ELECTRICAL ENGINEERING ADVANCED COURSE McGraw-Hill DookCompany PwSfisfiers offioofor Electrical World The Engineering and Mining Journal Engineering Record Engineering News Railway Age G aze ttx? American Machinist Signal Rnginoer American Engineer Electric Railway Journal Coal Age Metallurgical and Chemical Engineering Power ELECTRICAL ENGINEERING ADVANCED COURSE BY ERNST JULIUS BERG Sc. D. i PROFESSOR OP ELECTRICAL ENGINEERING UNION COLLEGE, SCHENECTADY, N. Y. AUTHOR OF "ELECTRICAL ENGINEERING," FIRST COURSE FIRST EDITION McGRAW-HILL BOOK COMPANY, INC. 239 WEST 39TH STREET. NEW YORK LONDON: HILL PUBLISHING CO., LTD. 6 & 8 BOUVEBIE ST., E. C. 1916 13 COPYRIGHT, 1916, BY THE McGKAW-HiLL BOOK COMPANY, INC. TUB MAIT-IC FKBSS T O H K PREFACE This volume contains abstracts of a series of lectures given to graduate students in electrical engineering at Union College. It is primarily intended to prepare the student to understand and to deal mathematically with phenomena which are incidental to abnormal or transient conditions in electric circuits. The first part is practically a reprint of a series of articles published by the author some years ago in the General Electric Review. These cover the simple transients in circuits containing concentrated inductance, capacity, and resistance, which have been treated by many authors, notably by BEDELL AND CREHORE in their " Alternating Currents," published 1893. The second part deals with the somewhat more difficult prob- lems of transients in circuits of distributed inductance, capacity and resistance. These were treated mathematically very fully almost thirty years ago by Heaviside in a series of papers on " Electromagnetic Theory/ 7 later published in book form. In 1909 Steinmetz's " Transient Phenomena" appeared. This book covered in a broad sense very much the same ground as that of the authors given above, but covered it in an essentially differ- ent way; introducing for the first time asfa-r as the author knows a really advanced book on practical electrical engineering problems. The third part of the book deals with problems in electro- statics. These again have been very fully treated almost fifty years ago by Maxwell in his famous books on "Electricity and Magnetism. " Since that time a large number of papers and books have appeared on the subject, notably by Heaviside, Kelvin, Gray, Jeans and Webster, and quite recently by Coffin in his interesting little book on "Vector Analysis." While the literature on this phase of engineering is thus very extensive, it has, for all purposes, been closed to the practical engineer because of his lack of sufficient mathematical knowledge. Dr. W. S. Franklin has, however, recently published a number of papers, which in a beautifully simple way have demonstrated that these advanced problems can be solved with simple mathematics. vi PREFACE The last part of the book gives an outline of the theory oi electric radiation. The mathematical theory was again given almost fifty years ago by Maxwell. Hertz's verification of Maxwell's theoretical work given twenty years later and pub- lished in his "Electric Waves" is today almost the last word in the theory of wireless transmission of energy. Yet it would be out of place to omit a reference to the recent excellent papers and books by Marconi, Lodge, Flemming, Pierce, Zenneck, Cohen, Austin and a score of others. It is evident then that the field covered in this volume is not new. Nevertheless, the book seems justified because it endeavors to give the theory in a way comprehensible to students who have had only the ordinary undergraduate course in electrical engi- neering. It is hoped that the volume will also serve a useful purpose in bringing to the attention of students a field of mathe- matics of extreme practical importance that is hardly known to them. The author is greatly indebted to one of his graduate stu- dents, MR. M. K. TSEN, who not only examined the manuscript in detail, but checked and elaborated upon the theoretical work. He is also indebted to DR. A. S. MCALLISTER, who kindly criticized the manuscript prior to its publication and offered valuable suggestions. CONTENTS CHAPTER PAGE INTRODUCTION 1 PART I TRANSIENT PHENOMENA I. CIRCUITS CONTAINING CONCENTRATED INDUCTANCE AND RESISTANCE 3 II. PROBLEMS INVOLVING MUTUAL INDUCTANCE 33 III. CIRCUITS OF RESISTANCE AND VARIABLE INDUCTANCE ... 56 IV. CHARACTERISTICS OF CONDENSERS 68 V. A CIRCUIT CONTAINING DISTRIBUTED RESISTANCE AND IN- DUCTANCE 106 VI. CIRCUIT CONTAINING DISTRIBUTED LEAKAGE CONDUCTANCE AND CAPACITY 110 VII. CIRCUIT CONTAINING DISTRIBUTED RESISTANCE AND CAPACITY 113 VIII. DISTRIBUTED INDUCTANCE AND CAPACITY 127 IX. DISTRIBUTED RESISTANCE INDUCTANCE LEAKAGE CON- DUCTANCE AND CAPACITY 143 X. PERMANENT CONDITIONS WHEN ONE OF THE FOUR CONSTANTS, R, L, G, AND C is NEGLIGIBLE 148 XL DISTRIBUTION OF FLUX OR CURRENT IN A CYLINDRICAL OR FLAT CONDUCTOR 150 PART II PROBLEMS IN ELECTRO-STATICS XII. FUNDAMENTAL LAWS 157 XIII. METHODS OF IMAGES, APPLIED TO THE PROBLEM OF POINT CHARGES + 10 AND 5, SEPARATED 5 CM 168 XIV. APPLICATION OF THE POTENTIAL FORMULA V = 2.* TO SOME MAGNETIC PROBLEMS 180 XV. DIVERGENCE OF A VECTOR, POISSON AND LAPLACE EQUATIONS. 186 XVI. LEGENDRE'S FUNCTION 189 XVII. DISTRIBUTION OF CHARGE ON AN ELLIPSOID 199 XVIII. CONCENTRIC SPHERES 209 XIX. CYLINDRICAL CONDUCTORS 218 XX. MUTUAL AND SELF-INDUCTION OF ELECTRO-STATIC CHARGES OR FLUXES MAXWELL'S COEFFICIENTS 232 XXI. TWO-CONDUCTOR CABLE 237 vii Vlll CONTENTS XXII. THE ELECTRO-STATIC EFFECT OF A THREE-PHASE LINE ON AN ADJACENT WIRE OR WIRES 249 XXIII. THE CURL OF A VECTOR 257 XXIV. THE EQUATION OF THE ELECTROMOTIVE FORCE 260 XXV. SOLUTION OF ALTERNATING CURRENT IN CYLINDRICAL CON- DUCTOR SKIN EFFECT '. 271 XXVI. ELECTROMAGNETIC RADIATION. . 278 APPENDIX I: PARTIAL DIFFERENTIATION . . APPENDIX II: ELEMENTS OF VECTOR ANALYSIS. 319 327 INDEX 331 ELECTRCAL ENIN-EERING ADVANCED COURSE PART I. TRANSIENT PHENOMENA CHAPTER I CIRCUITS CONTAINING CONCENTRATED INDUCT- ANCE AND RESISTANCE The study of transients in circuits of concentrated inductance and resistance involves as a rule a knowledge of the solution of linear differential equations of the first order. One example of such a differential equation is: 2+/i(*)=/i(*) (1) where fi(x) and fz(x) may be functions of x or constants, but must not be functions of y. For the sake of convenience fi(x) will be denoted by P and fz(x) by Q. P and Q in the most general case are then functions of x but not of y. Thus, equation (1) becomes % + Pdv = Q (2) A solution of this equation can be obtained, in several ways, all of which, however, involve "educated guesses." Let, for instance, y = uv (3) where u and v are unknown functions of x, which will be deter- mined in the most advantageous way. Since dy dv du y = uv, -7- = u -; \~ v-j- (4) 1 dx dx dx Substituting (3) and (4) in equation (2), dv du or Since u is entirely arbitrary, this expression can be greatly 4 ELECTRICAL ENGINEERING simplified, by selecting such a value of u as to make the coefficient of v or the parenthesis zero. Therefore let: , dx u .'. logu = - fPdx + C. Since the simplest possible function is sought, let that particu- lar one be chosen, which makes C = zero. Thus: log u - - fPdx, and u = e-SPdx ( 6 ) Substituting now this value in (5), there is obtained, .'. v = fef pd * Qdx + C. and since y = uv, y = t-fFd* [ffP** Qdx + C] (7) Special cases: First. Let P be constant, a; and Q be a function of x and y = e~ ax [fe ax Qdx + C] (8) Second. Let P be a function of x, but Q be a constant, b. and y = e -^ Fdx [bfef pd * dx + C] (9) Third. Let both P and Q be constants, a and 6 respectively, dy ' or > _ _ i n -ax Fourth. Let P be a function of x and Q be zero. and, y = < INDUCTANCE AND RESISTANCE 5 If P is a constant a, then y = Ce~ ax . Fifth. Let P be zero and Q be a constant, 6, . dy^ b and, y = bx + C. (12) Two useful integrals that can, of course, easily be solved but will frequently appear are given below for the sake of convenience. e ai cos ut dt = - -5 [o> sin ut -\- a COS co/1. a" + or rf sin at dt = - ~^\a sin co^ co cos co/1. a 2 + a; 2 / A study will now be made of the equation of the current flowing in such circuit when the impressed e.m.f. is steady and also when it varies with time. Referring to Fig. 1, it is evident that the following e.m.fs. exist: FIG. 1. First, the impressed e.m.f., E\ Second, the e.m.f. consumed by the resistance = ir; Third the e.m.f. consumed by the self-inductance = j^ --yr or di L di> where E is the impressed e.m.f. in volts, r the resistance in ohms, N the number of turns of the coil, L the inductance in henrys (assumed constant), -77 the rate of change of flux at a particular instant, t, and i the current in amperes at any particular instant. 6 ELECTRICAL ENGINEERING The e.m.f. consumed by self-inductance can be expressed as J* or L -jT- because the inductance by definition is : _AT0 = 10 8 i thus Nd _ di^ 10*dt " dt' The equation connecting these e.m.fs. is obviously: ' *-* + L (14) That is, at any instant the impressed e.m.f. E is numerically equal to the e.m.f. consumed by the resistance and the e.m.f. consumed by the inductance. Note that e.m.fs. consumed by but not e.m.fs. of resistance and self-induction are considered. The latter are: T di ir and L rr dt Equation (14) can be written: l+z'-f f Compare this equation with (2) and note that P = T an d Q = LJ E -j- are constant when the impressed e.m.f. is constant and not function of t. Thus the solution is found in equation (10) and * /~v 1 ' 1 "^ / -f t~*\ i = Ce L + - (16) The integration constant C is determined from the fact that time is required to impart energy, that is, in this case to produce or alter a magnetic field. Before the switch is closed, there is obviously no field sur- rounding the turns. Shortly after, however, there is a current and thus a field which appears simultaneously with the current. Thus since a magnetic field can not be produced instantaneously, no current can pass at the very first instant. Thus for t = 0, i = 0. Therefore = C.--L- +?, INDUCTANCE AND RESISTANCE but c = 1, therefore = C + ^, and C = - ^; and . Ef I = 11 r \ This equation shows, that as t increases, the current increases, and finally reaches a value, Assume now that after the current has reached this value, the circuit is disconnected from the generator, and at the same instant short-circuited. What can be expected to happen? The Dying Away of a Current in an Inductive Circuit. Re- ferring to Fig. 2, since the coil is surrounded by a magnetic field, and the field can not be destroyed instantaneously, and since the mag- netic field can not exist without a current, it is' evident that the cur- rent can not disappear instantan- eously, but must die away gradu- FIG. 2. ally. Referring to equation (15) which is the general equation of the current and remembering that the impressed e.m.f. E is zero, we have: the solution of which has been shown to be: i = Ce'i*' To determine the integration constant, it is remembered that at the very first instant when t = O, there was a definite current / in the circuit. Thus, i = I when t = 0, which substitued above gives: C = I, and the equation of the decaying current becomes: = - (is) 8 ELECTRICAL ENGINEERING If dW is the energy delivered during a short interval dt, then the rate of energy supply, or power is: _. dt The practical unit of power is the watt, which is work done at the rate of 1 joule per second. At any instant the power is the product of the instantaneous values of e.m.f. and current. Thus the power equation corresponding to equation (14) is: Ei = i X ir + iXL~ = i*r + Li j t (19) It is seen from this equation that when the instantaneous value of the current is i, energy is being dissipated at the rate of i 2 r joules per sec., or watts, in heat, and is being stored in the magnetic field at the rate of Li y- watts. The energy that has been supplied to the circuit t sec. after the switch is closed and the current started is: Eidt joules (20) f Jo The energy dissipated in heat = I Prdt (21) and the energy stored in the magnetic field , V (22) where 7 is the particular value of i when the time is t. In almost all calculations of transient phenomena, the ex- pression e~ ax is met with, e is the base of the natural logarithm. It has the numerical value of approximately 2.718. To calculate the numerical value of any particular expression, the ordinary logarithms are used. Thus, for instance, to find the value of y = c~- 2 , the method is as follows: log y = - 0.2 log e = - 0.2 X 0.434 = - 0.0868 + 0.9132 - 1, therefore y = 0.819, therefore INDUCTANCE AND RESISTANCE In Fig. 3 are shown the values of this function for a large number of values of the exponents. Since this curve is plotted on rectangular coordinate paper, it is rather unsatisfactory for small values of the exponent, and the table below has therefore been worked out. FIG. 3. X er x X e~ x X e~ x X e~ x 0.00 1.0 0.25 0.78 0.80 0.449 1.8 0.165 0.02 0.98 0.30 0.741 0.85 0.427 2.0 0.135 0.04 0.96 0.35 0.705 0.90 0.407 2.5 0.084 0.06 0.942 0.40 0.67 0.95 0.387 3.0 0.05 0.08 0.923 0.45 0.638 1.0 0.368 4.0 0.018 0.10 0.905 0.50 0.607 1.1 0.333 5.0 0.0067 0.12 0.887 0.55 0.577 1.2 0.301 6.0 0.0025 0.14 0.870 0.60 0.549 1.3 0.273 7.0 0.0009 0.16 0.852 0.65 0.522 1.4 0.247 8.0 0.00034 0.18 0.835 0.70 0.497 1.5 0.202 9.0 0.00012 0.20 0.819 0.75 0.472 1.8 0.165 10:0 0.00004 Example No. 1. A coil having 1000 turns and 5 ohms resist- ance is connected to a source of constant potential of 100 volts. 10 ELECTRICAL ENGINEERING (a) Show at what rate energy is being delivered to the entire circuit and to the resistance. Show at what rate it is being stored in the magnetic field as the current is increasing after the circuit is closed. (b) What is the rate of change of the flux when the current is 10 amp.? Referring to equation (13), * T? (23) therefore the rate of energy supply to the entire system is Ei watts, and Ei + 10 8 dt (24) The current will begin at zero value and finally reach a value of E i = I = = 20 amp. 2000 1800 1600 "1400 1200 c fc 1000 400 200 Rate of Energy Supply to Inductive Circuit Constants of Circuits e = 100 Volts r = 5 Ohms N = 1000 Turns 8 10 12 Current in Amperes FIG. 4. 14 1G 20 The rate at which energy is dissipated in heat is i*r and the rate at which energy is stored in the magnetic field is: 3? - w - * v (25) The three curves in Fig. 4 show these rates. It is interesting to note that energy is being stored at the greatest rate when the current is one-half of the the final value. This can readily be proven by differentiation of equation (25) and equating the result to zero, thus, E - 2ir = 0, INDUCTANCE AND RESISTANCE 11 therefore . E_ / ~ 2r ~ 2* The rate of change of the flux as the current changes is obviously d _ E - ir ~dt ~ F X IF 8 ' Therefore when the current is 10 amp. the rate of change is 5,000,000 lines per sec. The rate of change is greatest at first and becomes zero when the current reaches its final value. The determination by calculation of the inductance L of a circuit is usually very difficult, in fact almost impossible except in the very simplest cases, such as parallel long circular con- ductors. Approximations of one nature or another have almost always to be resorted to. Usually the inductive circuit contains iron, and in that case the reluctance (and hence the inductance) is not constant but changes with the degree of magnetization. Later in this volume the effect of the changing inductance in iron circuits will be considered, but at present it shall be assumed that L is a constant regardless of the value of the current. The inductance of the field circuit of a dynamo can readily be determined for any particular field current by experiment. All that is needed is to run the machine at some speed and to read the voltage and field current. These data in addition to those of the field and armature windings suffice. By definition, total flux X turns current X 10 8 The total flux per pole is determined from the voltage, speed and armature winding. Consider a 10-kw., two-pole, direct-current, 110-volt generator, having 2.5 megalines of flux per pole, and 1500 field-turns per pole. Assume that at normal voltage its field current is 3 amp. and that the field spools are connected in series. Thus T 2.5 X 10 6 X 1500 X 2 or L = 3 x 1Q8 = 25 henrys. Example No. 2. Figs. 5 and 6 represent the direct-current generator referred to above. M is the armature and F the field. If a voltmeter of 11,000 ohms resistance is connected as shown and switch S is opened without arc when the field current in ammeter A is 3 amp., what will be the effect on the voltmeter and will the ammeter and voltmeter read in the same direction as 12 ELECTRICAL ENGINEERING before the switch was opened? Before the switch is opened the current flow is as shown in Fig. 5. As the switch is opened the field flux can not die away instantaneously. The field cur- rent therefore can not die away instantaneously, but continues to flow through the only available path, which is that of the volt- meter. Since the resistance of the voltmeter is 11,000 ohms it is evident that the voltage across the instrument becomes at the very first instant very high. FIG. 5. x~i r i t FIG. 6. It tends to become ir = 3 X 11,000 = 33,000 volts. Thus the voltmeter will probably burn out as the needle swings to the opposite side of the scale. The ammeter needle will remain stationary for the first instant and gradually come down to zero. This problem gives an idea of the nature of the shock that is experienced where the field current of a generator is carelessly interrupted and permitted to pass through a person. Depending upon the nature of the contact the resistance of a body may be from 1000 to 10,000 ohms. If, therefore, a person touches both sides of the field winding when the field circuit is interrupted, he will experience a very severe shock. The energy stored is usually quite considerable. In this case it is J^L/ 2 = J X 25 X 9 = 113 joules. Since 1 joule is 0.74 ft.-lb., the energy available is 84 ft.-lb., i.e., that of a pound weight dropping 84 ft. It may bs asked, what would happen if the voltmeter were not connected across the field winding? Where would the initial rush of current, of 3 amp. flow, when the switch was opened? In reality it is impossible to open the field switch without an arc; therefore the current can not be interrupted instantaneously. INDUCTANCE AND RESISTANCE 13 Furthermore the circuit is more complex than assumed. The field winding has considerable capacity and therefore acts as if it were shunted by a condenser. A portion of the 3 amp. will therefore flow as condenser current, but a large portion will appear as secondary currents in the iron circuit of the poles. This phenomenon will be understood later from the investigation of circuits having mutual inductance. The problem is instructive in that it explains frequent burnout of voltmeters, and in that it teaches that the voltmeter should always be disconnected before the switch is opened, or otherwise be connected on the armature side of the field switch. It teaches also that in opening the field switch a relatively low resistance should be shunted across the field winding to prevent high vol- tage, and finally that it is well to open the field switch slowly. The importance of shunting the field circuit is best illustrated by a numerical example. j Example No. 3 (Fig. 7). Assume that the field circuit having a resistance of 36.5 ohms is shunted by a resistance of 50 ohms, and assume again, for the sake of simplicity, that the field current of it \ r = 36.5 L= 25 FIG. 7. 3 amp. is Interrupted without arc and that L is constant at 25 henrys. The total resistance in the circuit is then 50 + 36.5 ohms or 86.5 ohms. Determine the current in the field winding and the shunted resistance and the voltage across the field coils which is the same as the voltage across the resistance after the switch is opened. Referring to equation (18) = 3e -3.46 For t 0.05 0.10 0.20 0.5 1.0 -3.45< 1 0.84 0.71 0.50 0.18 0.03 i 3 2.32 2.13 1.5 0.54 0.09 iR 150 116.0 107.0 75.0 27.0 3.0 14 ELECTRICAL ENGINEERING It is seen that in this case the maximum voltage across the field coils, which, of course, occurs at the moment of opening the switch, is 150 volts, as compared with 33,000 when the voltmeter shunted the field coils. The field current i dies away very rapidly. In 1 sec. it has almost disappeared. The energy stored in the field is spent in heating as an i z r loss. Example No. 4. Prove that in discharging an inductive circuit all energy stored is spent in heat. The instantaneous value of the current was found to be: i = Ie~l l , therefore the energy expended in heat from time zero to infinite time is: 2r J"<-- /* i*rdt = Pr\ =o Jo [T 2r -i oo T2 r -*'*]. --^ It is of interest to study the rate at which the field flux, or what is equivalent, the field current, can build up when closing the field winding on a constant-potential busbar, and to see how much more rapidly the field current can be made to build up when a considerable resistance is inserted in series with the field coils. It will be assumed that use is made of the winding described in example 3, that is, one with a resistance of 36.5 ohms and inductance of 25 henrys. This circuit is connected to a direct-current busbar having a constant potential of 110 volts. Referring to equation (17), i = l ~^ =3[1 -6- 1 - 46 ']. The lower curve in Fig. 8 shows the result of this calculation. If, instead of exciting the winding from a 110-volt main, it is connected to a 220-volt circuit and sufficient resistance is inserted in series to keep the permanent current at 3 amp., the rise in current will be more rapid than in the first case, as shown in the upper curve of Fig. 8. There is an interesting mechanical analogy for the starting or stopping of a current in an inductive circuit. To bring a train up to speed a certain force is necessary; this force must overcome the friction and provide the necessary acceleration. INDUCTANCE AND RESISTANCE 15 Let F be the total force necessary, and fv the force of friction and wind resistance which, for simplicity's sake, is assumed to be proportional to the velocity v, and the mass M. Then F = fv + mass X acceleration or, dv f F^ dt ~~ M v " M' .2 .4 .8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 Time in Seconds FlG. 8. If the drawbar pull F as well as the coefficient of friction / be assumed constant during acceleration, F f t -* where C is the integration constant 16 ELECTRICAL ENGINEERING . If the train start from rest, then for t = O, v = O. F F .'. O = j + C, or, C - -, . -[,-.- By comparing this with the equation for the starting of a current Fr r ~~i in an inductive circuit, which is, i = 1^1 e~ L * J,it is seen that in electrical problems, the current corresponds to Velocity, the e.m.f. to the mechanical force, the ohmic resistance to frictional resistance and the inductance to the mass. The analogy can be carried further. The energy stored in the magnetic field, %LI 2 , corresponds to the kinetic energy of a moving body, %Mv 2 . The electromagnetic momentum LI cor- responds to the mechanical momentum Mv, etc. A problem involving mechanical as well as electrical transients will next be considered. Find the equation of the dying away of the field current in a direct-current self-excited shunt motor disconnected from the circuit and permitted to decelerate to standstill. Let the moment of inertia of the revolving part be /. Let the full speed be N revolutions per second corresponding to an angu- lar velocity of 0:0 radians per second. Let the power required to run the motor at full speed but at no-load be P hp., and assume that this power is represented by friction loss in the brushes and bearings, which is a very close approximation, particularly after a few seconds of deceleration, when the core loss becomes very small; and neglect the i 2 r loss. Assume that the saturation curve is a straight line, so that proportionality exists between the field current and the flux. Let the normal field current be J . Let the normal flux per pole corresponding to this current be <. Let the armature e.m.f. at full speed and flux be E, and the total field-circuit resistance be r, and let the motor have p poles and each field spool have n turns. Mechanical Calculations. 1. Determine the angular velocity o. It is, a = 2irN. 2. Determine the friction torque, or moment Q. We have X lb. INDUCTANCE AND RESISTANCE 17 3. Determine the stored energy. In general W = ^Mv 2 , in the case of a revolving wheel; if p is the radius of gyration, W = %M (2ir pN) 2 = %Ia 2 , where / = Mp 2 = moment of inertia. Thus, with revolving masses, / takes the place of M, and a. of v. During deceleration, no external force or torque is applied. Thus, = Q + I -JT = torque of friction + torque of deceleration. (For sake of simplicity the small power given electrically is neglected.) da = -jdt,or,a = j t + C; for t = 0, a. = a . .'. C = a - Q. . . a. = ao -j t. If T denotes the time at which the rotor stops, then for t = T, a = 0. Ci ^ T ' T* = a ~jT, ..7 - Q O . And, o,_ / t\ \ l "" T) a = Oi jfj t = o Check whether all energy is spent at t = T, neglecting the sup- ply of energy from the diminution of magnetic field and the con- sumption of energy in heat. The stored energy is 3^/a 2 . The energy consumed by friction during deceleration is: I force X vel. dt = \ Q2* Ndt = \ Qadt = Q substituting, ^ _ / ~ Q Q:O ' = Q L>| o - 7 Y Q2 a 2 = ~' Q ' E ' D ' Electrical Calculations. If the field current remained constant during the deceleration (which it obviously does not), the arma- 18 ELECTRICAL ENGINEERING ture voltage at speed a would be:ei = E, so that due to the loss of speed alone, the armature voltage is reduced from E to E. If the field current is reduced from / to i, the flux is reduced from to 2 + Qi - y T, or, log - 1 - I , ^2 + Q r, a a . e = Eo - j- = ir, OLQ IQ Q!o a . T di . . ir = IT + L --1 20 ELECTRICAL ENGINEERING s+f ('-=)- ao which, transformed, becomes: where 4=14- v _ and I +(!-'>-. but when = 0, z = 7o; rA .". C = If the problem given above is modified, so as to include a windage loss at full speed of 0.15 hp. as well as the bearing loss of 0.72 hp., the constants are: r = 40, I = 0.25, L = 2l, Q 2 - - 0.00525, o = 125.8, Qi = 3.16. T becomes 9.04 sec., and A = 5.71, ^ = 10.60, K = 0.021, _. 7 (-10.60* +504.6 -504.6e--2 1 . . & i 06 In curve 6, Fig. 9, is given the relation between the current and time in this case. INDUCTANCE AND RESISTANCE 21 The problems considered up to this point have all involved very simple integrations. Frequently, however, this is not the case, and to solve the differential equations, it is necessary to make algebraic transformations. The most important of these transformations is to separate fractions into partial fractions. 3.0 2.5 1,2.0 1.0 0.5 (b 5 6 7 Time, Seconds FlG. 9. 10 11 Almost any algebra deals with this; nevertheless it may be opportune to refer to it briefly here, although it is suggested that the student's memory be refreshed by reading, for instance, WILLSON'S " Advanced Algebra," from which the following is largely abstracted. If FJ-\ is a fraction, that is, the numerator is of lower degree than the denominator. It is known that F(x) can always be expressed as the product of linear factors, which are not necessarily real. If the factors are real, then F(x) can be expressed as the product of real linear and quadratic factors. Two cases will be here considered. First. No factors are repeated. Example. F(x) = (aiX + &i) (a& + & 2 ) (a$x 2 -f- & 3 # -f- C 3 ). Then A l -f F(x) aix + bi a 2 x -[- 6 2 a 3 x 2 + 6 3 a; + where AI, A 2 , A 3 , and B 3 are constants, which can readily be found, since if the expression: 22 ELECTRICAL ENGINEERING aox n -f aix n ~ l + a 2 x n ~ 2 + - i = b x n + to"- 1 + b 2 x n ~ 2 + holds for all values of x, then the coefficients of like powers of x must be equal, thus a = &o, i = &i> etc - Show that * 2 l 1 2 5 4z + 5 Second. Some factors of the denominator are repeated. F(x) = (aix + bi) 2 (a 2 x + 6 2 ) (3^ 2 + to + c 3 ) 2 Then + Prove that 2x* + b 2 7 15 (x - l) 2 (x - 3) 2 (x - I) 2 ' 2(x - 1) ' (x - 3) 2 2(x - 3) The application of this transformation is found in any transient phenomenon in which disproportionality between magneto- motiveforce and resulting flux exists. As an example the condition governing the self-excitation of a direct-current shunt-wound generator will be considered. (For a more detailed discussion see STEINMETZ'S " Transient Phenomena.") It will be assumed that the relation between the flux

. Then e c The e.m.f. e corresponding to any other flux

(K KHP) ek r k^e c (p k e c k r k\r e c k r Integrating each term we get after a slight transformation ~ - ~ '08 * - If at the time of closing the field circuit the remanent flux is

/ i x ke c log r log ; , 100e c (e c k r) L & e r & e c k r kie r l The voltage ultimately reached is e = e$ when t = . Thus , e c k r n/ioo loe; r i = oo thus P If - -- 7* A % -|P e c fc r e c /c r K\en = or eo = ; The greatest value of r which gives a positive value of e r = e c k. The condition of self-excitation is thus r ^ e c k. 24 ELECTRICAL ENGINEERING Up to this point, the problems have involved inductive circuits, on which a direct-current e.m.f. has been impressed. In case of alternating current the impressed e.m.f. varies from instant to instant and, while a harmonic e.m.f. is usually assumed, fre- quently the variation represented by a wave is much more com- plex. As long, however, as the e.m.f. is obtained from a dynamo of symmetrical poles, no matter how shaped, the e.m.f. wave can be expressed by a series of sine functions of odd frequencies. In the study of transient phenomena in connection with alter- nating current, the equations are derived for the fundamental wave only, that is, the instantaneous values of the e.m.f. are represented by e = E sin 6. If it is desired to know the result with distorted waves, the simplest method is to treat each harmonic independently and to add the instantaneous values so obtained. If the effective value is desired the square root of the sum of the squares of the effective value of each wave should be taken. As stated previously, the instantaneous value of the e.m.f. is generally expressed in two ways, either e = E sin coi or e = E sin 6, or the expression may be of more general form : e E sin (at -f a) and e = E sin (6 + a). In these expressions, e is the particular value of the e.m.f. at time t, or at phase angle 6, and E is the maximum value of the e.m.f. In the first case, the angle ut is expressed in radians, not in degrees, w is the an- gular velocity = 2 TT/, where / is the frequency. The relation between radians and degrees is 360 = 2ir radians, thus 1 radian is -77 = 57.3. To reduce equation e = E sin (cot + ) to ZTT degrees it should therefore be written e = E sin (57.3 ut + a), where in all cases a is expressed in degrees, as is customary. To reduce the expression to radians it should be written Note in connection with this that in the expression, y = sin x, x is expressed in radians, not in degrees. To bring it to degrees the equation becomes y = sin 57.3 x. In the development the value of the sine function Sin x = x - I + I - ^ + x is again expressed in radians. INDUCTANCE AND RESISTANCE 25 It is important to have this clearly in mind. It is well worth while to plot some curves of distorted waves from equations in- volving phase angle as well as radians. Example No. 5. Verify the e.m.f. wave in Fig. 10, e = EI sin ut + Es sin (3 ut + a) for E l =-- 10, E 3 = 5, a = 30 and the frequency 25 cycles. 1.2 1.0 8 G 4 2 2 4 G 8 1.0 1.2 / t f ~\ \ \ \ / / \ V 7 \ / \ / .4 .6 .* 1 JO 1 20 1 160 U80 200 220 2 Angle\in Degrees 10 2 2 50 3 30 3 20 3- ^3 \ / \ /^ / \ / / , V^ / \ t \ ^2 .01 .02 Time in Seconds FIG. 10. .04 Prove by integration that with a distorted wave : e = EI sin (ut -f i) + E 3 sin (3 cot + 3 ) the effective value is e// = "\/6i 2 e// + ^3 2 e// Thus in this instance, since the effective value of the funda- ~F 10 mental wave is ^ = 7= = 7.07, and that of the triple har- monic is 7= = T= = 3.53, the effective value of the wave V2 recorded by a voltmeter is e = \/7.07 2 + 3.53 2 = 7.9. Referring to Fig. 11: prove that ammeter A when placed in a circuit carrying 10 amp. direct current, 8 amp. 60-cycle current, and 5 amp. 125-cycle current reads 13.7 amp. 26 ELECTRICAL ENGINEERING Harmonic e.m.f. Impressed on a Circuit of Resistance and Inductance in Series. Let time be counted from zero value of the impressed e.m.f. and let the e.m.f. be rising. Thus e = E sin ut where e is the instantaneous value of the harmonic e.m.f. at time t. E is the maximum value, co = 2 TT/, is the angular velocity, / the frequency, r the resistance and L the inductance of the circuit. 10 Amperes, D.C. 8 Amperes, 60 Cycle 5 Amperes, 25 Cycle FIG. 11. If i is the instantaneous value of the current when the e.m.f. is e then: e E sin cot = ir + L -77 at or di r . E (27) (28) By comparing this equation with equation (2), it is seen that T V T = P and j- sin ut = Q. P is not a function of the independent variable t, but Q depends thereon, thus the solution is given in equation (8). It is (29) i = e L l \ I e + L Y sm ut dt + C The solution of this equation depends upon solving C + r -tE , E r ,r t I ^L jr sm wt dt = Y~ I L sm <*t dt. E. j- is a constant and can be left out of consideration at present. INDUCTANCE AND RESISTANCE 27 It is also convenient to substitute a single letter for Y' Let then L The immediate problem then is to solve ft at sin co dt. An integral involving exponentials or sine functions is usually easy to solve, because the differential of the functions are similar to the functions. If y = ax then ^ = ae ax . dx Similarly if y = sin ux, then -5- = co cos coz, or if y = cos ux, then - = co sin cox. Thus i at , _ j. at Ui f and / . , 1 , sin co* dt cos cot at. J w Fortunately for the engineer there are only very few methods of integration that need to be known. One of these is " Integration by Parts." That is: fudv = uv - fvdu (30) In integral ft* sin to* dt, let u = e at and dv = sin co* dt. .'. du = ae at and v = cos co* CO /. yV* sin co* dt = - cos cot + \ - e at cos co* dt (31) CO J CO This equation is indeed more complicated than the original. It is evident, however, that by again integrating the last term in 31, an integral results which contains an exponential term e at and a sine term instead of the cosine term. Thus the final expression will contain integrals of the same trigonometrical and exponential functions, which therefore can be solved directly. However, it is somewhat more convenient to use another method. Referring again to (30) let in this case: u = sin co* and dv = e at dt .'. du = co cos co* dt and v = - e at a. .'. ft at sin co* dt = sin co* - ( - e at cos co* dt (32) a J a 28 ELECTRICAL ENGINEERING By multiplying 31 by - and 32 by and adding the two equa- OL Ct> tions, it is readily seen that ./V" sin ( dt = ^-. (^ - ^-') (33) co- + a- \ co a I T Substituting = r and remembering that x t the reactance cor- responding to the inductance L is 2irfL = coL and that the impedance z = \/r 2 + x 2 . Then I L * sin wt dt = e + L -^,[ r sin coi x cos coi] (34) Let the angle of lag of current be ft thus tan ft = - and r = z cos ft (35) x = z sin ft (36) Substituting the values in 34: -t + r t L sin coi dt = e L - sin (cot ft) (37) z Referring to equation 29 E -L t i sin (cot ft) + Ce L (38) The integration constant C is determined from the particular problem under consideration. Assume that it is desired to find the value of the current at any instant after the switch is closed and the alternating e.m.f. is impressed upon the circuit, and that the switch is closed at time t = i b when the instantaneous value of the e.m.f. is e = E sin coii. Since, as has previously been discussed, it is impossible to establish a magnetic field instantaneously, the current can not flow at the first instant. Thus for t = t\, i = 0. Substituting these values in equation 38, then: 7? r = - sin (coii ft) + Ce~L tl , E Z l .'. C = e L sin (coii /3), Substituting this in (38) sin (coi ft) e * lfa) sin (uti ft) (39) i =-- ~\s INDUCTANCE AND RESISTANCE 29 It is often convenient, to eliminate t entirely from the expres- sion and to use the phase angle 8 only and to express 6 in degrees. That is, the e.m.f. is expressed as e = E sin 6. In that case B = ^t = 27T/7. The exponential term 6~ a ~' becomes ~^ (fl " 0l) = ~* ((? ~' l) T_ (9-fr) if 6 and 6\ are expressed in radians or & 57.3 if and 6\ are expressed in degrees. Thus when 6 and d\ represent degrees Er - r <-_*> -] i = -I sin (e - (3) - e * 57.3 sin (0i - /3) (40) The equation is, however, always written i = | [sin (0 - i?) - ~* "-' !) sin (0! - 0)] (41) and it is understood that the exponential term should be ex- pressed in radians. Equation (41) can, of course, be derived directly by using the phase angle instead of cat. Thus T di E sin ir = er + L -7- cfo may be written E sin - ir + x ^, (42) where x is the reactance. Thus, x = 2-irffj = coL and cot = 6. .". d# = codt or dt = - CO Prove that equation 41 is the solution of di E sin 6 ir -\- X~TC, The exponential term in equation (41), while of importance during the first second or so, ceases to affect the result very shortly after the switch is closed. Thus the equation for the current after the system is stable is i = E z sin (8 - 0) (43) The current lags behind the e.m.f., E sin 0, by an angle 0, whose x tangent is - 30 ELECTRICAL ENGINEERING The effective value of the e.m.f. is E = V5 and of the current j E ~~ V2Z It is of interest to note that the transient term is a maximum when sin (B l - /3) = 1, that is 0i - = 90, or 0i = 90 + /3. This value of 6\ also gives the maximum value of the perma- nent current. 