Southern Branch of the University of California Los Angeles Form LI JONES' OK STORE, . First St This book is DUE on the last date stamped below n I.-9-5iH-5.'24 ELECTRIC LIGHT ARITHMETIC. ELECTRIC LIGHT ARITHMETIC BY R. E. DAY, M.A. 1 LECTURER IN EXPERIMENTAL PHYSICS AT KING'S COLLEGE, LONDON Uontron MACMILLAN AND CO., LIMITED NEW YORK: THE MACMILLAN COMPANY 1897 Att rights reserved, RICHARD CLAY AND SONS, LIMITED, LONDON AND BUNGAY. First printed 1882. Reprinted 1883, 1887, 1888, 1889, 1891, 1893, 1897. TK PREFACE. SINCE the year 1878, when I undertook the direction of the Evening Classes in Physics at this College, I have sedulously encouraged the working out by the students themselves of numerous arithmetical ex- amples having a direct bearing upon the particular course of experimental lectures which, at the time, they are attending, and I find that by doing so they rapidly acquire a firm grasp of the principles of the subject, as well as a knowledge of its details, which it would be almost impossible for them to get by any other means. That the students themselves are conscious of this is, I think, established by the fact that within less than four years the attendance at our Evening Classes in Physics has increased more than fourfold. We have naturally had to pay a good deal of attention of late to the principles and practice of Electric Lighting, and it has been suggested to me that the arithmetical examples on this subject, which I vi PREFACE. have drawn up for the use of my own classes, might be of some service to a wider circle of students. In compiling these problems I have always had to keep in view the fact that the majority of our evening students are engaged during the daytime at other pursuits, so that they cannot devote much time to going deeply into the subject, and also that as a rule they have but a slight acquaintance with mathematics, so that no examples must be introduced which require for their solution anything beyond decimal fractions and elementary algebra. In the statements of the problems and in their solutions I have not thought it necessary to go into minute details about electrical formulas and theories because these will be found in all the recognised text- books on the subject, and this collection of examples is not intended to replace but to supplement whatever text-book on electricity the student may be using. R. E. DAY. KING'S COLLEGE; June, 1882. CONTENTS. ELECTRICAL RESISTANCE OF WIRES AND LAMPS . . I INTENSITY OF CURRENT IN A SIMPLE CIRCUIT . . 21 HEATING EFFECT OF THE CURRENT IN A SIMPLE CIRCUIT 30 WORK UTILISED IN A SIMPLE CIRCUIT 45 COMPOUND ELECTRICAL CIRCUITS 57 DISTRIBUTION OF ENERGY IN A COMPOUND CIRCUIT 69 TABLE OF SQUARES, SQUARE ROOTS, AND RECIP- ROCALS 83 TABLE OF THE BIRMINGHAM WIRE GAUGE . 88 ELECTRIC LIGHT ARITHMETIC. ELECTRICAL RESISTANCE OF WIRES AND LAMPS. THE electrical resistance (R) of a conductor varies directly with its length (L), and inversely as its sec- tional area (S), and may be expressed in terms of these quantities by the equation, where a is a constant quantity depending upon the material of the conductor and is numerically equal to the resistance of a wire of this material whose length and cross section are each unity. It is called the specific resistance of the material. Hence if we have two wires whose lengths are 4 and / 2 , their sectional areas s l and s 2 , their specific 2 ELECTRIC LIGHT ARITHMETIC. resistances T and # 2 , and their actual resistances R\ and R z , we shall have K! = af and R^ = 2 - 2 , S l S 2 whence by division we get and by means of this equation all problems connected with the relative resistances of wires in simple circuit can be readily solved. Example i. If the resistance cf one mile of a certain electric light cable be 3*58 Ohms, what is the resistance of 37 miles of the same cable? If we apply the data of this question to the above formula we find that ^i ^ ^2 ^and .'. the equation becomes L =37 f ^2 = 3-58 X 37 = 13-25 Ohms. *i-3-S8) Example 2. If the resistance of 700 yards of a certain cable be 0*91 Ohm, what will be the resistance of 1320 yards ? Ans. 172 Ohm. Example 3. The resistance of too yards of a certain wire is equal to 5 Ohms, what length of the same kind of wire would have a resistance of 1 3'2 Ohms ? ELECTRIC LIGHT ARITHMETIC. 3 In this case the quantity whose value is required is 4, and therefore transposing the equation (on page we have /2 = / lX x /?! > -f, But in the present case a^ a s and J, = s 2 and the equation reduces to and from the data of the question / t = 100, R v *= 5, ^2 = J 3'2. Therefore / 2 = 100 x l = 264 yards. Example 4. The resistance of a certain cable is found to be 4*55 Ohms and the resistance of a mile of this cable is known to be 1-3 Ohm. What is its length? Ans, 3*5 miles. Example 5. If the resistance of 130 yards of copper wire i-i6th of an inch in diameter be one Ohm, what is the resistance of the same length of copper wire I ~32nd of an inch in diameter ? B 2 4 ELECTRIC LIGHT ARITHMETIC. Since the areas of circles are proportional to the squares of their diameters we have also SE? and therefore the fundamental formula reduces itself to R z = 7?j x ^1 = 4 Ohms. J j Example 6. What is the resistance of a mile of copper wire which has a diameter of 65 mils if the resistance of a mile of copper wire 80 mils in diameter is 8-29 Ohms ? Ans. 12*56 Ohms. N.B. A ;//// is the one-thousandth part of an inch. Example 7. What is the diameter of a copper wire a mile long which has a resistance of 23 Ohms if a mile of copper wire 70 mils in diameter has a resistance of 10*82 Ohms ? Let d^ and d z represent the diameters of the wires in mils, then we have % = rttf * UJ ELECTRIC LIGHT ARITHMETIC. $ and ELECTRIC LIGHT ARITHMETIC. g Example 18. A wire one foot long and one mil in diameter has a resistance of 9-15 Ohms : find the specific resistance per cubic inch of the material of this wire. Ans. 0-5989 Microhm. Example 19. Find the resistance of 20 yards of platinum wire o - oi6 inch in diameter, the relative resistance of platinum with respect to copper being Referring to Example 14 we find that the resistance of 200 yards of copper wire 134 mils in diameter is o'34 Ohm, and therefore for the quantities in the fundamental formula we have A" 2 = '34 X X (^7\ X ^ = 26-95 Ohms. 2OO \ ID / I Example 20. What would be the resistance of 7 miles of iron wire 238 mils in diameter, the relative resistance of this particular iron with respect to copper being 7-5 ? Ans. 4979 Ohms. Example 21. What should be the respective lengths of two wires of silver and lead so that they may each offer the same resistance as 10 inches of copper wire of the same thickness, the conductivity of silver and lead with respect to copper being 1-0467 and 0-0923 respectively ? Ans. 10*467 and 0-923 inch. 10 ELECTRIC LIGHT ARITHMETIC. Example 22. What must be the relative thicknesses of wires of iron, silver, and platinum, of the same length, so that their resistances may be equal ? Ans. As 100 : 36 : 122. Example 23. If the resistance of a wire 3 metres long and weighing 3 grammes be 5 '88 Ohms, what is the specific resistance per cubic centimetre of the material, its specific gravity being 20-337 ? If $! represent the sectional area of the wire ex- pressed in square centimetres, then we have ^ X 300 X 20-337 = 3 300 X 20-337 of a square centimetre ; also / t = 300 ; s z = I and a z a \\ anc * therefore the fundamental formula becomes -i- X * = - >0 Ohrn 300 300X20-337 300x2033-7 = 9*638 Microhms. Example 24. Determine the specific resistance per cubic millimetre of a wire 437 millimetres long, which has a resistance of 0-1257 Ohm, and which weighs 0-411 gramme in air, and 0-365 gramme in water. ELECTRIC LIGHT ARITHMETIC. \ \ The weight of water displaced = -046 gramme, and therefore the specific gravity of the material - * - 8-935, 46 and then following the method of the last example we get R = 30*278 Microhms. Example 25. Find the specific resistance per cubic centimetre of the material of which the above wire is composed. Ans. 3-0278 Microhms. Example 26. A wire 874 "millimetres long and weighing 0-822 gramme in vacuo and 073 gramme in water, h'as a resistance of 0-1257 Ohm ; find the specific resistance per cubic millimetre of the material. Ans. 15-14 Microhms. Example 27. A piece of silver wire 3 feet long was found to weigh 7*2 grains, and its resistance was 0-3026 Ohm. What would be the resistance of a wire one foot long weighing one grain ? If a>! and w 2 be the weights of two wires of the same material whose lengths are /, and / 2 and their sectional areas s t and s%, then ^ = '*and... '-1 = 7 ' x ? 12 ELECTRIC LIGHT ARITHMETIC. Substituting this value for - in the fundamental J 2 formula, and remembering that a 1 = a z , we have = -3026 X (-Y X = o-24 Ohm. Example 28. What are the relative resistances of two wires, one of which is 30-48 centimetres long and weighs 35 grammes, while the other is 18*29 centi- metres long and weighs 10*5 grammes ? Ans. as 12 : lo/nearly. Example 29. Find the resistance at o C. of 20 metres of German silver wire weighing 52*5 grammes, having given that the resistance at o C. of a wire of this material I metre long and weighing I gramme is 1-85 Ohm. Ans. 14'! Ohms. Example 30. If the resistance of a foot of pure copper wire weighing one grain be 0-2 106 Ohm, and the resistance of a piece of ordinary copper wire 3 feet long and weighing 3-45 grains was found to be 0-5782 Ohm ; compare the conducting power of this sample of wire with that of a similar one of pure copper. ELECTRIC LIGHT ARITHMETIC. 