UNIVERSITY OF CALIFORNIA AT LOS ANGELES GIFT OF Mrs. Arthur C.V.'oodv.^rc THE ELECTEIC CIRCUIT WORKS BY THE SAME AUTHOR Published by McGRAW-HILL BOOK COMPANY The Electric Circuit Svo, xvi+229 pages. Cloth $2.00 (Sectmtt Etlttwn, entirely rewritten and enlarged) The Magnetic Circuit Svo, xviii+283 pages. Cloth $2.00 Published by JOHN WILEY & SONS Experimental Electrical Engineering Vol. I. 8vo, xix+469 pages, 328 figures. Cloth net, $3.50 Vol. II. Svo, xiv+333 pages, 209 figures. Cloth net, $2.50 Engineering Applications of Higher Mathe- matics Part I. MACHINE DESIGN. Small Svo, xiv+69 pages. Cloth net, $0.75 Published by FERDINAND ENKE, STUTTGART Ueber Mehrphasige Stromsysteme bei Ungleichmassiger Belastung. Paper Mk. 2.40 THE ELECTRIC CIRCUIT BY V. KARAPETOFF SECOND EDITION REWRITTEN, ENLARGED AND ENTIRELY RESET SECOND IMPRESSION McGRAW-HILL BOOK COMPANY 239 WEST 39TH STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E.G. 1912 COPYRIGHT, 1912, BY THE McGRAW-HILL BOOK COMPANY COPYRIGHT, 19 10, BY V. KAKAPKTOFF Stanhope jprcss F. H.GILSON COMPANY BOSTON, U.S.A. Engineering Lbruy -TK 15-3 PREFACE TO THE FIRST EDITION THIS pamphlet, together with the companion pamphlet en- titled The Magnetic Circuit, is intended to give a student in electrical engineering the theoretical elements necessary for cal- culation of the performance of dynamo-electric machinery and of transmission lines. The advanced student must be taught to treat every electric machine as a particular combination of electric and magnetic circuits, and to base its performance upon the fundamental theoretical relations rather than upon a separate " theory " established for each kind of machinery, as is often done. The first chapter is devoted to a review of the direct-current circuit, the next four chapters treat of sine-wave alternating- current circuits, and the last two chapters give the fundamental properties of the electrostatic circuit. All the important results and methods are illustrated by numerical problems of which there are over one hundred in the text. The pamphlet is not intended for a beginner, but for a student who has had an elementary de- scriptive course in electrical engineering and some simple labora* tory experiments. The treatment is made as far as possible uniform, so that the student sees analogous relations in the direct-current circuit, in the alternating-current circuit, in the electrostatic circuit and finally in the magnetic circuit. All matter of purely historical or academic interest, not bearing directly upon the theory of electric machinery, has been left out. An ambitious student will find a more exhaustive treatment in the works mentioned at the end of the pamphlet. The electrostatic circuit is treated in accordance with the modern conception of elastic displacement of electricity in di- electrics. No use has been made of the action of electric charges at a distance, or of the electrostatic system of units. The volt- ampere-ohm system of units is used for electrostatic calculations, 488205 yi PREFACE TO THE FIRST EDITION in accordance with Professor Giorgi's ideas (see a paper by Pro- fessor Ascoli in Vol. I of the Transactions of the International Electrical Congress, St. Louis, 1904). Those familiar with Oliver Heaviside's writings will notice his influence upon the author, particularly in Arts. 22 and 23,* where an attempt is made at a rational electrostatic nomenclature. Many thanks are due to the author's friend and colleague, Mr. John F. H. Douglas, instructor in electrical engineering in Sibley College, who read the manuscript and the proofs, checked the answers to the problems and made many excellent suggestions for the text. CORNELL UNIVERSITY, ITHACA, N. Y. August, 1910. PREFACE TO THE SECOND EDITION THE first edition of this book was issued as a pamphlet of some 85 pages which the author used for two years in his classes to supplement some other texts. In its present edition, the book is made independent of these texts, so that its size had to be more than doubled. The book has been practically rewritten, and completely reset in type. All the cuts are new. The topics are treated somewhat more in detail, and a large number of practical problems are provided. The new topics added are : the resistance of conductors of variable cross-section, the electrical relations in polyphase systems, performance characteristics of the trans- mission line, transformer and induction motor and the permittance (electrostatic capacity) of transmission lines. In the treatment of alternating currents by means of complex quantities, particular attention is paid to the trigonometric form E (cos 8 + j sin 0) of the expression for a vector. In fact, the transmission line, the transformer, and the induction motor to some extent, are treated in this trigonometric form. The author trusts that the reader will find this somewhat novel treatment more convenient in numerical applications than the usual form e+je'. * Chapter 14 in the second edition. PREFACE TO THE SECOND EDITION vii Since the appearance of the first edition, the author has been encouraged by some of his colleagues in his treatment of the electrostatic circuit in the ampere-ohm system of units, a treat- ment which involves the use of permittances in farads and elastances in darafs. He has extended this treatment to the calcu- lation of capacity of cables and transmission lines. The students grasp this mode of presentation much more readily than the old- fashioned way, based upon the law of inverse squares and elec- tric charges acting at a distance. The purpose of the present treatment is to impress them with the idea of a continuous action in the medium itself and with the role of the dielectric. Mr. F. R. Keller of the electrical department of Columbia University has read and corrected the manuscript and the proofs of the second edition, and checked the answers to the new prob- lems. The author wishes to express sincere appreciation of his painstaking, faithful and competent work. The author is also indebted to Mr. John F. H. Douglas for critically reading the galley proof of the second edition. CORNELL UNIVERSITY, ITHACA, N. Y. May, 1912. CONTENTS PREFACE v SUGGESTIONS TO TEACHERS xi CHAPTER I. FUNDAMENTAL ELECTRICAL RELATIONS IN DIRECT-CUR- RENT CIRCUITS 1 The volt, the ampere, the ohm, and the mho. Temperature Co- efficient. Resistances and conductances in series and in parallel. Electric power. CHAPTER II. FUNDAMENTAL ELECTRICAL RELATIONS IN DIRECT-CUR- RENT CIRCUITS (Continued) 13 Resistivity and conductivity. Current density and voltage gradient. Kirchhoff's Laws. CHAPTER III. CONDUCTORS OF VARIABLE CROSS-SECTION 22 Current density and voltage gradient at a point. The radial flow of current. The resistance and conductance of irregular paths. The law of current refraction. CHAPTER IV. REPRESENTATION OF ALTERNATING CURRENTS AND VOLTAGES BY SINE-WAVES AND BY VECTORS 31 Sinusoidal voltages and currents. Representation of a sine-wave by a vector. Addition and subtraction of vectors. Non-sinusoidal currents and voltages. CHAPTER V. POWER IN ALTERNATING-CURRENT CIRCUITS 45 Power when current and voltage are in phase. The effective values of current and voltage. Some special methods for calculating the effective value of an irregular curve. Power when current and voltage are out of phase. CHAPTER VI. INDUCTANCE, REACTANCE, AND IMPEDANCE 60 Inductance as electromagnetic inertia. Reactance. Impedance. Influence of inductance with non-sinusoidal voltage. The extra or transient current in opening and closing a circuit. CHAPTER VII. SUSCEPTANCE AND ADMITTANCE 75 Concept of susceptance. Concept of admittance. Equivalent series and parallel combinations. Impedances in parallel and admit- tances in series. CHAPTER VIII. THE USE OF COMPLEX QUANTITIES 82 Addition and subtraction of projections of vectors. Rotation of vectors by ninety degrees. Impedance and admittance expressed as complex quantities or operators. CHAPTER IX. THE USE OF COMPLEX QUANTITIES (Continued) 91 Power and phase displacement expressed by projections of vectors. Vectors and operators in polar coordinates. Vectors and operators expressed as exponential functions. x CONTENTS PAGE CHAPTER X. POLYPHASE SYSTEMS 99 Two-phase system. Three-phase Y-connected system. Three- phase delta-connected system. CHAPTER XI. VOLTAGE REGULATION OF THE TRANSFORMER 108 Imperfections in a transformer replaced by equivalent resistances and reactances. The vector diagram of a transformer. Analytical determination of voltage regulation. Approximate solution. Ana- lytical determination of voltage regulation. Exact solution. CHAPTER XII. PERFORMANCE CHARACTERISTICS OF THE INDUCTION MOTOR 122 The equivalent electrical diagram of an induction motor. The analytical determination of performance. Approximate solution. Starting torque, pull-out torque, and maximum output. CHAPTER XIII. PERFORMANCE CHARACTERISTICS OF THE INDUCTION MOTOR (Continued) 133 The secondary resistances and reactances reduced to the primary circuit. The circle diagram. The analytical determination of per- formance. Exact solution. CHAPTER XIV. THE DIELECTRIC CIRCUIT 143 The electrostatic field. A hydraulic analogue to the dielectric circuit. The permittance and elastance of dielectric paths. Per- mittivity and elastivity of dielectrics. Dielectric flux density and electrostatic stress (voltage gradient). CHAPTER XV. THE DIELECTRIC CIRCUIT (Continued) 157 Energy in the electrostatic field. The permittance and elastance of irregular paths. The law of flux refraction. The dielectric strength of insulating materials. The electrostatic corona. Dielec- tric hysteresis and conductance. CHAPTER XVI. ELASTANCE AND PERMITTANCE OF SINGLE-PHASE CABLES AND TRANSMISSION LINES 171 The elastance of a single-core cable. The elastance of a single- phasc line. The influence of the ground upon the elastance of a single-phase line. The equations of the electrostatic lines of force and equipotential surfaces produced by a single-phase line. The elastanre between two large parallel circular cylinders. CHAPTER XVII. EQUIVALENT ELASTANCE AND CHARGING CURRENT IN THREE-PHASE LINES 193 Three-phase line with symmetrical spacing. Three-phase line with unsymmetrical spacing. CHAPTER XVIII. DIELECTRIC REACTANCE AND SUSCEPTANCE IN ALTERNATING-CURRENT CIRCUITS 203 Dielectric reactance and susceptance. Current and voltage resonance. Voltage regulation of a transmission line, taking its dis- tributed permittance into account. Approximate formula; for the voltage regulation of a transmission line, considering its permittance concentrated at one or more points. APPENDIX 215 BIBLIOGRAPHY ' SUGGESTIONS TO TEACHERS (1) THIS book is intended to be used as a text in a course which comprises lectures, recitations, computing periods and home work. Purely descriptive matter has been omitted or only suggested, in order to allow the teacher more freedom in his lectures and to permit him to establish his own point of view. Some parts of the book are more suitable for recitations, others for reference in the designing room, others again as a basis for dis- cussion in the lecture room, or for brief theses. (2) Different parts of the book are made as much as possible independent of one another, so that the teacher can schedule them as it suits him best. Moreover, most chapters are written according to the concentric method, so that it is not necessary to finish one chapter before starting on the next. One can cover the subject in an abridged manner, omitting the last parts of some chapters. (3) The problems given at the end of nearly every article are an integral part of the book, and should under no circumstances be omitted. There is no royal way of obtaining a clear under- standing of the underlying physical principles, and of acquiring an assurance in their practical application, except by the solution of numerical examples. (4) The book contains comparatively few sketches, in order to give the student an opportunity to illustrate the important re- lations by sketches of his own. Making sketches, diagrams and drawings of electric circuits and machines to scale should be one of the important features of the course, even though it may not be popular with some analytically inclined students. Mechanical drawing develops precision of judgment, and gives the student a knowledge that is tangible and concrete. (5) The author has avoided giving definite numerical data, coefficients and standards, except in problems, where they are in- dispensable and where no general significance is ascribed to such Xll SUGGESTIONS TO TEACHERS data. His reasons are: (a) Numerical coefficients obscure the general exposition. (6) Sufficient numerical coefficients and de- sign data will be found in good electrical hand-books and pocket- lxx)ks, one of which ought to be used in conjunction with this text, (c) The student is likely to ascribe too much authority to a numerical value given in a text-book, while in reality many coefficients vary within wide limits according to the conditions of a practical problem, and with the progress of the art. (d) Most numerical coefficients are obtained in practice by assuming that the phenomenon in question occurs according to a definite law, and by substituting the available experimental data into the cor- responding formula. This point of view is emphasized through- out the book, and gives the student the comforting feeling that he will be able to obtain the necessary numerical constants when confronted by a definite practical situation. (6) The treatment of the electrostatic circuit is made as much as possible analogous to that of the electrodynamic cir- cuit. The teacher will find it advisable to make his students perfectly familiar with the juse of Ohm's law for ordinary electric circuits before starting on the electrostatic circuit. The student should solve several numerical examples involving voltages and voltage gradients, currents and current densities, resistances, re- sistivities, conductances and conductivities. He will then find very little difficulty in mastering the electrostatic circuit, and from these two the transition to the magnetic circuit, treated in the companion book, is very simple indeed. The following table shows the analogous quantities in the three kinds of circuits. Electrodynamic i Voltage or c.m.f. j Voltage gradient (or electric intensity) Electrostatic Voltage or e.m.f. Voltage gradient (or electric intensity) Magnetic Magnetomotive force M.m.f. gradient (or mag- netic intensity) I Electric current i Current density Dielectric flux Dielectric flux density Magnetic flux Magnetic flux density t Resistor < Resistance ' Resistivity Elastor Elastance Elastivity Reluctor Reluctance Reluctivity i Conductor < Conductance ' Conductivity Permittor (condenser) Permittance (capacity) Permittivity (dielec- tric constant) Permeator Permeance Permeability LIST OF PRINCIPAL SYMBOLS. The following list comprises most of the symbols used in the text. Those not occurring here are explained where they appear. When, also, a symbol has a use different from that stated below, the correct meaning is given where the symbol occurs. Page where defined Symbol. Meaning. or first used. a Radius of conductor of transmission line 176 a Radius of core of cable 171 A Cross-section 13 b Inter axial distance between conductors of transmission line 176 6 Radius of inner surface of cable sheathing 171 b Susceptance 75 C Constant 54, 72 C Permittance or electrostatic capacity 147 D Dielectric flux density 154 D max Rupturing flux density 165 e Electromotive force 1 e Instantaneous value of voltage 34 e Horizontal component of vector of e.m.f 83 e' Vertical component of vector of e.m.f 83 ei Local source of e.m.f 3 e t Terminal voltage 3 E Effective value of alternating voltage 48 E Vector of the voltage E 83 E m Maximum value of voltage 34 / Frequency of alternating current or voltage 33 F Mechanical force 60 g Conductance 2 g eq Equivalent conductance 8 G Voltage gradient or electric intensity 16 G max Rupturing voltage gradient 165 h Elevation of conductor above the ground 181 h Head of fluid 10 h Instantaneous value of harmonic 54 i Current 1 i Horizontal component of vector of current 87 i Instantaneous value of current 33 i' Vertical component of vector of current 87 7 Effective value of alternating current 48 / Vector of the current / 88 xiv LIST OF PRINCIPAL SYMBOLS Page where defined Symbol. Meaning. or first used. IL Primary load current 1 16 I m Maximum value of current 34 / Mesh currents in squirrel-cage rotor 133 j V^T 83. 85 k b Breadth factor of winding 133 A" Relative permittivity 151 / Length 13 log Common logarithm 172 L Inductance - 60 Ln Natural logarithm 171 m Mass 60 m Number of phases 133 p Number of poles 135 P Constant 73 P Input per phase of induction motor 123 P Power 10 Pate Average power 48 q Instantaneous displacement of electricity 193 q Rate of discharge of a fluid 10 Q Constant 73 Q Quantity of electricity 144 Q Quantity of heat '. 2 r Resistance 1 r tq Equivalent resistance 7 R Resistance 8 Ro Resistance at C 5 Rt Resistance at 1 C 5 Slip of induction motor 123 dl (31) The total voltage e between the terminals of the conductor is equal to the sum of these infinitesimal drops, or f Jo G-dl (32) Equation (32) is expressed in words by saying that voltage is the line integral of electric intensity (or voltage gradient). A clear understanding of relations (31). and (32) is of para- mount importance in the study of electrostatic and magnetic phenomena. This will be aided by recalling to mind the thermal analogy previously used. In the case of the flow of heat, G corresponds to the rate of change in temperature per unit length of the rod, while e represents the total difference of temperature between the ends of the rod. Equation (31) expresses the fact that, by taking the rate at a certain point and multiplying it by a very short element of the length of the rod, the actual difference of temperature between the ends of this element is obtained. Thus, for instance, let the drop in temperature at some point of the rod be equal to 2.5 C. per meter length. Then the actual drop in a very short element, say 0.1 mm., is 2.5 X 0.0001 = 0.00025 C. The element of length must be small, because by supposition the cross-section of the rod is not constant, and the rate of drop is consequently variable. For a short length the vari- able quantities can be assumed constant, or, more correctly, aver- age values can be used. Equation (32) thus states that the total difference of temperature between the ends of the rod is equal to the sum (or the integral) of the drops in the very small elements. CHAP. Ill] CONDUCTORS OF VARIABLE CROSS-SECTION 25 Similarly, in a pipe of variable cross-section the rate of loss of head per unit length is variable, so that it is only possible to speak of this rate G at a point. The total loss of pressure, or head e, is obtained by summing up the small losses of head in infinitesimal elements of the pipe. The loss of pressure for a length dl isGdl; the total head e is the integral of this expression, over the whole length of the pipe. This is expressed mathemati- cally by eq. (32). Relation (25) between G and U holds true for a non-uniform flow as well, because it merely gives a relation between the cause G and the effect U at a point, depending only upon the property of the material, as expressed by the factor 7 or p. This relation may be also considered as Ohm's law for an infinitesimal cylindri- cal conductor of length dl and cross-section dA, namely, Gdl = de= ( P dl/dA}UdA. Canceling dA and dl, relation (25) is obtained. Prob. 1. A current of 50 amp. is flowing along a cylindrical con- ductor 3 cm. in diameter. The resistivity of the material varies in con- centric layers in such a way that the current density is proportional to the cube of the distance from the axis. What is the current density at the periphery? Ans. 17.7 amp. per sq. cm. Prob. 2. A conductor of circular cross-section, 225 cm. long, has the form of a truncated cone, the diameters of the two terminal cross-sections being 1.2 cm. and 3 cm. respectively. The total drop at a certain current is 65 volts. What is the general expression for the voltage gradient G x at a distance x from the smaller end? Ans. Gx/G = [a/(a + x)] 2 , where a = 150 cm. is the distance from the smaller end to the apex of the cone, and G = 0.723 volt per centimeter is the voltage gradient at the smaller end. G is determined from eq. (32), namely, 65 - G f a*dx/(a + z) 2 . Jo Prob. 3. A non-linear irregular conductor, made of homogeneous ma- terial, has a current density U and an electric intensity G, varying from point to point in magnitude and direction. What is the general expres- sion for the power converted into heat? Ans. According to eq. (25a), /\ f* r y J J where dv is the element of volume to which G and U refer, and the inte- gration is extended over the whole volume of the conductor. The volume dv must be taken as a cylinder or parallelepiped, the length of which is in the direction of flow of the current, the cross-section being perpendicular to this flow. 26 THE ELECTRIC CIRCUIT [ART. 9. The Radial Flow of Current. The solution of problems involving a non-uniform flow of current usually requires con- siderable facility in the use of higher mathematics beyond ordinary calculus. An exception to this statement is the simple case of radial flow (Fig. 8) be- tween two concentric electrodes, cyl- indrical or spherical. The following exercises give an opportunity for prac- tice in the solution of problems of this kind. They serve to illustrate the concepts of current density and voltage gradient, and to prepare the student's mind for the solution of certain prob- lems on concentric cables, involving the dielectric and magnetic circuits. FIG. 8. Flow of current be- tween two concentric elec- trodes. Prob. 1. Calculate the resistance of a cylindrical layer of mercury MM (Fig. 8) of height h = 5 cm., between two concentric cylindrical terminals 7\ and T 2 , the radii of the contact surfaces being a = 10 cm. and b = 18 cm. The resistivity of mercury is 95 microhms per cubic cen- timeter. Solution: Take an infinitesimal layer of the mercury, between the radii x and x + dx; the resistance of this layer is P dx/(2irxh). The resistances of all the infinitesimal concentric layers are in series; therefore, r is obtained by integrating the foregoing expression between the limits a and 6, the result being r = [ P /(2^)].Ln(6/a) (34) Ans. r = 1.775 microhms. Prob. 2. In the preceding problem, when a current of 10,000 amp. flows through the mercury, what is the amount of heat generated per second per cubic centimeter of mercury, at both electrodes? Solution: The current density at the inner electrode is 10,000/(2x X 10 X 5) = 31.8 amp. per square centimeter. According to eq. (25a), the loss of power is (31.8) 2 X 95 X 10~ = 0.0958 watt per cubic centimeter. The heat loss at the outer electrode is 0.096 X (10/18) 2 = 0.0295 watt per cubic centimeter. Prob. 3. What is the curve of electric intensity G as a function of x in the preceding two problems, and what are the limiting values of Gl Ans. An equilateral hyperbola; Gi = 3.02, (7 2 = 1.677 millivolts per centimeter length. Prob. 4. A lead-covered cable, consisting of a solid circular conductor of A square millimeters in cross-section, is insulated with a layer of rubber c mm. thick, between the conductor and the sheathing. What is CHAP. Ill] CONDUCTORS OF VARIABLE CROSS-SECTION 27 the insulation resistance of I kilometers of such a cable, if the resistivity of rubber is p megohms per centimeter cube? Ans. [ P X 10~*/(2 nt)] Ln (1 + 1-772 c/V.l) megohms, according to eq. (34). Prob. 6. Show that by doubling the thickness of the insulation in the preceding problem, the insulation resistance is increased less than twice. Prob. 6. A current is flowing through a hemispherical shell of metal along radial lines. Express the resistance of the shell as a function of its radii a and 6, and the conductivity y of the material. Ans. (&-o)/(2irya&). Prob. 7. Apply the method of superposition and the result obtained in Arts. 60 and 63, to the calculation of the resistance of an unlimited conducting medium between two parallel cylindrical terminals. Such a case obtains, for instance, when a load resistor consists of two vertical pipes in a pond, the pipes being used as the terminals, the current flowing through the water. 10. The Resistance and Conductance of Irregular Paths. Let a conductor of irregular shape (Fig. 7) be connected to a source of constant voltage e. The power converted into heat in the conductor is e 2 /r, where r is the resistance of the conductor. This resistance depends upon the distribution of the current in the body of the conductor. The general law, demonstrated by all experiments, is that the distribution of the current is such as to make the dissipated energy a maximum. Since by supposi- tion e is constant (unlimited supply), the resistance r must be a minimum. Let now the same conductor be connected to a source of constant current for instance, an arc-light machine. The dis- tribution of the current in the conductor is such as to effect its passage with a minimum expenditure of energy, that is, minimum voltage at the terminals, or minimum iV. This again means that the resistance r is a minimum. The student is advised to consider similar cases in the flow of heat or of a fluid, in order to make the matter perfectly clear to himself. The general law of nature that of minimum effort or mini- mum resistance applies in all such cases, and is used in the cal- culation of the resistance of conductors of irregular form. The conductor is divided into small parts by means of stream lines and equipotential surfaces as shown in Fig. 7, drawing them to the best of one's judgment. These small cells are nearly cylindrical in form, so that their resistances or conductances are easily esti- 28 THE ELECTRIC CIRCUIT [Anx. 11 mated by using their mean lengths and average cross-sections. The resistance of the whole conductor is found by properly com- bining the resistances of these cells in series, and the conductances of the filaments thus obtained in parallel. Then the assumed shapes of the stream lines and of the equipotential surfaces are somewhat modified, and the resistance is calculated again, and so on. Thus, by successive trials, the minimum resistance, or the maximum conductance, of the given conductor is found, and this is the true value of resistance or conductance, as the case may be. The lines corresponding to this minimum give the true distri- bution of currents and voltages within the conductor. The work of the trials is made more systematic by following a procedure suggested by Lord Rayleigh, and further developed by Dr. Lehmann. This method is described in detail in Art. 54 below, in application to the electrostatic field, and also in Art. 41 of the author's Magnetic Circuit, in application to the magnetic field. The student will have no difficulty in applying the method to an electro-conducting circuit. The best way to make it clear to one's self is actually to draw a conductor of irregular shape '(in two dimensions for the sake of simplicity) and to calculate its resistance in the above-mentioned manner. 1 11. The Law of Current Refraction. The method out- lined above for the mapping out of stream lines and equipoten- tial surfaces applies only in a homogeneous conductor. When a current passes from one substance to another (Fig. 9), the stream lines suddenly change their direction at the dividing surface AB between the media, and in so doing they obey the law of cur- rent refraction, which is tan 0i/tan 2 = 71/72 (35) Here 0i and 2 are the angles of incidence and refraction, while 71 and 72 are the respective conductivities of the two media. This equation shows that the lower the conductivity of a sub- stance, the more nearly do the stream lines approach the direc- 1 In two-dimensional problems of this kind, the properties of conjugate functions may be used when the geometric forms involved can be expressed by analytic equations. However, the purely mathematical difficulties are such as to make this method applicable only in a comparatively few simple X Maxwell, Electricity and Magnetism, Vol. 1, p. 284; J. J. Thomson, Recent Researches in Electricity and Magnetism, chap. 3; Horace Lamb, Hydrodynamics, chap. 4. CHAP. Ill] CONDUCTORS OF VARIABLE CROSS-SECTION 29 tion of the normal NiN* at the dividing surface. In this way, the path between two given points is shortened in the medium of lower, and is lengthened in that of higher conductivity, by such an amount in each case that the total conductance of the composite conductor is larger with refraction than without it. Hence, the existence of refraction is a necessary consequence of the general law of least resistance. Mediums, of low conductivity Medium 1, of high conductivity FIG. 9. The refraction of a current, or of a flux. To deduce eq. (35), consider a tube of current between the equipotential surfaces ab and cd, and let the width of the path in the direction perpendicular to the plane of the paper be one centimeter. Let C/i and U z be the current densities in the tube, and let G\ and Gz be the corresponding voltage gradients. Two conditions must be satisfied, namely, (1) the total current through cd is equal to that through ab, and (2) the voltage drop along ac is the same as that along bd. These conditions are expressed by the equations Ui-ab = Uz-cd and d . bd = (? 2 ac. Dividing the first equation by the second and rearranging the terms gives Ui/Oi _ Ut/Gt bd/ab ac/cd 30 THE ELECTRIC CIRCUIT [ART. 11 But. according to eq. (25),_l/j/6ri = 71, and Ut/Gt = 72- From Fig. 9, bd/ab = tan 0i, and ac/cd = tan 2 . By substituting these values in the preceding equation, relation (35) is obtained. Thus, in mapping out an electro-conducting circuit in two media, the stream lines must be so drawn as to satisfy eq. (35), and the conductance must be a maximum for the combination, and not for each part separately. A similar law applies to elec- trostatic and magnetic fluxes (see Art. 55 below, and Art. 41a of the author's Magnetic Circuit). Prob. 1. Make clear to yourself the reason why the refraction of light follows a sine law, while in the case of the electric current it is a law of tangents. Prob. 2. Show that total refraction is impossible for an electric cur- rent. Prob. 3. Draw a set of curves giving values of X for different values of O t when the ratio of conductivities is 1, 2, 10 and 100. CHAPTER IV REPRESENTATION OF ALTERNATING CURRENTS AND VOLTAGES BY SINE-WAVES AND BY VECTORS 12. Sinusoidal Voltages and Currents. A large proportion of the electric power used for lighting, industrial purposes, and traction is generated in the form of alternating currents. Some of the advantages of the alternating current over the direct cur- rent are: (1) Alternating-current power can be easily converted into power at a higher or at a lower voltage, thus making possible the transmission of power over long distances; (2) the genera- tion of alternating currents is simpler than that of direct currents, the latter requiring a commutator, 1 which needs constant atten- tion in operation; and (3), by combining two or three alternating- current circuits into a polyphase system it is possible to convert electric into mechanical power, using motors of simple and rugged construction (induction motors and synchronous motors). Alternating voltage waves generated by commercial alter- nators are more or less irregular in shape, but for most engineer- ing calculations it is accurate enough to assume them to vary with the time according to the sine law (Fig. 10). This assump- tion simplifies the theory and calculations greatly; moreover, the results obtained with this assumption are comparable with one another, because they all refer to a standard shape of the voltage and current curves, instead of a particular form in each specific problem. If the curve of a voltage or current differs greatly from the sine-wave, it can be resolved into a series of sine- waves of different frequencies, so that even then the sine-wave remains the fundamental form (see Art. 15 below). Fig. 10 shows the well-known construction of a sine-wave, the instan- taneous values of the current or voltage being represented as 1 The homopolar machine, which is a direct-current machine without a commutator, has not proven, up to the present time, to be commercially suc- cessful. 31 32 THE ELECTRIC CIRCUIT [ART. 12 ordinates, against time as abscissae. Instead of actual time in seconds, the curve is sometimes plotted against some other quan- tity proportional to time for instance, fractions of a complete cycle. It is sometimes convenient to use as abscissae the angu- lar positions of a field pole of the alternator with respect to an armature conductor in which the electromotive force under, con- sideration is induced. Fia. 10. An alternating current represented by a sine-wave. To construct the curve of an alternating current or voltage, draw a circle the radius of which equals the maximum value of the wave. Divide the circle into a certain number of equal or un- equal parts, such as ab, be, etc., and mark on the axis of abscissae points a', b', c', etc., corresponding to the points of division on the circle. That is, a'b' is either equal or proportional to ab; b'c' is either equal or proportional to be, and so on. In general, an abscissa such as a'c' represents, to a certain selected scale, the central angle u corresponding to the arc ac. The lengtlTa'ra' represents to the same scale an angle of 360 degrees, or the time of one complete cycle of the wave. The ordinates of the sine- wave are equal to the corresponding ordinates of the circle. For example, the point c'" on the curve is obtained by transferring the ordinate cc" of the circle to the corresponding abscissa a'c'. The name "sine-wave" is derived from the fact that these ordinates are proportional to the sines of the abscissae, which represent to some scale the central angles of the circle of refer- ence. The equation of the curve expresses this property ana- lytically. Let the maximum value of the current, which is also CHAP. IV] SINE-WAVES AND VECTORS 33 equal to the radius of the circle, be denoted by 7 TO ; we have then from the triangle Occ" ....... (36) where the ordinate i = cc" = c'c'" represents the instantaneous value of the alternating current, at the moment of time corre- sponding to the angle u. The variable angle u is proportional to the time, because the radius Oc which generates the sine-wave is assumed to revolve at a uniform speed. Let time t be counted from the position Oa of this radius, and let T =a'mf be the in- terval of time necessary to complete one revolution of the radius, or the time of one complete cycle of the alternating wave. When t = 0, u = 0; and when t = T, u = 2ir. Therefore, in general, u = 2irt/T, ....... (37) because this expression satisfies the foregoing conditions. Sub- stituting this value of u into eq. (36), we obtain i = I m sm(2irt/T) ....... (38) For the values of t = 0, T, T, f T, etc., i = 0, as one would expect, because at these moments the current changes from positive to negative values, or vice versa. At t = j T, f T, T, etc., we have i = 7 m ; at these moments the current reaches its positive and negative maxima. Equation (36) is used when the sine-wave is plotted against the values of angle as abscissae. Equation (38) gives the same curve referred to time as abscissae. 'in practice, the rapidity with which currents and voltages alternate is not denoted by the fraction of a second T during which a cycle is completed, but, in a more convenient manner, by the number of cycles per second. Thus, instead of saying that an alternator generates current which completes a cycle within ^ of a second, it is customary to say that the frequency of the current is 60 cycles per second. Denoting the frequency in cycles per second by /, we have / = 1/T, ........ (39) and consequently i = / m sin2ir/ ........ (40) This is the usual expression for an alternating current having a frequency of / periods per second. Analogously, for an alternat- ing voltage we have e = E m sm2irft, ....... (41) 34 THE ELECTRIC CIRCUIT [ART. 12 where E m is the maximum instantaneous value, also called the amplitude, and e is the instantaneous value of the voltage at the time t. In numerical calculations, and when drawing sine-waves, the values of the ordinates for various values of u or t are obtained either graphically, as in Fig. 10, or from a table of sines. For approximate calculations, values of sines can be taken from a slide-rule. In the problems which follow, the student is advised to become familiar with each of the three methods of obtaining values of sines. In some cases one has to deal with two currents or voltages of the same frequency for instance, in two different parts of the same circuit. The two corresponding sine- waves (Fig. 11) usu- FIG. 11. Two alternating currents displaced in phase by an angle . ally differ in amplitude, and also pass through zero at different instants. Thus, in Fig. 11, when the current 1 is at a maximum, the current 2 is still growing, and passes through its maximum somewhat later. In other words, the current 2 lags behind the current 1, or, what is the same, the current 1 leads the cur- rent 2. The angle between the zero points (or between the maxima) of the two waves is called the angle of phase differ- ence, or simply the phase angle. If we regard the two waves as being formed by the revolving radii I m ' and 7 TO ", then is the angle between the radii at any instant. When the two waves are of different frequencies, there is no constant phase angle between them; but this angle varies peri- odically, so that at some intervals of time the two waves are nearly in phase, and at others they are nearly in opposition. The familiar method of synchronizing two alternators by means of incandescent lamps is based on this phenomenon. CHAP. IV] SINE-WAVES AND VECTORS 35 Prob. 1. An alternating current fluctuates according to the sine law between the values of 75 amp., making 6000 alternations per minute - (3000 positive and as many negative ones). Draw a curve of instan- taneous values of this current; mark on the axis of abscissae the time t in thousandths of a second, the angles u in degrees, and the same angles in radians. Prob. 2. What is the frequency of the current in the preceding probr lem, in cycles per second? Ans. 50. "^ Prob. 3. Plot on the same curve sheet with the curve obtained in problem 1 the sine-wave of a current the frequency of which is three times as great, and the amplitude, 52 amp. The curve is to be at its maximum when the first curve is at a maximum. Prob. 4. Supplement the preceding curves by one, the frequency of which is 50 cycles per second, the amplitude 63 amp., and which reaches its positive maxima at the same instants in which the first curve passes through zero. Show that with these data two distinct curves can be ^ drawn. Prob. 6. Draw on the same curve-sheet with the preceding curves a sine-wave representing a 50-cycle alternating current, the amplitude of which is 120 amp., and which lags by 30 degrees with respect to the current in problem 1. Prob. 6. The current mentioned in problem 1 is generated by a 12-pole alternator, that in problem 3 by a 14-pole machine. At what speeds must these machines be driven in order to give the required frequencies? . Ans. 500 and 1285 r.p.m. Prob. 7. Express the currents given in problems 1 to 5 by equa- tions of the form of eq. (36). Ans. i = 75 sin u; i = 52 sin 3 u; i = 63 cos u; i = 120 sin (u 30). Prob. 8. The angle u in the answers to the preceding problem is ex- pressed in degrees; rewrite the equations so as to have u expressed in radians, and in fractions of a cycle. Also represent the currents as func- tions of the time t. Prob. 9. Express by equations similar to eq. (41) the following sinus- oidal voltages of frequency /: (a) Amplitude E m volts, (b) Amplitude E m ' volts, lagging degrees with respect to the first curve, (c) Ampli- tude E m " volts, leading the second curve by radians, (d) Amplitude E m '" volts, lagging one nth of a cycle with respect to the curve (a). Prob. 10. The voltages required in the preceding problem are induced /' by four identical alternators, having p poles each, and coupled together. By what geometrical angles must the revolving or the stationary parts be displaced in order to give the required differences in phase? 13. Representation of a Sine-wave by a Vector. It is clear from the foregoing theory and problems that all sine-wave cur- rents or voltages are different from one another in three respects only, namely: (1) in amplitude; (2) in frequency; and (3) in relative phase position. In most practical cases, all the currents 36 THE ELECTRIC CIRCUIT [ART. 13 and voltages entering into a problem are of the same frequency, so that they differ from each other solely in their amplitudes and phase positions. In such cases it is not necessary to draw sine- waves, or even to write their equations; it is sufficient to indicate the radii /,' and I m " which generate these curves (Fig. 11), in their true magnitudes and relative positions. The rotating radius, at any instant, gives by its vertical projection to scale the magnitude of the alternating current or voltage at that instant. The absolute position of the radii is immaterial, because they are revolving all the time. It is their relative position which is permanent, and which determines the relative position of the sine-waves. The moment from which time is counted is arbi- trary in most problems; hence, one of the radii can be drawn in any desired position. Then, all other radii in the same problem are determined by their phase displacement with respect to this "reference" radius. It must be clearly understood that the foregoing representa- tion by vectors is true only when all the vectors are revolving at the same speed, that is, only with alternating quantities of the same frequency. When currents and voltages of different fre- quencies enter into a problem, the angle between the vectors varies all the time, and it is necessary to introduce an arbitrary zero of time for reference. In general, the graphical method of solution is unsuitable for such problems. In mathematics and physics, a quantity which has not only a magnitude, but also a definite direction in space or in a plane, is called a vector. Thus, for instance, in mechanics, force is a vector quantity, while volume is not. The radii which repre- sent sine-waves have both magnitude and direction in a plane. It is proper, therefore, to call them vectors. While the direction of the first vector is usually arbitrary, once it is selected, the directions of all the other radii become definite, so that with this limitation, the radii in alternating-current problems have definite directions and may be called vectors. While they must be imagined as revolving when generating their respective sine- waves, yet they revolve as a system, maintaining their relative positions unchanged. The required relations always depend upon the relative positions of the radii, so that the fact that they are revolving can be altogether disregarded, and the radii consid- ered as simple stationary vectors. CHAP. IV] SINE-WAVES AND VECTORS 37 Prob. 1. Draw the vectors of the currents in problems 1, 4 and 5 of the preceding article in their true magnitudes and relative positions. Prob. 2. A single-phase alternator has a terminal voltage of which the maximum instantaneous value is equal to 16 kilovolts. The maxi- mum value of the current supplied by the machine is 325 amp. The character of the load is such that the current wave lags behind the voltage wave by an angle of 37 degrees. Assuming both the voltage and the current to vary according to the sine Jaw, represent the foregoing condi- tions by two vectors. y Prob. 3. Draw a vector diagram showing the phase (star) voltages and currents of a 25-cycle three-phase system (Fig. 36), the amplitude of each voltage being 7235 volts, and each displaced in phase by 120 degrees with respect to the other two voltages. The current in the first phase is 30 amp., and lags behind the corresponding phase voltage by of a cycle. The current in the second phase is 47 amp., and leads its voltage by 18 degrees. The current in the third phase is 72 amp., and lags behind the corresponding phase voltage by 0.004 of a second. Note: In the foregoing three problems the student is supposed to draw the vectors equal in length to the amplitudes of the alternating waves. In practice, it is customary to draw vectors equal in length to the effective values of voltages and currents, and not to their amplitudes. For sine-waves the effective value is equal to the amplitude divided by \/2 (see Chapter 5). The difference is not important for our present purposes. The use of effective values would merely change the arbi- trary scale to which the vectors are drawn. 14. Addition and Subtraction of Vectors. There are many practical problems in which alternating currents or voltages have to be added, or subtracted one from another. For instance, when two or more alternators are working in parallel, the total current delivered to the station bus-bars is equal to the sum of the currents supplied by each machine. Or, to find the voltage at the receiving end of a transmission line, the voltage drop in the line is subtracted from the generator voltage. When the com- ponent quantities vary according to the sine law and are all of one frequency, the resultant quantity is also a sine curve of the same frequency. This curve may be found (a) graphically, by adding the component curves point by point; (b) analytically, by adding their equations; or (c) by adding the vectors of these curves. It must first be proved that the sum of two sine- waves of one frequency is also a sine-wave of the same frequency. Let the two currents to be added be represented by the equations i = I m sin (u + i' = /' sin (u + ' " (42) 38 THE ELECTRIC CIRCUIT [ART. 14 where u = 2 irft is the variable time angle, and and 0' are two constant angles characterizing the relative phase positions of the two waves with respect to some reference wave I m " sin u. The phase displacement between i and i' is 4>' 0. Expanding the foregoing sines of the sum of two angles, and adding the two equations, member for member, we obtain i e g = i + i' = (I m cos + Im cos ') sin u + (I m sin + I m ' sin 0') cos M, . . . . (43) where the constant coefficients of sin u and cos u are grouped together. The subscript eq stands for "equivalent." This ex- pression is of the form i eq = A sin u + B cos u, where A and B are constants. No matter what values A and B may have, the right hand side of this equation is reducible to the form i eq = I eqm sm(u + eq ) (44) Assuming this equation to be true, we equate the right-hand sides of eqs. (43) and (44), and expand sin (w + eg ). Equating the coefficients of sin u and cos u, we get I egm cos ca = I m cos + // cos 0'; 1 l eqm sin eg I m sin + I m ' sin 0'. ) These are two simultaneous equations with I eqm and e9 as the unknown quantities. Squaring and adding these equations, we obtain / e7m 2 = (/ sin + I m ' sin 0') 2 + (/ cos + 7 TO ' cos 0') 2 - (46) Dividing the second equation by the first gives tan C , = (I m sin + // sin 0')/(/ cos + 7 m ' cos 0')- (47) No matter what values I m , I m ', and 0' may have, the values of I eqm and e , determined from these equations are real. In other words, it is always possible to represent eq. (43) in the form of eq. (44). This proves the proposition, because we see from eq. (44) that i eq is a sine-wave having the same u = 2 irft for the variable angle, hence the same frequency as the component waves. The amplitude and the phase position of this resultant wave are determined by eqs. (46) and (47) . When two currents or voltages are represented by vectors, their sum or difference is also a vector, because, as proved before, CHAP. IV.] SINE-WAVES AND VECTORS 39 it is also a sine-wave of the same frequency. The problem is to find the vector of the resultant wave, knowing the vectors of the component waves in their magnitudes and positions. Any ordinate of the resultant wave must be equal to the sum of the corresponding ordinates of the component waves. Hence, the vector of the resultant wave must satisfy the condition that its projection upon the F-axis (Fig. 12) shall be equal to the sum of the projections of the component vectors on the same axis. This condition must be fulfilled at all instants of time, that is, during the rotation of the three vectors. To satisfy this requirement the resultant vector must be the diagonal of a parallelogram of which the other two vectors are the adjacent sides. FIG. 12. Addition of vectors. Let OA and OB be the given vectors to be added together. From the end B of the vector OB draw a line BC equal and paral- lel to OA. Connecting and C gives the resultant vector OC, in magnitude and position. It will be seen from the figure that the projection of OC upon the F-axis is equal to the sum of the projections of OB and BC upon the same axis. But BC is equal and parallel to OA, so that the projection of OC on the vertical axis is equal to the sum of the projections of the given vectors on the same axis. This construction holds true for any instant whatever. By drawing AC, the parallelogram OBCA is com- pleted, so that the construction is identical with that for finding the resultant of two mechanical forces. However, in practical applications it is not necessary to complete the parallelogram, because the resultant is perfectly determined by the triangle OBC. The resultant of two vectors obtained in this way is called their geometric sum. 40 THE ELECTRIC CIRCUIT [ART. 14 If the triangle were not closed, the condition of equality with the sum of the projections of the given vectors might be satisfied for one particular instant of the cycle, but would not be satisfied for other instants. Thus, for instance, assuming the line OC' to be the resultant vector, we see that for the instant shown in the sketch the projection of OC' upon the axis OF is equal to the sum of the projections of OB and BC upon the same axis; but the condition is not fulfilled when the vectors rotate. The rule for subtraction of vectors follows immediately from the preceding rule, because to subtract a vector means to add a vector with the opposite sign. Thus, let it be required to subtract the vector OA from OB (Fig. 13); this may mean, for instance, the subtraction of the voltage wave represented by OA from that represented by OB. From the end B of OB draw vector BC equal and opposite to OA. The resultant, OC, represents the difference of the two given vectors, in direction and magnitude, and thus determines the sine-wave of the resultant voltage. If it were required to subtract. OB from OA, it would be necessary to draw AC' equal and opposite to OB, thus obtaining the resultant OC', equal and opposite to the former resultant OC. This is in accord with the general alge- FIG. 13. Subtraction braic rule that A - B = -(B -A}. The preceding results with regard to the addition and subtraction of vectors are summed up in the follow- ing rule : Relations which are true algebraically for instantaneous values of sinusoidal currents and voltages, hold true geometrically for the vectors of these quantities. It is customary to provide vectors of currents with triangular arrows, as in Fig. 12; vectors of voltages are usually distinguished by pointed arrows, as in Fig. 13. This distinction enables one to see directly from the diagram whether a vector represents a current or a voltage, with- out reference to the text. Prob. 1. The currents generated by two alternators in parallel are 75 and 120 amp. respectively, the second current lagging behind the first by 30 degrees. Determine the magnitude and the relative phase position CHAP. IV] SINE-WAVES AND VECTORS 41 of the resultant line current by three methods: (a) point by point; i/ (b) analytically; (c) by means of vectors. Ans. 188.8 amp., lagging by 18 32' behind the first current. Prob. 2. Solve the preceding problem without the use of eqs. (46) ^/ and (47), simply by means of the theorem proved above, that the sum or the difference of two sine- waves is also a sine- wave. Solution: leg sin (u + eq ) = 75 sin u + 120 sin (u - 30). This equation is true for any instant, or for any value of u. It contains two unknown quantities, the amplitude and the phase position of the resultant curve. It is necessary, therefore, to apply this equation to two particular moments of time, in order to obtain two equations with two unknown quantities. It is most convenient in this particular case to choose u = 7I-/2 and u = 0. Substituting these values, two equations with two unknown quantities are obtained. This method is preferable in the solution of practical problems, because it is not necessary to remember eqs. (46) and (47), and also because the two values of u can be selected so as to give the simplest equations. Prob. 3. Two alternators, with the same number of poles, are coupled together so as to give voltages differing in phase by 27 degrees, the voltage of the second machine leading that of the first. The first alternator generates a voltage the amplitude of which is 2300 volts, the second 1800 volts. The two machines are connected electrically in series. Find graphically the vector of the resultant voltage in its magnitude and phase position. Find also the vector of the resultant voltage when the termi- nals of one of the machines are reversed. Ans. (1) 3988 volts, lead- ing the first by 11 49'; (2) 1074 volts, lagging behind the first by 49 32'. Prob. 4. An alternator, the terminal voltage of which is 6600, supplies its load through a transmission line. The conditions are such that the current lags behind the generator voltage by an angle of 35 degrees. The voltage drop in the line is 540 volts, leading the current ia -phase by an angle of 67 degrees. Find the receiver voltage by subtracting the voltage drop in the line from the generator voltage (geometrically) ; also determine the phase displacement between the receiver voltage and the current. Ans. 6149 volts; 32 20'. 15. Non-sinusoidal Currents and Voltages. When a current or voltage wave differs considerably from the pure, sine form, it is often convenient to represent it as the result of a superposi- tion of sine- waves of different frequencies (Fig. 14). No mat- ter how complicated a periodic wave may be, it can always be so represented with sufficient accuracy, by properly selecting the amplitudes and the phase relations of the component sine-waves, o'r harmonics, as they are called. Theoretically, an infinite num- ber of sine-waves is necessary in order to represent any given irregular wave exactly. In practice, however, a limited number of harmonics is sufficient. 42 THE ELECTRIC CIRCUIT [ART. 15 CHAP. IV] SINE-WAVES AND VECTORS 43 If the frequency of the given irregular wave is/, the frequencies of the harmonics are /, 2 /, 3 /, and so on. When, however, the given wave is symmetrical, that is, when the part above the axis of abscissae is identical with that below, all even harmonics (2/, 4/, etc.) drop out, and the wave consists only of the fundamental wave of frequency /, and the odd harmonics (3/, 5/, etc.). The student can easily convince himself of the truth of this statement by taking a fundamental sine-wave and adding to it a second harmonic and a third harmonic. In the first case the resultant wave will be unsymmetrical; in the second, symmetrical. Nearly all of the waves encountered in practice are symmetrical. Let the fundamental wave be represented by the equation 7/1 = Ci sin (u ai), the third harmonic by the equation y$ = Cs sin 3 (u as), etc. The meaning of Ci, 3, etc., and of ai, a 3 , etc., is clear from Fig. 14; is an arbitrary origin from which the angles are measured. The ordinates of the given composite symmetrical wave are represented by the equation y = Ci sin (u ai) + C 3 sin 3 (u a 3 ) + C 5 sin 5 (u 5 ) + etc (48) This expression is known as the Fourier series, and is of great importance in mathematical physics. In practice, the problem which presents itself is usually that of analysis; that is to say, it is often required to analyze or resolve a given irregular wave into its harmonics. In other words, know- ing y, one is asked to determine the values of C and a for one or more harmonics. This is a purely mathematical problem, and is not treated here, because the solution will be found in numerous textbooks, handbooks and magazine articles. 1 There are also mechanical wave analyzers on the market, by means of which any desired harmonic may be separated by tracing the given curve with a stylus, in a manner similar to the way in which a planimeter is used. It is of importance for an electrical engineer to train his eye in the discernment of prominent harmonics, without mathematical analysis. This training is afforded by exercises in wave synthesis, that is, in combining various assumed harmonics into irregular waves. 1 See, for instance, the author's Experimental Electrical Engineering, Vol. 2, p. 222. 44 THE ELECTRIC CIRCUIT [ART. 15 Take first a fundamental sine-wave and a third harmonic of a reasonable magnitude, say between 15 and 30 per cent of the fundamental. Combine these waves into one, with different rela- tive phase positions of the fundamental and the third harmonic. In this way a flat wave, a peaked wave and a one-sided "humped" wave will be obtained. Then change the magnitude of the third harmonic and construct similar waves, in order to see the influ- ence of this factor. After that, plot similar curves for the funda- mental wave with a fifth harmonic, a seventh harmonic, and so on. Finally, combine the fundamental wave with the third and the fifth harmonics simultaneously, and so on. After some prac- tice, the eye will easily discern prominent harmonics in a given irregular wave. Numerous oscillograms of irregular waves will be found in many current periodicals and in the transactions of the various electrical societies. Read in this connection Art. 30 of the Magnetic Circuit. Prob. 1. Draw two or three sets of curves suggested in the preceding paragraph; each set must comprise about six curves and each curve must have a harmonic with a different phase position. Prob. 2. Devise a simple apparatus by means of which harmonics can be combined mechanically, and the resultant waves observed, with- out actually plotting curves point by point. Prob. 3. Analyze a given irregular wave into its harmonics, using the method given in the reference above, or any other method found in the literature on the subject. CHAPTER V POWER IN ALTERNATING-CURRENT CIRCUITS V 16. Power when Current and Voltage are in Phase. Let a resistance r be connected across the terminals of an alternator, the voltage at the terminals varying according to the sine law. The current through the resistance also varies according to the sine law, because Ohm's law holds true for any moment of time, so that the curve of the current is in phase with that of the voltage. If the equation of the voltage wave is e = E m sin u, the equation of the current is i = (E m /r) sin u. Graphically, the current and the voltage are represented by two vectors of different lengths, but in the same direction for instance, like OC and OD in Fig. 13. Divide the time T of one cycle into a large number of small intervals A. Then the amount of energy delivered to the resist- ance r and converted into joulean heat during one of such inter- vals varies with the time position of the interval in the cycle, in other words, with the instantaneous values of the voltage and the current. This energy is practically equal to zero when the current and the voltage have values near zero, and it reaches a maximum with them. However, the dissipated energy, being in the nature of a frictional loss, never becomes negative, because whether the current flows in one direction, or in the other, the heat liberated, e i AZ = i*r A, is always positive. Since the voltage and the current vary with the time, the rate of liberation of energy, or the instantaneous power, is also variable. The expression P = ei = i z r represents the instantaneous power as with direct current. If e and i remained constant for one second, the energy liberated would be equal to i 2 r. As a matter of fact, e and i may be considered constant only during the infini- tesimal element of time dt, so that the energy liberated during the time dt is i 2 r dt. Nevertheless, it is proper to say that at the in- stant under consideration the energy is liberated at a rate equal to i z r per second, because (i z r dt}/dt = i 2 r. This is analogous to the 45 46 THE ELECTRIC CIRCUIT [ART. 16 way in which we speak of the instantaneous speed of a body during a period of acceleration or retardation. The speed varies from instant to instant, so that to say that the speed is v at a certain instant merely means that, if the body continued to move at this velocity for one second, it would cover a space equal to v. In the same sense, the instantaneous power indicates the amount of energy which would be developed per second, if the current and the voltage suddenly became constant. The total energy liberated in the form of heat during one complete cycle is n T r*T W = I i z r-dt = r I i 2 -dt. Jo Jo (49) When no local e.m.fs. are present, the same energy is repre- sented by the expression W = f T ei-dt ....... (50) Jo When there are local e.m.fs. in the part of the circuit under consideration, the total energy communicated to it during an interval of time is different from that dissipated as heat (Art. 4). According to eq. (19), we have W = r ei i-dt = r f Jo Jo Suppose, for example, that e\ is the counter-e.m.f. of a motor in the circuit, and therefore nearly in phase opposition to e t . Then the i'V loss on the right-hand side of the equation is the difference between the energy supplied to the circuit and that converted into mechanical work in the motor. The foregoing equations are true whether the current and the voltage vary according to the sine law or not. If they are sinus- oidal, the integration can be easily performed, and the energy per cycle evaluated by the following method. Let the current be represented as before by i = I m sin u. Substituting this value into eq. (49), we have W = I m z r f T sm z udt ...... (52) This expression is easily integrated by using the substitution sin'u = (1 cos2u). Or it may be evaluated by observing that its value remains the same if a cosine is substituted for the CHAP. V] POWER IN ALTERNATING-CURRENT CIRCUITS 47 sine. This is because the limits of integration are u = and u = 2 TT, and in summing up sines or cosines through 2 TT we take the same quantities, only in a different order. Hence we may write W = In?r f T cos?udt. (52a) Jo Adding the two expressions term by term, and remembering that sin 2 w -j- cos 2 u = 1, we get 2W = I m *r C T dt = I m z r.T, Jo or the energy converted into heat during one cycle is W = lI m *r-T (53) When there are no local e.m.fs. and the current is in phase with the voltage, we have E m = I m r, so that from eq. (53), and by analogy to eq. (18), we have W = %I m E m -T (54) and W = %(E m */r).T (55) The student must clearly understand that the phase relation between the current and the voltage is of no consequence in eq. (53), while eqs. (54) and (55) hold true only when the current is in phase with the voltage. Or else, E m in these latter expressions may be said to refer to that component of the total terminal voltage which is used up in Ir drop. Prob. 1. A sine-wave alternating current, which fluctuates between 75 amp., flows through a resistance of 10 ohms. Plot curves of instan- taneous values of the voltage and power; the frequency is 50 cy./sec. Ans. E m = 750 volts; max. power = 56.25 kw. Prob. 2. Determine the total energy liberated per cycle in the pre- \ ceding problem, by integrating graphically the curve of power. Ans. 562.5 joules (watt-seconds). Prob. 3. Prove analytically that the curve of power obtained in problem 1 is a sine-wave of double frequency, tangent to the axis of time. Proof: The equation of the curve is P = 7 m V sin 2 u. But from trigonometry . cos 2 u = cos 2 u sin 2 u = 1 2 sin 2 M. Substituting the value of sin 2 u from this equation into the expression for P, we get P = I Infr - i Ir,?T COS 2 U. The first term is constant, while the second represents a sine-wave of 48 THE ELECTRIC CIRCUIT [ART. 17 double frequency, because 2 u = 2 IT (2 f)t. The first term is never smaller than the second, so that P is always positive, and the whole curve lies above the axis of abscissae. The second term becomes equal to the first and P = 0, only when 2 u is a multiple of 2 IT. At these points the curve is tangent to the axis of abscissa;. Prob. 4. Deduce eq. (53) directly from (52), expressing sin u in terms of the cosine of the double angle, as in the preceding problem. Hint: . From eq. (37), dt = (T/2 TT) du, and the limits of integration are u = and 17. The Effective Values of Current and Voltage. In prac- tice, it is the average rate of delivery or dissipation of energy that is of interest, or, in other words, the average value of the variable instantaneous power. This is analogous to using in calculations the average speed of a machine, when the actual speed varies within certain limits. This average power is found by dividing the total energy developed during one cycle by the period T of the cycle. When the current varies according to the sine law, the total energy per cycle converted into heat is expressed by eq. (53). Dividing both sides by T, we find that the average power Pave = \ In?r ........ (56) It is convenient to use in eq. (56) a new value of the current, 7 = 7 m /v / 2 = 0.707 I m , ..... (57) instead of 7 m , because then the expression for the average power becomes identical with that in a direct-current circuit, namely, Pa* e = I 2 r ........ (58) Analogously, if we define E = E m /V2 = 0.707 E m , ..... (59) eqs. (54) and (55) become P ave = E'I ....... (60) and ....... (61) which are perfectly similar to the corresponding expressions in a direct-current circuit. E and 7, as defined above, are called the effective values of the alternating voltage and current respectively. We may say that by definition the effective value of an alternating (or variable) current is equal to such a constant current which, when flowing CHAP. V] POWER IN ALTERNATING-CURRENT CIRCUITS 49 through a resistance, dissipates the same average power as the actual variable current. This definition of an effective value applies to variable cur- rents of any form. It is used, for instance, in determining the temperature rise of electric railway motors. During the run of a car the current fluctuates within wide limits, but the heating of the motor windings is nearly the same as would occur with a certain constant current, which is called the effective value of the actual variable current. The condition for the same average i 2 r loss is Pr-T = r C where T is the interval of time for which it is desired to obtain the effective value. Hence C Jo (62) This equation expresses in mathematical language that 7 2 is the average value of i 2 , over the period of time T. Taking the square root of both sides of this equation, we can also define the effective value / as the square root of the mean square of the instantaneous values. This definition is true for any form of alternating or variable current. The effective voltage is defined by a similar expression, so that more generally f Jo (63) where y denotes an instantaneous value of current or voltage. Alternating-current ammeters and voltmeters are always cali- brated so as to indicate the effective values of current and voltage. When an irregular wave of current or voltage is given graphi- cally, its effective value is found by taking a sufficient number of equidistant ordinates (Fig. 15) and replacing the integration in eq. (63) by a summation. Let the half-wave be divided into k equal parts, where k is an even number, and let yo, yi, . . . yk be the corresponding ordinates. Then, according to Simpson's Rule, (63a) 50 THE ELECTRIC CIRCUIT [ART. 17 The larger the number of ordinates, the more accurate is the value of y,ff determined by this method. The value of y e /f 2 may also be found by plotting a curve of y 2 against u, as shown in Fig. 15, and determining its mean ordinate by means of a planimeter. When a large number of effective values must be determined, for instance, from the records obtained by a graphic ammeter during several runs of an electric train, the squaring of ordi- nates becomes a tedious process. Some practical methods, by means of which the necessity for squaring ordinates is eliminated, are described in the next article. FIG. 15. The effective and the average ordinates of an irregular half-v The effective value of a current or a voltage is also called the quadratic mean, to distinguish it from the arithmetical mean value defined by the familiar equation -i or for a periodic curve (64) (65) It will be noted that the upper limit of integration is TT and not 2 TT. It is evident that for a symmetrical wave the average ordinate over a whole cycle is equal to zero. The average value, therefore, always refers to a half-wave. CHAP. V] POWER IN ALTERNATING-CURRENT CIRCUITS 51 W = T>).J&~ "* ~ $ For the sine-wave sin u du = (2/ir)y m , - or y ave /y m = 2/7T = 0.637 (66) The ratio of the effective to the mean ordinate is called the form factor, because it gives an idea of the degree to which the curve is flat or peaked as compared to the sine-wave. For a sine-wave the form factor is (7/ m /V2)/(2i/ m /7r) = l.ll (67) For a perfectly flat-topped or rectangular wave, the maximum value, the effective value and the average value are all the same, so that the form factor is equal to unity. For very peaked waves, the influence of the high middle ordinates is more prominent in the quadratic mean, so that the effective is considerably higher than the mean value, and the form factor is larger than 1.11. Another ratio which helps in judging about the shape of a curve is the so-called amplitude factor, or the ratio of the maximum ordinate to the effective value. The author is not aware that either the form factor or the amplitude factor is used to any con- siderable extent in practice. Prob. 1. An electric heater was tested for power consumption on an alternating-current circuit, by having an ammeter in series with it, and a voltmeter across its terminals. Both instruments were calibrated to indicate effective values. The readings were 110 volts and 5.7 amp. Assuming the current and the voltage to have been in phase, which is nearly the case, what was the average power consumption of the heater, and what was its resistance? Determine also the maximum instan- taneous values of the current and the voltage, under the supposition of the sine law. Ans. 627 watts; 19.3 ohms; 155.56 volts; 8.06 amp. FIG. 16. A stepped curve of current or voltage. Prob. 2. Determine the average value, the effective value, the form factor and the amplitude factor of the curve shown in Fig. 16. Ans. 0.75 y m ; 0.791 y m ; 1.055; 1.264. 