1.4 1.2 1.0 .8 .6 .4 .2 -.2 -.4 -.6 -.8 -1.0 -1.2 -1.4 I 1 / z I /' 6 L Impressed Voltage ' Current when Swi ) " " ' ft t 1 v tch is Closed at 1 \ k & 270 79 Giving tli Value of - = = - .',' ,',' ,' e Permaaeu Current ^ > 7T- ~r s y S J Mill / y \ 1 LU= x A - s* / S x / ^s^ OS. s^ ~^^ \ / // / ^ s\% / \ k / / y / / s i: \\ \\ / w/ \ \s i / y , / / \ X \\ ^ / //' 7* / \ 5 / ^ / / / / \ \ \ \\ ^ / 420^ M 7 \ \ j W_ iKT 1201 1501 1 5Q 2 ^ ^ \\\ ,300 330 1 3U 390 l //// /^53 4i 5 5 w 5 6 An glei a Degrees \ Y ^U |y 1 \ y//// \ \ V V^k/ //' ^// ^ \ s \ _^^: -^// ^ > X. >s ^K t '// ^ X \ ' ^S s\ "v ? FIG. 12. The exponential term is zero, that is, there is no transient effect if #1 j8 = 6 or #1 f3 or, in other words, if the circuit is closed at such a time as would give zero value of the permanent current. Fig. 12 shows a series of such transient currents. Each curve corresponds to the closing of the switch at a particular value 0i of the phase of the e.m.f. Thus, for instance, curve D shows the starting current when the e.m.f. wave has a phase angle of +60, that is, when 0i = 60. These curves are calculated with the following constants E = 1 r = 0.196 x = 0.98. Problem No. 6. Check some curve in Fig. 12. It is of interest to study the rate at which energy is being sup- INDUCTANCE AND RESISTANCE 31 plied at any instant. This is equal to the product of the e.m.f. and the current: p = d = Es'm B X |[sin (6 - 0) - e~x (0 ~ dl} sin fa - 0)] (44) By simple transformations the equation becomes - cos (20 - 0) P : -^(0-00 -I - e * sin (0i 0) sin 0J (45) A -Impressed Voltage - Power Input C- Permanent Power FIG. 13. The first term in equation 45 must represent the power at any instant after the conditions have become stable. This power is expressed by E 2 P 1 = 22 [cos - cos (20 - ft] (46) It consists of two terms, on e a constant term ^ cos /3, the other a term which changes with double frequency; the net result of which over a complete period is zero, since the positive values are as large as the negative. Thus while the instantan- eous values of the power vary from instant to instant and may 32 ELECTRICAL ENGINEERING alternate from positive to negative values there is a definite average power delivered, which is E 2 P = 2z cos P The exponential part of the power, P 2 = - e ~ r x (e ~ 9l) sin (0! - 0) sin (47) is gradually decreasing in magnitude as well as oscillating at normal frequency. In Fig. 13 are given three curves; the first, A, is the wave of the impressed e t m.f.; the second, B, the power input; and the third, (7, the power curve after conditions are stable. These curves have been based upon the constants given in problem 6 and are well worth reproducing by calculation. The curves show that during the transient period the instant of maximum power is practically the same as that for permanent condition. They also show that the first rush of power is greater than that which corresponds to permanent condition, the reason being that the change of flux during the first part of the cycle is greater than during the corresponding time under stable condition. CHAPTER II PROBLEMS INVOLVING MUTUAL INDUCTANCE Up to this point the problems considered have dealt with cir- cuits of inductance and resistance only. However, in many circuits of commercial interest there are secondary circuits which are more or less closely coupled with the primary, and which influence the former materially. As instances of such circuits may be given the secondary winding of a transformer, the eddy currents in pole pieces of generators and motors, induced cur- rents in telephone lines running parallel to transmission lines, etc. Sometimes the secondary circuits carry currents by virtue of impressed e.m.fs., but frequently the currents are the result of the action of the primary currents. With a change of primary current obviously there is a change of the flux produced by the current and if this flux interlinks with the second circuit, e.m.fs. are induced therein, the values of which become higher as the interlinkage becomes more nearly perfect. While it is impossible to arrange two circuits so that all flux interlinking one will also interlink the other, the condition can be approached reasonably close under the most favorable conditions. The limiting case is, of course, perfect mutual induction, which condition will therefore first be considered briefly. Two Coils of Perfect Mutual Inductance. Assume then that it is possible to place two coils so close together that there is no leakage flux between them, that is, so that all flux that surrounds one coil also surrounds the other. Let the first coil, the primary coil, have NI turns and r\ ohms resistance, and the secondary coil N% turns and r 2 ohms resistance. Determine the open-circuit voltage of the second winding. When the first is connected to a source of constant potential E, we have obviously: F 'r r The rate of change of flux is thus d4> E i dt ~~ A 33 34 ELECTRICAL ENGINEERING Therefore the voltage of the second coil e 2 is N z d4> N 2 ~WTt~ ~Nl (1 At the instant of starting, when ii is zero, the secondary voltage is e 2 = - - -rp E, that is, it is proportional to the ratio of turns. 7^ When the primary current reaches its constant value J = the secondary voltage e 2 is zero. If the secondary winding has more turns than the primary, then at first the secondary voltage is higher than the impressed voltage. It decreases rapidly, however, and soon becomes zero. Prove that the two voltages are equal numerically when Assume that two coils, which, when considered alone, have resistances and inductances of r\, r% and LI, L 2 , respectively, are placed so close together that there is perfect mutual inductance between them (which of course is in reality impossible). Find the open-circuit voltage of the second coil if the first coil is connected to a source of constant potential. In the primary we have: E = i fl + L, J- The counter e.m.f. of self-induction of the primary coil is - LI -j7 and thus the voltage of the second coil is N 2 dii Check the values of the primary current and secondary voltage as given in full lines of Fig. 14. for E = 10 n - 0.10 L! = 2.5 NI = 10 r 2 = 0.50 L 2 = 10 N 2 = 20 In the case referred to above the primary current will rise from zero to a final value of 100 amp., while the secondary voltage decreases from 20 volts to zero. If when the primary current has reached its final value the coil is suddenly short-circuited, what will the primary current and secondary voltage be? PROBLEMS INVOLVING MUTUAL INDUCTANCE 35 The primary current will decrease according to equation: Check numerically the two dotted curves in Fig. 13. During the discharge of the primary the number of coulombs are | hdt = ( Jo Jo = 100 = r i 2500 coulombs. 1UU 90 80 70 S 60 r < 40 30 20 10 -2 -4 -6 s - s 3-10 p> -12 -14 -16 -18 20 \ I E - \ _^, -i ^ -^ s ^ ~- N / * * \ / Two Coils of Perfect Mutual Inductance A-Primary Current B- Secondary Voltage \ ^/ \ / - / / S N X / \ ^ / > / a / 1 ri 2 m ) V o or a s _4 50 \ e^ n S 2C ! \ J- ^ ^ ^ ^Z ? k^ ^ / / / \ / / / ~-~* / x j x^ / N / *s / ^ ^ ---, ^ / FIG. 14. Obviously, when connecting the primary to the source of supply, the number of coulombs required up to the time when the current becomes stationary is infinite, since it takes infinite time for the current to reach this value. Two coils of resistances and inductances of ri, r and LI L are connected in series and placed so close together that it is assumed that they have perfect mutual inductance. What will be the resultant resistance and inductance (a), if the coils are wound 3 36 ELECTRICAL ENGINEERING in the same direction; (6), if the coils are wound in opposite directions? The inductance of an air coil is subject to rigid mathematical determination, but the complete solution is very cumbersome. However, one of the best approximations, that of BROOKS and TURNER, published as an Engineering Experiment Station Bulletin by the University of Illinois, is: L = cm- IQfr + 12c + 2R 10 9 (6 + c + R) ^ 106 + lOc + 1.472 X 0.51og 10 (lOO + For coils which are not extremely thin or extremely long, this equation becomes approximately: L = 7 cm" (2) (b + c + Where L is expressed in henrys cm = centimeter length of wire b and c are the height and thickness respectively of the coil and R the outside radius, all in cm. 00000 OO( 00 f ooooo ooooo 00< oo< 00 00 ft ooooo ooc 00 ooooo oo< oo i R --> 1* ' ' FIG. 15. The maximum inductance is obtained when b = C and R (see Fig. 15). Then 2C L = 0.27Cm C X 10 9 henrys. It is seen that the inductance is proportional to the square of the total length of wire, which is, of course, proportional to the turns. Thus the inductance is proportional to the square of the number of turns, or L = KN\ (a) Coils in the same direction. PROBLEMS INVOLVING MUTUAL INDUCTANCE 37 Let N be the number of turns in the first coil, or, N = and NI the number of the turns in the second coil, or, JLl r The total number of turns in the two coils when considered as one coil (which is permissible when perfect mutual inductance is assumed) is The combined inductance is then Lo = KN 2 = (VL + VLi) 2 = L + Li + 2-v/LLL The resistance is obviously r = r + r\. (b) By similar reasoning it is found that if the turns are in opposite directions Lo = L + LI 2\/LLi and r = r + ^i- From the above it is evident that the equation for the starting current, for instance, is: Two Coils of Perfect Mutual Inductance Connected Simulta- neously to Sources of Constant e.m.f s. E and EI. Let r, r\ and L, LI be the resistance and inductances respectively, and assume that the circuits are closed at the same instant. Assume first that the coils are connected in the same direction, that is, in such a way that the permanent current in both coils will produce mag- netic fields of the same polarity. It is evident that in this case the impressed e.m.f. has to overcome not only the resistance and in- ductance drop due to the current in the coil, but also the e.m.f. which by transformer action is induced in one coil by the change of current in the other. Consider one coil alone, for instance the second coil: The counter e.m.f. of this coil is i~5T' If it has N\ turns, the voltage per turn is jj- rr' Since it has been assumed that /V i u/t 38 ELECTRICAL ENGINEERING there is no leakage field between the two coils, it is evident that this same voltage per turn is induced in the first coil by the cur- rent in the second. Thus the " transformer " e.m.f. in the first coil having N turns is -^- LI -rr, and similarly the transformer e.m.f. produced in the second coil by the current in the first is Ni T di ~W L dt But N IT. therefore the e.m.f. in the first coil caused by the mutual flux is L T dii _ ^/rr dil ' LI ^ ~ VLLl dt Thus it is seen how, when the mutual inductance usually denoted by M is perfect, M = \/LLi. In reality M is always smaller than \/LLi. The general equation dealing with e.m.fs. consumed by resistance, inductance and mutual inductance, are then To solve for instance i the following transformation is conven- ient, multiplying (3) by LI and (4) by M and add the equations so obtained. li is: LiE - ME, = Liir + LL l ~ at - Mi.r - M 2 ^ (5) Since with perfect mutual inductance M 2 = LLi (6) Liir - LiE + MEi dii _ LJT - (9) To determine the integration constant C, it would be a mistake to assume that the current i is zero when t = 0. All that is known is that the combined coil can not be surrounded instan- taneously by a flux it takes some time to produce or alter a magnetic field, because a transfer of energy is involved. It is possible that currents will flow the very first instant, currents which produce m.m.fs. of equal magnitude but in opposite direc- tion. One particular case of this would be where the currents were zero, but this is not a general solution. What is known, then, is that no flux will exist the first instant. Thus the m.m.fs. must be equal and opposite, and since the cross- section of the magnetic flux and the direction of the turns are assumed the same in both coils, it follows that for or N . . IL * i = -Ni iu "Ste Substituting this value in equation (7) LiE + (ID Mr, or L\EJ " Lri + for t = 0. i E ' r ~ _ MEir . . \j r(Lri ME 77 r(L Similarly ii is found to be . E MEjr + LErt , t , 10 , I = --- 77 = - r- 6 Lri+Lir (12) r r(Lr l + Lir) (13) i r 1 (Lr 1 +L 1 r) Problem No. 7. Prove by complete calculation that if the terminals of the second coil are reversed the following are the equations of the currents 40 ELECTRICAL ENGINEERING E LEr l - I = r r(Lr l r- 6 Ln+Lir Lri+Lir (14) (15) In the case that the two coils are excited from the same direct- current busbars when E = E\ the equations become : For coils wound in the same direction: E + i+Lir m Ln+Lir For coils wound in opposite direction: Lri Mr m t ll 1 Lri e Ln+Li 100 CO Starting Currents Two Coils of Perfect Mutual Inductance A - Coil No.l B - Coil No.2 D - Coil No'2 ) Wound PP site Directions Wound in same Direction (16) (17) (18) (19) FIG. 16. In Fig. 16 are given four curves showing the currents in two such coils of perfect mutual inductance, having the following constants : r = 0.10 ri = 0.50 L = 2.5 Li = 10.0 E = E l = 10 volts. PROBLEMS INVOLVING MUTUAL INDUCTANCE 41 It is assumed that they are connected in parallel to the same source of direct current at a constant potential of 10 volts. The full-drawn curves correspond to the condition in which the turns are in the same direction; the dotted curves to that in which the turns are in opposite directions. It is well to verify these curves by calculation. It is of interest to note from the full-drawn curves that, while the two coils are connected to the same source of con- stant potential, during the first few seconds the currents actually flow in opposite direction. The second coil having twice as many turns as the first, and therefore a smaller final value of current, has a current of negative value at the first instant of one-half the magnitude of the current in the first coil. Eventu- ally the currents become positive and are proportional inversely as the ohmic resistances. It is of interest to deduce the equations of the currents in the two coils when the first is connected to a source of constant poten- tial, and the second is short-circuited upon itself, as shown dia- grammatically in Fig. 17. FIG. 17. CO 40 20 Starting Current Two Coils of Perfect Mutual Inductance A - Coil No.l B - Coil No.2 FIG. 18. Prove that with the coils wound in the same direction: 11- - Lri r MEr Ln +Lir t TTr* 1 Ln+Lir (20) (21) 42 ELECTRICAL ENGINEERING In Fig. 18, which gives the values of the currents, it is of interest to note that the current in the second coil, under this condition, remains negative and approaches a value of zero. The initial values of the currents are twice as great as before. Thus the impedance is greatly reduced, as would be expected by the pres- ence of the short-circuited winding. As a further illustration consider: Two similar coils having perfect mutual induction and calcu- late the currents in the two coils when a sine wave of e.m.f. is impressed upon one coil while the other is short-circuited. Referring to Fig. 19 t , r, x The equations evidently become: di di\ di FIG. 19. E . di dii E - sin e and ^- - ^ = cos (22) Ex ~ and dii E dB 2x Zl = ~ 2~r COS i = - e ~ cos d0 -f- Ce - ~ - cos (6 - ?) + Ce L 9 /JQ I (23) where Z = 2 + (2x) 2 . The condition determining the integration constant is that when the switch is closed no flux exists, thus i ii. Let then the switch be closed when = 0^ Thus from (22) and from (23) E - 2ii = sin 0! -yr ~ cos (0! - ^) + JQ I (24) (25) PROBLEMS INVOLVING MUTUAL INDUCTANCE 43 From (24) and (25) = ^ X ^ [f5 COS (Sl ~~ ^ ~jr sin ' > """ -T 1 ] The transient term disappears when x sin 0i y- COS (01 (p) = Expanding and substituting it is readily seen that this occurs when tan 0i = -, that is when 0i = - 2)y = cosx (D \}u = cos x du -j -- u = cos x dx u = t x j*e~ x cos x dx + Cie x but e x j^e~ x cos x dx = % (sin x cos #) by simple integration (D - 2)y = y^ (sin x - cos x) + CV = X l dy (sin s - cos x) dx ~ 2y ~ ~^T Cie = Xl y = S*fe-**Xidx + C 2 e 2 *. The three integrals involved are readily solved and the result s cos a; snx - - 3 - In many cases a simpler integration is obtained by the following method which involves the breaking up of a fraction in partial fractions. 48 ELECTRICAL ENGINEERING It is known from algebra that if the equation given below holds for all values of x then the coefficients of the like powers of x are equal. a Q x n + aix n ~ l + a n = b x n -f- bix n ~ l + b n Thus a = b 0,2 = z Equation -j 2 + a -5 h fa/ = ^ can be written (D - mOCD - m,)?, = X or = But it is known from algebra that the fraction 1 A B (D - mi) (D - w 2 ) D - mi D - m 2 where A and 5 are to be determined. _ 1_ = A(D -m 2 ) +B(D -mi) (D - mi) (D - m 2 ) ~ (D - mi) (D - m 2 ) This equation shall hold for all values of D. Rearranging the equation we get: 1 D(A + B) - (Am 2 + (D - mi) (D - m 2 ) ~ (D - mi) (D - m 2 ) On account of the identity the coefficient for D must be zero and the constant terms must be equal, thus and Ani2 + Bm\ = 1 B = - - - and A = m 2 mi m 2 mi m 2 mi D mi D m 2 _JLJr_^__ i 1 mi m 2 LZ> mi D m 2 J Let y = u -\- v u = ^^ 1 D^ 1 (36) and 1 X ~ PROBLEMS INVOLVING MUTUAL INDUCTANCE 49 Equation (36) written out is du _ X dx u = Similarly v = - - -ft-** Xdx The general solution is thus y = Cie m i* + C 2 e m 2* + - ,m 2 x " <-*** Xdx. mi m z When X is a constant = K The solution evidently becomes: ^ = C^ mx + Cuc w * + y- When X is zero the equation is called the complementary func tion and its solution is found by making K = in the above. The solution of the equation, is y = Cie mlx + C 2 e w 2 a; . Before leaving the subject it is necessary to discuss the values of mi and m 2 which involve a square root in (33) and (34) which might be real, imaginary or zero. a 2 (a) The square root is real, that is -j > 6. We have shown then that mi and mz depend upon the auxiliary d^v du equation -j 2 + a -r- + by = and that the solution of this equation is, OL' (b) the square root is imaginary, that is b > - 50 ELECTRICAL ENGINEERING "fVlOTl / & . -4, / O \ / d . >* /. Ct \ ,, = rJ- + '\*-i> -4-c*(~2 -'\ 6 --4> where a = A/6 - -j' But e' ox = cos ax + j sin ax . ' . CV * = Ci cos ax -|- Cij sin ax . * . C 2 e~ ;oa: = C 2 cos ax Czj sin ax y = ~ a z [(d + C 2 ) + cos ax (Ci - C 2 ) sin axj]. In order that y shall be real it is necessary that C\ + C 2 and j (Ci C 2 ) shall be real, in other words, Ci C 2 must be im- aginary. Let A = Ci + C 2 and B = j(d - C 2 ) y = t~ 2 [A cos ax + B sin ax] (38) = Cie~f sin (ax + C 2 ) (39) where and tan C 2 = j} m (c) the square root is zero, that is -j- = 6, or m\ = m 2 This is not a complete solution of the complementary function because we have only one integration constant. The equation can obviously be written. (D - mi) 2 ?/ = = (D - mi}(D - mi)y = O (40) Let (D - m,}y = v (41) Then 40 becomes (D - mi)v = or dv -; m\v = O ax PROBLEMS INVOLVING MUTUAL INDUCTANCE 51 From (41) (D - mi)y = or f x - mi y = y = e m l a: [/V w l*C or i* [C lX + C 2 ] (42) Two Coils Having Resistance, Self -inductance and Imperfect Mutual Inductance Constant Impressed e.m.f. Let the constant e.m.f. impressed on the first coil be E and that on the second coil EI. Let their resistances and inductances be respectively r, TI and L, LI and let M < LLj. It follows that E = ri + L -j \- M -r^ (43) and dii , ,,di E 1 = ^ 1 rl + LI -j h M -r- (44) Differentiate (44). From (43) is found Substitute (46) and (47) in (45) and arrange the equation with reference to the derivatives. /72/ J 7 ' /. ^ (LLi - M 2 ) + -ft (Lir + Lri) + irr, = Er, (48) Or dH vLir + Lri~\ di dt 2 ' LLLi - M 2 J dt ' LLi - M 2 = zzr--ir 2 (49) E7 Similarly ii = ~ + BI*** + Brf- (51) 4 52 ELECTRICAL ENGINEERING where mi and m% are the roots of the equation. Lir -f Lri rri m + LL^W* m + LL^-W - (L,r + Ln) + VW-LrO' + Wrr, It is evident, from the factors under the square root sign, that in this case the two roots are real. Thus the solution is L\r -f- Lri - mi= 2(LLi - ,. and Lir + Lri , , 2 The integration constants AI, A 2 and J3i, 7? 2 are readily de- termined, since in this case (where the mutual inductance is not perfect) currents can not flow without producing some flux, and thus, since the establishment of flux requires time, the currents can not appear instantaneously. Therefore at t = 0, i = ii = 0. Referring to (50) and denoting the final current (where / == ^) (56) by = 7 + ^,4-^2, or A 2 = - (A l + 7) we get i = I + Aiei' - (Ai + I)e m * (57) and ^ = I I + B^" 1 * - (B l + 7i)e^ (58) These equations still contain the two unknown quantities A 2 and B 2 . To determine them, multiply (43) by LI and (44) by - M. L,E = L = - L^M (66) m z = - L T _ M (67) AI = #1 is found from equation (62) by substituting these values. E _ thus A 2 = B 2 = 0. Referring to equation (57) : (68) This shows that the mutual inductance acts as self-inductance. It is also evident that if the two coils are wound in opposite directions the circuit is almost non-inductive. It would be non- inductive if M = L; that is, with perfect mutual inductance. It is of particular interest to study the relations of the currents in two such identical windings inductively related when one is supplied with current from a source of constant potential and the other is short-circuited. 54 ELECTRICAL ENGINEERING It is well to deduce the equations from the two general ex- pressions: and . T dii . , f di = t,r, + L, ^ + M % However, it is evident that having once determined the general equations (57), (58), (62), (63), (64) and (65), it is possible to give the equations for the case in consideration by putting #1 = 0; that is, T _ I _ = n = i = I + Aie"* -. (Ai + 7)e m (69) and n = Bi m i* - #ie w 2< = B! ( i< - *) (70) Referring to equation (62) and substituting equations (66) and (67) and (71) Referring to equation (64) and making similar substitutions we get M' J (72) It is evident that these equations do not lend themselves to the limiting condition M = L, on account of the assumption made in determining the integration constants; that is, that leak- age flux exists between the two coils. To get these values, equations (20) and (21) should be used. In Fig. 20 are given some very interesting curves which show how the current in the short-circuited winding depends upon the leakage flux between the windings. These curves represent the conditions of two identical coils having a resistance of 0.10 ohm PROBLEMS INVOLVING MUTUAL INDUCTANCE 55 and an inductance of 2.5 henrys, placed at various distances apart so that the mutual inductance is M = L in curve a, M = 0.9L in curve b, M = 0.7L in curve C, M = 0.5L in curve d, and M = 0.1L in curve e. -4 2 -1 10 20 30 40 50 70 80 90 100 110 120 130 140 150 160 170 180 Time in Seconds FlG. 20. One of the coils is connected to a source of constant potential, e = 1 volt, while the other is short-circuited. Prove that the time for the maximum value of the secondary current is: Z^-M 2 L + M Iog L - M' CHAPTER III 10000 8000 CIRCUITS OF RESISTANCE AND VARIABLE INDUCTANCE In the discussions given so far it has been assumed that the inductance L has been constant. In almost all cases of interest to engineers this is, however, not the case because almost all magnetic circuits contain iron, and the permeability of iron is not constant but depends upon the magnetization. In other words the flux produced by a given current is not pro- portional to the current. Fig. 21 gives the saturation curve of an entirely closed iron magnetic circuit, as shown in Fig. 22. It is the familiar hysteresis loop, which shows how the magnetism lags behind the m.m.f. pro- ducing it. This particular sample has a remnant magnetism of 7600 lines per cm. 2 , so that this density corresponds to an exciting current of amp. The maximum density is 10,000, which corresponds to an exciting current of 4.5 FIG. 21. amp. If, after the maximum density is reached, the current is gradually reduced the rela- tion between existing current and density is found in curve a. The flux does not disappear until the current is 2.6 amp. in op- posite direction to the original 4.5 amp. If, instead of being entirely of iron, the magnetic circuit con- sisted partly of air circuit and partly iron (Fig. 23), the influence of the air circuit would as a rule be so much greater than that of 56 -8000 -10000 CIRCUITS OF RESISTANCE AND INDUCTANCE 57 FIG. 22. FIG. 23. the iron that the shape of the saturation curve would become materially modified. Thus the saturation curve of a dynamo, having a magnetic circuit largely of iron but also of at least a small air gap, can be represented by a set of curves similar to those in Fig. 24. If the air circuit is very small the two curves corresponding to a and b in Fig. 21 can be observed. If the gap is reasonably large the two curves merge into one as shown in the dotted line. FROLICH evolved an equation of such a saturation curve for a magnetic circuit consisting partly of iron and partly of an air gap; which, modified by KENNELLY, can be written thus ki : T+fo# where is the flux corresponding to an exciting current of i amp. 4 6 Current in Amperes FIG. 24. If the number of turns of the exciting winding is known then the inductance for any particular value of current i can be determined. It is N(f> -. where N = number of turns. :.L = A T /clO- 8 i + kj The general equation thus becomes: d , ^ di r NklQr 1 + kii dt iNkkMr* di (1 - kii) 2 dt (73) 58 ELECTRICAL ENGINEERING To solve this equation the variables are separated di or / *_ _ 10^ I 7 *\ 9 "\7*7 r J (e-i The integral is solved by breaking up the fraction into three fractions C 6 - ~ h i i T ~h The constants can be readily found and the integration carried out without the slightest difficulty. A becomes -= a 2 B becomes r a 2 and C becomes a The result is: Nk rr (1 rr l + /dz loe; e : h La e ir 10 8 a La " e - ir 1 + M J where a = r -\- eki. In this case a simple solution which is quite accurate is obtained if the last term in (73) be omitted since an inspection will show it to be small as compared with the second term. We have then di _ . JVMQ-8 di Separating the variables we get: di W*t ? (74) (e - ir) (1 + fet) Nk again 1 A B A +Be + i(Aki - Br) (e - ir) (I + Since the left-hand member does not contain the unknown i and since the constant term is 1, we get A + Be = 1 Aki = Br. CIRCUITS OF RESISTANCE AND INDUCTANCE 59 Thus A = _ _J1_ and B = fei r + efci r + eki The intergral is thus broken up into two simple integrals r_ <** r ^ _ , C- J (e- ir) (1 + kj) J (r + efci) (e - ir) ^ J ( r + *i)(H- ^ i fci fci r + eki * e ir If it is desired to find the value of i at any time after the circuit is closed then i = for t = (75) e ^r e The curve connecting z and t can conveniently be obtained by assuming different values of t and solving for the left-hand mem- ber of the equation. The value of i can then, of course, be easily determined. Curve a in Fig. 25 shows the relation between the exciting cur- rent and the time for the field current of a direct-current generator having the following constants: e = 100 volts = voltage impressed on the field. r = 100 ohms = field resistance. TV = 4000 = total number of field-turns in series. 01 = 1 megaline with 1 amp. excitation. 02 = 0.6 megaline with 0.5 amp. excitation. From FROLICH'S equation follows then: 1 = and 0.6 = ' k = L5 i .i ifci = 0.5. It is instructive to verify this curve. Curve b gives the corresponding values if the saturation curve had been a straight line, i.e., if the flux were 1 megaline for 1 amp. excitation, and 0.5 megaline for 0.5 amp. excitation. 60 ELECTRICAL ENGINEERING In that case the inductance L would be constant and would be 4000 X 1,000,000 L= 10' X 1 " and the equation e = ir + L -r would be 100 = lOOf + 40 -^ in which case i=l- e- 2 - 5 ' (76) ^ f * 7 s / ^ ^ ^ / .0 / ' / /a 7 / / / g . / 2J -6 7 / 3 7 g / / , // 3 .4 X ; / W A // K .2 j 1 / / n f 1.2 Time in Seconds FIG. 25. 2.0 It is interesting to see that the field flux builds up considerably slower than would have been the case if L had been constant. The reason for this is that, while at the final value i = 1, the inductance is the same in both cases, for all smaller values of current the inductance is greater because the flux is greater for the same current. When the saturation can not be expressed by a simple equation, there is no better method than to calculate step by step. Let Fig. 26 represent such a saturation curve. Determine the rise of current when a constant impressed e.m.f. of 100 volts is impressed on a coil 4000 turns having a resistance of 100 ohms. Thus N CIRCUITS OF RESISTANCE AND INDUCTANCE 61 Using differences instead of differentials: N A0 6:=ir + WAi or 10 8 (e - ir) M = 0.25 X 10 7 (1 - i (77) (78) 1.4 1.2 | 1-0 I * 8 c 5 .4 .2 .4 .8 1.2 1.6 Current in Amperes FIG. 26. 1.U .9 .8 .7 a Q> 3.6 ___ ^~~ f -^ z~-~ " ^ ^ "' ' ^ -<^ '"* ^. / / '/ /, ' /> ' /, /+ / f / 5 . p .4 / / / / w .3 .2 /,/ // /} / - / / } '/ "0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 1.1 1.2 1. Time in Seconds FIG. 27. If the values of current are determined every one tenth of a second A = 0.10. .'. A = 0.25 X 10 6 (1 - i). The actual flux at any time is of course SA< the corresponding relation as obtained by the use of the differential equation. 62 ELECTRICAL ENGINEERING The method of calculating is illustrated in the table given below and the results plotted in full-drawn lines in Fig. 27. First approximation Second approximation t A*> SAy> i> i - 1 A? SA

= .. , , . Assume that the resistance of the winding is r ohms, and that the impressed e.m.f. is a sine wave. At any instant the following relation exists: E sin ut = ir + L -r. -\- i -r:' dt at But L = j-fTjp where N = number of turns 1 i is read off the saturation curve or since in this particular case a satura- tion curve which can be expressed by FROLICH'S equation has been assumed for the relation i = 1.5 1.5 1 + 0.5* CIRCUITS OF RESISTANCE AND INDUCTANCE 63 again neglecting the last term. , di . . E sin ut = ir + - : : . -7- 1 + kii dt where a = JVfc 10~ 8 substituting for 1 -f- k\i = ; y y = e~ ~T d' r r + ki E sin o>< dt J 1 1 i (8o) dto* 1 d 2/i i?/ r r a dy k\y ki kiy dt dy kiy E sin ut = r ry a ~~ dy (r + fci E sin cup _ r_ dt+ y ~ a 'a r + ki E sin ut --J- . i E sin w< d< "] (82) the solution for i is found by a simple substitution. Unfortunately, however, the solution is not in a simple form and can not be simplified; and thus, while mathematically the problem is solved, practically it is unsolved. In cases like this it is necessary to proceed by a step-by-step method. Consider then the case of an alternating-current impressed upon a magnetic structure having a saturation curve of any shape. Let it, for instance, be expressed by ki = r+i^i The following relation exists at any instant : E sin orf = ir + g (83) 64 ELECTRICAL ENGINEERING where r is the resistance N /. E sin cot eft = ir dt + j^ d< (84) PI V 10 8 7> ID 8 *' d0 = sin ^ d * ~ "" d * (85) If, with full-load effective current I e the resistance drop is p per cent, of the rated voltage, then 7 e r = j^Q r=, and for any other value of / as i, ir = ^ ^- (86) ' 2 or since , cos coi , d = sin cot dt E X 10 8 r , cos co pidt d * = ~N L~ ~u TooT T ~ It is usually more convenient in alternating-current problems to introduce 6, the phase angle, instead of co. In that case = ut and dt = Referring to, (127) _ E X 10 8 pin 6 dB _ pidB 1 ~ ~~N~ co " I e 100 X/2COJ OT In most problems E, N, $ max and the frequency are known, so that numerical values can directly be substituted in the above equation. Since, however, there is a relation between them, one or more of the quantities may be unknown. The most general aspect of the problem is given by eliminating the numerical value of the impressed voltage, turns and fre- quency, and specifying the maximum value of the flux : maximum = $. We have the following well-known relation between $, N, E and co. 2irf 3>N 10 8 10 8 .'. - (88) CIRCUITS OF RESISTANCE AND INDUCTANCE 65 When in an inductive circuit the resistance is very small compared with the reactance so that the impressed and counter e.m.f. are equal numerically. Substituting this value in (87) 1.0 / NS / \ / \ / \ \ ' / \ / f ^ / v \ 1 l\ \ / / k ^ 1 I \ \ \ 1 / \ 7 / i \ \ \ 1 / / / / \ \ / / \ \ / / \ \ 7 / i \ \ \ \ j c ) $ 3 1( = i ,'HKV megalines. Assume that before the switch is closed the remanent magnetism is zero as is practically the case when the magnetic circuit con- tains an air gap. Assume further that under normal conditions CIRCUITS OF RESISTANCE AND INDUCTANCE 67 of operation the maximum flux is 1.4 megalines, that normal effective current is 1.7 amp., and that the resistance drop is 3.91 per cent. Then A = -- 1.4 [A cos + 0.00286i] - 1.4 A cos - 0.004?; The total flux is obviously ZA<. If the switch is closed when the e.m.f. passes through zero and is rising, the normal flux at that instant would be a maximum in the negative direction as shown in Fig. 29. As it has been assumed that the flux really is zero it is evident that there is a transient stage in the mag- netization before permanent condition is reached. It is evident also that if the switch were closed when the e.m.f. was a maxi- mum no transient condition would result, because the condition then demands zero flux, and the flux is assumed to be zero. In the numerical example it is assumed that the switch is closed when the e.m.f. wave passes through zero. The method of using the above equation is best shown by the table given below. No. 1 No. 2 No. 3 No. 4 No. 5 No. 6 No. 7 No. 8 No. 9 9 Cos 6 A cos - 1.4 A cos SA< i 0.004i SA< i 1.0 10 0.98 -0.02 0.028 0.028 0.01884 0.000075 0.027925 0.01875 20 0.94 -0.04 0.056 0.0839 0.0576 0.00023 0.08367 0.0573 30 0.87 -0.07 0.098 0.18167 0.1289 0.000516 0.1811 0.1288 40 0.77 -0.10 0.14 0.32115 0.2398 0.000959 0.3202 0.2395 50 0.64 -0.13 0.182 0.5020 0.402 0.001608 . 5004 0.4010 60 0.50 -0.14 0.196 0.6960 0.604 0.002416 . 6935 . 6020 70 0.34 -0.16 0.224 0.9175 0.881 0.00352 0.9139 0.8740 80 0.17 -0.17 0.238 1.1519 1.247 0.00497 1.1469 1.238 90 0.00 -0.17 0.238 1.385 1.712 0.00684 1.3781 1.699 Column No. 1, phase angle; No. 2, the cosine of the phase angle; No. 3, difference in the value of the cosine between two successive steps, for instance cos 20 cos 10; No. 4 is self-explanatory; No. 5, first approximation of the flux (sum of No. 8 of the preceding line and No. 4 on the line under consideration") ; No. 6, current as obtained from the saturation curve or the equation if such is given; No. 7, ohmic drop; No. 8, second approximation to the flux which takes into consideration the ohmic drop (the algebraic sum of No. 5 and No. 7) ; No. 9, current corresponding to the last approximation of the flux column, No. 8. CHAPTER IV CHARACTERISTICS OF CONDENSERS The charge q of a condenser is proportional to the voltage; or q = Ce, where C is the capacity the value of which depends upon the mechanical construction, dimensions, etc., of the con- denser, and 6 is the voltage. The charge q is expressed in coulombs or ampere-seconds. Thus the charge dq given in a time dt when the current is i amp. is: dq = idt. The capacity is expressed in farads, a very large unit; so large indeed that in actual practice it is never used. The capacities of condensers are almost always given in microfarads, that is, in a unit which is one-millionth of a farad. Nevertheless, in all formulae involving capacity, C stands for farads, not microfarads (m-f.) unless stated to the contrary. To give an idea of the capacity of condensers used in engi- neering, it may be of interest to know that the ordinary paraffine paper and tinfoil 500-volt blocks of the size of the average text-book have a capacity from 1 to 2 m-f. In a high-potential transmission line the capacity of one wire against neutral is about 0.016 m-f. per mile. The capacity of underground cables is relatively high. Depending upon the voltage and type of cable, etc., it must obviously vary much. It is usually less than 2 m-f. per mile and more than }/{ Q m-f. The capacity of an ordinary Ley den jar is extremely small a very small fraction of a microfarad. The fundamental equations for the condenser are as stated above q = Ce (1) and dq = idt (2) From these follow: q / Q \ = C dq = Cde (4) and . = | (5) 68 CHARACTERISTICS OF CONDENSERS 69 Substituting (4) in (2) dp Cde = idl ori = C ~ (6) (it or e, the voltage across the condenser = -~J*idt (7) The rate of energy supply or power is ei or from (6), Cde de or from (3) and (9), . _ ? . _ q_ dq ~ C ^ " C dt The energy stored in a condenser, which is the same as that required to charge a condenser to a voltage E or to a final charge Q, is therefore the rate of energy multiplied by the time. It is: f Jo Jo e o2 /~n , / or cedt = C - C 2C Equations (10) and (11) are obviously identical, since at any instant q = Ce thus for e = E when q = Q Q = CE, which, substituted in (11), gives CE 2 2C 2 As in the case of inductance, the calculation of the capacity of any but the simplest circuits is difficult. It will be dealt with in later chapters. Of particular interest to engineers, however, are a few simple forms of con- densers, the approximate capacity of which are given by equations which are ^ IG 39. well known. Thus the capacity between parallel plates, Fig. 30 is: c = ' in microfarads 70 ELECTRICAL ENGINEERING where K, the specific inductive capacity is approximately 1 for air, 2 for paraffin paper, 3 for rubber, 5 for mica and 6 for glass. A, the effective area is given in sq. cm. and d, the thickness of the dielectric, in centimeters. The capacity between concentric conductors (Fig. 31) is: 0.0386LK C log in microfarads io where the length I is given in miles of cable, K is the specific inductive capacity, D the inside diameter of the outside conductor, and d the diameter of the inside conductor. This is the capacity between the conductors, not the capacity to neutral or ground. The capacity of one conductor 1 mile long to neutral is twice as great. The capacity between transmission lines is: 0.0386? FIG. 31. C = in microfarads where I is expressed in miles and the capacity is that of one line against neutral. D is the distance between wires, center to center, and r the radius of wire. The charging current is thus . 2irfCe 10 6 where e is one-half of the line voltage in the single-phase system and 58 per cent, thereof in the three-phase system. Circuits Containing Concentrated Capacity and Resistance Consider at first the case of a constant e.m.f. E impressed upon a circuit of resistance r and capacity C, Fig. 32. After the cir- cuit is established a current flows and energy is delivered to the re- sistance and the condenser. In the resistance heat is developed and in the condenser an electrostatic field is produced. The energy given by the source of supply of power is fEidt. The energy supplied to the resistance is fftrdt FIG. 32. CHARACTERISTICS OF CONDENSERS 71 and the energy supplied to the condenser Thus fEidt = fi*rdt + Cq ~? . (12) Eidt = i*rdt + q ~l Ei = i z r + - Jj which is the power equation (13) and ,-, . q dq E = * r + Ci dt or substituting for dq = idt E = ir + -~ which is the voltage equation (14) C Obviously the voltage equation could have been derived directly, since ir is the e.m.f. consumed by the resistance and ~ the voltage across the condenser. The condenser voltage is thus ei = E ir; but or de, I = E dt ' Cr ? " * r Referring to equation^! = Ae c r -\- E (15) where A is the integration constant: The current is readily CA _JL< A _!, c cv' -e c, (16) r . ei found, since % = C ~r The charge g is = Cei - CAe~cr' + ^C. (17) Special cases: (a) Condenser charge. At time t = 0, e\ = 0. .'. referring to equation (15), A + E .'. A = E (18) Referring to equation (16) t---&' (19) 72 ELECTRICAL ENGINEERING Referring to equation (17) EC\l - (6) Condenser discharge. In this case the impressed voltage E Referring to equation (15) i and e = A i . (20) for t = 0, ei = e . = At cr l (21) 1 that is in opposite direction to charging current (22) 2000 Ohms. ""* -WV\A- FIG. 33. q = Ce e~cr (23) 2oom.f. Referring to the e.m.f. of the con- denser rather than to the impressed* e.m.f., the current becomes positive since the discharge current ,-.