13 If the second wire were of pure copper, then by the method adopted in the solution of Example 27 we should find that its resistance /^\ 2 i R = 0-2106 X (- ) X = 0-5494 Ohm. * l ' 3'45 Therefore Conducting power of this wire a similar one of pure copper = 5194= 91 nearly. 5782 100 Hence it appears that the conductivity of this specimen of copper wire is about 95 per cent, of that of pure copper. Example 31. It is found that 37 inches of a certain copper wire weighing 518 grains give a resistance ot 0*0041 Ohm ; find the conductivity of this copper compared with that of pure copper. Ans. 94 per cent. Example 32. Two samples of copper wire were brought to be tested, so a length of 20 feet was cut off each of them. The first weighed 150 grains and had a resistance of 0*613 Ohm, while the second weiehed 164 grains and had a resistance of 0-547 I 4 ELECTRIC LIGHT ARITHMETIC. Ohm. Find the percentage conductivity of each of these samples with respect to pure copper. Ans. 91-6 and 93-9 per cent. Example 33. The resistance of one mile of copper wire whose diameter was 0*065 mc h was found to be 1573 Ohms. The resistance of a wire of pure copper I foot long and o'ooi inch in diameter being 9-94 Ohms ; find the percentage conductivity of the copper of the first wire. Had the wire been of pure copper its resistance would have been R = 9'94 X 5280 X (~ ) 2 = 12-42 Ohms, \65/ and conductivity of the given wire a similar one of pure copper - ^ - 07896 1573 and therefore the conductivity of the sample is about 79 per cent, of that of pure hard-drawn copper. Example 34. An iron wire weighing 249 pounds per mile and having a diameter of '135 inch is found to have a resistance of 24*14 Ohms per mile. Compare the conducting power of this wire with that of a similar one of pure copper. Ans. ir8 per cent. ELECTRIC LIGHT ARITHMETIC. \ 5 Example 35. A certain kind of Brush lamp has a resistance of 6 Ohms, and there are 4 lamps in series which are separated from each other by spaces of 50 yards, and the dynamo-electric machine is 400 yards away from the nearest lamp. What must be the thick- ness of the leading wire if its conductivity is 96 per cent, of that of pure copper and the total resistance of the lead is 8 per cent of that of the lamps, no ground circuit being allowed? Total length of leading wire = 800+ 300 = 1 100 yards. resistance of lamps = 4 X 6 = 24 Ohms. leading wire = 24 X '08=1-92 Ohms; and by Example 33 the resistance of a wire of pure copper i foot long and i mil in diameter = 9-94 Ohms. But since d. z is the quantity whose value is required, the fundamental formula becomes where I'l^^lT - 0-96 and therefore -^aL-xS?! 1-92 x -96 i 1 6 ELECTRIC LIGHT ARITHMETIC. whence v 1*92 X '96 = -133 inch, or about No. 10 B.W.G. wire. Example 36. Ten Swan lamps, each of which had a resistance of 25 Ohms, were arranged in series, and the length of the leading wire was 50 yards. What was the diameter of the " lead," its conductivity being 80 per cent of that of pure copper, if its resistance was 2 per cent, of that of the lamps ? Ans. 19 mils, or about No. 26 B.W.G. Example 37. If the resistance of a certain incan- descent lamp, which is 12 feet from the supply main, is 80 Ohms, and the leading wires are of copper whose conductivity is 85 per cent, of that of pure copper, what should be the diameter of the wire so that its resistance may be '08 per cent, of that of the lamp ? Ans 66 mils, or about No. 16 B.W.G. Example 38. Five Brush lamps, each having a resistance of 6 Ohms, are to be worked in series, The lamps are 40 yards apart, and the nearest is 200 yards from the dynamo machine. If 10 per cent, be allowed for loss of energy in the " lead," which is of pure copper, what must be its diameter ? ELECTRIC LIGHT A Ri THME TIC. \ 7 By the question Resistance of (leading wire -j- lamps) _ 100 leading wire ~ To " Therefore the resistance of leading wire = i-gth of that of the lamps. But resistance of lamps = 30 Ohms. .. resistance of lead = Ohms. 9 and length of the lead = 400 -|- 320 = 720 yds. Hence if d z = the diameter of the leading wire in inches we have as in Example 35, -ooi x or about No. 14 B.W.G. Example 39. Thirty-eight Brush lamps, each of which has a resistance of 6 Ohms, are to be connected together and to the driving machine by an electric cable 3*5 miles long. If one-tenth of the total energy in the external circuit is allowed for loss in transit and the conductivity of the " lead " is 90 per cent, of that of pure copper, what should be its diameter ? A us. go mils. c iS ELECTRIC LIGHT ARITHMETIC. Example 40. The legal Ohm is a resistance equal to that of a column of mercury one square millimetre in section and 106 centimetres in length at the tem- perature of melting ice. What is the resistance between its ends of a cylindrical column of mercury 2 metres long and 2 millimetres in diameter ? R = I x 22? = o'6 Ohm nearly. 7T- IOO Example 41. Siemens' old unit of resistance was that of a column of mercury one metre long and one square millimetre in section. What was this in legal Ohms? Ans. 0-9434 Ohm. Example 42. The British Association unit of re- sistance being equal to 0-9889 of a legal Ohm, express a Siemens' unit in terms of a B.A. unit. Ans. 0-9539 B.A.U. nearly. Example 43. The resistance of the copper wire of a cable when the temperature was 12 C. was found to be 250 Ohms. What would be the resistance of this wire at 24 C., the variation in resistance of this sample of copper per degree Centigrade between these limits being approximately 0-39 per cent. ? If Rt and R? be the resistances of a certain wire at ELEC TRIG LIGH T ARITHME TIC. 1 9 the temperatures / and /' respectively, they may very approximately be connected by the relation ~Rt where K is a constant which has a particular value for each metal. In the present case we have R^ = i +.24X -0039 250 i + 12 X -0039' Ohms nearly. Example 44. The resistance of a certain length of copper wire at o C. being 100 Ohms, what would it be at 15 C. ? Ans. 105-85 Ohms. Example 45. Some resistance coils made of plati- num-silver alloy are correct at 12 C. What amount of error will arise from using them when the tempera- ture is 30 C, the variation in resistance of this alloy for i C. being about 0*03 per cent. ? Here fa = L+3o_xjooo3 = roog , & /v ]2 i -j- 12 x "0003 roo3o Hence the error due to change of temperature amounts to an increase of resistance of o - 53S per cent. C 2 20 ELECTRIC LIGHT ARITHMETIC. Example 46. The mean summer and winter tem- peratures in Shetland being 12 C. and 4 C. re- spectively, what would be the limits of error due to variation of temperature which might occur in using a set of platinum- silver resistance coils, accurately adjusted for the mean annual temperature ? Ans. o'i2 per cent. Example 47. Between what limits of temperature could this set of coils be used without introducing temperature corrections, if the margin of error is not to exceed 0-25 per cent. ? Let x be the number of degrees above or below 8 C., for which the temperature correction would not be greater than 0^25 per cent. Then we should have x X '03 = -25. /. x = 8-3 nearly. Hence the required limits of temperature are ap- proximately 1 6 C. and o C. Example 48. What would have been the approxi- mate limits of temperature if these coils had been made of platinoid wire, for which the temperature coefficient is o'O2i per cent.? Ans. + 20 C. and - 4 C. approximately. ELECTRIC LIGHT A Rl THME TIC. 2 1 INTENSITY OF CURRENT IN A SIMPLE CIRCUIT. All problems connected with this part of the subject are worked out by the aid of a formula which is known as Ohm's Law ; namely '=1 in which 7 represents the intensity of the current ; E the resultant of all the electro-motive forces in circuit ; and R the total resistance of the circuit. Example I. The internal resistance of a certain Brush dynamo machine is io - 9 Ohms, and the external resistance is 73 Ohms ; the electro-motive force of the machine being 839 Volts. Find the strength of the current flowing in the circuit. 1 In this case we have E = 839 ; R = 73 + 10-9 = 83-9 Ohms, and therefore substituting these values in the aboi'e equation we get / = ^2 =10 Amperes. 1 In a rotating armature the resistance appears to vary with the speed, on account of the possession by every electric current of a property sometimes called electric inertia, and sometimes self-indnction. In the following examples the resistance of the dynamo is to be understood to depend upon the particular conditions of the experiment. 