINCIPLES OF ALTERNATING CURRENTS Although it is possible to neutralize the mutual-induction between a transmission line and a telephone line by proper transposition of the lines, in practice complete neutralization of the mutual-induction is often difficult. In many cases the telephone line is not exactly parallel to the transmission line. Under such conditions the transposed sections of the lines are at different distances from each other, and complete neutralization of the mutual-induction cannot be obtained unless the lengths of the transposed sections are varied or are made very short, neither of which is practicable. Moreover, in certain cases taps may be taken from the transmission line at certain points, making the currents unequal in the different transposed sections. Voltage Induced in a Telephone Circuit by a Three-phase Transmission Line which is Parallel to the Telephone Line. It is not necessary to find the linkages with each conductor of the telephone line. All that is required is the flux linkages with the loop formed by the two conductors of the telephone circuit. Refer to Fig. 114, page 367. Let the currents which are there marked i A , i B and i c be read as I A , I B and 7 C respectively, i.e., let them be vectors. In the figure, the distances from the conductors 1, 2 and 3 of the three-phase transmission line to one conductor i of the telephone line are D A , D B and D c . Let one of the unlettered circles be the other conductor of the telephone line. Let the distances between this latter conductor and the three con- ductors of the three-phase line be D A ', D b ' and D c r . The total linkages with the telephone circuit per centimeter length of telephone line, due to the currents J A , I B and /<, in the conductors of the transmission line, are F.L. (circuit) = 21 A log, -yf + 2I B log, -yf- + 2I C log, -=p- (47) DA DB D c The voltage induced by these flux linkages per unit length of telephone circuit is V = 27r/(F.L.) f- CD/") 2 (D '} 2 (D '} 2 \ = 2tf I A log. Vr^V + /. log, Tff4 + Ic log. Trfw (48) If the currents in equation (48) are expressed in abamperes, the voltage will be in abvolts. If the currents are expressed in amperes and equation (48) is multiplied by 10~ 9 , the voltage will be in volts. REACTANCE OF A TRANSMISSION LINE 371 The voltage induced per mile of telephone circuit, using common logarithms, is V = 2.33/{ I A logio 4jf^7 + IB log 4^4 + i c logio my^} 10 ~ 3 volts - ( 49 ) The currents in equation (49) must be expressed in amperes and in their complex form. It is convenient in equations (48) and (49) to put the 2 ; s, which are coefficients of the logarithms in equation (47), as exponents of the D's. This simplifies the use of the equations. In most cases in practice, the distances between the conductors of adjacent transmission and telephone lines are not directly known, but they can easily be found from the known horizontal and vertical spacing of the conductors and the height of poles and distance between poles of the two lines. Putting the 2 ; s in the equations as exponents of the D's, instead of as coefficients of the logarithms, is convenient as it avoids taking the square roots which would be necessary in finding the distances between the conductors. The square of the distances between the con- ductors is readily found by taking the sum of the squares of the horizontal and vertical distances between them. It must not be forgotten that equations (47), (48) and (49) are vector equations and that all terms in them must be added vectorially. The currents should be expressed in their complex form, referred to some conveniently chosen axis. Voltage Induced in a Telephone Circuit by an Adjacent Three- phase Transmission Line which Carries an Unbalanced Load. Equation (49) holds for either balanced or unbalanced loads. However, when a careful study or analysis of the inductive effects of a transmission line on a telephone line is to be made, it is best to resolve the unbalanced load currents carried by the transmission line into direct-phase, reverse-phase and uniphase components. (See Chapter XII.) The uniphase components, when they exist, cause by far the most trouble, since they are all in phase. They are usually called the " residuals" by telephone engineers. Since the residuals are all in phase, no amount of transposition of the transmission line will diminish their inductive effect on an adjacent telephone line. To get rid of the inductive 372 PRINCIPLES OF ALTERNATING CURRENTS effect of the residuals, the telephone line itself must be transposed. Since the vector sum of the direct-phase components and also the vector sum of the reverse-phase components are zero, they would produce no inductive effect on an adjacent telephone line if the conductors of the transmission line could be equidistant from , 5ft , each conductor of the telephone line. This state- _g[ g ment would not apply to the residuals, since they are in phase. Jj Example of the Calculation of the Voltage Induced in a Telephone Line by Electromagnetic o Induction. Two three-phase, 60-cycle, transmis- 1. sion lines, a, b, c and a', &', c', and a telephone line d, e are carried on the same poles with spacings between conductors shown in Fig. 115. The conductors of the transmission line are No. 000 wire and have a diameter of 0.41 inch. The transmission lines are paralleled at the generating station, with corresponding conductors connected to the same phase. The phase order is a, 6, c with the current in phase a leading the current in phase b. Find the voltage induced electromag- netically in the telephone line, per mile, when FIG. 115. eacn f ^ ne three-phase transmission lines carries a balanced load with 200 amperes per conductor. The power-factors of lines a, b, c and a', b', c' are unity and 0.85 (lagging current) respectively. Also calculate the voltage induced in the telephone line when each transmission line is operated with an inductive load of 0.80 power-factor and 200 amperes per conductor. The square of the distances between the conductors of the transmission lines and the conductors of the telephone line are given below. (bd) 2 = (b'eY (b e y = (Vdy = (c'e) 2 = (c'd) 2 (22) 2 + (3.5) 2 = 496.3 feet squared. 5) 2 = 486.3 feet squared. 5) 2 = 268.3 feet squared. 5) 2 = 258.3 feet squared. S) 2 = 112.3 f eet S q Uar ed. 5)2 = 102.3 feet squared. (22)2 + (1 (16) 2 + (3 (16) 2 -f (1 (10) 2 (lO) 2 + (S + (1 REACTANCE OF A TRANSMISSION LINE 373 Let (p with subscripts a, 6, c and a', &', c' be the flux linking the telephone line, per centimeter length of telephone line, due to the currents in the conductors a, 6, c and a', b', c r respectively. The magnitudes of these fluxes are: log* = 20 X 2.303 X 0.00884 =0.407 CL/ log. = 20 X 2.303 X 0.0165 = 0.760 2 = 20 X 2 ' 303 X - 0405 = 1-866 The resultant flux through the telephone circuit per unit length of line is equal to the vector sum of the component fluxes

    , w and (p c >. a a' ~\- c ~

    lags (p c by cos" 1 0.85 = 31.8 degrees. (p a leads ^ by 120 degrees and leads

    by 240 degrees. Use the flux (p a as an axis of reference. Then the resultant flux linking the telephone circuit, per centimeter length of the telephone circuit, is ^o = p a (cos - j sin 0) - ^ <(cos 31?8 - j sin 31?8) + <^(cos 120 - j sin 120) - *v(cos 151?8 - j sin 151?8) + ^ c (cos 240 - j sin 240) - *v(cos 2718 - j sin 2718) = 0.407 -j 0.000 - 0.346 +J 0.214 - 0.380 -; 0.658 + 0.670 +j 0.359 -0.934 +j 1.616 -0.059 -j 1.865 = -0.642 -j 0.334 l. Hence the equi- potential lines in any plane perpendicular to the axes of the fila- ments are circles whose centers lie on the axis of X, i.e., on a line through the axes of the filaments. The actual equipotential surfaces due to the filaments are circular cylinders whose axes are parallel to the axes of the filaments. When = 1, the radius of the equipotential line becomes infinite, equation (8), and the line is a straight line passing through b perpendicular to the line joining the axes of the fila- ments. The corresponding equipotential surface is an infinite plane passing through b and perpendicular to the line joining the axes of the filaments. Since the surface of any conductor must be equipotential, a conducting cylinder may be introduced in the field due to the charges on A and B with its surface coincident with any equi- potential surface, without altering the field. If the equipotential surface in question is one of the group which surrounds A, the charge on the filament A may be considered as existing on the surface of such a cylinder. As the surface of any conducting body must be equipotential, the charge will distribute itself in such a way .as to make the surface of the cylinder equipotential. 