52 THE ELECTRIC CIRCUIT [ART. 18 Prob. 3. Check sonic of the values of the form factor and the ampli- tude factor given in the table in the Standard Handbook (see Index under " form factor"). This will afford practice in calculating effective values of curves when they are given by analytic equations of the form y = f(t), using eq. (63). Prob. 4. Plot an irregular wave, taken from an available oscillograph record, and calculate its average and effective values by the point-by- point method, or by using a planimeter. v 18. Some Special Methods for Calculating the Effective Value of an Irregular Curve. As is mentioned in the preceding article, squaring a large number of ordinates in order to find the effective value of a curve is a tedious process, and methods are available which sometimes lead to the end more quickly. It must be admitted, however, that for one who has to do this work only occasionally, the plain point-by-point method described above is probably the quickest and the most reliable. FIG. 17. An irregular curve (Fig. 15) and the equivalent sine-wave, plotted in polar coordinates. (a) Fleming's Method. The given curve (Fig. 15) is replotted in polar coordinates (Fig. 17), so that equal polar angles O/fc cor- respond to equal distances ir/k upon the axis of absciss. The ratio between an abscissa u in Fig. 15 and the corresponding polar angle in Fig. 17 is of no consequence; in other words, it makes no difference what total central angle fl corresponds to the total distance TT in Fig. 15. The area of an infinitesimal triangle sub- tended by a polar angle du is \ i/ 2 do>, because y is the base of the CHAP. V] POWER IN ALTERNATING-CURRENT CIRCUITS 53 triangle, and y dco is its altitude. Thus, the total area of the curve in Fig. 17 is S-|/V, at right angles or in quadrature with the voltage, is called the reactive component. 1 Instead of resolving the vector of the current into two com- ponents, it is sometimes preferable to resolve the voltage E into 1 The older name for this reactive component is wattless current. CHAP. V] POWER IN ALTERNATING-CURRENT CIRCUITS 57 the components E cos < and E sin <, in phase and in quadrature with the current. In this case, eq. (77) is expressed in words by saying that the average power is equal to the current times the component of the voltage in phase with it. These components of the voltage are also called the energy component and the reactive component respectively. The two components of power, the true power El cos , and the reactive power El sin 0, stand in the same relation to the apparent power El as the two sides of a right triangle bear to the hypotenuse; that is, (El)* = (El cos <) 2 + (El sin ) 2 (78) Let now the current and voltage curves be different from pure sine-waves, and also different from each other in form. The fundamental equation Pa* e = (1/T) f T ei-dt (79) Jo holds true in all cases, so that if the curves are given graphically, the energy per cycle is found by multiplying the corresponding instantaneous values of e and i, and using the planimeter on the resultant curve. The average ordinate of this curve gives the average power. Of course, the parts of the resultant curve below the axis of abscissae must be evaluated separately from those above it, and the difference of the two taken to represent the total energy. If the two waves are given in the form of Fourier series, an expression for the average power may be obtained in terms of the effective values of the harmonics. Substituting the expan- sions for e and i into eq. (79), two kinds of terms are obtained, those containing products of two harmonics of the same frequency, and those containing products of two harmonics of different fre- quencies. The terms of the first kind, after integration, give re- sults of the same form as for the fundamental wave; that is, for the nth harmonic \ E n l n cos <, where E n and / are the amplitudes of the nth harmonics, and n is the phase displacement between them. The terms of the second kind give zero after integration, the proof of this being analogous to that in problem 3 of the pre- ceding article. Thus Pave = vEJi cos 0i + \EJ Z cos < 3 + etc. . . (80) In other words, each harmonic contributes its own share of power, as if it were acting alone. 58 THE ELECTRIC CIRCUIT [ART. 19 Let E and 7 be the effective values of some non-sinusoidal periodic voltage and current, measured, for instance, by means of hot-wire or dynamometer-type instruments. Let P ave be the average power according to eq. (80), or measured by a dyna- mometer-type wattmeter. Then the ratio P ave /EI is called the power-factor of the system, the same as with sinusoidal curves. This ratio is also often denoted by cos <, meaning by the phase angle between the equivalent sine-waves of voltage and current, as defined in the preceding article. With the use of this angle and of the equivalent sine-waves, vector diagrams may be constructed and the corresponding calculations performed with currents and voltages deviating considerably from pure sine-waves, though of course such calculations check only approximately with the actual measurements. Prob. 1. Assuming the line current in problem 4, Art. 14, to be 452 effective amperes, calculate the average power delivered by the alternator, and the power received at the opposite end of the line. Ans. 2444 kw.; 2350 kw. Prob. 2. Referring to problem 1, Art. 17, a wattmeter was connected into the heater circuit, and the true power was found to be 598 watts. Assuming all the three instruments to be in calibration, calculate the power-factor and the angle of displacement between the current and the voltage in the heater; also the energy component and the reactive com- ponent of the current. Ans. 95.4 per cent; 17 30'; 5.39 amp.; 171 amp. Prob. 3. Deduce expression (77) for power by direct integration. Solution: Let the current be expressed by /, sin u; also let the voltage he leading by an angle , and therefore expressed as E m sin (u + <#>). Substituting these values into eq. (79), we get C T l\ve = (E,,J m /T) I sin u sin (u + $) dt, Jo = (E m I m /2 *) / sin u [sin u cos + cos u sin ] du, = (E m I m /2 *) [cos tt> I sin 2 u du + sin < / sin u cos udu]. Jo Jo From a table of integrals we find that the value of the first integral is *-, and that of the second is zero. Substituting these values, and introduc- ing the effective values of voltage and current, formula (77) is obtained. Prob. 4. Plot a sine-wave representing an alternating voltage of 500 effective volts, and a current of the same frequency, of 20 effective amperes, lagging behind the voltage by 30 degrees. Plot on the same curve sheet the sine-wave of the instantaneous power, and check the CHAP. V] POWER IN ALTERNATING-CURRENT CIRCUITS 59 average ordinate of this curve with the value obtained by formula (77). Explain why the power is negative during a part of the cycle, remember- ing that there are local electromotive forces in the circuit. Prob. 5. Prove that the curve of power consists of a sine- wave of double frequency, plus a constant term, the latter representing the aver- age power. Compare with problem 3, Art. 16. Suggestion: ie = I m E m sin u sin (u + }. Use the trigonometric transformation, 2 sin A sin B = cos (A - B] - cos (A + B). Prob. 6. A non-sinusoidal voltage is represented by the equation e = 270 sin u + 62 sin 3(u + 15) + 16 sin 5(u - 25); the correspond- ing line current is i = 18 sin (u - 30) - 7 sin 3 (u + 50) + 2.5 sin 5(w + 10). Calculate the true average power and the power-factor of the system. Ans. P ave = \ (4860 cos 30 - 454 cos 75 - 40 cos 5) = 2027 watte; the power-factor is 75 per cent. CHAPTER VI INDUCTANCE, REACTANCE AND IMPEDANCE 20. Inductance as Electromagnetic Inertia. Experiment shows that an electric current in a variable state behaves as if it possessed inertia; there is an opposition to any change in its magnitude and direction. This opposition is manifested in the form of an "induced" electromotive force in such a direction as to tend to counteract the change in current. Thus, if an external e.m.f. tends to increase the current, the induced e.m.f. is in a direction opposite to that of the current; but when, on the other hand, the current for some reason decreases, the induced e.m.f. is in the same direction as the current, and therefore tends to strengthen it. These reactions of the current are similar to those exerted by a moving body; for instance, the water in a pipe, when its motion is accelerated or retarded. In practical applications, it is convenient to consider, not the reactions themselves, but the external forces necessary to overcome them. Thus, in the case of a moving body of mass m, the external force necessary to com- municate to it an acceleration dv/dt is F = m dv/dt. Here F is positive when the acceleration is positive, and vice versa. Similarly, to increase a current at a rate of di/dt, an external e.m.f. is necessary of the magnitude e = Ldi/dt, (81) where L is a constant which characterizes the circuit and is analo- gous to the mass m in the mechanical motion. The coefficient L is called the inductance of the circuit, and depends upon its shape and proportions, the presence or absence of iron, the number of turns which the conductor makes, and some other factors, which it is not necessary to discuss here. In most books the right-hand side of eq. (81) is written with the sign minus, because e is under- stood to mean the induced e.m.f. or the reaction of the circuit; while in our case e designates the external voltage, equal and 60 CHAP. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 61 opposite to this reaction. The form of the equation used here is preferable, because in practice one deals with components of the applied voltage rather than with the induced counter-e.m.f . ; moreover, the minus sign is apt to confuse a beginner. The inertia effect of the electric current is brought about through the mechanism of the magnetic field produced thereby. When the current varies, the flux embraced by the electric circuit also changes, and according to Faraday's law of induction this flux induces in the circuit an e.m.f. Thus, postulating the exist- ence of electromagnetic inertia, and stating the law of induced e.m.f., are perhaps but two different ways of expressing the same physical phenomenon, the true nature of which is at present un- known. Any arrangement of the circuit which increases the flux linked with it, also increases its inductance L or the inertia effect. The inductance of a given electric circuit can be calculated with more or less accuracy, 1 or it can be measured experimentally, using eq. (81). For our present purposes we shall assume L to be a constant quantity, which characterizes the inertia of a given electric circuit, according to eq. (81), without any reference to the nature of the magnetic flux which produces it. Mechanical inertia is used in physics and in engineering as a fundamental entity, without explaining it in any other terms, while the mystery as to its cause is just as deep as that surrounding the electro- magnetic inertia. Some modern physicists even believe that all inertia is of an electromagnetic nature. The fact that a body resists acceleration, together with the law of conservation of energy, leads to the conclusion that a mov- ing body possesses a certain amount of stored energy. The external work done upon a body while it moves through a dis- tance ds is F ds = m (dv/dt) ds, or, since ds = v dt, we have F ds = mv dv. The total work done upon the body while accel- erating it from rest to a velocity v is therefore W = C S Fds = r mv dv = \ mv 2 . According to the law of conservation of energy, this work is stored in the moving body as its kinetic energy. " The electrical work done in increasing a current against the induced electromotive force, during the time dt is d W = ei dt, or 1 See the author's Magnetic Circuit, Chapters 10 to 12, 62 THE ELECTRIC CIRCUIT [ART. 20 substituting for e its value from eq. (81), dW= Lidi. The total energy supplied to the circuit from the external source of power, while the current increases from zero to a certain value i, is W = r Lidi = $Li* (82) Jo This does not include the energy required for supplying the i-r loss. According to the law of conservation of energy, expression (82) represents the energy stored in the circuit as long as the value of the current remains the same. When the circuit is broken, this energy is converted into heat. Analogously, when a non- elastic moving body is stopped, its accumulated energy is con- verted into the heat of impact. Inductance can be defined from either eq. (81) or (82); and for most purposes the two definitions are identical. Similarly, in mechanics, mass may be defined either as the ratio of F to dv/dt, or as a ratio of the kinetic energy to \v z . The unit of inductance in the ampere-ohm system is called the henry. According to eq. (81), a circuit has an inductance of one henry when one volt is necessary in order to increase the current at a rate of one ampere per second. This one volt does not include, of course, the e.m.f. necessary for overcoming the resistance of the circuit. The henry being rather a large unit, inductance is frequently measured in millihenrys. Substituting into eq. (81) the physical dimensions of the voltage in the ampere- ohm system, we get [IR] = [LI/T] or [L] = [RT]. In other words, the henry stands for the "ohm-second." For this reason, one instrument for measuring inductance has been called by its inventors "the secohmmeter. " All actual circuits which possess inductance, at the same time, have some resistance, however small it may be. Therefore, the total instantaneous voltage applied during a variable state is e = ir + L di/dt (83) Ohmic resistance may be compared to mechanical friction, so that eq. (83) can be interpreted by reference to the mechan- ical analogy used above, in the following way; namely, the force necessary to accelerate a body must be augmented in practice by the amount required for overcoming the inevitable friction. CHAP. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 63 Prob. 1. A circuit which possesses an inductance of 12 millihenrys and carries a direct current of 150 amp. is broken within one-fifth of a second. What is the average voltage induced in the circuit during this interval of time? Ans. 9 volts. Prob. 2. Calculate the electromagnetic energy stored in the circuit of the preceding problem while the current is steady. Ans. 135 joules. Prob. 3. The current in a coil is made to vary at a uniform rate of 250 amp. per second. At the instant when the current is equal to 150 amp., a voltmeter connected across the terminals of the coil reads 295 volts; when the instantaneous current is 100 amp. the voltmeter reading is 230 volts. From these data calculate the resistance and the induc- tance of the coil. Ans. 1.3 ohms; 0.4 henry. 21. Reactance. It is natural to expect the inductance to exert a considerable influence upon the voltage and current rela- tions in an alternating-current circuit, because the Current is varying in magnitude all the time. The influence of inductance in this case is analogous to that of the inertia of the moving parts in a reciprocating engine; i.e., energy is stored during the periods of increase in velocity (or in current), and is returned to the source of power during the intervals of time when the velocity (or the current) decreases. There is no net gain or loss of energy for a complete cycle, although the instantaneous values of current and voltage may be considerably affected. Consider first a part of a circuit which has inductance only, the resistance being negligible. Let the current vary according to the familiar law i = I m sin (2 irft a). Substituting this value into eq. (81), we get e = 27r/L7 m cos(27r/Y-), .... (84) which means that the voltage necessary to force a sinusoidal current through an inductance also varies according to the sine law, and is in leading quadrature with the current. The ampli- tude of the voltage E m = 2 -irfLI m , or the relation between the effective values of voltage and current, is E = 2irfLI (85) It will be seen from this relation that, in alternating-current cal- culations, the quantities / and L always appear as a product. It is therefore convenient to introduce, for the sake of abbrevia- tion, a new composite quantity x, defined by the relation . . (86) 64 THE ELECTRIC CIRCUIT [ART. 21 The quantity x is called the reactance of the circuit, and always refers to a stated frequency /. Equation (85) becomes then E = xl, (87) from which it follows that reactance is measured in ohms, like resistance. This does not mean, however, that the two quanti- ties are similar in their physical nature. r Let now some resistance be con- . nected in series with the inductance E ||l (Fig. 18), or let the coil which pos- sesses inductance have also an appre- ciable resistance. Substituting the expression for i, given above, into eq. (83), we get e = rl m sin (2 irft - a) + xl m cos (2 */-).. . (88) Since the sum of two sine-waves is also a sine-wave (see Art. 14), the total voltage e varies according to the sine-law. These com- ponent sine-waves of voltage, their sum, and the current wave are shown in Fig. 19. The student is advised to study this figure FIG. 18. Resistance and re- actance in series. FIG. 19. The instantaneous current and voltage relations in the circuit shown in Fig. 18. very carefully, because it represents one of the most important fundamental relations in the whole theory of alternating currents. The same relations are represented vectorially in Fig. 20, and the two figures may be conveniently examined together. One de- scribes the phenomenon from instant to instant; the other gives CHAP. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 65 the salient features in a symbolic form. The vector E consists of one component Ir in phase with the current, and another Ix in leading quadrature with the current. The first component ir i FIG. 20. The current and voltage relations in the circuit shown in Fig. 18, represented vectorially. serves to overcome the ohmic resistance; the second, the reac- tance of the circuit. From the triangle of voltages we have (89) For the phase displacement between the current and the voltage we have tan< = Ix/Ir = x/r, ...... (90) or the power-factor cos = r/Vr 2 + z 2 ....... (91) The hydraulic analogue shown in Fig. 21 may make these relations clearer. ACDGA represents a closed pipe circuit in which water is made to oscillate to and fro by means of the piston B. The water is assumed to be devoid of inertia, and the inertia of the whole circuit is concen- trated in a heavy mass F, which moves freely with the water. The force upon the piston rod H is analogous to the alternat- ing voltage E in Fig. 18; the velocity of the water is analo- FIG. 21. A hydraulic analogue to Fig. 18. gous to the alternating current, the friction in the pipes represents the ohmic resistance r, and the inertia of the heavy mass F stands for the inductance L. To make the analogy closer, we assume that the piston is forced to perform a simple harmonic motion; 66 THE ELECTRIC CIRCUIT [ART. 21 so that the velocity of the water varies with the time according to the sine law, and may be represented by the curve for i in Fig. 19. The force upon the piston B consists of two parts, that re- quired for overcoming the friction in the pipes, and that necessary for accelerating and retarding the mass F. These two compo- nents of the force can be represented by the curves for ir and ix in Fig. 19. The frictional reaction is at a maximum when the piston is in the middle of its stroke, because there the velocity of the water is the greatest. On the other hand, the acceleration is zero in this position, so that the mass F exerts no reaction. At the ends of the stroke the acceleration or retardation is at a maximum, so that the force necessary for constraining the mass F to the prescribed motion is at a maximum; however, the fric- tional resistance is equal to zero. Adding the two sinusoidal com- ponents, we find the resultant force upon B, corresponding to the curve e in Fig. 19. It will be seen that e reaches a maximum before the center of the stroke; this gives a phase angle between the force and the velocity that is analogous to the phase angle between the voltage and the current. The student can easily deduce that the force leads the velocity in phase, and that the displacement is greater the larger the mass F, as compared to the frictional resistance; in other words, the greater the reactance as compared to the resistance. It may be shown also that the inertia reaction of the same mass F is greater for a higher fre- quency of oscillation, because the acceleration and retardation are proportionately larger. Prob. 1. The inductance of a coil is 0.2 henry; its ohmic resistance is negligible. Draw a curve giving the voltage necessary to maintain a current of 12 amp. through the coil, at frequencies ranging from zero to 100 cycles per second. Ans. A straight line through the origin; at / = 100, E = 1508 volts. Prob. 2. A reactive coil without iron draws a current of 75 amp. when connected across a 110- volt 25-cycle circuit. What current would it draw at 60 cycles and at the same voltage, provided that the effect of its resistance can be neglected? Plot a curve of current at intermediate frequencies. Ans. 31.25 amp.; equilateral hyperbola asymptotic to both axes. Prob. 3. The reactive magnetizing current of a 2200-volt, 600-kilo- volt-ampere, 50-cycle transformer must be not over 2.5 per cent of the full-load current. What is the lower limit of its no-load reactance and inductance? Ans. 322.5 ohms; 1.027 henrys. CHAP. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 67 Prob. 4. The coil considered in problem 1 is connected in series with a 100-ohm resistance; it is required to maintain a current of 12 amp. through the two, at various frequencies. Supplement the curve obtained in that problem with curves of voltage drop across the resistance, and the total voltage across the combination. Plot also the corresponding values- of power-factor. Determine the ordinates of the curves graphically, and check a few points analytically. Ans. E r = 1200 volts, independent of the frequency. At / = 0, E M ai =E r , and cos 0=1. At /= 100, E tota i = 1927 volts, cos = 62.25 per cent. Prob. 5. Three simultaneous instrument readings in a power house are: 7520 kw.; 66 kv.; 147 amp. The power-factor meter shows that the current is lagging behind the voltage. What are the readings at the same instant at the receiving end of the line, if its resistance is 45 ohms and its reactance 83 ohms. Hint: Draw the vectors of the generator voltage and current in their true relative position. Subtract the ohmic drop in phase with the current, and the reactive drop in quadrature with it. The result will give the receiver voltage in its true magnitude and phase position. Ans. 6547 kw.; 53.4 kv. Prob. 6. In order to determine the power input into a single-phase 110- volt motor, without the use of a wattmeter, the motor is connected in series with a non-inductive resistance across a 220-volt circuit. The resistance is adjusted so that the voltage across the motor terminals is 110, when the motor is carrying the required load. Under these condi- tions the voltage across the resistance is found to be 127, and the current through the motor 23 amp. From these data determine graphically the power-factor of the motor, and calculate its power input. Ans. 72.3 per cent; 1826 watts. Prob. 7. Referring to the preceding problem, calculate cos trigo- nometrically, from the triangle of voltages, instead of determining it graphically. 22. Impedance. When a reactance is connected in series with a resistance, eqs. (89) and (91) indicate that the current and voltage relations are determined, not by the value of the reac- tance alone, but by a composite expression z = Vr* + x\ (92) The quantity z has the dimension of a resistance, and is called the impedance of the circuit. It can hardly be called a physical quantity, but rather an abbreviation for a certain combination of the physical properties of a circuit; in other words, an abbre- viation for the radical in eq. (92). Introducing the value of z into eqs. (89) and (91), we obtain E = zl (93) and cos = r/z. . (94) THE ELECTRIC CIRCUIT [ART. 22 Impedance may be defined from eq. (93) as the ratio of the voltage to the current in a circuit containing resistance and reactance. In a non-inductive circuit the impedance is simply equal to the total resistance, while in a purely inductive one the impedance is equal to the reactance. It must be clearly understood that eq. (93) gives only the relation between the magnitudes of the vectors. The phase relation is given by Fig. 20, or by eq. (94). The three quantities r, x, and z form a triangle of which z is the hypothenuse (Fig. 20). This triangle is similar to the tri- angle of voltages, but the quantities r, x, and z are not vectors in the same sense as currents and voltages are. From the impe- dance triangle we have the following useful relations: r = z cos (95) and x = z sin (96) When two impedances are connected in series (Fig. 22), the voltage and current relations are as represented in Fig. 23. The total terminal voltage E is less than the arithmetical sum of the voltages EI and E 2 across the two impedances, and is equal to their geometric sum. FIG. 22. Two impedances in series. The resultant phase angle has a value intermediate between the phase angles i and < 2 of the two component impedances. It will be seen from the triangle ABC that the resultant voltage Fia. 23. The current and voltage relations in the circuit shown in Fig. 22. is the same as that required by an impedance which consists of a resistance TI + r 2 and a reactance Xi + z 2 . In other words, CHAP. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 69 the resultant impedance is z = (n + r 2 ) 2 + ( Xl + z 2 ) 2 , .... (97) and the resultant phase angle is determined from the equation tan = (si + z 2 )/(ri + r 2 ) ..... (98) These equations show that two impedances are added in series by adding the resistances and the reactances separately. An impedance of 5 ohms in series with one of 7 ohms is not equal to an impedance of 12 ohms, but as a rule is less. The relations shown in Figs. 22 and 23, and eqs. (97) and (98), are easily ex- tended to any number of impedances in series. Dividing all the voltage vectors in Fig. 23 by the value of the current 7, the dia- gram of voltages is converted into one of impedances, as in Fig. 20, the relations being represented by eqs. (97) and (98). It must be borne in mind, however, that from a physical point of view the latter relations are not vectorial in the same sense as are those of the voltages. Prob. 1. The impedance of a coil is 7.5 ohms at 60 cycles; the re- sistance measured with direct current is 6 ohms. What is the inductance? Ans. 11.9 millihenrys. Prob. 2. Two impedance coils are connected in series across a 292- volt line. The voltages across the coils are 152 and 175 respectively; the current is 7.3 amp. Knowing that the resistance of the first coil is 10 ohms, determine graphically the resistance of the second; also the impedances of both coils. Ans. r z = 23.8; z\ = 20.82; z 2 = 23.97, all in ohms. Prob. 3. When a certain non-inductive resistance is connected across a source of alternating voltage, a current / flows through it. When an inductance, containing negligible resistance, is connected across the same source of voltage, the current is /'. What are the current and the phase displacement when the resistance and the inductance are connected in series across the same source? Solution: Let the unknown voltage be E. The unknown resistance is r = E/I; the unknown reactance x E/I'. When the two are connected in series, the impedance z = [(E/I) 2 + (E/I'Y\*. Consequently, the current is E/z = 7/'/(/ 2 + / /2 )* ; tan = x/r = I /I'. 23. Influence of Inductance with Non-sinusoidal Voltage. (a) Let an alternating voltage e of an irregular form, such as is shown in Fig. 14, be applied at the terminals of a pure resistance r (non-inductive). The current through the resistance is at any instant equal to e/r, and consequently has the same wave form as the voltage. 70 THE ELECTRIC CIRCUIT [Aux. 23 (I)) Let now the same voltage be applied at the terminals of a pure inductance L (without resistance). It may be said a priori that the current wave will be different from that of the voltage, and will approach more nearly a sine-wave. This follows from the very concept of inductance as the inertia of the circuit; the high- frequency harmonics in the voltage are unable to produce currents of the same magnitude as at lower frequencies, because the react- ance offered to each harmonic is proportional to its frequency. This property of an inductance of choking higher harmonics is useful in some applications. Let the voltage across an inductance be given in the form of a Fourier scries, e = E t sin (2 irft - i) + #3 sin 3 (2 irft - 3 ) + etc. Substituting its value in the fundamental eq. (81), we get E l sin (2 Trft - + E 3 sin 3 (2 Trft - 3 ) + etc. = L di/dt. Multiplying both sides of this equation by dt and integrating gives - (#!/2 TT/) COS (2 Trft - i) - (#3/6 TT/) COS 3 (2 Trft - a 3 ) etc. = Li + const. The constant of integration is equal to zero, because the current cannot have a unidirectional component without a commutating device or electric valve of some sort. Therefore i = - (Ei/2 7T/L) cos (2 Trft - - (#3/6 7T/L) cos 3 (2 Trft - 3 ) - etc., (99) which means that each harmonic in the e.m.f. produces its own current, as if this harmonic were acting alone. The total current is the sum of such harmonic currents. The reactance of the coil for the nth harmonic is n times as great as for the fundamental wave; therefore, the higher harmonics in the current are rela- tively smaller than those in the voltage wave. (c) Let now a non-sinusoidal alternating voltage be impressed at the terminals of an impedance coil, and let it be required to determine the wave form of the current. The result to be expected will be intermediate between those derived for a pure resistance and a pure inductance; viz., the current wave will be more nearly of sine form than the voltage wave, but not to the same extent as in the case of a pure inductance. Substituting the above given expansion for the voltage wave CHAP. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 71 into the fundamental eq. (83), we obtain a differential equation for i, which equation some readers may not be able to solve. We choose, therefore, the opposite way; that is, we assume the current wave to be given, instead of the voltage wave, and determine the corresponding voltage wave from eq. (83). This procedure is much simpler, because it involves differentiation instead of inte- gration. Let the current be given in the form i = /i sin (u i) + 7 3 sin 3 (u 3 ) + etc., where u = 2 irft. Substituting this value into eq. (83) and rearranging the terms, gives e = [Iir sin (u i) + 2 irfLIi cos (u ai)] + [/ 3 r sin 3 (u a 3 ) + 6 Tr/L/g cos 3 (u -> 3 )] + etc. . . . (100) This result shows that each harmonic of the current requires a corresponding harmonic of the voltage, as if it were flowing alone. The total voltage is equal to the sum of the harmonic voltages. Therefore we conclude that, conversely, if the voltage were given, the current would be equal to the sum of the harmonic currents produced by the respective harmonics in the voltage. If the impedance to the first harmonic is z\ = \/r 2 + x 2 , that to the third harmonic is z 3 = Vr 2 + (3 x) 2 , and in general the impedance to the nth harmonic is z n = Vr 2 + (nx) 2 . The phase displace- ment between the corresponding harmonics of current and voltage is determined from the condition, tan < = nx/r, or cos n = r/z n . The general conclusion reached is as follows: When the applied voltage contains higher harmonics, the total current is found by summing the harmonic currents due to each harmonic of the voltage acting alone. Prob. 1. The effective value of the fundamental wave of an e.m.f. is 110 volts; it has a pronounced third harmonic, of 24 per cent of the fun- damental wave. This voltage is applied across a pure reactance, equal to 5 ohms for the fundamental frequency. Calculate the current. Ans. 22.07 amp. Prob. 2. An alternating voltage is represented by the expression 170 sin 250 t + 62 sin (1250 t + 2.3). It is applied to an impedance coil having an inductance of 45 millihenrys and a resistance of 7 ohms. Show that the current in amperes is equal to 12.82 sin (250 1 1.015) + 1.09 sin (1250 + 0.853). 24. The Extra or Transient Current in Opening and Clos- ing a Circuit. Since an electric current possesses inertia in 72 THE ELECTRIC CIRCUIT [ART. 24 the form of inductance, no current can be established or broken instantly, unless the applied electromotive force be infinitely large. Thus, when a large electromagnet is connected to a source of continuous voltage, the current increases during an appreciable interval before it reaches its final value. Again, when the circuit is broken, the current continues in the form of an arc through the air for an appreciable time. In a majority of cases these transient phenomena at the opening and closing of a circuit are of no practical importance, yet there are circumstances under which they must be taken into consideration; for instance, in switching on and off large amounts of energy, in high-frequency oscillations, in highly inductive circuits, etc. We shall consider here two simple cases of such extra currents; namely, when a circuit pos- sessing resistance and inductance is connected to a source of (a) continuous voltage and (b) sinusoidal alternating voltage. (a) Direct Voltage. When e in eq. (83) is constant, one value of i which satisfies this equation is i = e/r, because in this case di/dt = 0. However, this is not the most general solution, because it is possible to select an exponential expression in addi- tion to the constant i, which will satisfy the equation. Put i = /r + CV-'* f (101) where e is the base of natural logarithms, and C and T are certain constants. Substituting this value of i into eq. (83), we get Cre-'/' - (L/r)Ce-^ = 0, or T = L/r. Besides, i = when t = 0, so that expression (101) becomes = e/r + C, from which C = -e/r, and consequently i = (e/r) (1 - -*/*) (102) In other words, when a direct-current circuit is closed, the current increases at first rapidly, then more and more slowly; and theo- retically it reaches its final value of e/r only after an infinite time. In reality, the current becomes practically constant after a frac- tion of a second, unless the inductance is exceedingly large. The factor T = L/r is called the time constant of the circuit; it deter- mines the rate of the initial rise in current, and has the dimension of time. CHAP. VI] INDUCTANCE, REACTANCE AND IMPEDANCE 73 (b) Sinusoidal Voltage. If the voltage follows the law e = E m sin 2 irft, one solution of eq. (83), as we have seen before, is i = (E m /z) sin (2 irft $), where cos< = r/z. But this is not the most general solution, because it is possible to add to it an expo- nential term of the form Ce~ t/T , and to select the time constant r in such a way that this term will cancel in eq. (83). Since the sine term of the current alone satisfies the equation, we will find as before T = L/r. The constant C is determined by the condi- tion that i = when i = 0, or = - (E m /z) sin + C, from which C= (E m /z}sm4>=.E m x/z\ Therefore the current i = (E m /z) sin (2*fl - <) + (E m x/z 2 )<--. From this triangle we obtain two useful relations, g = y cos 4> b = y sin 0. When there are several susceptances and conductances in parallel (Fig. 26), the reactive and the energy components of the current I ^ and (116) FIG. 26. Susceptances and conductances in parallel. must be added separately (Fig. 27). Therefore, the amperes per volt in phase or the conductances, and the amperes per volt in E&3 FIG. 27. The voltage and current relations in the circuit shown in Fig. 26. quadrature or the susceptances, must also be added separately, so that the equivalent admittance y = V( gi + 2 + etc.) 2 + (61 + 6 2 + etc.) 2 , . (117a) and tan c/, = (&! + 6 2 + etc.)/(0i + g* + etc.). . . . (117b) The student should compare Fig. 27 with Fig. 23 in order to see the similarity of procedure and the difference in the physical phenomena in the two cases. With a series connection, it is the current that is common to all the parts of the circuit, while the partial voltages are added geometrically. In a parallel combina- tion, the voltage is common to all the branches, while the com- ponent currents are combined in their proper phase relations. 