-- -C^ dt dt _ C Cr (24) Cr * " ' r - (25) In order fully to understand the action of condensers it is not sufficient to follow the equations given above, but it is essential and indeed necessary to figure a number of numerical examples. h 2| ||.05 S.2.04 JU If 02 jfiOl 3 oo .1 .2 3 .4 .5 .6 .7 .8 .9 1.0 1.1 Time in Seconds FIG. 34. 90 1.8 80 1.6 6o|l.2| 40^ .8 300 .6 20 .4 10 .2 For this reason Figs. 33 and 34 are given. The curves shown there should be checked numerically by every student. They CHARACTERISTICS OF CONDENSERS 73 are calculated under the assumption that a constant impressed e.m.f. of 100 volts is impressed on a circuit of 2000 ohms resist- ance and 200 m-f. capacity, as shown in Fig. 33. An interesting problem in connection with the charging and discharging of condensers, is to consider the flow of current be- tween two Ley den jars of different capacity and voltage (Fig. 35) . The energy stored in condenser A at volt- age E is J^CE 2 . The energy stored in condenser A at voltage e is The energy stored in condenser B at FIG. 35. voltage Ei is %CiEJ. The energy stored in condenser B at voltage e\ is J^Ciei 2 . While current flows between the two condensers, a readjustment of energy takes place. The energy equation is obviously: 0.5CE 2 + 0.5Ci#i 2 - 0.5Ce 2 - O.SCi^ 2 = fi z rdt. By differentiating this equation, the following results: - Cede - Cieidei = i*rdt (26) As it is assumed that the voltage of A is higher than that of B, the latter being charged; thus where e\ is the voltage of B at any time. Equation (26) contains three variables, e, e 1} and i, which, however, are dependent upon each other. At any instant the following relation exists between the e.m.fs. e = ir -f- ei Thus de_ di^ dei _ d 2 ^ dei dt ~~ T dt "" dt dt 2 ~ ~dt Substituting in (26) ClCir-rr 4-ei) (Ctf-tt + -37) dei-j^ = Ci 2 r (-57) \ eft / V di 2 d^ / rfi \ dt I or 6 , dei ei \ n ei lr r ei j 757 ' Cl ~di( Cir ~dt + ei or dei \ / dei rr d 2 ei deA ) l Cl ~di + CCir W + c ~dt) = 74 ELECTRICAL ENGINEERING Since Cir -57 + 61 can not be zero at + C-0 (29) Integrating (157) or dei C + Ci K Kl ei = K 1 + K 2 e- (3D where CCi / /-? i /^ C + Ci The integration constants KI and X 2 are determined from the initial condition that for t = o, ei = EI and e = E .'. Ei = KI + K 2 or K 2 = Ei - K l i+ (E, -Kfc c (33) l dt~ C r" but e = ir + e\ /.-- ^(^i - Xi)e " ^ + K! Co for t = o e = E :. E =- ^(E l - K,) + K Co .'. K, = E, + ^ and Co The problem can be solved in a simpler way if it is realized that the total charge in the system, is not changed after the switch is closed. Thus Qo = EC + E, d = q + qi (36) Where q and qi are the charges at any time in jars A and B respectively. CHARACTERISTICS OF CONDENSERS 75 In that case e = ir + e\\ or since q = eC and qi = e\C\, Assuming E>Ei, then jar A is being discharged thus dq or g i dt ' Cdr q ~ where ^ CC\ Co ~ C + C for J Since condenser B is being charged K EC si , si Co(E Since condenser A is being discharged ,__* + z*j.-,-. at r The voltage across condenser A, which is being discharged is for t = o e = E n , , ( 76 ELECTRICAL ENGINEERING = O = O C (E E } ~ ~Cor 1 TJ C (E EI 1 E - E^ /"* {s QT< Ci r i for t = o, and C , _. ,, x C With a slight modification of this equation it is seen that for t = oo the final voltage between the coatings of the Ley den jars is E = Q Numerical example: condenser A has a capacity of 1 m-f. and is charged to 1000 volts; condenser B has a capacity of 2 m-f. and is charged to 500 volts; the resistance is 10,000 ohms. Find the current after the switch is closed. The original charge in A is then 1000 X T = 0.001 cou- 500 X 2 lomb; the charge in B is I-^G = 0.001 coulomb also. E - EI = 500 2 X 10 6 2 3 X 10 12 3 X 10 6 and 1 C + Ci = 3 X 10- 6 c V c = 0.667 c _^ Ci = 0.333 *' = i 600 1.060 S 000 .140 .120 10 1000 .100 I ! 8 5 800 | .08 > S i 6.SCOO-J .06 ir* -^ ^ 4- . 4 3 400 g .04 ? 2 200 O .02 > -2 -200 -.02 -4 -400 \ Tim Case No.2 Voltages H -500 )00 .005 .010 .015 Time in Seconds FlG. 36. 025 .025 For t = oo , e Q = e i 0.667 volts which is the final voltage of the two jars. Fig. 36 gives the result of these calculations. Harmonic E.m.f. Impressed upon a Circuit of Resistance and Capacity in Series. Let e = E sin cot be the impressed e.m.f., r the resistance, C the capacity and q the charge at any instant. 78 ELECTRICAL ENGINEERING Then, referring to Fig. 37, E sin wt = ir -{- Differentiating = ir -\- -^ \ i - \ idt di u cos u>t = r-r + -^ di i Eos cos cot (1) E Sin FIG. 37. Thus, _ i t [ C +- 1 -tEu J i = e Cr \ I Cr cos cotdt + K The integration is readily made and the result is: E i = -y sin (cot + j8) + -KV where and Z = r' a^ r tan /8 = r The voltage across the condenser is: = ~ sn fldt + ^ fe c At the moment of closing the switch e c = 0. Thus for t = t 1} e c = 0. (2) (3) thus, or t = | [sin (8 cos cos (4) (5) CHARACTERISTICS OF CONDENSERS 79 As an interesting application of these equations and the corre- sponding equation for inductive circuits, consider the nature of the current supplied to a tuned circuit, Fig. 37, when the resist- ance is small. x = - x C jZ = \/r 2 -}- x' 2 = -\/r 2 -\-x c 2 The line current at any time is the sum of the currents in the two circuits. Q (0 - 8) + sin (0 + 8} - e ~x (e ~ ei} sn 1 - - = ^| 2 sin cos j3 e r ' ' cos (0i /?) XC. ""I r e ' cos (0i -f )8) (7) The line current is a combination of a sine wave of form - ~ sin and two exponential or logarithmic curves. Since r is ^_ fa a \ small compared with x or x ej one of the logarithmic curves e * M *^ c fa a sin (0i ft) dies down at a slow rate, whereas the other -V 7 l cos (0i -f- (3) dies down with extreme rapidity. x In the limiting case when - is large and ft therefore approaches 90 the permanent term disappears since cos = 0, and Er -J-(g- gi ) x c _ x f(0_0,) . ~| = '7? I * COS 0i H r sin 0i ZL r J for # r = OLto = ~~ cos QI x Thus the interesting situation occurs that if an alternating voltage is impressed on a tuned circuit as shown in Fig. 37 and if the resistance is zero the line current is a steady unidirectional current having the value: . _ E cosd i x If the switch is closed when 0i = 0, then the direct current is a maximum and is . At any other time it has a smaller value. x 80 ELECTRICAL ENGINEERING In Fig. 38 is shown a series of curves which illustrate this in the case where the resistance is considerable and the circuit is closed when 0i = 0. The constants for the circuit are: E = 1, r = 0.05, x = x c = 1 z'o is the total line current, the dotted sine wave is the impressed e.m.f., IA the current in the condenser circuit, and IR the current in the inductive circuit. As is seen, i Q is a unidirectional current l.U ( ^ Q ^ x ^^, __,, ---' ~~ ^ 'o 5 K. M. i 1 . * ^^ ^' f \ / \ / ^ / ( 1 1 \ i * 11 a 1 i \ -* -~- A i^ -^ <"4 ) Q 1 ) 11; ^ ?, () ^- ^ i() 4 g (1- - , ft ) _^- 1 -.1 \ / \ .2 ( .3 \ / N J 1 N , / ^ / \ ' t- A g \ / \ 1 7 \ / \ 1 g 1.9 E M F, 5 1 8 /" > \ 4 1 7 \ 2 V l r / / \ \ / ^ 1 fi t \ c / \ /' 2 / \ / \ \ / / \ \ / 1 4 \ \ / / V , 72 1 3 91 ;/ li 9 2 70 p 30 4 / b ip \ Gl / 1 9 V \ 1 \ N 1 / o 1 1 A \ \ / q 1 q N I 7 A / 4 9 / A i ^ -5 / \ 1 \ \ 6 7 / \ / \ \ 7 g / \ / \ 1 \ 5 J ^i K \ I ._ / 9 4 N \ / y / 1 3 . \ \ ;' r / / \ \ / \ I \ / N \ /y \ i / 12 ) 9 1 ^p 2 70 \ 3 30 / 4 oO 5 IP 6 \ I 2 V f \ \ i Q x^ ^} \ 4 \ / \, / \ , ^^ e -.6 FIG. 38. of slight pulsation, slowly decreasing in magnitude. After a number of cycles it would become a small alternating current, as shown in the curve marked Final i . This feature of a tuned circuit might be of practical importance in connection with problems of rectification charging of storage batteries from an alternator by occasional interruption of the CHARACTERISTICS OF CONDENSERS 81 current and starting it at the time when normal current in either branch would be a maximum. Circuits of Inductance and Capacity. While practically such circuits can never exist they offer much interest from a theoretical point of view since their study represents an introduction to oscil- lating circuits, which are of much importance in electrical engineering. In Fig. 39 is shown such a cir- cuit. In practice the condition there indicated is approached FIQ when a very low resistance over- head transmission line supplies power to a cable net work, which case, however, is fully treated in a subsequent chapter. The following relation exists at any time. where E is the instantaneous value of the impressed voltage and e\ the voltage across the condenser. But . _ n dei - dt thus di _ d*ei = dl* thus LC^ + ei = E or d*ei , _fi_ J^ dt 2 ~^ CL~ CL The solution of this equation has been given, it is ei = E + Aie mi ' mi and ra 2 are the roots of equation; m. - f i . , f> / -'" '* V 1 1.2 .8 .4 / 7-) \ SI 7 v / s \ / \ . -.4 \ / s i - -1.2 ^ ^**s < ^T=2if\TcI. Seconds > The frequency of the alternating current / = - ~/TT * s c the natural frequency. 1 It is almost always much higher than the frequencies used in commercial alternating-current circuits. The Discharge of a Condenser through an Inductance. Re- ferring to Fig. 41. Let E be the voltage of the condenser before the circuit is closed and i the current at any instant after the switch is closed. Then But the discharge current e = L thus and or i = - C di di dt de ( dt FIG. 41. L dt CL dt 2 eo = -CL' 1 di 2 CL the auxiliary equation becomes : m 2 + -== =0 or m = 1 In case of a transmission line the natural frequency will be shown to 1 84 thus and ELECTRICAL ENGINEERING cos cos The integration constants A and B are determined from the fact that at t = 0, i = 0, and e = E . Thus: EQ = A sin B = - C D cos B '. cos B = and B ^ and .'. sin B 1 and A t = En sm cos 2 E I = En COS sn s It is seen that the discharge frequency is the same as the frequency at charge, and that the maximum value of the current P ft E \L- As another application of this will be considered the condition when a short-circuit is suddenly opened and the large current in- stantly interrupted. This condition is diagrammat- ically illustrated in Fig. 42. S represents a switch which short- circuits the condenser and is opened at the instant under consideration. (In practice this switch may represent a short-circuit across the cables opened by the magnetic effects of the current.) The current I in the short-circuit is evidently the same as the current in the inductance ; therefore the energy stored in the magnetic field is 0.5LJ 2 . At any time after the switch is opened the current i flowing through the condenser, inductance and generator (all assumed as FIG. 42. CHARACTERISTICS OF CONDENSERS 85 having zero resistance) is governed by the condition that the energy stored in the condenser and inductance is the same as the original energy. .'. 0.5Li 2 + 0.5Ce 2 = 0.5L/ 2 LI' 2 - Li* = Ce* but . _ ~de 1 ^7. dt thus /de ,- 2 = C2 /^V \dt) :.LP - LC*(~} 2 = Ce\ Differentiating or/^2 ^ 2g ~ 2LL dt 2 dt = . d*e I -+ 6 '- and . / t % e '-'- A sin + B ) for t = Q,i = I,e = Thus = A sin 5; and sin 5 = 0, and B = 1C = i = I cos It is interesting to note that while at the instant of opening the switch, or the short-circuit, the voltage e across the condenser is zero, one-quarter of a period later (period being here the natural period which is extremely short) the voltage is a maximum and is 6max = These equations are instructive in that they show that the maximum voltage obtained in opening a short-circuit in a cable 86 ELECTRICAL ENGINEERING or transmission line is independent of the length of the line and depends only upon the constants of the circuit per unit length and the current at the time the circuit is interrupted. They also show that when the circuit is closed on a transmission line of considerable inductance and capacity, the maximum rush of current is also independent of the length of the line and depends only upon the value of the e.m.f. and the circuit constants. Harmonic E.m.f. Impressed upon a Circuit of Inductance and Capacity but Negligible Resistance. This strictly theoretical condition is chosen for two reasons. The solution of the equa- tions introduces some mathematical operations which have hitherto not been considered and the problem from the electrical point of view illuminates in a relatively simple way what happens in the extreme case in switching high-potential circuits. The general equation obviously becomes: E sin ei (1) where e\ is the voltage across the condenser, Fig. 43. But ^ dt .'. E sin thus L -7 at CL or FIG. 43. d 2 6i 61 E W + CL = CL sln (2) It is seen that the right-hand member of the equation is a function of L To solve such equation the solution of the comple- mentary function is first found; that is, zero is substituted for the right-hand member: V ' r CL ~ i 2 +-== = or m = j .J CL/ (3) CHARACTERISTICS OF CONDENSERS 87 Ai +A/ cE' + AvT^m,' = A^ at + A#- iat (4) where The equation, as has been shown previously, can be written 61 = A sin (at + #) Thus the complementary function is e Q = A sin (at + B) (5) The next step is to eliminate the sine function from the general equation (2) by two successive differentiations: Substituting the value of Ea 2 sin coi from equation (2) and arranging the equation in the order of the derivatives: (7) The complete solution of (7) is obtained in the usual way: m 4 + (a 2 + a> 2 )ra 2 + coV = W 2 (m 2 + a 2 ) + a> 2 (ra 2 + 2 ) = or (m 2 + w 2 ) (m 2 + a 2 ) = .*. m = + jb) m = + jo. Thus d = Ai sin (at + #0 + A 2 sin (a + 5 2 ) (8) By referring to (5) it is evident that A 2 and 5 2 must be the same as A and B. Thus i = e Q + AI sin (o> + BI) The integration constants AI and BI are determined from the fact that the expression A\ sin (co + BI) must be a particular solution of (2). Thus i = AI sin (o> + 1) -7- = AICO cos (&t -}- BI) d z d - = -- Aiw 2 sin (cot + BI) 88 ELECTRICAL ENGINEERING Substituting these values in (2). Thus - Aico 2 sin (ut + Bi) + Ai 2 sin (coZ + BJ = Ea* sin ut (9) or A^a 2 - co 2 ) sin (at + BI) = Ea 2 sin ut. Thus equating the coefficients of similar terms: Ea 2 = Ai(a 2 - co 2 ) * --^.-5%; (10) or or c a: and 1 = /. from (8) (11) ei = EXc sin ut + A sin (erf + 5) (12) 3?c 32 The second term in this equation may in this case be more advantageously written : AQ sin at + Bo cos a /. ei = - - sin coi + A sin a^ + Bo cos at (13) 3?c E = 7a: c sin coi + A sin a^ + Bo cos at (14) Where / is the maximum value of the permanent current, that is, / - ~^ (15) X c - X i = C 3 - = Ix c u cos ut + A Q a cos at B a sin CK (16) at Considering the problem of starting a current in such a circuit : when t = ti, i = and e\ = 0. If these values are substituted in equations (13) and (16) it is readily found that BQ = Ix c sin ati cos coi sin at\ cos cd\ (17) AQ = Ix c sin ati sin wt\ -\ cos at\ cos co^i \ (18) Substituting these values in the equations for the current and e.m.f. across the condenser we find: ei = - sin at cos coZi sin a(t ti) Xc X L a - sin coii cos a(t - fi)l (19) CHARACTERISTICS OF CONDENSERS 89 and i = C -TT = ~ cos at cos coti cos a(t ti) at Xc x L 1C 1 + x c ^Ij- sin coti sin a(t - fc) I (20) While as a rule equations of the form given in (2) having a sine function or exponential function on the right-hand member, can be solved by the method given above, it may be opportune here to call attention to another well-known method of more general application. The differential equation is: fJZf, *' (25) These four integrals are solved independently below : First, s 3 .*f. - 2 ' J f sn ( la sm co* co cos co *\ EC , jat J I A co + a 2 / J co 4 +a 2 (ja sin co* + co cos co*) (26) Second, r ~ iat cos co^ dt = a* I co" a* 2 - I -jat u s ^ n ^ ~ J a cos ^ \ ^ a?co \ C " co 2 + a 2 / : ~ J co 4 -a 4 (a cos co* jco sin co*) (27) Third, r _. at _ _. at r _ 2 . a< O 1C I C (Zt' vy l6 I C ttt' J J " 2ja e ' a< ~ 23at = 2ja e ~ } ' at = C ^ at (28) Fourth, C 2 e>' at (29) The last two terms can be written C 3 e-' a < + C 2 e>< = C 4 sin (a* + C 6 ) (30) In the general equation it is seen that the second integral is negative thus: co 2 sin co + a 2 sin co^ jaco cos co^ (31) = C 4 sin (at + C 6 ) H -- ri jaw cos + co 2 si n - / * = C 4 sin (at sin ( This equation is identical with (12). = C 4 sin ( + C B ) + -- sin co* (32) X c X CHARACTERISTICS OF CONDENSERS 91 As an application of these formulae will be considered a 100- mile 60-cycle transmission line supplying power to a cable net- work of 50 miles. For the sake of simplicity and for the sake of later instructive comparison the resistance of the cable and of the overhead line will be neglected in this particular investigation and it will also be assumed that the inductance of the cables and the capacity of the overhead lines are so small as to be negligible when compared with the inductance of the overhead line and the capacity of the cables. While the inductance of a line, of course, depends upon the size of the conductors and the distance between them, in reality it is not subject to a great deal of variation in ordinary lines. It is about 0.002 henrys per mile of single conductor. The capacity of a cable system is, however, subject to great variation, depending upon the nature of the cables. Assume that in this case it is 2 m-f. per mile of single conductor, when referring to neutral voltage: C = 4 farads and L = 0.2 henrys = 26.4 ohms x = 2irfL = 75 .4 ohms = 27T/ = 377. If the circuit is closed when the impressed e.m.f. is zero, that is, when ti = 0, then equations 209 and 210 become: ei = - 0.54#[sin 377* - 1.69 sin 223*] and i = - 0.0204#[cos 377Z - cos 223*] The time for one complete cycle of the fundamental wave is gg = 0.0166 sec. If, therefore, the circuit is closed when the impressed e.m.f. is a maximum, that is, when h = t = 0.00416 then the equations become: d = - 0.54#[sin 377* - cos 223(* - 0.00416)] i = - 0.0204#[cos 377* + 0.59 sin 223( - 0.00416)] 92 ELECTRICAL ENGINEERING These curves are shown in Figs. 44 and 45 when the impressed e.m.f. is 100 volts. The curve e\ in Fig. 44 shows the e.m.f. across the condenser, the curve i the current when the switch is closed at zero value of 4 ^-~, / \, / \ 2100 ^_ / \ / N. ^ > / ^^ s c i Y "X ' V , m / / / V ? V \ / / \ V / \ s N \ 55 1 50 / / / \ \ \ v/ cY / \ / / \ \ ^0 *0 / .X \ \ \ / \ / \ / "- . \ o > i ^0 \ 2 70 \f 30 ^ oO b ty 6 JO \ I 8 LO / yu a 1 50 \ \ i A / a. t A A / \ \ $/ \ / / \ / \ x i > v \^ ^ \ / \ / / U 2 100 \ V / \ / V. ^x \ / 4 s^ FIG. 44. 2 200 150 .S oO 50 1 100 150 2 200 90 isp 270 \ /450 720 9(0 FIG. 45. the impressed e.m.f. The corresponding lines in Fig. 45 show the same quantities when the switch is closed when the impressed e.m.f. is a maximum. In both figures the dotted sine wave is the impressed e.m.f. CHARACTERISTICS OF CONDENSERS 93 Circuits Having Resistance, Inductance and Capacity. Con- stant Impressed E.m.f. Let E, Fig. 46, be the constant impressed e.m.f. r, the resistance L, the inductance and C, the capacity in farads. L c: Then FIG. 46. T di , Differentiating di d*i i = T + L + or dH > r dii _ dt* + L dt + CL " The solution of this equation has been shown previously to be : Where mi and ra2 are the roots of the auxiliary equation. where and _ 2L 4L 2 CL CL 2L ~ = = - - 4L 94 ELECTRICAL ENGINEERING Three conditions are possible: (a) r 2 -T^ is positive. (b) r* - -~r is negative. 4L. (c) r 2 -fT is zero. Considering first the Case (a), ^ ~^r (positive). Then i = ^[A*? + A*-'* 1 }. The e.m.f. across the condenser is: ei = E -ir -L d ~ = E - re- at [A^ 1 + A*'* 1 ] - L[((3 - a The integration constants for starting the current in this circuit are determined from the fact that when t = 0, i = and e\ = 0. Thus E Ai = Ai and AI = ^j ZLp 2L/3 2 4L \ r '^ :=>S > If then . El -l'-=^)i -f'J^\ t \ *-( - ) By differentiating i and substituting its value and that of i in the equation of the voltage of the condenser: e\ = E ir Jl -j-.j dt we get [1 / r ~ s t r+s t\~] i if._i_ cr\ o7~ i (^ Q\ f 9r~ t \ 1 ^.^ I (r +- o) 2L IT ^Jc *^ ^o \ / J The equation for the discharge current can readily be proven to be exactly the same as that of the initial current except for reversed sign. CHARACTERISTICS OF CONDENSERS 95 During the discharge the voltage is r-f S where E is the voltage of the condenser before the discharge. Case (6).- r - 4 negative. In this case + r 2 ~- can be written j -\\-~ r 2 \ C \ C Si where S x == - r 2 It has been shown previously that this equation can be written : (1) where A and 5 are integration constants. The e.m.f. across the condenser is Substituting the values of i and - as obtained from equation (1) Counting time from the instant of closing the circuit, then for t = 0, i = 0, and e\ = 0. From equation (1) it is found that A sin B = 0; .'. B = 0, 2# and from equation (2) A = -~- 01 . 2E * t . Sit t = ST 6 2L Sm 2L where tan 7 = In a similar way is found the equation for the discharge current which will be identical with the charging current, and the voltage across the condenser, e a = E Q e ~2L l sin ( ^F + T ) > where EQ is the original voltage across the condenser. 96 ELECTRICAL ENGINEERING Case (c). 4L r 2 In this case, as has been shown previously, i = e m i* (A + Bx) r i = e 2L (A + Bt) or = E re ~2L l (A + Bt) - LBe 2L l + ^~- If the time is counted from the instant of closing the switch, then for t = 0, i = 0, and ei = A = and - " - ^ hi E - LB, or B and The equation for the discharge current is the same as that of the initial charging current, the voltage across the condenser during discharge being: As an application of these equations will be considered the case of starting a direct current at 500 volts in a 20-mile concentric cable having the dimensions given in Fig. 47. In this instance it will be assumed that the capacity of the line can be represented as that of a condenser at the end of the line taking one-half of the charging current. It will be assumed that the specific inductive capacity is 3, the diameter of the inside conductor 0.5 in., the inside diameter of the outside conductor 0.7 in., and the' out- side diameter 0.86 in. The resistance of the 20 miles of cable is 8.8 ohms. FIG. 47. / CHARACTERISTICS OF CONDENSERS 97 The capacity per mile of concentric cable previously given is n o^sft Jf Cmf = - ~ = 0.795 m-f. per mile. ^og- Thus the capacity of the 20 miles of cable is 15.9 m-f. and the equivalent capacity at the end of the line is 7.95 m-f., or 7.95 by 10~ 6 farads. Since the determination of the inductance of a concentric cable involves the general method applied to other systems, it will be given below, although such determinations do not come within the scope of this treatise. The inductance is recollected to be numerically equal to the interlinkages of the turns and flux per unit current. In general if the m.m.f. acting in a circuit is M then the flux 4irM X area of magnetic circuit produced is , -r * ~r- = ^ The interlinkage length of magnetic circuit factor is that fraction of the total current which is enclosed by the flux and L = y-2J flux X turns X interlinkage factor. Consider first the flux in the inside conductor due to the assumed uniform distribution of the current. At a distance x from the center see Fig. 47, the m.m.f. is ^ / where / is the total current. The area of the flux per centimeter of length of conductor is dx and the length of the magnetic circuit is 2irx. x 2 dx x ... dv , 1 = 4 ,_ / _ ==2J _ & ( TTX 2 This flux interlinks with ^ of the total current, and hence the x 2 interlinkage factor is ^. 1 C r x 3 .'. LI = I 27 dx = % (assuming /x = 1) Between the conductors the flux interlinks with the whole current, and hence by a similar reasoning we get 98 ELECTRICAL ENGINEERING The current in the inner conductor interlinks with the entire flux which is inside of the outer conductor but which is caused by the difference in m.m.f . in the inner and outer conductor. At a distance XQ the m.m.f. is thus _ R 2 -R 2 * *R Q 2 -R 2 The interlinkage of this flux with the current in the inner con- ductor is of course unity; therefore L * = I J ~*o flo 2 ^!^ dx = R 2 - R 2 log ^ " " l The inductance of the outer conductor should be added to give the total inductance of the cable. The m.m.f. is _ r RQ ~ X Q Ro 2 - 1 The interlinkage factor is xp 2 - R 2 -R 2 ,2 _ ' L * - - 7/ z (Ro 2 -R 2 ) ^ R Q 2 - R 2 + (R 2 -R 2 ) 2 log ~R The total inductance L = LI -j- L% + L s + LI which is readily proven to be: This inductance is expressed in the absolute system of units. By dividing by 10 9 the inductance is expressed in henrys. The combined inductance L = 0.0039 henrys; thus r = 8.8, L = 0.0039, and C = 7.95 m.f. r 2 = 77.5 1960 is thus negative. c Therefore this problem comes under the second case and t = ^/%- - r 2 = 43.4. ^ = 5550 ~ = 1130 tan 7 = - = 4.93r = 78.5 LU ZLi T = 1.37 radians /. i = 23e- 1130 ' sin 5550* and e c = 500[1 - 1.02 e - 1130( sin (5550* + 1.37)]. CHARACTERISTICS OF CONDENSERS 99 The frequency of the oscillation is 5550 27T = 885 cycles, and the time for one oscillation 0.00113 sec. The maximum value of the current is determined by differen- tiation. It occurs when t = 0.000246 sec. When 5550^ = 78.5, the current is 17.1 amp. The next maximum value occurs when t = 0.000246 + 0.00113 = 0.001376 sec. The maximum value of the voltage across the condenser is also determined by differentiation. It occurs when t = 0.000565 sec., when e c = 763 volts. The next high value occurs obviously at t = 0.001695 sec. These curves are shown in Fig. 48. 0004 Starting V; Condenser Volt; K urrent m- .001 odd! 0012 0014 0016 .0018 400 200 ,002 FIG. 48. It is of interest to note that for a given distance of transmission the capacity, and therefore the charging current, is several times as great in the case of the concentric cable as in the case of the cable with parallel wires. Similarly the inductance is several times as great in the case of an overhead line as in the case of the cable. As a second numer- ical application of these equations will be considered: 100 miles of overhead transmission line supplying energy to a cable network 50 miles in length. It will be assumed that the cable system consists of a large number of short cables projecting in different directions from the terminal substation, as would be the case when a high-tension 7 100 ELECTRICAL ENGINEERING line supplies energy to a city lighting network. The resistance of the cable system can therefore be neglected. It will be assumed that the high-potential line is three-phase and consists of No. 00 B. & S. wire, having a resistance per 100 miles of 40 ohms and an inductance of approximately 0.2 henry. Hence the capacity of the overhead line is very small compared with that of the cable and it will be neglected. The problem is to determine the values of the current and vol- tage across the condenser when a steady e.m.f. of E volts is applied at the generating station. E = 100 r = 40 L = 0.2 C = 0.0001 farad. /. r 2 = 1600 -Q = 0.8 X 10 4 = 8000. 4L. .'. r 2 = -^ is negative. Therefore there is an oscillation when the switch is closed, and the constants are to be obtained from case (6). = 2= 80 ' 7 = !o = 2 " tan ^ = 2 and r 40 Si 80 V - 6 3.5 2L = ^4 =: 10 > 2L = 04 = :.i = 0.025#e- 10 * sin 160* and a = E[l - e" 100 ' 1.12 sin (160* + 63.5)]. The time for a complete cycle is T^T = 0.0392 sec., corresponding to a natural frequency of ^ = 25.5 cycles per sec. It is inter- esting to see that the effect of the resistance is to lower the natural frequency, since if the resistance is neglected it would be 7= = 35.5 cycles. CHARACTERISTICS OF CONDENSERS 101 Circuit Containing Resistance, Inductance and Capacity in Series. Harmonic E.m.f. Impressed. Fig. 49. From previous discussions it is evident that the general equation is di dt E sin cot = ir -\- L ^- -f- ~- I idt (1) r E sin = ir + x~ + x c (idd (2) FIG. 49. The latter form is preferable when dealing with alternating- current phenomena, but of course it must be remembered that x and x c refer to the impressed frequency and not to the natural frequency of the system; that is, x = 2-nfL and x c = The solution of (2) can best be obtained by differentiating twice, di d 2 i E cos 6 = r -TT + x -7T- 2 + x c i E sin 6 cttf d z i d*i di dl ' idB (3) Differentiating (3) and rearranging the equation: d*i dH d*i di] Xcl = The auxiliary equation is: xm* + rra 3 + (x c + x)m 2 + rm + x c = ra 2 (zra 2 + rm x c ) + xm 2 + rm + x c = /. (m 2 + 1) (xm 2 + rm + x c ) = Vr 2 - Let r 2x 102 ELECTRICAL ENGINEERING then mi = + j W2 = j m 3 = a + ft 1714 = -- a ft and i = A, sin (0 + A 2 ) + ^'(A*-* 8 + A,^ e ) (6) The integration constants A \ and A 2 could be found by methods outlined in the chapter on circuits of inductance and capacity. It is possible, however, for students familiar with elementary electrical engineering to determine them at once. Apparently the first term represents the permanent and the second the transient condition. In permanent operation the current leads or lags behind the impressed e.m.f. by an angle which depends upon the numerical values of the two reactances. The final value of the current becomes, then, -T7 i = ^ sin (0 + 0) ^o if tan 4 = ^ and Z == vV + (x c - x) 2 (7) .'. i = ~ sin (6 + )+ ) + AiT'to sin (0J + j) (11) ^0 where CHARACTERISTICS OF CONDENSERS 103 and , A B As + A 4 7 = tan" 1 -r- = tan" 1 7-1 A 6 (A 3 - A 4 ) a S t = ~- sin (0 + ) + ~"*[A 6 cos 0i0 + A 6 sin faO] (12) (c). r* - If (12) is true, then m 3 = w 4 , and we do not have a general solution; that is: - a + = - - since = (13) A general solution is obtained by letting w 4 = ra 3 + h (14) where h is very small. Here w 3 = a and ra 4 = a + h then (see also equation 42, Chap. II), E i = -yr sin (0 + 0) + A 3 e~' ' -f A 4 e ( ~ (15) which may be written: i = sin (0 + 0) + e~ ae [A 8 + A 9 0] (16) where A 8 = A 3 + A 4 and A 9 = A 4 /i. Each case will be considered independently. Case (a). r 2 4xx c (positive). Since r 2 4xx c is positive, is a real number and the solution of the differential equation (4) is : i = ^- sin (0 + 0) + A 3 e~ ( * 6 -\~ A 4 e~ ( (17) Let ( - 0) = K and (a + 0) = K (18) By differentiating equation (17) and substituting in the equation d = E sin ir X-J-Q, (19) the voltage across the condenser is: 1?7 ei = E sin - -^ sin (0 + + ) + A 3 e~ J ^ lfl (^ia: - r) (20) where .,. x (21) (22) 104 ELECTRICAL ENGINEERING If the problem is to find the current and the condenser voltage at any instant after the circuit is closed, and if the circuit is closed when = 0i, then i = and ei = 0. Substituting these conditions in equation (23), it may be written : A s e~ Kei + A,*-* 1 ' 1 = -- |^sin (0! + 0) (24) Also equation (20) can be written: EZ Ax' 1 (Kx - r) + A 4 e~ J 9l (Kix - r) = -^- sin (0i + + ^) - E sin 0i (25) Solving equations (24) and (25) for A 3 and A 4 we have: E A 3 = K01 [Z sin (0i 4- + ^) + (Kix r) sin (0i -f 0) Z sin X ] (26) and, E ^r e Kl01 \Z sin (0i + d> + ^) + x - r) sin (0i + 0) - Z sin Oi] (27) Case (6). As stated before the expression \A* 2 xx c is imaginary and from equation (28) Equation (11) may be written: E7 z = ^- sin (0 + 0) + e~ a0 [A 5 cos |8i0 + A 6 sin j3i0] (29) -^0 where A 6 = A 3 + A 4 and AG = j(A 3 A 4 ). If the switch is closed at = 0i, e\ = and i = 0. From these conditions A 5 = [- f- sin (0! + ) - A 6 e- afll sin /3 1 1 ]-^- (30) L Z/o J COS Piwi From these relations and the equation (19) can be written - - r) cos 0i0] + A 6 [- $ix cos 0i0 + (ax - r) sin di e\ = E sm zr x -> EZ = E sin - -=r- sin (0 + + ^) + e~ a0 [A 5 [/3iZ sin CHARACTERISTICS OF CONDENSERS 105 where Ecosp,d } ae [Z . A 6 = - TToT- e r sm 0i - cos (01 + 0) - 01 and [# sin (^i + 0)e a) - a sin (0i + ) (33) Z/o L ^ A 8 = - ~ e^ 1 sin (19! + 0) - A 9 (34) CHAPTER V A CIRCUIT CONTAINING DISTRIBUTED RESISTANCE AND INDUCTANCE An aerial transmission line with negligible capacity and leakage conductance is an example of such circuit. Fig. 50 represents an aerial transmission line with negligible capacity and leakage conductance and with a load having an impedance of \/Ri 2 + Li 2 co 2 . Measuring x from the receiving end, consider an element of the line dx. Let the resistance of the line be R ohms per unit length of the conductor and the inductance L henrys per unit length. Then the resistance of the element is Rdx and its inductance Ldx, and the voltage across the element is de. Therefore, or de = Rdxi de TJ di MX df di As the same current flows in all parts of the line i is not a function of x, thus equation (1) is readily integrated, it is: 106 RESISTANCE AND INDUCTANCE 107 where T di x = 0, e = R]i + Li -T , . = di f n . . T di\ r di . . e = (/fa + L--i- iz -f 72 1* + Lrr (2) \ d< / d/ where x = I, e = E or E sin o>Z, depending upon whether the generator voltage is constant (a) or alternating (6), therefore, (a) E = Ri + LJJ I + R,i + Lr = (fl/ + R,)i + (JH+Zu)^ (3) (6) sin co* = (Rl + 720* + (^ + LI)^ (4) /?? + Ri is the total resistance and LI -f ^i the total inductance of the circuit. Hence, neglecting capacity and leakage conduct- ance, a circuit of distributed resistance and inductance may be considered as if the resistance and inductance were concentrated as far as the determination of the current is concerned. Case (a). Unidirectional voltage impressed on the circuit. For the current, solution of (3) gives E f Rl + Ri\ (1 LI+LI* (see equation (17), Chap. I) (5) i \ / P 7 , P til -f rt Substituting (5) in (2) n \ From (5) and (6) it is seen that at the moment of closing the circuit, that is, when t = 0, i = and e = TJIT 1 E. In the case of non-inductive load (when LI = 0), e = jE and e = for x = 0. It is interesting to note, that, while resistances consume no voltages when i = 0, inductances do consume voltages when i = 0, provided -^ = 0. When t = , that is, when the current and the voltage reach their permanent conditions, 77? J?T \ 7? *' = ~ e = " E ' the expected results - 108 ELECTRICAL ENGINEERING Problem. Assume reasonable values for the constants, and plot a series of curves for the voltages at various values of t and x. Case (b). Alternating voltage impressed on the circuit. From equation (39) in Chap. I, the solution of (4) is found to be: E r Rl + Rl < t , n i = | [sin (co* - 13) - sin ( w *i - /3)e " T+T7 (t (7) where ti is the time at which the circuit is closed. Z = V(M + Ri) 2 + (LI + LJW and _ , (LI + Li) co (Rl + Ri) Differentiating (7), cos (/ - 0) + fj^-f^ sin co/! - |9)e " ^TiT w ' J (8) Substituting (7) and (8) in (2), or, sin (ft' 0) sin (wti 0) (9) where -Li) 2 co 2 , and j8 r = tan -1 Referring to (7) and (9) it is seen that the current is the same as if the resistance and inductance were concentrated, but the voltage is different at different points, being modified in magni- tude, and displaced in time phase. It is noticed that no transient component in the voltage or current exists at any point of the line, if the circuit is closed at R ti = -, or in other words when sin (wti (3) = 0. CO If sin (co/i |8) is not zero, the transient voltage appears at all values of x except x = I, for ft' can not equal (3, or ^ ZTT? (LI + I/O co can not equal to P7 , 5 unless x = I. 11 + K\ When t becomes large, that is, many cycles after the circuit is closed, the exponential term approaches zero and the whole cir- cuit becomes free of the transient, and (7) and (9) become : i = | sin (ut - |8) (10) EZ' e = --sm (co* + j8' - 0) (11) RESISTANCE AND INDUCTANCE 109 Z f where x = I, that is, at the generating end. -^ = I L &' = ]8, therefore, e = E sin ut, as assumed. The voltages at other values of x can readily be computed from (11) and it is seen at once that the amplitude is proportional 7' to -;= and the phase is ($' /3) radians leading the impressed e.m.f. The effective value of the voltage at any value of # is: (12) where e e /f = the effective value of the voltage at x, and E e f/ that at generator, that is, at x = I. When the line is open, that is, when Ri = oo , then (7) and (9) become : i = 0, and e = E sin co for all values of x. No current flows, which is to be expected. Grounding the receiving end of the line, Ri = and LI = 0. .'. -jj = -y and &' = j8, hence (7) and (9) become: / i i = 277[sin (ut - 0") - sin (wti - /3")e v "'J (13) and Ex . e = - sin at (14) where ., / , ,.,, _, Lo> A = \/ R 2 -f- L 2 co , and p = tan "5" It is interesting to note that in this case the voltage has no transient component and is in time-phase throughout the line. PROBLEMS 1. Assume reasonable constants of the circuit for equation (9) and plot e against t for (a) x = 0, (6) x = ^' 2. When an accidental ground occurs on an aerial transmission line the voltage 10 miles away from the generator station is found to be 60 per cent, of the generator voltage. Determine the point of grounding. CHAPTER VI CIRCUIT CONTAINING DISTRIBUTED LEAKAGE CON- DUCTANCE AND CAPACITY A low-voltage cable may be considered as an approximate representation of such a circuit, since it contains distributed leakage conductance and capacity but usually low resistance and inductance. Since the resistance and inductance are considered negligible as a limiting case, it remains to consider a system of parallel conductance and capacity. The voltage may be con- sidered the same at all points of the circuit, that is, independent of x. Let i in Fig. 51 be the current at x, then i + ---dx is the current at x + dx. Let C be the capacity in farads per unit length of the conductor against the ground or neutral, and G the conductance 'i -hdz -> - -,* 1 1 > 1 J 1 r\ = : i - -- : _ < > ~ < < * > - > 1- > < ! i > ^ := > - rr \ ' 1 ' FIG. 51. in ohms per unit length of the conductor to the neutral. Hence de the current in the path of the capacity is Cdx-?-, and the current in the path of the conductance is Gdxe. The difference in cur- rent between the two sides of the element dx is dx. Therefore dx T- dx = Cdx -JT + Gdxe f or (1) 110 DISTRIBUTED LEAKAGE 111 This equation is similar to (1) in Chap. V with i for e, e for i, C for L, and G for R. As e is independent of x, equation (1) integrated gives: K (2) It is of no interest to consider short circuit of the cable, since the resistance and inductance are neglected, for it would mean a dead short-circuit on the generator. Therefore consider the case of switching the generator on the open cable. Thus, where x = 0, i = 0. .'