380 PRINCIPLES OF ALTERNATING CURRENTS The effect at any point outside of the cylinder will be the same as if the charge were concentrated along the filament A. It follows that any two parallel, straight conductors of circular cross-section having equal and opposite charges per unit length may be replaced by two charged filaments at the points A and B, Fig. 116, page 376, provided the charges are not influenced by the presence of any other charges. The positions of the points A and B are determined by the radii of the cylinders and their distance apart in accordance with the expressions given in equa- tions (8) to (11) inclusive. The equipotential surfaces due to two parallel, straight, charged, conducting, circular cylinders are shown in Fig. 117. f FIG. 117. The solid lines in this figure are the intersections of the equi- potential surfaces with a plane perpendicular to the two charged cylinders. The solid heavy circles are the projections of the cylinders on this plane. The dots in the circles represent the projections of the filaments. The lines of electrostatic force due to the charges on the cylinders are shown dotted. These lines must be everywhere perpendicular to the corresponding equipotential surfaces. It may be shown that the lines of force are also circles. CAPACITANCE OF A TRANSMISSION LINE 381 Remembering that the two cylinders have equal and opposite charges per unit length, it follows from equation (4), page 377 that the difference in potential between B and A is T B ' where r A f and r B ' are the distances for the cylinder B correspond- ing to the distances r A and r B for the cylinder A . (See Fig. 1 16, page 376.) Since the capacitance of any two conducting bodies, which are removed from the influence of all other charged bodies, is equal to the ratio of the charge on either to the difference of potential between them, the capacitance of two parallel straight cylinders or conductors of circular cross section, which are removed from all other charged bodies, must be c = v 2q TBTA ' (13) 2 lo < C is the capacitance in electrostatic units per centimeter length of the conducting cylinders. Equation (13) is not in a convenient form for use since r A , r B , r A and r B ' are all variables. These may be replaced in terms of the radii, r a and r^ of the cylinders and the distance between their centers. The radii of the cylinders are given by equation (8), page 378, d radii 1 - This is positive when = k is less than unity. It is negative when - - = k is greater than unity. For the equipotential sur- ' B face corresponding to the outer surface of the conductor surround- ing A , = k is less than unity and the expression for its radius * B is positive. Therefore, if r a is the radius of the conductor," d 382 PRINCIPLES OF ALTERNATING CURRENTS and r A d d 2 For the equipotential surface corresponding to the outer surface of the conductor surrounding B, -, = k\ is greater than unity B and the expression for its radius is negative. Therefore, if r b is the radius of the conductor, d and 7 r A r d fc ' = ^ = 27> Substituting in equation (13) the values of and from equations (14) and (15) respectively gives for the capacitance K_ i / . . \ / . . \ (16) Equation (16) gives the capacitance in electrostatic units per centimeter length of line. Since d, r a and r b enter only in ratios, it is immaterial in what units they are expressed, provided they are all expressed in the same unit. The distance, d, between the filaments may be easily found in terms of the radii, r a and r b , of the conductors and the distance, D, between their centers. Refer to Fig. 116, page 376. Since - is constant and P'Pc is a circle r B r_ A T B Ac + AP 1 Ac AP' " c ~ P 7 /? Ac - AP' cB + P'B 2r a cJ5 - P'B 20 A 2r a OA 2r a + 2P'B r a r a + d - (r - OA) r a = d(OA) + ( OA- - + + (17) CAPACITANCE OF A TRANSMISSION LINE 383 OA is the offset of the filament A from the center of the con- ductor about A. The offset, O'B, of the filament B from the center of the con- ductor about B may be found in a similar manner. It is O'B = - + r b *+ (18) Then D = d + OA + O'B Solving equation (19) for d gives , 2 _ > 4 + (r a 2 - TV*)* - 2Z> 2 (r a 2 -f r 6 2 ) - 2(r a 2 -f r> 2 ) (20) Equation (20) may be put in the following form d z = (D + r a -f r b ) (D + r a r b ) (D r a + r b ) (D r a r b ) . . By calculating the numerical value of d from either equation (20) or (21) and substituting this and the values of the radii in equation (16), the capacitance of two parallel, straight conductors of circular cross-section may be found in electrostatic units per centimeter length of the conductors. Equations (16) and (21) may be combined and reduced to the following form C = l 2 lo ft ( + V = 1_ 2 cosh- 1 a Where Z) 2 -r a 2 - r 6 2 (23) (24) 2 r a r b When the distance, D, between the conductors is great com- pared with their radii, r a and r b , d will be very nearly equal to D. (See equation (20).) The conductors of a transmission line are of equal radius and are in air. Therefore, for a transmission line K = 1 and r a = 384 PRINCIPLES OF ALTERNATING CURRENTS r b = r. When the radii of the conductors are equal, d, from equation (21), is d = \/(D + 2r)(D - 2rj = Substituting this value of d and r a = n = r in equation (16), page 382, 2 log. *+w 1 4 cosh~ D 2r (25) (26) Since, for transmission lines, D is large compared with r, //D\2~ /D \ V \2r/ "~ W ^ k e ver y ne arly equal to I - ) and may be assumed equal, especially as it enters in a logarithm. Putting D in equation (25), C = I 4 log, (27) for the capacitance of a single-phase transmission line in electro- static units per centimeter length of line. It is convenient to reduce equation (27) to electromagnetic units, either per thousand feet or per mile of line. It is also convenient to have the capacitance expressed in terms of common logarithms. From the dimensions of the electrostatic and electro- magnetic units of capacitance it may easily be shown that the ratio of the electromagnetic unit of capacitance to the electrosta- CAPACITANCE OF A TRANSMISSION LINE 385 tic unit of capacitance is equal to the square of a velocity. This velocity has been shown to be the velocity of light or 3 X 10 10 centimeters per second. Therefore, to reduce a given capacit- ance expressed in electrostatic units to its equivalent capacitance in electromagnetic units, it is necessary to divide by the square of the velocity of light or by (3 X 10 10 ) 2 . Hence, for a single-phase line consisting of two equal, parallel, straight conductors of circular cross-section, to get the capaci- tance in microfarads per 1000 feet of line and in terms of common logarithms, multiply the capacitance given by equation (27) by X (2.540 X 12 X 1000) X (3 X 10 10 ) 2 2.303 X (10 9 X 10 6 ) = 14.706 X 10- 3 r 14.7 X 10- 3 _ 3.68 X 10~ 3 (t> . ^microfarads J) J) 4logio logio- microfarads per 1000 feet of line. Per mile of line this becomes 19 - 4 X 10- 3 microfarads TN microfarads per mile of line. Equation (29) would be exact if the surface densities of the charges on the cylinders were uniform. In this case the positions of the filaments on which the charges may be considered con- centrated would coincide with the axes of the cylinders. The distribution of the charges on coaxial cylinders would be uniform. In all cases occurring in practice, where it is necessary to determine the capacitance or the charging current for a trans- mission line, the distances between the conductors and between the conductors and the earth are so great compared with the diameters of the conductors that the distribution of the charges on the conductors may be assumed to be uninfluenced except by the shape of the conductors themselves. Since the conductors of transmission lines are nearly always circular in cross section, the charges on the conductors of transmission lines may be assumed to be concentrated on the axes of the conductors. Equation (29) may be applied in the case of all aerial transmis- sion lines without sensible error. 25 386 PRINCIPLES OF ALTERNATING CURRENTS Example. A certain two-wire transmission line has the con- ductors spaced ten feet on centers. The conductors have a radius of 0.285 inch. What is the capacitance of the line between conductors in microfarads per mile of line? r _ 19.4 X IP" 3 D logio- 19.4 X IP" 3 10 X 12 logl -0285" = 19.4 X 10~ 3 2.6243 = 7.39 microfarads per mile. Capacitance between the Earth and a Straight Conductor Parallel to Its Surface. The capacitance between the earth and a straight conductor parallel to its surface may be calculated on the assumption that the earth is an equipotential surface of zero potential. Since the earth is a conductor, the electro- static lines of force must enter it perpendicular to its surface. It can be shown that the electrostatic field between an infinite conducting plane and a charged straight conductor which is parallel to this plane is the same as would be produced between one of two parallel, straight, charged conductors and the equipo- tential surface midway between them. The surface of the earth may be assumed to be the equipotential surface ab, Fig. 117, page 380. According to this assumption, the capacitance between the earth and a conductor parallel to it at a perpendicular distance h above its surface may be calculated by assuming a hypothetical conductor or image parallel to the actual conductor and situated at a distance h below the surface of the earth. This hypothetical conductor or image will have a charge equal and opposite to the charge on the actual conductor. The charge which is actually on the surface of the earth is assumed to be on the hypothetical conductor or image. Since the surface of the earth corresponds to the equipotential surface a&, Fig. 117, between the actual conductor A and its image B, the difference of potential between the earth and the actual conductor will be one-half as great as that between the conductor and its image. Therefore since C = -, the capaci- CAPACITANCE OF A TRANSMISSION LINE 387 tance of the conductor with respect to the earth will be twice that given by equations (28) and (29), page 385, or 7 36 X 10~ 3 C to earth = - - microfarads per 1000 feet. (30) , Zifl logio 38.8 X 10~ 3 . . = - ^r microfarads per mile. (31) I log io When applying the formulas for capacitance to earth, it should be remembered that the earth in the neighborhood of the con- ductor is assumed to be a level plane of perfectly conducting material. The assumed distribution of the electric field and therefore the calculation of the capacitance may be considerably modified by the presence of buildings, trees and poles, especially steel poles. It will also be influenced by the variation in the height of the conductor above the ground, as well as by the fact that the earth, in certain localities or in dry weather, may differ considerably from