78 THE ELECTRIC CIRCUIT [ART. 27 The following table gives the quantities defined in this and the preceding chapter, in their proper relations. Friction Effect Inertia Effect Connection Result Unit Resistance r Reactance x Series Impedance z Ohm Conductance y Susceptance 6 Parallel Admittance y Mho - Prob. 1. What susceptance must be connected in parallel with a resistance of 0.2 ohm, in order to bring the power-factor of the combina- tion down to 80 per cent? Also, what is the value of the resultant ad- mittance? Ans. 3.75 mhos; 6.25 mhos. Prob. 2. Two electrical devices are connected in parallel to a line of voltage E. One device consumes a current /i at a power-factor cos i', the total line current is 7, lagging behind the voltage by an angle <. Show how to determine graphically the susceptance and the conductance of both devices. 27. Equivalent Series and Parallel Combinations. Let a resistance r a be connected in series with a reactance x a ; also let another resistance r p be connected in parallel with a reactance x p . If the values of the resistances and reactances are so selected that the series combination, when connected to the same source of supply, will let through the same current at the same power- factor as the parallel combination, then the two combinations are called equivalent. It is sometimes convenient to replace a given series combination by an equivalent parallel combination, and vice versa. For instance, when some parts of a circuit are in parallel and others in series, it is convenient for numerical cal- culations to replace them all by an equivalent parallel or series combination. The problem is to find the relation between the four quanti- ties r,, r p , x, and x p , if these quantities form two equivalent com- binations. According to the above-given definition, the angle is the same for both, and besides, according to eqs. (93) and (114), y-i/, (us) where y refers to the parallel combination and z to the equiva- lent series combination. Combining now eqs. (116), (95) and (96), we have lAp = 9 = y cos = (l/z) (r./z) = r 8 /z 2 ; CHAP. VII] SUSCEPTANCE AND ADMITTANCE 79 or r.r p = z*= l/ (121) and 6 = x t /z z (122) On the other hand, if g and b are given, r. = g/y 2 ; (123) x. = b/y* (124) The reciprocals of r, and x, are of no practical importance, and are not used in this work. Prob. 1. An impedance coil has a reactance of 7.5 ohms; the resist- ance of the winding is 2 ohms. What are the susceptance and the con- ductance of the equivalent parallel combination? Ans. 124.3 and 33.2 millimhos. Prob. 2. Check the answer to the foregoing problem by actually cal- culating the current and the power-factor of the series and the parallel combinations at some assumed voltage. Prob. 3. Show that r p and x p are always larger than r t and x, re- spectively. Hint: In eqs. (119) and (120) replace z z by r a 2 + x, 2 , and solve for r p and x p . Prob. 4. An apparatus takes 25 amp. and 2000 watts at 110 volts, the current being a lagging one. What are the equivalent conductance and susceptance of the device? What are the resistance and reactance in series equivalent to this apparatus? Ans. 0.165 mho; 0.156 mho; 3.2 ohms; 3.04 ohms. Prob. 6. In adjusting a measuring instrument, a non-inductive re- sistance of 120 ohms was used in parallel with a choke coil. The imped- ance of the coil was 75 ohms, its resistance 16 ohms. In the regular manufacture of the instrument it is desired to use a resistance and a re- actance in series. Determine their values, either graphically or analyti- cally. Ans. r a = 38.0 ohms; x a = 44.3 ohms. 80 THE ELECTRIC CIRCUIT [ART. 28 28. Impedances in Parallel and Admittances in Series. In the preceding chapter we have learned how to add impedances in series, and in this chapter how to add admittances in parallel. Let now two or more impedances be connected in parallel, and let it be required to find the equivalent impedance. This is done by replacing each of the given impedances by an equivalent parallel combination, and then adding their admittances in par- allel, according to the rule developed above. Conversely, let sev- eral admittances be connected in series, and let it be required to find the equivalent admittance. To solve this problem, each parallel combination is replaced by an equivalent series combina- tion, and then the impedances are added in series. The student understands, of course, that the addition in both cases is geo- metric, and that only like components can be added algebraically. Problems of this kind occur, for instance, in the theory of trans- mission lines, transformers, and induction motors; for this reason it is important that the student understand the equivalent com- binations, and that he acquire facility in changing from a series to a parallel combination, and vice versa, as is explained in the preceding article. Prob. 1. The load of a single-phase, 6600-volt generator is estimated to consist of 1200 kw. of lamps, practically non-inductive, and of 800 kw. of motors, working at an average power-factor of 75 per cent. What will be the expected generator output, in amperes, and the power-factor? Solution: The energy component of the motor current is 800/6.6 = 121.2 amp.; the reactive component is 121.2 tan ^ = 106.8 amp. The lamp current is 1200/6.6 = 181.8 amp. The total energy component of the generator current is 121.2 + 181.8 = 303 amp. Consequently, the total generator current is (303 2 + 106.S 2 )* = 321.3 amp.; the power-factor is 303/321.3 = 94.3 per cent. Prob. 2. Check the solution of the preceding problem graphically. Prob. 3. Three resistances of 2, 5 and 10 ohms, and two reactances of 4 and 2.5 ohms, are all connected in parallel across a 250-volt alternating- current line. What are the total current and the power-factor of the com- bination? Ans. 258 amp.; 77.5 per cent. Prob. 4. Three impedance coils, having ohmic resistances of 2, 3 and 4 ohms respectively, and inductances of 13, 10 and 22 millihenrys, are connected in parallel across a source of 220-volt, 60-cycle alternating voltage. Calculate the total current and the power-factor. Check the solution graphically. Ans. 110 amp.; cos = 0.495. Prob. 6. Solve the preceding problem for a frequency of 25 cycles per second. Construct the vector diagrams of the currents in both problems to the same scale, so as to see the influence of the frequency. CHAP. VII] SUSCEPTANCE AND ADMITTANCE 81 Prob. 6. In problem 4, let the total current be given in magnitude, but not in its phase position; assume the inductance of the third coil to be unknown. Show how to determine analytically and graphically the vector of the current in the third coil, and the position of the vector of the total current. Prob. 7. The admittance of a winding is 0.2 mho; the current through the winding lags by 34 degrees with respect to the voltage at its terminals. Determine the resistance and the reactance of the winding. Ans. 4.145 ohms; 2.796 ohms. Prob. 8. A coil having a resistance of 2.3 ohms and a reactance of 5 ohms is connected in parallel with another coil, for which r = 3 ohms and x = 4 ohms. Calculate the resistance and the reactance of the equivalent series circuit. Ans. 1.36 ohms; 2.255 ohms. . Prob. 9. The coils given in the preceding problem are connected in parallel across 55 volts. Calculate the total current, its energy and re'ac- tive components, and the power-factor of the combination. Ans. 20.85 amp.; 10.78 amp.; 17.88 amp.; cos = 0.5165. CHAPTER VIII THE USE OF COMPLEX QUANTITIES 29. Addition and Subtraction of Projections of Vectors. With the explanation given in the preceding four chapters, the stu- dent is enabled to handle, by means of vector diagrams, problems involving resistances and reactances in alternating-current cir- cuits. A number of problems in transmission-line calculations and in the theory of alternating-current machinery may be solved by the use of such vector diagrams. The disadvantages of the graphical method are: (1) Results are usually obtained which hold for one specific case only; an analysis of the effect of various factors is often difficult. (2) Some vectors may be many times smaller than others; for instance, the voltage drop in a transmis- sion line, as compared to the line voltage itself. Therefore, the diagram must be drawn to a very large scale, or else the results are not sufficiently accurate. In addition to these drawbacks, some engineers object to graphical methods in general, as involv- ing the use of drawing instruments, which may not be convenient. On the other hand, vector diagrams are quite convenient in some practical cases; moreover, they are helpful for the under- standing of general relations in a circuit, without reference to particular numerical values. Again, in some problems, the un- known vectors can be calcu- lated from the vector diagram trigonometrically , without the necessity of actually drawing it to scale. It is possible to treat vectors analytically, using their projections on two axes, FIG. 28. A vector and its projections. as in analytic geometry (Fig. 28). A vector, such as E, can be denned either by its magnitude and phase angle 0, or by its projections e and e' upon the axes of coordinates. 82 CHAP. VIII] THE USE OF COMPLEX QUANTITIES 83 If E and 6 are given, the projections are calculated from the expressions e = Ecosd',} e' = Esme.l' If the projections are given, the vector itself is determined in magnitude and position from the equations E = (e 2 + e' 2 )*; (126) tan 6 = e'/e (127) In numerical computations it is more convenient first to calculate tan from eq. (127) and then to determine E from one of the eqs. (125), using trigonometric tables. This does away with the necessity for squaring the projections and extracting a square root. The fact that e and e' are components of the vector E along two perpendicular axes is expressed symbolically thus: E = e+je' (128) Here j is a symbol which indicates that the projection e' refers to the vertical axis. This symbol must not have any real value; for the time being, it may be considered merely as an abbrevia- tion of the words " along the vertical axis." The sign plus in eq. (128) denotes the geometric addition. The dot under E sig- nifies that by E is meant not only the magnitude of the vector, but its direction as well, the latter being defined by the projec- tions. When the magnitude only is meant, the dot is omitted. The foregoing notation has been introduced by Dr. Charles P. Steinmetz, and is now universally used in this country. Much credit is also due to Dr. Steinmetz for developing the analytic method, used below, of dealing with alternating currents and voltages by means of their projections. The addition and subtraction of vectors are reduced simply to the addition and subtraction of projections. According to Fig. 12, the projection of a vector on any axis is equal to the sum of the corresponding projections of its component vectors on the same axis. Thus, if a current is represented as a vector by its projec- tions 50 -f- j 70 amp., and another current by 100 + j 40 amp., the vector sum of these currents is 150 + j HO amp. Or, the resultant of two voltages, EI e\ + jei and E 2 = e 2 + jez', is E eq = EI -\- E z = (ei + e = 7520/(66^X 147) = 0.775; sin = 0.632. The generator voltage is 66 X 0.775 +j 66 X 0.632 = 51.15 +.7 41.71 kilovolts. The voltage drop in the line is 147(45 +j 83) = 6615 +j 12,200 volts. Hence, the load voltage is (51.15 - 6.61) + ./(41.71 - 12.2) = 44.54 +j 29.51 kilovolts. The nu- merical value of the load voltage is (44.S4 2 + 29 .51 2 )* = 53.43 kilo- volts. Prob. 2. Determine analytically the resistance r 2 required in problem 2, Art. 22. Prob. 3. A voltage equal to 180 + j 75 produces a current of 7 + j 1.5 amp. What is the impedance of the circuit? Ans. 26.78 + j 4.97 ohms. Prob. 4. Power is transmitted from a single-phase alternator to a load consisting of a resistance of 1.17 ohms in series with a reactance of 0.67 ohm. The generator voltage is 2300, and the impedance of the transmission line is 0.085 -f j 0.013 ohm. Determine (a) the line cur- rent; (b) the voltage drop in the line; (c) the receiver voltage. Take the generator voltage as the reference vector. Ans. (a) 1413.6 -j 769.6 amp.; (b) 130.1 -j 47 volts; (c) 2169.9 + j 47 volts. Use the admittance operator to obtain the current, and the impedance operator to calculate the line drop. Prob. 5. A voltage, e + je', is impressed across the impedances r\ + jxi and r 2 + jx z in parallel. Find the total current. Solution: The total conductance is g = n/Zi 2 + r 2 /2 2 2 , and the total susceptance is b = Ji/zi 2 + z 2 /z 2 2 . Hence, the current i + ji' = (e + je') (g jb) = (eg + e'b) + j (e'g - cb). Prob. 6. Extend the solution of the preceding problem to the case in which more than two impedances are in parallel. Ans. i + jV = [e s(r/) + e'z(z/*)] + j [e'z(r/s) - e S(z/z 2 )]. Prob. 7. Two impedances, r\ + jx\ and r 2 + jx 2 , in parallel, are connected in series with a third impedance r + jx. Show how to deter- mine the total voltage, knowing the total current i + ji' ; or, how to find the expression for the total current when the total voltage e + je' is given. Prob. 8. Show how to solve the preceding problem when both the current and the voltage are given, but either the impedance n + jxi or the impedance r + jx is unknown. CHAPTER IX THE USE OF COMPLEX QUANTITIES (Continued) 32. Power and Phase Displacement Expressed by Projec- tions of Vectors. Let an alternator supply a current / = i -f- ji' at a voltage E = e -f je', and let it be required to calculate the power output of the generator. The expression for the average power is P = El cos <, where < is the phase displacement between E and /. The angle < is the difference between the angles d e and which the vectors E and / respectively form with the reference axis. Hence, we have P = El cos = El cos (0. - 6i) = E cos d e I cos 0i + E sin d e / sin 0. Remembering that E cos d e , Esin6 e , etc., represent the projec- tions of the given vectors on the axes of coordinates, we have simply P = ei + e'i' (143) Another way of deducing expression (143) is to resolve the given vectors of current and voltage into their components along the axes of coordinates, and to consider the contribution of each pro- jection to the total power. The projections e and i, being in phase, give the power ei. Similarly, the projections e' and i' give the power e'i'. The projection i' of the current gives zero average power with the projection e of the voltage, the two being in phase quadrature. For the same reason the average power resulting from e' and i is equal to zero. Thus, ei + e'i' represents the total average power. To find the phase displacement, or the power-factor of the output, we write N tan 6 e tan 0j tan = tan (0, - 00 = ^. ^- , 1 + tan e tan 0< ....... Knowing tan , its cosine is found from trigonometric tables. 1 1 Or else the power-factor, cos = cos (O e Oi), can be found from the relations O e = tan- 1 e'/e, and ^ = tan- 1 i' /i. 91 92 THE ELECTRIC CIRCUIT [ART. 32 Power-factor can also be determined directly from the expres- sion cos = P/EI = (ei + eY)/[(e 2 + e' 2 ) (t 2 + i' 2 )]*, (145) but the calculations are more involved than when formula (144) is used. The power calculated by means of formula (143) sometimes comes out negative, if some of the projections of E and / are negative. The interpretation is that the phase displacement between the current and the voltage is over 90 degrees, so that power is being supplied to the machine, instead of being delivered by it. In other words, the machine acts as a motor and not as a generator. Tan in formula (144) may also be negative, which means either that the current is leading, or that it is lagging by an angle larger than 90 degrees. The question is decided by reference to the sign of the power. For the reactive power (Art. 19) we have P r = EI sin = EI sin (0 e - { ) = E sin d e l cos t - - E cos e e l sin 0,, or P T = e'i - ei' (146) The apparent power is P a = (# + i'rf (e 2 + e' 2 )* (147) However, it is sometimes more convenient to determine the appar- ent power from the relation P a = P/ cos 0, (148) where P is calculated from eq. (143), and cos is found from trigonometric tables, knowing tan from eq. (144). Prob. 1. The terminal voltage of an alternator is 5370 +.7 735; the line current is 173 - j 47 amp. Calculate the output of the machine and the power-factor of the load. Ans. 894.5 kilowatts; 92 per cent. Prob. 2. In the preceding problem, what must be the projection of the current upon the 7-axis in order that the power shall become zero? Ans. i'= -1264 amp. Prob. 3. Let the line current in problem 1 be 58+jl2 amp. Explain the negative sign of the power and the plus sign of tan . Draw the vectors of the current and voltage. Prob. 4. A synchronous machine generates a voltage equal to 2300 - j 50 volts, and supplies a current, through an impedance of 5 + j 50 ohms, to another synchronous machine generating a counter-e.m.f. of 2300 + j 50 volts. What is the power output of the first machine? Is the current leading or lagging? Make clear to yourself the physical meaning of the answer. Ans. 4.55 kw.; = 173 3' lagging. CHAP. IX] THE USE OF COMPLEX QUANTITIES 93 Prob. 6. A current of 350 j 75 amp. is maintained through an im- pedance, the power output being 952 kw. at a power-factor of 86 per cent lagging. Find the voltage across the impedance. Hint: Solve eqs. (143) and (144) together for the unknown projections e and e'. Ans. 2930 + j 987 volts. Prob. 6. Solve problem 5 by calculating the value of the impedance, and multiplying the impedanee by the current. Hint: power = I-r; x = r tan . Prob. 7. Solve problem 5, using the expression for the reactive power. ^/ 33. Vectors and Operators in Polar Coordinates. Instead of representing a vector by its orthogonal projections, as in eq. (128), it is sometimes more convenient to express the same vector as a complex quantity in terms of its magnitude and direction. Sub- stituting the values of e and e' from eqs. (125) into eq. (128), we obtain # =#(cos0+./sin0) (149) Similarly, a current in phase with this voltage is expressed as 7 = 7(cos0+.7'sin0), (150) while a current lagging by an angle behind the voltage E is represented by the equation 1=1 [cos (0 - 0) + j sin (0 - 0)]. . . . (151) When the vectors of currents and voltages are expressed in the trigonometric form shown above, it is convenient to use the operators Z and Y in a similar form. Substituting the values of r and x from eqs. (95) and (96) into eq. (135), we get Z = z (cos + j sin 0) (152) In a similar manner, using eqs. (116) in eq. (138), gives Y = y(cos0 -jsintfO (153) When calculating the voltage drop IZ or the current E/Z it is necessary to find the product or the ratio of two complex expressions of the form cos + j sin 0. By actually performing the multiplication and separating the real from the imaginary term we find that (cos 0+ j sin 0) (cos 0+ j sin 0) = cos (0+0) + j sin (0+0). (154) This gives a simple rule for the multiplication of two or more complex quantities in the trigonometric form. In order to deduce 94 THE ELECTRIC CIRCUIT [ART. 33 a similar rule for division, we observe that cos<> jsm = cos ( 0) + jsm( <). . . (155) This relation is easily verified by multiplying the numerator and the denominator of the left-hand side of the equation by cos - j sin <{>, so as to get rid of the complex quantity in the denomi- nator. Equation (155) leads to the following rule for the division of complex quantities in trigonometric form : (0-/)). (156) Thus, for instance, if the current given by eq. (150) flows through an impedance expressed by eq. (152), the required terminal volt- age is E = IZ = Iz [cos (0 + 0) + j sin (0 + )], . (157) which result simply means that the voltage is equal to Iz and leads the current by the angle 0. The operator given by eq. (152) multiplies a vector by z and turns it by the angle < in the positive direction. Hence, the operator (cos + j sin 0) simply turns a vector by the angle <, without changing its length. Thus, if it be required to turn a vector A = a + ja r by an angle a in the positive direction, the projections of the new vector are found from the following expres- sion: (a + ja'} (cos a + j sin a) = (a cos a a' sin a) + j (a' cos a + a sin a) ..... (158) Of course, the same result could be obtained by first calculating the angle 6 which the vector A forms with the reference axis, from the relation tan 6 = a' /a, and then determining the new projec- tions A cos (6 + a) and A sin (6 + a). Voltage Regulation of a Transmission Line. 1 As an example of the use of complex quantities in the trigonometric form, let us consider the voltage regulation of a single-phase transmission line. Let the resistance and reactance of the line, and the gen- erator voltage EI, be given; and let it be required to determine the receiver voltage E 2 for a given current I and a given power- 1 The electrostatic capacity of the line is disregarded here; a complete treatment of the regulation of a transmission line, taking into account the capacity and leakage, is given in Arts. 68 and 69 at the end of the book. CHAP. IX] THE USE OF COMPLEX QUANTITIES 95 factor of the load cos 0'. In the symbolic notation we have E^Ei + IZ, (159) where the impedance Z of the line is known, and is expressed by eq. (152). The phase angle refers to the line, the angle 0' to the load. When actually solving an equation such as (159), it is highly important to select the reference axis in the most advantageous way, so as to simplify the calculations as much as possible. In the case under consideration, it is convenient to select the refer- ence axis in the direction of E 2 , because then E 2 is determined by its magnitude alone, the direction angle being equal to zero. The generator voltage is expressed by EI (cos 6 + j sin 0), where the magnitude of E\ is given, but the angle 6 is unknown. The current lags by the angle 0' behind E 2 , and therefore is expressed by the formula / (cos 0' j sin 00- Thus eq. (159) becomes E! (cos 6 + j sin 0) = E 2 + Iz [cos (0 - 00 + j sin (0 - 0')]- (160) Equating the real and the imaginary parts gives # 1 COS0 = # 2 + /ZCOS(0-00; - (161) EI sin 6 = Iz sin (0 - 0') (162) From eq. (162) sin e = (Iz/EJ sin (0 - 0') (163) Knowing 9, we find from eq. (161) E 2 = #1 cos 0- Iz cos (0-0'). . . . (164) In practice, one is usually required to determine the voltage regulation of the line. According to the definition adopted by the American Institute of Electrical Engineers (Standardization Rules, Art. 187), per cent regulation = 100 (E 2 - Ez)/E 2 , . . (165) where E 2 is the value of E 2 at no load. But here E 2 = EI, be- cause the electrostatic capacity of the line is neglected. It is pos- sible to determine the difference EI E 2 directly from eq. (161), by substituting for cos the expression 1 2 sin 2 0. We obtain then A# = EI - E 2 = Iz eos (0 -0')+2 E! sin 2 \ 6. (166) Equation (165) becomes pr cent regulation = 100 kEKEi - A#). . (167) 96 THE ELECTRIC CIRCUIT [ART. 33 When it is required to calculate the voltage regulation for several loads, the computations are conveniently arranged in a table of the following form: I...-L.I current / Line drop u Load power- factor coa' Angle -*' sin <*-*') C08(0-<') sin. Angle e sin? i 6 A Per cent regula- tion In practice, the voltage regulation is usually required for a certain load of P 2 watts, so that, strictly speaking, the current / is not known. But, since E 2 is not much different from Ei, it is an easy matter to estimate the current with sufficient accuracy. Or else a curve of voltage regulation is plotted against the load as abscissae, so that the regulation may be read off at any desired load. It is possible to solve the problem exactly, by using for / its expression P 2 /(E 2 cos <') in eqs. (161) and (162). In this case the equations are squared and added, so as to eliminate 6. This gives a biquadratic equation for Ez, from which the receiver voltage can be computed. This problem can also be solved when the complex quantities are expressed in the orthogonal form, instead of the trigonometric form here used. The student is urged to work out the details, in order to become thoroughly familiar with complex quantities in both forms. Prob. 1. A vector 72 + j 53 must be turned by 25 degrees in the nega- tive direction. What are its new projections? Ans. 87.65 + j 17.6. Prob. 2. A single-phase aluminum line is to be built from a power house, at which a voltage of 11,500 is maintained at a frequency of 50 cycles, to a point 25 km. distant. When a current of 60 amp. at 80 per cent lagging power-factor is delivered at the receiver end, the power loss in the line must not exceed 10 per cent of the useful power. What must be the size of the conductor, and what will be the per cent voltage regula- tion at this load? The spacing between the wires is to be 61 cm. Ans. No. 0000 B. & S.; 11.4 per cent. Prob. 3. Check the answer to the preceding problem graphically. CHAP. IX] THE USE OF COMPLEX QUANTITIES 97 Prob. 4. Explain the theory of Mershon's diagram found in various electrical handbooks and pocketbooks, and check by means of it the answer to problem 2. Prob. 6. Show how to determine the voltage regulation of a trans- mission line when the receiver voltage is given. Prob. 6. Show how to calculate the receiver voltage E 2 from eq. (159), using the orthogonal projections of the vectors and operators. Discuss the relative advantages and disadvantages of the rectangular and polar coordinates in this case. 34. Vectors and Operators Expressed as Exponential Func- tions. 1 Expressions (149) to (153) are sometimes written in the exponential form, using the identity cos e + j sin 6 = e#, ..... (168) where e is the base of natural logarithms. This important equa- tion follows from the well-known expansions for sin 9, cos and e 9 , obtained by Maclaurin's Theorem in calculus; namely, The last series, when j d is substituted for 9, becomes Substituting these values into eq. (168), it is found to be an iden- tity. Thus, we have E = E (cos 6+j sin 9) = E& ..... (169) Similarly, the impedance operator becomes Z = z (cos + j sin 0) = ze 1 '*, .... (170) and the admittance operator Y = y (cos - j sin <) = y-*+. . . . (171) If, for instance, a current is given as / = Ie }0 and if it flows through an impedance Z = Zf?*, the required voltage is found by multi- plying these two expressions, or E = IZ = IzeW+< ...... (172) 1 This article may be omitted if desired, without impairing the conti- nuity of the treatment in the rest of the book. <)8 THE ELECTRIC CIRCUIT [ART. 34 This shows that the absolute value of the voltage is Iz and that it leads the current by the angle 0. Equation (172) corresponds to eq. (157) in trigonometric notation. The projections of the current vector are / cos and / sin 6; while the projections of the voltage vector are Iz cos (6 + ) and Iz sin (6 + ). Thus, it is always possible to change from the exponential form to the trigonometric form, and finally to the orthogonal projections, or vice versa. The exponential form is more concise, and possesses marked advantages in the solution of some advanced problems relating to alternating currents and oscillations. 1 However, for the simple problems treated in this book, the plain algebraic nota- tion a + ja' and the trigonometric notation A (cos a + j sin a) are amply sufficient. It is deemed advisable to explain the expo- nential notation here in order to enable the student to read books and magazine articles in which it is employed. 1 See for instance J. J. Thomson, Recent Researches in Electricity and Magnetism. CHAPTER X POLYPHASE SYSTEMS 35. Two-phase System. The student knows from his ele- mentary work that the induction motor operates on the principle of the revolving magnetic field, and that such a field is produced by a combination of two or more alternating currents differing in phase. An electric circuit upon which are impressed two or more waves of e.m.f. having definite phase displacements is called a polyphase system. A large majority of the alternating-current circuits used in practice hi the generation and transmission of electrical energy are polyphase systems; it is therefore essential that the student become familiar with the current and voltage relations in such circuits. Theoretically, the simplest polyphase system is a four-wire two-phase system (Fig. 30), although it is not the most econom- Phase 1 N g o -c Generator Receiver Phase 2 > FIG. 30. A four-wire two-phase system with two independent circuits. ical one hi practice. The two generator windings are independ- ent, and are relatively displaced by ninety electrical degrees. The two alternating voltages induced in these windings are there- fore displaced in time phase by a quarter of a cycle. Each phase may be used separately, for instance for lighting, or both phases may be combined in the windings of a synchronous or induction motor, for the production of a revolving magnetic field. Each phase can be treated separately, as if it belonged to an independ- ent single-phase system. 100 THE ELECTRIC CIRCUIT [AST. 35 Some economy in line conductors and insulators is achieved by combining two conductors belonging to different phases into one return conductor (Fig. 31). Such a system is called a three-wire o Generator Fid. 31. A three-wire two-phase system. two-phase system. The current and voltage relations for a balanced load and a lagging current are shown in Fig. 32. The vectors EI and EZ represent the voltages induced in the two generator or transformer windings from the point to the points A and B respectively; in other words, they are the voltages between each phase wire and the return wire. The vector Eiz* is the geometric difference of the two, and represents the voltage between the two phase wires. That "12 is the differ- ence and not the sum of EI Fiu. 32. A vector diagram of currents in- > \ ,\ and voltages for the two-phase system and E * 1S P rOVed ^ the shown in Fig. 31. following reasoning: Let the wire OGG'O' be permanently grounded, so that its potential is zero. Let the potential of the wire A A' at a certain instant be for example 100 volts above the ground, and that of the wire BB' 60 volts above the ground. Then the difference of potential, or the voltage between A A' and BB', is 40 volts. The same reasoning applies to every instant, so that the vector of the voltage between A A' and BB' is the geometric difference between EI and E*, which are the vectors of the voltages between the points A and 0, and B and re- * Pronounced E one two, and not E sub twelve. CHAP. X] POLYPHASE SYSTEMS 101 spectively. That is, the voltage between A and B is represented in phase and magnitude by the vector connecting the ends of the vectors EI and E z . Numerically, E IZ = E 1 V2 = E 2 V2 (173) The currents in the conductors A A' and BE' are represented by the vectors l\ and 7 2 , lagging by an angle with respect to the corresponding voltages. The current in the return conductor is the geometric sum of the two phase currents, and is represented by the diagonal vector /i 2 . It will thus be seen that the common return current is A/2 times as large as each component current, or In = Ii V2 = 7 2 V2 (174) If it is desired to have the same current density hi each of the three conductors, the cross-section of the return wire must be V2 times that of each of the other two wires. The two phases hi Fig. 30 are sometimes electrically inter- connected at their middle points, as shown in Fig. 33 at the left. FIG. 33. A star-connected quarter-phase system to the left, a mesh-connected system to the right. This is done in order to fix the difference of potential between the two phases. If the voltage between A and B is E, then the volt- age between the common point and each wire is \ E, and the voltage between the two phases is equal to \ E V2 = E/V2. For example, the voltage E da between D and A equals E a E d . These relations are shown vectorially in Fig. 34. This circuit is some- times called the star-connected quarter-phase system. The four windings of a generator or motor are sometimes con- nected in mesh, as indicated in Fig. 33 to the right. With the 102 THE ELECTRIC CIRCUIT [ART. 35 star connection, the star voltages OA, OB, etc., are induced directly, while the mesh voltages AD, DB, etc., are established by the combination of the star voltages. With the mesh connection of the windings, however, the mesh or line voltages are induced directly. The mesh and star voltages are shown in Fig. 34. Electrically the two arrangements are equivalent, provided that the proper numbers of turns are used in the windings. The line and mesh currents are indicated in Fig. 34. The line currents and those in the star-connected windings are represented FIG. 34. A vector diagram of currents and voltages in the quarter-phase system shown in Fig. 33. by the sides of the square, each current lagging by the angle with respect to the corresponding star voltage; the angle of lag depends upon the character of the load. In Figs. 33 and 34 the voltages are taken in the cyclic order AC, CB, BD, DA; hence, it is natural to take the positive direction of the current in the same way. With the arrows in Fig. 33 showing the positive directions of the currents, each line current is the difference between two adjacent mesh currents. Hence, in the vector diagram the mesh CHAP. X] POLYPHASE SYSTEMS 103 currents are represented by the radii from the center to the vertices of the -current square. It will be seen that the angle between the mesh currents and the mesh voltages is also equal to . While the mesh voltages are V2 times as large as the star voltages, the mesh currents are 1/V2 times the star currents. This condition is necessary in order to have the same power per phase in the mesh and star-connected systems. 36. Three-phase Y-connected System. This system is shown in Fig. 35; the current and voltage relations are repre- Fio. 35. A three-phase Y- or star-connected system. sented in Fig. 36. OA, OB, and OC represent three generator windings; O'A', O'B', and O'C' are the windings of a receiving apparatus for instance, an in- duction motor. The three gener- ator windings are placed on the armature core at angles of 120 electrical degrees with respect to each other, so that the alternating voltages induced in these windings are displaced in phase by one- third of a cycle, the positive di- rection in the windings being outward. They are represented r, 3 r, Fio. 36. The line and star voltages, by the vectors E a , E b , and E c in and the ljne currentg in the y . connected system shown in Fig . 35 . Fig. 36. The motor windings are similarly displaced, so that the whole system is symmetrical with respect to the three phases. The currents lag behind the voltages by an angle depending upon the relative amounts of resistance, reactance, and counter- e.m.f. in the circuit. 