22 ELECTRIC LIGHT ARITHMETIC, Example 2. The resistance of the dynamo machine being r6 Ohm, and the external resistance 25-4 Ohms ; calculate the strength of current in the circuit, the electro-motive force being equal to 206 Volts. Ans. 7-6 Amperes. Example 3, Three arc lamps in series have a resistance of 9-36 Ohms,, while the resistance of the leading wires is ri Ohm and that of the dynamo machine is 2 '8 Ohms. Find what must be the electro-motive force of the machine when the strength of the current produced is I4'8 Amperes. In this case we have R = 2-8 -f 9-36 + ri = 13-26 Ohms, / = 14-8 Amperes ; and therefore substituting these values in the equation E = /X R it becomes E = 13-26 x 14-8 = 196-3 Volts. Example 4. A certain arc lamp has an electrical resistance of 2-5 Ohms, while that of the leading wires is "5 Ohm, and that of the machine 0*5 Ohm. What must be the electro-motive force of the machine so as to send a current of 25 Amperes through three of these lamps and the leading wires ? Ans. 213 Volts nearly. ELECTRIC LIGHT ARITHMETIC. 23 Example 5. Calculate from the following data the average resistance of each of three arc lamps which are arranged in series. The electro-motive force of the machine is 244 Volts and its resistance is 37 Ohms, while that of the leading wires is 2 Ohms, and the strength of current flowing through each lamp is 21 Amperes. If x represent the average resistance in Ohms of each lamp, then the total resistance of the circuit is R = 3.r + 2 + 37. But by Ohm's law 5'7 -^- = n'6i Ohms. . . x = i '97 Ohms nearly. Example 6. In an experiment with a Gramme machine working on a simple circuit it was found that the electro-motive force developed by the machine was 81-58 Volts, the strength of current 29*67 Amperes, and the external resistance 1-14 Ohm. What was the internal resistance of this Gramme machine? Ans. r6i Ohm. Example 7. In a similar experiment with another dynamo machine its resistance was known to be equal 24 ELECTRIC LIGHT ARITHMETIC. to 4-58 Ohms, while the electro-motive force was 158-5 Volts, and the current 17-5 Amperes. What was the resistance of the exterior part of the circuit ? Ans. 4*48 Ohms. Example 8. Three Maxim incandescent lamps were placed in series. The average resistance, when hot, of each lamp was 39*3 Ohms, and that of the dynamo machine and leading wires 1 1 '2 Ohms. What electro- motive force was required to maintain a current of I '2 Ampere through the circuit ? In this case we have R = 3 X 39'3 + ii'2 = 129-1 Ohms, and I ** i'2 Ampere; and therefore by Ohm's law E = 7 X R = i'2 X 129-1 = 154*9 Volts. Example 9. Two incandescent lamps, each having a hot resistance of 73 Ohms, were joined up in series to the poles of a dynamo machine, and the total resistance of the circuit was 158 Ohms. What was the electro-motive force of the machine when a current of i Ampere was circulating through it? Ans. 158 Volts. ELECTRIC LIGHT A AY Til ME TIC. 2 5 Example 10. The resistance of the arc of a certain Brush lamp was 3*8 Ohms when a current of 10 Amperes was flowing through it. What was the electro-motive force between the two terminals ? By Ohm's law E = I X R = 10 X 3'8 = 38 Volts. Example n. The electro-motive force between the terminals of a Serrin lamp was 31'! Volts, and there was a current of 35-8 Amperes flowing through it. What was the resistance of the arc? Ans. 0-869 Ohm. Example 12. Twenty-five exactly similar galvanic cells, each of which had an average internal resistance of 15 Ohms, were joined up in series to one incan- descent lamp of 70 Ohms resistance, and produced a current of o'H2 Ampere. What would be the strength of current produced by a series of 30 such cells through 2 lamps, each of 30 Ohms resistance ? The data of the first part of the problem enable us to determine the average electro-motive force of each cell of the battery. Let this be represented by E, then we have 25 E = I x R = -112 X (25 X 15 + 70) = -112 X 445 - ' II2X 2 = 2 Volts nearly. 2 6 ELECTRIC LIGHT ARITHMETIC. Then from the data in the second part of the problem we have by Ohm's law / = 3QX2 = 160 = Q . ii8 A re _ 30 X 15 + 2 X 30 510 Example 13. Two incandescent lamps whose resist- ances were 16-9 and 32 Ohms respectively, were joined up in series with a battery of 40 similar voltaic cells, the total resistance of which was 20 Ohms, and the strength of the current produced was ri6 Ampere. What will be the strength of current produced by a battery of 60 such cells through a series of 4 lamps whose respective resistances are i6'9, 32, 20, and 16 Ohms ? Ans. 1-043 Ampere. Example 14. What would have been the strength of current in Example 12 if the area of each of the battery plates had been doubled, all else remaining the same? The resistance of a battery cell varies inversely with the area of the plates, and therefore we should have had 30 y + 2 x 30 ELECTRIC LIGHT ARITHMETIC. 27 Example 1 5. What would have been the strength of current in the case of the four lamps in Example 13, if the size of the battery plates had been doubled ? Ans. 1-2 Ampere. Example 16. At the Mansion House three Crompton arc lamps were joined up in series with a Biirgin dynamo machine. The resistance of each lamp was 2 Ohms, which was also that of the leading wires. The electro-motive force of the dynamo was 244 Volts, and the current 2 1 Amperes. What was the internal resistance of the dynamo machine ? If x = resistance of machine in Ohms, then the total interpolar resistance being 8 Ohms the resistance of the circuit was R = x -f- 8, and substituting these values in the equation we get r+8 = ^ = 11-62 .'. x = 3'62 Ohms. Example 17. The current from a certain Brush machine, whose electro-motive force was 839^02 Volts, was sent through a series of 16 arc lamps each of which had a resistance of 4/5 1 Ohms. The resistance 28 ELECTRIC LIGHT ARITHMETIC. of the leading wires was o'8 Ohm, and the strength of the current was io'O4 Amperes. What was the re- sistance of this Brush machine ? Ans. io'6i Ohms. Example 18. What was the difference of potential between the battery terminals in the case of the first battery described in Example 13 ? As in the solution of Example 10 we have P.D. = I x R = 4S'9 X ri6 = 56724, /. P.D. = 567 Volts nearly. Example 19. What would be the P.D. between the terminals in the case of the second battery mentioned in Example 13? Ans. 88-55 Volts nearly. ELECTRIC LIGHT ARITHMETIC. 29 HEATING EFFECT OF THE CURRENT IN A SIMPLE CIRCUIT. Whenever an electrical current flows through a con- ductor, heat is invariably produced, and it can be proved theoretically, as also Joule has shown experi- mentally, that the total quantity of work converted into heat per second in a circuit may be expressed by the formulae W = P- X R = ~ = E X /ergs if the units be absolute = E Xf X io 7 ergs if the units be practical, where W is the quantity of work converted into heat, and /, E, and R have the same meanings as before. So also if we consider any particular portion of a circuit, the resistance of which is r, and of which the difference of potential at the two ends is e, then the quantity of work converted into heat in this portion of 30 ELECTRIC LIGHT ARITHMETIC. the circuit in the unit of time may be expressed by the formulae = *?/ergs if the units are absolute = el x io 7 ergs if the units are practical. The heat equivalent of the quantity of work ex- pended in t units of time in a circuit, the resistance of which is /?, by a current of strength /, is given by the equations 4-2 X io 7 if the units are absolute /2A>/ calories if the units are practical. ELECTRIC LIGHT ARITHME TIC. 3 1 Example i. One portion of an electrical circuit was composed of 10 feet of copper wire 7 mils in diameter, and another portion of 5 feet of iron wire 148 mils in diameter. If the resistances of copper and iron are to one another as 112 to 825, compare the heating effects of the current in these two portions of the circuit. Since the same current flows through both wires the value of / is the same for both, and for their relative resistances we have = . nearly. 121 Therefore Heat developed in the copper section ii iron _ Resistance of copper .__ 121 iron I Example 2. The interpolar portion of an electrical circuit consisted of two different wires joined end to end. One was of copper 3 feet long and 65 mils in diameter, 3 2 ELECTRIC LIGHT A AY TffAtE TfC. while the other was of platinum 6 inches long and 35 mils in diameter. The resistances of copper and platinum being to one another as 112 to 1243, com- pare the quantities of heat developed by the current in these portions of the circuit. A ns Heat in platinum _ 638 copper ioo* Example^. The internal resistance of a certain dynamo machine is 5 Ohms, that of the lamp is 10 Ohms, and of the leading wires 2 Ohms. Compare the quantities of heat produced in the lamp and in the machine, and find what proportion of the total energy in the circuit will appear in the lamp. Ans. Heat in lamp = p = 2 machine 5 i ' Energy in lamp Resistance of lam p Total energy in circuit Resistance~of^holTcT?c^it Example 4. Two Swan incandescent lamps each of which has a resistance of 35 Ohms are connected jp ELECTRIC LIGHT ARITHMETIC. 