. K = 0, and (2) becomes: (3) Case (a). Unidirectional voltage impressed on the cable. In this case, it is assumed that e = E from i = to t = , but just before t = 0, e = 0. Therefore it is assumed that = co just before t = 0, and -77 = 0, just after t = 0; that is, equivalent to assuming that the fictitious condensers were charged with an infinitely large current during an infinitely small period. In reality, the rise of the impressed e.m.f. takes time, though extremely short, and the resistance and inductance of the circuit limit the initial value of the current and lengthen the period of charging. These assumptions thus do not allow a study of the transient condition. The equation indicates simply that i = <*> , for t = 0. de For the permanent condition we have -rr = 0, and (3) becomes: i = GEx (4) which is expected. Case (b). Alternating voltage impressed on the cable. Let the impressed e.m.f. be e = E sin wt, and the time of apply- ing to the cable be fa, thus e = E sin u>(t ti). Hence, de - -J7 = Eu COS 0>(2 ti). Substituting these in (3) i = [Cw cos u(t - ti) + G sin w(t - ti)]Ex, 112 ELECTRICAL ENGINEERING or i = #zVC 2 u> 2 + G 2 sin (ut - wi + 0) (5) where ft = tan- g^ This equation represents the permanent values of the current, and shows that the current is proportional to x at all values of t, and leads the impressed e.m.f. by /3 radians at all values of x. The transient component does not appear in this equation, as explained in case (a). The transients will be studied in the following chapters, where the capacity and inductance are considered. CHAPTER VII CIRCUIT CONTAINING DISTRIBUTED RESISTANCE AND CAPACITY In the study of the problems involving distributed inductance and capacity, and the simpler problems involving the penetra- tion of current or flux in conductors, etc., where alternating cur- rent of sine shape is assumed, a certain differential equation, given below is met. Its solution is of importance to the engineer and deserves consideration. The equation is ^ - * ^ (i) dx*~ ' at A general solution which can readily be verified differentia- tion is: y = A Q + 2Ae ax+bt sin (ax + # + 7) (2) or y = A + Aie 0lX+6 " sin ( ai x + ft it + 71) + A 2 e a * x+M sin (a 2 x + fat + 72) + . . . The evaluation of the different constants is accomplished partly from the known conditions at some points of the system, and partly by solving for the constants by differentiation and substitution. In most problems, y or its derivative is known at some point, where for instance after permanent condition has been reached y = Y sin ut. If the point happens to be where x = x\, then equation (2) becomes Y sin ut = A + SAe ox i +w sin (aXi + fit +. T) = Ao + SA'c M sin (#+/) = Ao + A\e blt sin fat + T 'i) + A'*** sin fat + y'*) +. . . 113 114 ELECTRICAL ENGINEERING Writing the equivalent of sines in terms of e Jut _ f -jwt j(0it + y'i) _ ' Y- J(P*t + y',) _ -j(ft*t + y'd -- -+. . it - j(0,t + y't) + . . . Thus, since the left-hand member contains the imaginary only, and the right-hand member a constant and the complex imagin- ary and the two sides must be equal for all values of t it is evident that AQ and the b's are separately equal to zero. .'. y= 2Ae ax sin (ax + ft + 7) (3) ^ = 2Ae ax [a sin (ax + ft + 7) + a cos (ax + ft + 7)] oX *~ = 2Ae ax (a 2 - a 2 ) sin (ax + ft = 7) + 2aa cos (ax + ft + 7) oX ^f = 2Ae ax p COS (ax + # + 7) at Substituting these values in (1) and equating the coefficients for similar trigonometric terms, a* - a 2 = 0, and 2aa = k z fi (4) .*. (a -\- a)(a a) = 0, thus a + a = 0, or a a = 0, or both. For a + a = 0, or a = a, the second equation of (4) gives, 2 2 = fc 2 /3, which is evidently impossible. Thus there remains only a a = 0ora = a. Then 2a 2 = k 2 P, or, a = a = fc -v/^ where a and a must have like signs. The general solution then becomes: y = A,e ax sin (ax + ft + 71) + A 2 e~ ax sin ( - ax + ft + 72) (5) and, ~ = A ie ax a[ sin (ax + # + 71) + cos (ax + ft + 71)] - A z e~ ax a[ sin ( - ax + ft + 72) + cos ( ax + ft + 72)] = V2 a [A ic * sin (ax + ft + 71 + | ) - Aae'^sin (- ax + # + 72 + ) (6) RESISTANCE AND CAPACITY 115 Application of these equations will be found in the case of a circuit of distributed resistance and capacity but negligible inductance and leakage conductance, such circuit being approxi- mately represented by the cable in Fig. 52. Let R and C be the resistance and capacity respectively per unit length of the cable. Let the distance be counted from the receiving end of the line. Let the voltage at B be e, and the voltage at A e -f- dx. ox .'. the voltage consumed in the line element is: . de j de j e + dx e = dx. dx dx di Let the current at B be i and the current at A be i + dx. ox Receiving End of Line A FIG. 52. Thus the difference in current on each side of the line element is . i -f dx i = dx. ox ox de or, .'. dx = iRdx, dx de = ^R dx (7) The difference in current on each side of the element is the charging current of the element. or . di , nj de . . dx = Cdx dx dt di = de dx dt (8) 116 ELECTRICAL ENGINEERING di 1 d 2 e From equation (7) we get: - = >, ~~ 2 oX it oX l_ d 2 e _ d^ ' ' R dx 2 ~ di' or, d 2 e de Referring now to the general equation, it is seen that k 2 = CR, [RC0 and y = e, and a = a = \l~cT- .'.e = A^s'm&x + ft + 71) + A 2 e~ ax sin(- ax + ft + 72) (10) and, 1 de A/2a r . / TT\ * = Rdx= -ir[^ a *^(+ fl + Ti + y)- A 2 e-*sin(- ax + ft + 72 + ^)] (11) Case (a). Alternating current supplied to a circuit of distrib- uted resistance and capacity. Example No. 1. If the voltage at the generator end of the line (x = e) is e = E sin cot, and if the cable is open at the receiving end, then i = for x = and all values of t, and e = E sin ut for x = e. From (10) E sin ut = Aie al sin (al + ft + 7i) + A 2 ~ al sin (- al + at +72) = [Aie al cos (al + 71) + A<*r al cos (- al + 72)] sin ft + [Aie al sin (aZ + 71) + A 2 ~ al sin (- al + 72)] cos # .'. Aie al cos (aZ + 71) + A 2 e~ al cos ( al + 72) = 1? and |8 = w, and Aie a/ sin (aZ + 7i) + ^2- z sin (- al + 72) = (12) For x = 0, i for all values of t from (11), = Ai sin (# + 71 + j) - A 2 sin (# + 72 + | Since this must hold for all values of i, Ai = Az = A and 71 = 72 = 7 Then from (12) we have: E t al cos (al + 7) + e~ al cos ( - al + 7) and, e al sin (al + 7) -f e~' sin ( al + 7) =0 or, RESISTANCE AND CAPACITY f = tan" 1 I al _ al tan al , and 117 A = e = A [e ax sin (ax and. = ^-~ A [e a * sin ( E (13) 2al + -2aZ + 2 (COS 2 ttZ - SHI 2 (rf) -f- 7) + e-*sm (- a* + a>* + 7)! (14) (15) I ax + a>c + 7 + e~ ax sin f ax + a>2 + 7 + Where A and 7 are given in (13) and o- +^/^ If the voltage at the receiving end of the line were known rather than the voltage at the generator then : e = EQ sin ut for x = 0, EQ being the maximum value of the voltage at the receiving end of the line. Thus, EQ sin ut = AI sin (/ft + 71) + A 2 sin (/ft + 72) = sin /ft(Ai cos 71 + A 2 cos 72) -(- cos /ft(A x sin 71 -f A 2 sin 72) .'. AI cos 71 + A 2 cos 72 = EQ, fi = u and, AI sin 71 + A 2 sin 72 = (16) For x = 0, i = for all values of t (assuming again an open line). .'. AI sin \8t -\- 7i + T) = A 2 sin (/ft + 72 + j) AI sin /ft cos (71 + ^rj -f AI cos fit sin (71 + T) = A 2 sin fit cos (72 H- jj + A 2 cos fit sin (72 + jj In order that this shall hold for all values of t, the coefficients of the similar trigonometric terms of t must be the same. .*. AX cos (71 + T) = A 2 cos (72 + T-J and, or. AI sin (71 + jj = A 2 sin (72 + - .'. tan (7! + j) = tan (72 + | j , 7i = 7? = 7- (17) 118 ELECTRICAL ENGINEERING Then from (16) (Ai + A 2 ) cos 7 = EQ and (Ai -f A 2 ) sin 7 = .'.7 = and AI + A 2 = #o. From (17) A , . A - -- ^-1 /12 A . . A - ~ Therefore 771 e = __[ a* gm ( aa . _f_ ut ) + -ax gm (_ a:C and, where a = ~\~ \l~~n & n d .E'o is the maximum value of the e.m.f. at the receiving end. In the examples, both 1 and 2, the current leads the voltage by 45 at all points of the line. Let CQ be the voltage at the receiving end and e\ that at the generating end. From (14) 2E sin (co + 7) e Q = 2A sm (wt + 7) = , V>' + *-** + 2 (cos 2 al - sin 2 al) and ei = E sin From (18) e Q = EQ sin and TJ ci = ^[c ai sin (al + + e~ a ' sin (- aZ + )] TTf = -n[(e al + ~0 cos aZ sin co^ + (c ai e~ al ) sin ai cos = 2^o e 2 ^ + ~ 2 ^ + 2 (cos 2 aZ - sin 2 al) sin (co* - 7). Hence both equations show that, (a) the voltage at the re- ceiving end leads the generator voltage by an angle 7, and (6) the maximum voltage at the receiving end is: _ 2 _ V* 201 + e- 20 ' + 2 (cos 2 aZ - sin 2 aO times the maximum generator voltage. In fact examples (1) and (2) refer to one phenomenon, but one terminal condition already known and one terminal condition to be determined are interchanged in the statements of the examples. RESISTANCE AND CAPACITY 119 Example No. 3. The same phenomenon may be studied in still a different way, namely, measuring x from the generator end, that is, x = I refers to the receiving end of the line. When the generator is taken as the point from which the dis- tance is measured, then, as the voltage and current decrease as x increases, we have: dx = iRdx, C/Jv and di , de - dx = Cdx dx dx which by a similar transformation, also resolves in the differ- ential equation : *_!?. _ rp de - dx* ~ K dt .*. e = A^ ax sin (ax + &t + 71) + Ax~ ax sin (- ax + pt + 72) and, For x I. i = for all values of t. .'. A** 1 sin (al + fit + 71 + j) = Ax~ al sin (- al + pt +. 72 + | or, = A! ai [sin pt cos (al + 71 + ^) + cos # sin (+ al + = A 2 e- ai | sin ^ cos ^ a^ + 72 + ^j + cos pt sin f a/ + 72 + 4) I As this must hold for all values of t, .'. Aie al cos (al + 71 + ^) = A*-" 1 cos (- al + 72 and, Ai6 fll sin (a/ + Ti + |) = A 2 c- ai sin (- al + 72 + .'. tan (al + 71 + ^j = tan ( al -f- 72 -f- |j /. 72 = 7i + 2aZ, and it follows that, At = Aie (20) 120 ELECTRICAL ENGINEERING For x = 0, e = E sin ut ; .*. E sin cot = Ai sin (fit + 71) + A 2 sin (j3t + 72) = (A i cos 71 + A 2 cos 72) sin pt + (Ai sin 71 + A 2 sin 72) cos fit. In order to make this hold for all values of t, |8 = CO AI cos 71 -f- A 2 cos 72 = E, and Ai sin 71 + A2 sin 72 = 0. From (20) Ai sin 7! + Ai 2oZ sin (71 + 2al) = .*. sin 71 = 2aZ (sin 71 cos 2al + cos 71 sin 2aZ), - e 20 * sin 2aZ ' ^ = tan 1 + e-cos^ Let 7 = 71 + al, then 71 = 7 al, and 72 = 7 + aZ. And let A = Aie z = A 2 e- z . Then, pi e~ al cos (7 - al) + c aZ cos (7 -f aZ) = -T (22) From (21) and 71 = 7 a, we have: sin (7 al) e 2aZ sin 2al tan 71 = tan ( T - = COS(Y _ oi) - j + cos 2al > .'. 7 = tan i r-?l- ( -l _ e0( + e - ol which is the same as that in example (1) Substituting the value of 7 in (22) V ^ al + e~ 2ai + 2(cos 2 al sin 2 aZ) also the same as that in example (1). Hence, e = A[e~ a(l ~ x} sm( al x + wt + y) + c^^^sin (aZ x + ut -f 7)] sn a which are identical with the equations obtained in example (1), only with (Z x) in the place of x. RESISTANCE AND CAPACITY 121 It is noticed that at any particular point of the line the current and e.m.f. waves are sine waves. The wave length X is found when, x \^~ = .*. X = X i = = 2ir\hnr CRa) \CRu mr/~ \fCR' The time required for the wave to go one complete wave length is H- Thus the velocity of propagation is: distance / w I Ifw /2o> Thus the velocity of propagation is proportional to the square root of the frequency. Higher harmonics travel faster than the fundamental. The third harmonic travels 73 per cent, faster, etc. But while the higher harmonics travel faster than the funda- mental their attentuation is greater as will be seen. When the wave has traveled one complete wave length, that is, when x = X = ^\J~ ( The exponential term becomes: 2 CRu = e~ 2v = 4- = 0.0019. That is, the wave is only 0.2 per cent, of its original value. It has reached = = 0.368 of its original value when CRu ~~ = l r x ~- = 2 / 2 /I cT^ = VCRW = \Cte Thus the third harmonic has decreased to 37 per cent, of its original value in a distance which is only 58 per cent, of that required by the fundamental to be reduced to 37 per cent, of its original value. 122 ELECTRICAL ENGINEERING To find the time for the wave to decay to - of its original value, we have: distance / 1 //TT time = ; r? = -\/ ~^ -j- 2 A/T^: = velocity \CRirf \CR Thus the time required for a given decay varies inversely as the frequency. The third harmonic requires only one-third of the time of that of the fundamental. Instance. A concentric cable 100 miles long. Assume a capacity of 1 m-f. per mile to the neutral. Using the mile as the unit of distance, C = 10 6 . Assume the cable to have a resistance (of one conductor) of 1 ohm per mile. Then R = I At 60 cycles, / = 60 and co = 377. /2 X 377 .'. Velocity of propagation = A/ 6 = 27.500 miles per sec. The velocity of the triple frequency wave would be, \/3 X 27,500 = 47.500 miles per sec. The main wave is reduced to 37 per cent, of its original value after = = 0.00265 sec.; and the triple frequency wave is reduced to the same fraction in one-third of the time or 0.0009 sec. In the first case the wave has traveled 73 miles; in the case of the triple harmonic 42 miles. Problem. Develop the equation of the voltage and the cur- rent in a closed cable under alternating impressed e.m.f. Case (6). Direct current supplied to the cable. Example No. 1. Consider the line open at the receiving end (x = 0). Assume, e = K + 2A ax+bt sin (ax + ft + y) where x = 1, e = E for all values of t, This is evidently only possible if 2Ae al+bt sin (al + pt + 7) = and K = E. RESISTANCE AND CAPACITY 123 As this must hold for all values of t, p = 0, and al + 7 = nir. 1 7T It is found convenient to let 7 = ~, that is, to make z cos al = 0, and al = mr + = i^L+ii*: where 7 TT STT STT 7?r ai== 2 ' "2" T T' etc " Thus, e = E + ~SAt ax + l " cos f) f* . 2 = a ae ax cos ^ _ aae ax sn ot Substituting these values in the general equation d*e de .W Tt and equating the coefficient of similar trigonometric terms we get: a 2 -a 2 ~CR~ and ax = or a = 0, since a. can not equal zero. 'eosia+M? (23 ) 1 sin v ' ' (24) K dX K n = o 2i When * = 0, x < 1, e = n = 1 In this equation appear several constants, some of which are determined by the terminal conditions, others by mathematical transformations. It is, of course, possible to do a certain amount of choosing as long as the choice satisfies the differential equation as well as the known conditions which exist in the problem. So, for instance, we may assign an arbitrary value of 7 and carry the calculations through when we may find that the final expres- sion is simple or too complicated to be of practical value. It is reasonable that in the first trial 7 may be assumed as zero. When the problem is worked out on this basis it is seen that the answer is not sus- ceptible to a simple equation. The trial will suggest another value, most likely 7 = - |. This is therefore used. 124 ELECTRICAL ENGINEERING In order to determine the values of A n , multiply both sides of (24) by cos - dx and integrate between and e, thus C Jo e 7r(l + 2k}x n =~ . 7r(l + 2n)z, cos ^ J - 2, A n cos- -7^7 - dx 21 n =o 21 Each term on the left-hand side equals zero except that one which has n = k, and hence this particular value is used, and we have A, cos * * :. A l = LkA B (16) A 2 = - LkA 7 (17) and, -A 9 = LAA 6 (18) -A 4 = - LkA b (19) For x = I, i = for all values of t, .' . = A 5 sin al sin kat + A 6 sin al cos fcotf + A 7 cos sin fccx -f A. 8 cos cos fc (B) .'.A 5 sin a? + AT cos a/ = .'.A e sin al + A 8 cos al = .'.A 7 = - A 5 tan Z (20) /.A 8 = - A 6 tan al (21) .'. e = LkAs sin ax sin fcotf LfcAr sin ax cos fcZ - LkA 6 cos ax sin kat + LfcA 5 cos ax cos fcaL For a; = 0, e = E sin ut .'. # sin ut = LkA & sin kat LkA$ cos /ca^, W .' . E = LkA & , or A 6 = yr-' A 5 = 0, and co = ka. From (19) and (19) A 3 = E and A 4 = From (20) and (21) A 7 = W and, A 8 = yy tan al From (16) and (17) A l = 7 tan aZ and A 2 = Therefore, A^ = E tan al A 2 = A 3 = # A 4 = A 6 = A A 7 = A 8 = TT ^ an a ^ i>/C e = E tan a? sin ax sin fcatf + E cos ax sin INDUCTANCE AND CAPACITY 131 ' i = -- v-r sin ax cos kat + -f-p tan at cos axkat. LK LK Substituting, , _ 1 _ _ " VLC * e = #[tan ul-\/LC sin co\/LC sin co + cos u\/LC x sin coj], rr v-~ or, e = E sin w<[tan ul^/LC sin u\/LC x + cos u\/LC x], / _ _ i = E-l~ cos co^ftan coZ\/LC cos u\/LC x sin w\/EC a; rr t = v-~[tan wl-\/LC cos w\/LC x cos co sin u^/LC a; cos coZJ, or, cos co \-LC (Z a;) e = # sin coi - , (22) cos co v LC t C ^sin cos coi - (23) The voltage at the end of the line is : E sin co^ Example. If the receiver voltage, instead of the generator e.m.f. is known and if the distance had been counted from the receiving end of the line, then i = 0, for x = and, de _ di d~x~ ' L dt Thus the signs for A 5, A 6 , A^ and A*, in equation (A) would have been reversed. :.Ai = - AsLk. A 2 = + A 7 Lk. A 4 = Equation (B) would have been: = A 7 sin kat + A 8 cos kat, .'. A 7 and A 8 = .'. e = LkAs cos ax sin kat LkA & cos ax cos fc For c = 0, e E Q sin ut. .'. EQ = sin co = LkA 6 sin kat .'. EQ = LkA 6) and A 5 = and o .". e = EQ cos ax sin fca Z = s * u ax cos ^ a ^ 132 ELECTRICAL ENGINEERING or, e = EQ cos to\/LC X sin coi (25) i = Eo+Lr sin u\/LC X cos WT (26) \L Therefore the generator voltage is: e = cos u-\/LC IE Q sin cot = E sin co, cos co \/LC I and (25) becomes: cos co \/LC x e = E sin coi - , ? cos co \/LCl which is obviously identical with (23) as obtained before. It is seen at once that the receiver voltage is 1 cos co \/LC I times that of the generator e.m.f. As the cosine is always less than unity except I = 0, the receiver voltage is always greater than the generator e.m.f. Therefore the receiver voltage would approach infinity, when 2wfVLCl = I , J_ _1 JL _ 4(Ll)(Cl) 4L Co that is, when the natural frequency of the line and the frequency of the impressed e.m.f. coincide. The wave length is X = / WV.L/O Co X 2 ^/ 1 Thus the velocity of propagation = = /, - == /, ^ 1 coV^o^o V-^o GO If the inductance inside of the conductor is negligible, then the velocity becomes that of light = 188,000 miles per second. In reality it is somewhat less. So for instance in a transmission line consisting of No. B. & S. wires, 18 in. apart, L = 1.6 X 10~ 3 henrys per mile. C = 0.019 X 10~ 6 farads per mile. -4= = 182,000 miles per sec. INDUCTANCE AND CAPACITY 133 For short distances, sin co -\/LC x = u-\/LC x cos co \/LC x = I .*. e = Eo sin ut Ic i = Eo -^1 co -\/LCx cos jC/o cos co = cos co X c where x c is the capacity reactance of length a of the cable. It is seen that the current in time phase leads the voltage by 90. Transient Condition. When a steady voltage is impressed upon the circuit. DR. FRANKLIN in his book on waves and his paper before the A. I. E. E. of April, 1914, has approached the subject from a most simple and instructive point of view and has been able to make some generalizations which are of great value. He shows that whatever the distribution of the current or e.m.f. in a travelling wave along a transmission line there must be a fixed ratio between the instantaneous values, which ratio is C Y when the line resistance and leakage reactance are negligible, and it can be represented by a somewhat more complicated ex- pression when they are taken into consideration. His reasoning is briefly as follows: If the current in an element of the line is i the magnetic flux in the area a, b, c, d, Fig. 53, is Lidx. If the current wave progresses toward the right with a velocity V the time required for the flux to sweep past be is -y] thus the e.m.f. induced along be is ^ x = LiV. y Similarly if e' is the voltage in the line element then the charge on ab is e'Cdx. This charge flows past the point in time y, where V is the velocity of propagation of the e.m.f. distribution; thus ., _ Q _ Cdx 1 t = '' dx :=CeV ' V 134 ELECTRICAL ENGINEERING In order then that these distributions shall sustain each other, i = {', e = e r and V = V. * .'. e = LiV and i = CeV or and Ce' - Li- or,^ = + - e ' \L The -f sign belonging to outgoing waves, arid, the sign to the reflected waves. . i 1C , i' 1C i i' . . - = + A / T and -SB - /-- or-= - e \L e \L e e where index ' refers to the reflected waves. When the line is open at the receiving end the sum of the in- coming and reflected current waves must be zero, thus i -\- i' = , ., e i . c' = + e . . e' = i' -. = -f- -e = + e . i i When the line is short-circuited at the receiving end, e + e' = .'. e f = e. Thus i' = i. When the receiving circuit is noil -inductive and of resistance/^, e + e' = R(i + i'} but e 1L e substituting these values above, then, (R - a) . a - R e = e ~^r- t and i = i ^* R + a a + R It is seen that the reflected current and e.m.f. waves may be positive, zero, or negative, depending upon the relative values of R and .*/- In DR. FRANKLIN'S American Institute Paper (April, 1914) is given a very full discussion of the nature of these reflected waves and some highly instructive diagrams are shown. For example, when the receiving circuit is inductive the line acts at the instant of reflection as if it were open circuited, since the current can not rise instantaneously in an inductive circuit. After some time the condition becomes that of a non-inductive INDUCTANCE AND CAPACITY 135 receiving circuit, discussed above (since we are dealing with direct-current voltage). Between the two periods of time the current and e.m.f. change according to a simple exponential law. Of special interest is the condition of the waves when the line constants change. DR. FRANKLIN illustrates this condition in the case of an overhead line connected to a cable system. Let i y i r and i t be the instantaneous values of the outgoing current, the reflected current and the transmitted current, and let e, e r and e t be the corresponding values of the e.m.f. Then e + e r = e t i + i r = it e e r e t * - = - = a ~ = b + I Ir It From these equations are found b -a It is of interest to apply these simple relations numerically. Assume that the inductance and the capacity of a cable sup- plying power to an overhead line are: L = 0.0002 henrys and 8 C = ~Q farads per mile, and that the corresponding constants for the overhead line are LI = 0.0015 henrys and C\ = ~T^ farads per mile. dllCl -, ; \J*\J\J1.U /, ~ v_ i If therefore such cable-overhead line combination is connected to a source of steady e.m.f., e, the voltage at the junction as the 548 wave reaches it will be e t = oo? o = 1-88 times that at the .6OC7.O 136 ELECTRICAL ENGINEERING generator. Should the overhead line be open at the receiving end the voltage will be doubled as the reflected wave starts on its journey back. Thus as a maximum at the junction the volt- age would equal 3.76 times the impressed value. The mathematical solution of the problem is given in equation (11) which can be written in the following way: e = K + ZA sin ( ax + kat + 7) (27) where = + a applies to the waves issuing from the generator and a to those going toward it. From the expression ax + kat, it is seen that the waves of all frequencies travel with the same velocity, +k or k where the signs indicate the direction of motion. It will be shown that in the case of an open line connected to a source (of negligible resistance) of undirectional voltage, four waves have to be considered before the cycle repeats itself. First the outgoing rectangular wave of value E which begin- ning at the generator progresses toward the open end of the line. Second the reflected wave also of strength E which returns from the open end toward the generator which with the initial wave gives a wave of double voltage. Third a negative wave of strength E which progresses from the generator toward the open end of the line, which wave is necessary in order to maintain the generator voltage E. Fourth the reflected wave of the negative wave which is of strength E and which progresses toward the generator. Consider now what happens at a point located say at one-fourth of the length of the line from the generator. If the time required for the wave to reach the end of the line is T, it is evident that during Y T there is no voltage at the point. After that time the voltage remains constant at a value E until the first reflected wave arrives. This occurs evidently when t = 1%T. Thus between t = ^ and t = 1.757 7 the voltage at the point is E. From that on it has a value of 2E until the negative gener- ator wave reaches the point which occurs when t = 2T -f- T - = 2.25 2 T . After that time the voltage has a value of 2E - INDUCTANCE AND CAPACITY 137 E = E until the reflected wave of the negative wave arrives T which is when t = 4T - -j = 3.75 T 7 . Then the voltage = 2E - 2E = 0, and it remains zero until a time t = 4.25T 7 when the voltage again equals E and the cycle is repeated. The result is the wave shown in Fig. 54. E.M.F, Wave E ',., 5 1 2 .0 1 5 2 2(1 -X) 2 5 3 2 3 X .5 J.O 2(1 -x) 4 5 k A V fc Q. r,i A k A train of waves would pass the point indefinitely since we have neglected the energy loss in resistance. The wave length is evidently four times that of the open line. Consider now the current wave of Fig. 55. As successive equal elements of the line are being charged to voltage E a constant current has been shown to flow from the generator while the voltage wave progresses toward the end of Current Wave l-x Jc 1+ X k 3i- X ** t Ti 31 4 X k -J FIG. 55. the line. At the end the current must be zero, therefore the reflected current wave must be equal but opposite to the incoming wave. The reversed current reaches the generator after a time 2T, when the current becomes zero. After that time the genera- tor supplies E voltage and a negative wave of current flows until it also is neutralized by the reflected current which occurs when t = 4T. Consider the current at the particular point mentioned above. 138 ELECTRICAL ENGINEERING T From t = to t = -j no current flows. After that the current is constant until the reflected current reaches the point (at t = 1.75T) when it drops to zero. It remains zero until the negative current issuing from the generator reaches the point fat t = 2T T\ -f -T) . Then it becomes negative and remains negative until the negative reflected current now positive reaches the point It = T\ 4T -j] t when it again is zero and so forth. In general centering our mind on a particular point x, from the receiving end of the line there is no e.m.f. or current at that point until t = JT . After this the voltage is E, the generator voltage for some time. At t = T, the waves reach the end of the line and I + x reflect, therefore after t = j- , e = 2E for a period of time. At 21 t = -j-, the waves return to the generator end. In order to keep the voltage at the generator end constant at E the generator must now begin to supply E. Therefore after time t = ]T~~> e at x becomes 2E E = E] and after t = -W , e at x becomes 2E 41 2E = 0. At t = -r the generator reverses its voltage from E to +E again, and the voltage at x repeats its cycle again and again. Referring now to equation (27) e = K + SA sin (ax + kat + 7) + SA' sin (- ax + kat + 7') (28) when x = I, e = E for all values of t. :. 2A = SA' and sin (al + 7) = - sin (al + 7') *v' 'Y ~4~ TlTT = sin (mr al + 7') thus a = - ^ where n = is an odd number (29) and K = E :. e = E + SA[sin(aa? + A;a/ + 7)+sin(- ax + kat + y')] (30) INDUCTANCE AND CAPACITY 139 At the receiving end of the line (x = 0). e = for all values of t which are less than T. But when t = T the voltage is 2E. /c Thus t = is a transition point, a point of discontinuity, e is either or 2E. Substituting the two values in equation (30) we get: + E = 2,4 [sin (al + 7) + sin (al + 7')] respectively (31) At the open end of the line where there is complete reflection the incoming and outgoing waves are identical .'. 7 = 7'. Thus from (29) a = -~j where n is an odd number. .'. (31) becomes + E = 22A sin ( + 7) (32) In the development of the trigonometric series it is found that : 4 n=c sin nO + 1 = - - S - where n is an odd number (33) where the negative sign refers to values of 6 between ?r and 2w and the positive sign to values between and TT or TT < 6 < 2ir for negative sign. < 8 < TT for positive sign. See "BYERLY'S FOURIER'S Series and Spherical Harmonics" (page 51). Comparing equations (32) and (33) it is evident that: _ E 4 1 An ~ 2 IT n and nw y + T = nfl. It remains to determine the value of 6. The two series have TT and 2-K or in common. It remains to choose the proper value of these. If TT were chosen the signs in (32) and (33) would be reversed, if however, or 2ir is chosen, the signs are satisfied. Thus mr nir + 7 = 2n?r or . . 7 == - 140 ELECTRICAL ENGINEERING Thus equation (30) becomes: e = E + j- -S^Fsin ^ (x + kt - 1) + sin ~ (- x + kt - I)] (34) a Tt H L 1 &l J and since de di^ dx ~ dt 2 i[sin ^ (a + fc - - sin ^ (- s + & - Z)] (35) The curves drawn in Figs. 54 and 55 may be verified as follows : = for t < x = T Z -x k x -\- kt I < x -\- I x I or smaller than .'. 6 in the trigonometric series lies between TT and 2ir. Therefore S - sin (a; + kt - Z) = - . Tl 4 Consider with the second term in (34), x + kt I < x+l 31 x I, thus smaller than 2x or smaller than 2 + -r or 1.51 thus 6 is again negative and the series of the second term in (35) - -7 kt l = X-\-l adds up to - .'. e = E -\- ^ - ( -7 jj =0 which agrees with the curve. When k = -- = = we are in the first quadrant and - x + kt - I = - x + I - I = - x = - 6 lies in the fourth quadrant .*. 2J- sin ( x -\- kt 1) = 2" .'. e =1 which agrees with the curve. 07 F or j = _ k + kt-l=*x + 2l l = x + l= 1.751. 6 lies in the second quadrant- thus: INDUCTANCE AND CAPACITY 141 and - x + kt - I = - x + I = 0.25Z. 6 lies in the first quadrant, thus : Z^sin (-x + kt - Z) - + TV 71 .'. e = 2# which agrees with the curve. The current wave may similarly be checked. When for instance t = T, it is readily seen that the algebraic sum of the trigonometric terms become j E 4 1C TT 1C i = ?r ~\ IT o = E \ T which agrees with the curve. A 7T \-L/ A \ Li It is thus seen that when considering the outgoing waves only the i JC relation between the current and e.m.f. waves must be - = \ T> e \L the equation also shows that when considering the reflected waves The effect of the line resistance is to taper the waves so that instead of their being represented as a ribbon of parallel sides the sides slant toward each other; thus the reflected e.m.f. wave is not as great as the original wave, and the line soon reaches a state of permanent condition. In reality the wave front is not vertical but slants and the corner is rounded off, due to the skin effect of the conductors. The higher harmonics of the current meet a much higher resist- ance than do the lower, and hence the resistance is not a constant quantity but different resistances should be assumed in connec- tion with the different harmonics. The mathematics involved becomes, however, altogether too complicated for any practical application. The important point is that if the values of the waves are determined in a cir- cuit having no resistance, the most pronounced variations in current and e.m.f. are discovered. A circuit having no resistance and no leakage is said to produce pure waves the characteristics of which are, as has been shown, such that 142 ELECTRICAL ENGINEERING That is the electric energy is always equal to the magnetic energy. The wave may, however, be pure even if there is resistance and leakage but in that case the energy dissipated in heat per unit length of line must be equal to the energy dissipated by leakage in the electric field. e 2 .'. i 2 R = 77- where R i is the leakage resistance per unit length. /?/? e2 L ' ' RHl == i* C A line in which this relation exists is called a distortionless line. For a full discussion of such circuit the reader is again referred to DR. FRANKLIN'S book on waves. CHAPTER IX DISTRIBUTED RESISTANCE, INDUCTANCE, LEAKAGE, CONDUCTANCE AND CAPACITY Let R, L, G and C, Fig. 56, be the line constants per unit length of the line, K being expressed in ohms, L in henrys, G in ohms and C in farads. The voltage equation is evidently or, or, de . di dx = Rdx i -\- Ldx -- ox ot de . di dx dt dx = Gdxe 4- Cdx TT ox dt (1) (2) R L FIG. 56. Differentiating (1) partially with respect to x and (2) with respect to t and combining the results with (1) and (2) we get: ^ = LC d ~ + (RC + GL) ~ + RGe OX" Ol" Ol and, (3) (4) where a' and & may be positive or negative, real or imaginary, simple or complex. 143 The general solution of these equations is 144 ELECTRICAL ENGINEERING Substituting the general solution in (3) or (4) and equating the coefficient, we get a' 2 = LCp'*(RC + GL)ff + RG (5) Substituting a + ja for of and b + JP for ff in (5) and separat- ing the real and imaginary terms, we have a 2 - a 2 = LC (6 2 - 2 ) + (RC + GL) b + RG (6) and 2aa = 2LC6/5 + (RC + GL)/? (7) A slight consideration shows that the exponential solution given above can be written e = k + ZAt* ax bt sin (pt ax + 7) (8) If now for the sake of simplicity only the permanent condition is considered we get e = k + SAc ax sin (pt a* + 7) (9) If as a further limitation the current and e.m.f. are assumed to be simple sine functions, depending in time upon the impressed frequency, then p has only one value /, then E = sin ut = A 1 [e al sin (pt + al + 71) + t~ al sin (pt - al + 7^] which by simple transformation becomes E sin at = A i (sin pt [e al cos (al + 7i) + e~ al cos (- al - 71)] + cos pt[e al sin (al + 71) + e~ al sin (- al + 71)]} + c- 20 ' + 2(cos 2 Z - sin'aZ) sin (pt + B) DISTRIBUTED RESISTANCE 145 where , _ _ e al sin (al + 71) + e~ al sin ( al + 71) al cos (al + 71) + e~ al cos ( al + y^ Thus 13 = co and = # = V^M- = Biae ax sin (cot + ax 4- ^>i) Bia*~ ax sin (co a ox + Biae ax cos (co^ + 0:0; + <^?i) Biat~ ax cos (coi ca z sin (co ax + <^ 2 + pyf? [ e az sin (w^ + aa; + 71) e~ oa; sin (coi ao: + V^ + e~ 2al + 2 (cos 2 al - sin 2 ^) 146 ELECTRICAL ENGINEERING where [ e al f al ^r+7^ tan a 1 a = tan" 1 a . Ceo

be the flux in the tubular element of radius x, and + d is the increment of flux in the tubular element, as x increases from x to x + dx, but the total flux in the tubular element is 0. is the result of the external m.m.f . and the m.m.f. (demagnet- izing) due to the current between x and XQ', + dcf> is the result of the external m.m.f. and the m.m.f. due to the current between x + dx and XQ. Therefore d is caused by the decrement of demagnetizing m.m.f. due to the current between x and x + dx, i.e., within dx. Let i be the current density at x, then the current within dx is ildx, and the m.m.f. due to it is also ildx, as the number of turns is unity (7 being the length of the cylinder). Let B be the flux density at x then dB the increment of flux density as x increases from x to x -\- dx. Thus d(j> = 2irxdxdB. m.m.f. Since flux = 0.4 TT and the reluctance in this case is -= reluctance I 2ir X dxu, * OAirildx We get d