104 THE ELECTRIC CIRCUIT [ART. 36 The diagram of connections shown in Fig. 35 is also called the star connection, and the points and 0' are called the neutral points of the system. The voltages between the line conductors and the neutral points are called the star or phase voltages, as distinguished from the line voltages, or voltages between any two line conductors. A line voltage, for instance between the points A and B, is equal to the geometric difference between the voltages OA and OB, as has been shown above in the case of a two-phase line. Consequently, the line voltages are represented in Fig. 36 by the three vectors E^, E bc , and E ca , which connect the ends of the vectors of the phase voltages. It will be seen that the line voltages are V3 times as large as the phase voltages. When the three phases are perfectly balanced and the currents are nearly sinusoidal, the two neutral points and 0' may be connected by a wire, as shown by the dotted line, or grounded, and very little current will flow through this connection. The reason is that the algebraic sum of the three currents flowing towards or from the neutral points is equal to zero at all instants, because sin u + sin (u -f- f TT) + sin (u f TT) =0. (175) This identity is easily proved by expanding the left-hand member, using the expression for the sine of the sum of two angles. It will also be seen from Fig. 36 that the geometric sum of the three current vectors is equal to zero, because these vectors when added form a closed triangle. In practice, there are transmission lines on which one or both neutrals are grounded, although in some installations both neutrals are insulated from the ground. To prevent large currents with unbalanced loads or during short- circuits, the neutrals are often grounded through protective resistances. The question of grounded vs. ungrounded neutrals is still in a somewhat controversial stage. The power developed in the generator windings and available at the generator terminals is 3 lyEy cos 0, where Ey is the phase voltage. Since the line voltage E = EY V3, we have P = 3 / Y #Y COS = 7 Y #A V3 COS . . . (176) In practical calculations of three-phase transmission lines and electrical machinery, only one phase is considered; that is, the three-phase circuit is reduced to an equivalent single-phase cir- cuit. Let it be required, for example, to calculate the cross-sec- CHAP. X] , POLYPHASE SYSTEMS 105 tion and per cent voltage regulation of a three-phase 66,000-volt line, to transmit 50,000 kw. at 80 per cent power-factor, and at a loss of 10 per cent of the useful power; the spacing to be 1.8 m. First of all we find that the voltage between each wire and the neutral is 66,000/V3 = 38,100 volts, and that the power per phase is 50,000/3 = 16,700 kw. Hence, the problem is reduced to the following one: Determine the cross-section and per cent voltage regulation of a single-phase 38,100-volt line, having a spacing of 1.8 m., the i 2 r loss in one conductor being 1670 kw., and the resistance of the return conductor being negligible. The solution of this problem is given in Art. 33 above. A drop of say 5 per cent in the phase or star voltage means also a drop of 5 per cent in the line voltage, because of the fixed ratio 1/V3 between the two. Prob. 1. Assuming the reference axis in Fig. 36 to be horizontal, the line voltage equal to 44 kv., the current per phase 73 amp., and the angle equal to 15 degrees, write down the complex expressions for all the cur- rents and voltages. Ans. E b = 22 - j 12.7 kv. ; E* = 22 - j 38. 1 kv. ; / = 18.9 + j 70.5 amp. Prob. 2. A three-phase 60-cycle line is 16 km. long; the spacing be- tween the wires is symmetrical and is equal to 61 cm., the conductors con- sisting of copper wire of 14 mm. diameter. It is required to maintain a voltage of 6700 between the conductors at the receiver end of the line. What is the generator voltage when the load is equal to 1000 kw. at unity power-factor? Ans. 7040. Prob. 3. Show that a three-phase transmission line may be treated as a single-phase line which transmits one-half the power at the same voltage. The three-phase line requires three conductors of the same size as the single-phase line, with the same spacing (25 per cent saving in material) . Prob. 4. When the phase currents have higher harmonics, show that equalization currents must flow through the neutral connection, even though the phases are perfectly balanced. What happens when the neutrals are insulated from each other? Prob. 5. Show that the line voltage cannot have the third, the ninth, the fifteenth, etc., harmonics, even if these harmonics are present in the phase voltages. 37. Three-phase Delta-connected System. This method of three-phase connection is shown in Fig. 37, one end of the line being connected, for instance, to an alternator, the other end to a motor or to three transformers. Fig. 38 represents the current 100 THE ELECTRIC CIRCUIT [AHT. 37 and voltage relations. The currents in the windings are differ- ent from those in the line. With the positive directions of the FIG. 37. A three-phase delta- or mesh-connected system. currents indicated in Fig. 37, each line current is equal to the difference between the two adjacent currents in the " delta," Fia. 38. The voltages and currents in the delta-connected system shown in Fig. 37. Hence, in the vector diagram the line or " star " currents are represented by a triangle, and the delta currents by the rays from the center to the vertices of the triangle. It will be seen that the delta currents are equal to 1/V3 of the line .currents, and are displaced in phase by 30 degrees with respect to them. This also follows from the identity / sin u - I sin (u + 120) = \/3 / sin (u - 30). . (177) CHAP. X] POLYPHASE SYSTEMS 107 While there are no neutral points with a delta system, one or more of them may be artificially created by connecting three resistances or reactances in star as shown in Fig. 37 by dotted lines. The .Y- voltages between this neutral and the line conductors are shown in Fig. 38 by the vectors E a , -E& and E c . The delta volt- ages are \/3 times as large as the star voltages, while the star or line currents bear the same ratio to the delta currents. This is necessary in view of the power relation P = 3 #A/A COS = 3 # Y /Y COS <. . . . (178) In design and performance calculations one phase only is considered, the three phases being identical when the load is balanced. As far as the line is concerned, the delta-connected generator and load may be replaced by equivalent star-connected windings to give the same line currents and voltages. Then the line is designed and its performance calculated the same as in the preceding article. As a matter of fact, for line calculations it is only necessary to know the power, the voltage, and the power- factor of the load. The fact that the generator or the load is delta- or Y-connected has no bearing upon the line performance with a balanced load. Prob. 1. A 2000-kw. 6600-volt induction motor is fed from a 66,000- volt three-phase line through three step-down transformers, the high- tension windings of which are connected in Y, the low-tension windings in delta. What are the currents in these windings when the motor is carrying a 25 per cent overload? It is estimated that at this overload the power-factor is 90 per cent and the efficiency 92 per cent. The mag- netizing current of the transformers is negligible. Ans. 26.4 and 153 amp. Prob. 2. Show that, while the instantaneous electrical output of a single-phase alternator varies at double the frequency of the current, the output of a polyphase machine is practically constant as long as the load remains constant. Show that the same is true for motors. Note: For the electrical relations in two- and three-phase systems with unbalanced loads, and also for the theory of the V and T connections, see the author's Experimental Electrical Engineering, Vol. 2, Chapter 25. A more exhaustive treatment will also be found in his investigation entitled Ueber mehrphasige Stromsysteme bei ungleichmdssiger Belastung (published by Enke, 1900). See also the chapters on polyphase systems in Dr. Steinmetz's Alternating-current Phenomena. CHAPTER XI VOLTAGE REGULATION OF THE TRANSFORMER 38. Imperfections in a Transformer Replaced by Equivalent Resistances and Reactances. The reader is familiar in general with the construction and operation of the constant-potential transformer (Fig. 39). It consists of an iron core upon which two windings are placed as closely as possible to each other. When one winding is connected to a constant-potential alternat- ing-current source of power, an alternating magnetic flux is excited in the iron core and an alternating voltage is induced in the other winding. If this latter winding is connected to an electrical load, an alternating current flows through it, and causes a correspond- ing flow of current through the first winding, in order that power may be transmitted from the primary into the secondary circuit. The constant-potential transformer is one of the most perfect pieces of electrical apparatus, in that its efficiency (in medium and large sizes) is nearly one hundred per cent, and its voltage regula- tion with varying load is quite close. On the other hand, the requirements for voltage regulation are quite exacting, there being no provision in the apparatus itself for adjusting the voltage, like the field rheostat in a generator. Therefore, the pre-determina- tion of the voltage regulation of a transformer is of considerable .practical importance. Numerically, the 'regulation of a transformer is expressed in a manner similar to that given in Art. 33 above for the trans- mission line. Let, for example, the rated secondary voltage of a ten-to-one transformer be 220 volts, and let us suppose that a primary e.m.f. of 2280 volts is necessary in order to have the rated secondary voltage at the rated load. Let now the second- ary circuit be opened ; the secondary voltage will rise to prac- tically 228 volts, provided that the primary voltage is kept con- stant. Then, by definition, the regulation of the transformer at this load is 100 (228 - 220) /220 = 3.64 per cent. Let, in general, 108 CHAP. XI] REGULATION OF THE TRANSFORMER 109 the secondary terminal voltage at a certain load be E 2 , that at no load EM Then, by definition, per cent regulation = 100 (#02 - Ed/E* . . (179) The difference between the no-load voltage and that at full load is due to slight imperfections in the transformer itself. There- fore, in order to be able to calculate the voltage regulation at a given load, it is necessary to learn the nature of these imperfec- tions; for purposes of computation, it is convenient to replace these imperfections by certain resistances and reactances, as shown in Fig. 39. In an ideal transformer the ratio of the primary to the second- ary voltage is equal to the ratio of the numbers of turns in the corresponding windings. The same relation is very nearly true in any good transformer at no load. This follows from the fact that the two windings are linked with the same magnetic flux, and hence the voltage induced per turn is the same in both. Hence, denoting the primary and secondary induced voltages by En and Eiz, and the corresponding numbers of turns in series by n\ and HZ, we have E il /E i ^ = n l /n z ....... (180) Furthermore, in an ideal transformer 7ini = / 2 n 2 , . ...... (181) that is, the currents are inversely as the numbers of turns, or the primary ampere-turns are equal and opposite to the secondary ampere-turns. This is because an ideal transformer is supposed to have no reluctance in its magnetic circuit, so that no ampere- turns are required to maintain a magnetic flux in it. Consequently, any secondary current required by the load automatically draws a compensating primary current from the source of power, of such value that eq. (181) is satisfied. In a real transformer the primary ampere-turns are slightly different from the secondary ampere- turns, and the difference between the two is just sufficient to main- tain the alternating flux through the reluctance of the core, and to supply the core loss. Multiplying eqs. (180) and (181) term by term, and canceling n\ and n 2 , we find that This simply means that an ideal transformer transmits power from the primary into the secondary circuit without loss. 110 THE ELECTRIC CIRCUIT [ART. 38 (a) The Ohmic Drop. One of the causes of the internal volt- age drop in a transformer is the ohmic resistance of its windings. Because of the resistance of the primary winding, the primary terminal voltage E l (Fig. 39) is slightly larger than the induced I, Iron Cure Windings FIG. 39. Imperfect ions in a transformer represented by resistances and reactances. counter-e.m.f. EH which balances it. The secondary resistance causes a voltage drop, so that the secondary terminal voltage E 2 is smaller than the secondary induced e.m.f. E i2 . Thus, the effect of the internal resistances upon the terminal voltages is such as to make the ratio E*/Ei smaller than the ratio n 2 /ni. The windings themselves may be thought of as devoid of resist- ance, but corresponding resistances r\ and r 2 may be placed out- side the transformer, as shown in Fig. 39. 1 (b) The Reactive Drop. Another imperfection or cause of internal voltage drop is the so-called leakage reactance of the windings. The total magnetic flux in a loaded transformer may be considered as consisting of three components; viz., the useful flux linking with both the primary and the secondary windings, the primary leakage flux linking with the primary winding only, and the secondary leakage flux linked with the secondary winding only. In an ideal transformer the two last-named fluxes are absent because the two windings are supposed to be perfectly interwoven, so as to leave no room for the leakage flux. The primary leakage flux, being produced by the primary current, is in phase with it, and induces an e.m.f. in lagging quadrature with 1 The equivalent resistances TI and r 2 must replace not only the true ohmic resistances of the windings, but should also account for the eddy- currcnt loss in the conductors. In low-tension windings made of heavy conductors, this latter loss may be at least as great as the theoretical iV loss. In new transformers the eddy-current loss can only be estimated; in actu- ally built transformers it is calculated from the wattmeter reading on short circuit. CHAP. XI] REGULATION OF THE TRANSFORMER 111 this current. This e.m.f. must be balanced by part of the applied primary voltage, so that either this voltage or the induced e.m.f. EH must be different from that in an ideal transformer. The effect of the secondary leakage reactance is similar, in that it absorbs part of the secondary induced voltage E i2 , and makes the secondary terminal voltage E z different from that in an ideal transformer. It is shown below that, with a load of lag- ging power-factor, the reactive drop in both windings lowers the secondary terminal voltage. With a leading secondary current, the reactive drop is hi such a phase position as to raise the sec- ondary voltage. For purposes of computation, the transformer windings are assumed to produce no magnetic leakage fluxes, but imaginary reactance coils are connected in series with the wind- ings (Fig. 39). The reactances x\ and Xz of these coils are such as to cause the same reactive voltage drop as that due to the actual leakage fluxes in the transformer. 1 (c) The Exciting Admittance. Having thus made the windings of the transformer perfect by placing their impedances outside, we still have the problem of making the magnetic circuit ideal also. As stated before, the primary and secondary ampere-turns are not quite equal, because of a certain number of ampere-turns necessary to magnetize the iron. This means that a current must flow through the primary winding even when the secondary circuit is open. This current is called the no-load or magnetizing current of the transformer. Its amount depends upon the reluct- ance of the magnetic circuit and upon the core loss (hysteresis and eddy currents). For purposes of computation the iron core may be assumed to be of zero reluctance, and to have no core loss; but we may imagine a fictitious or equivalent susceptance 6 and a conductance g (Fig. 39) connected across the primary winding to draw a current equal hi phase and magnitude to the exciting current of the transformer. Both go and 6 are shown connected across the induced voltage EH, because both the mag- netizing current and the core loss depend upon the value of the flux and consequently upon the value of En, which is proportional to the flux. Let the calculated or measured core loss be equal to Po watts; then g is determined from the equation P = En*go (182) 1 For further details in regard to the leakage reactance of transformers, see the author's Magnetic Circuit, Art. 64. 112 THE ELECTRIC CIRCUIT [ART. 39 The pure magnetizing current, without the core-loss component, is in phase with the flux which it produces, and therefore is in quadrature with the induced voltage En. For this reason, it is represented as flowing through a pure susceptance. Knowing the pure magnetizing current IQ ', the susceptance b is determined from the equation l bo = I 7En (183) Neither the core-loss component nor the pure magnetizing current are proportional to the flux or to the voltage EH, so that strictly speaking both b and g are functions of the counter-e.m.f. EH. However, in practice, EH varies so little with the load that it is admissible to assume g and &o to be constant quantities. More- over, the influence of the magnetizing current upon the voltage regulation is negligible in most cases. The magnetizing current is mentioned here only for the sake of completeness, so as to make the transformer core absolutely perfect. We shall see in the next chapter that 6 and g are of considerable importance in the per- formance of the induction motor. Thus, by the foregoing reasoning, both the core and the wind- ings of the transformer are made ideal, and all the imperfections are replaced by external resistances and reactances. Having done this, the performance of a transformer can be readily treated either graphically or analytically, as explained below. Prob. 1. Draw a diagram similar to Fig. 39 for a transformer with several secondary windings supplying independent load circuits. Prob. 2. Draw a diagram similar to Fig. 39 for an auto-transformer. 39. The Vector Diagram of a Transformer. Having re- duced the transformer to an equivalent electric circuit (with a perfect magnetic link), the current and voltage relations at a certain load may be represented by a vector diagram (Fig. 40). In order to make the relations clearer, the voltage drop and the losses are greatly exaggerated. For this reason, the graphical treatment is more suitable for purposes of explanation than for numerical computations. For actual calculations the analytical method given in the next article is preferable. The calculation of the core loss and of the magnetizing current belongs properly to the theory of magnetic phenomena, and is treated in detail in the author's Magnetic Circuit, Arts. 19, 33, and 34. Here the values of P and /o' are supposed to be known. CHAP. XI] REGULATION OF THE TRANSFORMER 113 Let the secondary terminal voltage and the load be given, so that the vectors E 2 and 7 2 can be drawn in magnitude and rela- tive phase position. The secondary induced e.m.f. E i2 is found FIG. 40. The vector diagram of a transformer. by adding to E 2 the ohmic drop 7 2 r 2 in phase with 7 2 , and the reactive drop I 2 Xz in leading quadrature with 7 2 . The primary induced voltage En is in phase with E iz , because both are induced by the same magnetic flux. The magnitudes of the two voltages are as the respective numbers of turns; see 114 THE ELECTRIC CIRCUIT [ART. 39 eq. (180). The vector marked En in Fig. 40 is in reality equal and opjx>site to En, and represents the part of the primary terminal voltage that balances EH. Without the primary drop, the total applied primary voltage would be equal to En. But, on account of the primary drop in the transformer, the applied voltage Ei is obtained by adding to En the resistance drop /ir x in phase with the primary current /i and the reactive drop IiXi in leading quadrature with I\. In order to be able to construct the vectors /in and IiXi it is necessary to know the vector of the primary current /i in magnitude and phase position. In an ideal transformer, the primary current is in exact phase opposition to the secondary current, and the ratio of the two currents is inversely as the ratio of the respective numbers of turns; see eq. (181). In the actual transformer, the primary current, in addition to this component 7 2 (nz/ni) " transmitted into the secondary," has a magnetizing component 7 , which serves to maintain the alternating flux in the core, and which is not transmitted into the secondary circuit. The total primary current is the geometric sum of the two com- ponents, and can be constructed if the magnetizing current 7 is known. The magnetizing current itself consists of two components, as explained in the preceding article, under (c). One component, / t /dt =-2 ^ m rh. cos 2 irft, . . (185) the second sine-wave lagging by 90 degrees behind the first. CHAP. XI] REGULATION OF THE TRANSFORMER 115 It is assumed in the construction of Fig. 40 that the primary and secondary inductive drops can be calculated separately. Such is, however, not the case with our present state of knowledge; both theory and experiment enable us to determine only the total re- active drop, including primary and secondary. Therefore, when it is desired to use the vector diagram for actual computations, it is customary to ascribe one half of the Ix drop to the primary and the other half to the secondary circuit. Usually, the magnetizing component of the primary current can be neglected; then it does not make any difference how the inductive drop is distributed. It will be shown in the next article that in such case the voltage regulation depends only upon the total impedance drop, either calculated or determined from a short-circuit test. When the internal voltage drop is given in per cent, it is understood to refer to the no-load voltage of each particular circuit. For instance, if the reactive voltage drop in a 20/1-kv. transformer is said to be 5 per cent, this means that the secondary drop is 2.5 per cent of 1000 volts, or is equal to 25 volts, and that the primary drop is 2.5 per cent of 20,000 volts, or is equal to 500 volts. Prob. 1. What is the regulation of a 600-kw., 2200/220- volt, 25-cycle transformer at the rated current and at 80 per cent power-factor (lag- ging)? The total reactive drop is 10 per cent, the primary ohmic drop is 2.2 per cent, and the secondary ohmic drop 2.8 per cent. The mag- netizing current may be neglected. 1 Ans. 10.1 per cent. Prob. 2. Determine the per cent voltage regulation of the trans- former specified in the preceding problem at the rated load and at 80 per cent power-factor, leading. Ans. 1.4 per cent. The negative sign indicates a rise in secondary voltage, instead of a drop. Prob. 3. Correct the vector diagram of problem 1 for the magnetiz- ing current, knowing that the core loss amounts to 20 kw., and that 8500 effective ampere-turns are necessary to maintain the flux, without the iron loss. The number of turns in the secondary winding is 64. Prob. 4. Adapt the diagram shown in Fig. 40 to an auto-transformer. 40. Analytical Determination of Voltage Regulation. Ap- proximate Solution. As explained above, it is preferable to calculate the voltage regulation of a transformer analytically, 1 An excessive internal drop is selected purposely to enable the student to construct an accurate vector diagram to a convenient scale. The losses and the magnetizing current in problem 3 also are too high for a standard transformer. 116 THE ELECTRIC CIRCUIT [ART. 40 because the vectors of voltage drop are very small as compared with those of the primary and secondary voltages. The relations shown in Figs. 39 and 40 are expressed analytically by the equations E t =En-!&, (186) E^Eu+ItZi (187) Since our purpose is to find the relation between E t and E%, it is necessary to eliminate from these equations EH and E i2 . The relation between EH and E iz is given by eq. (180); therefore, we multiply eq. (186) by wi/n a and subtract it from eq. (187). The result is E l - (ni/n 2 )# 2 = IiZj. + (wi/n 2 )/ 2 Z 2 . . . (188) The correct relation between /i and 7 2 is (Fig. 39) 1 1 = 1 2 (n 2 /ni) + I = IL+IO, . . . (189) where /L = /t(ni/nO (190) is the primary load current, or that part of the primary current which is transmitted into the secondary circuit. In a great majority of practical cases the magnetizing current is only a few per cent of the total primary current at the rated load. The voltage drop in the primary winding is also but a few per cent of the line voltage EI. For these reasons, it is permissible in Fig. 39 to transfer the exciting admittance Y from the place MN to the primary terminals AB. The voltage drop in the transformer is then caused only by the load current, so that for the purpose of calculating regulation we may use the approximate relation /, = /L=/(n,/n 1 ) (191) Substituting for /i and 7 2 their values from eq. (191) in terms of I L, we finally obtain ^i-^L=/L[Zi + (ni/n0^d (192) In this equation, the quantity E L = (ni/n 2 )# 2 (193) is called the primary load voltage, or the secondary terminal volt- age reduced to the primary circuit. The expression (ni/n 2 ) 2 Z 2 is called the secondary impedance reduced to, or transferred into, the primary circuit. The quantity Z = Z l + (n!/n 2 ) 2 Z 8 (194) CHAP. XI] REGULATION OF THE TRANSFORMER 117 is called the total or equivalent impedance of the transformer reduced to the primary circuit. Using in eq. (192) the abbreviated notation introduced in eq. (194), we get E 1 -E L =I L Z (195) Equation (195) corresponds to the simplified equivalent diagram of the transformer shown in Fig. 41. This diagram differs from Fig. 39 in two respects: (1) The magnetic link is omitted, the primary circuit being connected directly to the modified second- ary circuit; (2) the exciting admittance is connected across the primary terminal voltage instead of across the induced voltage. The latter change makes the equivalent diagram only approxi- mately correct, but simplifies computations greatly. Equation (195) is identical in form with eq. (159), Art. 33, for the voltage drop in a transmission line; both are solved," and the per cent voltage drop determined, in the same way. In fact, without the exciting admittance Y , the equivalent diagram shown in Fig. 41 reduces the performance of a transformer to that of a transmission line. Expression (194) for the equivalent impedance shows that resistances and reactances can be transferred from the secondary B N D FIG. 41. The approximately equivalent diagram of a transformer or an induction motor. circuit into the primary, and vice versa, by multiplying them by the square of the ratio of the numbers of turns. For instance, in a 10,000/1000-volt transformer, a 1-ohm resistance in the low- tension circuit causes the same per cent voltage drop as a 100- ohm resistance in the high-tension circuit. This is easily verified as follows: Let the current in the low-tension circuit be 20 amp.; then in the high-tension circuit the current will be 2 amp. The drop in the 1-ohm resistance is 20 volts, or 2 per cent of the secondary voltage. The drop in the 100-ohm resistance is 200 volts, which is 2 per cent of the primary voltage. In other words, the same reduction in the load voltage will be produced by using 118 THE ELECTRIC CIRCUIT [ART. 40 cither a resistance of one ohm in the secondary circuit or 100 ohms in the primary circuit. The secondary resistance r 2 transferred into the primary cir- cuit is denoted in Fig. 41 by r 2 ', where r,' = r, (m/nO 1 ....... (196) C or respo ndingly a^aiCni/n,) 2 ....... (196a) The equivalent impedance Z consists of the quadrature sum of the equivalent resistance .... (197) and the equivalent reactance x = x, + xz' = X! + (ni/n 2 ) 2 :r 2 .... (197a) The equivalent resistance is easily calculated, knowing the re- sistances of the two windings and the voltage ratio of the trans- former. Or else it is calculated directly from the i 2 r loss measured by a wattmeter in a short circuit test. The equivalent reactance is calculated from the terminal voltage in the short-circuit test, making a proper allowance for the known resistance drop. To illustrate, when the secondary circuit is short-circuited, eq. (195) becomes E l = I L Z ........ (198) Ei and IL are measured directly, so that Z can be calculated. Knowing the equivalent resistance r = P/Ii?, the reactance is calculated from the expression x = Vz 2 r 2 . For a new trans- former, the total equivalent leakage inductance is estimated with sufficient accuracy by means of various semi-empirical formulae; l or else the total impedance drop I L Z is taken as a certain per- centage of the rated voltage, from previous experience with similar transformers. Prob. 1. Check analytically the answers to problems 1 and 2 in the preceding article. Prob. 2. The high-tension winding of a 2000-kva., 33/1 1-kv. trans- former was short-circuited, and the voltage on the low-tension side ad- justed so as to circulate the rated current through the windings. The instrument readings were 470 volts and 30 kw. Calculate the per cent ohmic and reactive drops in the transformer. Ans. 1.5 and 4 per cent. Prob. 3. Deduce a formula similar to (192), but referring to the secondary circuit. 1 See for instance the author's Magnetic Circuit, Art. 64. CHAP. XI] REGULATION OF THE TRANSFORMER 119 Prob. 4. Show how the voltage regulation of a transformer can be estimated, using Mershon's diagram given in various electrical handbooks and pocketbooks. Prob. 5. The primary voltage of a given transformer is kept constant at a known value. Determine the percentage internal drop (Ei EL)/ Ei for a given impedance of the load. 1 Solution: Let the load impedance, reduced to the primary circuit, be ZL', then the load current is IL = Ei/(Z L + Z), where Z is the equivalent impedance of the transformer itself, supposed to be known. The load voltage is E L = E,- ZI L = EiZ L /(Z L + Z) = #!/[! + (Z/Z L )]. Having expressed all the known and unknown quantities in the complex form, in either Cartesian or polar coordinates, the magnitude and direc- tion of EL can be determined, by using the general method, i.e., equating the real and the imaginary parts on both sides of the equation. Prob. 6. The equation for EL given in the preceding problem leads to involved numerical computations. Moreover, the difference EI EL cannot be accurately determined in this way when EL differs but little from #1. Show how to simplify the numerical work, by taking advan- tage of the fact that Z is small compared with ZL- Solution : When a quantity a is small compared to unity, we have by division 1/(1 + a) = 1 a + a 2 etc. We have accordingly E L = E 1 [l- (Z/Z L )] approximately, or EL = Ei - E^/ZL ..... . ..... (A) Let EL be the vector of reference; consequently EI = EI (cos 9 + j sin 0). Let also ZL = ZL (cos $L + j sin + j sin <). Then according to eqs. (154) and (156), E.Z/ZL = E.Z/ZL [cos (e + 4> - to) + j sin (e + - to)], or, denoting EiZ/Zi by A^i and L by 0, we have A#j = A#! [cos (e + 0) + j sin (e + 0)]. Equation (A) may now be written in the form Ei (cos e + j sin e} = EL + A#i [cos (0 + 0) + j sin (e + ft)], where is a known quantity, as well as the angle /3 = # $L. Separating the real and the imaginary parts, we get E 1 coso = E L + &EiCos(e + p), ......... (B) Ei sin e = &Ei sin (0 + 0) = &Ei sin cos + A#i cos 6 sin 0. (C) 1 The conditions in this problem differ from those in the text above in two respects: (1) The primary voltage is given instead of the secondary; (2) the load is given by its impedance instead of the current and power- factor. 120 THE ELECTRIC CIRCUIT [AKT. 41 From eq. (C), dividing both sides by cos 6, we find tan = AEi sin P/(Ei - A#i cos 0), . . . . (D) from which can be calculated. Using in eq. (B) the transformation cos e = 1 2 sin 2 e, the same as in Art. 33, we get, after division by Ei, (Ei - E L ) /EL = ( A#,/#,) cos (e + 0) + 2 sin 2 J 0. . . (E) While the derivation of formulae (D) and (E) may seem somewhat tedi- ous, the results are in the form most convenient for numerical work. 41. Analytical Determination of Voltage Regulation. Exact Solution. 1 The approximation made in the preceding article consists in shifting the exciting admittance Y so that it is con- nected across the primary terminal voltage EI, instead of across the primary induced voltage EH (compare Figs. 39 and 41). Retaining the exciting admittance in its correct place, we obtain the equivalent diagram shown in Fig. 42. The secondary im- B N FIG. 42. The correct equivalent diagram of a transformer or an induction motor. pedance is reduced to the primary circuit as before, by being multiplied by the square of the ratio of turns (ni/n 2 ) 2 . This pro- cedure is strictly correct, the magnetic link being by assumption perfect. Equations (186) and (187) hold here as before, but instead of using the approximate eq. (191) we shall use the correct rela- tion (189). The magnetizing current is Jo=#iiF , (199) so that the total primary current is /!= /L+^-iFo (200) Expressing 1^ and 7 2 in eqs. (186) and (187) through the load current / L , and eliminating EH and E i2 as before, we obtain $1 ~ Z!/ L )/(I + ZiFo) = E L + I L (m/n,) 2 Z,. (201) This equation takes the place of the approximate eq. (192). The two equations become identical when F = 0. 1 This article may be omitted if desired. CHAP. XI] REGULATION OF THE TRANSFORMER 121 The complex quantity 1 + ZiY which enters into eq. (201) may be called a correction factor, and may be represented in the form K= l+Z 1 F = A;(cosa+jsina). . . . (202) Since the ratio of two complex quantities is also a complex quan- tity, eq. (201) may be expressed in the form E cl =E L + I L Z c , (203) where the corrected primary voltage Ed = Ei/K = (Ei/k) [cos (6 - a) + j sin (8 - a)], (204) and the corrected equivalent impedance Z c = Zi/K + (n./n^Z, = z c (cos ^ + j sin ^) . . (205) Equation (203) is of the same standard form as eqs. (195) and (159), and can be solved by the method given in Art. 33. l This problem can be solved also by keeping the complex quan- tities in the orthogonal form. The student will profit by working out the details for himself. Sometimes it is desired to know the voltage regulation of a trans- former over a certain range of loads, the results being represented in the form of a curve. In such a case, it makes no difference for which particular loads the regulation is actually calculated, provided that these loads are selected within certain limits. If the primary voltage is given and is constant, it may be more convenient to perform the calculations (according to Fig. 42), not for an assumed current /2, but for an assumed load imped- ance Z L . Combining the impedances in series and the admit- tances in parallel, the whole circuit connected at the primary terminals is finally reduced to one impedance. Dividing the primary voltage by this impedance gives the primary current, and consequently the drop in the primary impedance Zi. Thus, the voltage EH becomes known, and the current 7 can be calculated. After this, the current I L and the drop / z,Z 2 are calculated. This drop, being subtracted from En, gives the desired secondary voltage Ei, reduced to the primary circuit. 1 Namely, when I L = 0, E L = E c i, so that per cent regulation is equal to lQO&E/(E c i &E), where AE = E c i E L (algebraically, not geometrically). Consequently, eq. (166) and the table on page 96 are directly applicable. CHAPTER XII PERFORMANCE CHARACTERISTICS OF THE INDUCTION MOTOR 42. The Equivalent Electrical Diagram of an Induction Mo- tor. The student is supposed to be familiar with the general (quali- tative) explanation of the performance of a polyphase induction motor, and with the general shape of the load characteristics. 