a perfect conductor. It has been shown experimentally that the equipotential plane corresponding to the surface of the earth actually lies below the earth's surface. Experiments carried out at the Massachusetts Institute of Technology in Cambridge have shown it to be approximately coincident with the mean tide-water level. Charging Current of a Transmission Line. Consider a three- phase transmission line with all three conductors in a single plane which is parallel to the surface of the earth. Assume the surface of the earth to be a perfectly conducting plane. According to this assumption the image of any conductor, as for example conductor 1, Fig. 118, will be at a distance below the surface of the earth equal to the height, h, of the conductor above the surface. A line joining any conductor and its image must be per- pendicular to the surface of the earth, since this surface forms the equipotential plane of zero potential. (See Fig. 117, page 380.) Let the distances between the middle conductor and each of the outside conductors be equal. Call this distance D. Let h be the height of the conductors above the earth. Assume the distances between the conductors and also the height of the conductors above the earth to be large compared with the radii. Under these conditions the charges on the conductors may be 388 PRINCIPLES OF ALTERNATING CURRENTS considered as concentrated along the axes of the conductors, so far as any effect outside of the conductors is concerned. Figure 118 shows the conductors of the line and their images. All charges and voltages will be expressed as vectors. Let the charges on the conductors per centimeter length of line be Q it Q 2 and Q 3 electrostatic units. The corresponding charges on the images will then be Qi, Q? and Q 3 . Let the radii of the conductors be r. *(&-- C ! < k--^ fe. 1 1 1 1 h \< D ^ kd-^ < D > --f/c-B h' h' ! i E a, th | i i ////////////////// ////////// i i i ' i i : 1 ! i : _ _ 1' 2' 3' FIG. 118. Consider first the potential difference, Fi 2 , between conductors 1 and 2 due to the three actual charges on the conductors and the three apparent charges of opposite sign on the images. The difference of potential between conductors 1 and 2 is equal to the algebraic sum of the potential differences which would be pro- duced by the three charges and the three images each acting separately. Fi2 = 2^! log, ^ + 2Q 2 log, ^ +' 2Q 3 log, ^ -Qi) log. 2h + 2( -Q,) log. 2( -Q 8 ) log. (2D) (32) CAPACITANCE OF A TRANSMISSION LINE Similarly, on n r F 28 = 2Q, log, -p + 2Q 2 log, 7 + 2Q 3 log, ^ 389 log, ~ + 2Q 2 log, + 2Q 3 log, 1 (33) (34) The computation may be somewhat simplified by placing the 2's before the logarithms as exponents of the quantities under the logarithms. Making this change, = i log, X log. X v 4 X = Ci log, T x D log, x 4/i 2 2 ^2 I x -- __ + g 3 iog, x (35) (36) (37) F 3 i = < Also Fi2 + F 23 + Fsi = (38) Qi + Q 2 + Qs =0 (39) Knowing Fi2, F 23 and F 3 i in complex, Qi, Q 2 and Q 3 may be found in complex. Equations (35), (36) and (37) hold only for electrostatic units. The ratio of the electromagnetic unit of charge to the 390 PRINCIPLES OF ALTERNATING CURRENTS electrostatic unit of charge is equal to the velocity of light or 3 X 10 10 centimeters per second. The corresponding ratio for the units of voltage is the reciprocal of the velocity of light or 10 reciprocal centimeters per second. Therefore, if o X J-U volts are used in equations (35), (36), (37) and (38), the resulting expressions for Qi, Q% and Qs derived from these equations must be multiplied by 10 " X X 10 | = 9 X 11 (40) in order to get the charges in coulombs. Inspection of equations (35), (36) and (37) will show that the effect of the images on the potentials and hence on the charges is negligible for ordinary heights and spacings of the conductors of transmission lines. For example, consider the first term in equation (35). If the image Qi is neglected, this term becomes D 2 Suppose D is 10 feet and the height h is 40 feet. Let the radii of the conductors be 0.5 inch. Then the first term actually is { 100 *X 144 X 4 X 1600 j 025- < 4 x 1600 + 100 2.303 Qi logio {57,600 X 0.985} = 2.303 Qi X 4.7537 If the image be neglected, the first term becomes 2.303 Q! logiof 57,600} = 2.303 Q l X 4.7604 Neglecting the image in this case makes an error of only 0.14 of one per cent in the first term of the equation for Viz- Neglect- ing the images will produce errors of the same order of magnitude in the other terms. If the effect of the images is neglected, equations (35), (36) and (37) become V 12 = Qi log, ~ + Q 2 log, ~ + Q 3 log, \ (41) F 23 = Qi loge 4 + Q 2 log. ~ + Q 3 log, ^ (42) 2 4D 2 -j + Q, 3 log, -3- (43) CAPACITANCE OF A TRANSMISSION LINE 391 The charging current may be obtained from the charges in the following manner, q = Qm sin a>t dq ' ~ i = -^ = wQ m cos wt ... (44) The charging currents in amperes per conductor per centimeter length of line may be found by multiplying the root-mean-square values of the charges per conductor per centimeter length of line by co = 2irf. If the drop in potential along the line is negligi- ble, the total charging current may be found by multiplying the charging current per unit length of line by the length of the line. Although the solution of the preceding equations for the charges on the conductors of a three-phase line is a simple matter when the conductors are in a plane parallel to the surface of the earth, the equations do not take into account the effects of the poles and steel towers or the effect of adjacent lines. They also neglect the effect of adjacent trees and buildings, irregularities of the earth's surface, etc. They assume that the equipotential surface due to the earth is actually coincident with the earth's surface, an assumption which may be considerably in error, especially in very dry sections of the country. This last source of error, however, is of little consequence since the effect of the images on the charges is so small as to be negligible under ordinary conditions. An exact solution taking into account all the factors just mentioned is obviously impossible. These considerations, together with the fact that with ordinary spacing of the con- ductors of a transmission line and the height of the conductors above the earth the charging current is due almost entirely to the capacitance between the conductors themselves, make it possible to neglect the effect on the charging current of the earth and other adjacent bodies. Example of the Calculation of the Charging Current of a Three-phase Transmission Line. A 1 10,000-volt, three-phase, 60-cycle transmission line has conductors 1.14 inches in diameter, which are in a horizontal plane and are spaced 10 feet between the middle and each outside conductor, at a height of 40 feet 392 PRINCIPLES OF ALTERNATING CURRENTS above the earth. Assuming there is no drop in voltage along the line, find the charging current in each conductor per mile of line. The line will be assumed not to be transposed, and the capacitance of the line to earth will be neglected. Balanced voltages will be assumed. The arrangement of the line and the notation used will be the same as in Fig. 118, page 388. Take Fi 2 as the axis of reference. Then 7i2 = 110,000(1+^0) = 110,000 + jO 7 23 = 110,000 - * -j = -55,000 -j95,260 = 110,000 - + = -55,000 + j95,260 From equations (41), (42) and (43), page 390 LO,OC f JO = 2.303 | Q l Iog 10 ^^~-- + 2 lo glo 1 -55,000 - j'95,260 = 2.303 X 9 X 10 11 (45) (10 X 12) 2 logio 4 + Q 2 logio (o^7)s~- + 3 lo glo X 9 X 10U (46) -55,000 + J95,260 = 2.303 lo glo - X 9 X 10" (47) 110,000 +jO = 96.31 X 10 n Qi -- 96.31 X 10 n Q 2 - 12.48 X 10 U Q 3 (48) -55,000 - j95,260 = 12.48 X 10 n Qi + 96.31 X 10 U Q 2 - 96.31 X 10 n Q 3 (49) -55,000 + j95,260 = -108.8 X 10 u Qi + 108.8 X 10 U Q :} (50) Qi + 2 + Q 3 = (51) CAPACITANCE OF A TRANSMISSION LINE 393 Substitute in equation (51) the value of Q 2 from equation (48). -- [96.31 X IQ 11 -^ 12.48 X IQ 11 -^ 110,000 | -- Vl-| ~ 196.31 X 10 llVl 96.31 X lO 11 ^ 3 96.31 X lO 11 !"^ 3 " Q l + 0.4352 Q 3 - 0.571 X 10~ 8 = (52) Substitute in equation (50) the value of Qi from equation (52). -55,000 + j95,260 = -108.8 X 10 n (-0.4352Q 3 + 0.571 X 10~ 8 ) + 108.8 X 10 11 Q 3 Q 3 = 45.63 X 10- 11 + J609.8 X lO' 11 coulombs per cm. (53) Substitute in equation (50) the value of Q 3 from equation (53) . -55,000 + j'95,260 = -108.8 X 10 ll Qi + 108.8 X 10 U (45.63 X 10- 11 + J609.8 X 10~ n ) Qi = 551 X 10~ n - J265.7 X 1Q- 11 coulombs per cm. (54) Substitute in equation (51) the values of Q 3 and Qi from equa- tions (53) and (54) respectively. 551 X 10- 11 - J265.7 X 10- 11 + Q 2 + 45.63 X lO" 11 + j'609.8 X 10- 11 = Q 2 = -597 X 10- 11 - J344.1 X 1Q- 11 coulombs per cm. (55) The numerical values of the charges are Qi = V(551) 2 + (265.7) 2 X 1Q- 11 = 612 X 10~ n coulombs per cm. length of line. Q 2 = V(597) 2 + (344. 