jr X-d X dB = - , thus dll = . ,^ If p is the spec, resistance then the resistance that the current within dx meets is I 7.7 and the e.m.f. consumed by the resistance = ildx I yr~ = 2-jrpxi. Let e be the e.m.f. induced in the circle of radius x, and e + de that in the circle of radius x + dx. As no external e.m.f. is applied around the circle of radius x the sum of the consumed and the induced e.m.fs. is zero, thus: e + Zirpxi = (2) Substituting (2) in (1) UlJ v/.i/iyuc/ /Q\ " - ^r W 152 ELECTRICAL ENGINEERING e is induced by all the flux within the circle of radius x, and e -f- de by all the flux within the circle of radius x -f- dx, thus de is induced by all flux in the tubular element 2irxdx, which is 1 d according to our notation. Hence de = ^c s -77 or using partial differentials, de - 10 8 dt or or, 0.2-n-xdxdB de 10 8 dt 2irx dB io 8 l>i (4) Equation (3) may be written: = Differentiating with respect to x, d 2 B dB X r -f - = dx 2 dx Combining (4) and (5), dividing by x d 2 B I dB 0.47rju dx 2 " + x dx = : T0 8 P Q.47TM de +x> dx dt (5) (6) A long thin flat bar may be considered as a cylindrical bar of infinitely small curvature or infinitely large radius, thus x considered as the radius becomes infinity andeauation (6) becomes: d*B dx 2 0.47T/Z dB 10 8 P dt Wmdmg while dx and x take the meanings as shown in FIG - 58 ' the Fig. 58. Equation (7) may be directly derived from consideration of a flat bar in place of a cylindrical bar. Distribution of Current in Cylindrical and Flat Bars. Reason- ing as in the previous paragraph, but considering the current and flux interchanged in their places, not only similar but also iden- tical equations will be derived for the distribution of current. DISTRIBUTION OF FLUX 153 Let B in Fig. 59 be the flux density at x, i be the current density at x, and i + di the current density at x + dx. The flux in the tubular element dx is Bldx. The reluctance of the flux path is r-r- 0.47T m.m.f . Ihus since rlux = reluctance 2irBx m.m.f. = m = j^-r 2 ATT p. (8) (Current I 1 Impressed i 1 E.M.F. 1 FIG. 59. As x increases from x to x + dx, the m.m.f. increases from what is within the circle of radius to that of radius x + dx dm = 2irxdxi, or -T- = 2irxi where i is the current density at distance x Differentiating (8) dm dx Equating (2) and (3) 27T 0.47TM (+') f. dx , o. 4 ,,,- x (9) (10) (11) As x increases from x to x + dx, the increment of current density is di, and the increment of current in the tubular element is 2Trxdxdi. , The resistance of the material that this increment / Therefore the increment of the of current traverses is consumed e.m.f. is: 2irxdx 2irxdxdi 2-irxdx P Ui. 154 ELECTRICAL ENGINEERING Hence the decrement of the induced e.m.f., de is pldi. This -de is caused by the flux in the tubular element, viz., Bldx. Therefore de= pldi = -- ^ -r.Bldx, J. \J Ct'L using signs of partial differentials, and re-arranging, we have : dx = W~p dT Differentiating (11) with respect to t and (5) to x, d 2 B I dB di + - = 0.47T/*- (13) (14) a* ~ lov at The solution of equation (15) is somewhat difficult and is therefore delayed until later (see Skin Effect in Cylindrical Conductors). Equation (16), however, which shows the flux distribution in a lamination is readily solved when the impressed m.m.f. and there- fore, at least in non-magnetic materials the flux density is a sine function of time. Let B = B m sin cot - The effective value of the first expression may be represented by vector OA = B, and that of B A the second expression in the derivative by OM = FIG. 60. dxdt ' x dt " dt and d 2 i 1 d 2 B dx 2 10 8 p dxdt Substituting (-12) and (14) in (13) d 2 i \ di 0.4717* di dx~ 2 ~^~ x dx = 10 8 p ~di For long or thin flat bars, di Thus dealing now with effective values we can write: DISTRIBUTION OF FLUX 155 where P 10 8 To solve this equation we write, d 2 B - v 2 B = /. m 2 - v 2 = .'. m = v and B = A^ vx + A^e~ vx . Since the density must be the same at equal distances from the center line A^ vx -f A 2 e~ vx = Ai~ vx + Aze Lvx which requires that A i = A 2 = A 2 _ 04^ plO 8 ' It is readily seen that if v = (1 + j)a V 2 = 2ja 2 2j P 10 8 - oa: ~ Jaz or B = A[e ax e jax + e- oa: c Substituting trigonometric expressions and combining the real and imaginary terms we get: Since e :iax = cos ax j sin ax B = A[(e ax + e~ ax ) COS ax + j(e ax e ax ) sin ax]. If B i is the effective value of the surface density then B = BI for x = d. If, furthermore, it is assumed as an approximation that e~ a& is small compared with e aS then, B l = Ae a5 (cos ad + j sin a 8) and flic"*' cos a8 + j sin ad t * + e " X) COS + " - ") n ] 156 ELECTRICAL ENGINEERING and B Bie~ a V e 2ax + e~ 2ax + 2 cos 2 ax where as given above <0.27r2/^7 : \ p io 8 For iron p is approximately IO 5 and /* may be anything up to 18,000. 1 For copper p is approximately and i = 1. PART II. PROBLEMS IN ELECTROSTATICS CHAPTER XII FUNDAMENTAL LAWS Coulomb's Law. The fundamental law upon which our know- ledge of electrostatic or electromagnetic phenomena rests was found experimentally by COULOMB. It is similar to NEWTON'S law of gravitation and is: F = c^orF = c^ (1) iv>i ntZ where F is the force acting upon the point charges Qi and Q 2 , or the magnet poles of strength mi and w 2 , c is a constant depending upon the system of units employed, and r is the distance between the charges or magnet-poles, respectively. In the electrostatic system of units, as well as in the electro- magnetic system of units, c is taken as unity when the medium is air, or rather vacuum, .:r-3&.,**r-*F- . (2) where F is expressed in dynes. Thus two unit charges or two unit magnet-poles repel each other with a force of 1 dyne when separated 1 cm. If the medium has a specific inductive capacity K, then 1 = K r* If the magnetic medium has a permeability /z, then The strength of unit poles is then measured assuming that it were possible by the repulsion between two similar poles. When the force is 1 dyne and the distance is 1 cm., the poles have unit strength. 157 158 ELECTRICAL ENGINEERING Field Intensity. Surrounding electric charges or magnet-poles is a field, and the intensity of the field at a point is defined as the property of the space, which is measured by the force exerted by the field on unit charge or unit pole located in that point, when electric and magnetic fields, respectively, are considered. Because of this definition, it must not be inferred that the intensity of the field is a force; it is not a force, but merely a space function just as the gravitational field intensity is a space function. The force acting on a certain mass at a certain point may have any value, depending upon the particular mass used in the experiment. Important Theorems. While COULOMB'S law forms the basis on which the theories rest, the progress in the art would probably have been slow were it not that a number of theorems have been worked out more or less directly from that law. These theorems are: GAUSS'S theorem, the divergence theorem, GREEN'S and STORE'S theorems, etc., all having important bearing on prac- tical problems. Surface Integral of a Distributed Vector. As a preliminary to these theorems the surface integral of a distributed vector will be defined. It will be assumed that an electric field exists due to some charge and that lines of force or tubes of force radiate from the charge in all directions. It is desired to find the number of lines that go through a surface, say a cap that is placed in the field. In Fig. 61, AB may be assumed to be, for instance, the inter- section of the plane of a loop of wire, over which the cap is made, with the plane of the paper. If the surface of the cap were divided up into a number of elements and the direction and the intensity of the field at every point were known, then it obviously would be possible to calculate the total number of lines (the flux) that crosses the cap or the surface. The sum of the fluxes normal to each element of the surface is called the surface integral of the normal field intensity over the cap, or the total outward flux through the cap. (The normal to the elementary surface is always understood to be drawn out- ward from the surface. On account of the sign of trigonometric function a normal drawn inward will lead to a negative surface integral.) FUNDAMENTAL LAWS 159 If another diaphragm or cap (Fig. 62) were stretched across the wire loop AB, it is evident that a certain amount of flux would enter the space between the diaphragms and a certain amount leave it. It will be shown in this case that the total normal outward flux from the space enclosed by the two caps will be zero, as long as no charged particles are enclosed by the diaphragms. Consider then a distributed vector field (Fig. 63), and let R be the value of the vector at the small surface element dS, making an angle 6 with the normal to the surface element. R represents the electrostatic or electromagnetic field intensity. The field intensity along the normal is then R cos 6, and the flux going through dS is: dif/ = R cos 6dS, i.e., df = R cos (N t R)dS. .'. \l/ = total outward flux through the surface, RcosddS, where N is the component of the field intensity normal to dS, i.e., N = R cos 0. This can also be expressed in rectangular coordinates by vector analysis, provided that dS represents not a surface dS, but a vector at right angles to dS of size dS. (See appendix for dot product.) Let and R = X(x,y,z)i + Y(x,y,z)j + Z(x,y,z)k dS = dS x i + dS y j + dS z k. 160 ELECTRICAL ENGINEERING Then, ffR-dS = ff(XdS x + ~ ,~ ' ^,N / //T?J j where obviously dS z = dydz. YdS y + ZdS.) =ff(Xdy-dz _ y + Zdxdy), Another way of expressing the surface integral of a distributed vector field is: ff(Xl + Ym In these equations X, Y andZ are the components of the vector along the three axes and I, m, and n, the direction cosines of the normal to the surface. Thus: I = cos (N,x), and, m = cos (N,y), n = cos (N,z). .'. IdS = dydz, and, mdS = dzdx, ndS = dxdy. .'. ff(XdS x + YdS y + ZdS.) = ff(Xl + Fw + Zn)dS. Gauss's Theorem. According to GAUSS'S theorem the total normal outward flux from a closed surface containing a charge Q is = 47rQ. Let N be the component of the field intensity R normal to an elemental surface of the bag dS. The theorem can be expressed mathematically by: ffNdS =- 47rQ. Let dw, Fig. 64, be the solid angle at A corresponding to dS or dSi, which is perpendicular to R or, dSi = r 2 w. ffNdS = SfRcosedS = ffRdSi = f fRr*dvu. But COULOMB'S law gives: _QQ or since by definition F _i r 2 F = R, when FUNDAMENTAL LAWS Qi = unity. 161 (1) FIG. 64. the integral to be taken around the entire surface, that is over the complete solid angle, which is 47r, Thus, = 47rQ. ffNdS = 47TQ, or if there are many charges in the envelope, ffNdS = 4irSQ. It is seen that the total flux radiating from a point charge Q or a magnet pole m is

M = N = --> and H = a^ a^/ a^ an and, 7=2; I J dS = - 47rZra, if the envelope contains magnetic particles; and I I - dS = 0; if the envelope does J J on not contain magnetic particles. 166 ELECTRICAL ENGINEERING Equipotential Surfaces shown in Dotted Lines Around Two Point Charges Separated 5 cm.; Q l = +1Q.; Q 2 = - 5. FlG. 68. Lines of Force shown in Fine Lines and Equipotential Surfaces shown in Heavy Lines Around Two Point Charges Separated 5 cm. ; Q t = + 10 , Q 2 = - 5 FIG. 69. FUNDAMENTAL LAWS 167 In these equations, H is the intensity of the magnetic field, L, M and N are the components of the intensity, H, along the three axes. V Application of the Formula, V = 2 - In Fig. 68 is shown the equipotential surfaces between two point charges, Qi = +10 and Qz = 5, separated 5 cm. The potential at any point, P, is obviously : V P - ' + $. ri r 2 The lines of force can not well be shown in a plane, but a fair idea of their shape can be gained from Fig. 69. The direction of each line of force is obtained by combining 72 1, the intensity at a point due to Qi, and R 2 , the intensity due to Q 2 . CHAPTER XIII METHOD OF IMAGES, APPLIED TO THE PROBLEM OF POINT CHARGES +10 AND -5, SEPARATED 5 CM In plotting the equipotential surfaces of the problem given above, it is readily seen and proven that the surface of zero potential, Fig. 70, is a sphere, and that the following relations obtain: (1) a = FIG. 70. 25 _ 50 cm. and p = - Substituting, we get a = =~ X 5 = 3.33 cm. It is evident that the field distribution will not be affected if a grounded metallic sphere of radius, p, at a distance (D + a) from the positive charge is surrounding the negative point charge at B. And it is also evident that the potential will be the same ( = 0), if the charge at B is removed altogether. 168 METHOD OF IMAGES 169 It is thus evident, that, reversing the line of argument, the potential distribution in a system involving a point charge at A and a grounded sphere of radius p with center at a distance L from the point charge, can be determined without the laborious deter- mination of the distribution of the induced charge on the sphere, simply by using two point charges at A and B. The location of B and the charge which must be assumed at the non-existing point B can be determined from the following rela- tions which are easily proven: L P (L - D) = P 2 , or D = L - p *> L and ~ p Potential Distribution between a Point Charge and a Metallic Sphere. While it is evident then that the field can be determined without much labor in the case of a grounded sphere, the problem becomes quite involved if the sphere is insulated and kept at a certain potential, V, which is not zero. To calculate the potential distribution in that case, it is neces- sary to study the distribution of the surface charge. FIG. 71. Consider first the case of the grounded sphere. The intensity of the field at a point is the resultant of the intensities due to the charges at A and 5, the so-called inverse points. It must be expected that the direction of the resultant field is perpendicular to the surface at all points, thus we can draw the 170 ELECTRICAL ENGINEERING diagram in Fig. 71. Remembering that R } the intensity is the vectorial sum of -4 and v or of Ri and ri 2 r 2 The intensities R\ and R 2 are resolved along the radius CP ana along a line parallel to AB. It is seen that PE and PG are equal, and cancel each other, so that the resultant intensity is in line of CP and is PF + PH, algebraically. Let the radius of the sphere in Fig. 71 be p. By similar tri- angles, PF PC PC p. ~R~ = PA ' =/*/! p^ : = *ti , and PH _ PC _ T PC Po " P#' " But and And it has been shown that Q 2 = Qi j-' Li Since La = p 2 , from the figure, it is seen that, n L p ' C 2 72\ C1\ p7? (p - L) By Coulomb's theorem, 47r(T = .R, where o- is the surface density of charge, or charge per square centimeter, - 2 ) (2) METHOD. OF IMAGES 171 When the radius is very large, the surface of the sphere ap- proaches a plane, Fig. 72, and a approaches p. Thus, if d, in Fig. 72, is the distance of the point charge from the plane of zero potential, we have: L = P + d, which, substituted in (1), gives: R = 3 (p2 ~ (p =: ( ~ 2pd ~ or, since d- is small compared with 2pd, and the surface density of charge is: Qid 27rri 3 FIG. 72. The surface density of charge decreases inversely as the cube of the distance from the point. Assume now that the sphere is insulated and without charge, it will then have some potential not zero. It was shown, that, when the sphere is at zero potential, it acts as if it had a charge Qz = Q\Y ^ ^ ne inverse point B of point A. In order that its charge shall be zero, we have to apply mathematically, somewhere in the sphere, a charge = Q 2 = _|_Qiy. Then the total charge obviously is zero. Li Since the resultant potential of the external charge Qi and the internal charge Q\j gives zero potential of the spherical surface, in order to maintain a uniform potential V all over the sphere, the assumed charge must be applied in the center of the sphere. Thus we deal with three charges, which combined cause the external field. First. The field due to the external point charge Q\. Second. The field due to the charge Q 2 at tne inverse point. Third. The field due to the charge Q 2 in the center of the sphere. 172 ELECTRICAL ENGINEERING The charge Q 2 gives a uniform surface density of Q 2 (?i 47T P 2 *~ 47T P L The combined effect of Qi at A and Q 2 at 5 has been shown to give a surface density of = Thus the actual surface density is: Equation (3) then gives the distribution of the surface charge on an insulated sphere without any independent charge. The equation must, and does show, that a is positive on one side and negative on the other side, in order that the total charge be zero. The potential of the sphere is obviously, charge _ Qip _ Qi radius pL L This is of interest, in that it shows that the potential of a sphere due to a point charge Qi situated L cm. from the center is -j^- This can be proven in a more general way as follows : Assume that a non-conducting sphere be placed in an electric field caused by a number of point charges, a, 6, c, etc. Let the potential of a small element of the sphere be V. The value of V changes from point to point of the surface of the sphere. The average value of the potential V is: VdS where dS is an element of the surface. Referring to Fig. 73 : dS = r sin 6d4>rdd .'. V m ^^ffVs and the average potential gradient along the radius is: * Cf irr 2 J J dS. 4irr 2 J J dr Since is the intensity as well as the gradient it follows that METHOD OF IMAGES 173 dS is the flux diverging from the sphere. This is zero as we have assumed that no charge exists in the sphere. Thus -~^ and V m is a constant for all values of r. We conclude then that the average potential of a sphere is the same as the potential at the center. rdO FIG. 73. Suppose now that the insulated sphere had a charge Q Q . In order that the surface of the sphere shall be an equipotential surface, this charge also should be considered as placed in the center, and its surface density should be added to those given above. L n 3 The potential of the sphere will obviously be the sum of the potentials due to its own charge Q and due to the point charge Qo , Qi or, Usually V is known rather than It is of interest to find the attractive or repulsive force between the point charge at A and the sphere. 174 ELECTRICAL ENGINEERING The force is, by COULOMB'S law, proportional to the product of charges and inversely proportional to the square of the distance. The following conditions therefore exist: First. A charge + Qi at A. Fig. 74. Second. A charge Q\ y- at B. Li Thisd.A charge + #1 ~ + Q at C. Li -< D ^ r V - - : FIG. 74. Thus the force between the sphere and the point charge is: F = L 2 p 2 But it has been shown that D = j > Lt p = ~ Qi 2 L 2 jo Qi 2 p QoQi " (L 2 -p 2 ) 2 " L~* L 3 L 2 ' which, by transformations, becomes : QoQi ,p 3 (2L 2 -p 2 ) L 2 " Vl L 3 (L 2 - p 2 ) 2 Example Qi = 1, Q Q = 10. L = variable, p = 10. For L = 100, 10 1000 20,000 - 100 1 10 4 ' 10 6 (10,000 - 100) 2 " * 1000 yn For L = 11, 10 1000 (242 - 100) F - 121- 1330 (121 -10Q) 2= - 0.158 dyne attraction. It is thus seen that a lightly charged particle may be repelled, if far away from a charge of the same sign, and may be attracted when near. If, however, the charges are of opposite signs, the charges attract always. METHOD OF IMAGES 175 Problem. Construct the equipotential surfaces between an insulated charged sphere and a point charge, when p = 10 cm., L = 20 cm., Qi -- i, and _ 21 ~ 20' Potential Distribution between Two Spheres. Let sphere A in Fig. 75 have a pot. V and a radius R] and sphere 5 have a pot. FI and a radius FIG. 75. Calculate first the charges at A and B and the location of these charges, when A is at potential V and B is at zero potential. Then reverse the operation, and calculate the charges at A and B and the location of these charges, when B is at a potential Vi and A is at zero potential. Then add the charges and potentials respectively, and the desired solution is obviously obtained. (1) Calculation of the charges on A and B when the potential of A is V and that of B is zero : The first approximation is obtained when the potential of A alone is considered. We have then, since in general Q = VR, a charge in the center of A of value Q = VR, and we may, for completeness, say that its distance a from the center is zero. VR This charge affects B by giving B a potential, which is -y Since, however, the potential of B must be zero, it is necessary VR to supply B with a charge which gives a potential - -j-. This charge, which may be called Q'i> has previously been shown /7? \ to be Q'i = VR(-J-}, shall not be placed in the center of 176 ELECTRICAL ENGINEERING the sphere, but at a distance 61, which is obtained by the relation previously proven: (radius) 2 ~ distance from charge to center of sphere \T~D~D But the charge Q'i or -- j at 61 affects the potential of A, so that its potential is no longer V, but V + f -- r -j -s- (distance from charge Q'i to center of A) V- VRRl L(L - &0* To bring the potential of A back to V, A must be supplied with a charge, which is: As far as the external action of the charge is concerned, it is located at a i} where as before This charge at a\ affects sphere B and induces a potential which is VR 2 Ri L(L-bi) (L-ai)' In order to bring the potential back to zero, a charge Q'% has to _ T/P2P be added to B, which gives a potential of r /r _ ^ \ (T~I )' and this charge, as far as external influences are concerned, is located at a point 62, where , RS 62 = Continuing the process, the necessary additional charge on A to balance the effect of Q'% at 62 is found to be : Qi= _^_^_ = and Again, n , v 3 L(L - 61) (L - ai) (L - 6 2 ) (L - at)' and , R\ 2 3 " (L-a 2 )' METHOD OF IMAGES 177 The total charge on A is Q A = Q + Q l + Q 2 + . . . ; The total charge on B is Q B = Q'i Q'z + Q'z + . . . . But, it must be remembered that in order to find the intensity of the field at any point, the position of the charges has to be considered. , (2) By an identical method, a new set of charges are obtained, when A is kept at zero potential and B at its potential V. The total charges on A and B are the sum of all the charges so calculated. Assuming, for instance, that the potentials of A and B are both positive. The first set of calculations will then give a number of positive charges in A, all of which, except the first, located at points, riot its center, the charges in B will all be negative, and all be located at points not its center. The second set of calculations (not shown above) will result in a series of negative charges in A, all of which are located at points not its center, and a set of positive charges in the sphere B } the first of which is at its center. Thus the total charge in either A or B is a sum of a series of positive and negative charges. Simple Case. For two similar spheres, one at zero potential and the other at a potential, V, we have: On the sphere of pot. V On the sphere of pot. zero Q = VR a = o' VR2 Ql = T h R2 bl== ~L o VR3 o' VRi L(L - 6x) R 2 L(L - bJ(L - a x ) t ^ ai ~ (L - 60 (L - at) Q VR * Q's - VR 6 L(L - 6 X )(L - ai )(L - 6 2 ) R 2 L(L-6 1 )(L-a 1 )(L-6 2 )(L-a 2 ) b> - RZ * (L - 6 2 ) (L - a,) 178 ELECTRICAL ENGINEERING The total charge on the sphere of potential V is : QA = Qo + Qi + Q 2 + . . . ; and that of the sphere of zero potential is : QB - Q'l + Q' 2 + Q'z + . . To study the sphere gap, the following problem has been solved to show more particularly, that, while the difference in potential between two gaps may be the same, one gap may break down with considerably lower potential difference than the other. Air at atmospheric pressure appears to sustain, as a maxi- mum, a density of about 100 lines per sq. cm., or a potential gradient of 100, electrostatic units or in practical units 30,000 volts per cm. If, therefore, the potential to ground is high, the air may well break down around the spheres, even though the potential difference between the spheres may be comparatively low. When the air breaks down, corona appears. Then the effective dimensions of the spheres are increased and the gap length correspondingly lowered. The following three cases are calculated, and the results are tabulated below. Diameter of the spheres, 25 cm. Distance between surfaces, 14 cm. Potential difference 1000 electro static units or 300,000 volts. In the first case, sphere A has a potential of 1000 and B is at zero potential, in the second case the spheres are at potentials +500 and 500 respectively, and in the third case they are at potentials +1500 and +500 respectively. In the example the potential gradient G is calculated at the surface of the sphere of highest potential on the center line between the spheres al- though it may, of course, be greater at some other points. In general G = S .,. . 2 - The gradients due to the two spheres should obviously be added if the charges are of opposite potential. Since the intensity of the field is in the same direction at the point considered. METHOD OF IMAGES 179 Summary of the first case: For sphere A, a = Co = 12,500 ttl = 4.46 Ci = 1,430 a 2 =4.53 Q 2 = 186 a 3 =4.53 Q 3 = 24.4 For sphere B, 60 = Q'o = b l = 4 Q'i = - 4,000 62 = 4.5 Q' 2 = - 516 6 3 = 4.53 Q' 3 = -- 67.3 6 4 = 4.53 Q\ = - 9 In general G = ^ .'.G = - S ~ = - 114.6, or, - 34,500 volts per cm. Thus the sphere probably begins to glow. Summary of the second case : ao = 6 = o Qo = Q'o = 6,250 Ol = bi = 4.01 Qi = Q'i = 2,000 a 2 = 6 2 = 4.45 62 = Q' 2 = 714 a 3 = 6 3 = 4.51 Q 3 = Q's = 258 a 4 = & 4 = 4.53 Q 4 = Q\ = 93.5 a 5 = 6 6 = 4.54 Q 6 = Q' 6 = 34.2 (^ = -100.2 or about -30,000 volts per cm. The spheres ought to be just about on the point of glowing. Summary of third case : Qo = 18,750 Q'o = 6,250 Q l = - 2,000 Q'I = - 6,000 Q 2 = + 2,140 Q't = + 714 Q s = - 258 Q ; 3 = - 775 #4 = + 280 e ; 4 = + 93.5 Q b = - 34.2 Q' B = - 102.6 The a's and 6's are the same as above. G = -128 or -38,400 volts per cm. Thus the spheres glow undoubtedly, and if "ground" is under the spheres the potential gradient may be slightly higher below the line connecting the centers of the spheres. 12 CHAPTER XIV APPLICATION OF THE POTENTIAL FORMULA V = 2 - TO SOME MAGNETIC PROBLEMS The magnetic potential at a point in a magnetic field is, as has already been stated, the work done in ergs in bringing a unit pole from infinity, or a point of no magnetic field, to the point under consideration. By GAUSS'S theorem the outward normal flux from a pole of strength m is 4irm. Thus the intensity of the magnetic field, H, at a distance, r. from the pole is | ^ '> or, H = m- f r 2 tlll(l) . -|-T flv 7 "V. or in general, V = 2 Obviously, a magnetic pole can not exist alone; there is always a north pole and a south pole in every magnet. Thus to get the potential at a point, at least two poles of opposite signs must be considered. The potential of a small magnet at distance large compared with its dimension is: V = - ' S > where 6 is the angle the axis of the magnet makes with the radius vector to the point. This is readily seen, if the magnetism be assumed as con- centrated at the poles of the magnet. Referring to Fig. 76, the potential at P is: _ m m m V = AP + ~BP~ m , (1) 180 THE POTENTIAL FORMULA 181 If r is large, compared with I then V = m m A/r 2 + lr cos \/r 2 Ir cos The square root can be expanded by the binomial theorem. We have, = |l -- J^ - cos -h . . . .' 1 J - cos . . .1 = % cos 6 (approximately) (2) Aba. FIG. 76. It is seen from (1) that the magnetic potential at P is in times the difference in -, as we go from one pole to another, where r is the distance from a pole to the point P. Let I = ds, then the rate of change of - along ds, is: - (-) . thus the total difference is (-)ds, ds W ds W . dr If I', m', and n f are the direction cosines of the magnetic particle at (x, y, z), we can then also write, >""[> I +'sC') + i OK Magnetic Shell. A thin piece magnetized at right angles to its surface is called a magnetic shell. It can thus be assumed as 182 ELECTRICAL ENGINEERING made up of a large number of small magnets as shown in Fig. 77. Let the total pole strength in Fig. 78 be m and the area S, then the pole strength per unit area is -. Let the thickness of the shell be I, then the potential at P due to the shell is from equation (2). v S where is the solid angle at P subtended by the surface of the shell. (Recollect that the solid angle, doi = -y cos 0.) FIG. 77. FIG. 78. ml is called the magnetic moment, the strength of a mag- netic shell or the moment per unit area is usually denoted by 0, ml and, V = grco. WEBER proved experimentally that a small circuit in a plane carrying current produces the same kind of a field as a magnet, and that the potential at a point depends upon the area A of the coil, the current 7, and the distance to the point, by a relation : _. KAI cos V = y > r 2 from which the electromagnetic unit of current can be deter- mined by making k unity. AI cos ml cos thus, and 7 = r- = g, the strength of the magnetic shell sur- rounded by the circuit or coil. THE POTENTIAL FORMULA 183 Since we have proven that V = #co, we get the following simple relation between magnetic potential and current: V = 7co, where co is the solid angle subtended at the point by the surface of the coil. It is evident then, that, as long as we do not tread the circuit, and as long as we return to the starting point, the work done in moving a pole in the field is zero. To illustrate this, the potential at a point on the axis of a cir- cular wire carrying I abs. amp. will be determined. First let the point be at the center of the coil, Fig. 79, then co = 27r, and, V = 2?r/. FIG. 79. FIG. 80. If the point is on the axis, but a distance x from the face of the coil as shown in Fig. 80, then the solid angle is: co = 2ir(l - COS a) = 27r( 1 and, - / x ' The magnetic field intensity along the z-axis, which is the direction of the magnetic field, is: dx ~ (R 2 + x*)* ' and the force in dynes on a pole of strength m is : 2TrR 2 m = (R 2 + X*)* ' for x = 0, that is if the point is in the plane of the coil and in its center, H = 27T/ R ' The work done in bringing unit pole once or several times through a loop carrying a current / will now be investigated. 184 ELECTRICAL ENGINEERING Referring to Fig. 81, before the journey starts, the potential at P has been shown to be 7 gence of the vector is -T- It is written div. R, div. (X, Y, Z) or V where V stands f or + ^r + ^ dx dy dz differential parameter. It is evident, from what has been said above, that unless some charges are enclosed in the small volume, there can be no divergence. If there are as many units of positive charge as of negative charge in each small volume, there can also be no divergence, i.e., div. R = 0. The divergence is positive, if there is an excess of positive charge; it is negative (sometimes called convergence), if there is an excess of negative charge. The presence of divergences involves the presence of charges. In hydraulics the presence of divergence means either the presence of some source of fluid in the element or some change in density. Consider a small volume represented by a cube, in Fig. 83 for the sake of simplicity. This cube is assumed to be a small part 185 T Axis FIG. 83. 186 ELECTRICAL ENGINEERING of the total volume enclosed by the envelope that contains the charges. Let X, Y and Z be the components of the field intensity R parallel to the coordinate axes and at the center of the surface a, b and c. If R is a continuous function, which depends upon the space coordinates only, and if the edges of the cube are dx, dy, dz then the value of the ^-component of the field intensity at Ci = Zi = Thus, the incoming flux at c is: Zdxdy, the outgoing flux at is [Z + -T- dz\ dxdy. Consequently, the difference is 'dZ (dz}dxdy; Similarly, for the other sides, and / - dy } dzdx. dy The total diverging flux is thus: Hence by definition div. R = V - R~ = V-R. dv If p is the charge per unit volume or the volume density, then the outward normal flux is 4?rp. ax ar az dx dy dz' A vector field is said to be solenoidal, if there is no divergence. Such a field is, for instance, the electric field in free space or the field of force of gravitation in free space. The divergence theorem connects the surface and volume inte- grals and states that the surface integral of the normal outward flux of a distributed vector is equal to the volume integral of the divergence taken throughout the volume. It is one of the forms of Green's theorem. ELECTRICAL ENGINEERING 187 It is ffR cos SdS = Using the notation of vector analysis, we get: ffn-RdS = fffRdv, where n is the unit normal vector. This theorem is subject to rigid mathematical proof, but can be understood without advanced mathematics, if the volume enclosed by the surface is assumed to be divided up into a large number of small volumes, each fitting tightly against the others. As we add the normal outward fluxes of the different elemental volumes, all will cancel, except those on the very outside surface, since every wall separating two elements is integrated over twice with normals in opposite direction. The outward normal flux is J* J* R cos Ods. Since the excess of outgoing flux over the incoming flux in the element of volume, dxdydz, is: -- h ~^~ + -Q-J dxdydz, it follows that the total outgoing flux is: III f-r + -.J h -Q-} dxdydz, which is equal to I I R cos 6dS. Poisson's equation is: d*V\ " = " p - This becomes: dX dY dZ If X } Y and Z are gradients or intensities of a scalar point function V, so that X = - - V - Y = - d ~- Z = - d# ' dw ' 62 ax ay -ay = az 62 = dz 2 ' 188 ELECTRICAL ENGINEERING and dx* " dy 2 " dz* = 47rp > where p is the density of electrification or charge per unit volume. This equation then applies, when the region of the electrostatic field under consideration contains positive or negative charges, or sources and sinks as some writers call them. Laplace's equation is: VV VV dF _ dx* ~ dy 2 " dz 2 ' or, as it is often written, V 2 F = 0, (Read del square V = zero) and refers to a region in which there are no charges, or to a solenoidal field. By means of LAPLACE'S equation it is possible to determine the potential at any point in the dielectric surrounding a charged body. If the body is unsymmetrical in every way the equation becomes very involved, but if, as is almost always the case in practice, there is some axis of symmetry and particularly if the body has circular symmetry then the potential distribution can usually be calculated fairly , easily, especially if a table of LEGENDRE'S coefficients is available. CHAPTER XVI LEGENDRE'S FUNCTION The potential at points outside of the bodies having circular symmetry, such as circular discs, circular rings, etc., can be determined very readily by means of a certain function, viz., LEGENDRE'S function, which has been worked out and is tabulated much in the same way as trigonometric functions. LAPLACE'S equation g-O (1) dx 2 can be used as has been shown in exploring the space surrounding charged body. With circular symmetry of the charged body it is obviously advantageous to express the equation in spherical coordinates (see Appendix heading Partial Differentiation). Thus, rd 2 (rV) _J_J)/. dF\ 1 d 2 V dr 2 sin 6 dd \ Sm BO/ sin 2 6 d The left-hand term is a function of r only, the right-hand term of 6 only. In order then that this shall hold for all values of r and 0, each term must not only be a constant, but must be the same constant. Let this constant which is entirely arbitrary, be a 2 , -'-0 (6) and 1 d /sin 6 Equation (6) becomes; rS ^ + 2r dr 2 dr dr The solution of (8) is readily found, it is : where and (9) It is evident then that r m and -^i are particular solutions of equation (6). If we choose for a 2 a value which is: a 2 = m(m + 1), then equation (9) is satisfied, since LEGENDRE'S FUNCTION 191 It has been shown that r m is a particular solution of R', thus using this solution at first, we get V = r m B\ Substituting this in equation (3) we get, and, 36 Equation (11) can be solved for 0'. We have, d I . a0'\ . a 2 0' Let 80' a: = cos 0, there sin = \/l x 2 '. rift' In equation (12), is to be determined -- and - ^-' ou do 80' 86' 8x 86' . 86' ^ = " smd== " \j ^/v \j \J v/**/ / a a0'\ ax _ r_a_ / a0_' \ax a0 / a0 ~ Lax \ ax a 2 0' a0 2 (14) U^l/ r) f) f Substituting the value of T~T- from equation (13) and the value ofj of j from (14) in (12), we get: a Thus equation (11) becomes: r) 2 /?' m(m + 1) 8' + ~ (1 - x ^-(x + x) = 0, + !)' = (16) 192 ELECTRICAL ENGINEERING This equation, which very important, is called LEGENDRE'S equation. It can also be written: [(1 - x) ~] + m (m + 1) tr - (17) since> Assume now that 0' can be expressed in whole powers of x multiplied by constant coefficients, that is, 9' = 2a n x n = a + aix' + a 2 z 2 + a 3 z 3 + . . . (18) Referring to equation (17), ~\nf (1 - x 2 ) - = ai - aiz 2 + 2a 2 x - 2a z x 3 + 3a 3 x z - 3a 3 a: 4 (1 - x 2 ) = - 2aix + 2a 2 - and, m (m + 1) 6' = m (m + I)a + m (m + 1) aiz' + m (m + 1) a 2 ^ 2 + m(m + 1) a 3 z 3 + . . . Collecting the coefficient for similar powers of x we get: [2a 2 + m(m + I)a ] is the constant term; [6a 3 2ai + m(m + l)oj is the coefficient of x 1 ', [ 6a 2 + 12a 4 + m(m + 1) a 2 ] is the coefficient of x 2 ; Since, from equation (17), each of these coefficients is zero, we get: m(m -f- l)ao m(m -\ 2 *** a a i > D m (m -f- 1) a 2 + 6a 2 _ m (w + 1) ~ 6 It is seen that if a = 0, all the even terms disappear; if ai =0, the odd terms disappear. LEGENDRE'S FUNCTION 193 The coefficients are related in a comparably simple manner, as follows: I ON L'\"* or, (fc+l)(fc m(m + 1) - (fc +!)(*- From (20) it follows, that, if fc = m - 2, (m - 2+ l)(m - 2 + 2) m(m - 1) ~ (m - m + 2)(m + m - 2 +1) am ~ 2(2m - 1) m(m - 1) (m - 2) (m - 3) _ 2.4(2m- I) (2m - 3) m(m l)(m 2)(m 3)(m 4)(m 5) x i _ . L- _ ft . rff> 2 4 6 (2m - l)(2m - 3) (2m - 5) It is thus possible to express equation (18) as follows: 6' = 2 a n z n ; if the highest power of x is x m , then we get: , 2 (2m - 1) W ( m - 1) (m - 2) (m - 3) ] 2 4(2m - 1) (2m - 3) where a m is entirely arbitrary, and it is convenient to choose a value, (2m - 1) (2m - 3) (2m - 5) ... 1 a m = 7 because, for this value of a m , 6' 1 when x = 1. . , = (2m - 1) (2m - 3) (2m - 5) ... 1 r _ m(m - 1) m _ 2 m! 2(2m - 1) ^ m(m-l)(m-2)(m-3) 1 2-4(2m - 1) (2m - 3) Since 6' is a function of #, and contains no higher power of x than >x m , it is customary to write, instead of 6', P m (x), or since x was cos 0, P m (cos 6). Before enumerating some values of 6', recollect that (factorial 0) = 1, or 0! = 1, or I? = 1; and since |1 = 1, 0=1 = 1. This is readily seen In 1 since j n _ l = n] forn = 1, = = 1, .*. |0 = 1. 194 ELECTRICAL ENGINEERING Example. Find P 3 (cos 0). m = 3, .'.P 3 (z) = (6 - 1) (6 - 3) (6 - 5) 1-2-3 r 8 3 3-1 , -i 2 ' 6 - l x . Note that only three terms can be used in the numerator in front of the parenthesis, -since the last term must end with 1 as is shown in equation (22). The parenthesis contains only two terms, because the next term would give a negative exponent, and we have assumed that the powers of x are positive integer numbers. Thus, for m = 3, P 3 Similarly, for m = 2, (4 - 1) (4 - 3) 1-2 For m = 1, Pi(s) = ^ x = x or, V = A rP (cos 0) + (23) For m = 0, But we assumed as a particular solution: V = r"0' .', V = 2A m r m P m (x), or 2A m r m P m (cos 6) (24) (cos 0) + A 2 r 2 P 2 (cos 0) + A 3 r 3 P 3 (cos + . . . (25) Referring now to equation (10), we see that there is also another particular solution, namely: or, _ A 2 P 2 (cos^) (26) Before applying these equations to some practical problems, it may be of interest to note that the LEGENDRE'S function can be LEGENDRE'S FUNCTION 195 obtained by expanding ^ where R' is the distance between two points (Fig. 85). and R' = \r 2 + ri 2 - 2rri cos 0. If n > r, l f ; Ur 5 t __ 2r .T^ A where A = (1 + h 2 - where T p = cos e. FIG. 85. Expanding A by the binomial theorem, we get: hp Po + - 3p) + /i 2 P 2 + /i 3 P 3 + . . (27) The similarity between (23) and (27) is obvious. Returning now to the problem of a circular wire carrying current, we have shown that the potential at a point on the axis, that is, r coincides with i/-axis and 6 = 0, is : where r and R are shown in Fig. 86. If R > r, see Fig. 86, then FIG. 86. 196 ELECTRICAL ENGINEERING Remembering that when -^ = K is a fraction Since equation (23) holds for all values of 0, it also holds when 6 = 0. Thus we can readily determine the coefficients A , Ai, A 2 , etc., which are: Ao = 2*1, A - R ' A, = 0, _ .,_, 1 A 4 = 0, A 6 = 0, AT- + y = 27r7[l -- ^P! (cos 0) + M ^3^3 (cos 0) - M ^5 (cos 0) + ^ P 7 (cos 0) + . . .] (28) If r > R, then, r ^ !. 3B 4 l-3-5fl 6 -i 2.7 [1 - 1 + K - - - + ^- Q - + [7? 2 P^ z?6 n >i^-M^+K 6 |+ ] (29) From equation (26) we get: + P l *e + A l *& .*. Ao = 0, R2 1 = 2?r/ 'T' LEGENDRE'S FUNCTION 197 A 2 = 0, A 3 = - 2irl A, = 0, A 6 = /. V = 2irl [^^1 (cos 0) - % ^ 4 P S (cos 6) + 7?6 -, . ;J (30) As a second application of the use of the LEGENDRE'S function, the following problem will be considered. Find the potential at points outside of a thin circular disc, Fig. 87, charged to a certain potential, V. It will be proven that the distribution of the surface charge is : where Q is the total charge, that is, the charge on both sides. FIG. 87. FIG. 88. We first calculate the potential at a point PI on the axis (Fig. /"Vo = /"*ro = 2Q Q 1 r 2 - = 2K cos ^T^ as can be readily found by simple integration. This expression then must be expanded in a power series. This can not be readily done, but its derivative with respect to r becomes a simple expression, which can be expanded, the resulting series can be readily integrated. Thus, ^. ["-*?- -i r2 - R2 1 = Q drl2R C r 2 + R 2 1 R 2 + r 2 If R > r, then Q ^H ,J1 J 6 , 1 fi a + r 2 B 2 L R 2 ~*~ R 4 R & ^ ' ' ' J 198 ELECTRICAL ENGINEERING Integrating, R R -&- -+ c } OH J For r = 0, i.e., on the disc, and the potential of the disc will be proven, to be ^ ' D"' r - - " 2 7 -if-- - -4- 4- 1 " R 12 R + 3^ 3 + 5^ 5 " when r > R, it is found in a similar way that: '-[?-+-+] Equation (31) is similar to: 7 = A r P (cos 0) + Air ! Pi(cos 0) + A 2 r 2 P 2 (cos 0)+ . . . '' Ao = '' Al = ~'' A2 = > As = '^ )A ^ ' etc ' * V = |[I P (COS 6) " i Pl ( cos ^) + 3^^2(cos 0) Equation (32) is similar to: _ A P cos0 ' AiPi (cos 0) A 2 P 2 (cos 0) f l f 2 r 3 - ' /. A = 5- B, A! = 0, A 2 = - ^ y, A 3 = 0, A 4 = ^ y,e CHAPTER XVII DISTRIBUTION OF CHARGE ON AN ELLIPSOID If an ellipsoidal thin shell is formed by two similar, similarly situated ellipsoids, and the charge per unit volume, p, is constant in the shell, then the force at any point inside the ellipsoid is zero, that is the poten- tial is constant. The outer surface is an equipotential surface. 1 To prove this, consider the attraction at o of the two masses at A and B, Fig. 89. FIG. 89. The volume at A is r z du dr .'. charge, q = pr 2 dudr. The volume of B is n 2 du dr .'. charge q' = pr^ dudri .'. The attraction of A at is -^ = pdu dr. The attraction of B at is 2 = pdu dri. But from geometry it is known that with two ellipsoids, one of axes a, b and c, and the other of a (1 + a), b(l -f a) and c(l-f ), that is, with two similar, similarly situated concentric ellipsoids, dr must always be equal to dri. Thus the attraction at must be zero. In the case of a conducting ellipsoid charged with electricity, the charge is confined to the surface and the distribution will be shown to be such as is represented by the thickness of the shell in Fig. 89. It is greatest where the curvature is greatest and least on the flat point of the surface. The problem then is to express the thickness of the shell in terms of a variable surface charge, cr. The volume of the shell is evidently = %irabc [(I + a) 3 1]; considering uniform volume charge, the total charge is: 1 NOTE. See "Analytical Statics," vol. II, by ROUTH. 