1 It will be shown here how to predetermine the performance char- acteristics of a given induction motor by reducing it to an equiva- lent electric circuit, similar to that of a transformer. The following experiment shows the possibility of such an equivalent diagram. A brake test is performed on the motor, and the primary current and the power-factor are plotted against the output as abscissae. Then the rotor is blocked, and variable non-inductive resistances are inserted into its phase windings. If the rotor has a squirrel-cage secondary, resistances must be inserted in series with each bar, or into each section of the end- rings between consecutive bars. The motor is thus reduced to a polyphase transformer, the inserted secondary resistances repre- senting the load. A load test is performed on this transformer, and the curves of primary current and power-factor are plotted against the total fir loss in the external resistances. These curves are found to coincide very closely with the curves obtained from the brake test, provided that the brake power and the fir power are plotted to the same scale, one representing the mechanical, the other the corresponding electrical output. Some difference in the curves is due to the fact that the stationary transformer has no friction loss; this is, however, partly or wholly compen- sated by a greatly increased secondary core loss. The theoretical reasons for this equivalence of an induction motor to a polyphase transformer will become clear by consider- 1 See, for instance, the author's Experimental Electrical Engineering, Vol. 1, chap. 17. 122 CHAP. XII] THE INDUCTION MOTOR 123 ing the electrical relations in the rotor under the following three headings: (a) The Relationship between the External Resistance and the Slip. Let the input into the rotor be P watts per phase of the secondary winding, and let the motor be running at a slip s. For instance, s = 0.05 means that the speed of the rotor is 5 per cent lower than the synchronous speed, or the speed of the revolving field. Then, sP watts are converted into heat in each phase of the secondary winding, and (1 s)P watts are available on the shaft as the output (including friction and windage). This is because the tangential electromagnetic effort is the same on the surface of the stator as it is on the surface of the rotor. But while the gliding magnetic flux travels at synchronous speed, the rotor travels at (1 s) times the synchronous speed. The elec- tromagnetic coupling between the stator and the rotor is simi- lar to a friction coupling between two shafts, having a certain amount of slip. If the speed of the driven shaft is say 5 per cent below that of the driving shaft, on account of the slip in the coupling, 95 per cent of the power is transmitted and 5 per cent is lost in heat in the coupling. With the rotor blocked, let R be the external resistance per phase of the secondary, and let J 2 be the secondary current per phase. For a slip s we must have the condition s (R + r 2 ) 7 2 2 = r 2 / 2 2 , or s = r 2 /(fl + r 2 ) (206) If the slip is given, the required external resistance is R = r a (l - s)/s (207) (b) Equal Secondary Current and Phase Displacement with the Rotor Running or Blocked. Let the reactance of the secondary winding per phase be Xz ohms, at the primary or synchronous frequency. With the rotor running at a slip s, the frequency of the secondary currents is only equal to s times the primary fre- quency, so that the reactance per phase is sx^. Therefore, the phase displacement 2 between the induced secondary voltage and the current is determined by the relation tan 2 = sz 2 /r 2 (208) With the rotor blocked and provided with external resistances satisfying condition (206), the total resistance of the secondary 124 THE ELECTRIC CIRCUIT [ART. 42 circuit per phase is R + r 2 = r a /s. The secondary frequency is equal to that in the primary circuit, so that tan < 2 = Xt/(r 2 /s} = sx 2 /r z ; thus the phase displacement is the same as that given by eq. (208). With the same revolving magnetic flux in both cases, the currents are also equal. While with the stationary rotor the induced secondary e.m.f. is larger in the ratio of 1: s, because of a higher speed of cutting the secondary conductors, yet the total secondary resistance R + r 2 is also larger in the same ratio of 1 : s, according to cq. (206). The secondary reactance is also larger in the same ratio on account of the higher frequency. Thus, with the rotor blocked, both the e.m.f. and the impedance of the secondary circuit are larger in the ratio of 1 : s than when it is running at a slip s. Hence, the current, which is equal to the ratio of the e.m.f. to the impedance, is the same in both cases. (c) The Reaction of the Secondary upon the Primary Circuit is the Same with the Rotor Running or Blocked. The magneto- motive force of the revolving rotor is the same as that of the stationary rotor with the resistance R in series, provided that in both cases the magnetomotive force is considered with respect to the stationary primary circuit. In the latter case the frequency of the secondary currents is equal to that of the supply, so that the resultant magnetomotive force due to all the secondary phases travels in the air-gap at synchronous speed, the same as the resultant magnetomotive force of the primary currents. The two magnetomotive forces ^form one resultant magnetomotive force which produces the revolving flux. With the revolving rotor, the frequency of the secondary currents is s per cent of that of the supply, so that the secondary magnetomotive force glides relatively to the body of the rotor at a speed equal to s per cent of the synchronous speed. But the speed of the rotor itself is the (1 s) part of the synchronous speed. Hence, the velocity of the secondary magnetomotive force with respect to the stator is s + (1 ~ s) = 1, or is equal to the synchronous speed, and is the same as the velocity of the primary magnetomotive force. We have seen above that the secondary currents and their phase relation are the same in the two cases, so that the secondary magnetomotive force is also the same in phase and magnitude. Consequently, with the same flux, determined by the applied volt- CHAP. XII] THE INDUCTION MOTOR 125 age, the primary magnetomotive force is also the same in both cases. This means that the primary current and power-factor are the same with the stationary rotor loaded electrically as with the revolving rotor loaded mechanically. We have thus proved theoretically, as well as experimentally, that the performance of an induction motor may be reduced to that of a stationary transformer. But we know from the pre- ceding chapter that a transformer can in turn be replaced by an equivalent electric circuit, either approximately (Fig. 41) or accu- rately (Fig. 42). Thus, the same equivalent diagrams can be used in the predetermination of the performance of an induction motor. All the quantities which enter into these diagrams are understood to be per phase of the primary circuit (usually per phase of Y in a three-phase motor). When the number of the secondary phases and the method of connections are different from those in the primary circuit, the secondary winding is replaced by an equivalent one of the same number of phases, and with the same kind of connections as in the primary circuit; see Art. 45 below. 1 Prob. 1. Explain the principle of the speed control of an induction motor by means of adjustable external resistances in the secondary cir- cuit. Prob. 2. Explain the principle of the direct and differential cascade connection of two induction motors. Prob. 3. Show how an induction generator can be reduced to an equivalent electric circuit. 43. The Analytical Determination of Performance. Ap- proximate Solution. The problem is to calculate the perform- ance characteristics of a given induction motor in other words, against the output as abscissae, to plot the following curves; viz., primary amperes, kilowatts input, primary power-factor,' slip, torque, and efficiency. The resistances and the leakage reactances of both windings, reduced to the primary circuit, are supposed to be known, so that each primary phase of the motor can be replaced by either the approximate or the exact diagram (Figs. 41 and 42). The approximate diagram only is considered here, because it is sufficiently accurate for most practical purposes. The exact solu- tion is given in Art. 47 below. The iron loss, friction, and the 1 The values of the leakage reactances of the windings are supposed here to be known; for their calculation from the dimensions of the motor see the author's Magnetic Circuit, Art. 66. 126 THE ELECTRIC CIRCUIT [ART. 43 magnetizing current are also supposed to be known, so that the exciting admittance Y is known. While in reality it varies some- what with the load, in the approximate solution it is considered to be a constant quantity. The problem is solved similarly to that of the voltage regula- tion of a transmission line or of a transformer, treated above; that is, a load current I L is selected, and the circuit is solved in the complex notation. In order to make the treatment independ- ent of the other chapters, a complete solution is given below, with some minor changes which simplify the numerical work. As in the transmission line and in the transformer, we have Ei=*E L + ZI Lt (209) or, expanded, Ei (cos + j sin 0) = E L + I L (r + jx) . . . (210) Here the direction of the unknown load voltage EL is again selected as the reference axis. The current IL is in phase with EL, because by assumption the external resistance R is non-inductive. Sepa- rating the real and the imaginary parts, we get Ei cos 9 = E L + I L r; (211) Ei sin 6 = I L x (212) When plotting the curves, it is immaterial which values of the load are selected for computation. We assume, therefore, a series of reasonable values for I L , and from eq. (212) calculate the corre- sponding values of sin 6. Then from eq. (211) we find the values of EL, and finally determine the outputs per phase from the equation PL = ! L E L (213) Knowing I L , E L , and the angle 6, the rest of the values for the performance curves are calculated as follows: (a) The Slip. The external resistance, reduced to the primary circuit, is R' = E L /I L - (214) and the slip is found from the equation s = iV/GR' + r,'), (214a) which is identical with eq. (206), except that r 2 ' and R' are second- ary quantities reduced to the primary circuit. (b) The Primary Current and Power-factor. The total primary current per phase is !i= !*+!L, (215) CHAP. XII] THE INDUCTION MOTOR 127 where the magnetizing current / is known, and can be repre- sented with respect to the terminal voltage E : as /o = / (cos o jsin^o) ..... (216) The load current, in its phase relation with respect to the terminal voltage, is I L = I L (cos9- jsinfl), ..... (217) so that !i = ii-jii = (!L cos 0+/o cos ) J(!L sin 0+/o sin ). (218) Knowing the projections ii and i\ of the primary current with respect to the terminal voltage E\, the primary phase angle 0i is found from the equation tan 0i = *YAi, ...... (219) and then the primary power-factor, cos 0i, is taken from a trigo- nometric table. The current itself, 7i = ii/cos0i ....... (220) (c) The input per phase is Pi = Eil i cos 0i = Eiii ..... (221) The efficiency is equal to the ratio of the output to the input. (d) The useful torque in synchronous watts, or the input into the secondary, is equal to the output plus the secondary copper loss. The tangential effort per phase, in kilograms at a radius of one meter, or the torque per phase, in kg.-meters is T = 973.8 (P L + 0.001 7 L V 2 ')/(synchr. r.p.m.), . (222) PL being expressed in kilowatts, so as to avoid large numbers in numerical applications. Sometimes the performance data are desired for one particular load, PL, only; for instance, at the rated output of the machine. The method outlined above may in this case be somewhat tedious, because one has to find by trials the proper values of IL and EL which give the desired output. It may lead more quickly to the desired end to solve eqs. (211), (212), and (213) as three simul- taneous equations for the unknown quantities EL, IL, and 6. Squaring the first two equations and adding them together, the angle 6 is eliminated, and we get Substituting for EL its value from eq. (213), gives a quadratic equation for z/L 2 , namely (zI L *Y -2 (z7 L 2 ) (i tfi 2 - P L r}/z + PL* = 0. . . (223) ^4. CVt 128 THE ELECTRIC CIRCUIT [ART. 43 The solution of this equation is 2/L 2 = (i #i 2 - PiA/z - (l Ei* -Pir)*/* -Pi*. The minus sign only is retained before the radical, because it gives a smaller current. It can be shown that the solution with the plus sign corresponds to the unstable region of operation of the motor. For numerical computations the preceding equation is put in the form zlj} = 1000 (Q - VQ2 _ p L 2), . . . (224) where, for the sake of brevity, we introduce the notation, Q= (5QOES- P L r)/z (225) In the last two equations P L and Q are in kilowatts, so as to avoid large numbers, and EI is in kilovolts. The student is reminded that EI is the phase or star voltage, and not the line voltage, and that PL is the output per phase. When PL is small compared to Q, formula (224) represents the difference of two quantities of nearly equal value. The result is inaccurate, and it is better to expand the expression [1 (Pz,/Q) 2 ]*> according to the binomial theorem. This gives zlj} = 1000 Q ft (P L /QY + i (PL/QY + ^ (PL/QY + etc.]. (226) The latter formula is much more convenient for numerical applica- tions than eq. (224), because the second term in the brackets is small as compared to the first, and the third term can usually be neglected. Knowing IL, the rest of the values are determined as before. Prob. 1. Plot complete performance curves of a three-phase, 25- cycle, 150-kw., 6-pole, 2200-volt, induction motor between no load and 25 per cent overload, from the following data: Total no-load input (for all three phases) is 10.5 kw.; the no-load current per phase of the line is 13 amp. With the armature blocked, the input is 230 kw., the cur- rent per phase being 227 amp. 1 The resistance of the primary winding per phase of Y is 0.60 ohm. Hint: Follow consistently the approximate diagram, Fig. 41; that is, do not correct the no-load reading for the primary i-r loss, and assume the magnetizing current with the armature locked to be the same as at no load. 1 The data with the armature locked refer to the rated voltage; they are obtained by extrapolating the curves taken at lower voltages. It would not be practicable to apply the full line voltage to a large motor with the armature blocked. CHAP. XII] THE INDUCTION MOTOR 129 ABS. At the rated output the primary current is 49 amp.; the power-factor is 90.8 per cent; the slip, 3.5 per cent; the efficiency, 88.5 per cent; and the torque, 302 kg.-m. Prob. 2. Check the answer to the preceding problem, using eq. (226). Prob. 3. Extend the theory and the formulae given above to the performance characteristics of an induction generator. 44. Starting Torque, Pull-out Torque, and Maximum Out- put. When judging the performance of a given induction motor, or designing a new motor, the following features are of importance : (a) The starting torque, either in its absolute value, or in its ratio to the torque at the rated load. If the motor is to be started by means of resistances in the secondary circuit, one may be required to. calculate the values of these resistances necessary for a prescribed starting torque, or for a maximum starting torque. (b) The pull-out torque, or the torque at which the motor reaches the limit of stable operation, and comes to a stop. This torque is usually given through its ratio to the full-load torque. (c) The maximum output of the motor, in kilowatts. This output takes place at a smaller slip than that at which the motor pulls out. The output is a maximum when the product of torque times speed is a maximum, but not when the torque itself is greatest. The three^ quantities mentioned above can be determined by using the equations deduced in the foregoing article. (a) The Starting Torque. In the general formula (222), P L = at start, because the motor supplies no mechanical output, the speed being equal to zero. In the equivalent electrical diagram (Fig. 41) this corresponds to a short curcuit of the load, or R = 0. Hence, IL = E\/z; substituting this value into eq. (222), we find that the starting torque per phase, in kg.- m. T 8t = 0.9738 #i 2 r 2 '/ (z 2 X synchr. r.p.m.). (227) It will be seen from this expression that the starting torque is proportional to the square of the line voltage. This fact permits one to determine the initial torque when starting a motor on a lower voltage, by means of auto-transformers. The same equa- tion shows that the starting torque increases with the secondary resistance. It is not quite proportional to it, because r 2 ' is also implicitly contained in z. 130 THE ELECTRIC CIRCUIT [ART. 44 When the motor has a phase-wound secondary and is started by means of resistances in the secondary circuit, r* in formula (227) includes this secondary resistance. It is thus possible to calculate the external resistance required for a given starting torque. For example, when the starting torque is given, the ratio z 2 /r 2 ' in eq. (227) is a known quantity, or */r t ' = [(n + r/) 2 + *W = c, . . . (228) where c = 0.9738 #i 2 / (desired torque X synchr. r.p.m.). (229) Solving the quadratic (228) for r/, we obtain *). (230) In applications, the minus sign only is retained before the radical because one would naturally use the smaller of the two resistances which give the same torque. The value of r 2 ' determined from this expression comprises both the resistance per phase of the rotor proper and the starting resistance per phase, if any is used, both reduced to equivalent primary values. To obtain their actual values, see Art. 45 below. If the external resistance is to be so selected as to give a max- imum starting torque, c in expression (228) must be a minimum. Equating to zero the derivative of c with respect to r/, and solving for r 2 ', we get r,' = (z 2 + ri)*, (231) that is, r-i is very nearly equal to x. This value comprises the resistance of the rotor proper and the starting resistance, both per phase of the primary circuit. If a resistance is selected which is either less than or greater than that determined by eq. (231), the motor does not develop its full starting torque. This checks with eq. (230), which shows that the same torque can be obtained with two different values of starting resistance. (b) Pull-out Torque. According to eq. (222), the torque is a maximum when EJ L + IM = max (232) Here EL and IL are functions of the independent variable 6. Expressing them through from eqs. (211) and (212), and omit- ting the constant factor #i 2 , we obtain (1/z) sin 9 [cos - (r/z) sin 0] + (r 2 '/z 2 ) sin 2 = max., CHAP. XII] THE INDUCTION MOTOR 131 or, after simplification, x sin 2 - ri (1 - cos 26) = max. . . . (233) Equating to zero the derivative of this expression with respect to 6, gives x cos 2 6 - ri sin 2 6 = 0, or tan 2 6 = x/ri (234) Knowing 6, the values of EL and IL are calculated from eqs. (211) and (212), and then the torque is determined from eq. (222). It is of interest to note that the angle 0, at which the motor pulls out of step, is independent of the secondary resistance r 2 '. Neither does this resistance enter into eq. (233). Hence, the maximum torque which a motor is capable of developing is inde- pendent of its rotor resistance. This resistance determines only the speed at which the maximum torque takes place. The higher the secondary resistance, the lower the speed at which the motor pulls out of step. By using an external starting resistance, and a rotor winding of low resistance, two maxima of the torque are obtained, one at the start, with the external resistance in, and the other near synchronism, with it out. (c) Maximum Output. The problem is to find the values of EL and IL for which the product EL!L is a maximum. Again expressing E L and I L through the angle 6 from eqs. (211) and (212), and omitting the constant factor Ei*/x, we have sin 6 [cos (r/x) sin 6} = max. Equating to zero the derivative of this expression with respect to 0, gives * = **, (235) where tan = x/r. Knowing the angle 0, the values of IL and EL are calculated from eqs. (211) and (212), and then their product EL!L is determined. Prob. 1. The motor specified in problem 1 of the preceding article is designed to be started at a reduced voltage. What per cent tap should be used on the auto-transformers in order to get a starting torque of about 30 per cent of the full-load torque? Ans. 60 per cent of the line voltage. Prob. 2. The same motor is provided with a phase-wound secondary and is to be started by using resistances in series with the rotor windings. 132 THE ELECTRIC CIRCUIT [AaT. 44 What external resistance is necessary in order to obtain a starting torque equal to 1.5 times the full-load torque? Ans. 0.78 ohm per phase, in terms of the primary circuit. Prob. 3. What starting resistance in the preceding problem would give the maximum starting torque? Ans. About 5.7 ohms. Prob. 4. Show that the motor specified in problem 1 of the preceding article pulls out of step when the torque exceeds 2.45 times the rated full-load torque. Prob. 5. Check the answer to problem 4 by using the answer to prob- lem 3. Prob. 6. Show that the maximum output of the same motor is equal to 2.15 times the rated output. Prob. 7. Show that the input into an induction motor is a maximum when 6 = 45. Hint: IL cos 6 = max. Prob. 8. Show how to calculate the per cent slip at which the motor pulls out of step, and also the speed at which the output is a maximum. CHAPTER Xin PERFORMANCE CHARACTERISTICS OF THE INDUCTION MOTOR (Continued) 45. The Secondary Resistances and Reactances Reduced to the Primary Circuit. It is proved in Art. 40 that in a trans- former the secondary resistance and reactance can be transferred into the primary circuit by being multiplied by (rii/w 2 ) 2 . The same rule holds true for the induction motor, provided that the number of phases is the same in the primary and in the secondary windings, and that the two windings are of the same type (the same number of slots per phase and the same winding pitch). This is hardly ever the case, and with a different number of phases and different types of winding in the primary and second- ary, the following formula holds true : r 2 '/r 2 = (roi/rog) (fc 6i n 1 //c 62 n 2 ) 2 , .... (236) and analogously for the reactances, x 2 '/x z = (mi/m 2 ) (fcmi/fan,)*. . . . (237) In these expressions, m stands for the number of phases, n is the number of turns per phase, and fa is the so-called breadth factor which characterizes the winding. The subscripts 1 and 2 refer to the primary and secondary windings respectively. The quan- tities r 2 and z 2 are the actual resistance and reactance per phase of the secondary circuit; r/ and Xz f are the equivalent quantities per phase of the primary circuit. 1 When Wi = ra 2 and k bi = fc& 2 , the preceding formulae become identical with eqs. (196) and (196a) for the transformer. Equations (236) and (237) refer to the resistances and react- ances per phase, with the understanding that the windings of each phase are all in series, both in the stator and in the rotor; and that the connections are either both star, or both mesh, even if the number of phases be different. Otherwise, the actual con- 1 For a proof of these formulae, see the author's Magnetic Circuit, Art. 44; the values of kb will be found in Arts. 27 to 29 of the same book. 133 134 THE ELECTRIC CIRCUIT [AKT. 45 ncctions must be taken into consideration and the values of r 2 ' and x 2 ' further modified, keeping in mind the fact that the total tV loss must be the same in the equivalent winding as in the ac- tual one. As a simple illustration, let the stator be three- phase Y-connected, and the rotor three-phase delta-connected. If the equivalent secondary resistance calculated by means of eq. (236) is r/, then only ^ r 2 ' must be used in the equivalent diagram, per phase of Y. This is because the current per phase of Y is v3 times as large as that per phase of delta, hence, for the same &r loss, the resistance per phase of Y must be only one- third of that per phase of delta. The same relation holds true for reactances, because the stored electromagnetic energy is also FIG. 43. A squirrel-cage rotor and the FIG. 44. The vectorial relation between star resistances equivalent to the end the star and mesh currents in a sym- ring. metrical m-phase system. proportional to the square of the current (Art. 20), and the equivalent diagram must express correctly the cyclic exchange of energy between the primary and the secondary circuits. In the most general case, let an ra-phase symmetrical system be given, for instance a two-pole squirrel-cage rotor (Fig. 43), and let it be required to find the relation between the mesh resist- ances r m and the star resistances r, such that the i z r loss per phase shall be the same in both. The relation between the vectors of the star currents and those of the mesh currents is shown in Fig. 44, the star currents /, forming an m-sided polygon, and the mesh currents I m being the radii of the polygon. This diagram is correct because it satisfies the following conditions : (a) the star currents CHAP. XIII] THE INDUCTION MOTOR 135 are displaced in phase relatively to each other by equal angles of 2 TT/W, over the whole range of 2 TT; (b) the same is true for the mesh currents; (c) each star current is the geometric differ- ence of the two adjacent mesh currents; (d) the geometric sum of the star currents is equal to zero (Kirchhoff's first law). From the geometry of the figure we have H, = / m sin(7r/m) ...... (238) The condition 7 8 2 r g = 7 m 2 r m leads to the ratio m). . . . (230) A similar relation holds for the reactances. When m = 3, we find as before that r A /ry = 3. Equation (239) finds its practical application in the calcula- tion of the equivalent resistance and reactance of a squirrel-cage rotor. The sections of the end-rings between the bars are mesh- connected, while the bars themselves may be considered as parts of a star-connected w 2 -phase system, where w 2 is the number of bars per pair of poles. Let r& be the resistance of each bar, includ- ing the two contact resistances between the end-rings and the bar; let r r be the resistance of a section of an end-ring between two consecutive bars. The resistance of the rings can be replaced by added resistances in series with the bars, so as to change the connections to a pure star system (Fig. 43). According to eq. (239) we find that the new resistance per bar must be equal to r& + 2 r r /[4 sin 2 (7r/m 2 )]. If the motor has p poles, or \ p pairs of poles, there are \ p bars in parallel belonging to the same phase, so that the total resistance of the secondary winding per phase is only the 2/-p part of that of one bar. Hence, assuming that the primary winding is star-connected, and that all the coils in each phase are in series, the value of r 2 to be used in eq. (236) is . . . (240) Analogously, . . . (241) Since there is only one bar per phase, and one bar is equivalent to one-half of a turn, the value n 2 = ^ and fc& 2 = 1 must be used hi eqs. (236) and (237). Prob. 1. A two-phase induction motor has the primary winding ar- ranged for two independent phases; the secondary is three-phase Y-con- 136 THE ELECTRIC CIRCUIT [AiiT. 46 ncctcd. When the rotor is stationary and its circuits arc open, 440 volts impressed at the primary terminals produce 97 volts between the slip- rings. The calculated starting resistance, per phase of the primary circuit, is 14 ohms. What is the actual resistance to be used in series with the rotor windings? Hint: Consider the primary circuit as ajour-phase star-connected system, so that njtbi/njtbz = I X 440/(97/ V3). Ans. 0.34 ohm. Prob. 2. A six-pole, three-phase, Y-connected induction motor has a squirrel-cage rotor of 80 cm. diameter with 73 bars; the resistance of each bar is 120 microhms (including the contact resistance). In order to have a certain required torque and slip, the equivalent rotor resistance per phase of the primary circuit must be equal to 1.07 ohms. What must be the actual resistance of each end-ring per centimeter of its length? There are 100 turns per phase of the primary winding, and kb\ = 0.95; for the squirrel-cage winding k b2 is always equal to unity. Ans. 5.8 microhms. 46. The Circle Diagram. Let the values of primary current obtained from a brake test on an induction motor be plotted as vectors at proper phase angles with respect to the vector EI of the primary voltage (Fig. 45). The locus of the ends of the current vectors is found to be very nearly a semicircle. This i? i;/ FIG. 45. The circle diagram of an induction motor. locus, together with some auxiliary lines, is called the circle dia- gram or the Heyland diagram of the induction motor. A similar diagram holds true for the transformer, although it is hardly ever used in practice. The importance and the convenience of the circle diagram lie in the fact that the complete performance of an induction motor can be predicted if the diameter of the semicircle and its posi- CHAP. XIII] THE INDUCTION MOTOR 137 tion with respect to the voltage vector EI are known. The semi- circle is usually determined by the vector 7 of the no-load current, and the vector of the current I 8 obtained when the rotor is locked (the subscript s stands for starting or short circuit). At any other load, the extremity of the current vector 7i lies between those of 7 and 7 8 . The Heyland diagram is simply a graphic representation of the current and voltage relations in the approximately equivalent circuit diagram shown in Fig. 41. The exciting current and the no-load losses are assumed to be constant at all loads from no- load to standstill. The no-load current is resolved into a loss component / (5) The total impedance between the primary terminals "is Z eg = (TMN + n) + j (XMN + #1). (6) The corresponding admittance Y eq = g eq - jb eq is calcu- lated from eqs. (121) and (122). (7) The primary current is /i=#iF eg (245) (8) The voltage across MN #1 = tfi - /xZi = ^(1- ZiF.,). . . . (246) (9) The load current is I L = /!- 7 = /j-tfaFo, or, substituting the values of /i and EH from eqs. (245) and (246), IL = Ei [Y eq (1 + FoZi) - Fo]. . . . (247) (10) The load voltage EL = lift' (248) CHAP. XIII] THE INDUCTION MOTOR 141 The rest of the quantities are calculated in the same manner as in Arts. 43 and 44. In numerical work, it is convenient to take E\ along the ref- erence axis. Having determined the value of Y eq , computations are begun with the composite admittance in the brackets in eq. (247). Either the orthogonal expressions of the form r + jx or the polar expressions of the form z (cos + j sin ) may be used, according to one's preference or familiarity with one or the other form. The student ought to be familiar with both forms. The trigonometric form is convenient for multiplication and division, while the Cartesian form is preferable in addition and subtraction. It may be advisable to use both forms in the same problem. The calculator should avoid long algebraic expressions, jper- forming numerical operations step by step. Much time is saved by arranging the consecutive steps in a table, so as to repeat the same operations mechanically for different values of R'. An irregularity of the values in a column is a sure indication of a numerical error. Much time is also saved by intelligently discriminating be- tween the principal terms and small correction factors in an ex- pression. For instance, in eq. (247) Y eq is large as compared to Yo and to Y eq YoZi. It would be a waste of time to figure out the latter expression accurately, when, in all probability, the principal term will be affected only by its first significant figure. On the other hand, the principal term, Y eg , must be calculated to a degree of accuracy at least equal to that desired in the result, if not to a higher degree. Considerable skill, experience, and judg- ment are necessary to determine the proper accuracy of computa- tions in engineering problems. This is an art which grows by intelligent exercise, and it is never too early to begin practicing it. The rewards are time and mental energy saved for better things, while obtaining an accuracy which is commensurate with the desired result. 1 1 For a complete set of final formulae for induction motor characteristics, see Arnold's Wechselstromtechnik, Vol. 5, part 1 (1909), pp. 65-78. A very slight inaccuracy is introduced there in the beginning, by neglecting the imaginary part in a complex quantity. See also Dr. Steinmetz's Alternating- current Phenomena, under " Induction Motor." 142 THE ELECTRIC CIRCUIT [ART. 47 Prob. 1. Make out a table showing in detail the order of computa- tions for a complete set of performance characteristics of an induction motor, according to the method developed above. Prob. 2. Mark on the curve sheet obtained in problem 1, Art. 43, a few points determined according to the exact equivalent diagram, in order to see the inaccuracy resulting from the use of the approximate method. CHAPTER XIV THE DIELECTRIC CIRCUIT 48. The Electrostatic Field. 1 In the following discussion, it is assumed that the student knows the fundamental phenomena of electrostatics from his study of physics. The purpose of the treatment given here is to deduce the principal numerical relations which are of importance in electrical engineering. The electro- static field is considered in this book from Faraday's point of view, viz., as consisting of displacements of electricity, and stresses Battery Condenser FIG. 46. A plate condenser completing a direct- current circuit. in the dielectric. This is different from the older theory of the action of electric charges at a distance. Let a source E of continuous electromotive force (Fig. 46) be connected to two parallel metallic plates A and B, the combina- tion of which is commonly known as a condenser. Let the plates 1 See the footnote at the beginning of Chapter 3. 