1) 2 X 10~ n = 689 X 10~ u coulombs per cm. length of line. X 10~ u = 612 X 10~ n coulombs per cm. length of line. To get the charging current in amperes per mile of line, the charges in coulombs per centimeter length of line must be multi- plied by 27r/(2.540 X 12 X 5280) = 377 X 160,900 = 6.066 X 10 7 /i = 612 X 10- 11 X 6.066 X 10 7 = 0.371 amperes per mile of line. 7 2 = 689 X 10- 11 X 6.066 X 10 7 = 0.418 amperes per mile of line. 7 3 = 612 X 10- 11 X 6.066 X 10 7 = 0.371 amperes per mile of line. 394 PRINCIPLES OF ALTERNATING CURRENTS The currents 7i and 7 3 are equal since the conductors 1 and 3 are symmetrically placed. If the line were transposed, as all transmission lines are in practice, the average charging current per mile or per unit length of line would be the same for all conductors. This assumes balanced voltages. Capacitance of a Balanced Three-phase Transmission Line with Conductors at the Corners of an Equilateral Triangle, Neglecting the Effect of the Earth. When the effect of the earth and adjacent bodies is neglected, the l Q Q i expressions for the charging currents and / \ for the capacitance of a balanced three- / \ phase transmission line, with conductors D jR at the corners of an equilateral triangle, / \ become very simple. Refer to Fig. 119. Q / \ 2 Let the distances between the con- ( ----- D ~~o^ ductors be D, assumed to be large FIG. no. compared with the radii. Let the charges on the conductors 1, 2 and 3 be, respectively, Q 1; Q 2 and Q 3 electrostatic units per centimeter length of conductor. Consider the conductors 1 and 2. The charge on conductor 3 cannot produce any difference of potential between conductors 1 and 2 since it is the same distance from each. The potential difference Fi 2 due to Q 3 is = (56) The resultant potential difference between conductors 1 and 2, due to all three charges, is that produced by charges Qi an d Q* only. - Q 2 ) log e j (57) in a vector sense, as is indicated in the equation. For balanced condition of impressed voltages, since the line is symmetrical, Qi and Q 2 must be equal in magnitude and must differ by 120 degrees in time phase. The magnitude of the vector difference of any two equal vectors which differ in time phase by 120 degrees is equal to the magnitude of either vector multiplied by the CAPACITANCE OF A TRANSMISSION LINE 395 square root of three. Hence for balanced conditions, the alge- braic form of equation (57) may be written V=2V3Q log, y (58) or Q = - -r (59) Equation (59) gives the magnitude of the charge Q per con- ductor per centimeter length of line in electrostatic units in terms of the magnitude in electrostatic units of the voltage V between conductors. Since i = -TT, the charging current per conductor per centimeter length of line is 2x/F / = - ~. statamperes (60) where V is expressed in statvolts, i.e., in electrostatic units. Both I and V are root-mean-square values. If V is expressed in volts, the right-hand member of the equation must be multiplied by Q 1 Q 11 to give the current in amperes. /. r . = J Qi per conductor per centimeter amperes 2-v/31og 1 9 X 10 11 length of line (61) T j It will be seen from equation (57) that the voltage Viz is in phase with the vector difference of the charges Qi an d Qz, i.e., in phase with (Qi Q 2 ). Similarly Vzz is in phase with (Q 2 Qz) and Vz\ is in phase with (Q 3 Qi). For balanced conditions, the three charges Q i} Q 2 and Q 3 must be equal in magnitude and must differ by 120 degrees in time phase. The only way in which the charges can be equal in magnitude and differ in phase by 120 degrees, and at the same time have their vector differences taken in pairs in phase with the line voltages, is for the charges to be in phase with the wye voltages of the system. D D Viz = VIQ - V Z o = 2Q!log, - 2Q 2 Jog y (62) 396 PRINCIPLES OF ALTERNATING CURRENTS where the voltages Fio and Vzo are the voltages between the neutral and conductors 1 and 2 respectively. The phase rela- tions between the voltages and charges will be made clearer by referring to Fig. 120. This figure shows the phase relations between the vectors in equation (62). The charges on the conductors per unit length of line are the same as would exist on three equal condensers connected in wye across the three-phase line, each condenser having a capaci- tance of -Q-. = l statf arads (63) -^ _ &e - 2 log, - V3 r r Equation (63) gives the equivalent capacitance, in electrostatic units, per phase to neutral, per centimeter length of a balanced three-phase transmission line, with con- ductors at the corners of an equilateral triangle. It gives the capacitance for each of three equal condensers which, if connected in wye across the line, would take the same charging current as the line actually takes. Equation (63) holds only for balanced conditions. It should be noticed that the equiva- FIQ. 120. l 611 ^ capacitance to neutral of a balanced three-phase transmission line, with con- ductors at the corners of an equilateral triangle, is equal to twice the capacitance between conductors of a single-phase line with the same conductor spacing. (Equation (27), page 384.) It is equal to the capacitance to neutral of the single-phase line. In terms of common logarithms and microfarads per mile of line, equation (63) becomes (see equation (29), page 385). r 38.8 X 10- 3 . C =r microfarads per mile of line (64) logio - The charging current per conductor per mile of a balanced three-phase line with conductors at the corners of an equilateral triangle, neglecting the drop in voltage along the line, is CAPACITANCE OF A TRANSMISSION LINE 397 27r/F n X 38.8 X IP" 9 D logic - 2 44 F f X 10~ 7 r -- amperes per conductor per mile of line. (65) V n is the voltage to neutral. Equations (57) to (65) inclusive neglect everything except the line itself. They hold only for balanced conditions. When the conductors of a balanced, transposed transmission line are not at the corners of an equilateral triangle, the average capacitance of the line to neutral may be calculated by using the spacing of the equivalent equilateral arrangement. As an approximation, this spacing may be taken equal to D = ^D, XD 2 XD 3 (66) where DI, D z and D& are the actual distances between the conductors. Example of the Calculation of the Average Charging Current per Conductor per Mile of a Transposed Line by the Use of the Equivalent Equilateral Spacing. A 1 10,000-volt, 60-cycle, three-phase, transposed transmission line has its conductors arranged in a horizontal plane 40 feet above the earth, with 10 feet between the middle conductor and each outside conductor. The radius of the conductors is 0.57 inch. What is the average charging current per conductor per mile of line. The equivalent equilateral spacing will be used in calculating the charging current. D = ^/D, XDtXDs I = = v"lO X 10 X 20 = 12.60 feet. 2.447 n / X 10- 7 - 244X 110>OQQ X 60 X 10- 7 V3 9.30 X 10- X 10- 7 / = - log] 12.60 X 12 logio 265.3 0.57 = 0.384 amperes per conductor per mile of line. 398 PRINCIPLES OF ALTERNATING CURRENTS The actual charging current per conductor per mile of the com- pletely transposed line may be found from the charges per centi- meter length of each conductor when the line is not transposed. These charges were found to be (see page 393). Qi = 551 X 10- 11 - J265.7 X lO' 11 Q 2 = -597 X 10- 11 -J344.1 X 1Q- 11 Q 3 = 45.63 X 10- 11 + J609.8 X 1C)- 11 coulombs per conductor per centimeter length of line, where Qi, Qz and Q 3 are the charges on the three conductors with the conductors in the positions shown in Fig. 118, page 388. If the line is transposed so that the conductors from left to right, Fig. 118, are 3, 1, 2, the charge Qi on conductor 1 would be the same in magnitude as Q 2 given above, but it would not be in phase with it since the middle conductor would now be connected to phase 1 instead of to phase 2. Since the cyclic order^of the voltages used in finding the charges Qi, Q 2 and Q s was Viz, ^23, Vsi with Vzs lagging Viz, the charge Qi on conductor 1 when it occupies the middle position, i.e., the position 2, Fig. 118, may be found both in magnitude and in time phase by multi- plying Q 2 by the operator which produces a rotation of +120 degrees. If Q ~ Qi is the charge on conductor 1 per centimeter length of line when this conductor occupies the position shown in Fig. 118, the charges on it, per centimeter length of line, when it occupies the positions 2 and 3 may be found by rotating the charges Q z and Q 3 as given above through +120 and 120 degrees respectively. The charges on conductor 1 per centimeter length of line when the conductor occupies the three positions necessary for complete transposition are Position 1 Q = (551 - j265.7)10- n Position 2 Q = (-597 -j 344.1) (-0.500 + J0.866)10- n = (596.5 - J344.9)10- n Position 3 Q = (45.63 + j609.8)(- 0.500 - j0.866)10- n = (505.3 - j344.4)10- n CAPACITANCE OF A TRANSMISSION LINE 399 Average O = I { (551 + 596.5 + 505.3) + o j(- 265.7 - 344.9 - 344.4) JIG" 11 = (551 - j318)10- u Average Q = V(551) 2 + (318) 2 X 10~ n = 636 X 10~ u coulombs per cm. length of line. The actual charging current per conductor per mile of the com- pletely transposed line is / = 2-nf X 636 X 10- 11 X (2.540 X 12 X 5280) = 0.386 amperes This differs by less than one per cent from the value found by the use of the equivalent equilateral spacing. Voltage Induced in a Telephone or Telegraph Line by the Electrostatic Induction of the Charges on the Conductors of an Adjacent Transmission Line. A telephone line is on the same poles as a three-phase transmission line, whose conductors are in a plane parallel to the surface of the earth and at a distance h above it. Let the distance between the middle and each of the outside conductors be D. Let the two telephone conductors, A and B, be at a distance 2d' apart and each at a distance h' above the surface of the earth. Also, let the telephone conductors be at equal distances from a line joining the middle conductor of the transmission line and the image of this conductor. Refer to Fig. 118, page 388. Let the charges on the conductors of the transmission line per unit length of line be Qi, Q 2 and Q 3 electrostatic units. The potential difference between conductor A of the telephone line and the earth, due to the charges +Qi on conductor 1 of the transmission line and Q\ on the image of conductor 1, is V AO ' = 2Q l \og, V(h h , )2+(D _ dv = statvolts (66) h , )2 + (D _ Similar expressions hold for the voltages VAO" and VAO" between conductor A and the earth due to the charges on conductors 2 and 3 and their images. The resultant potential difference 400 PRINCIPLES OF ALTERNATING CURRENTS between