199 200 ELECTRICAL ENGINEERING But a = pd, where 5 is the variable thickness of the shell, :.Q- or 3QS J s QS FIG. 90. But the thickness of the shell 5 can be ex- pressed as the distance between two parallel planes going through any point of the shell. We have from geometry (see Fig. 90) that the distance from the center of an ellipsoid to a tangent plane is : P = ~ (1) / g * \ a 4 "" fe 4 "*" c 4 Neglecting infinitesimals of higher order than the first, d = p(l + a) - p = pa. Qp . . a = 4irabc y + a + l) ; or at the limit = 0, (T = 4:irabc Consider now a very thin flat elliptic disc in the x y plane (c is small) we have from (1) Q o when c approaches zero, 47Ttt& / x 2 V J - * - v As a consequence for a circular disc, Q a = - r DISTRIBUTION OF CHARGE ON AN ELLIPSOID 201 where R is the radius of the disc and r the particu- P lar distance from the center, where a is the surface density on the disc. To find the potential of the circular disc, we calcu- late the potential at a point on the axis, Fig. 91. FIG. 91. rdrQ A = 2Trrdr2 rA0; 202 ELECTRICAL ENGINEERING the volume of an element of thickness Ar is : r 2 sin 0A0A0Ar. If p is the charge per unit volume, then the charge on the small volume is: q = pr 2 sin 0A or, (2) (3) 206 ELECTRICAL ENGINEERING The potential at P on the axis due to the ring-shaped element surface is: (4) (r - Substituting (2) and (3) in (4) we get: the potential at P due to the whole ellipsoid Qdz r, 2cV(r-z) 2 From the equation of the ellipsoid, substituting in (5), "= (5) f_ c 2V - (a 2 - c 2 ) z 2 - 2rc 2 z + c 2 (a 2 + r 2 ) , (a 2 - c 2 )z + re 2 1 c , sin" 1 - . = = 2V a 2 c 2 caVa 2 - c 2 + r 2 J _ c Q f. . a 2 -c 2 + rc . 1 -a 2 + c 2 + rc] / . sm" 1 x = sin" 1 . : (6) 2V a 2 - c 2 [ aVa 2 - c 2 + r 2 a-ya 2 - c 2 + r 2 J To find the potential at a j>oint, like PI, which is not on the z-axis, LEGENDRE'S function may be employed, and the equation (6) is to be expanded into a series in the terms of r. In order to obtain an expression which may be easily expanded, differentiate (6) with respect to r, expand the result into a series, and then integrate the series. Thus differentiating (6), dV p Q (7) dr (a 2 - c 2 + r 2 ) Expanding (7), dV P -Q r r 2 r 4 _r^_ l dr a 2 -c 2 L a 2 -c 2 ~ i (a 2 - c 2 ) 2 " (a 2 - c 2 ) 4 ' 'J 1 when c < r < \/a 2 - c 2 (8) V Q -[_ c , y( _ c V _ " 2 2 2 2 \Va 2 - C V DISTRIBUTION OF CHARGE ON AN ELLIPSOID 207 For a point on the surface, i.e., when r = c, .-. C = V~ + tan-' -7= (10) ^ V a 2 c 2 Since F Pl is a function of V P and /, the solution for V Pl takes the following form: V Pl = Ao + AiriPi(cos 5) + A 2 ri 2 P 2 (cos 0) + A 3 ri 3 P 3 (cos 0) + .... When = 0, n = r, Pi = P 2 = P 8 = - - . 1, and F Pl = F P . ^ + - - tan- 1 - 7 J= Q QVa 2 - c 2 \/a 2 - c 2 J a 2 - c 2! = 0; . U-- - (a 2 - cV 1 c A P^cosfl) tan L 7= = I 7^- ^r (a 2 c 2 ) P 3 (cos e) , P 5 (cos 0) , , P 7 (cos 6) 7 _ I v / ~* 3 * L- M 5 I x y , 7 _i_ I , I O / 9 9\ 9*1 F* / *> 9\ *? ' 1 I^ T / 9 9\ A ' 4 I I 3 (a 2 c 2 ) 2 5 (a 2 c 2 ) 3 7(a 2 c 2 ) 4 which is applicable, when When expanding (7), Whenr = c, V P = 0, .'. (7 = 0. And _ Q , ii) , A 2 P 2 (cos0) , 3 3 , I n" " r! 2 n 3 4 - 208 ELECTRICAL ENGINEERING When = 0, PI = P 2 = P 3 = . . . = 1, r l = r, and V Pl = V P) /. ^o = Q; A! = 0; 4 Q(a-c). "T"" ; A* = 0; = 0; a2 - c2 ) Ps ( cos ^) . (a 2 - c 2 ) 2 P 4 (cos 0) ~~ ~ (a 2 - c 2 ) 3 P 6 (cos 6>) 7ri 7 - which is applicable, when ri >Va 2 -c 2 . (12) (Two similar series can be derived for an oblong ellipsoid. For this and the potential at a point inside an ellipsoid, see W. E. BYERLY'S "Series.") CHAPTER XVIII CONCENTRIC SPHERES Fig. 97 represents a system of concentric spherical shells. It is desired to find the potential at any point in the medium (which is assumed free from charge). Since we are dealing with spherical bodies and since the body is symmetrical, indeed a sphere, LAPLACE'S equation in spherical coordinates becomes : d*V a PP endlx ) (1) FIG. 97. To solve this equation, one first ascertains if the relation dV = A dr ~ r 2 is satisfactory. (We may well assume this solution, since it can be expected that the intensity or force on unit charge varies inversely as the square of the distance.) Then, d^V _2A dr 2 ~~ r 3 ' Substitute in (1) to see if the solution satisfies the equation -~ + l = o,Q.E.D. Thus, 37 _ A ~dr ~ 7* 209 210 ELECTRICAL ENGINEERING satisfies the equation (1). o r ,V=- + B (2) Or we might have solved the equation as follows : Let _ dV . dW _ dy y ~~ dr' ' dr 2 dr .-.*+*.- - J*-dr -log r2 A A . . v = Ae = Ae = , 1) -$ e log r* r 2 " Or again we might have developed the equation directly, without using LAPLACE'S equation, by assuming a positive charge Q on the inside sphere. The intensity of the field at a point in the medium at a distance r is then by GAUSS'S theorem: A~n n R = .'. V = - \ Rdr = + ^ + B (3) an equation of the same form as (2). Referring to equation (2), let Vi be the potential of the inner sphere of radius r\ and Vz that of the outer, then, and ' F 2 = - + B. r 2 r rir 2 r 2 r 2 - ri r where ri < r < r 2 CONCENTRIC SPHERES 211 To determine the meaning of B assume that the outer shell is grounded, or which is the same, at zero potential, then and from (4), . v tfi D . K r l . . V 2 = ~ ~~ - -T jD. . . > = - - V i. r 2 - ri r 2 - n From (5), y = Vi n r 2 _ _TI __ r 2 - ri r r 2 - ri * = _FVi_ rra _ 1 1 == JV r 2 ri Lr J r 2 - r 7*1 r If the outside sphere is very far off so that r 2 approaches infinity and F 2 zero, then, V z = 0, r 2 = oo ; 7 2 = o = - + 5, /. 5 = 0. r 2 ? r r The potential gradient in the space between the conductors is: 7.-F. It is the greatest at the surface of the inner sphere, where r = ri. r 2 The potential gradient at the inner surface of the outer conductor is evidently : r 2 ri r 2 Referring to equation (7), R = 4^ri = ^J equating to (8), Qi _ 7i-7 2 ra. r 2 - 14 212 ELECTRICAL ENGINEERING Example. Calculate the average potential gradient in the space between two concentric spheres separated by a distance of 2 cm. Assume that the potential gradient at the surface of the inside conductor is 100 electro-static units per centimeter, that is, just about on the point of glowing. Consider a concentric sphere, Fig. 98, the inner sphere of which has a charge Q\ and the outer a charge Qo = $2 + Qs- FIG. 98. Evidently, Qo = Q 2 + Q 8 . Since all tubes of force beginning at the surface of the inner conductor terminate at the inner surface of the outer conductor, it is evident that the charge Qz = Qi- .'. Qo = - Qi + Q 8 . The potential at a point outside of the outer conductor is thus, from (6), T7 Qo Qs-Qi , y = ^_ = ^ *_, w here r = r 3 . r T Since the capacity of an electric field is the ratio between the charge on the positive boundary and the potential difference between the boundaries, c - r i-F 2 Thus horn (9), c== Ii^Z_ 2 . rir2 . 1 r 2 ri Fi K 2 / 2 / 1 The capacity of the inside sphere alone is ri. Capacity of concentric spheres _ r^ Capacity of inner sphere r 2 CONCENTRIC SPHERES 213 If the thickness of the dielectric is small compared with the radius, then: C = r , ^^ = T 4' where S = r, - r,. 6 d 4?rri 2 _ area of sphere as a limiting case, where TI = r^ = we get parallel plates, and, area on one plate 47r(distance between them) The capacity is expressed in cm. not in farads. To get the capacity in farads divide C by 9 X 10. u The energy input to a condenser is: W = Thus the energy stored in the field between two concentric spheres, is: Infinite Parallel Planes. LAPLACE'S equation applies in this case so long as there are no charges between the condenser plates, _ d*V dx 2 " dy 2 ~~ dz 2 = Since the field depends upon the distance between the plates only, that is, upon one of the coordinates only, we get, w _ . dv __ "T "^ U, . . j L/o dx 2 dx and V = C x + d. , If the charge on plate A (Fig. 99) A ifc^T | Ql is Qi and the potential FI; and the , charge on plate B is $2 and the poten- B - - 1 - Q * tial Vz', and if the distance between the plates is d] We have: Vi = + Cj, and F 2 = C d + Ci. Subtracting, Vi - V z = - Cod, or, C = - (Fl "T V ^. 214 ELECTRICAL ENGINEERING :.V -^ -x + Ci; or, since V 1 = d, d The potential gradient, that is the potential drop per cm. is: =-=^ It is constant all through the dielectric. The total outward flux from A is 47rQi, one-half of this enters the space between the plates. The inward flux to B is ^irQi, and one-half of this is added to the flux from A. Thus the total flux in the space between the plates is: But the charge on A, Qi, must be numerically the same as that on B, Q%, since all tubes of force leaving A enter B, thus Qi = $2, numerically, but of course of opposite sign, which, however, is taken care of in the above discussion. Thus the total flux in the gap is 4?rQ, where Q is the charge on one of the plates. 47T0 .'. R, the intensity of the field, is j- where A is the area of one side of the plate. And ' G = R = ^; or from (1), 4rQ Fi- F 2 . r _ Q A. A d " ' ' L = Fi - 7 2 " This could have been calculated in still another way. Since D V = - fRdx = - X For x = 0, V = 7i; /. Ci = 7i. for , ,, T7 . T7 T7 oj = d, V = 7 2 ; . . V z = Vi -- T d, CONCENTRIC SPHERES 215 1s . 2 .. If the plates are separated by uniform insulation of specific ., v ,, ., . KA KA inductive capacity, A, the capacity is -r, cm., or . , Q 1()ll farads. If the dielectric consists of several layers of different specific inductive capacities then one can consider that the condenser is made up of a number of condensers in series and the capacity of each is: KiA Ci = -7r> etc. 47rdi The total capacity is obtained from the well-known relation: 1 1 C ~ Ci ' 1 f 7T + . , or, C 2 1 1 , 1 , irdi 4ird 2 Ci F c 2 + V * V * ' AI A 2 All these formulae are approximate, however, since no allow- ance has been made for the effect of the edges, but the plates were assumed to be infinite. Concentric Cylinders. LAPLACE'S equation can again be used if it is assumed that there are no charges between the cylinders. Moreover since we are dealing with cylinders, it is best to put LAPLACE'S equation in cylindrical coordinates. Thus we have: ~dr* + r ~fo ~~ let y = -r-> then (1) becomes -j- H y = 0. The solution of this equation is -f*L = A = A y ^^ -. log r " M I dV A 216 ELECTRICAL ENGINEERING To determine the integration constants, let V = Fi, r = r, (Fig. 100) and V = F 2 , r = r 2 . Then, Vi = A log (n) + B, and T r 2 = A log (r 2 ) + 5. .'. Fi - F 2 = A (log ri - log 7- 2 ) = A log and, ><* FIG. 100. The potential gradient or the intensity of the electrostatic field is: dV V, - V Z /1\ 47TW 2^ where Qi = charge per unit length of conductor, and Z = length of conductor. per centimeter length of conductor. The potential gradient is the greatest at the surface of the inner conductor, where it is: 1 7, - 7i Graded insulation between the conductors. In order that G may be constant at all points of the dielectric it is evident that the specific inductive capacity must be the high- est at the inner conductor, and be inversely proportional to the distance from the inner conductor. CONCENTRIC SPHERES 217 Let the specific inductive capacity be expressed by the follow- ing formula: K = ~, where a is a constant. With a charge Q on the inner conductor, the flux per centimeter length is 4irQ, thus the force on unit charge is: _ 2Q K2irr ~ Kr =- C^dr=- (^dr=- f* J Kr J ar J a dV 2Q G = = -- = constant. dr a The same result could have been obtained directly from (2), which, in the general case when Kl, becomes: R 4,0 2Q K2irr Kr a K = -> r Substituting R = G) 20 R = - - = constant, Q.E.D. CHAPTER XIX CYLINDRICAL CONDUCTORS Line Charge. Assume that the conductor which is perpendicu- lar to the page is infinitely long and its diameter so small that it may be considered as line, and let the charge per unit length beQ. The electric field is then represented by radial lines in planes parallel to the page or, which is the same, at right angles to the axis of the conductor. The intensity of the field at a point P, Fig. 101, is obviously: 2Q And the difference in potential between two points PI and P is: - C^ dr Jhi r - 2Q [log n - log fcj = 2Q log (1) FIG. 101. Two equal but opposite line charges separated by a distance 2hi: Let A and B (Fig. 102) be the locations of the line charges. The difference in potential between midways between the charges and P, due to the charge on A alone, is and has been shown : V p - V. = 2Q log (2) The difference of potential between o and P due to the line charge Q on B is obviously, V p - V, = - 2Q log J- 1 (3) TZ 218 CYLINDRICAL CONDUCTORS 219 Thus the difference of potential between and P due to both line charges is: V 9 - V = (2Q log - log ) = 2Q log . (4) Referring to equation (2) or (3), if P lies midway between A and B, so that r\ = r 2 = hi, then: V p - V = 0, thus as long as the charges are equal and opposite, the potential at is zero, which would, of course, have been concluded without proof. V = 2Q log ^ (5) where V is the potential of P due to the charges on both lines. From (5), follows T2 -- = 2Q = a = a constant (6) for all surfaces of potential V. Equation (6) represents a circle, defined by the following relation : ~OA X OB = R 2 (7) referring to Fig. 103, where is the center of the circle, A and B Fig. 103) are called the inverse points, and O f the center of inversion. FIG. 103. FIG. 104. To prove that equation (6) represents a circle refer to Fig. 104. 4- or. O, 220 ELECTRICAL ENGINEERING which is the familiar equation of a circle having a radius of o __ , "1-C 2 and its center at a point whose coordinates are: "1-C 2 from A; = 0, /. OA X OB = from B. = R 2 ; (1 - C 2 ) 2 thus, equations (6) and (7) are proved. The ratio, > can be expressed by a simple equation involving h, the distance of the center from the neutral plane, and the radius, R. FIG. 105. Referring to Fig. 105. R 2 = OA X OB = (h - hi)(h + hi = or, i = 2 - R 2 But triangles OPB and OP A are similar, since OP 2 = OA X OB; - _T?_ Tl , 'OP ~ OA' r, = OP = R = B ri ~ OA ~ h hi ~ a (8) (9) CYLINDRICAL CONDUCTORS 221 Substituting (8) in (9), rs = R = R(h + ri ~ h- We can then determine the potential of a circle, or, which is equivalent in this case, a cylinder, whose center is h cm. from the neutral plane and whose radius is R, as Similarly the potential at a circle around the negative charge 7 2 = - 2Q log R (12a) /. V = V l - 7 2 , that is, the potential difference between the two cylinders is: h + 4Qlog R (13a) For the sake of convenience, will be added other expressions for Vi, Vz and V, involving hi, and R instead of h and R. From (8), h 2 = R 2 + /U 2 , which, substituted, gives V l = 2Q log R - -2Qlog- R and, = 4Qlog- 4- V hi 2 + (126) (136) It is now evident how we can go from line charges to charges on actual conductors. It has been proven that the equipotential surfaces around the line charges are cylinders and hence if circular cylinders be substituted for the circles, the distribution of the field is not affected. The capacity per centimeter length of two such metal cylinders (that is, of the double conductor) is : 4 log h + = cm. (14) R 222 ELECTRICAL ENGINEERING CW--- -f- -/=== farads (15) 9 X 10" 4 log * ^~ or, C m _/. per 1000 ft., of circuit (double conductor) m-f. (16) R 'where logic means the ordinary logarithm not the natural logarithm h is half the distance between conductors, and K the specific inductive capacity. If E is the effective value of the alternating-current line voltage, then the charging current per 1000 ft. of double con- ductor is readily proven to be: C m . f . The capacity to neutral is obtained directly from (lla) and is: c= 1 n , h + Vh 2 - R 2 2 log - gr- It is thus seen that the capacity to neutral is twice as great as that between the lines. This results, of course, in the same charging current as in the E first case, since in this case the voltage is -^- Thus the capacity of 1000 ft. of one wire to neutral or ground is: C m -f. = ~ j-- , = m-f. per 1000 ft. of transmission. logio - ~/j>~ Two Parallel Cylindrical Conductors of Different Diameters but Equal and Opposite Charges. Since OA X OB = Ri 2 and (FB X WA = R 2 2 , we have a(a + 2hi) = Ri 2 , or a hi + \/h\ 2 + R\ 2 (1) and -'- ' 2hi) = Rz 2 , or = -hi + V/^F^T 2 (2) and ^ F,= -2Qlog^= -2Qlog| 2 (3) CYLINDRICAL CONDUCTORS 223 Substituting (1) and (2) in (3), Vi - V 2 = 2Q log and C = Q (- hi '+ Vfti 2 + #i 2 ) (- fti + \//ii 1 2 log RiR: - hi + Vfti 2 + Ri 2 ) ( - fti + Vfti 2 + #2 2 ) To obtain an expression in terms of h and R, instead of hi and R, from Fig. 106 we have: (3 = 2h - 2hi - a (4) .'. /? + 2hi = 2h - a (5) FIG. 106. Substituting (4) and (5) in (2), (2h - Solving (1) for 2hi and substituting it in (6), a (6) or, or, (2h = 2ha 2 + a)(2h a) R% , - #i 2 - 4/i 2 ) a + 2hRS = 0, - V(Ri 2 - R* 2 + 4/i) 2 - 16ft 2 4ft where the sign in front of the radical is minus not plus, because a = when Ri = 0. Similarly, 4ft .'. C = 2 log _ __ (RiR*\ 224 ELECTRICAL ENGINEERING 2 log 4/i 2 - (Ri 2 + R 2 2 ) - V 16/i 4 - which becomes: C = - 4 log h + \A 2 - R 2 R if R is substituted for both RI and R z , a result obtained before. Construction of Equipotential Surfaces around a Cylindrical Conductor, Charged to a Certain Potential, V. Let the distance be- tween the center of the conductor, Fig. 107, and ground be h, and the distance of the equivalent line charge above ground be hi. Since the ground is an equipotential surface, it is evident that the problem will in no way be affected, if a second conductor with a charge Q be placed equidistant below the ground surface, and the equipotential surfaces around A be considered as due to a positive charge, Q at A, and an equal but opposite ("image") charge Q, at the inverse point A'. Suppose that it is desired to draw the equipotential surface through a point P, distant d from the ground. The first step is to locate the equivalent line charge in the original conductor of radius R and distance h from ground. We have, /ii 2 = h 2 - R 2 , .". hi = Vh 2 - R 2 (1) from A, the (2) (3) (4) FIG. 107. h 2 - R 2 To find the radius of a circle whose center is location of the equivalent line charge, we have, But from the figure we have, hi + ttl = fa + d .*. i = RI + d hi. CYLINDRICAL CONDUCTORS 225 Substituting (4) in (2), (Ri + d- /ii) (2/i! + -to - hi) 2d (5) The potential of the circle of radius Ri, which goes through the point, P, is: Fl = 2 Q log ^ = 2Q log *' But V, the potential of the conductor, is: log Knowing the radius from (5), and the center is Ri + d above ground, the equipotential surface through P can be drawn, and the potential of that surface is given by (6). Potential of a Cylinder due to External Charges. In order to determine the potential due to a number of charged cylindrical conductors, it is necessary to calculate the potential of one cylinder due to charges on other cylinders placed in the vicinity. FIG. 108. Consider a line charge Q at B in Fig. 108 and determine the average potential due to Q on a non-conductive cylinder A. The potential at P is, as has been shown : V = 2Q log -> but from the triangle OPB, r 2 = d 2 - C 2dr l cos - -~ cos ELECTRICAL ENGINEERING r = d V/c 2 + 1 2k cos , 226 or, where Thus the average potential of A is = TT I *TrjQ log r Jo log (1 - cos - cos /"2r where 2Q \oghi-2Q\ogd- I log (a-b cos a = 1 + fc 2 and, b = 2k. Evaluating the definite integral (see PIERCE'S "Table of Integrals") we find that the last term is zero. Thus, V A = 2Q(log h - log d) = 2Q log -j 1 (D Thus, the average potential is independent upon the radius of the conductor. But equation (l)has been shown previously to be the potential at a point distant d from a line charge distant hi above ground. Thus to determine the potential of a cylindrical conductor A, due to a line charge at B distant d, the diameter of the conductor does not enter as long as, with metallic conductors, the field can be assumed not disturbed by the conductor. ^-^ ?> / >\ I d z Wa _X h* \ FIG. 109. Referring to Fig. 109, The potential of A due to B is: log - CYLINDRICAL CONDUCTORS 227 The potential of A due to C is : F 2 = 2Q 2 log |- .'. V = Fi -}- F 2 = 2Qi log ^ + 2Q 2 log ^ ai a 2 Lines of Force between Parallel Cylinders. Let s-s (Fig. 110) be a part of a line of force, and N-N a line at right angles to it. Thus the projection of G\ on the normal is Gi 1 = Gi cos a, where G\ is the intensity at P due to the line charge at A . Simi- larly the projection of (7 2 on the normal is G 2 cos 0. The sum of the projections must be zero, since N-N is perpendicular to the line of force. cos a -f G 2 cos |8 = (1) But and cos cos a = Similarly, L ds -2Q TT as 15 228 ELECTRICAL ENGINEERING Substituting in (1), de l + de = o or 0i + 6 2 = constant. This equation represents a family of circles through A and B, with center on the line 0-0. Construction of Lines of Force. Referring to Fig. Ill, as P is in the center line, n = G l cos or, FIG. 111. Knowing the values of x and the fixed points, A and B, the lines of force, being circles, can be readily constructed. Problem. Draw equipotential surfaces around a line charge placed 10 cm. above the neutral plane, when the charge is 1 electro-static unit per centimeter of conductor. Find the radius of the conductor containing the line charge whose potential is 2000 volts. Draw surfaces corresponding to 400, 800, 1200 and 1600 volts. Draw lines of force whose intensities at the neutral plane are 120, 110, 100, 90 and 80 volts per centimeter. Solutions. First. Radius of conductor: Since 2000 volts corresponds to 6.67 electro-static units, we have: 6.67 = 2Q log hi + Vhi 2 + R 2 . 10 + \XI66~+ fl 2 - = 2 log - R R .. 10 + VlOO +^ n^Q/ix/QQ 1 AA^ . . logio ^ = 0.434 X 3.3 = 1.445. R . 10 + V 100 + R 2 R = 28.05 .*. R = 0.72 cm. By a similar process the radii corresponding to 1600, 1200, 800 and 400 volts are found. CYLINDRICAL CONDUCTORS 229 These being calculated, the corresponding values of A, the distances from the neutral plane, are found by the relation h = VV + R 2 - Second. To find the intersection between the neutral plane and the line of force of intensity 100 volts per centimeter or 0.333 electro-static units, we have: - 1 = 10 ).333 X 10 -1 = 10X0.447 = 4.47 cm. Capacity of Two Cylindrical Conductors, when the Effect of the Proximity of the Earth is Considered. Consider, for the sake of simplicity, the case of two cylinders of equal radii, and charges Q and Qi respectively. FIG. 112. Referring to Fig. 112, it has been shown that the potential of A due to its own charge, Q, and the charge on its image, A' is: 2Qlog R (1) It has also been shown that the potential of A due to the Qi, on conductor B is: V, = 2Q 1 log ^ (2) Similarly, the potential of A due to the image of B is : V, = 2Qi log - 1 (3) 230 ELECTRICAL ENGINEERING Thus the total effect of conductor B on A is: V z + V, = 2Q l log ~ (4) And the resultant potential of A is: " V A =- V l + F 2 + V 3 = 2Q log Similarly, ^-^log^f^-H*]** (6) Special Cases. Two wires in parallel at same distance from ground. Thus h = H, Q = QL .'. V A = F B == 7. Thus the capacity per centimeter of each wire is: Q l 2 log [I" R J and the capacity of the two wires taken together, is: c = - -]> > + -i/M^fiT (8) \d h + vft 2 # 2 108 U' ~R~ "J In the case of a transmission line, ft is large compared with R, and d f is approximately 2ft. 1 1 It has been shown that the capacity of a single wire to neutral is: approximately. (10) 2 log " ' - " " - ' 2/i JLl/ Xl/ Thus the proximity of the other wire has reduced the capacity of each wire, so that the combined capacity of the two in parallel is usually not more than 25 to 30 per cent, greater than that of a single wire. CYLINDRICAL CONDUCTORS 231 As an instance, let R = 0.5 cm., h = 1,000 cm., and d = 20 cm. 2000 (log -- + log = 0.0388 cm. per centimeter; .*. 2Ci = 0.0776 cm. per centimeter and the capacity of one single wire alone is C """2000 = 0-0603 cm. per centimeter. 21 six I =, o.oi. '* C = 2 log 400,000 = 0< 388 ' C' = s-^ -777 = 0.1352. 2 log 40 .'. K L1 = 0.087, Ki . 2 = - 0.0482, f^- 1 = - 1.806. A 1.2 Discussion. To show the application of these coefficients, the following problems will be considered. A. Compare the capacities between a wire and ground, (a) when the wire is alone; (b) when an adjacent wire is grounded. B. Compare the charging currents for the same applied voltage between the two conductors when the two wires are insulated, and when one is grounded. In the latter case, give the relative proportions of the current in the grounded wire and in the ground itself. The numerical case will be: R = 0.5 cm.; h = 1000cm.; and, d = 20 cm. The problems will be best solved by the use of the MAXWELL'S equations, viz. Qi = #i.iFi + Kt.iVt + Ki. t V*, Q2 = #J.lFi + #2. 2 F 2 + #2.3^3, and, Q 3 = K Z . 1 V 1 + MUTUAL AND SELF-INDUCTION 235 In these equations, index 1 refers to conductor No. 1, index 2 to No. 2, and index 3 to the ground. Since the potential of No. 3 is zero and since we assume two similar and similarly placed conductors, V 3 = 0, KM = #2.2 and K lft = K,. 3 . /. Qi = K 1 . 1 V 1 + #!. 2 F 2 (7) Q* = #i. 2 F! + #i.iF 2 (8) and Q 3 = #i. 3 Fi + K^V* (9) Case A. (a) It has been shown that with a single conductor suspended above ground, the capacity is: C = -- ^r = 0.0601 cm. per cm. (10) r i ""' 21og fl Thus if V is its potential the charging current is: ri = 0.0601 ^ (6) since No. 2 is grounded, T 2 = 0. Thus from (7), Qi = ^i.iFi .'. capacity = KI.I = 0.087, and ri = 0.087 ^ The capacity of wire No. 1 is increased 45 per cent, by the proximity of the grounded adjacent wire No. 2. Case B. Under normal conditions, Qz = - Qi and KI.I = K 2 .z, .'. Qi = tfi.iFi + X!. 2 F 2 , Thus the capacity between the conductors is: C = g " ~ g " = 0.0676. i If FI F 2 = F, and if z'i is the current in conductor No. 1, then dV dV dV i l = C^ = y 2 (#!.! - #,.,) ^- = 0.0676^- 236 ELECTRICAL ENGINEERING When No. 2 is grounded, Vz = 0. .'. Qi = Ki.iVi = Ki.iV, thus the capacity, C" = = 1.285. *1 JV.1.1 ~ A- 1.2 The charging current in conductor No. 1 is increased 28.5 per cent, by the proximity of the adjacent grounded wire. The charge in conductor No. 2 is: Qz = K 2 .iVi -f #2.2 V z = Ki. 2 V, since 7 2 = 0. But Xi.t = - 0.135 + 0.087 = -- 0.048 Thus ^2= -0.048 f The current carried in the ground is obviously - 2 3 = (0.087 - 0.048) -^ .: ,-,-- 0.039 If the current in No. 1 after grounding No. 2, is taken as 1 amp., then wire No. 1 carries 1 amp., No. 2, 0.554 amp. and the ground, 0.446 amp. Problem. Assume three similar horizontal conductors of R = 0.5, h = 1000 and d = 20. Give the relative values of the charging current between No. 1 and No. 3 if No. 2 is indulated, and if it is grounded. Also give the charging current if No. 2 is removed entirely. Consider the current in the last case to be unity. CHAPTER XXI TWO-CONDUCTOR CABLE Since the conductors as well as the lead covering are of metal, the surfaces of each are equipotential surfaces. In order to simplify the calculations it is desirable to substitute for the sheath and each conductor a system of conductors, i.e., the conductor, and its image, which will give the same distribution of potential. Consider first the system of Fig. 114 consisting of A, its image A' and the lead sheath. It is necessary to determine the position of the line charges at distance hi from the neutral plane, so that the conductor A and the sheath are equipotential surfaces. From what has been shown previously, it is evident that the following relations exist: and hi 2 = h 2 r 2 , when considering the conductor; hi 2 = (h + a) 2 r*i 2 , when considering the sheath. i 2 - r 2 - a 2 Having determined h from (1), hi is determined, as hi = \/h 2 r 2 (1) (2) FIG. 114. Referring to Fig. 114, it is evident that the potential of A is due to its own charge and the charge on its image, and the charges on B and its image. TVf) It is also recollected that the latter potential is: 2Q log = mp 237 238 ELECTRICAL ENGINEERING if we neglect the shortening of the lines of force from m to p in going through conductor B, where np is the distance between the line charge in B and the center of A, and mp is the distance be- tween the line charge in B' and the center of A. .'. np 2a -f- h hi, and mp = 2a -f- h + hi. 21 r' + VW-r'^a + h-h! to neutral (3) i" r* za -f AI Aii \ Approximation. Frequently, in fact almost always, the follow- ing approximation can be made : h = hi. (4) 2 log - "' r a If furthermore r 2 is small compared with ri 2 a 2 , and is small compared with A 2 , then, h -4- A//? 2 r 2 2h ri 2 a 2 - /I . 2a r r ar thus, c = _^ __ ^ the capac j ty to neutral (5) 2 log ( 1 , \r r^ + Thus, the capacity between the two conductors is approximately C " /aa'r.. - (6) 4 lo g (7- " ^rf or, in microfarads per 1000 ft. of cable, (7) , log l To determine the capacity of the two conductors in parallel against the sheath, the two conductors are given positive charges, +Q, and hence the charges on the images are Q. TWO-CONDUCTOR CABLE 239 The potential of A due to its own charge and the charge on its image is: V' A - 2Q log *X^Ei The potential of A due to the charges on B and its image is: or, using the same approximations as before, (9) The potential of the sheath, if insulated, due to the charges in A and its image is: V. = 2Q log *' Similarly, due to B and its image is : y". = 2Qlog^- r\ .-. F. = 4Q log (10) Tl Using the same approximations as before, V 8 =- 4Q log ^ (11) Thus the potential difference between the sheath and either of the conductors (when they are connected in parallel) is approximately : V -V A -A. = 2Qlog-~- - 2Qlog (12) Thus the total capacity between the two conductors in parallel and the sheath is: log 240 ELECTRICAL ENGINEERING In connection with this it may be of interest to determine the capacity between the conductor and the sheath in a single con- ductor, eccentric cable, Fig. 115. The potential of A due to its own charge and the charge on its image is: n + VV + r 2 V A = 2Qlog FIG. 115. The potential of the sheath due to the charge on A and its image is: 7 8 = 2Q log hi + Vhi 2 + .'. C = II (13) Denoting the conductor A with 1, B with 2 and the sheath with 3, the values of KI.I, KI.% and KI. S , are respectively identical with K 2 .2, K 2 .i and ^2.3. To determine them we have and, Q 3 = K Z . 1 V 1 + ^3. 2 7 2 + # 3 . 3 7 3 . If we are concerned with the distribution of currents in the conductors and lead sheath, it is convenient to consider the sheath grounded, that is, 7 3 = 0. and, TWO-CONDUCTOR CABLE 241 If then FI = F 2 = F, that is, if both conductors are given the same positive charge, then Qi V(# 1;1 + #,. 2 ) /. C = #1.1 + #1.2; but C has been determined in (12) which gives, (14) If the two conductors have potentials FI and FI, respectively, then: Qi = #1.1^ + # 1>2 F 2 = F^L! - #i. 2 ), .'. C = #1.1 ~ #1.2- This capacity has been given in (5), which is: 1 C = 2a r! 2 + a From equations (14) and (15) the values of KI.\ and KI.Z are readily obtained. Consider finally that when the two conductors are in parallel, that is, at the same potential and the charging current returns over the grounded sheath, we have, Qi + Q 2 + Q 3 = 0, and Vi = F 2 = V. .'. (Ki.i + Ki.2 + KM + #2.2 + #3.1 + #3.2) F = 0, or, 2#!.! + 2Ki. 2 + 2#!. 3 = 0. .*. #1.3 = - (#1.1 + #1.2) (16) Problems. Find the charging current under the conditions shown in Figs. 116-120, when r t = 4r; a = 2r; .'. h = 2.75r; hi = 2.55r; KI.I + #1.2 = 0.4; #1.1 - KI.I = 0.57 .'. #1.1 = 0.485; #1.2 = 0.085 and #1.3 = 0.4 (using no approximations). (a) (Fig. 116.) F! = F 2 , F 3 = 0. Q 2 = Fi(#i. 2 -f- #2.2). .'. ii (#1.1 + #1.2) -jj, dV and iz = (#1.1 + #1.2) ~ 9 242 ELECTRICAL ENGINEERING the total charging current is 1 "*" l '^ ~dt ' ' dt' (6) (Fig. 117.) 'i.iVi + Xi. 2 F 2 , .-.2 + ^1.0 = 0; = (X 2 .i + X 2 . 2 + K 2 . 3 + X 2 . ) 7, .'. #1.1 + 2X 1 . 2 + XLO = 0; = (X 8 .i + #3.2 + K 3 . 3 + X 8 .o)7, /. #1.1 + 2#!. 2 + Xi. = 0; any one of these equations gives: #1.0 = - (#1.1 + 2Xi. 2 ) (21) Thus XI.Q is determined. Problem. Verify the equations of the charging current under the conditions given below (Figs. 122-130) and apply the follow- ing numerical values: TI = 4r, a = 2r. (Fig. 122) i = (Fig. 123) i = 2 FIG. 123. ; ^ - 0<826 - Xi.i / dt dt 248 ELECTRICAL ENGINEERING (Fig. 124) i = 3(/M.i + 2/M.j) ^ = 0.903 Ki.i 2 - #i.2 2 dV (Fig. 125) t dt M* dV dt . _ ZKi.z* to ~Tr~ A. 1.1 -IV 1.1 Si = FIG. 128. (Fig. 126) ti = FIG. 129. K dV l * = K ~di i K dV 13 ~ Kl -*~di = l ' St = dt (Fig. 127) i = 2(i.i + K lm - ~K" K" A (Fig. 12J (Fig. 129) i dV i = 0.608 = - 0.418 = 0. dV dt' FIG. 127. v v FIG. 130. = 0. = 1.488 dV dt' ~dt' (Fig. 130) Three-phase: z = (K lfl - K^) - = 0.744 CHAPTER XXII THE ELECTROSTATIC EFFECT OF A THREE-PHASE LINE ON AN ADJACENT WIRE OR WIRES The potential of the wire W, Fig. 131, due to A, B, and C and their images is obviously: V = 2Q A log g + 2 Q B g + 2Q C log g 2Q B b, + 2Qcd (1) where . r 2 01 = log > 61 = log p and, , r 6 Cl = log - If C is the average capacity of the three lines against neutral. then: Q A = Ce\, QB = Ce z , and Q c = Ce s , where e\ t e 2 and e s are the instantaneous values of the Y voltages. e 2 6i -f e 8 ci) 2CE[ai sin 6> + bi sin (^ + 120) + d sin (^ + 240)]. - 60 + b^ - cO + Ci( Cl - ai) (2) where 7 is the maximum value of the Y voltage, that is, of the voltage to neutral. To determine the average capacity of the three wires: The potential of A, Fig. 132, due to its own and the other charges is evidently, V A = 2Q A log ^ + 2Q B log g + 2Q C log ||- If the average value of R 2 , R* and RQ is #!, and the average 249 250 ELECTRICAL ENGINEERING values of R 3 and R$ is D, then the potential of A can be reason- ably well expressed as: V A = 2Q A log y + 2Q B log ^ + 2Q C log ~ (3) WA log - 1 + 2 (Q* + Q c ) log ^ 7 ; r = Radius of Conductor FIG. 131. But Q B + Q c = -Q A , thus 7 A = 2Q A log (~ j^ = 1 FIG. 132. /. C = where D is the average distance between the conductors. E '.' Vmax. = T?\/l 61) + 61(61 Ci) + Ci(Ci - (4) (5) Problem. Prove that the maximum value of the induced potential on a telegraph wire placed under a three-phase trans- mission line of 100,000 volts (effective) between the lines is A THREE-PHASE LINE 251 approximately 3100 volts, when H = average height of trans- mission wires above ground = 1500 cm., D = 300 cm., and r = 0.5 cm. The telegraph wire is 800 cm. above the ground, and 50 cm. to the left of the center line of the pole. It is seen that when the three-phase line is operating under normal conditions, the voltage induced in an adjacent wire is only a few per cent., in this case only 3 per cent, of the line voltage. If, however, one of the three-phase lines is grounded, so that the system is unbalanced electrostatically, then very considerable voltage is induced as will be shown. If ei = E sin e, e 2 = E sin (e + 120), and, 6 3 = E sin (0 + 240) are the Y voltages or phase voltages, then it is well known that the line voltages are : V 1 - F 3 = EV3 sin (6 + 30), and, V 2 - F 3 = EV3 sin (e + 90). Therefore, if phase No. 3 is grounded or at zero potential, then we have the relation between the line voltage as shown in Fig. 133. The line voltages differ 60 in time phase, when one phase is grounded. For the sake of simplicity, let : Vi - 7 3 = V3 sin e = V a ; ' V 2 - V, = EV3 sin (6 + 60) = 7 6 ; 7 3 = = V c . or, V a = #o sin 0; where E G = E\/3> V b = E sin (6 + 60) ; FIG. 133. Using MAXWELL'S equation, applying index e for ground, and remembering that V c = V e = 0, we have, Qa = Ki.iVa 4~ Ki.zVb, Q b = # 2 .iF + K*. 2 V b , Qc = #3.lF + #3. 2 F 6 , and, Q. = K..iV a + K e . 2 V b . 252 ELECTRICAL ENGINEERING But KI.I X 2 .2, Xi.2 = Xi.s and K e .i = X e . 2 approximately. * Qa Xi.iFa H- Ki.zVb, Q b = Xi. s 7. + Xx.iFi, Q c = K^V a + Xi. 2 F 6 , and, Q e = Xi.,7. + Xi. 6 F 6 . '. Qa = Xi.i# sin + #1.2 EQ sin (0 + 60) Xi.i + 0.6X1.2) sin + - # Xi. 2 cos (6) sin 6> + Xi.i ^ sin (0 + 60) = #0(^1.2 + 0.5Xi.i) Sin + --KLI cos (7) Qc = Xi. 2 (F + Vb) = E Ki 2 (1.5 sin 6 H pr cos 6) (8) t and, Q e = Ki. e (V a + Vb) = E Ki. e (1.5 sin d + -g cos 0) (9) Assuming for the present that the values of the MAXWELL'S coefficients are known, it is then possible to obtain, in a manner similar to that used for the balanced system, the potential of the telegraph wire. While in this case we deal with four charges, the effect of the charge of the earth is not felt at the telegraph wire, because the earth may be considered as an infinite cylinder, enclosing all wires; thus the effect of its charge on any point inside it, re- sults in no potential. The potential of the wire is now readily obtained from equation (1). The charging current in the three wires and the earth is found from equations (6) to (9), remem- bering that = cot. . . dQ a r \/3 . I ^a = ~TT = L(\CO\ (/Vi i H~ U.O/Vi 2) COS COl ~ Al 2 Sin 001 , at L 2 J r \/3 i 4 = EQCO\ (Xi.2 + O.SXi.i) cos cot ~ XLI sin cot ', i e = E OJ\ Xi.2 (1.5 COS Cot ^~ SHI CoZ ', l e = E Q CO\ Ki. e (1.5 COS Cot ^~ Sm W j It remains now to determine the values of the MAXWELL'S coefficients. Give each of the three conductors the same charge Q, and assume average values of the distance between the conductor (10) A THREE-PHASE LINE 253 and ground as H and the distance between conductors as D. Then we have approximately the following relation: OI7 9T/ 977 V. = 2Q log + 2Q log ^ + 2Q log ^ = 2Q log ' ' V We have also, .'. #1.1 + 2# lt2 - - -^ (11) 2 log Wr Give now three-phase charges to the three conductors, then, QA = ^l.l^a + #1.2^6 + K-l-zVc = Ki.iVa ~ #1.2^6 = V a (#!.! - #x. 2 ). Thus #1.1 #1.2 is the capacity of one of the three lines against the neutral, which has been shown to be: 1 . . #1.1 #1.2 ~ T^- (12) 21ogf From (11) and (12), the numerical values of Kn and #i. 2 can be determined, as well as #1.3, so that all the coefficients are known. It may be of interest to consider the problem from another point of view. By grounding one conductor, while the potential difference between the conductors is not changed, the potential of the system of three conductors has been changed. It should be possible, therefore, to calculate the charge Q , which should be given to each conductor, in order that the new potential distribution shall exist. The charge should ob- viously be such that the potential of C shall be reduced to zero. Before grounding, the potential was +V C , and hence Q Q should be such as to give C a potential of V c . :. -V c = 2Q log y 2 + 2Q log |^ + 2Q log j* 254 ELECTRICAL ENGINEERING SH 3 = 2Q log -j, using the approximations. Since V c = E sin (cot + 240), the maximum value of the TjJ charge is Q = - The charges on the conductor A after grounding the conductor C are therefore, QA + Qo = E sin cot Similar expressions are of course readily written for the charges on the conductors B and C. The potential of the telegraph line after grounding is thus, V = 2 [(Q A + Q )ai + (Q B + Qo)6i + (Qc + Qo)CJ. By applying these equations to the numerical example given previously, it will be found that the induced potential of the telegraph line will be 25 per cent, of the phase voltage or 14.5 per cent, of the line voltage. In the case of an insulated balanced system, it was found about 5 per cent, of the phase voltage or about 3 per cent, of the line voltage. The Effects of a Grounded Horizontal Wire on the Distribution of Electricity in the Atmosphere. It has been observed that frequently considerable potential difference exists between successive layers of the atmosphere. A potential gradient of 600 volts per m., or roughly 200 volts per ft., is not unusual. It is of interest then to see how much the potential at a given height may be reduced by a grounded overhead line such as is used in high-potential transmission systems. Assume that the gradient, not far from the earth, is 2 electro- static units per m. (600 volts per m.). It is readily seen that the distribution can be quite closely represented by the effect of a charged cylindrical conductor, say 300 m. or more above the surface of the earth. The conductor then represents whatever cause there was for the potential gradient. The charge per centimeter length of the fictitious conductor is determined by the fact that the potential at a certain height A THREE-PHASE LINE 255 is known. Thus according to the assumption, the potential at 15 m. above the ground is 30 electro-static units. Thus referring to Fig. 134, 01 C V = 2Q log = 30 .'