33 series by thick wires with a series of 40 Grove cells ; the average resistance of each cell being o'5 Ohm. Compare the quantity of heat developed in each lamp with that developed in each cell of the battery, and find what proportion of the total energy yielded by the battery appears in each lamp. . Heat in lamp 70 Ans. - r. = '_ ,, cell i Energy in one lamp _ 7 ^ of battery 18 Example 5. The internal resistance of a voltaic battery is equal to that of 3 metres of a particular wire. Compare the quantities of heat produced both inside and outside the battery when its poles are connected by one metre of this wire with the quantities produced in the same time when they are connected by 37 metres of the same wire. Taking the resistance of one metre of the given wire as our unit and I v 7 2 as the intensities of the current in the two cases, we have /, = and 7 2 = ; whence - = 10, 4 40 7 2 and therefore for the heat developed in the battery in the two cases 34 ELECTRIC LIGHT ARITHMETIC. For the heating effects outside the battery 100 J_ _ 100 "i" 37 ~ I?" Example 6. A Grove cell whose electro-motive force is 1*9 Volt and internal resistance o - 4 Ohm has its poles connected (i) by a wire of 3 Ohms, and (2) by a wire of 30 Ohms resistance. Compare the amounts of heat developed in the cell in the two cases. Ans. As 80 : i. Example J. A dynamo machine the resistance of which is 6 Ohms has its poles connected (i) by a circuit of '5 Ohm resistance, and (2) by a circuit of 500 Ohms resistance. Compare the quantities of energy which are dissipated as heat in the machine assuming that the driving engine works always at the same horse-power. Ans. As 78 : i nearly. Example 8. The interpolar portion of a circuit consisted of a piece of copper and a piece of platinum wire. The platinum was I foot long and 28 mils in diameter, while the copper was 20 yards long and 65 mils in diameter. Compare the quantities of heat developed, per running foot, in each of these wires. Ans. As 60 : I nearly. ELECTRIC LIGHT ARITHMETIC. 35 Example 9. The difference of potential between the ends of the two carbons of a certain Brush arc light is 36 Volts, and a current of 10 Amperes is passing through it. Find the quantity of heat developed per second. When expressed in the Centimetre-Gramme-Second (C.G.S.) system of units One Volt = io 8 C.G.S. units of potential, One Ohm = io n C.G.S. electro-magnetic units of resistance, One Ampere = io- 1 of the C.G.S. unit oi current. One water-gramme degree Centigrade = 4/2 X io 7 ergs. In the present case we have W =. E x I 36 x io 8 x io x io- 1 = 36 X io 8 ergs per second, Example io. An Edison incandescent lamp has a resistance of 125 Ohms, and the difference of potential between the ends of the carbon filament is no Volts. Find the strength of current flowing through it and the quantity of heat developed in the lamp per second. Ans. Current = o - 8S Ampere. Heat = 23*05 w.g.d. C. 36 ELECTRIC LIGHT ARITHMETIC. Example II. The house-main wire for supplying 20 Edison lamps, each of which requires a current of o'S Ampere is of copper of 86 per cent, conductivity, and 65 mils in diameter. What is the amount of heat developed per second in every foot of this wire ? Following the method of Example 33, page 14, we shall find that the resistance of one foot of this wire is 0-00274 Ohm, and then substituting the numerical values in the equation Q = R X /- = 2 74 x io- 3 x 10 x ('08 X 20)2 ergs per second. Example 12. A current of one Ampere circulates through a wire whose resistance is 0*9536 Ohm. Find the amount of heat developed in this wire in 5 minutes. Ans. 68 w.g.d. C. Exatnple 13. A current of 075 Ampere was sent for 5 minutes through a column of mercury whose resistance was 0-47 Ohm. The mass of the mercury was 20-25 grammes and its specific heat 0-0332 ; find the rise of temperature, assuming that no heat escapes by radiation. By the method adopted in the previous examples we find that the number of units of heat developed in 5 ELECTRIC LIGHT ARITHMETIC. 37 minutes by the passage of a current of 075 Ampere through a resistance of 0*47 Ohm is 18*88 units, and by the question this quantity of heat will raise 20*25 grammes of mercury through x C. of temperature. Cut the amount required for this purpose is = 20-25 x '33 2 X x = 1 8*88 units by the question. Hence 18-88 20-25 X 0-0332 28 3 C. Example 14. A coil of very fine wire which had a resistance of 46-64 Ohms was placed in 1000 grammes of water at o C. The current from a series of 50 voltaic cells, each of which had an electro-motive force of I Volt and a resistance of 6 Ohms, was sent through the coil for 10 minutes. Find the rise of temperature of the water. Am. i -39 C. Example 15. A small coil of German silver wire was placed in one of Lenz's alcohol calorimeter?, and a steady current of i"8 Ampere was sent through it. The resistance of the coil was 1*2 Ohm, and it was found that the temperature of the alcohol rose i C. in 23 seconds. Had the resistance of the coil been 38 ELECTRIC LIGHT ARITHMETIC. equal to i Ohm while the current strength was the same, what would have been the time required ? Ans. 27*6 seconds. Example 1 6. A wire of pure copper, 0*165 centi- metre in diameter, has a current of 10 Amperes flowing through it. 1 What will be the limiting temperature of the wire ? The specific resistance of pure copper per cubic centimetre is 1-65 Microhm, and therefore the resistance of one metre of the above wire will be 7'8 X lo- 3 Ohm. Therefore the quantity of heat developed per second in one metre of this wire is _ 7 8 X 10 _ .jg thermal unit. 4-2 X 10' The surface of a length of one metre of this wire is = loo X 7T X 0-165 = S 1 '^ square centimetres. Hence the development of heat in this wire is at the rate of =^ or 0^00359 of a thermal unit per square centimetre of surface per second. But the rate of loss of heat by radiation and convection from an unpolished surface of copper (or other solid material) is about l~4oooth per square centimetre per degree of excess of temperature above that of the surrounding medium. 1 This assumption is true only in certain exceptional cases ELECTRIC LIGHT ARITHMETIC. 39 It follows therefore that the temperature of this wire will go on rising until it has reached such a value that the loss per second by radiation and convection is equal to 0*00359 of a thermal unit. Hence if x be this temperature JL. = -00359 an d .'.*= I4*36 C. 4000 Hence, if this wire be freely exposed to the air, its limiting temperature will be i4'36 C. above that of the surrounding air. Example 17. Taking account of the loss ot heat by radiation, calculate what would be the rise of tempera- ture of the wire referred to in Example 11. Ans. 42"3 C. Example 18. A short piece of lead wire is included as a safety catch in a circuit, and it is required to find what its diameter should be so that a current of 7-2 Amperes may just fuse it, assuming that the specific resistance of lead per cubic centimetre is 19-85 Microhms, and that the melting point of lead is 335 C. Let x diameter of the wire in centimetres. The resistance of a lead wire one centimetre long and x centimetres in diameter is = 19-85 X 10 -6 X Ohms, 4o ELECTRIC LIGHT ARITHMETIC. and therefore the quantity of heat developed in one second in one centimetre of this wire . (72)' X 19-85 X io 3 X4 n d GO ITX* X 4'2 X I0 7 But the area of the surface of one centimetre of this wire is irx square centimetres, and therefore the development of heat per square centimetre of surface is at the rate of (7^XL?l5X_irf X4 thermal units per second . TrV X 4'2 X io 7 Now while the current is flowing the wire will go on rising in temperature, but the rate of increase will be slower and slower as the temperature approaches the limiting one at which the gain of heat per second is just balanced by the loss by radiation and convection, and if this limiting temperature be equal to or higher than that at which lead fuses then the wire will melt and the current will be interrupted. If therefore the diameter of the wire is to be such that the wire shall just melt with the above current \ve have the equation 335 ^ (72) 2 X 19-85 x io 3 X 4 4000 n-x 3 X 4'2 X io 7 whence ^ = 4000 X (72)" X 19-85 X io 3 X 4 335 X 7T 2 X 4'2 X io 7 ELECTRIC LIGHT A RlTHME TIC. 4 1 and .