conductor A and the earth, due to all three charges Qi, Q 2 and Q z and their images Qi, Qz and Q 3 , is (h -h f y + (D - + < + <2* log. statvolts (67) If VBO is the resultant difference of potential between the conductor B and the earth, found in the same manner as the potential difference VAO, the potential difference between the conductors A and B of the telephone line is VAB == VAO ~ VBO (68) In order to evaluate VAO and VBO, it is necessary first to calciK late, by the method already given, the charges Qi, $2 and Q 3 in complex from the voltage, height and spacing of the conductors of the transmission line. Substitute these values of the charges in equation (67) for VAO and in the corresponding equation for VBO- The voltage VAB then may be found from equation (68). In most cases it is not necessary to consider the effect of the images. Transposition, when it can be properly carried out for either the power line or the telephone line, will eliminate the resultant voltage in an adjacent telephone line due to eletromagnetic in- duction. With the power line transposed, the induced voltages neutralize in such a length of the telephone line as is exposed to a complete transposition of the transmission line. With the telephone line transposed, the voltages induced in adjacent sec- tions of the telephone line are equal and opposite and neutralize. When either the telephone line or the transmission line is trans- posed, the voltage induced by electromagnetic induction in any section of a telephone line will be too low to be troublesome or dangerous. Transposition, when it is properly carried out for either the tele- phone line or the transmission line, will eliminate the resultant voltage in a telephone line due to electrostatic induction of an adjacent transmission line, i.e., it will make the resultant voltage acting along the line zero. Transposition, however, will CAPACITANCE OF A TRANSMISSION LINE 401 not eliminate the danger from the voltage induced by electrostatic induction, as the magnitude of this voltage is independent of the exposed length of the telephone line. The voltage in a telephone line due to the electrostatic induction of an adjacent transmission line, and especially a single-phase transmission line with ground return, may not only be troublesome but also dangerous to We. This situation may be met by providing a shunt, of low reactance for transmission frequency, between the conductors of the tele- phone line and between these conductors and the earth. This ;\ P Equations (5) and (6) are the current and the potential gradients, respectively, at a distance L from one end of the line. Differentiating equations (5) and (6) with respect to L gives d 2 ! - dV ~dT* ~ dL Combining equations (7) and (6) and equations (8) and (5) gives (9) - yzV = (10) Equations (9) and (10) are the differential equations for current and voltage of a transmission line. Equations (9) and (10) are linear differential equations of the form p = e mq is a solution of equation (11). The complete solution of equation (11) is p = A l miq + A 2 e m ' fl a l m 2 e mq + a 2 me mq + a s e m(I = aim 2 + a 2 m + a- 6 = From equations (9) and (10) a\ = 1, a 2 = and a 3 = yz. m 2 yz = m = \/P The solutions of equations (9) and (10) are, therefore, (13) SERIES-PARALLEL CIRCUITS 409 Equations (12) and (13) apply only to the steady state, i.e., after any transient has disappeared. The constants of integration, AI, A 2 , BI and B z must be determined from assumed terminal conditions either at the load or at the source of power. Let them be determined from the assumed terminal conditions at the load. For this case, L is the distance from the receiving end of the line to the point at which the voltage and current are desired. It must be reckoned positive when measured from the load end of the line to the source of power. Let the assumed current and voltage at the receiving- or load-end of the line, i.e., for L = 0, be I R and VR respectively. Differentiate equation (12) with respect to L and put L = and V = VR. This gives (1 Wp r^_ -WP - yV y V K = AI \/$z - A 2 Vp (14) Putting L = and V = VR directly in equation (12), J R = A! + A 2 (15) Combining equations (14) and (15), yVa = (In - A 2 ) \/yz - A 2 \/p From which _ I R \/yz y V R (16) * \ \ 'A I and (17) The constants BI and B z may be found in a similar manner from equation (13). Substituting the values of the constants AI, A 2 , BI and # 2 in 410 PRINCIPLES OF ALTERNATING CURRENTS equations (12) and (13) gives the current and voltage at a distance L from the receiving end of the line. 7 R coshL-v/P + "F*A/ sinh Z/Vp (22) ,_. . ; + '\2V i (24) = y K cosh L\/yz + /a A/- sinh L\/yi (25) ' c/ 7 and F in equations (22) to (25) are the current and voltage at a distance L from the receiving or load end of the line. The expressions \/yz, * and ^ are the fundamental constants of a transmission line. If L is measured from the generator end of the line, i.e., is positive from the source of power towards the load, equations (22) and (25) become = I G cosh L P - VG- sinh Vp (26) - 7 G J| si * c/ sinh L Vp (27) where 7 and F are the current and voltage at a distance L from the generator end of the line. I and VG are the current and voltage at the generator end of the line. . /- is called the surge impedance and is denoted by Z Q . + _ is called the surge admittance and is denoted by y Q . Both the surge impedance and the surge admittance are complex quantities. is known as the propagation constant and is denoted by a. SERIES-PARALLEL CIRCUITS 411 It follows from Chapt. I, pages 21 and 22, that (28) cos (B z + By) + j sin = (B 2 + B v ) (29) L & = a \Ba = r+ j +ia * ) may be written e La > jLa \ I = \ (la + yoV*) e La ' e iLa * + (/* - 2/o V) e~ La > -*" (39) For increasing values of L, z.e., in going from the load to the generator, the first term of the second member of equation (39) is a vector which increases in magnitude logarithmically with L, on account of the factor e La '. This first term advances in phase at a rate which is directly proportional to L, due to the factor J 1 "*. This term may be written I Direct = (In + 2/0 V R ) e L wave reflected at the load which travels from the load to the generator. SERIES-PARALLEL CIRCUITS 413 The two terms of the second member of equation (23) may be considered to represent direct and reflected voltage waves. (42) V Reflected = (V R ~ 2 /ft) (43) where Z R is the impedance of the load and 7* is the load current. Equations (41) and (43) show that reflection of the current and voltage waves will not occur at the load when y R = y Q or 2ft = 2 , the equality being considered in a vector sense. In general, reflection will always occur at a point in a circuit where there is a change in the electrical constants of the circuit. The direct and reflected waves combine to produce standing waves with maximum and minimum points, i.e., loops and nodes, equally spaced along the line. As these loops and nodes must occur a quarter wave-length apart, they will not show on lines of present commercial length and frequency, since a quarter wave length, even for sixty cycles, is about 775 miles. A more complete discussion of direct and reflected waves will be found in works dealing specifically with long-distance power transmission. 1 Propagation Constant, Attenuation Constant, Wave-length Constant, Velocity of Propagation and Length of Line in Terms of Wave Length. It was stated on page 410 that is known as the propagation constant. The real part of the propagation constant, i.e., i = Vy~z cos ^ (By + 0.) (44) determines the attenuation of the traveling waves. It is known as the attenuation constant. (Equation (39), page 412.) 1 Electric Phenomena in Parallel Conductors, F. E. Pernot. 414 PRINCIPLES OF ALTERNATING CURRENTS The coefficient of the imaginary part of the propagation constant, i.e., 2 = Vyz sin 2 (B y + 0.) (45) determines the amount of phase retardation of the component waves in their direction of propagation. It is known as the wave-length constant. (Equation 39, page 412.) The wave length, X, of the traveling waves is X = ^ (46) <*2 where 2 is in radians. The unit in which the wave length, X, is expressed depends on the unit of length selected for the con- stants y and z. If they are per mile of line, X will be in miles. If they are in any other unit, X will be in the same unit. The magnitude of the product yz is independent of the units chosen for admittance and impedance, provided corresponding units are used, i.e., mhos and ohms or abmhos and abohms. X= ^ - (47) \/yz sin ^ (y + #*) (Equation (32), page 411.) The length, L, of a transmission line, expressed in terms of the wave length for the impressed frequency, is The velocity of propagation is X/ = 2 ^ (49) where / is the frequency. yz = (g + jb)(r + jx) = (i + jo? 2 ) 2 (rg xb) + j(rb + gx) = i 2 + j c loL\a.