143 144 THE ELECTRIC CIRCUIT [ART. 48 be separated from each other by air, or by some other non-conduct- ing material. When the key K is pressed upwards, a certain quantity of electricity, Q, flows from the battery to the plate A, and the same quantity flows from the plate B back to the battery. This quantity can be measured by the ballistic galvanometer shown in the circuit. Within a very short time the difference of potential between the plates becomes equal and opposite to that of the battery and the flow of current stops. Since electricity behaves like an incompressible fluid, the same quantity, Q, is displaced through the whole circuit, including the layer of insulation or dielectric between the condenser plates. This displacement is accompanied by a stress in the dielectric, similar in some respects to a mechanical stress in an elastic body. The directions of the electric stress and of the lines of displacement of electricity through the air are shown in the figure by dotted lines. These stresses produce a counter-electromotive force, which finally balances that of the battery. When the key is opened, the condenser remains charged, since the stress and the displacement can be relieved only in a closed circuit. To dis- charge the condenser, its plates must be connected by a conductor; this is done by pressing the key down. The deflection of the ballistic galvanometer during the discharge is equal and opposite to that during the charge, and the electric energy stored in the condenser is dissipated by the current in the form of heat. The difference between a dielectric and a conductor is that the resistance of the former to the passage of electricity is of an elastic nature; that is, the stress can be relieved and the stored energy returned to the circuit. On the contrary, the resistance to the flow of electricity in a conductor is of the nature of friction. The energy is converted into Joulean heat and cannot be restored. The modern electronic theory of electricity is not sufficiently advanced at this writing to give a clear account of the true nature of these displacements and stresses in a dielectric. It is therefore preferable for our purposes not to specify the mechanism by which these stresses and displacements are produced. We shall simply assume, as a matter of fact, the structure of dielectrics to be such that an e.m.f. across a layer of such material produces a displace- ment of a certain quantity of electricity, which is proportional to the e.m.f. When the e.m.f. is removed and a closed circuit is provided, the stresses within the dielectric are relieved, and the CHAP. XIV] THE DIELECTRIC CIRCUIT 145 displacement disappears. The analogy to an elastic body sub- jected to external mechanical forces naturally suggests itself. Experiment shows that, with given metallic plates (Fig. 46) and the same applied e.m.f., the value of the electric displacement depends upon the nature of the dielectric. With solid and liquid insulating materials, such as glass, oil, mica, etc., the same e.m.f. produces larger displacements of electricity than with air as the. dielectric. These materials are therefore said to possess higher permittivity than the air (some writers use the word inductivity) . When an alternating voltage is applied at the terminals of a condenser, the displacement of electricity in the dielectric varies continually in its magnitude and periodically reverses its direc- tion; consequently, it gives rise to an alternating current in the conducting part of the circuit. This is called the charging or capacity current. This current leads the alternating voltage in phase by 90 degrees, as may be seen from the following consider- ations: When the voltage has reached its instantaneous maximum the charging current is zero, because at the crest of the wave the voltage and the displacement remain practically constant for a short period of time. As soon as the voltage begins to decrease, the current begins to flow in the direction opposite to that of the applied voltage, because the elastic reaction of the dielectric is now larger than the applied electromotive force. At any instant, the current, or the rate of flow of electricity, is proportional to the rate of change of the applied voltage. But if the applied voltage varies according to the sine law, the rate of variation is also represented by a sine function differing in phase by 90 de- grees from the original function, because d (sin x)/dx = cos x = sin (90+ x); see also Art. 66 below. That there must be a dis- placement of 90 degrees between the voltage and the current follows also directly from the assumed elastic structure of the dielectric. The energy is supposed to be periodically stored in the dielectric and given up again without any loss; hence, the average power must be zero, and the current must be reactive. 49. A Hydraulic Analogue to the Dielectric Circuit. The hydraulic analogue shown in Fig. 47 may assist the student in the understanding of the electrostatic circuit. A is a pump which corresponds to the source of electromotive force in Fig. 46. The pipes B and C represent the leads to the condenser, or the metallic parts of the circuit. The cylinder D corresponds to the condenser, 146 THE ELECTRIC CIRCUIT [ART. 49 and the elastic partition K is analogous to the dielectric. Let the pipes and the cylinders be filled with water, and let the piston in A be in its middle position, the partition K not being stressed. Let the stopcock M be open, and the stopcock N closed. When a Stop Cocks FIG. 47. A hydraulic analogue of a dielectric circuit. pull to the right is exerted upon the piston rod and it is forced to move, the water in the system is displaced, and the elastic partition K is strained, as shown in the figure. With a given pull, or a given electromotive force, the movement stops when the pull is balanced by the elastic reaction of the partition. The charge, or the total displacement, is represented by the amount of water shifted; it can be measured by the water-meter W, which thus takes the place of the ballistic galvanometer. If the pipes are frictionless, and the inertia of the piston and water is assumed negligible, the analogy can be followed still further; namely, the phase difference in time between the pull and the velocity of the water is equal to 90 degrees, the velocity leading the pull. Assuming the motion of the piston to be har- monic, the velocity of the flow of water is at its maximum when the piston is at the center of its stroke. The required pull is equal to zero at this moment, because the elastic partition is in its middle, or unstrained position. At the end of the stroke the velocity is zero, but the pull is at its maximum, because the partition is strained to its extreme position, and exerts its maxi- mum elastic reaction. Thus the pull lags behind the velocity. Substituting another partition, made of a more yielding CHAP. XIV] THE DIELECTRIC CIRCUIT 147 material (material possessing higher permittivity), a larger dis- placement is produced with the same pull; this corresponds to the case in which some solid or liquid dielectric is substituted for the air. Closing the stopcock M corresponds to breaking the electric circuit of the condenser. It will be seen from analogy that the condenser remains charged. To discharge the condenser, the stopcock N must be opened; this equalizes the pressure on both sides of the elastic partition. Since, in reality, water possesses some inertia, the partition does not stop in its middle position during the discharge, but the momentum of the water carries it beyond the center. The electromagnetic inertia of the electric current produces a similar effect, and we thus have a simple explanation of the oscillatory character of the electric discharge. During this discharge, the energy is alternately transformed into the potential energy of dielectric stress, and into kinetic energy of the magnetic field. The oscillations of the partition are gradually damped out by the frictional resistance of the pipes. In the electric circuit, oscillations are damped by the ohmic resistance of the conducting parts of the circuit. The student can follow this analogy still further, and exp^iin free electrical vibrations, current and voltage resonance, also the effect of a resistance in series and in parallel with a condenser, etc. 50. The Permittance and Elastance of Dielectric Paths. Let Q (Fig. 46) be the total displacement of electricity in the dielectric, measured in ampere-seconds or coulombs, and let E be the voltage impressed across the condenser or " permittor." Experiment shows that up to a certain limit Q is proportional to E; this is similar to the behavior of an elastic body, in which the strains are proportional to the applied forces until the limit of elasticity has been reached. Thus, we may write Q = CE, (249) where the coefficient of proportionality, C, is called the permittance of the condenser. The older name for C is electrostatic capacity. When E is in volts and Q in coulombs, permittance is measured in units called farads. A condenser has a permittance of one farad when a displacement of one coulomb is produced for each volt applied at its terminals. The farad being too large a unit for practical use, permittances are usually measured in micro- 148 THE ELECTRIC CIRCUIT [ART. 50 farads, one microfarad being equal to one millionth part of a farad. The larger the permittance of a condenser, the larger is the displacement of electricity with the same voltage; hence C is a measure of the ease with which an electric displacement can be produced in a given condenser. In this respect the concept of permittance is analogous to those of electric conductance and magnetic permeance. In some cases it is convenient to speak, not of the degree of ease, but of the difficulty with which an electric displacement can be produced in a given condenser. For this purpose, a coefficient of proportionality, the reciprocal of C, has to be used; and eq. (249) becomes E = SQ, (250) where S = C- 1 (251) is called the elastance of the condenser. Elastance is thus analo- gous to electric resistance and to magnetic reluctance. When permittance is measured in farads, the unit of elastance is the reciprocal of the farad, and may therefore be properly called the daraf. This is a name derived by spelling the word farad back- wards, that is, in the same way in which mho is derived from ohm. 1 A condenser has an elastance of one daraf when one volt of pres- sure is required for each coulomb of displacement within it. The farad being too large a unit for practical use, the daraf is con- sequently too small a unit. Therefore, in practice, elastances should be measured in megadarafs, one megadaraf (= 10 6 darafs) being the reciprocal of one microfarad. 2 When two or more permittances are connected electrically in parallel, the resultant permittance is larger than that of any of the component condensers, because a larger path is offered to 1 It may be of interest to mention in this connection a similar derivation of the name for a unit of magnetic reluctance. The henry being the natural unit of magnetic permeance (or inductance) in the ampere-ohm system, the author has proposed calling the corresponding unit of reluctance the yrneh, a word derived by spelling the word henry backwards. See his Magnetic Circuit, Art. 5. 2 For a complete rational nomenclature of electric and magnetic quanti- ties, see the table on page xii at the beginning of the book, and also the one in the Appendix. CHAP. XIV] THE DIELECTRIC CIRCUIT 149 the displacement. The relation is similar to that of conductances or permeances in parallel. Let d, C 2 , etc., represent permit- tances connected in parallel across a source of constant voltage E, and let Qi, Q 2 , etc., be the corresponding electric displacements through these condensers (or permittors). Then, according to the definition of permittance, we have Qi = C,E (252) The equivalent permittance, C eq , must be such as to allow of a displacement equal to the sum of the partial displacements, with the same voltage; hence, SQ = C eq E (253) Adding eqs. (252) together gives Qi + Q 2 + etc. = E (Ci + C z + etc.), or, by comparison with eq. (253), C., = 2C (254) In other words, when permittances are connected in parallel, the equivalent permittance is equal to their sum. When condensers (or elastors) are connected in series, it is more convenient to use their elastances. Since electricity behaves like an incompressible fluid, the displacement through several elastances in series is the same in all of them. Let this displace- ment be denoted by Q, and let the voltages across the terminals of the individual elastors be E if E z , etc. Then, (255) where Si, S z , etc., are the elastances of the separate condensers. The equivalent elastance must allow of the same displacement Q with the same total voltage, or 2E = S eg Q. ....... (256) Adding eqs. (255) together gives Ei + E z + etc. = Q (Si + S 2 + etc.), 150 THE ELECTRIC CIRCUIT [Aux. 51 or, by comparison with eq. (256), S eq = 2S. . (257) In other words, when elastances are connected in series, the equivalent elastance is equal to their sum. The analogy to the addition of conductances in parallel and resistances in series is self-evident (see Art. 3). Prob. 1. A condenser, which has a permittance of 10 microfarads, is connected to a direct-current magneto, the speed of which is increased at a uniform rate, so that the voltage rises at a rate of 1.7 volts per second. Calculate the charging current. Ans. 17 microamperes. Note: This is the principle of an appara- tus used for measuring the acceleration of railway trains. Prob. 2. An elastance of 10 kilodarafs is connected across a 220- volt, 50-cycle line. Show that the effective value of the charging current is 6.91 amp. Solution: The maximum displacement in the dielectric is 220 V2/(10 X 10 3 ) = 22 V2 X 10~ 3 coulombs. This displacement is reduced to zero within aU of a second; hence, the average charging current is 4.4 \/2 amp. The effective value, assuming a sine-wave of current, is 4.4 v 7 2 X (-**/ V2) - 6.91 amp. Prob. 3. Show that with two condensers in parallel the ratio of the displacements equals that of the permittances or is inversely as the ratio of the elastances. What is the analogous relation for conductances and resistances? Prob. 4. When two condensers are in series, show that the ratio of the voltage drops across them equals that of the elastances, or is inversely as the ratio of the permittances. What is the analogous relation for resistances and conductances? Prob. 6. A sectionalized condenser, such as is used for calibration and exact measurements, is built up of the following permittances: 0.5, 0.2, 0.2, 0.05, and 0.05 microfarads. What is the extreme range of permittances and elastances possible by combining these sections in series and in parallel? Ans. From 1 to 0.0192 mf., or from 1 to 52 mgd. Prob. 6. Referring to the preceding problem, the sections of the con- denser are connected as follows: 0.2, 0.05, and 0.05 mf. are in scries, and the combination is shunted by 0.2 mf. Then the whole is put in series with 0.5 mf. Show that the resultant permittance is equal to 0.154 microfarads. 51. Permittivity and Elastivity of Dielectrics. Experiment shows that the permittance of a sample of any dielectric varies with its dimensions in the same way that the conductance of a metal or the permeance of a magnetic path in a non-ferrous medium does; namely, the permittance is proportional to the cross-sec- tion of the layer and inversely proportional to its length in the CHAP. XIV] THE DIELECTRIC CIRCUIT 151 direction of the lines of force. By increasing the cross-section of the path perpendicular to the lines of force (Fig. 46), the displace- ment is increased in the same proportion. On the other hand, the displacement is found to be inversely proportional to the thickness of the dielectric, since the distance through which the voltage must act is greater if the thickness is increased. These relations follow directly from the laws deduced in the preceding article for the addition of permittances in parallel and elastances in series. Thus, by analogy with eq. (21), Art. 5, we put C = KA/l, (258) where K is called the permittivity of the dielectric. It is analogous to the conductivity of a conducting material, or to the perme- ability of a magnetic medium. Permittivity may be defined as the permittance of a cubic unit of dielectric, when the lines of displacement are straight lines perpendicular to one of its faces. For air the permittivity is K = 0.08842 X 10" 6 microfarads per cm. cube. . (259) For other dielectrics, liquid and solid, the permittivity is higher than that of air; that is to say, they are more yielding to an electro- motive force. It is convenient to express their permittivities in terms of that of the air; for instance, we may say that the per- mittivity of a certain transformer oil is 2.1 times that of the air. The relative permittivities of some important insulating ma- terials are tabulated in Art. 56 below, merely to indicate their order of magnitude. For accurate values, the reader is referred to various published physical tables and engineering handbooks. The older name for relative permittivity is specific inductive capacity (or dielectric constant). It is more convenient in prac- tice to use relative than absolute permittivities, because the necessity of tabulating small quantities like * in eq. (259) is avoided. Besides, the data are more readily comparable with one another, and with the permittivity of air, which is a standard dielectric. This procedure is analogous to tabulating the con- ductivities of various metals in terms of that of pure copper, taken as 100 per cent. The absolute permittivity of a material is obtained by multiplying the absolute permittivity of air by the relative permittivity of the dielectric in question. Equation (258) thus becomes C = K Ka A/I, (260) where K stands for the relative permittivity. 152 THE ELECTRIC CIRCUIT [ART. 51 The clastance of a prismatic piece of dielectric, with the lines of displacement parallel to one set of its edges, is expressed by analogy with eq. (20), Art. 5, as S = ffl/A, (261) where a = K~ l (262) is called the elastivity of the dielectric. Elastivity is analogous to the resistivity of a conducting material or to the reluctivity of a magnetic medium, and may be expressed for practical purposes in megadarafs per centimeter cube. For air, the absolute elastivity is, according to eq. (259), ffa = Ka -i = 11.3 x 10 6 megadarafs per cm. cube. . (263) The concept of relative elastivity could be introduced if necessary, in which case its values would be equal to the reciprocals of the relative permittivities tabulated in Art. 56. However, it is suf- ficient to use the relative permittivity, even when dealing with elastances, so that eq. (261) becomes S = (ffa/K)l/A (264) The nomenclature used above is due to Mr. Heaviside; 1 it is consistent and uniform with the nomenclature used in the electro- conducting and magnetic circuits, and is suggestive as to the nature of the phenomena. The electrostatic nomenclature now in general use comprises but three terms; namely, condenser, capacity, and specific inductive capacity. It is hoped that the more rational and complete nomenclature used here will help to a clearer understand- ing of the dielectric circuit, and will simplify engineering calcula- tions relating thereto. 2 Note: The author considers the above-given value of , eq. (259), to be an experimental coefficient, in the same sense in which other prop- erties of materials are characterized by experimental coefficients. For an engineer, the volt and the ampere are arbitrary units established by an international agreement, no matter what their relation to the so-called absolute units. The value of can be calculated theoretically, assuming the ratio between the electrostatic and the electromagnetic units to be known. In the absolute electrostatic system of units, with air as the dielectric, a plate condenser having an area of A sq. cm. and a distance between the plates equal to I cm., has a capacity equal to A/ (4*1). The 1 O. Heaviside, Electromagnetic Theory (1894), Vol. 1, p. 28. 1 See the author's paper " Sur Quelques Calculs Pratiques des Champs Electrostatiques," in the Transactions of the Congresso Internazionak delle Applicazioni ElcUriche, Turin, 1911. CHAP. XIV] THE DIELECTRIC CIRCUIT 153 factor 4 w enters on account of an unfortunate selection of the expression for Coulomb's law, which should have been gig 2 /4 Trr 2 , instead of qiq^/r 2 . In the absolute electromagnetic units the same capacity is equal to (A/4 Ti-Z) (3 X lO 10 )" 2 , where 3 X 10 10 is the velocity of light in centimeters per second. To obtain the result in microfarads, the foregoing expression must be multiplied by 10 15 . On the other hand, the same capacity expressed in the rational -units defined above is K a A/l. Equating the two expressions gives Ka = lQ- 5 /(9 X 4 *) = 0.08842 X lO" 6 microfarads per centimeter cube. The fact that K a can be expressed through the velocity of light does not make *, the less an empirical coefficient, because the velocity of light itself is determined experimentally. As a matter of fact, one of the ways in which the velocity of light is determined consists in calculating it indirectly from the value of a obtained from measurements. Prob. 1. Show that in the English system * a = 0.2244 X 10~ 6 micro- farads per inch cube. Prob. 2. A condenser (Fig. 46) consists of two metal plates, 50 by 70 cm. each, in contact with a glass plate 3 mm. thick between them. When a continuous voltage of 2400 is applied to the condenser, the ballis- tic galvanometer shows a charge of 17.1 microcoulombs. What is the relative permittivity of the glass? Ans. 6.9 Prob. 3. A 0.5-mf . mica condenser is to be made out of sheets of mica 12 by 25 cm., 0.3 mm. thick, and coated on one side with a very thin film of silver. How many sheets are required? The relative permittivity of the mica is about 6. Ans. About 96 sheets, 48 sheets in parallel per terminal. Prob. 4. Let the dielectric in problem 2 consist, instead of glass, of three layers of different materials. Let the thicknesses of these layers be 1.2, 0.7, and 1.1 mm., and let the corresponding values of relative per- mittivities be 2, 3, and 5. What is the capacity of the condenser? Hint: Calculate the equivalent elastance as the sum of three elastances in series. Ans. 2.94 X 10~ 3 mf. 52. Dielectric Flux Density and Electrostatic Stress (Voltage Gradient). Referring again to the uniform electrostatic field (Fig. 46), consider a cube of the dielectric, one square centimeter in cross-section, and one centimeter long in the direction of the lines of force. Let a quantity of electricity Q be supplied by the battery, as shown by the ballistic galvanometer; then the same quantity of electricity must be displaced in the dielectric. Neg- lecting a small displacement at the edges and at the outside sur- faces of the plates, the whole quantity Q is uniformly displaced between the plates. Therefore, if the area of each plate is equal to A square centimeters, the displacement through the cube under consideration is equal to D = Q/A (265) 154 THE ELECTRIC CIRCUIT [ART. 52 Since Q is the total electrostatic flux, D is naturally called the dielectric flux density. If Q is measured in coulombs, D is ex- pressed in coulombs per square centimeter. In practice, Q is measured in microcoulombs, and D is expressed in microcoulombs per square centimeter. The dielectric flux density is analogous to current density U (Art. 6) and to magnetic flux density B. When an electrostatic field is non-uniform (Fig. 48), it is con- veniently subdivided by lines of force and equipotential surfaces perpendicular to the same. The procedure is similar to that used in Art. 8. In this case, the total flux or displacement divided by the area of an equipotential surface gives only the average flux density through the surface. The actual density varies from point to point, and it is therefore proper to speak of the dielectric flux density at a point. Take a tube of infinitesimal cross-section formed by lines of force, and let dQ be the displacement of electric- ity through this tube. The displacement is the same through any normal cross-section of the tube, because electricity behaves like an incompressible fluid. Let dA be a particular cross-section of the tube; then the flux density at this cross-section is D = dQ/dA, (266) D being usually expressed in coulombs (or microcoulombs) per square centimeter. Since the cross-section of the tube is infinites- imal, D is the density at the point corresponding to the position of dA. If the flux density in a uniform field is given, the total displace- ment is Q = DA (267) In a non-uniform field, the flux density must be given as a function of the coordinates of the field; so that Q = f DdA, . (268) Jo the integration being extended over the whole area of an equi- potential surface, or over the part of this area through which the flux is to be calculated. The electromotive force impressed at the terminals of a condenser is balanced in the whole thickness of the dielectric; that is, each small length of path in the dielectric produces its own counter-electromotive force. Therefore, it is possible to speak of the voltage drop per unit length of the path in the CHAP. XIV] THE DIELECTRIC CIRCUIT 155 dielectric, the same as in Art. 6. This voltage gradient, or electric intensity, in a uniform field is expressed by G = E/l, (269) and is measured, as in the conducting circuit, in volts per centi- meter, kilovolts per millimeter, or in other suitable units. In a non-uniform field, the electric intensity, or voltage gradi- ent, varies from point to point. Let the voltage between two infinitely close equipotential surfaces MN and M'N' (Fig. 48) FIG. 48. A non-uniform electrostatic field, represented by lines of displacement and equipotential surfaces. be dE, and let the distance mn between the surfaces, along a certain line of force HH', be dl. Then the voltage gradient along mn is G = dE/dl (270) The length of the line mn being infinitesimal, G is the intensity at any point between m and n. When the voltage gradient is uniform, we have for the total voltage across the field E = Gl (271) In a non-uniform field, G has to be given as a function of I, so that E = (272) 156 THE ELECTRIC CIRCUIT [ART. 52 the integration being performed between any two points on the equipotential surfaces between which the voltage is to be deter- mined. Imagine a uniform field existing in a dielectric, and consider a unit cube of the material. The total displacement through such a cube is equal to the flux density D, and the voltage across it is equal to the voltage gradient G. The permittance and the elas- tance of the cube are respectively equal to the permittivity and the elastivity of the material. Thus, applying to the cube eqs. (249) and (250), we have D = K G, (273) and G = dq, . Jo Jo (275) where T is the total time of charging, and dq = i dt is the infini- tesimal charge or displacement added to the condenser during the interval of time dt. The quantities dq and e can be expressed 1 The voltage and the charging current rise gradually, even though the key K be closed suddenly. This is on account of an ever-present magnetic inductance which acts as a kind of electromagnetic inertia. 157 158 THE ELECTRIC CIRCUIT [ART. 53 through the instantaneous flux density D t and the stress G t ', namely, from eq. (267), dq = A dD t , and from eq. (271) e = G t l. Performing the substitution, and taking the constant quantities A and I outside of the sign of integration, we get = Al C T G t dD t ...... (276) Jo In order to integrate this expression, D t must be expressed through G t , or vice versa. The relation between the two is given by eq. (273). Eliminating D t , we obtain W W = K Al f T G t dG t = \ K VG\ . . . (277) where V = A I is the volume of the dielectric, and G is the final value of the stress, at the time T. Hence, the energy stored per unit volume of the dielectric, or the density of energy, is W' = W/V = \ KG 2 = \ G*/a ..... (278) Using relations (273) and (274), the preceding formula can also be written in the following forms: W' = i GD = i 2 /K = \ aD* ..... (279) The analogy to the corresponding formula in Art. 69 of the Mag- netic Circuit is apparent at once. The total stored energy can be expressed through the per- mittance or elastance of the dielectric. We have from eq. (249) dq = C'de; substituting in eq. (275) and integrating, we get W = \ CE* = i E 2 /S ...... (280) Since the final charge, or total displacement Q equals CE or E/S, the energy can be represented also in the following forms: W = i QE = \ Q*/C = \ Q*S ..... (281) These formulae are analogous to the corresponding expressions in Art. 57 of the Magnetic Circuit. Let now the dielectric and the field be of an irregular form as shown in Fig. 48. The stress G and the displacement D are different at different points, so that it is necessary to consider infinitesimal layers of the dielectric between consecutive equi- potential surfaces, and infinitesimal threads of displacement be- tween the electrodes. Consider an infinitesimal volume mnqp of the dielectric, comprising the part of a tube of displacement EH' between two equipotential surfaces MN and M'N'. The sides CHAP. XV] THE DIELECTRIC CIRCUIT 159 mp and nq can be provided with infinitely thin metal films, because these sides lie in the equipotential surfaces, and therefore no current would flow along these metal coatings. Then the element of volume under consideration is converted into a small plate condenser; the flux density and the stress within this element can be considered as uniform, so that formula (277) holds true, and we have V ....... (282) Differentials are used because both the volume and the stored energy are infinitesimal. The density of energy W = dW/dV = !G 2 , ..... (283) and has the same expression as in the case of a uniform field; but its numerical value is different from point to point, because G is variable. The other expressions for the density of energy, eqs. (278) and (279), also hold true for the points of a non-uniformly stressed dielectric, provided that proper values of D and G are used for each point. The total energy stored in a non-uniform electrostatic field is W = \ f V K&dV = \ C V GDdV = \ f V D*dV/ K ; (284) Jo Jo Jo two more expressions may be written in which I/a- is used in place of K. In order to perform the integration G and D must be given as functions of coordinates, and the integration extended over the whole space occupied by the field. Equations (280) and (281) are true for condensers of any shape, because in the deduction of these formulae no assumption is made as to the particular form of the dielectric or the electrodes. The expressions for the electrostatic energy of the field, derived above, are analogous to the corresponding ones for the potential energy of stressed elastic bodies; and this is consistent with the assumed behavior of dielectrics. Consider the work necessary per cubic centimeter to strain mechanically the elastic fibers of a given material. The external mechanical force being applied gradually (so as to avoid oscillations), the stress varies from zero to its final value G. Let G t be some intermediate value of the stress, and let D t be the corresponding strain. The same symbols G and D are used here to denote the mechanical quantities analogous to electric stress and displacement. While the strain 160 THE ELECTRIC CIRCUIT [ART. 54 increases from D t to (D t + dD t ), the stress G t may be considered constant; the infinitesimal work done is therefore equal to G t dD t . The total work of deformation is W But, according to Hooke's law of elasticity, strains are propor- tional to stresses, so that a linear relation exists between D t and G t , similar to eq. (274). We thus arrive again at the result that the work necessary to strain one cubic unit of an elastic material is equal to | from r to r' is equal to this elastance multiplied by the electric displacement Q' per centimeter length of the cable. Or E rr ' = S f r r' - Q' = (ffQ r /2 TT) Ln (r'/r}. . . (290) This formula finds its important application below in the calcula- tion of the permittance of single-phase and polyphase transmission lines. It is absolutely essential to agree in regard to the signs in eq. (290). In the applications that follow, Q' is taken with the plus sign when the positive displacement is directed from the con- ductor, and with the minus sign when it is directed towards the conductor. It is also important to write the distances r' and r in the order given, because interchanging r' and r in eq. (290) changes the sign of E rr '. If the insulation consists of two or more concentric layers of different materials, the elastances of the layers are calculated separately, according to formula (288), and then added in series. The permittance of the cable as a whole is the reciprocal of this resultant .elastance. The same formulae apply to a concentric cable without sheathing, the outside conductor taking the place of the sheathing as far as stresses in the dielectric are concerned. With two cylindrical conductors side by side the elastance is cal- culated as shown in Art. 63 below. With three conductors the theory is rather difficult; as is also the case when the conduc- tors are not of circular cross-section. Those interested will find 1 This simple derivation of the formula for the capacity of a single-core cable demonstrates in a particularly striking manner the usefulness of the concept of elastance. CHAP. XVI] ELASTANCE OF CABLES AND LINES 173 extensive literature on the subject in the European electrical magazines and proceedings of electrical societies. In practice, the permittance of such cables is usually determined by test. The distribution of the electric stresses in a single-core cable is of considerable practical importance. The total displacement Q being the same through every concentric layer of the dielectric, the flux density and consequently the stress is a maximum at the surface of the inner core. For a layer of radius x we have Q = D x '2-n-xl = const., (291) where D x is the density of displacement through that layer. Hence, D x x = const., (292) which means that the density of displacement is inversely pro- portional to the distance from the center. Since displacements are proportional to stresses (with a uniform insulation), we also have G x x = const (293) A useful relation between the total applied voltage E and the stress G x at a given point in the dielectric can be deduced from eq. (293). We have G x = const, /x; and if we multiply both sides by dx and integrate between a and b, remembering that voltage is the line integral of intensity, we obtain G x dx = E = (const.) Ln (6/a). Eliminating the constant between these two equations, gives G x = E/[x Ln (6/a)] (294) Equations (292) and (293) show that a homogeneous dielectric is fully utilized with regard to its dielectric strength only at the surface of the core, the stress gradually decreasing toward the periphery. This condition could be helped by gradually increasing the elastivity of the material toward the sheathing, so as to in- crease the voltage drop and the stresses there. If the elastivity of each layer could be made exactly proportional to its radius, the stress G x would be the same throughout the dielectric. Such a condition would be an ideal one, with regard to economy in 174 THE ELECTRIC CIRCUIT [ART. 59 material, provided that^ the dielectric strength of the " variable insulation " were constant. This purely theoretical conclusion leads to the important practical question of the grading of insulation of cables. With high-tension cables, in which the thickness of insulation is large, it pays to provide two or more layers of different materials, utiliz- ing their permittivities and ultimate strengths in the most ad- vantageous manner. The problem is primarily to relieve] the stress near the inner core, and this is done by placing near it a layer of insulation of high permittivity, so as to cause a low volt- age drop. One case where the opposite arrangement would be advantageous is in a low-voltage cable in which it is desired to keep the total permittance as low as possible (for example, to obtain small capacity or low charging current at high frequen- cies). In this case the layer surrounding the core must have as high an elastance as possible, because it is this layer that contrib- utes most to the total elastance of the cable. With a clear understanding of these principles, the student will be able to design a graded insulation for given conditions, if he knows the properties of the available materials. 1 Prob. 1. A single-core cable receives a charge of 1.18 millicoulombs per kilometer when a continuous voltage of 12 kv. is applied between the core and the sheathing. The core consists of a solid conductor the diameter of which is 5 mm. ; the insulation is 9.5 mm. thick. Determine the value of the relative permittivity of the material of insulation, and the extreme values of the dielectric flux density. Ans. K = 2.78; D maz = 0.00750; D min = 0.00156 microcoulombs per sq. cm. Prob. 2. The insulation used in the cable specified in the preceding problem breaks down at a flux density of 0.062 me. per sq. cm. Show that the critical voltage for the cable is about 70 alternating kilovolts. Prob. 3. What is the ratio between the maximum and the average stress in the insulation in problem 1? Ans. 2.42. 2 Prob. 4. Deduce formula (290) from the fact that the voltage is the line integral of the electric intensity. 