-. x = -106 centimetre, or about No. 19 B.W.G. For the sake of simplicity it has been assumed in this solution that the resistance of the lead does not vary with the temperature, and also that the rate of loss of heat by radiation follows the same law at comparatively high temperatures. The student should also notice that the value of the specific heat of lead only affects the time which would elapse from the commencement of the flow before the wire fuses, and does not affect the practical question as to whether or not fusion will ultimately take place. Example 19. A leaden safety catch is to be inserted in a circuit which will fuse if the current strength exceeds 20 Amperes. What must be its diameter ? N.B. A safety catch is a short piece of wire whose thickness and material are so chosen that if it be inserted in a circuit it will fuse and so interrupt the circuit if the strength of the circuit exceeds a given value. In the present case we shall find by the method adopted in Example 18 that x = '209 centimetre, or the wire is about No. 14 B.W.G. Example 20. A copper wire is to be inserted in a circuit as a safety catch for a current of $00 Amperes. Taking the melting point of copper at 1050 C. and its 42 ELECTRIC LIGHT ARITHMETIC. specific resistance per cubic centimetre equal to r652 Microhm ; find what must be its diameter. Ans. '53 centimetre, or about No. 6 B.W.G. Example 21. If a piece of zinc rod is to be used as a safety catch for a current of 500 Amperes, find what must be its diameter, having given that the melting point of zinc is 422 C. and its specific resistance per cubic centimetre is 5-689 Microhms. Ans. I'ogi centimetre, or about No. ooo B.W.G. Example 22. A piece of platinum wire 0^55 milli- metre in diameter is inserted in a circuit. If the resistance of one metre of platinum wire one milli- metre in diameter be o'ii66 Ohm and the melting point of platinum be 1700 C, calculate the strength of current which will be required to fuse the above wire. The resistance of a length of one centimetre of this wire is = ' II66 100 X ( V Ohm. v- 5S / and therefore the quantity of heat developed per second by a current of x Amperes is V X 'Il66 X io 9 _J^ w.g.d. C. 100 x ('55) 2 X 4'2 x io 7 ELECTRIC LIGHT ARITHMETIC. 43 But the surface of one centimetre of this wire =?rX '055 square centimetre, and therefore the heat developed per second per square centimetre of surface is ' w.g.d. O. loo x ( - 55) 2 X n- X '055 x 4'2 and if the limiting temperature be that of the melting point of platinum this expression must be equal to -^ and therefore 4000 jr a = 17 X IPO X C55)' 2 X 4'2 X ?r X -055 40 x '1166 and .'. x = 8-9 Amperes nearly. Example 23. A lead wire which is o'Si millimetre in diameter is to be used as a safety catch. What strength of current will it just bear without fusing ? Ans. 476 Amperes. Example 24. Twenty of Edison's incandescent lamps, each of which requires a current of o - 8 Ampere, are to be supplied from one main wire. If a margin of 50 per cent, excess of current be allowed, find the thickness of a lead safety catch which should be inserted in the supply wire. The fusing current is in this case one of 24 Amperes 44 ELECTRIC LIGH T A RITHME TIC. and then working the question out as before we find the diameter required = 0-236 centimetre. Example 25. A piece of lead wire of the same diameter as the copper conductor is inserted in an electrical circuit and a strong current is sent through the circuit until the lead wire fuses. Assuming that there is no loss of heat by radiation, find the tempera- ture of the copper conductor at the moment when this occurs. If we take the specific resistance of lead per cubic centimetre to be 19*85 Microhms, and that of copper 1*65 Microhm, the ratio of the resistances of similar wires of lead and copper will be very nearly as 12 : I, and therefore the quantities of heat developed in equal lengths by the same current in one second will be as 12 to t. But the specific gravity of lead is 1 1 -38, and that of copper is 8-89, while the specific heat of lead is 0-0314, and that of copper is 0-0948. Therefore Heat required to raise a given volume of lead rC. ,, ,, same copper 11- 0-0948' and consequently while the temperature of the copper wire is being raised I C. the temperature of the same ELECTRIC LIGHT ARITHMETIC. 45 length of lead wire of the same diameter through which the same current is flowing will be raised 8-89 X 0-0948 12 = 2go c 11-38 X 0-0314 I But the melting point of lead is 335 and therefore by the time the lead is heated to 335 C. the copper will have had its temperature raised by 28-3 The student will understand that this result is only approximately true, because no notice has been taken of the change of resistance with increasing tempera- ture, nor of the loss of heat by radiation and convection. WORK UTILISED IN A SIMPLE CIRCUIT. Example i. The difference of potential between the electrodes of a certain Swan lamp was 31'! Volts, and the current flowing through it was 1*22 Ampere. Calculate the energy, in horse-power, absorbed by the lamp, and also the amount of heat developed. By Joule's formula we have IV = PR = I X E, 46 ELECTRIC LIGHT ARITHMETIC. and expressing the values of the Volt and Ampere in absolute C.G.S. measure we have in the present case W = X 31-1 X io s ergs 10 = 37942 X lo 8 ergs per second. But one horse-power is equivalent to 7-46 X lo 9 ergs, and therefore the energy absorbed by the lamp per second . 37942 * 10* _ l thofaHiR 7-46 X I0 9 20 Again, one water-gramme-degree Centigrade is equivalent to 4-2 X io 7 ergs, and therefore the amount of heat developed in this lamp per second is - 37943 X io 8 = d ^ 4-2 X io' Example 2. The strength of current flowing through a certain lamp was found to be 2-42 Amperes and the electro-motive force between the terminals was 57 Volts. What amount of horse-power was being expended in the lamp? Am. 0-185 H.P. Ex-ample 3. What horse-power is required to drive a current of 4'i6 Amperes through a resistance of 21*4 Ohms ? ELECTRIC LIGHT ARITHMETIC. 47 By Joule's formula we have W = /2 X K = 04i6) 2 X 21-4 x io 9 ergs = (-416)2 x 21-4 X io 9 R p 7-46 X io 9 = l - H.P. nearly. Example 4. The resistances of three incandescent lamps which are placed in series are 60, 32, and 50 Ohms respectively. What horse-power will be required to send a current of one Ampere through them, neglecting the resistance of the connecting wires? Ans. 0-19 H.P. Example 5. A certain dynamo machine has an internal resistance of 3 Ohms, and that of the lamps and leading wires is io Ohms. What horse-power is expended in driving a current of 15 Amperes through this circuit ? Ans. 3^9 H.P. Example. 6. Find the amount of horse-power required to maintain a current of io Amperes through 8 arc lamps, each of which has a resistance of 6 Ohms, the resistance of the leading wires being 8 per cent, of that of the lamps. Ans. 7 H.P. nearly. Example 7. Suppose that we have 1 6 Brush lamps in series, that each lamp has a resistance of 4*4 Ohms, 48 ELECTRIC LIGHT ARITHMETIC. and that the resistance of the rest of the circuit is I2'65 Ohms. If 60 per cent, of the total energy expended appears in the lamps, find the amount of, horse-power required to maintain a current of io v Amperes. Ans. 1573 H.P. Example 8. A certain size of Swan lamp has, when hot, a resistance of 30 Ohms. There are 30 of these lamps in series and the resistance of the remainder of the circuit is 15 Ohms. What horse-power will be required to maintain a current of 1*5 Ampere through this series of lamps? Ans. 276 H.P. Example 9. An electric circuit is composed of 5 arc lamps, each of which has a resistance of 2*5 Ohms, the armature of the machine which has a resistance of 2 Ohms, and the leading wires which have a resistance of 1*5 Ohm. Find what proportion of the total energy expended in the circuit appears in the lamps. The total resistance of the circuit = 16 Ohms, and the ratio of the work expended in any part of the circuit to the total work is equal to the ratio of the resistance of that part to the resistance of the whole circuit. Hence in the present case the proportion of the total energy which is developed in the lamps = -r = 78 or 78 per cent. 10 ELECTRIC LIGHT ARITHMETIC. 49 Example 10. If the internal resistance of a dynamo machine be 3 Ohms and that of the external circuit 1 8 Ohms : what is the ratio of the external or useful work of the current to the total work expended ? Atts. 6-7ths. Example 11. The resistances of two incandescent lamps arranged in series are 50 and 100 Ohms respec- tively, and the total resistance of the circuit is 200 Ohms. What fraction of the total work is expended in each of these lamps ? Ans. j and . Example 12. In an experiment with a certain dynamo machine the resistance of the armature was o-o 1 6 Ohm, and that of the external circuit 0757 Ohm The power required to turn the armature at a certain rate in the magnetic field was 7 '604 H.P. and this produced a current of 837 Amperes. Calculate the duty of the generator and also its commercial efficiency. The total resistance of the circuit = 0757 + 0*016 = 0773 Ohm, and therefore by the method of Example 3 we find that the electrical energy in the current - ( 8 '37) 2 X Q773 _ ? . 