^ CL\/LC (52) where co = 27r/ and L and C are the inductance and capacitance per unit length of line. For a line without transmission losses, the wave length would be 2ir 27r 1 , KO , X = = - -7-= - (53) From the dimensions of inductance and capacitance, it may be shown that /=-^ has the dimensions of velocity. This velocity has been shown to be the velocity of light, or 3 X 10 10 centimeters per second. It is the limiting velocity of propagation for electric waves. The actual velocity of propagation is slightly less than this on account of the terms zy and rg in the expression for 2 (Equation (51)). The difference between the actual velocity and that found by assuming no transmission losses is not great as a rule. Solution of the Transmission Line. If a table of Hyperbolic Functions of Complex Quantities is available, equations (22) and (25) or (37)_and (38), pages 410, 411 L should_be used for cal- culating I G and Vo at the generator when 7and Va at the load are known. For calculating I R and VR at the load when 7 and Vo at the generator are known, equations (26) and (27) should be used. The open-circuit voltage at the load is found by putting I R equal to zero in equation (25), page 410, or in equation (38), page 411 and then solving for VR. The charging current of the line, i.e., the no-load current, may be found for fixed generator voltage by substituting, in equation 416 PRINCIPLES OF ALTERNATING CURRENTS 4 (22), page 410, or in equation (37), page 411, the no-load volt- age at the receiving or load end of the line, as found above, and then solving for 7 G , IR being zero. The length of the line in terms of the wave length of the impressed voltage is found from equation (48), page 414. The efficiency of transmission is PR = VR!R COS B R P G V G I G cos e G where P R and P G are the power at the load-end and the generator- end of the line respectively. The angles R and 6 G are the power-factor angles at the load-end and the generator-end of the line respectively. The power may be found as indicated, but it may be more convenient to find it from the complex expressions for current and voltage. Evaluation of the Equations for Current and Voltage when Tables of Hyperbolic Functions of Complex Quantities are not Available. From equation (30), page 411. cosh La = coshL(i + ja 2 ) _ 1 (cos La 2 + j sin L 2 ) - + -e~ LQr ' (cos La z j sin La 2 ) + j\(e La > -e- 1 ""') sin L 2 cosh La\ cos La: 2 + j sinh La\ sin Loti (54) It may be shown, in a similar manner, that sinh La = sinhLai cos La 2 + j cosh Lai sin La 2 (55) Then Vo = VR cosh La + 7#2o sinh La = Fftfcosh Lai cos L 2 + j sinh Lai sin La 2 \ Lai cos La 2 + j cosh Lai sin La 2 [ (56) SERIES-PARALLEL CIRCUITS 417 where zo = 2 cos (^ - e v ) +j sin(0 2 - IG = IR cosh La + V R yo sinh La = IR( cosh Lai cos La 2 + j sinh Lai sin La 2 j + F2/o{sinh Lai cos La 2 + j cosh Lai sin La 2 } (57) where 1 1 cos ^(By e z ) + j sm -(e v B z ) j I COS ^(By + Bz} 1. = Vyz sin ^(By + 9.) Fo, F/z, /o and 7 a must be in complex. If VR and / are known, the solution of equations (56) and (57) will give VG and J G in complex. If VG and I G are known, the solution of these equations will give VR and I R in complex. The open-circuit voltage at the load or receiving end of the line is found by putting 7 B equal to zero in equation (56). F*(open circuit) = - = ^V -y r-^ (58) cosh Lai cos La 2 + J sinh Lai sin La2 When J R is zero, the first term of the second member of equation (57) is zero. If VR' is the open-circuit voltage at the load end of the line, found from equation (58), the charging cur- rent is given by J (charging) = V R 'y {cos \(B V - 0.) +jsin^(0 v - 0.) 1 L & X {sinh Lai cos La 2 + j cosh Lai sin La 2 j (59) When a transmission line is not uniform, due to loads being taken off at intermediate points or to a change in conductor spacing or in size of conductors, it must be considered in sections of such length that the principles of uniform lines are applicable. The ends of the sections will be at the points where loads are applied or where the conductor spacing or the size of conductors changes. 27 418 PRINCIPLES OF ALTERNATING CURRENTS Example of the Calculation of the Performance of a Trans- mission Line for Steady Operating Conditions from its Electrical Constants. A 150-mile, 25-cycle, 110,000-volt, 3-phase trans- mission line has its conductors placed at the corners of an equi- lateral triangle. The conductors are 10 feet apart on centers. Each conductor is a 19-strand copper cable with an overall diameter of % inch and an equivalent cross section of 300,000 circular mils. The skin effect at 25 cycles for conductors of the size used for this line is very small and may be neglected. The line operates below its corona voltage, i.e., below the voltage at which the loss due to corona discharge between conductors appears. The line insulation is assumed high. The corona loss and the leakage between conductors over line insulators, etc. will be assumed to be negligible. Calculations will be based on an average line temperature of 20 degrees centigrade. What are the voltage, current, power and power-factor at the power station or generator end of the line for an inductive load at the receiving end of 30,000 kilowatts at 110,000 volts and 0.90 power-factor. What is the efficiency of transmission? If the voltage at the generator end of the line remains constant at the value required for the given load conditions, what will be the charging current of the line and the open-circuit voltage at the receiving or load end of the line? What is the length of the line in terms of the wave length of the current and voltage waves? The resistance of the copper conductors will be assumed to be 10.58 ohms per mil-foot at 20 degrees centigrade. The mile will be used as the unit of length. coon r = 10.58 X ^~ v" A = 0.1862 ohm per mile. jUUU g = negligible (assumed zero). 38.8 X 10- 3 38.8 X C (to neutral) = - , 10 X 12 2.584 logio 5/ 71 Q = 15.0 X 10~ 3 microfarad per mile. b = 2irf X 15.0 X 10~ 3 X 10- 6 = 157 X 15.0 X 10- = 2.355 X 10~ 6 mho per mile. SERIES-PARALLEL CIRCUITS 419 x = 27T/ (741 lo glo 1U ^ e 12 + 80) X 10- 6 = 157(741 X 2.584 + 80) X 1Q- 6 = 0.3130 ohm per mile. z = \/r 2 + x 2 = V(0.1862) 2 + (0.3130) 2 = 0.3642 ohm per mile. 0.3130 tan ' = OJ862 = L681 e z = +59.25 degrees per mile. z = 0.36421+ 59?25 hm per mile. y = vV + V = V(O.OOO) 2 + (2.355 X 10~ 6 ) 2 = 2.355 X 10~ 6 mho per mile. 2.355 X 10- 6 By = +90.0 degrees per mile. y = 2.355 X 1Q- 6 |+90?0 ohm per mile. a = ^/^z = \/2.355 X 1Q- 6 X 0.3642 = 0.9261 X 10~ 3 per mile. _ B y + 6 Z _ 90?0 + 5925 6Q = 63,510 volts. V3 r / N 30 >00 X 1000 1P __. n I R (per conductor) 3 x 63?51Q x Q 9Q - 175.0 amperes. cos B R = 0.90 B R = 25.84 degrees. Take VR as the axis of reference. VR = 63,510 |0 f 0000 I* = 175.0|-25?84 cosh aiL = 1.001 sinh aiL = 0.03684 1 Qf) 2 L = 0.1340 X - 7T = 7.675 degrees, sin a 2 L = 0.1336 cos a 2 L = 0.9912 cosh aL = cosh a\L cos a^L + j sinh a\L sin aju = 1.001 X 0.9912 + J0.03684 X 0.1336 = 0.9922 + J0.004920 0.004920 = V(0.9922) 2 + (0.004920) : = 0.9922 1 +0?284 tan" 0.9922 SERIES-PARALLEL CIRCUITS 421 sinh aL = sinh a.\L cos a 2 L + j cosh ctiL sin a^L = 0.03684 X 0.9912 +j 1.001 X 0.1336 = 0.03651 +j 0.1337 0.1337 V(0.03651) 2 + (0.1337) ; tan- 1 0.03651 = 0.13861+ 74 72 V = 7 fl cosh oL -f- H R z Q sinh oL = 63,510 lOOQQ X 0.9922 1 +0?28 + 175.01 -25 84 X 39&3 H1538 x 0. 13861+ 74 ?72 = 63,000 +0?28 + 9538i+3350 63,000(1.000 + J0.00495) + 9538(0.8338 + J0.5521) 70,950 + J5578 5578 V (70,950) 2 + (5578) : tan~ 70,950 = 71,170J+4495 volts to neutral. V G = 71,170 X \/3 = 123,300 volts between conductors. . 123,300-110,000^ Line regulation for the given load = - Tin 000 ~ = 12.1 per cent. The no-load voltage at the load with 123,300 volts maintained at the generator is V R (no load) = -^= V (cosh ctiL cos azL) 2 + (sinh a\L sin a^L) 2 123,300 = 0.9922 = 124,300 volts between conductors. The percentage rise in voltage at the load-end when the load is removed, the generator voltage being maintained at 123,300 volts, is Io = IR cosh ^Jj -f Vxyo sinh L = 175.0|-2584 X 0.9922 1 +0284 + 63,510 10000 X 0.002543 jf!5?38 x 0.1386 1 +74 72 = l73.6'-2556 + 22.38 1+9010 - 173.6 (0.9022 - J0.4314) + 22.38(-0.0017 + j 1.000) 422 PRINCIPLES OF ALTERNATING CURRENTS = 156.6 - J52.51 = \/(156.6) 2 + (52.51) 2 tan' -52.51 156.6 = 165.2 1 - 18 ?54 amperes. &G = By 01 = 4495 - (-1854) = 23.03 degrees. Power-factor at generator = cos 6 G = cos 23 ?03 = 0.9205 P G = V G I G cos e G = 71,170 X 165.2 X 0.9205 X 10~ 3 = 10,820 kilowatts per phase. By complex method P G = (70,950 X 156.6 - 5578 X 52.51) X 10~ 3 = 10,820 kilowatts per phase. 10 000 Efficiency of transmission = - ^ X 100 = 92.5 per cent. When the generator voltage is maintained at the value required to give 110,000 volts between conductors at the load, the load being 30,000 kilowatts at 0.9 power-factor (inductive), the no- load charging current is 7 G (charging) = I G = V R 'yo sinh aL where VR is the voltage at the load on open circuit. V R ' IG = ^r (V R y sinh V R The magnitude of (FR#O sinh L) has already been found for a voltage at the load of V R = -- ^= volts to neutral. (See calculation of the generator current for load conditions.) It is the second term of the second member in the expression for the generator current. The charging current at no load varies di- rectly with the voltage. _ 123,300 " 110,000 > = 25.1 amperes. SERIES-PARALLEL CIRCUITS 423 Length of the line in wave lengths is 2 X 3.142 where X is the wave length of the current and voltage waves. Length of line, neglecting the transmission losses, is T Length of line in miles X frequency v L x = , T i ., , .. , . . TJ - X wave length Velocity of light in miles per sec. 150 X 25 " 186,000 X = 0.02016X INDEX Active component, of current, 58; of voltage, 62. Active, current, 59; power, 60; voltage, 63. Addition, of admittances, 216; of currents or voltages, 40, 43, 110; of imped- ances, 200. Admittance, 213, 243; addition of, 216; complex expression for, 215; polar expression for, 217. Air -core transformer, induced voltage in windings of, 187; vector diagram of, 189. Alternating current, definition of, 27; ampere value of, 34, 82; calculations based on sinusoidal waves, 29; measurement of, 37, 79; strength of, 33. Alternating voltage, definition of, 27; generation of, 31, 69, 245; measure- ment of, 37, 79; volt value of, 35, 84. Ampere value, of an alternating current, 34, 82. Amplitude factor, 106. Analysis, of non-sinusoidal waves into fundamental and harmonics, 89, 92. Apparent power, 55. Attenuation constant, 413. Axis, imaginary, 6; real, 6. Balanced, Y-connected load connected across a three-phase circuit having balanced voltages, 274, 277; Y- and A-connected loads in parallel, 277. Braking, an inductive circuit, 127. Capacitance, 118; effect of, 119, 136, 137, 140, 142, 150; of two, straight, parallel conductors, 376; of two, straight, parallel conductors, per 1000 feet of line, 385, per mile of line, 385; between earth and a straight conductor parallel to its surface, 386; between earth and a straight conductor parallel to its surface per 1000 feet of conductor, 387, per mile of conductor, 387; of a balanced three-phase transmission line with conductors at corners of an equilateral triangle, 394, per mile, 396; of a three-phase transmission line from its equivalent equilateral spacing, 397. Charging current, of a transmission line, 387, 391, 395. Circuit, containing constant resistance and capacitance in series, 136, 137, 140, 142, 149; containing constant resistance and inductance in series, 119, 121, 125, 127, 128, 134; containing constant resistance, inductance and capacitance in series, 150, 151, 157, 161, 168; coupled, see coupled circuits; inductive, breaking of, 127; filter, 229; see three-phase, four-phase and six-phase. Coefficient, of electromagnetic coupling, 180; of leakage, 178; of leakage- induction, 183; of mutual-induction, 116, 174, 176, 180, see mutual inductance; of self-induction, 115, 180, 185. 