. Q Q = 155. Suppose now that it is desired to find the change in a grounded overhead wire of radius r = 0.5 cm. placed 15 m. above ground. Since the potential of A, Fig. 135, is zero, it is evident that the potential of A due to its own charge and the charge on its image plus the potential of A due to the charge on the fictitious conductor and its image must be zero. nr. TT i TL n~L Thus 2Q log - : + 2Q log JT~ = = 2Q log - + 30. .'. Q = - 30 = - 1.72 E.S.U. T P Abs FIG. 134. FIG. 135. FIG. 136. The potential at a point P, Fig. 136, distant hi, from the ground is then: V = 2Q log -TT + 2 Q log , but 2Q log -^ is, according to KI TI III the first assumption of uniform gradient, 0.02/ii (hi being given in centimeters). Thus the potential of P is : V P = 0.02/i! - 3.44 log - (1) The effect of two ground wires A and B on the potential at a point P in the vicinity of the wires : 256 ELECTRICAL ENGINEERING The potential of A or B due to the fictitious and the two actual conductors and their images must be zero. The potential of A, Fig. 137, is: 2Q log |4i + 2 <3 lo S y + 2Q log ^ = = or 0.02ft + 2Q log y ^ = 0, /. Q = fj^ (2) If the wires are 2 m. apart and 15 m. from ground then r 4 = 3010 cm. and r 3 = 200 cm. :. Q = -1.31. T The potential at a point P, Fig. 138, is then: V p = 0.02^! - 2.63 log T -~ (3) It will be seen that by means of a single ground wire above a transmission line the potential is reduced by some 30 per cent., and when two ground wires are used by some 40 to 50 per cent., and that there is little gain in using ground wires of large diameter. CHAPTER XXIII THE CURL OF A VECTOR In vector representation, the curl of a vector is represented by the cross-product of the differential operator V and the vector. It is: V X R = curl R = dZ dY i j k dx dy dz X Y Z dX dx) /dY _ dX\ ~ dill \dx dy iCx + jCy + kC,. The curl of a vector is thus a vector and its components along the axes are C x , C v , and C z . It is important to analyze the meaning of this new vector. dy c ^Z, dz FIG. 139. Consider a small rectangle in the y-z plane, Fig. 139. Let the component of R along the ^/-axis be Y and let it change to FI, as we move along the z-axis from a to b. dz dz Similarly, 257 258 ELECTRICAL ENGINEERING The line integral around the rectangle is then : dL = Ydy + Zidz - Y^dy - Zdz idZ dY\ , , = I - - ) dydz. \dy dz j Extending this to all three planes, we get: the line integral around dS, -}dzdx + dz dx dy = C cos adS, where a is the angle between curl C and the normal to the surface dS. The z-component of the curl C x is then seen to be the limit of the ratio between the line integral of the vector around a small element in the y-z plane and the area of the element. Since it is the ^-component, it is, of course, at right angle to the surface, dydz. In general, r. i r AL dL Curl = lim -TO = -TO, AS dS where surface dS is normal to the vector C. Stokes's Theorem. STOKES'S theorem states that the line integral of a vector R around any closed contour is equal to the surface integral of the curl of the vector over the surface or cap enclosed by the contour. The theorem holds always when transforming from the line integral to the surface integral, but applies in the transformation from the surface to the line integral only when -r + h -T = 0, that is, only when the curl has no divergence. Depending upon the system of notations used, it is written in either of the following ways: In vector notation, it is: fR dr = f f (V X R) ' NdS, which is to be read: The line integral of the electric field in- tensity along the circuit is equal to the surface integral of the curl of the vector over any surface (any cap) bounded by the circuit, where N is the unit, outward drawn normal to dS. THE CURL OF A VECTOR 259 Obviously, the theorem may also be written: fRds cos (Rds) = f(Xdx + Ydy + Zdz) dZ dY\ dX dZ BY d where I, m and n are defined below. The theorem can best be proven by calculus of variations, but may be understood without mathematics by the following reasoning. Refer to Fig. 140, which shows the cap divided up into a number of small elements. It is evident that the sum of the line integrals around all these small areas resolves itself into the line integral around the contour, 'since all lines, except the contour, are traced in two equal and opposite directions. Thus if dL is the line integral around one of the small areas, then 2dL = fR cos (Rds)dS. But it has been shown, that fdZ dY\. dX dZ. /BY dL = " \. - dx) dzdx cos adS, where a is the angle between the curl C and the normal to the surface dS. (2) .*. dL = C cos adS' but dydz = IdS, where I = cos (Nx)' } dzdx = mdS, where m = cos (Ny) ; dxdy = ndS t where n = cos (Nz); by substituting these values in (2), equation (1) is proved. 17 CHAPTER XXIV THE EQUATION OF THE ELECTROMOTIVE FORCE It has been shown that the potential difference between two points in an electric field is the line integral. V = f(Xdx + Ydy + Zdz) = fGds (1) where X, Y and Z are the components of the field intensities or gradient along the x, y and z axes and V is expressed in electro- static units. It will be shown later that the conversion factor between the electro-static units and electromagnetic units of potential is the velocity of light, v = 3 X 10 10 cm. per sec. The e.m.f. in the electromagnetic system of units is v times that in the electro-static system of units. Equation (1) should be written: V = vf(Xdx -f- Ydy + Zdz) in electromagnetic units (2) Experiments have also shown that the e.m.f. in electromagnetic units in a circuit is equal and opposite to the product of the turns enclosing the magnetic flux and the rate of change of the flux. If L, M and N are the components along the x, y and z axes of the magnetic field intensity, and if I, m and n are the direction cosines of the normal to the surface dS, and if /z is the permea- bility then the flux is: = ffpQL + mM + nN)dS = ffpH - dS Then the e.m.f. induced per turn is: V = - ~ = - ^ [ff(lL + mM + nN)dS] (3) combining (2) and (3), and assuming M constant, But from STOKES'S theorem, we can write: f(Xdx + Ydy + Z*i) =//[Kf - ) + (f - g 260 THE ELECTROMOTIVE FORCE 261 Equating (4) and (5), we get: dY\ /dX dZ\ /dY dX\-\ m \~d~z - dx) + n (to " a J = r ~ v dt (6) If the circuit be closed, a conduction current will flow, and its magnitude will depend upon the resistance. NOTE. If the circuit is inductive, this applies equally well, since in these equations the total variation in flux is considered. Let I, with components u, v and w be the current density along the x, y and z axes, and p be the resistivity of the ma- terial. Let ds with components dx t dy and dz be an element of the circuit, and A x , A y and A z be the projected areas of an elemental surface dS, then the resistance along the z-axis is - dx and dV = (resistance X current) = -- T uA x = pudx, A x A x but V dV Y X = ~~dx =pu > - X = pu - Similarly, Y = pv, and; Z = pw. It should be noted that X, Y and Z are expressed in electro- static units. Thus by transforming the relations to electro- magnetic units, we get: pu = vX] pv = zY', pw = vZ. The Equations of the Current. Let the components of the current density along the three axes be u, v and w, in electromag- netic units. Let I, m, and n be the direction cosines of the normal to surface dS' t then the total current is: + mo + nw)dS. 262 ELECTRICAL ENGINEERING It was shown by AMPERE that the work done in carrying unit pole around an element carrying current i was 4iri. The work done is J* (Ldx -f Mdy + Ndz), where, as usual, L, M and N are the components of the magnetic field intensity. .*. f(Ldx + Mdy + Ndz) = 4x7 = 4*ff(lu + mv + nw)dS. But by STOKES'S theorem, f(Ldx + Mdy + Ndz) = ff(lC x + mC y + nCJdS. + mv + nw)dS. dN dM = C x = -r -- -5 a?/ a^ r aL aAr = C y = - - , and, 4 _ c ._ 5M _ aL " dz dy Energy of the Electric Field. Consider a small cube-shaped volume dxdydz, Fig. 141, in the electric field, and let the po- tential difference between the two sides dxdy be V. FIG. 141. The capacity of the field enclosed by the cube has been shown to be: M J M J 4ird 4irdz The energy stored in the field is J^CF 2 , and the potential dv V is Zdz, where Z = _ Kte*y ^ _ KZVxdydz 4irdz Sir KZ* , KZ* . - dv, or the energy per unit volume = -^ , when only the ^-component of the field is considered. THE ELECTROMOTIVE FORCE 263 If the components of the electric field intensity R, are X, TT Y and Z, then, the total energy per unit volume = Wo = ^ O7T (X 2 + F 2 + Z 2 ). Similarly, it is proven that the energy stored per unit volume in the magnetic field is: W = ^ (L 2 + M* + TV 2 ). Thus the total energy per cubic centimeter in space occupied by magnetic and electric field is: W = ^ [ M (L 2 + M 2 + JV 2 ) + K(X* + F 2 + Z 2 ).] There appears to be no limit to the possible intensities of the magnetic field, but for the electric field in air at atmospheric pressure, experiments indicate a maximum possible gradient, or field intensity of 30,000 volts per cm., or 100 electro-static units of potential per cm. Thus in the electric field the maximum amount of energy at normal pressure is: 100 2 W max . = -~ - = 400 ergs per cu. cm. or 0.00004 joules per cu. cm. Maxwell's Displacement Current. MAXWELL assumes that when a potential difference exists in any part of a dielectric, an electric displacement, or a displacement of electricity has taken place along the lines of electric intensity (force). The greater the displacement, the greater the difference in potential. The displacement, however, is resisted by the electric elasticity of the medium, which, for the lack of a more satisfactory analogy, can be thought of as being in a way similar to that existing in an elastic body, against which a particle is pressed. For a given potential difference, the displacement is greater the greater the specific inductive capacity; for example, if the dielectric be glass, the displacement may be five to six times as great as would be true with air or vacuum. A metal may be considered to have zero capacity, in other words, energy can not be stored into it, but electricity would continue to pass through it as long as a potential difference existed. Dielectrics, on the other hand, would permit electricity to flow up only to a certain distance, and the flow ceases when the 264 ELECTRICAL ENGINEERING force causing the electricity to flow is exactly equal to the opposing force due to the elasticity of the dielectric. The displacement of electricity is in the direction of the lines of electric force; since the displacement has magnitude as well as direction, it is a vector quantity. According to MAXWELL'S theory an electric current is a time rate of change of the displacement of electricity. The charge on a body is a measure of the displaced electricity. Indeed, MAXWELL states that a charge Q on a body causes a displacement of Q units of electricity out from the body, and he has defined the displacement D as the charge per unit area. It is then numerically equal to a 1 , the charge per unit area, but while or is a scalar quantity, D is a vector. D can be expressed as a function of the intensity R and the specific capacity K. T ^ j. *. i? xi_ /> i j flux ^ 4?r C T In air the intensity of the field is - - = j In area area A other dielectric of specific capacity K, P 1 47TQ ARK R = K ~T " Q ~- ^T Q ARK RK The surface charge = T = - A A 4: Thus the displacement D is also, The displacement of electricity is in the direction of the field. Thus if /, g and h are the components of the displacement, and X, Y and Z are the components of the electric field intensity, then, KX g 47T ' ^^ In these equations, the units are in the electro-static system, and, , KZ h = -T. 4?r The amount of electricity displaced is the product of current and time, or considering current per square centimeter or current density, the displacement is the product of current density and time. THE ELECTROMOTIVE FORCE 265 Let Ud, v d , and w d be the components of the current density, then: u d dt = df, Vd dt = dg, and, dt = dh. It has been shown that the conduction current density in electro-static units was: X and, Z w = , where p is the specific resistance. Thus the total current density along the x-axis is: similarly, = X ,df = X JtdX p dt p 4ir dt ' Y , dg Y KdY v + v d = h -7.- = h T- -7:7 p at p 4?r at Z dfc Z X dZ Thus, applying AMPERE'S relation, that in electromagnetic units the curl of the magnetic field intensity is 4?r times the current density, we get : 47T , - (w similarly, and, 1/47TZ 9\ p dX\ _ lr47T K dt) := vlp - dt dM dz dx dM dL (16) L p dtJ dx dy where v at present is the unknown ratio between the units. The corresponding equations for the e.m.f. were shown to be jT *\ rr *\ ~\7 fj, (JLJ O j O I ~ v dt dz ~ dx 266 ELECTRICAL ENGINEERING v dt = f ~?x ^ _ p cW _ dY _ dX ~ v dt = dx " dy By combining equations (1) and (2), it is possible to arrive at equations of the electric and magnetic field intensities in any medium conductor or non-conductor. Differentiate (la) with respect to t, 7v 7" ~dt + K ~W = dydt ~ ~dzdt Differentiate (2c) with respect to y, IJL d 2 N d 2 Y d 2 X * ' ~~ ~v ~didy = dxdy ~ 'dy 2 Differentiate (26) with respect to z, fi d 2 M d 2 X d 2 Z " ~v Htdz == 'dz 2 ~ dxdz ^ 2 ~y Substitute (4) and (5) in (3), and add and subtract ^ = T ( -^ } , the following equation results: 47TM dX 3 2 X rd*X .^X.^X d_ p dt ' AM dt 2 '' V Idx 2 " dy 2 " dz 2 ' dx which is the most general equation. If there is no divergence, that is if we are interested in medium having no charges, then the equation becomes: It is readily seen that exactly similar equations not only result for the Y and Z components of the electric field intensity, but also for the components of the magnetic field intensity, L, M and N. Special Cases. (a) In a dielectric, p = , thus the equations become : (8) THE ELECTROMOTIVE FORCE 267 or in general, -Qp = a 2 V 2 ?7, where a 2 = ~, and U stands for either X, Y, Z, L, M or N. This is the well-known equation of the propagation of any disturbance at finite speed. The velocity of the propagation is a = - . In air. k = 1 and M = 1, thus the velocity of propagation of the electric and magnetic field is v. This value has been measured and found to be that of light, thus the conversion factor is the velocity of light. Thus v = 3 X 10 10 . ' This important fact was deduced by MAXWELL in 1865. (6) In a conductor, the specific inductive capacity may be assumed as zero, thus we get: or, d*U , d 2 U 47r dU . d^ + ^ = ^-ar in rectangular coordinates, and, , , 1 dU . ^ a0T + -^ + r -^T == > ln cylindrical coordinates. ; Assuming, as an application, that it is desired to determine the current distribution at any time in a cylindrical conductor at any distance from the origin and any distance from the center of the conductor. If the practical system of units is used, v 2 = 1; and on account of circular symmetry, the term involving disappears. Thus the equation becomes: d z i d z i 1 di 4?r di , , a7 2 + a^ 2 + r dr = 7 Hi Distribution of current in a cylindrical conductor: If it is of interest to find the distribution along a radius only, the equation becomes : dH 1 di _ 47r di , . dr 2 + r dr ~ p dt 268 ELECTRICAL ENGINEERING It will be of interest to verify this equation directly. It has been shown that the work done in ergs, in taking unit pole once around a conductor carrying current 7 is 4?r7, where 7 is the current enclosed in the path. Consider, for the sake of simplicity, a cylindrical conductor, Fig. 142. Let the instantaneous values of the current density at distant r from the center be i, and that at r + dr be i + dr. FIG. 142. Let the magnetic field intensity at distant r be H] and at riff distant r + dr, be HI = H -f dr. The work done on unit pole in going from a to b is: #i(r + dr)6 = (H + ^ dr) (r + dr)8 = (\TT v Hr + #dr + r -~- dr\ , neglecting the term which in- volves (dr) 2 . The work done in going from b to c, or from d to a, is zero, because we travel on an equipotential surface. The work done in going from c to d is Hrd. :. W = e( Hdr + rdr = edr And by definition given above, W = 4irir0dr, neglecting the term which involves (dr) 2 . H dH /1 x (1) THE ELECTROMOTIVE FORCE 269 The ohmic drop in voltage along 1 cm. of the conductor at the outer edge of the segment, that is, at r + dr from the center, perpendicular to the paper, is: (i + dr) p, where p is the specific resistance. The drop along the inner edge is ip; thus the difference in the e.m.f. at the two edges is: de = This must be then the e.m.f. which is consumed by the self- induction due to the flux in the element. The flux in the element is = pH(dr X 1 cm.) == Hdr (3) AA. AH (4) (5) From (2) and (4), di dH di _ dH Differentiating (1) with respect to t, di __ 1 dJH d*H ' dt r dt + drdt Differentiating (5) with respect to r, Substitute (7) in (6), di 1 dH p dH .*. 4?r = _- + - z Substitute the value of --- from (5) in (8), di _ 1 p di p dH dt r n dr IJL dr 2 or, M dH . 1 di _ p dt ~ ar 2 "" r dr in electromagnetic system of units. Equation (9) is very important in connection with problems of heat as well as electricity, it has been studied by great mathe- maticians, notably, MAXWELL and LORD RAYLEIGH. It is to be noted that the right-hand member of equation (9) is LAPLACE'S equation transformed to cylindrical coordinates, 270 ELECTRICAL ENGINEERING when the cylinder has circular symmetry. Thus we could have written : Special Case. Flat bar: Referring to Fig. 69, in the case of flat bar, r approaches infinity, and (9a) becomes: Si FIG. 143. The distribution of flux in a cylindrical conductor surrounded by an energized solenoid is determined in a similar way. Fig. 143 shows the path of the current and flux. The dots represent the current, and the lines around the current, the flux. The result for a cylinder is identical with equation (9), if H is substituted for i. Similarly, for a flat bar equation (10) is applicable with the same substitutions. CHAPTER XXV MATHEMATICAL SOLUTION OF EQUATION 11, PAGE 267, DEALING WITH ALTERNATING CURRENT DISTRIBUTION IN CIRCULAR CYLIN- DRICAL CONDUCTOR The general equation is as has been shown: dH 1 di 47TM di dr 2 + r dr ~ p dt Since we are dealing with sine waves, let: i = i\ cos cot + z*2 sin cot (2) where ii and i Zj the current densities, are functions of r but not of t. Substitute first, i = i\ cos cot, di dii - =cosa^-> - 2 and > fU -a* sin erf. at dH\ 1 dll 4:TTfJL . .'. cos ut T-^- H cos (^t T = - iico sin cot (3) Similarly, for i = i% sin co, J a 2 l2 , 1 dll +47TM . sin ut T 2 - + - sm o)t = - izoj cos cot (4) Adding (3) and (4), ' COS M ^ + - fr- H| and d 2 i z 1 5i 2 W+r dr = 271 272 ELECTRICAL ENGINEERING + 6 2 r 2 Assume : i\ == a<) ~\~ air and Then: v- 1 = 01 + 2a 2 r + 3a 3 r 2 -=-- = 2a 2 + 6a 3 r Let + n(n b n r n .n-l + HI' (8) (9) (10) (11) (12) (13) (14) (15) '. ai = 0;4a 2 = m 2 6 ;9'a 3 = 7^ 2 6 1; in general, n 2 a n = m 2 6 n _ 2 (16) By similar substitutions in (14), we have: 1 = Q; 46 2 = m 2 a ; 96 3 = m 2 aij in general, (6) and (7) can be written: dr and, dr dr Substituting (9), (10) and (11) in (13), 2a 2 r + 6a 3 r 2 + + n(n l)a n r n ~ l - + a! + 2a 2 r + 3a 3 r 2 + + na n r n ~ l 6 n = - m 2 a n _ 2 , or (n - 2) 2 & n _ 2 = - m 2 a n _ 4 Combining the last equations in (16) and (17), m 4 a n = From (17), n 2 (n - 2) (17) (18) (19) Since ai = 0, and 61 = 0, from (18) and (19) all the a's and 6's with odd indices separately equal to zero. And those with even indices are as follows: SOLUTION OF ALTERNATING CURRENT 273 a c = a m 4 w 4 as = fi-FToS a and so forth m 42 . 6 2 * m 4 m 8 8 2 .^02 6 = + 42"7g2Tg2~ and so forth (19)] 60 =+ 7^02 [See (16)] a m 10 2 2 -4 2 -6 2 -8 2 7o 2 o and so forth m 2 8 2 m 2 m 6 4 2 -6 2 -8 2 -(10) 2 -(12) 2 and so forth Therefore, i\ = a (l ~ 02742 + "02. 42.^2.02 ~ ' * ' ) H 9 , " t " and 52 2 2 -4 2 -6 2 2 2 -4 2 -6 2 -8 2 -(10) 2 ~2"2~ ~ 2 2 -4 2 -6 2 + ***] + m 4 r 4 ) (21) LORD KELVIN has denoted the first series in (20) by ber (mr) and the second in (20) by bei (mr), thus: oer(x) 2 2 -4 2 -6 2 -8 2 , and /v.2 ^.6 /rlO ,.XN *> ^ | t' = 2*~ 2 2T 4 2T 6 2 "*" 2 2 " T 4 2 6 2 8 2 (lO) 2 And 4 ii = a for (mr) + 2 a 2 6et (mr), and, 4 t' 2 = ao 6ei (mr) + 5 a 2 6er (mr) (22) (23) These functions, 6er and 6^', have been worked out and appear frequently in books on mathematical physics. 274 ELECTRICAL ENGINEERING Therefore, i = a her (mr) -\ - 2 a 2 bei (mr) cos ut + \ a z ber (mr) a bei (mr) I sin cot (24) The constants a and a 2 are determined from the fact that the extreme outside layer is not surrounded by any flux. (We consider only the flux in the wire in this calculation.) Thus the sine term is zero at all values of t. Let Jo = maximum value of the current density at the surface, then, and 7 = a ber (mR) -\ ^ bei (mR), 2 = ^ her (mR) a bei (mR) (25) Equations (25) are readily solved and give: 4a 2 _ JQ bei (mR) m? ~~~ ber 2 (mR) -f bei 2 (mR)' IP ber (mR) ber 2 (mR) + bei 2 (mR) {[ber (mR) ber (n and, bei (26) bei ber 2 (mR) + bei 2 (mR) (mr)] cos ut + [bei (mR) ber (mr) ber (mR) bei (mr)] sin co} (27) Thus the square of the effective current density I r)P7*^ I ? rr I r>^7*^ ( 'YyiT'i -j~\\ TO [t/C'/ \^ffvL\j) Ut/l \ifvl ) ber 2 (mR) bei 2 (mr) + bei 2 (mR) bei 2 (mr) + bei 2 (mR) ber 2 (mr)] 7 2 [ber 2 (mr) + bei 2 (mr)]. 2[ber 2 (mR) + bei 2 (mR)] V2 (mR) + bei 2 (mR) (28) At the center of the conductor, r = 0, .'. ber (mr) = 1, bei (mr) = 0. 1 leff. = _ A/2 Vber 2 (mR) + bei 2 (mR) at r = 0. SOLUTION OF ALTERNATING CURRENT 275 With very low frequency, the current density approaches the direct current case where it is normal and is: thus the ratio of the alternating current density at the center. to that of the direct current is: _ __ 1 __ \/ber 2 (mR) + bei 2 (mR) For copper, fj. = 1, = and p = 1600, If the radius is 1 cm., and the frequency is 60, mR = 1.72, [ber 2 (mR) + bei 2 (mR)]~* = 0.87. .'. the current density at the center is 87 per cent, of that at the surface, and also 87 per cent, of what it would be with direct current. If the conductor had a diameter of 50 cm., the current density at the center would only be 25 per cent, of that at the surface. Actual watts consumed in heat are : '72 [ber 2 (mr) + bei 2 (mr)]d(r 2 ) -(29) C 2 I Jo 2[6er 2 (mR) + bei 2 (mR)] W = (ohmic resistance) (total eff. current) 2 (30) Ohmic resistance = - (31) Total current = irK i2irrdr = ber 2 (mR) + bei 2 (mR) ber (mR) \ ber (mr)d(r 2 ) + bei (mR) \ bei (mr)d(r 2 ) r C R C R i Cosut-\-\bei(mR) I ber(mr)d(r 2 ) ber(mr) I bei(mr)d(r 2 ) sincoZ L Jo Jo J 272 .'. (total eff. current) 2 = or , , . PN , , . 9 . ^r\- 2[ber 2 (mR) + bet 2 (mR)] \ C R I 2 \ C R 1 2 1 I ber (mr)d(r 2 ) + j bei (mr)d(r 2 ) (32) U Uo 276 ELECTRICAL ENGINEERING 7rp/o 2 j|J o 6er(mr)d(r 2 )J 2 +|J o bei (mr)d(r 2 ) \ 2R 2 [ber 2 (mR) + bei 2 (mR)] The coefficient of skin effect = w act . ' C R 2 C R Jo Jo 2 (mr)d(r 2 ) \R 2 J W jf ber (mr)d(r 2 )\ 2 + \ i I 2 bei (mr)d(r 2 ) (33) (34) (mr) = 1 - 312 + 25,500 210,100% (mr) + bei 1 (mr) == 1 + 313 + 3900 I [6er 2 /mr\ 12 21,700 Q + (mr) 3100 (35) C R \ ber J ' (mr)d(r 2 ) = I I SOLUTION OF ALTERNATING CURRENT 277 Substituting (35) (36) and (37) in (34), K = 10 (38) The following tables give the coefficient of skin effect at various values of mil and the values of m for copper, aluminium and iron. mR K mR K mR K 1 3.0 1.32 6.0 2.39 0.05 1.0001 3.5 1.49 8.0 3.10 1.0 1.005 4.0 1.68 10.0 3.79 1.5 1.026 4.5 1.86 15.0 5.57 2.0 1.08 5.0 2.04 20.0 7.32 2.5 1.17 5.5 2.22 Material M p in. e.m.u. m Copper. . 1 1700 at 20 C. 216 A/7 Aluminium 1 3,000 162 A/7 Iron 300 to 1200 10,000 . 09 A/M? The value of /x for iron is usually taken as 300, but experi- ments on iron wires used as transmission lines seem to give values of M as high as 1200. LORD RAYLEIGH has shown that when the penetration is so slight that the above table can not be used a close approximation of the " effective thickness" in centimeters of the surface layer which causes the current is: == 7 where K is the specific conductivity. 6.6 This formula becomes 6 = 7= for copper approximately. 8 8 ^= for aluminium approximately. 16 Vrf for steel approximately. CHAPTER XXVI ELECTROMAGNETIC RADIATION Introduction. The laws governing electromagnetic radiation were stated by MAXWELL fifty years ago. The experimental verification was presented twenty years later by HERTZ in a series of most extraordinary papers, which were later published in book form. The practical application was made by MARCONI. An extensive literature is now available, notably FLEMING'S "The Principles of Electric Wave Telegraphy and Telephony," and ZENNECK'S "Wireless Telegraphy." In writing this chapter the author has drawn extensively upon the information which is given in these books. Since it is likely that students who have not read what preceded this chapter will want to understand the principles of wireless transmission it has seemed wise to built up the theory from the fundamental laws even though this procedure necessarily involves some repetition of what has been given in previous chapters. Fundamental Conceptions. Surrounding any body charged with electricity is an electric field. The intensity of the field usually varies from point to point, but, at any point it is propor- tional to the charge, that is, the amount of electricity on the charged body. To charge a body we connect it to a source of potential when a current momentarily flows from the source to the body, the cur- rent stopping when the potential of the body is the same as the potential of the source. If i is the current flowing during an interval of time dt then the resulting charge on the body is dq = idt, or, i - ^ 1 ~ dt For reasons that will appear later, it has been assumed that the outward field of flux from a body charged with Q units of electricity is \l/ = 4wQ lines of electric force. 278 ELECTROMAGNETIC RADIATION 279 If the lines are uniformly distributed over a closed envelope of area A sq. cm., then the density of the electric field is By the introduction of the constant 4ir in the flux formula this density becomes in space the same as the force in dynes per unit charge which is numerically the same as the intensity R of the electric field at the particular point considered. This is easily seen from COULOMB'S law, which states that the repulsive force between two charges Q and Qi is f _QQi J - Kr 2 where r is the distance between them and Vi = 1. In the ideal case the charge is confined to a point and the flux is distributed uniformly in every direction. . R = __ = ^Q = Q area of sphere 4?rr 2 r 2 where r is the distance from the point to the point charge. or, Q = Rr* /. / = j3 rQ! = RQ,. If, therefore, Qi ==!,/= R. The potential difference between two points in an electric field is by definition numerically the same as the work done in moving unit charge from one point to the other. Thus, if X represent the intensity of the electric field in a cer- tain direction, say a direction parallel to the x-axis in a rectangu- lar coordinate system, then the potential difference across a short element dx is dV = Xdx = force on unit charge at dis- tance x, or, Y _dV = dx' Similarly v dV = dy and _ dV J 7 * dz Y and Z being, respectively, the electric intensities along, or parallel to, the s&-a&d-y axes. * * 280 ELECTRICAL ENGINEERING If we desire to find the potential difference between the ends of a wire bent in a small rectangle in the x-y planes and the inten- sities along the x and y axes are X and X\, Y and FI, then refer- ring to Fig. 144, dV = Xdx + Yidy - X,dx - Ydy (3) dy dx FIG. 144. For y = y = dy X = X X = X, The rate of change of X as we travel along the ?/-axis is thus the total change in distance dy is : dX, -r dy dy y Similarly, Substituting (4) in (3) we get dX (4) It is one of the properties of the electric field alone when free from charges that the above potential difference is zero in a closed circuit. If, however, an e.m.f. is induced in the rectangular circuit by change of flux treading through the circuit, then we get: dt dN 1 1 -^dxdy ELECTROMAGNETIC RADIATION 281 where N is the density of the magnetic field perpendicular to the plane of the electric circuit. By a similar reasoning we get then the following three impor- tant equations: dX dy = ar dx dX dz dy _ dx dY_ dz dN dt dM dt dL dt (5) where X, Y, Z, L, M, N, are respectively the electric intensities and magnetic intensities in the same system of units parallel to the x, y and z axes. Note that in air the densities are 'the same as the intensities. The next consideration is in re- lation to the magnetic effect of a current. Let A, Fig. 145, represent the end view of a wire A carrying a certain current dl, perpendicular to the plane of the paper. Let the curved line be in the plane of the paper. The magnetic field intensity at P is then // and this is defined similarly to R as numerically the same as the force on unit pole. Let, therefore, a pole of unit strength be carried along the curved path, Fig. 145. The work done per unit pole in completing the journey once is evidently W = fll cos 6ds = ( - - cos 0ds J r FIG. 145. but rda rda -3 =. cos 6 , . as = cos 6 r a -2* :. w : Jtx = 2dlda = The work is independent of the position of the current element and the path. Thus if there are a number of filament currents inside the path, I = Zdl W = 47rJ (6) 282 ELECTRICAL ENGINEERING This quantity is by physicists called ^ the magnetomotive force, around the circuit, whereas, engineers would call I Ll it 47r X m.m.f . d Consider now a small rectangular surface, Fig. 146, in the x-y plane of a magnetic field, and let L and LI, M * an d MI be the components of the FIG. 146. magnetic field intensities, along the x and y axes respectively. Then the line integral, or work on unit pole around the element is Ldx -\- Midy Lidx Mdy. The rate of change of L as we travel along the y-axis is ^- thus the total change is dy, thus L l= L + - dy. Similarly dM dL idM dL\ - - dxdy = ( d - - ~) dxdy. From (6) it is seen that dy where I z is the total current flowing through the rectangle per- pendicular to dxdy. Depending upon the medium, this current may be the ordinary conduction current such as flows in a wire or the charging current which is incident to a change in the electric field, or indeed, the sum of the two currents. In this analysis it will be assumed that the air surrounding the oscillator is free from ionization, so that its resistance is infinite; thus the only currents considered are the " displacement, or charging currents." MAXWELL assumed that surrounding a charged body is an electric field, the strength of which is proportional to the charge, and that the intensity of the field is a measure of what he calls ELECTROMAGNETIC RADIATION 283 displaced electricity. The displacement of electricity is in the direction of the field intensity, and is thus a directed quantity. Numerically a charge Q displaces Q units of electricity outward from the body. Since dQ = idt, it follows that the displacement current or, as engineers say, the charging current is proportional to the time rate of change of the electric field intensity. Or, dR where R is the intensity and a a constant to be determined. MAXWELL worked out his theory on the basis that the dis- placement is numerically the same as the charge per unit area. Thus area But the outward normal flux from a charge Q is \j/ = 4-n-Q ; thus the intensity of the field is R area area r> .'. R = 4:ird or d = 7- in air. i = ~ 4?r dt where i is the current per unit area or current density. If, therefore, u, v and w are the components of the displace- ment current densities along the x } y and z axes and X } Y and Z, the components of the electric intensities then : 1 dX 1 dY 1 dZ u = -7 w = -: and w = - A TT (8) 4-7T d 4;r ^^ 4.w dt everythir^g being given in electro-static units. From (7) and (8) it is evident that one can write /dM dL\ . dZ _ ( ^r I dxdy = 4:irwdxdy = dxdy. \ dx dy] dt or, dM _ dL dZ dx ' dy == dt' 284 ELECTRICAL ENGINEERING dN dM dX dy dL dz dN dt dY z dx dt dM dL dZ dx dy dt 1 By a similar reasoning are obtained the following three relations. (9) everything being given in the same system of units. The simplest form of oscillator, or rather that form which lends itself to the simplest mathematical treatment, is that used by HERTZ. The oscillator consists of two large spheres separated by a considerable distance and connected by wires through the spark gap to the source of energy as shown in Fig. 147. o Magnetic Field FIG. 147. O It will be assumed that the electric field is due to the spheres alone, and the magnetic field to the linear conductor. It will be assumed that the axis of the oscillator is the 2-axis. Thus the magnetic field which is in the form of rings around the conductor has, in the x-y plane, no component in the direction Z and therefore no e.m.f. can be induced in the x-y plane. However, e.m.fs. will be induced in the direction of the Z-axis. Whatever the potential distribution in the x-y plane it must thus be due to the charges on the spheres alone, that is, due to the electric field alone. The distribution of potential around an electric double, that is, ELECTROMAGNETIC RADIATION 285 around two spheres given equal but opposite charges. Referring to Fig. 148, since dV = - Rdr, V = -fRdr = -J^ r = f. The potential at P is (Fig. 149) : FIG. 149. It is when r is large compared with dZ (see note). NOTE. Proof: (10) a (q ^r 3/ 9 cos 6 and (2 by the use of the binomial theorem it is easily seen that this becomes: 2qdz cos 6 -^2 Equation (10) may be written where /i, one-half of the length of the oscillator is substituted for dz. NOTE. Equation 11 is not limited to spheres but is quite general as long as the distances dealt with are long compared with the length of the oscil- 286 ELECTRICAL ENGINEERING lator. Suppose, for instance, that we are dealing with a linear oscillator. We assume then that the potential at a point P can be expressed as due to two point charges located at some points on each rod (not the end of the oscillator) which will give the same potential as the linear conductor actually gives at distances far away from the oscillator. While this assumption is quite justified when dealing with points in space far away from the oscillator, it is obviously not at all permissible at points near the oscillator, because it is readily seen that the potential distribution at the surface of the two halves of the oscillator must be such that the surfaces themselves are equi- potential surfaces and two point charges, no matter where located, can not give such equipotential surfaces. Fortunately, we are for practical purposes interested in only what happens far away from the oscillator, where equation 11 applies. The subsequent equations can indeed be used with such linear oscillator if instead of letting Q or 7 represent the charge and current respect- 2 2 ively, we use the average value along the oscillator which is - Q and - 7. 7T 7T The ratio between X the wave length and h the height of the sending antenna is in such case, theoretically 4, but in reality due to various effect nearer 4.8. When P is far away from the oscillator the electric condition is not due to the instantaneous value of the charge q at the oscillator but due to the value of q which existed somewhat earlier in time. Thus the charge causing the electric field at P is not q = Q sin ut but q = Q sin u(t At) where At is the time required for the distribution to reach P. If v is the velocity of the propagation which is that of light, then T vAt = r 'or At = - v .'. q = Q sin f ut j If X is the wave length then 27T .*. q = Q sin I cot rj = Q sin (oot mr) = Q sin (mr cot) .'. V = - 2Qh ^ dz r . _ dF 6 2 sin (mr - ut) (13) v dV d 2 sin (mr - ' ay dydz ELECTROMAGNETIC RADIATION 287 H M Z the component of the electric field intensity perpendicular to the x-y plane cannot be obtained from V alone as discussed above. We shall now consider some of the properties of the magnetic field intensities. Consider the x-y plane (Fig. 150). It is obvious that since the lines of force are circles, the sum of the projections of the components of the magnetic field intensities along the x and y axes on a radius vector must be zero. Let L and M be the components of H along the FIG. 150. x and y axes, we have, but and or Since L itself is negative in the position shown, L cos a + M sin a. = cos a = - P sin a = - P '. Lx + My = 0, - - but x 2 + y 2 = x L M .'. xdx + ydy = 0. Thus L_ dx M dy or Ldy Mdx = This is satisfied as long as L = du du and M = dy dx (14) (15) where u is any function of x and y N the component along the 2-axis is obviously zero. From equations (9) and (15) dt ~dt "aT dy dL dz dz dz dxdz dL _ d 2 u dz ~~~ dydz _ dM dL _ /d z u d*u\ '' ~dx "~ fry " dx (16) 288 ELECTRICAL ENGINEERING Referring now to (13) and differentiating X with respect to t - <2Qh - ( d * sin (mr ~ w ' dt " 2Qh dt(dxdz ' ~~r It is evident by comparing (16) and (17) that, d sin (mr wQ Substituting this value in (16) we get: dX .d 3 (nsM. sin (mr dt dtdxdz or v on , A = ZQfi F = 2Q/i N = where n = sin (mr - cop It is now a simple matter to get the different derivatives of n. An inspection of the several terms will readily show that some are much larger than others. There is little object in investi- gating conditions close to the oscillator by these equations even if all terms are used without considerable caution, because an approximation was made in the assumption that the electric field emanated from two point charges. The derivatives contain trigonometric terms having coefficients of raV 2 mr and unity. The terms containing w 2 r 2 are so much larger than terms involving mr and unity that the latter can be neglected. Making these approximations and placing P in the x-z plane we get for distances involving several wave lengths, ELECTROMAGNETIC RADIATION 289 7 = Z = - -- sin (wr 7, = .. rr M = // = sin 2 . . sin (rar cot) sin v (19) (20) Everything is given in electro-static units at present since all terms involve Q the charge which is expressed in such units. FIG. 151. R the intensity along the surface of a sphere through r is (from Fig. 151): R = Z sin B X cos 6 = - sin (mr at) sin (sin 2 + cos 2 0) or R = sin (mr cot) sin (21) It is of interest to compare equations (20) and (21) 47T 2 since and We can write X 2 2ir ma) = - 47T 2 X2T H = R (22) when the charge is given in electro-static units. The electric and magnetic intensities perpendicular to each other in space are in time phase. Thus the product of the two represents power. (This is the case only at some distance from the oscillator, near the oscillator the large part of the fields is in quadrature.) 