259 HiR 7-46 The duty of the generator is the ratio of the total electrical energy developed to the amount of energy E 50 ELECTRIC LIGHT ARITHMETIC. expended in turning the armature in the magnetic field. In the present case it is The commercial efficiency of the generator is the ratio of the amount of electrical energy which appears in the external circuit to the total energy which is expended in driving the machine. In the present case the commercial efficiency _ 757 X 7-259 _ 773 X 7'6o4 Example 13. The internal resistance of a certain dynamo machine was 1-5 Ohm, and the external re- sistance 27 Ohms. An expenditure of 3*87 H.P. in turning the armature developed a current of 25-5 Amperes. Calculate from these data the duty of the generator and also its commercial efficiency. Ans. Duty = 0^95 ; Commercial efficiency = 061. Example 14. In an experiment with a Brush machine the work expended in driving the armature was 15-3 H.P., the internal resistance was 10-5 Ohms, the external 73 Ohms, and the strength of the current was 10 Amperes. Find the commercial efficiency of this machine. Ans. 0-64 ELECTRIC LIGHT ARITHMETIC, 5 1 Example 15. The external resistance of an electrical circuit being 0-923 Ohm, and the internal resistance of the Gramme machine 1-077 Ohm, an expenditure of 4 H.P. in driving the machine produced a current of 32 Amperes. What was the commercial efficiency of this machine ? Ans. 0-32 nearly. Example 16. What was the duty of the machine alluded to in the last example ? Ans. o'6<) nearly. Example 17. The resistance of a certain Gramme machine is' 1-67 Ohm, and that of the leading wires ox>3 Ohm, while that of the lamp is 2 Ohms. If 3*5 horse-power is expended in driving the armature, and the duty of the machine is 072 ; find the quantity of energy which will be expended in the lamp. The electrical energy in the circuit = 3'5 X 0-72 = 2-52 H.P. Total resistance of circuit = 1-67 + -03 + 2 = 37 Ohms. Resistance of lamp = 2 Ohms. Hence the quantity of energy expended in the lamp = 1-36 H.P. If we wish to know the value in Amperes of the current flowing through the lamp, we can calculate it from the formula E 2 5 2 ELECTRIC LIGHT ARITHME TIC. I M/ W = PR whence / = / v ^ and in the present case W = 1-36 X 7-46 X io 9 ergs. R2Xio ) electro-magnetic C.G.S. units of resistance. therefore / = / r36 X 7 ' 46 = 2-25 C.G.S units of current, V 2 = 22-5 Amperes. Example 1 8. Three arc lamps, each of which had a resistance of i'5 Ohm, were joined up in series with a dynamo machine whose resistance was 2 '8 Ohms. The resistance of the leading wires was o p 22 Ohm, and the duty of the machine 0-95. Find the quantity of energy developed in each lamp when 5*45 H.P. was expended in driving the armature. Ans. 1-03 H.P. Example 19. What was the strength of current flowing through each lamp ? Ans. 22'6 Amperes. Example 20. A certain Brush machine had an internal resistance of 5'o6 Ohms, and it was connected up by wires of 1*3 Ohm resistance with a series of 16 lamps, each of which had a resistance of 3'8 Ohms. ELECTRIC LIGHT ARITHMETIC. 53 The duty of the machine was 073. What was the amount of energy developed in each lamp when 14 H.P. was being expended in driving the armature? Ans. 0-58 H.P. Example 21. What was the strength of current, in Amperes, flowing through each lamp ? Ans. 1 0*67 Amperes. Example 22. The internal resistance of a Gramme machine provided with a permanent steel magnet of constant strength was o - 4 Ohm. When the terminals were insulated the power required to drive the bobbin at a certain rate was o - oi H.P. When the terminals were connected by a wire of 3'6 Ohms resistance, the power required to drive the bobbin at the same rate as before was o - i H.P. What power would be required to drive the bobbin at the same rate if the terminals were connected by a wire of 0*2 Ohm resistance ? If this wire be immersed in one pound of cold water through how many degrees per minute would its temperature be raised, the mass of the wire being negligible ? Since the machine is provided with a permanent magnet of constant strength, and since the rate of revolution of the armature is constant, it follows that the electro-motive force of the machine is constant. 54 ELECTRIC LIGHT ARITHMETIC. But by Joule's equation W / X E = = - where a is a constant. R R Now by the question o'oi H.P. is used in overcoming friction, and therefore when the total energy expended is o'i H.P. and the resistance R =^ 0-4 -f 3-6 = 4 Ohms, we have - = O'l - o'oi = 0-09 H.P. ; .'. a = '36. 4 Let x represent the horse-power required in the second case, then since W-'& R we get IV = 3 = 3'6 = . 6 H

ELECTRIC LIGHT A RITHME TIC. 6 1 we have 45 = ^X 1-2 (^ + -01) = I'2 X 35 + A' X '012 and .*. x = 250 lamps. Example 10. Find the number of lamps which can be driven in simple multiple arc by a dynamo machine whose resistance is 0-032 Ohm, and its electro-motive force 55 Volts, if each lamp has a hot resistance of 28 Ohms and requires a current of r6 Ampere. Ans. 199 lamps. Example 11. The internal resistance of a certain dynamo machine is i Ohm, and its electro-motive force is 484 Volts when the interpolar portion of the circuit consists of 200 incandescent lamps arranged in 20 series of 10 each. If the resistance of each lamp be 30 Ohms, what is the strength of current flowing through it ? The resistance of each series of lamps . . . = 30 X 10 = 300 Ohms. The resultant resistance of the 20 series . . . = ^ = 1 5 Ohms. 20 Total resistance of the whole circuit . .= 154-1= 16 Ohms. 62 ELECTRIC LIGHT ARITHMETIC. Hence the current flowing through the machine = i-i = 30-25 Amperes, ID and therefore the strength of the current through each one of the 20 series is _ 30^ = ^^ Ampere. Example 12. Thirty-five incandescent lamps, eacli of which has a hot resistance of 35 Ohms, are arranged in 5 series of 7 each, and are then connected in multiple arc with the poles of a dynamo machine whose internal resistance is 2 '5 Ohms and its electro-motive force 309 Volts. What is the strength of current through each 'lamp? Ans. i '2 Ampere. Example 13. The resistance of a certain dynamo machine is 2 Ohms, and it is required to arrange So incandescent lamps, each of which has a hot resistance of 50 Ohms, in such a way that the resistance in the external circuit shall be 20 times that of the machine How is this to be done ? Let x = number of series. Then = number of lamps in each series, and ., = the resistance of the external circuit, = 40 Ohms, by the conditions of the problem. ELECTRIC LIGHT ARITHMETIC. 63 Hence 80 x 50 x 2 -- - 100; .-. x = TO rows 40 Example 14. A set of 120 incandescent lamps, each of 307 Ohms resistance, are to be arranged in such a way that their resultant resistance shall be about 16 times as great as that of the dynamo machine, whose resistance is i 6 Ohm. How is this to be done ? Ans. 12 series of 10 each. Example 15. A certain kind of incandescent lamp requires a current of 2*5 Amperes with an electro- motive force of 73 Volts to work it. There are 60 of these lamps, arranged in 20 series of 3, and the resis- tance of the machine is 0*27 Ohm. What must be its electro-motive force ? The resistance of each lamp = ^ = 29-2 Ohms external circuit = = 4-38 Ohms. Total resistance of circuit = 4*38 + '27 = 4'6s Ohms. But since there are 20 series, and each requires a current of 2-5 Amperes, the current through the dynamo machine = 20 X 2-5 = 50 Amperes, and therefore the requisite electro-motive force of the machine = 50 X 4'65 = 232-5 Volts. 64 ELECTRIC LIGHT ARITHMETIC. Example 16. The current required for a certain incandescent lamp is i'2 Ampere, with an electro- motive force of 63-6 Volts. A hundred of these lamps are joined up in 50 series of 2 each to the poles of a dynamo machine whose internal resistance is 0^14 Ohm. What must be its electro-motive force ? A us. 135-6 Volts. Example 17. Forty-five incandescent lamps, each of which has a resistance of 98 Ohms, are arranged in 15 series, and a current of 0-67 Ampere is sent through each. If the resistance of the dynamo be 1-23 Ohm. what must be its electro-motive force ? Ans. 209-3 Volts. Example 18. Find the number of Daniell cells which would be required to construct a voltaic battery having the same electro-motive force and internal resistance as a certain Brush machine whose electro- motive force is 839 Volts, and its resistance is 10-55 Ohms, having given that the electro- motive force of a Daniell cell is one Volt, and its resistance equal to 5 Ohms. Let x number of cells in one series, and / the number of series; then, since the EMF of a Uaniell - one Volt, we have x = 839. ELECTRIC LIGHT ARITHMETIC. 65 The resistance of the battery = - 9 x 5. = , -^ by the question ; ...,.1^1. 397-63, also total number of cells required = 397'63 X 839 = 333612 cells. Example 19. A certain dynamo machine has an electro-motive force between its terminals of 1 10 Volts, and the resistance of the whole circuit is 0*1 1 Ohm. If a Grove's cell has an electro-motive force of i - 97 Volt, and on the average a resistance of 037 Ohm, find the number of such cells which would have to be grouped together so as to produce a current equal to that of the above machine through a short thick wire. Ans. 10487 cells. Example 20. A Gramme machine, whose resistance is o'i Ohm, is employed to send a current through 16 incandescent lamps, each of which has a resistance of 25 Ohms. The electro-motive force of the machine is proportional to the speed at which it is driven. If the driving engine work always at the same horse-power, and if the armature of the Gramme make 1200 revolu- tions per minute when all the lamps are in series, find the number of revolutions it will make when all the 66 ELECTRIC LIGHT ARITHMETIC. lamps are (i) in multiple arc, (2) arranged in 4 series of 4. Let E represent the electro-motive force of the machine, and / the strength of the current flowing through the machine. Then since the amount of energy converted into electricity is constant we have E X / = a constant quantity, and . . E = - where m is some constant. Again by Ohm's law we have But R = B + L, where L resultant resistance of the lamps and B the resistance of the machine. When all the lamps are in series, L = 16 X 25 = 400 Ohms ; .