425 426 INDEX Complex, method, 3; quantities, 3, 23, 135; expression for admittance, 215; expression for impedance, 134. Components, active, energy, or power of a current or voltage, 58, 61, 62; direct-phase, of an unbalanced three-phase system, 338, 341, 345, 346; imaginary of a vector, 6; reactive, wattless, or quadrature, of a current or voltage, 58, 61, 62; real of a vector, 6; reverse-phase, of an unbal- anced three-phase system, 338, 341, 345, 346; uniphase components of an unbalanced three-phase system, 338, 341, 345, 346. Conductance, 213, 243. Connections, delta, 255, 292; four-phase, 262, 302; mesh, 257, 262, 302, 306; six-phase, 305, 306; star, 257, 262, 302, 306; wye, 255, 289. Constants of a transmission line, attenuation, 413; propagation, 413; wave-length, 414. Copper, relative amounts required for power transmission with different numbers of phases, 320, 321, 322. Copper loss, in an unbalanced three-phase circuit, 352. Crest factor, 106. Coupled circuits, differential equation for, 181; effect of short-circuiting one of two inductively coupled circuits on the apparent self-inductance of the other, 182; leakage coefficients of, 178; leakage inductance of, 183; magnetic leakage of, 178. Coupling, close, 181; coefficient of electromagnetic, 180. Current, at one end of a transmission line in terms of line constants and current and voltage at other end, 406, 410, 411; active, energy or power, 58, 61; active, energy or power component of, 58, 61; alternating, defini- tion of, 27; ampere value of, 34; average value of, 35; continuous, defini- tion of, 27; charging, of a transmission line, 387, 391, 395; direct, definition of, 27; effective value of, 33, 34, 82; instantaneous value of, 27; measurement of effective or root-mean-square value of, 37, 79; oscillating, definition of, 27; positive direction of, 249; pulsating, definition of, 27; ratio of average and effective values of, for sinusoidal waves, 36; reactive, wattless or quadrature, 58, 61; reactive wattless or quadrature components of, 58, 61; simple harmonic, definition of, 29; strength of, 33, 34, 82. Current, non-sinusoidal, analysis of, 89, 92; effective oj root-mean-square value of, 82; effective or root-mean-square value of, from polar plot, 88; equivalent sine waves, 107; measurement of effective or root-mean- square value of, 79; representation of by a Fourier Series, 72. Currents, addition of, 40, 43, 110; polyphase, generation of, 245; positive direction of, 249; relative magnitudes of line and phase, for a balanced three-phase circuit, 258, 266, 291, 292; for a balanced four-phase circuit, 262, 266, 304, for a balanced six-phase circuit, 266, 309; for a balanced twelve-phase circuit, 266, for a balanced n-phase circuit, 264. Cycle, 27. Delta connection, 255, 292; equivalent wye connection, 280, 281, 282, 283, 284, 294; relative magnitudes of line and phase currents and voltages and INDEX 427 their phase relations, 258, 289, 292; third harmonic short circuited in, 292; with unbalanced A-connected load, 268; with single-phase load, 261. Deviation factor, 107. Direct, current, definition of, 27; voltage, definition of, 27. Direct-phase components, of currents and voltages of an unbalanced three- phase circuit, 338, 249, 252; determination of, 341, 345, 346. Double -subscript notation, 248, 250. Effective, reactance, 237; resistance, 234; value of an alternating current or voltage, 33, 34, 82; value of current or voltage, measurement of, 37, 79; value of current or voltage from a polar plot, 88. Electromotive force, see voltage. Electrostatic induction, voltage induced by, in a telephone or telegraph line by the charges on the conductors of an adjacent transmission line, 399, 401, 403. Energy, component of current, 58; component of voltage, 62; of electro- magnetic field, 116, 124; of electrostatic field, 139. Equations, vector, solution of, 12. Equivalent, equilateral spacing of a three-phase transmission line, 397; phase difference, 107; resistance and reactance, 240; sine waves, 107; Y- and A-connected circuits, 280, 281, 282, 284, 294. Equilateral spacing, equivalent, of a three-phase transmission line, 397. Factor, amplitude, peak or crest, 106; deviation, 107; form, 37, 106. Filter circuits, 229. Fisher -Hinnen, method of analysing a periodic non-sinusoidal wave, 92. Flux, mutual, of inductively coupled circuits, 186; producing mutual induc- tance, 185; producing self -inductance, 185; producing leakage induc- tance, 185. Form factor, 37, 106. Four -phase, alternators, 245, 262; circuits, connection of, 262, 302; har- monics in, 299, 301; relative magnitudes of line and phase voltages for, 262, 266, 303. Fourier Series, for a non-sinusoidal current or voltage, 72, 75; determination of constants of, 89, 92; for rectangular and triangular waves, 78; for symmetrical and non-symmetrical waves, 75. Frequency, 28, 247. Generation, of an alternating electromotive force, 31, 69, 245. Harmonic, 73; simple, current or voltage, 29. Harmonics, 73; effect on power, 85; effect on power factor, 86; even, effect of on wave shape, 75; in balanced three-phase circuits, 287; in balanced four-phase circuits, 299; in balanced six-phase circuits, 304; third, in neutral connection of a three-phase Y-connected circuit, 291; third, in neutral connection of a four-phase star-connected circuit, 304; third, short-circuited in A-connected circuit, 292; third, not short-circuited in four-phase mesh-connected circuit, 304. Henry, 116. 428 INDEX Imaginary, components, 6; axis, 6. Impedance, 243; of a circuit containing constant resistance and self -induc- tance in series, 134, complex expression for, 134, polar expression for, 136; of a circuit containing constant resistance, and capacitance in series, 148, complex expression for, 149, polar expression for, 150; of a circuit containing constant resistance, inductance and capacitance in series, 168; complex expression for, 169, polar expression for, 170. Impedances, equivalent Y- and A-connected, 280, 281, 282, 283, 284; in parallel, 211, 215, 243; in series, 196, 199, 243; in series-parallel, 224. Inductance, of a single-phase transmission line, 359; leakage, 183, 185; mutual, 116, 173, 174, 176, 180, 185. Induction, electrostatic, between a transmission line and an adjacent telephone or telegraph line, 399, 401, 403; mutual, between transmission lines or a transmission line and a telephone line, 367; voltage, electro- magnetic, induced by, in a telephone line by a three-phase transmission line, 370, 371, 372; voltage, electrostatic, induced by, in a telephone line or telegraph line by the charges on the conductors of a transmission line, 399. Instantaneous, value of current or voltage, 27. j, operator which rotates a vector through ninety degrees, 4, 5, 10. Kirchhoff's Laws, 271, 272. Leakage, coefficient, 178; flux, 178, 185; inductance, 183; magnetic, 178. Magnetic leakage, 178. Magnitudes, relative, of currents and voltages in polyphase systems, see delta, wye, four-phase, six-phase, twelve-phase and n-phase. Measurement, of current and voltage, 37, 79; of power (active), 57, 79; of reactive power, 63. Mechanical system, having constant friction, mass and elasticity, 170. Mesh connection, 257, 262, 264, 302, 306. Mutual flux, of inductively coupled circuits, 186. Mutual induction, 116, 173, 174, 180, 185; between transmission lines or between a transmission line and a telephone line, 348, 367; of two coupled circuits in a medium of constant permeability, 176; differential equation for a coupled circuit having constant mutual and self-induc- tance, 176; flux corresponding to, 185. Neutral, third harmonics in, for a three-phase Y-connected system, 291; for a four-phase star-connected system, 304. Non -sinusoidal current or voltage waves, addition and subtraction of, 110; effective or root- mean-square, value of, 82; representation of, by a Fourier Series, 72. Non-sinusoidal voltage, impressed on a circuit containing constant imped- ances in series and in parallel, 232. N -phase system, power in, 314; relative magnitudes ii}, of line and phase currents for sinusoidal waves, 264. INDEX 429 N -wattmeter method, for measuring power in an n-phase system, 326. (N-l) -wattmeter method, for measuring power in an n-phase system, 326. Operators, (cos a j sin a), 7, 8; exponential, 15; j, 4; polar, 19; powers of, 9, 24; product of, 9, 20; reciprocal of, 11; roots of, 9, 20, 24; successive application of, 9; which produce uniform angular rotation, 11, 18. Oscillation, 