290 ELECTRICAL ENGINEERING It is remembered that in the ordinary electric circuit involv- ing capacity and inductance the magnetic and electric field intensities are in time quadrature and, therefore, the product represents " wattless power or better reactive power." Energy Radiated. Equations (20) and (21) can be trans- formed to read, and 2Ihm . , H = - sin (mr co/) sin 21 hm 2 . , K sin (mr co/) sin 6 co r (22) Since dq q = Q sin (co/ mr) and i = -37 = Qco cos (co/ mr) dt .'. I = QcoorQ = Qco cos (mr co/) If 7 is expressed in amperes and R in volts per centimeter, then and But H = 0.21 sin (mr - co/) sin 6 21 Hm 2 R = 300 X 10 - - sin (mr - co/) sin 6 m = and co = 2irf = 2?r -r- .', A A 0.47T/ h H = - - - sin (mr co/) sin 6 r A 1207T/ h . R = - r- sin (mr co/) sin 6 r \ (23) (24) From what has been shown, it is remembered that H and R are perpendicular to each other in space. The e.m.f. in a circuit is proportional to R, the current is proportional to H and the power to HR sin a, where a is the angle between H and R. Since these fields are perpendicular to each other in space the energy radiated in time dt eidt is proportional to HR, or, W = kHRdt and it remains to determine the value of k. ELECTROMAGNETIC RADIATION 291 The voltage per centimeter is R\ thus e = R when considering 1 cm. of circuit. The m.m.f. that produces a magnetic in- 4:iri tensity H is - where I is the length of the magnetic circuit in air. . . HI ~ 0.47T or, if we consider 1 cm. length of magnetic circuit, i = H 0.47T* Thus the energy transmitted through a square centimeter area is: RHdt W = eidt = 0.47T Thus the energy radiated through the whole sphere of radius r enclosing the oscillator is (from Fig. 152) : FIG. 152. s*e = TT rt = T r>Tj W = I 7r-r- 2wr J e = Jr = 0.47T sin Brdddt h 240 7T 2 I 2 sin 2 (mr - ut) sin 3 OdOdt but - o sin 2 (mr <*t)dt = -~ approximately 19 (25) 292 ELECTRICAL ENGINEERING and sm vav = y% :. w = 1600^ W h 2 .'. watts = ^ = 1600 ^- 2 I 2 . If I is given in effective current then, h 2 watts = 3200 ^I 2 (26) In the case of wireless transmission the radiated power corre- sponds to one-half of the area of the sphere thus, h 2 Watts = 1600 ^- 2 / 2 (27) where I is the effective value of the current. The " Radiation" resistance is obviously R = 1600 ^ (28) It is noted that in the case of wireless telegraphy the energy radiated is greatest along the equatorial plane, that is, near the surface of the earth. Since the receiving antenna is near the earth this result is, of course, very desirable. MARCONI'S improvement upon HERTZ'S oscillator resulted from his connecting the lower end of his oscillator through a spark gap to ground, by which he was able not only to obtain the maximum energy, where it was most useful, but also to make use of half the length of oscillator for the same distribution of the magnetic and electric field above ground. This will be evi- dent at once if it is considered that the earth being a perfect con- ductor, its surface is an equipotential surface. It is easily proven from the equations given that the energy received near the surface of the earth through unit surface is 1.5 times the average value of the energy per unit surface. It is also of interest to note that with an " ideal" simple antenna where X = 4h and the current is zero at the top at all 2 times and therefore the average value of the current is - 1 that 7T the power radiated in watts is 40/ e 2 or the radiation resistance is R r = 40 ohms. ELECTROMAGNETIC RADIATION 293 In this connection it is of interest to add that MAXWELL'S general equation of propagation of electromagnetic waves in space free from electric charges or magnets has been shown to be : where u is any of the components of the electric and magnetic intensities. In the case of spherical waves it is readily proven by trans- forming the equation in spherical coordinates that any function of r vt divided by r satisfies the equation. Thus r The function used so far was _ sin (mr coQ r which satisfies the above since mr ut = r = m(r vt). m In the case of sustained oscillations the function chosen was obviously most suitable. In the case of damped oscillations we would naturally choose TT = ~ e - a(ut - mr} sm(mr - ut) r where A and a depend upon the amplitude and damping of the circuit. Of special interest is the magnetic intensity // near the surface of the ground and the electric intensity R perpendicular to the surface but near the ground. Equation (21) gives, R = sin (mr wt) sin = 2h~ ' - sin (mr ut) sin = r co r 4?r - / - sin (mr cot) sin r A where I is expressed in electro-static units. If the current be expressed in amperes and the potential gradient in volts per centimeter 4wlh V X T7* X 300 sin (mr ojt) sin 10 = 377 - sin (mr - ut) sin (29) r A 294 ELECTRICAL ENGINEERING Thus the maximum value of the potential gradient near the surface of the ground is Rmao. = 377 - r- volts per cm. (30) / A If, therefore, the height of the receiving antenna is hi cm. the maximum value of the potential difference between earth and top is F 377 h h I El = T~ x hj or the impedance of the receiving antenna is Z Ei 377 h , 2 ~- 7- ~7~ x * onms - ( 31 ) and the effective value of the voltage is : E. = I f Z 1 (32) where E f is the effective value of the voltage across the receiving antenna and J e is the effective current in the sending antenna. In a simple antenna the current is a maximum at the gap and is zero at the top. Thus the current is not uniform as is the case in the HERTZ oscillator. 2 The average value of the current is - 7. With such simple 7T antenna the wave length X should be 4/i if there were no disturb- ing effects. Substituting these values we get as the impedance of the receiv- ing antenna in an ideal simple antenna Z\ = 60 ~ ohms (33) The magnetic field intensity H (equation 10) is similarly modified to H =* 2 -h sin (mr ut) sin B = - r- sin (mr wt) sin 6 0) T T A where I is in abamperes, or if / is expressed in amperes rather than abamperes h . . . . /Q ., H = - r- sin (mr cot) sin 8 (34) T A It is of interest to note that this agrees with the intensity due to an infinitely long conductor if h - JL X ~ 27T* ELECTROMAGNETIC RADIATION 295 If the sending antenna were a simple rod then the current at the top would always be zero and the average value of the current 2 would be - 7. In that case the wave length would be 4/i. 7T Substituting these values we get : 0.27 H = - - sin (mr ut) near the surface of the earth. Thus the approximation sometimes made in writing 27 H = - sin(rar to/) sin 6 is not very far from right and is correct in the case of an " ideal simple antenna." It should again be emphasized that equations (29) and (34) give the values of the electric and the magnetic intensities several wave lengths away from the oscillator. It can very readily be proven by carrying out the differentia- tions in equation (18) that near the oscillator the magnetic intensity decreases inversely as the square of the distance and the electric intensity inversely as the cube of the distance. Power Factor and Logarithmic Decrement. Prior to the use of high-frequency alternators for the production of radiation the trains of waves were oscillating, with decaying current and e.m.f. in the antenna and the word decrement had therefore a very significant meaning. When alternators or oscillating arcs are used the current and the e.m.f. at the antenna are sustained, and therefore " decrement" ceases to have any meaning. It is, therefore, appropriate to discuss the power factor rather than to try to treat of the decrement in such circuits. If RQ is the sum of the radiation resistance and the effective resistance of the wires and the ground connection, then the power consumed in the circuit is P -= PR , where 7 is the effective cur- rent. If E is the effective voltage, then I = 2irfCE = coCE where C is given in farads, thuS Pf = a>CR (35) 296 ELECTRICAL ENGINEERING Numerical application : Let C = 003 m-f. - ~i farads. h = 50 m. X = 3000 m. .'. Radiation resistance = 1600 (^7.7^;) 2 = 0.5 ohm approxi- , XoUUU/ mately W = 27T/ = 27T = 27T10 5 . A Let the effective resistance of the wires and ground be 2 ohms, then R = 2.5 ohms. .'. Pf = 27rl0 5 j^ 2.5 - 0.0047 or approximately one-half of 1 per cent. The radiated energy corresponds in this case of course to only one-fifth of this amount. The product of the current and the e.m.f. is 200 times as great as the power consumed in heat and radiations and 1000 times as great as the power radiated. Determination of the "Logarithmic Decrement." If a con- denser is discharged in a circuit of negligible resistance an alter- nating current will flow indefinitely, and no energy will be ex- pended since the energy is transferred alternately between the magnetic and the electric field. When the current is a maximum (either positive or negative) the e.m.f. across the condenser is zero; when on the other hand the current is zero, the e.m.f. is a maximum. Thus twice in each cycle the magnetic energy w m = is transferred to electric energy W e = The total amount of energy in joules surging during a cycle is then W = L/ 2 where / is the maximum value of the current, or, W = CE 2 where E is the maximum value of the voltage. ELECTROMAGNETIC RADIATION 297 If, however, the circuit contains resistance, 'the current will not alternate indefinitely but will die down gradually, the rate of decay being greater the greater the energy consumption. During these oscillations energy is also transferred between the electric and magnetic field but each pulse of energy is smaller, than the preceding by the loss of energy in the resistance. Ultimately all energy stored in the condenser becomes dissi- pated in heat or radiated away. The energy stored in a condenser is %CE 2 joules where E is the voltage and C the capacity in farads. Thus if the condenser is charged and discharged N times per sec. the sum of the energy N converted to heat and radiated away is -^CE 2 joules per sec. Zi or watts. FIG. 153. Thus Wi = ^CE 2 watts (36) z In a circuit of concentrated inductance and capacity it is shown in the elementary theory of alternating current that the oscillating current can be expressed quite accurately by the following equation: i = I 6 - at smut. IT* Where a = ^ and R Q and L are assumed constants which how- iLi ever is not the case in ironclad inductors and in arcs. 298 ELECTRICAL ENGINEERING The ratio g as is readily proven. The logarithmic decrement is (37) Incidentally 5 is also the ratio between the energy absorbed by the resistance and the surging energy per cycle. Thus I 2 RoT R T -~ which agrees with (37). In the case of the HERTZ oscillator or the umbrella type of antenna the inductance is confined largely to the linear con- ductor and the capacity to the spheres or superstructure; thus we may consider the inductance and capacity as separated rather than distributed, thus (38) (39) The resistance in the above formula is the sum of the radiation resistance, the resistance of the wires (taking into consideration the skin effect), the ground and the radiation resistance. When an arc is used the resistance of the arc should also enter. Unfortunately the latter is not a constant but depends upon the current carried, and hence the decrement is not logarithmic. However, for the purpose of this article the arc resistance may be assumed constant at say 5 ohms. For a very complete dis- cussion of this whole subject the reader is referred to FLEMMING'S " Principles of Electric Wave Telegraphy." Equation 39 contains the inductance and capacity as well as the resistance. The inductance is usually very difficult to determine since at different wave lengths more or less inductance is added to that of the antenna proper. The capacity of the antenna is however, usually not changed but it depends upon the construction of the aereal. The complexity of the structure is, however, such that its value can hardly be calculated except in the very simplest cases rarely used in practice. ELECTROMAGNETIC RADIATION 299 FLEMMING expresses the approximate capacity of a vertical wire of radius, h cm. long as: C v = - - farads, 2 log - X 9 X 10 11 when, as is the case in wireless stations the lower end of the wire is near ground, the capacity may, however, be say 10 per cent, greater. He also expresses the capacity of a horizontal wire placed h\ above ground as I C h = 2 log- 1 X .9 X 10 11 where I is the length in centimeters and h the height above ground. Thus the capacity of a T-shaped antenna may be approxi- mated as: c = c v + c h obviously the total capacity is not at all proportional to the number of wires connected in multiple. It is only slightly increased as the number of wires is increased. If the value of the capacity is difficult to calculate accurately it is measured relatively easily and will therefore be assumed as known. It ranges according to ZENNECK approximately as follows : 0.001 m-f. in torpedo boat antenna. 0.002 m-f. in battleship antenna. 0.007 m-f. in BRANTROCK station. 0.18 m-f. in NAUEN high-power station. The capacity of the antenna of the experimental installation at Union College is 0.0012 m-f. When the wave length is considerably more than four times the height of the antenna the current distribution is fairly uni- form in the conductor, and, the circuit can be treated as consist- ing of "bunched" rather than distributed inductance and capacity when the following relation obtains. T = 27T/LC /. L = ~ 300 ELECTRICAL ENGINEERING thus .C -f (40) \ Numerical application: Union College station with an antenna having a capacity of 0.0012 m-f. sending out waves of 700 m. length. Assume R = 10 ohms. (By far the greater part of this is the ground and spark resistance.) Then 5 = 2 o 10 _!2_ 3 X _ 10 10 70,000 In the case of the simple antenna it has been shown that the 2 radiation resistance assuming X = 4/i and I avo > = - / is 40 ohms. Thus the radiation decrement is : 3 X 10 10 = 2 log 9 X 10 10 log In reality due to the proximity of the earth and other causes the wave length is nearer 4.8 than four times the antenna height, 2 and the average value of the current is nearer 0.7 and -. 7T Substituting these values we get : R = 34 ohms instead of 40 ohms and the radiation decrement for the simple antenna is : ,_*!!* 8 X16*-^ (41) 2 log ^9X10"' log" Abraham gives s 2.45 o - r* i ] g ~ General Conclusions. Since the power radiated from an antenna is: W = 1600 ^ 7 2 it is evident that at a given voltage as the capacity of the super- structure is increased the current and the wave length are in- creased. Since, however, the energy is proportional to the square ELECTROMAGNETIC RADIATION 301 of the current and the wave length is proportional to -\/C it follows that by adding capacity to the superstructure and there- fore increasing the wave length the radiated energy is increased. Therefore, if the capacity is made four times as great, the cur- rent 2 is 16 times as great and X 2 is only four times as great, and hence, the radiated energy for the same antenna height is in- creased four-fold. Unfortunately, however, there is hardly a practical way of increasing the top capacity without decreasing the effective height so that the gain is not as great as indicated and if the umbrella is carried to an extreme, the effective height may be so much decreased that the energy radiated may eventually begin to decrease. With a given construction of the antenna the wave length may be increased by the introduction of inductance. In this case the energy radiated is, however, reduced. It is noted that for a given current the radiated power is greater the higher the frequency. This does, however, not necessarily mean that the power received is greater, since as will be shown later the absorption of energy in space is much greater with short wave length than with long. At times it is necessary to send at two widely different fre- quencies. The natural wave length may be say 600 m. and it is desired to communicate at a wave length of 300 m. In that case a condenser may be connected in the series with the antenna. Since two condensers in series have a smaller capacity than each and thus the frequency is increased. The relation between the effective value of the antenna current and the maximum instantaneous value of the current and e.m.f. If the damping is not excessive the discharge current of a condenser of voltage E can be represented by the following equation : _^o i = EuCe 2L sin u = Ie~ at sin co where / = EuC and = g = |- R Q being the total resistance in the circuit which is assumed constant, not depending upon the current. 302 ELECTRICAL ENGINEERING The rate at which energy is being converted to heat and radiated is then: Ro P -* a t sin 2 coZ. The energy developed in one train of waves then is, R I 2 e~ 2at sin 2 utdt = approximately (43) If the antenna is charged and discharged N times per sec. then the power is W _ A/- L (AA\ 43/ If I c is the effective value of the antenna current as read by a hot-wire instrument, .:I = I e ^jy (45) and since / Substituting JRo and r 1 L = we get l2R C Je , (47) These equations connect the instantaneous max. values of the antenna current and e.m.f. with the effective current read by a hot-wire instrument. Numerical application : At the Union College station R = 10, 12 X = 700 m., C = -, N = 500. Using the small sending set. I = 2.5 amp. 3 X 10 10 70,000 0.43 X 10 6 ELECTROMAGNETIC RADIATION 303 I = 2.527T.4310 6 = 46 amp. E = 14,400 volts. Relation between E.m.fs. Frequencies and Coupling in In- ductively Connected Circuits. Let e\ in Fig. 154 be the voltage across the primary capacity, e 2 in Fig. 154 be the voltage across the secondary capacity. Then neglecting resistance we get: ^7 = but, dt " dt ~dt l ~df - = and (1) (2) FIG. 154. Substituting the current values of equation (2) and equation (1), and writing ei = Ei sin ut e z = EI sin (ut + a). Thus we get, (3) (4) 304 ELECTRICAL ENGINEERING From (3) 61 = F^rfc^ 2 (5) Substitute (5) in (4) and assume that C\Li = C 2 L 2 = CL, that is, assuming that the circuits when independent are tuned to the same wave lengths, then, (6) - 2LCco 2 - co 4 (L 2 C 2 - lf 2 CiC 2 ) - 1 k " CL(1 - where 7 M k = . (7) &nu\j \ j. t\j~ \ a. A/~ or, (8) where /o is the frequency of each circuit when alone. It is seen from these equations that two frequencies exist in the circuit and that they become nearer and nearer alike as the coefficient of coupling is decreased, that is the less the value of the mutual induction as compared with the self-induction. In the case of transformation by ordinary transformer where the mutual induction is almost perfect, only one frequency will appear, namely, / 2 the forced frequency which in that case is / 2 = / -7=- 1 other words the radiated frequency has only one V2 value and that value is 70 per cent, of that of each circuit when alone. Since /I _ \2 /2 Xi' It follows that two different wave lengths are transmitted and that fT^fc X 2 - ' '- ELECTROMAGNETIC RADIATION 305 Wave meters are used to show the wave length, and hence, if the wave length is known the coefficient of coupling can be deter- mined, it is: X 2 2 - Xa 1 (10) k = (11) A 2 -f AI* It is evident from the above that the current and voltage in two such circuits must be expressed as functions of two frequencies. Let 61 = AI cos coi 4~ Bia) 2 t e 2 = A 2 cos coit -f- B 2 cos ai 2 t These equations are justified since the resistance is negligible, and hence, no appreciable phase displacements exist between the two voltages. For t = 0, ei = E lt and e 2 = 0. /. E! = A, + B 1 | = A 2 + B 2 \ Consider then the two waves separately. We have from (5) (12) 1 - and where 1 A 2 B l (13) a. = ft = Thus from (12) 1 jLido? 2 2 Ai = E 1 - B 1 A 2 = - B 2 and A, = E l - Bi .'. A! = - aEi (14) a. 306 But and ELECTRICAL ENGINEERING E l jCl 1 A .'. A 2 a Thus From (13) from (6) - a wi- C02 2MC - a (16) Equation (16) shows the relation between the maximum voltage and the capacity when the circuits have negligible resistance. When damped oscillations are considered the equations become more involved. FIG. 155. Consider the simplest case when the primary of the exciting transformer, Fig. 155, is supplied with power from an alternator or other source of sustained oscillations. Due to the mutual induction between the primary and second- ELECTROMAGNETIC RADIATION 307 ary circuits an e.m.f. EI sin co^ will be impressed upon the secondary. The differential equation of the secondary circuit is thus : EI sin &it = iRz + L^rrr + ^2 at where e 2 is the voltage across the secondary condenser. But . ~ de 2 .-. E 1 sin Wl = C,R + C 2 L 2 2 + e 2 (1) The sine term can be eliminated by two successive differentia- tions and the result will be a well-known linear differential equation of the fourth order the solution of which is: e 2 = E f sm (wi$ + 2 2 - a) 2 + (2eoia 2 ) 2 When the secondary circuit has the same natural period as the primary impressed frequency the secondary current becomes a maximum. Thus for 0> 2 = COi I = I r since a 2 4 is very small compared with 4coi 2 a 2 2 . This is readily shown to be .*. The square of the effective value of the secondary currents is L 2 - C0 2 2 - 2 2 ) 2 or 2 2 is small compared with co 2 2 , thus I _ 2a 2 a>i T r = tW //I \/(l \ \ but and B 2 and " 2L 2 / 2 ' 2ZT 2 = 52j ELECTROMAGNETIC RADIATION 309 r* (-i-Y I 2 \ irfi / .'.(I -o; 2 ) 2 where * or _ - and If x is near unity then 1 x 2 = (1 + x) (1 x) = 2(1 x) and = 27r(l - x) rjL \ir 2 - r If the secondary current J is read by a hot wire instrument then since the effective values are proportional to the maximum values, or, er If the frequency of the secondary is so adjusted that J 2 = -- z then we get a formula which is used estensively in connection with wave meter measurements. Inductively Coupled Oscillating Circuits Having Considerable Damping. We have shown (equation 5) the following relation between the effective secondary current J r at resonance and the induced e.m.f. EQ when the primary is supplied from a source of sustained power 310 ELECTRICAL ENGINEERING The corresponding equation when both the primary and second- ary circuits are oscillating has been worked out by BJERKNES and others who found that as long as the decrements are small the following relation obtains: 16/ 3 L In this equation J max * is the maximum possible value of cur- rent read by a hot wire instrument in the antenna circuit. N is the number of condenser discharges per second. EQ is the maximum value of the e.m.f. induced in the antenna circuit. L 2 is the inductance of the antenna circuit in henrys, / is the frequency and di and d z the logarithmic decrements in the primary and antenna circuits per full period. 7i = E = <**MCiEi and E 2 _ _ = " 16/ 3 " Similarly, In the case of sustained primary power. The maximum instantaneous value of the antenna current is from (45) remembering that in these equations the decrements per full period is used. T 2 A f S T 2 T 2 " max. ^ J "2 A "max. f? ~w~ i\r ~A^ 52 The maximum instantaneous value of the antenna voltage is /2 (ID Numerical Examples. Union College small set. E l = 5000 volts 120 . Ci == JQ-TO farads farads N = 500 ELECTROMAGNETIC RADIATION 311 and or X = 700m.; /. / = 0.43 10 6 >! = 0.05 5 2 = 0.10 k = 0.10 .'. did 2 (d! + d 2 ) = ~- 6 i o v 1 20 1 5 . 2 = 50,000 0.43 10 6 ^f^ 25 10 6 0.01 ~~ = 20 10 75 T _ jr *J max. ~ Tt.tJ 4 \/ OH - 0.43 10 6 X 0.10 = 1375 % 7 2 = 37 amp. 10 10 278 X 37 = 11,400 volts. 2x0.43 10 6 12 BJERKNES has shown how with a slight modification equation (6) can be used to determine the decrement of the secondary circuit which may, for instance, be the antenna circuit by means of a third tuned circuit which is called a wave meter: This expression is: d + 5i = 27r (l - } (12) where 5 is the decrement of the circuit being tested and <5i is the decrement of the meter. The formula is limited as is the case of equation (6) to the condition that J r and J being the effective values of the current in the wave meter. It is also limited to the condition that both d and Si are small and that di is considerably smaller than 8 and that finally Xi and X 2 do not differ more than, say, 5 per cent. Referring to Fig. 156, W is the wave meter which is a calebrated closed circuit of known inductance, capacity and therefore of known natural period. The resistance is FlG - 156 - made as low as possible so that the decrement of the meter is small. The value of the current or the (current) 2 is frequently deter- mined by means of a low resistance heating element, actually a thermal couple, which supplies a direct current to a galva- nometer G. 312 ELECTRICAL ENGINEERING In that case the galvanometer deflection is obviously pro- portional to the square of the current value. The procedure is as follows. The meter is loosely coupled to the antenna and the capacity of the wave meter is varied until the largest galvanometer deflection G> is obtained and the corresponding wave length \o is read. Then the capacity is changed so that the deflection of the C 1 galvonometer is -~ when the meter reads shorter wave length. We have then from (12) 5 + d l = 27T (l - ^] To determine the decrement of the meter it is desirable to insert in the meter circuit such non-inductive resistance that at resonance, that is when the wave meter reads Xo, the galvanom- C* eter deflection is ~ The capacity is then varied until the galvanometer deflection C 1 -^ when the wave length is X 2 . We have then if 5 2 is the decrement due to the added resistance, It has been shown in equation (8) that the relation between the effective values of the resonance current with different decrements are related as follows: J/ 2 dd* (d l + da) In our case Jr' 2 Gr> 2 d\ = d = decrement of the antenna. d'z = di + 5 2 = decrement of the wave meter in second test. di = 8 = decrement of the antenna. <2 2 = 5i = decrement of the meter in the first test. ELECTROMAGNETIC RADIATION 313 2 = ( g * + *)(* + fr + frJ 5i(5 + 5 2 ) 1-^ Si + 5 2 Xo Si + 5 2 Si } Xi Xo - 8 = 2 I ( Xl ~ Xa ) ( x ~ x ^) * * \ \ i \ o\ AQ AO "T~ *2 ^Ai Numerical Example. X = 500 m. Xi = 485 m. \z = 475 m. .-. + .! = 2ir (l -^) = 0.189. - ^ ( J - 500) - ' 314 - 0.125. 27T 10 X 25 = 0. 500 5 d = 0.126. Conditions Affecting the Receiving Station. It has been shown that at some distance from the sending antenna the maximum value of the potential gradient in volts per centimeter near the equatorial plane is G== l20irh I (1) where / is the maximum value of the current at the sending antenna, the current being assumed the same at all points of the conductor. The dimensions are given in centimeters. A more general formula would be * ( . E 2 = j- (2) where E 2 is the maximum value of the voltage across the whole receiving antenna, a is a correction factor for the current dis- 2 tribution which is - for a simple antenna and unity for an antenna 7T in which the height constitutes only a fraction of a quarter wave, as is most frequently the case in actual practice. h\, hz, X and r may be given in any units as long as they are the same, hi and h 2 are the heights of the sending and receiv- 314 ELECTRICAL ENGINEERING ing antenna, X the wave length and r the distance between hi and h z . In order to be applicable to wireless transmission this formula needs to be elaborated in several respects. (a) The voltage is actually greater due to the concentration of energy as the waves sweep over the surface of the earth. (b) The voltage is smaller on account of the energy which strays away from the curvature even if the surface of the earth is assumed to be perfect of conductivity. (c) The voltage is reduced on account of the energy absorption of the earth current which effect is prominent near the sending conductor where the concentration of current is greatest. (d) The voltage is sometimes increased, but more often re- duced, due to reflection, absorption, etc., depending upon the condition of the atmosphere. FIG. 157. Conditions (c) and (d) have not been studied theoretically, but a considerable amount of data has been given from actual tests, notably by AUSTIN and FULLER. l The Effect of the Curvature of the Earth. Assume that the sending antenna is at A and the receiving antenna at B, Fig. 157. The distance between A and B is -=-. In the case of a plane z wave the receiving antenna for the same distance would then be at C where, A -C = \R. Thus in this latter case the energy would be spread over a AUSTIN, Bulletin, Bureau Standards, 1914. FULLER, Proc., A. I. E. E., April, 1915. ELECTROMAGNETIC RADIATION 315 circumference wnereas due to the curvature of the earth the circumference is only 2irR. There is, therefore, a concentra- tion of energy which can be represented by a coefficient k' = and since the intensity of the electric field is proportional to the Venergy, the concentration coefficient for the electric field at a distance r under the condition given above is R0. .'. Energy Let distance AC, Fig. 157, be equal to AB per unit length of circumference at C is E Energy per unit circumference at B is _E_ 2irR sin 6 2irR6 6 2irR sin sin B or = (3) FIG. 158. The effect of the straying of power on the potential gradient due to the curvature of the earth is included in the equation according to theoretical works done by SUMMERFIELD and ZEN- NECK by the introduction of a divergence factor. 0.0019r 316 ELECTRICAL ENGINEERING AUSTIN'S experiments indicate, however, that with continuous waves this coefficient is: 0.0915r and FULLER'S experiments show 0.0045r = (4) AUSTIN'S equation gives values which lie between ZENNECK'S and FULLER'S and has the advantage of being simpler than the other two. Thus the general formula for continuous waves becomes : , , #2 = kki -- (5) Note, however, that in equation (4) the dimensions are ex- pressed in kilometers. The maximum value of the antenna current in the case of Tjl sustained oscillations is evidently ^2 TT where R% is the total /l2 resistance of the antenna that is the radiation resistance, the effective resistance, ground resistance, and resistance of the receiving device. The equation 01 the current in the case of damped oscillations is slightly different. It has been shown that if an e.m.f., EQ, is impressed on a tuned circuit the following relations obtain: T 2 = ~ where EQ is the voltage induced, which in our case is E%. di and d-2, are the decrements in the two circuits. Thus di and d% are in this case the decrements of the sending and receiving circuits respectively. Equation (7) may be written: / di\ ,,* ^i + ^ But the decrement of the receiving antenna is 2-L 2 / ELECTROMAGNETIC RADIATION 317 where / M20 h = . , = 4/di wnere j i is in Tt 2 N m g*Ji' but /i 2 = Ji 2 ^ where Ji is the effective value of the sending antenna current. 4/fl2 2 (l+JW RS (l+J) \ 2/ \ 2/ and J2 = _ ^/^ (g) The effective value of the voltage across the receiving antenna TT ^ AsaJi (10) _ where 2 is the effective value of the receiving voltage and Ji is the effective current at the base of the sending antenna. It is evident from the above that the ratio between the effect- ive values of the received e.m.f. with sustained and with damped oscillations is: damped if the decrements of the sending and receiving antennas were the same then the ratio would be V2. Method of Determining Power Received. AUSTIN based his determinations on the fact that if in two circuits in parallel we know the power in one we can calculate the power in the other and the total power from the relations of the resistances Rys' in the circuits. The total power supplied is FT F( E + E \ E R + S - ~ E \R + S)~ RS The power of circuit R is P -^ Fr ~ R %l - R + S P - R + S " P r ~ RS S D I Cf or the total power = P r . 318 ELECTRICAL ENGINEERING The minimum power P r required for distinguishing between dots and dashes of resistance R is determined experimentally by ob- serving the current in the receiving antenna under conditions that can be conveniently controlled. Knowing P r and R and the resistance S which is shunted across the telephone receiver enables one to determine the total power received. In FULLER'S experiments at Honolulu this minimum power was found to be 3.2 X 10 10 watt, when dealing with sus- tained oscillations. APPENDIX I Partial Differentiation. The complete differential of a func- tion V of several independent variables r, / d\ z = r cos 6. x 2\K = - = sin cos v? 2 dx y = ^ = sm sin dy (x 2 + t/ 2 + z*)* r dr _ 2 _ z dz ~ (x 2 + y 2 + )W ~ r = I^A.( X 2 I ^2)^ /2 ^ r y J X dx x 2 + y* x* + y 2 + z* " ( x * Z Z COS COS ^ COS

] dy* = dy (sin sin ^ = = r [1 ~ si av d n a0 i ^=^cos0 , -sm0- = + -sin'0 d 2 6 cos 6 cos

= - = cos cos " cos 1 . 2 (cos cos

H cos sm r r sm = I -j sin ^ cos ^ cos 2 <^ + sin 9 cos cosV cos . "I r r Sin 2 <^> sm d 1 1 r = ^- - cos sm $? = = - -I 2 sin cos sin 2 cos i - COS

r 1 ( sin (p cos r 2 sin fa r sin 2 cos < d r sin dx sin <^g dr 1 sin

sin

sin 2 cos 2 + sin 2 sin 2

161> given by the diagonal mn as shown in Fig. 161. Products of Vectors. There are two kinds of vector products : I. The dot product which is denned as, a dot b = a b = ab cos (a, b) where a and b are the two vectors to be multiplied together, and a and b are the numerical values of the vectors. II. The cross product which is denned as: a cross b = a X b = e ab sin (a, b) . 327 328 ELECTRICAL ENGINEERING where e denotes that the product is a vector. It is the unit vector perpendicular to the plane formed by a and 6. The above names have been introduced by WILLABD GIBBS and they are used principally by American writers. The reader is familiar with the resolution of vectors into com- ponents which can be treated according to the laws of ordinary algebra. The great advantage of vector analysis is that it deals with vectors directly. It is found useful, however, to resolve vectors into their components and in such case a vector a is defined in terms of its magnitude along any direction, say x, times a unit vector i along x. For convenience rectangular coordinates are used and the unit vector along the z-axis is denoted by i, the unit vector along the y-axis is denoted by j and the unit vector along the z-axis byfc. Thus a = a x i -f- a v j -f- a x k a = A/a* 2 + a v 2 + a 2 2 a = a(i cos a -f- j cos /3 -f k cos 7) where a, and 7 are the direction cosines. Now it will be easily seen from the definition of the dot product that: i - i 1 i - j = j-j = l t-fc = k-k = 1 j -k = a - a = a 2 It is also clear that the condition of perpendicularity of two vectors is that their dot product shall be zero. The dot product is also called (by HAMILTON) the scalar product, because the product is a scalar. The cross product is called the vector product, because it is a vector. . a a X 6 gives a vector c, Fig. 162, whose mag- nitude is (ab) sin (a,6); its direction is along the normal to the plane of the vectors a and 6, and finally the sense of c is taken so that as one goes from a to 6 he follows a right-hand screw. In other words from a to b we follow the threads of a corkscrew whose direction of progress determines the sense of 6. This is, of course, the well- APPENDIX 329 known rule of MAXWELL for the relation between the direction of flux, the motion of a conductor, and the e.m.f. thereby generated. It is clear from the definition of a cross product that in Fig. 163 i X j = k = - j Xi j x k = i = - k X j k X i = j = -- i X k i X * = j X j = k X k = 0. FIG. 163. The cross product of two vectors can also be obtained in terms of the components and the unit vectors i, j and /b; only it is evident that care should be taken not to invert the order of factors, since a Xb = 6 X a. Exercise. Prove that if a x a v a 2 , b x b v b z are the rectangular components of a and b. a X b = (a v b z - aj) v ) i + (a f b x - ajb,) j + (a x b v - a v b x )k or in determinant form, * j a X b = a x a v b x by b z Exercise. Prove that the absolute value of a X b which is written a X b = (a) (b) sin (a, b) a X 6= b v 2 + b f 2 ) - (a x b x + a v b 330 ELECTRICAL ENGINEERING Now it will be noticed that in the last exercise, a 2 x + a 2 y + a z 2 is simply equal to a a. Thus: First term = a a Second term = b - b Third term = (a 6) 2 = (ab cos a) 2 where = < a, b so that a X & = V(a )(& &) - (ab cos a) 2 a X 6 = "a 2 & 2 - a 2 6 2 cos 2 a = a& \ 1 cos a = ab sin a The product of a X 6, must be the normal to the plane of the vectors a and b is seen as follows : Assume c to be the vector and find a c = a. (a X b) also b-c = b. (a X b) Multiplying these out in the ordinary way we find a c = 6 c = 0, i.e., ac cos (a, c) = be cos (bj c) = which is satisfied when c is normal to the plane ab. The above are intended to cover the very small part of vector analysis used in this book. For further information the reader should consult special treatises written on the subject. HEAVISIDES' " Electromagnetic Theory;" ABRAHAM and FOPPL'S "Theory of Electricity and Magnetism" can be recom- mended highly. An excellent short treatise on the subject is "Elements of Vector Analysis" by BURALLI-FORTI and R. MAREOLONGO, and a somewhat larger work is that of WILLARD GIBBS, edited by WILSON. Finally COFFIN'S "Vector Analysis" may be men- tioned among works of reference, it appears indeed as best suited for the introduction to vector analysis. INDEX Attenuation, 121 Austin, 316 Auxiliary equations, 46 B Ber and bei function, 273 Bjerknes, 313 Capacity between concentric con- ductors, 70 between parallel planes, 68 between transmission lines, 70 of antenna, 299 of a single wire, 230 of concentric cable, 96 of isolated spheres, 163 of two cylindrical conductors, 224 of two wires in multiple, 230 Charge distribution on an ellipsoid, 199 Circular symmetry, 188 Complete differential, 164 Complimentary function, 46, 86 Concentric cylinders, 215 spheres, 209 Condenser, capacity of, 68 characteristics of, 68 charged, 71 discharged, 72 energy supplied to, 71 Coulomb's law, 157 Coupling, effect on frequency, 303 Curl of a vector, 257 Current, equation of, 261 Curvature of earth, 314 Cylindrical bars, 150 conductor, current and flux distribution, 268 conductors, 218 D Differential equations, higher order, 44 operator, D, 45 Differentials and differences, 61 Direct-current generator, field cir- cuit, 59 Displacement current, 263, 264 Distortionless line, 142 Divergence of a vector, 185 theorem, 186 E Electric doublet, 284 field, energy of, 262 intensity, 290 Electromagnetic radiation, 278 Electromotive force, equation of, 260 F Field intensity, 158 Flat conductors, 152 . Flux and current distribution, 152 Forces between point charges and spheres, 175 Froelich's equation, 22 Fuller, 316 G Gauss' theorem, 160 Graph of function y =