-. R = 400-1 Ohms. If x = number of revolutions per minute, then E = nx where n is a constant ; and substituting this value for E in equation (i) we get m = - - .. x 2 = R X q where q is some constant. A ELECTRIC LIGHT ARITHMETIC. 67 But when R = 400'!, x = 1200 ; When all the lamps are in multiple arc, R = o-i + 1-563 = 1-663 Ohms ; .'. ** 3599 X 1-663, whence x = 77 revolutions. Again, when the lamps are arranged in 4 series of 4, R = 25' i, and then * 3599 x 25' 1 ; ' x 3 01 revolutions. Example 21. Sixty accumulators are arranged in series to feed a number of glow-lamps arranged in parallel. Each accumulator has an internal resistance of '005 Ohm and a discharge electromotive force of 2 Volts, while each lamp requires a current of \ an Ampere and 1 1 1 volts potential difference between its terminals. How many lamps can be thus fed ? Let x = number of lamps, then the current through the accumulators is - Amperes. Since the total fall of potential is equal to the sum of the fall of potential in the accumulators and of that in the lamp circuit, we have 68 ELECTRIC LIGHT ARITHMETIC. - X 'COS X 60 + I* 1 = 60 X 2 /. x = 60 lamps. Example 22. How many lamps could be fed by 30 of the above accumulators, if each lamp required one Ampere of current and 50 volts of electrical pressure ? Ans. 66 lamps. Example 23. If each lamp required 90 Volts and 2*25 Amperes of current, how many lamps could be run in parallel by 47 such accumulators in series ? Ans. 7 '6 lamps nearly. That is, 8 lamps all a little below normal brightness, or 7 above Example 24. How many of these lamps could be run in parallel with 48 of these accumulators in series ? Ans. 1 1 lamps all a little above normal brightness, 12 ., below Example 25. How many lamps similar to those described in Example 21 could have been run in parallel by 61 accumulators of the kind mentioned in that example ? Ans. 72 lamps all a little above normal brightness. Example 26. Each of 30 glow-lamps arranged in ELECTRIC LIGHT ARITHMETIC. 69 parallel requires 100 Volts and 75 of one Ampere. How many accumulators arranged in series will be required to run them if each accumulator has an internal resistance of 0-0059 Ohm and an electro- motive force on discharge of i'8 Volts? Let x be the number of accumulators required; then, as in the solution of Example 21, we have x x '0059 x 22-5 + 106 = x x i'8 ; .'. x = 60 very nearly. Example 27. If each lamp required not less than 50 Volts and i\ Amperes of current, and there were 20 lamps in parallel, how many of the above accumulators would be required ? Ans. 31 very nearly. DISTRIBUTION OF ENERGY IN A COMPOUND CIRCUIT. Example I. Two coils of platinum wire, whose resistances are as 5 : 3, are placed in vessels of water forming calorimeters, and are connected in series in an electric circuit, when it is found that the quantities of heat produced in 10 minutes are 72'22 and 43*33 units respectively. The coils are then placed in a multiple arc, and the quantities of heat produced in 10 minutes are then found to be 2'6 and 4*33 units respectively. 70 ELECTRIC LIGHT ARITHMETIC. Deduce from these experiments Joule's law as to the heating effect of an electric current. (i) When the coils are in series. Let HI represent the heat developed in ist coil, and /?i its resistance. Let HI represent the heat developed in 2nd coil, and R% its resistance. Then we have = 2522 and % = 5 H* 43-33 X>2 3 whence R-jHi = 5 X 43 '33 = RiHi 3 x 72-22 H, (2) When the coils are arranged in multiple arc. In this case the current will divide itself between the two coils in the inverse ratio of their resistances, and therefore the current in the first coil will be 3-8ths, and that through the second coil 5-8ths of the main current. Let /! and 7 2 represent the strengths of the currents in the two coils, then we shall have ELECTRIC LIGHT ARITHMETIC. 71 A 2 X A -H* 9 X 5 A 4'33 7 a 2 X R z HI 25 x 3 X 2-6 _ I2 99 _ Therefore * X R z I? X Rj. and from (a) and (/3) we get Joule' s equation H = KRI Z where K is some constant Example 2. A current of 2*4 Amperes is sent through two lamps in series, each of which has a resistance of 60 Ohms. The lamps are then placed in multiple arc and the same current is divided between them. Com- pare the quantities of heat developed in each lamp in the two cases. Ans. As 4 : I. Example 3. The resistance of a certain dynamo machine is 2 Ohms, and when it is making 820 revo- lutions a minute it can drive a current of 2*5 Amperes through a single extrapolar resistance of 30 Ohms- The extrapolar part is then made up of 5 incandescent lamps, placed in multiple arc, and each having a resistance of 150 Ohms. Assuming that the electro- motive force is proportional to the velocity of rotation, 72 ELECTRIC LIGHT ARITHMETIC. find the rate at which the armature must be made to turn so as to send a current of i Ampere through each of these lamps. Under the conditions stated in the first part of the problem, the electro-motive is found directly by Ohm' s law EI = Ji X RT. = 2-5 X (2 + 30) - 80 Volts. In the second arrangement the resultant resistance of the circuit is R 2 = 2 + - J -i = 32 Ohms, and the current flowing through the dynamo is equal to 5 Amperes, hence its electro-motive force is E 9 = 5 x 32 = 1 60 Volts, and since the electro-motive force is proportional to the velocity of rotation, the velocity required is = ~ X 820 = 1640 turns a minute. Example 4. When the armature of a certain machine is making 1200 turns a minute, it can send a current of II Amperes through an external arc-lamp resistance of 12-5 Ohms. The arc lamps are then replaced by 10 incandescent lamps in simple multiple arc, each of ELECTRIC LIGHT ARITHMETIC. 73 them having a resistance of 125 Ohms. The resistance of the machine is 0*5 Ohm, and it is required to find at what rate the armature must turn so as to maintain a current of i'2 Ampere through each of these lamps. Ans. 1309 turns a minute. Example 5. A current of 10 Amperes is sent through a series of 10 arc lamps, each of 3-8 Ohms resistance, by a dynamo machine whose armature is making looo revolutions a minute. The arc lamps are then replaced by 8 incandescent lamps, each of 120 Ohms resistance and arranged in 4 series of 2 each. The armature is then made to turn at 1044 revolutions a minute. The resistance of the machine is 3 Ohms. Assuming that the electro-motive force developed by the machine is proportional to the velocity of rotation, find the strength of current in each lamp. In the case of the arc lamps the total resistance of the circuit = 3'8 X 10 + 3 = 41 Ohms, and therefore, by Ohm's law the electro-motive force of the machine is EI = 41 X 10 = 410 Volts. In the second arrangement the electro-motive force, which is assumed to be proportional to the velocity of rotation of the armature, is - x 4IO = 42 g VoltS) 74 ELECTRIC LIGHT ARITHMETIC. and the resultant resistance of the circuit is = 2 X I2 + 3 = 63 Ohms, and .'. the current flowing through the machine = I? 8 - =6-8 Amperes, 63 and the current through each lamp = =17 Amperes. 4 Example 6. Fifteen incandescent lamps, each of 120 Ohms resistance, are arranged in simple multiple arc between the poles of a dynamo machine whose resistance is I Ohm, and which is making 1395 revolutions a minute. When the extra polar circuit consists of 3 arc lamps, each of 3 Ohms resistance, a velocity of 1550 revolutions develops an electro-motive force of 200 Volts. Find the strength of current in each arc lamp, and also in each incandescent lamp. Ans. Current in arc lamp = 20 Amperes. Current in incandescent lamp = 1*3 Ampere. Example 7. The poles of a certain dynamo machine whose resistance is 0-08 Ohm, are in one case con- nected by 4 incandescent lamps in series, and in another case by the same 4 lamps arranged in multiple arc. The resistance of each lamp is 50 Ohms, and ELECTRIC LIGHT ARITHMETIC. 