155, 159, 162; period of, for a circuit containing constant resis- tance, inductance and capacitance in series, 208; time of, for a circuit containing constant resistance, inductance and capacitance in series, 156, 159, 162; of a mechanical system, having constant friction, mass and elasticity, 170. Parallel, impedances in, 211, 243; resonance, 220. Peak factor, 106. Period, 27; of oscillation of a resonant circuit containing constant resistance, inductance and capacitance in series, 208. Periodic, time, 27. Phase, 2, 30; balancer, 350; equivalent difference in, for non-sinusoidal waves, 107. Phase order of harmonics, in a balanced three-phase circuit, 287, 289; in a balanced four-phase circuit, 301; in a balanced six-phase circuit, 307. Polar, expression for admittance, 217; expression for impedance, 136, 150, 170; form of operator to produce rotation, 19. Polyphase currents, generation of, 245; see mesh and star connections, three- phase, four-phase, six-phase, n-phase, delta and wye. Potential, see voltage. Power, active, 60; absorbed and delivered by a circuit, 48; average, 50; average, when current and voltage waves are sinusoidal, 51, 52, 53; average, when current and voltage waves are non-sinusoidal, 85; calculation of, from complex expressions for current and voltage, 66; component of current or voltage, 58, 62; in polyphase circuit, 314; in a balanced three-phase circuit, 314; in an unbalanced three-phase circuit, 349; instantaneous, 49; measurement of, 57, 79; measurement of, in a three-phase circuit, 323, 324, 326 ; measurement of, in an n-phase circuit, 326; reactive, 60; reactive, calculation of, from complex expres- sions for current and voltage, 66; reactive, measurement of, in a single- phase circuit, 63; reactive, measurement of, hi a balanced three-phase circuit, 336; transfer of, by mutual induction among conductors of a three-phase transmission line, 365; virtual, 55. Power component, of current or voltage, 59, 63. Power factor, 56, 61, 86; of balanced polyphase, circuit, 314, 317; of balanced three-phase circuit, 316; of balanced four-phase circuit, 317; of unbal- anced polyphase circuit, 317; determination of, for a balanced three- phase circuit, 316, 333; measurement of, 58; when current and voltage waves are non-sinusoidal, 86, 316, 333. Powers, of operators, 9; of vectors, 24. Propagation constant, 413. 430 INDEX Quadrant, 116. Quadrature component, of a current, 58; of a voltage, 62. Reactance, average, of a completely transposed, three-phase, transmission line, 360, 363, average, of a completely transposed, three-phase, trans- mission line whose conductors are at the corners of an equilateral triangle, 363, 364; average, of a completely transposed, three-phase, transmission line whose conductors are at the corners of an isosceles triangle, 363; average, of a completely transposed, three-phase, trans- mission line whose conductors are in the same plane, 364; capacitive, 148, 168; effective, 237, 243; equivalent, 240; function of current, 230; in terms of conductance and susceptance, 216; of a single-phase trans- mission line, 355, per 1000 feet, 359, per mile, 359. Reactive, component of current, 58; component of voltage, 62; current, 59, 61; factor, 57; power, 60; power, measurement of, in a single-phase circuit, 63; power, measurement of, in a balanced three-phase circuit, ,336; power, measurement of, in a balanced or an unbalanced three- phase circuit, 327a; voltage, 63; volt-amperes, 60 ; volt-amperes, measure- ment of, 63. Real, axis, 6; components, 6. Reference point, changing the position of, 77. Relative magnitudes of line and phase currents and voltages, in a balanced three-phase system, 266, 289, 292, 294; in a balanced four-phase system, 266, 303; in a balanced six-phase system, 266, 308, 310; in a balanced twelve-phase system, 266. Resistance, effective, 234, 243; equivalent, 240; function of current, 230; in terms of conductance and susceptance, 216. Residuals, or uniphase components in an unbalanced three-phase system, 337, see uniphase components. Resonance, parallel, 220; series, 204. Reverse-phase components, in an unbalanced three-phase circuit, 338, 249, 252; determination of, 341, 345, 346. Root-mean-square value, of an alternating current or voltage, 33, 35; measurement of, 37, 79; relation between root-mean-square and average values, 36. Roots, of operators, 9, 20. Secohm, 116. Self-inductance, 115, 180, 185; effect of, 117, 119, 121, 125, 127, 128, 150; function of current when magnetic material is present, 125. Self-induction, coefficient of, see self-inductance. Series circuit, containing constant resistance, and inductance, 119, 121, 125, 127, 128, 134; containing constant resistance and capacitance, 136, 137, 140, 142, 149; containing constant resistance, inductance and capaci- tance, 150, 151, 157, 161, 168. Series-parallel, impedances, 224, circuits, 405. Series resonance, 204. Simple harmonic, current or voltage, 29. Six-phase circuits, harmonics in, 304, 307; relative magnitude of line and phase voltages for, 266, 308. INDEX 431 Skin effect, 236. Star connection, 257, 262, 302, 306. Strength, of current, 33. Subscript, double, notation for, 248, 250. Subtraction, of non-sinusoidal waves, 110. Susceptance, 213, 243. Summary, of conditions in series and in parallel circuits, 243. Third harmonics, in three-phase circuits, 287, 289, 292, 297; in four-phase circuits, 299; in six-phase circuits, 304; in neutral connection of aY- connected three-phase circuit, 291; in neutral connection of a star- connected circuit, 304; in a mesh-connected circuit, 304, 309; short- circuited in a three-phase A-connected circuit, 292. Third -harmonic current, in neutral connection of a Y-connected three- phase circuit, 291; in a mesh-connected four-phase circuit, 304; in mesh-connected six-phase circuit, 309; in neutral of four-phase star- connected circuit, 304; short circuited in A-connected three-phase circuit, 292. Three-phase, alternators, advantage of, 248; circuits, delta connection of, 292; circuits, harmonics in, 287, see harmonics and third harmonics; circuits, wye connection of, 289; circuits, relative magnitude of line and phase currents and voltages for, 266, 289, 294. Three-phase connection, see three-phase circuits, delta and wye connection; of generator, advantages of, 248. Three-wattmeter method, for measuring power in a three-phase circuit, 323, 324. Time, periodic, 27. Transfer of power, by mutual induction, among conductors of a three-phase transmission line, 365. Transformer, see air-core transformer. Transmission line, calculation of performance of, 418; capacitance of, see capacitance; direct and reflected waves in, 411; equations for, 406, 410, 411, 415, 416; inductance of, see inductance, induction and mutual- induction; reactance of, see reactance. Two -wattmeter, method for measuring power in a three-phase circuit, 326; relative magnitude of the wattmeter readings for balanced load, 330. Uniphase, components of an unbalanced three-phase circuit, 337, 338, 341, 249, 252. Vector, algebra, 2; diagrams of air-core transformer, 189; equations, solu- tion of, 12; oscillating, 25; representation by use of operator j, 4; representing a simple-harmonic current or voltage, 39; representing a voltage rise or fall, 64", 249; roots and powers of, 20, 24; rotating, which decrease in magnitude with time, 26. Vectors, product and ratio of, 20; addition of, 12, 40, 43. Virtual power, 55. Volt, value of an alternating voltage, 35, 82. 432 INDEX Voltage, active, 62; active component of, 62; alternating, definition of, 27; analysis of, into its fundamental and harmonics, 89, 92; at one end of a transmission line in terms of line constants and voltage and current at other end, 406, 410, 411; continuous, definition of, 27; direct, defini- tion of, 27; effective value of, 35, 82; effective value of, from a polar plot, 88; energy component of, 62; equivalent sine wave, 107; fall x 48, 49, 50, 64, 249; generation of, 31, 69, 245; induced in alternator, 31, 69, 245; induced by electromagnetic induction in a telephone circuit by a three-phase transmission line, 370, 371, 372; induced by electro- static induction in a telephone line by the charges on the conductors of a transmission line, 399; induced in the windings of an air-core trans- former, 187; instantaneous, definition of, 27; measurement of effective or root-mean-square value of, 37, 79; non-sinusoidal, representation of, by Fourier Series, 72; pulsating, definition of, 27; quadrature component of, 62; reactive component of, 62; rise, 48, 49, 50, 64, 249; simple harmonic, 29; wattless component of, 62. Voltages, equivalent wye and delta, 294; relative magnitude of line and phase, for three-phase circuits, 258, 266, 290, 294; relative magnitude of line and phase, for four-phase circuits, 266, 303; relative magnitude of line and phase, for six-phase circuits, 266, 308. Volt-amperes, 55, 61. Wattless component, of current, 58; of voltage, 62. Wattmeter, 57, 79; two-wattmeter method, 326, 330; three-wattmeter method, 323, 324; n-wattmeter method, 323, 326; (n-1) -wattmeter method, 326. Wave, analysis of, 89, 92; equivalent sine, 107, 108; non-sinusoidal, 29, 75, 76; non-sinusoidal, representation of by a Fourier Series, 72; shape or form, 28; sinusoidal, 29. Wave form, 28; determination of, 81; of alternators, 31, 69; similar, 77. Wave length, for transmission line, steady state, 414; constant, 414. Wye connection, 255, 289; equivalent delta connection, 280, 281, 282, 294; relative magnitude of line and phase voltages, 258, 266, 289; third har- monics in neutral of, 291. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below, JUN1 INTER LIBRARY LOAN IE MONTH AFTER RECE1P <'2 1985 LD 21-95w^ll,'50(2877sl6)476 re L UNIVERSITY OF CALIFORNIA LIBRARY