75 the same amount of energy is expended in each case by the driving engine. Compare the amounts of heat developed in the machine. If H be the total amount of energy developed in the circuit which is the same for both cases, then when the lamps are in series the heat developed in the machine is H = 0-08 X H H 200 -f 0-08 ~ 2501 In the second case the resistance of the extra polar circuit is = = 12-5 Ohms, and therefore the amount of heat developed in the machine is o-oS x H H H 12'5+0'OS IS7-25 * nearly. N.B. The student will notice from this result that a dynamo machine should never be let run on short circuit, as in that case all the energy developed by the machine is dissipated as heat in the machine itself. Example 8. In one case 100 incandescent lamps, each of 120 Ohms resistance, were joined up in 25 series of 4 each to the poles of a dynamo machine whose resistance was 2 Ohms. In another case the 76 ELECTRIC LIGHT ARITHMETIC. same lamps were joined up to the same machine in 50 series of 2 each. If the same horse-power was expended in each case by the driving engine find what proportion of the total heat is dissipated in the machine in each case. Ans. 0-18 and 0-29 respectively. Example 9. An Edison lamp of 120 Ohms resistance has a current of ex 8 Ampere flowing through it. A wire of 20 Ohms resistance is then placed across its terminals as a shunt. If the same current be now divided between the lamp and shunt, by how much will the heat developed in the lamp be diminished ? The current flowing through lamp = - x 0-8 = Ampere, 120 + 20 35 and if H l and H 2 be the quantities of heat developed in the lamp before and after the addition of the shunt, then Example 10. Two Swan lamps, one of which had a resistance of 30 Ohms and the other one a resistance of 20 Ohms, were arranged first in series, and then in multiple arc, and a current of 1-2 Ampere was main- tained between the extreme terminals. Compare the ELECTRIC LIGHT ARITHMETIC. jj amounts of heat developed in each lamp in the two cases. For the first lamp ^ 2 = 1 ; Hi 2S For the second lamp ? = H\ 25 Example 11. A dynamo machine has an internal resistance of '0046 Ohm, and it supplies a current of 9 Ampere to each of 1000 incandescent lamps arranged in simple multiple arc. If each lamp has a resistance of 120 Ohms, find the amount of power consumed in the machine and in each lamp, and also the amount of heat developed. The current flowing through the machine = looo X '9 = 900 Amperes. .'. Work consumed in the machine = PR = (goo) 2 X -0046. = 3726 Watts. Work expended in each lamp = (*9) 2 X 120 = 97' 2 Watts. The amount of heat developed in the machine and in each lamp can be readily expressed in terms of the new electro-magnetic unit of heat which has been suggested by Dr. C. W. Siemens. It is the amount 7 8 ELECTRIC LIGHT ARITHMETIC. of heat which is generated in one second by a current of i Ampere flowing through a resistance of i Ohm and is called a Joule. In the present case Heat developed in machine per second = 3726 Joules. each lamp = 97^2 Joules. Example 12. The mechanical equivalent of the amount of heat which is necessary to raise the temperature of one gramme of water i C. is 4'2 X to 7 C.G.S. units or ergs. How many grammes of water would be raised i C. by the amount of heat represented by one Joule ? One water-gramme degree Centigrade = 4-2 X io 7 ergs. One Joule = io 7 ergs. ... One Joule - .-. _ 0>238 w g d> C o Example 13. The poles of a certain dynamo machine, whose internal resistance is '125 Ohm, are connected in simple multiple arc by 50 incandescent lamps, each of 100 Ohms resistance, and a current of one Ampere is sent through each lamp. Find the amount of heat expended in each lamp and consumed in the machine per second. Ans. Heat in lamp = 100 Joules. Heat in machine = 6*25 Joules. ELECTRIC LIGHT ARITHMETIC. 79 Example 14. Two incandescent lamps, each of 120 Ohms resistance, are connected in multiple arc with the poles of a dynamo machine whose resistance is 4 Ohms, and a current of I Ampere is sent through both. The lamps are then placed in series and again a current of i Ampere is sent through both. What amount of heat is developed in the machine in each case? Ans. 1 6 and 4 Joules. Example 15. The resistance of a certain battery is 13*5 Ohms, and its electro-motive force is 98-5 Volts. Its poles are first short circuited by a thick wire and then are connected by a lamp and leading wires of 1 1 Ohms resistance. Find the quantity of heat developed in the battery in each case. Ans. 7187 Joules and 218-2 Joules. Example 16. The resistance of each of 1320 Edison lamps, arranged in simple multiple arc, is 140-5 Ohms and that of the armature of the dynamo machine is 0^0042 Ohm. The resistance of the leading wires is o'oi Ohm and that of the field coils, which are arranged in parallel circuit with the lamps, is 7*067 Ohms. If the machine can convert 142 H.P. into electrical energy, find the amount of heat developed in the field coils. The resistance of the lamp circuit = o-oi + ^-$ = 0-1164 Ohm. 1320 8o ELECTRIC LIGHT ARITHMETIC. The resistance of the field coils = 7 '067 Ohms, .'. resultant resistance of external circuit 71Q67XO-1 164 g 7-067 +0-1164 and the resultant resistance of whole circuit = 0*0042 + o'ii45 = 0-1187 Ohm. .'. Energy expended in external circuit But energy expended in the field coils external circuit resistance of lamp circuit resistance of (lamps + field coils) 0-1164 _ 0-1164 7-067 + 0-1164 ~ 7'i834 .. Energy expended in the field coils . = 2-2195 H.P. But the heat equivalent of one H.P. is 746 Joules and .'. the amount of heat developed per second in the field coils of this dynamo machine is = 2-2195 X 746 = 1656 Joules. ELECTRIC LIGHT ARITHME TIC. 8 1 Example 17. Find the amount of heat developed in the revolving armature of the above machine. The energy expended in the armature - -42 x - 7 = 5-024 H.P .. Heat developed in armature = 5 -024 X 746 = 3748 Joules. Example 18. Find the amount of heat which would have been developed in the armature if its resistance had been 0-0041 Ohm instead of 0*0042 Ohm, all the other resistances as well as the total amount of electrical energy in the circuit being the same as before. Ans. 3662 Joules. N.B. The difference in the amount of heat developed in the armature shows that, in the construction of a dynamo machine for the develop- ment of strong currents, it is important that the resistance of the armature should be as small as possible. Example 19. The armature of a certain Edison dynamo machine has a resistance of 0-2 Ohm, and G 82 ELECTRIC LIGHT ARITHMETIC. the field coils have a resistance of 41-5 Ohms and are connected in parallel circuit with 70 lamps, each of 140 Ohms resistance, arranged in simple multiple arc. Neglecting the resistance of the leading wires, find the amount of heat developed in the armature and in each of the lamps when the machine is converting 8 H.P. into electrical energy. Ans. Heat in armature = 566-2 Joules. Heat in each lamp = 73-6 Joules, ELECTRIC LIGHT ARITHME TIC. 8 3 TABLE OF SQUARES, SQUARE ROOTS, AND RECIPROCALS. X *!x I X" I I I -000 roooo 2 4 1-414 0-5000 3 9 1732 0-3333 4 16 2'OOO 0-2500 5 25 2-36 0-2000 6 36 2-449 0-1667 7 49 2-646 0-I429 8 64 2-828 0-I250 9 81 3-000 O'lIII 10 IOO 3-I62 O'lOOO n 121 3-3I7 0-0909 12 144 3'464 0-0833 13 I6 9 3-606 0-0769 H I 9 6 3742 0-0714 IS 225 3-873 0-0667 16 2 5 6 4'OOQ 0-0625 17 28 9 4-I23 0-0588 18 324 4-243 0-0556 19 36l 4-359 0-0526 20 400 4-472 0-0500 1 84 ELECTRIC LIGHT ARITHMETIC. Jt X* ^ X i X 21 441 4-583 0-0476 22 484 4-690 0-0455 23 529 4796 0-0435 24 576 4-899 0-0417 25 625 5-000 0-0400 26 676 5-099 0-0385 27 729 5-196 0-0370 28 784 5-292 0-0357 29 841 5-385 0-0345 30 900 5-477 0-0333 31 961 5-568 0-0323 32 1024 5-657 0-0313 33 1089 5745 0-0303 34 1156 5-831 0-0294 35 1225 5-916 0-0286 36 1296 6-000 0-0278 37 1369 6-083 0-0270 38 1444 6-164 0*0263 39 1521 6-245 0-0256 40 1600 6-325 0-0250 ELECTRIC LIGHT ARITHMETIC. 85 X ^ V* i X 41 1681 6-403 0-0244 42 1764 6-481 0*0238 43 1849 6-557 0-0233 44 1936 6-633 0-0227 45 2025 6-708 0-0222 46 2116 6782 0-0217 47 2209 6-856 0-0213 48 2304 6-928 O'O2o8 49 2401 7-000 0'0204 So 250x3 7-071 O'O2OO Si 2601 7-141 0-0196 52 2704 7'2II 0*0192 53 2809 7-280 0-0189 54 2916 7-348 0-0185 55 3025 7-416 0-0182 56 3136 7-483 0-0179 57 3249 7-550 0-0175 58 3364 7-616 0-0172 59 348i 7-681 0-0169 60 3600 7-746 0-0167 86 ELECTRIC LIGHT ARITHMETIC. y * V* i X 61 3721 7-810 0-0164 62 3844 7-874 0-0161 63 3969 7'937 0-0159 64 4096 8-000 0-0156 65 4225 . 8-062 0-0154 66 435o 8-124 0-0152 67 4489 8-185 0-0149 68 4624 8-246 0-0147 69 4761 8-307 0-0145 70 4900 8-367 0-0143 71 5041 8-426 0-0141 72 5184 8-485 0-0139 73 5329 8-544 0-0137 74 5476 8-602 0-0135 75 5625 8-660 0-0133 76 5776 8718 0-0132 77 5929 8-77.5 0-0130 78 6084 8-832 O'OI'2S 79 6241 8-888 0-0127 80 6400 8-944 0-0125 ELECTRIC LIGHT ARITHMETIC. 87 X X s VF X Si 6561 9'ooo 0-0123 82 6724 9-055 O'OI22 83 6889 9-110 O'OI2O 84 7056 9-165 O'OIig 85 = 7225 9*220 O'OIlS 86 7396 9-274 o-oi 1 6 87 7569 9-327 0-0115 88 7744 9-381 0-0114 89 7921 9-434 0-OII2 90 8100 9-487 o-oiii 9i 8281 9-539 o-oi 10 92 8464 9-592 0-0109 93 8649 9-644 o-oioS 94 8836 9-695 0-0106 95 9025 9747 0-0105 96 9216 9798 0*0104 97 9409 9-849 0-0103 98 9604 9-899 0-0102 99 9801 9-950 o-oioi 100 10000 lO'OOO O'OIOO 88 ELECTRIC LIGHT ARITHMETIC. TABLE OF THE BIRMINGHAM WIRE GAUGE. B. W. G Diam. Diam. Circum. Area No. Inches. Centims. Centims. Sq. Centims. I circ i oooo 000 oo o I 2 3 4 I'OOO 454 425 380 340 300 284 259 238 2-540 I-I53 I -080 965 864 762 721 658 604 7-980 3'622 3-393 3-032 2714 2-394 2-265 2-067 1-898 5-067 1-044 916 731 586 456 4 08 "340 287 I 9 '22O 203 180 165 148 '559 '457 419 376 1756 1-621 1-436 1-316 riSr 245 20 9 l6 4 -I 3 8 'III IO II 12 "'34 120 109 340 305 277 1-068 958 870 9 I 073 060 H 095 083 241 "211 757 663 046 15 16 072 065 I8 3 I6 5 '575 518 026 021 18 058 049 147 124 462 "390 017 OI2 20 21 22 23 24 h 042 035 032 028 025 'O2 2 107 089 08 1 071 063 055 314 280 254 223 198 173 00 9 006 00 5 004 003 'OO2 University of California SOUTHERN REGIONAL LIBRARY FACILITY 405 Hilgard Avenue, Los Angeles, CA 90024-1388 Return this material to the library from which it wss borrowed. STATE jn Universit South* Libra