1 For a theoretical treatment of the grading of insulation, and for the bibliography of the subject, see H. S. Osborne, Potential Stresses in Die- lectrics (1910), a thesis presented to the Massachusetts Institute of Tech- nology for the degree of Doctor of Engineering. * There is a tendency in practice to deal with average stresses even when the field is far from being uniform. The answer to this problem shows that one has to be careful in using an average value, unless its ratio to the maxi- mum stress is known. CHAP. XVI] ELASTANCE OF CABLES AND LINES 175 Prob. 6. Show by actual calculation that in the foregoing cable the maximum stress in the dielectric is reduced by increasing the diameter of the conductor to 7.5 mm., with the same diameter of the sheathing. This is in spite of the fact that the insulation becomes thinner, and consequently the average stress greater. Prob. 6. Referring to the preceding problem, show that it is of ad- vantage to make the ratio b/a about equal to , where e = 2.71828 . . . is the base of the natural system of logarithms. If the diameter of the conductor be further increased, so that the ratio b/a becomes less than e, the maximum stress does not continue to decrease, but increases in- stead. Solution: The stress at the core is G a = E/[aLn (b/a)] according to eq. (294) . As a varies, G a reaches its maximum when dG a /da = 0. Differentiating, we get dG a /da = E [1 - Ln (6/a)]/[a Ln (b/a)] 2 = 0; whence, 1 Ln (6/a) = 0, or b/a = t. Prob. 7. Explain the following deduction from the theorem stated in the preceding problem. In a concentric cable subjected to an excessive voltage, if the insulation is quite thick, the layer around the inner core is first gradually destroyed or charred up to a certain thickness, and then the rest of the insulation suddenly breaks down. With a thin layer of insulation no such phenomenon is observed. Prob. 8. A cable is provided with several concentric layers of insula- tion, the external radii of which are 61, 6 2 , etc., and the relative permit- tivities, Ki, K 2 , etc. Show that the elastance of the cable is expressed by the formula S = (41.45/0 [Kr l log (bi/a) + Kr l log (6 2 /6,) + Kr l log (63/62) + etc.]. Prob. 9. Show that in a single-core cable the density of energy stored in the dielectric varies inversely as the square of the distance from the center. Prob. 10. A conductor 2 a cm. in diameter is surrounded by a con- centric metal cylinder of 26 cm. inside diameter. What alternating volt- age can be allowed between the cylinder and the conductor at a factor of safety k against the formation of corona? Ans. E = 18.4 a (D c X 10 3 /&) log (6/a) effective kilovolts, where D c is the flux density in microcoulombs per sq. cm., at which corona is formed. Prob. 11. Show that the elastance of the dielectric between two con- centric spheres of radii a and 6 is equal to (o- a /4 *-./?) (I/a 1/6) mega- darafs. Prob. 12. Show that with two concentric spheres the equation corre- sponding to (294) is G x = E/[x* (a~ l - &-)]. Prob. 13. Apply the formulae given in the text above to the theory of a condenser-type terminal. 1 1 See A. B. Reynders, "Condenser Type of Insulation for High-tension Terminals," Trans. Amer. Inst. Electr. Engrs., Vol. 28 (1909), p. 209. 176 THE ELECTRIC CIRCUIT [ABT. 60 60. The Elastance of a Single-phase Line. The general character of the electrostatic field between two infinite parallel conductors is shown in Fig. 50. The lines of force are arcs of circles extending from one metal surface to the other; the equi- potential surfaces are circular cylinders eccentric with respect to the conductors (see Art. 62 below). It is required to calculate the elastance of the air between the two conductors, for a unit axial length of the line. Knowing this elastance, the charging current of the line can be calculated for a given frequency. This elastance, or its reciprocal, the permittance, is used in the pre- determination of the regulation of a transmission line (Art. 68 below). We shall consider in this article the usual practical case in which the radius a of the conductors is small as compared with the interaxial distance b. It is shown in Art. 63 below how to determine the elastance when the diameters of the cylinders are comparatively large. For purposes of analysis it is convenient to consider the field shown in Fig. 50 as the result of the superposition of two simple radial fields similar to that in Fig. 49. Consider the conductor A , together with a concentric cylinder of an infinitely large radius, as one electric system. Let the conductor B with a similar con- centric cylinder form another independent system. Let the con- ductor A be connected to the positive pole of a battery of voltage E, the conductor B to the negative pole, and the two cylinders at infinity to the middle point of the battery. In the first concentric condenser the displacement of positive electricity is from the con- ductor A to the infinite cylinder, while in the second system the positive displacement is from the infinite cylinder toward the con- ductor B. The displacements due to the two systems are equal and opposite at the two infinite cylinders, and the cylinders them- selves coincide at infinity, because the distance AB between their axes is infinitely small as compared with their radii. Hence, the two displacements at the cylinders cancel each other, and the combina- tion of the two cylindrical condensers is electrically identical with the two given parallel conductors A and B. In a medium of constant permittivity the resultant stress or voltage gradient, produced at a point by the combined action of two or more independent electric systems, is equal to the geometric sum of the stresses produced at the same point by each system. CHAP. XVI] ELASTANCE OF CABLES AND LINES 177 This principle of superposition can be considered either as an ex- perimental fact or as an immediate consequence of the fact that in a medium of constant permittivity the effects are proportional to the causes. This principle being true for electric intensities, the component flux densities at a point are also combined accord- ing to the parallelogram law, because they are proportional to the intensities. Hence, the resultant electrostatic flux can be re- garded as the result of the superposition of the fluxes created by FIG. 50. The electrostatic field produced by a single-phase trans- mission line. the component systems. Furthermore, the actual voltage be- tween any two points in the dielectric is the algebraic sum of the voltages due to the component systems, because each voltage is the line integral of the corresponding voltage gradient, and the principle of superposition is valid for these gradients. This line integral is a function only of the positions of the two points, and is independent of the path along which the integration is per- formed. This latter fact is very convenient in applications of the principle to the solution of problems. 178 THE ELECTRIC CIRCUIT [ART. 60 In order to be able to apply the formulae deduced in the pre- ceding article, it is essential that the diameters of the two wires be small as compared with the distance between them. The reason is that each component system is supposed to possess a radial field, in spite of the presence of the other conductor. This is practically true when the second conductor is so small, or so far distant from the first, that the infinite permittivity of its material does not appreciably distort the radial field. To be more precise, the dis- tortion of the radial component field originating from each con- ductor, caused by the presence of the other, must be negligible. It is sufficient to calculate the elastance of that part of the system between one of the conductors and the neutral plane of symmetry 00', the total elastance being equal to twice that value. This we can do by computing the voltage needed to produce a displacement Q' per unit length of the line. The volt- age between the surface of the conductor A and the point N in the plane of symmetry 00' is equal to \ E. On the other hand, the .same voltage can be expressed as the sum of the voltages due to the two component systems. Referring to eq. (290), let the distance r' refer to the point N, and let r refer to a point on the surface of the conductor A. Then, as far as the first component system is concerned, the voltage between A and N is equal to (aQ'/lv} Ln (\ b/a), where Q f is the actual displacement per unit length of the line. In the second system, the voltage between the same two points is (ffQ'/2v) Ln (\b/b}. The minus sign is due to the fact that the displacement in the second system is toward the conductor B, and hence must be considered as negative if that at the first conductor is regarded as positive. The ratio r' IT for the second system is more accurately equal to \ b/(b a), but a, being by supposition small as compared to 6, is neglected in the denominator. Equating the sum of the preceding two expres- sions for the voltage between A and N to the actual voltage E, we get \E = (aQ'/2 TT) Ln (b/a) (295) Hence, the elastance between one of the conductors and the neutral plane 00', for a unit of axial length, is S 1 = (C")- 1 = \ E/Q' = (a/2 T) Ln(6/o), . (296) or, with air as the dielectric, S' = (C')- 1 = 41.45 log (6/a) megadarafs per kilometer. (297) CHAP. XVI] ELASTANCE OF CABLES AND LINES 179 The corresponding permittance is C' = (S')~ l = 0.0241/log (b/a) microfarads per kilometer. (298) When using these formulae, one must not forget that the per- mittance is proportional to the length of the line, while the elas- tance varies inversely as the length of the line. The total elastance for a unit length between the two conductors is equal to 2 S f , the corresponding permittance being \ C'. Prob. 1. For a few standard spacings and sizes of conductor, check the values of permittance given by eq. (298) with those tabulated in an electrical pocketbook. Prob. 2. For some assumed values of a, 6, and E, corresponding to an actual transmission line, plot a curve of values of the voltage gradient along the line A B, and also draw the horizontal straight line represent- ing the average gradient E/b. Hint : At a distance x from A the intensity due to the system A is aQ'/ (2 TTZ); that due to the system B is oQ' / [2 * (6 - a;)], both inten- sities being directed from left to right. Prob. 3. In Fig. 50 let A and B be small spheres, instead of cylinders. Show that the elastance between one of the spheres and the neutral plane 00' is equal to ( or 1 + BC'/AC',' ..... so that the equipotential line can be easily drawn for a given C or C'. Let it be required to calculate the elastance of the slice of die- lectric between the neutral plane 00' and the equipotential sur- face passing through a given point C. It is sufficient to find the expression for a unit axial length, knowing that the elastance is inversely proportional to the length of the conductors. Write the expression for the voltage between the points N and C, using again eq/ (290). For the systems A and B we have E NC = (*Q'/2ir) [Ln (AC /AN) - Ln (BC/BN}}, or, since AN = BN, E NC = ( = const. According to the principle of superposition, the equation of a line of force in the resultant field is w w' = const., the minus sign being due to the fact that A' is negative if A is positive.' Let the point of intersection of two lines of force w = C and ' = C" be the starting, point for draw- ing a line of force in the resultant field. Then the point of intersection of the next lines = C + w/n and J = C' + v/n also belongs to the same line of force in the resultant field, because for both points u u' = C C'. In other words, the lines of force in the resultant field are diagonal curves with respect to the lines of force in the component fields, and may be drawn from intersection to intersection. A similar construc- tion holds for equipotential surfaces. The student is strongly urged to try this construction for some assumed data, because the method of diagonal curves is generally applicable when a given field can be resolved into two simpler fields. 188 THE ELECTRIC CIRCUIT [ART. 63 63. The Elastance between Two Large Parallel Circular Cylinders. The formulae derived in Art. 60, for the elastance and permittance of a homogeneous medium between two parallel cylinders, hold true only when the diameters of the cylinders are small as compared to the interaxial distance, for the reason there explained. When the diameters of the "cylinders are compara- tively large, the elastance is derived by reducing the conditions to those obtaining in Art. 60. Let A and B (Fig. 50) represent as before two conductors of very small diameter, and let a difference of potential of 100 volts be maintained between them by means of a battery. Let the volt- age between the conductor B and the equipotential surface CPC' be 20 volts. Place an infinitely thin metal sheet so as to coincide with this surface, and connect this sheet to a point of the battery such that the voltage between it and the conductor B still remains equal to 20 volts. These changes do not affect the electrostatic field either inside or outside the surface CPC', the displacement being normal to this surface. Now remove the conductor B al- together, leaving a difference of potential of 80 volts maintained by the battery between the conductor A and the cylinder CPC'. The field outside the cylinder is not affected; that inside of it has entirely disappeared. We have now a field between the cylinder A of very small diameter and the cylinder CPC' of a compara- tively large diameter. Take now another equipotential surface, for instance KMK', symmetrical with CPC', place a metal cylinder so as to coincide with it, and connect it to a tap on the battery, so that the same difference of potential of 20 volts remains be- tween this cylinder and the conductor A. The field is not altered by this connection, and now the conductor A may be removed. Thus, we finally obtain a field between two cylinders of compara- tively large diameter. The difference of potential between the cylinders is only 60 volts, while the original difference of potential between the conductors A and B was 100 volts. Conversely, let the cylinders CPC' and KMK' be given, and let it be required to find the shape of the field between them, and the elastance of this field. The problem is reduced to that of finding the positions of the infinitely small eccentric conductors A and B, with respect to which the given cylinders are equi- potential surfaces. Then the field is mapped out according to the formulae given in the preceding article, leaving out the space CHAP. XVI] ELASTANCE OF CABLES AND LINES 189 inside the cylinders. The elastance between one of the large cylinders and the plane 00' is calculated by using formula (313). This method is applicable whether the two cylinders are of the same radius or not, and whether one is outside or inside of the other. It is always possible to find the positions of the lines A and B with respect to which the given cylinders represent equi- potential surfaces. The details of the calculation are given below. Consider first the case of two cylinders CPC' and KMK' of the same diameter d; let the distance between the centers p and q of these cylinders be equal to c. In order to use eq. (313), it is necessary to express AC and BC through the given quantities c and d. According to eqs. (308) and (309), we have AC/BC = AC'/BC f (314) All the quantities which enter into this equation can be expressed through one unknown length, for instance BC. We put BC = AK = x] then AC = CK + AK = (c - d) + X-, AC' = x + c; - (315) BC' = d - x. Substituting these values into eq. (314), and solving the resulting quadratic equation for x, we obtain, retaining the positive value only, = %d[-(a-l) + \/ 2 - 1], (316) where the ratio of the interaxial distance to the diameter is de- noted by a, or a = c/d (317) By substituting this value of x into the expression for AC m eqs. (315), we find AC = i[(c - d) + Vc 2 - d 2 ] = id [(a - 1) + W - 1], (318) so that AC/BC = [(a - 1) + V 2 - !]/[ _ ( - 1) 4- Va 2 - 1]. This expression can be simplified by multiplying both the numer- ator and the denominator by the value of the numerator, so 190 THE ELECTRIC CIRCUIT [ART. 63 as to get rid of the square root in the denominator. The result is AC/BC = a + V 2 - 1 (319) The expression (313) for the elastance between one of the cylinders and the plane of symmetry, per unit of axial length, becomes S' = (0/2*) Ln [a + Vo 2 ^!]. . . . (320) Those familiar with hyperbolic functions will notice that the pre- ceding equation can be simplified into S' = (cr/271-) Cosh- 1 ** (321) Since tables of hyperbolic functions are readily available, the evaluation of elastance is simpler in this form than it is if eq. (320) is used. 1 When the diameter of the conductors is small as compared to the interaxial distance, a is a large quantity, and unity under the radical sign in eq. (320) may be neglected. This equation be- comes then practically identical with eq. (296). For large values of a, the term (1 I/a 2 )*, obtained by factoring in expression (320), is conveniently expanded according to the binomial theorem, the result being S' = (a/2 TT) Ln (2 a - i a" 1 - | or 3 - T V a~ 5 - . . .). (322) With the exception of 2 a, all of the terms in parentheses are small corrections to the result. Let now the diameters of the two given cylinders be different. In addition to relation (314), we also have BK/AK = BK'/AK' (323) It is necessary in this case to introduce two unknown quantities, BC = x and AK = y. Equations (315) are modified accordingly. All of the quantities in eqs. (314) and (323) are expressed through x and y, and then these two equations are solved together for x and y. After this, the elastance between each cylinder and the plane 00' is expressed by using eq. (313). 1 Dr. A. E. Kennelly, " The Linear Resistance between Parallel Conduct- ing Cylinders in a Medium of Uniform Conductivity," Proceedings Amer. Philosophical Soc., Vol. 48 (1909), p. 142; also his article on " Graphic Rep- resentations of the Linear Electrostatic Capacity between Equal Parallel Wires," Electrical World, Vol. 56 (1910), p. 1000. See also his book on Ap- plications of Hyperbolic Functions to Electrical Engineering (1912). CHAP. XVI] ELASTANCE OF CABLES AND LINES 191 In some cases it is required to calculate the dielectric flux density at a point in the field between the cylinders, or at the sur- face of one of the cylinders. Let P (Fig. 52) be a point in the field between two parallel cylinders, small or large ; the flux density at P is the geometric sum of the densities due to the systems A and B. The flux density due to the system A is while that due to B is These component densities are directed as shown in Fig. 52. The resultant density D is directed along the tangent to the line b FIG. 52. Dielectric flux density at a point, determined by the method of superposition. of force through P. From the preceding two equations, we have the relation so that the triangles APB and Pmn are similar. The correspond- ing sides are marked with one, two, and three short lines respec- tively. From these triangles we can write D : D, = b : r 2 , or, substituting the foregoing expression for DI, D = Q'6/(2W 2 ) ....... (324) From this expression, the flux density can be calculated at any point in the field or on the surface of one of the cylinders. Mul- tiplying the flux density by the elastivity of the medium, the cor- responding dielectric stress is obtained. It must be kept well in mind that b, r i} and r 2 refer to the points A and B, and not to the centers p and q of the actual cylinders. 192 THE ELECTRIC CIRCUIT [ART. 63 Prob. 1. Take two equal cylinders at a comparatively short distance apart, and (a) calculate the permittance per meter of the axial length; (b) divide the field into 10 equal elastances in series and into 10 equal permittances in parallel; (c) plot a curve of the flux density distribu- tion on the surface of one of the cylinders. Prob. 2. Show that on an equipotential surface surrounding A, and consequently on the corresponding metal surface, the flux density varies inversely as r, 2 . Prob. 3. Show how to calculate the permittance between a large cylin- der and a given infinite plane. Prob. 4. Show that A and B are inverse points with respect to any equipotential circle; this means that the radius qC is the geometric mean between the distances qB and qA, and the radius pK is the geometric mean between the distances pA and pB. This is true whether the radii qC and pK are equal or not. Prob. 5. Extend the theory given in this article to the calculation of the elastance and flux density distribution between two large spheres. Consult the chapters on electrostatics in some standard work on the mathematical theory of electricity and magnetism. CHAPTER XVII EQUIVALENT ELASTANCE AND CHARGING CURRENT OF THREE-PHASE LINES 64. Three-phase Line with Symmetrical Spacing. Consider an unloaded three-phase line, and let the three conductors be denoted by A, B, and C respectively. There is a displacement of electricity between each pair of conductors, and since the three in- stantaneous voltages are different, the displacements between the three pairs of conductors at any instant are also different. The three sets of lines of force are relatively displaced and the flux density varies from instant to instant, so that there is produced in reality a revolving electrostatic field. Let the instantaneous displacements which issue from the three conductors per unit of axial length be denoted by qi, q 2 , and q 3 , where the subscripts 1, 2, and 3 refer to the conductors A, B, and C respectively. To be consistent with the notation used before, these symbols should be provided with the " prime " sign, but this sign is omitted in order not to obscure the formulae. The displacements are considered positive when they are directed from the conductors into the die- lectric. Since electricity behaves like an incompressible fluid, as much of it as is displaced at any instant out of one conductor must be displaced into the other two conductors, so that at all times the following relation holds, namely, ffi + ? 2 + ? 3 = (325) The three q's vary with the time according to the sine law. With a symmetrical spacing of the wires, and symmetrical voltages forming an equilateral triangle (Fig. 53), the effective values of the three q's are equal, and the corresponding instantaneous values are displaced in time phase by 120 degrees. The charg- ing current per unit length of a conductor is equal to the rate of change of the corresponding displacement with the time, or i = dq/dt (326) 193 194 THE ELECTRIC CIRCUIT [AitT. 64 But, with sinusoidal voltages, the displacements vary also ac- cording to the sine law, or 5 = Q ro sm27r/f, ...... (327) where Q m is the maximum value of the displacement from one of the conductors. Substituting this value of q in eq. (326), we find i = 2 7rfQ m cos 2 TT/Y ...... (328) Consequently, the amplitude of the charging current ....... (329) and the same relation holds true for the effective values of the displacement and current. It is to be noted that the charging current leads the flux by 90 electrical degrees. Thus, knowing the displacement, the charging current can be calculated from eq. (329). If Q m is expressed in microcoulombs per kilometer, I m is in microamperes per kilometer. The actual charging current which flows through a cross- section of the conductor, is equal to that necessary to supply the displacement between this cross-section and the receiver end of the line. In other words, the charging current varies along the line, from a maximum at the generator end to zero at the receiver end. If the effective voltage along the line were constant in phase and magnitude, the amplitude of the charging current would vary according to a straight-line law. In reality, the volt- age varies along the line, due to its resistance and inductance, so that the variations in phase and amplitude of the charging cur- rent along the line follow a much more complicated law. The influence of the permittance of the line upon its voltage regulation is treated in Arts. 68 and 69 below. The problem here is a preliminary one; namely, with a given size and arrangement of conductors in a three-phase line, to find the permittance per kilometer of the equivalent single-phase line, for which the volt- age regulation is usually calculated. The problem is solved by applying again the principle of superposition. Each conductor is considered as forming a condenser with a concentric cylinder of infinite radius, the three phases being star-connected and the three cylinders grounded. The vectors of the star and delta volt- ages are shown in Fig. 53, the subscripts 1, 2, 3 referring again to the conductors A, B, and C respectively. CHAP. XVII] ELASTANCE OF THREE-PHASE LINES 195 Applying eq. (290) for the voltage between the conductors A and B, we have, for instantaneous values, en = ((79i/2 7r)'Ln (6/a) + TT) Ln (a/6), (330) where, as before, the spacing is denoted by 6, and the radii of the conductors by a. The first term on the right-hand side of this equation represents the action of system A, the second term that of system B. The action of the system C is equal to zero, be- cause, on applying eq. (290) for this system, it is observed that r = r', on account of the symmetrical spacing. In other words, FIG. 53. Electric displacements in a three-phase line with symmetrical voltages and symmetrical spacing. for system C the conductors A and B lie on the same equipoten- tial cylindrical surface. The preceding equation is simplified to ev, = (91 #>) S r , where S' is the elastance expressed by eq. (296), that is, the elastance between one of the conductors and the plane of symmetry 00 r , as if the third conductor did not exist. Owing to symmetry, the other two equations are similar; thus we have to -fa -,) e 23 = (92-93)5'; (331) 631 = (q*- qi )S'. 196 THE ELECTRIC CIRCUIT [AKT.64 This result is interpreted graphically by Fig. 53, remember- ing that relations which hold true algebraically for instantane- ous values of sinusoidal quantities, hold true geometrically for the corresponding vectors of these quantities. According to eqs. (331), the instantaneous values of (qi - g 2 ), (92 - Qs), and (q 3 - 1) are in phase with the corresponding voltages e i2 , e 23 , and e 3 \. For this reason, the vectors (Qi - Q 2 ), (Qz - Qa), and (Q 3 - Qi) are drawn in phase with the vectors E n , E 23 , and E 3 i. In regard to the quantities Qi, Q 2 , and Q 3 , we know that, for reasons of symmetry, they are equal numerically and are displaced in phase relatively to each other by 120 degrees. Therefore, they must be repre- sented by vectors from the center to the vertices of the triangle MNP. The condition is then fulfilled that each side of this tri- angle is equal to the difference of two vectors from the point 0. We see now that the three electric displacements Qi, Q 2 , and Q 3 are in phase with the corresponding star- or Y-voltages of the system; also, from the similarity of the triangles, we have Eu/Ei = (Qi QaVQi, with corresponding relations for the other two phases. Consequently, eqs. (331) are reduced simply to E l = Q'; E 2 = QzS'; (332) E 3 = Q 3 S'. We thus arrive at the following important conclusion: The dis- placement (and consequently the charging current) per phase of a three-phase line with symmetrical spacing and symmetrical voltages is equal to that in a single-phase line with the same conductors and the same spacing, provided that the star voltage of the three-phase line is equal to that between one conductor and the neutral plane 00' in the single-phase line. As explained in Art. 36, an equivalent single-phase line is obtained by taking one conductor of the three-phase line and assuming the transmission voltage to be equal to the star voltage of the actual transmission line; the return conductor is supposed to be devoid of both resistance and inductance. The preceding rule gives a simple method for finding the permittance of the equivalent line; namely, the permittance of the equivalent single- phase line is equal to that between one of the conductors of the actual line and the plane of symmetry between it and one of the other con- ductors, as if the third conductor did not exist. CHAP. XVII] ELASTANCE OF THREE-PHASE LINES 197 The calculation of the charging current with an unsymmetrical spacing of conductors is much more involved, and is explained in the next article. Fortunately, however, the spacing between the conductors affects the value of the charging current but little, with the usual ratios between size of conductor and spacing. The student can easily verify this fact by consulting any available table of capacities or charging currents of transmission lines. The reason for this is that the principal part of the elastance be- tween two small conductors occurs near the conductors, where the flux density is comparatively high. Consequently, it is pos- sible in practice to estimate the permittance per phase of a three- phase line with unsymmetrical spacing, by finding the limits of the permittance with symmetrical spacings. For instance, let two conductors be placed on a cross-arm and the third on top of the pole, forming an isosceles triangle. Let the spacings be 2 m. and 1.6 m. respectively. The charging currents are differ- ent in the three conductors, but the average value is larger than with a symmetrical spacing of 2m., and smaller than with a sym- metrical spacing of 1.6 m. Having found the charging currents or the equivalent permittances for these two spacings, one can assume an intermediate value by interpolation, or else take one of the two limits, whichever gives the more unfavorable operating conditions of the line. It is rather a tedious problem to estimate the influence of the ground upon the charging currents in a three-phase line. The theory is simple, the ground being replaced by the images of the three conductors, as in Fig. 51; but the formulae are long and in- volved, because the effects of six separate systems must be super- imposed. See problem 3 in the next article. Prob. 1. Show that when one of the conductors in a three-phase line fails, the charging current in the other two conductors drops to 86.6 per cent of its former value. Solution: Let C be the permittance between one of the conductors and the plane of symmetry between it and one of the other conductors. Then the charging current with the three phases alive is kC(E/V^3), where E is the line voltage, and k is a coefficient of proportionality with which we are not concerned here. Operating single-phas^, the charging current is kC(%E). The ratio of the two is 0.5/ (I/ V3) = 0.866. Prob. 2. A three-phase, 140-kv., 25-cycle transmission line consists of conductors 2 cm. in diameter; the spacing is symmetrical and equal to 3.5 m.; the length of the line is 250 km. What is the total reactive 198 THE ELECTRIC CIRCUIT [ART. 64 power necessary to keep the line alive, and what are the voltage and the permittance per kilometer of the equivalent single-phase line? Ans. 7270 kva.; 80.8 kv.; 0.00947 mf. per km. Prob. 3. A three-phase transmission line consists of conductors 18 mm. in diameter, suspended all three in the same vertical plane, at a distance of 2.4 m. between the adjacent conductors. What are the limits of the clastance of the equivalent single-phase line? Ans. 100 and 113 megadarafs per km. Note: The proximity of the two limits shows that it is sufficient for practical purposes to consider the symmetrical spacing only, as far as the dielectric and magnetic effects are concerned. Mr. J. G. Pertsch, Jr., has called the author's attention to the fact that, with certain simplifying assumptions, and when the three wires are transposed, the equivalent spacing for inductance and capacity is equal to the geometric mean of the three actual spacings, or b eq = ^6,2623631. In the case under consideration the equivalent spacing is 3.02 m., and the corresponding elastance equals 105 mgd. per km. Prob. 4. Extend the treatment given in this article to the case in which the three delta voltages are different (Fig. 54), and show that the point coincides with the center of gravity of the triangle, a sym- metrical spacing of the conductors being presupposed as before. Solu- tion: Equations (331) hold true as before, and the sides of triangle MNP are parallel to those of 123, but the point cannot be deter- mined in this case from the sym- metry of the figure. Any point within the triangle 123 gives a set of star voltages Ei, E 2 , and E 3 , which will produce the given set of delta voltages; but there is only FIG. 54. Electric displacements in a one point from which the rays three-phase line with unsymmctrical ^ t- ne vertices of triangle MNP voltages and symmetrical spacing. satisfy condition (325). Since the displacements in the equivalent single-phase lines must be proportional to the voltages, condition (325) requires that the geometric sum of Ei, E t , and E 3 shall equal zero. The parallelogram 021'3 gives the geometric sum of E t and E 3 equal to Or. If is the correct point, 01' must be equal and opposite to 01, and Om = JOl. Similarly, the condition must be fulfilled that On = \ 02, and Op = \ 03. It is known from elementary geometry that the three bisectors of a triangle divide each other in the ratio of 2 to 1, and that the point of their intersection is the center of gravity of the triangle. Hence the point is the center of gravity of the triangle 123. CHAP. XVII] ELASTANCE OF THREE-PHASE LINES 199 From eqs. (331) we again derive eqs. (332), and finally arrive at the same conclusion as that printed in italics after these equations. The three star voltages being different, one from another, the three charging currents are also different, each leading the corresponding Q by 90 degrees. The permittance and the voltage of the equivalent single-phase line are also different for each phase, in spite of the symmetrical spacing of the conductors. Prob. 5. For a given three-phase line with symmetrical spacing and voltages, draw the electrostatic field for the instant when one of the delta voltages is equal to zero; also make three drawings of the field at the ends of intervals A, A and T s z of a cycle later. Use the principle of superposition explained in Prob. 5, Art. 62, and apply it to the three component systems, A, B, and C, keeping in mind the relative magni- tudes of the instantaneous displacements. 65. Three-phase Line with Unsymmetrical Spacing. 1 As is mentioned in the preceding article, the calculation of charging currents in a three-phase line with unsymmetrical spacing is much more involved than with symmetrical spacing, and is not of much practical importance at present. An outline of it is given here in order to fix more firmly in the student's mind the general principle of superposition, and the method by which the results are derived in the preceding article. Moreover, the in- fluence of the dielectric is becoming more and more important, as the transmission voltages and the lengths of transmission lines are increased. The time may come when the exposition given in this article will be of assistance in the solution of practical prob- lems. Let the three spacings be denoted by 612, 623, and 631 respec- tively. Equations (325) to (329) inclusive hold true as with a symmetrical spacing, but eq. (330) now becomes en = ( = 1.377/0.6667 = 2.065; cos = 43.58 per cent (leading). Total current = 0.6667/0.4358 = 1.53 amp. Prob. 3. The condenser and the resistance in the preceding problem are connected in series, instead of in parallel. What is the equivalent parallel combination? Ans. C p = 1.387 mf.; r p = 926 ohms. Prob. 4. The voltage at the receiver end of a 25-cycle, single-phase transmission line is 45 + j'57 kv.; the load current is 178 + j 69 amp. The series magnetic impedance of the line is 32 + j 68 ohms, and its capacity is 4.24 mf. Calculate the generator current and voltage. For purposes of calculation, one half of this capacity can be assumed to be CHAP. XVIII] DIELECTRIC REACTANCE 205 connected across the generator end of the line, the other half across the receiver end. Solution: The dielectric susceptance at the receiver end of the line is -2 * X 25 X 2.12 X KT 6 = -0.333 X 10~ 3 mho. The corresponding charging current is j 0.333 X 10~ 3 (45000 + .; 57000) = - 19 + j 15 amp. Consequently the total line current is 159 + j 84 amp. The line drop is (159 + j 84) (32 + j 68) = -624 + j 13500 volts. The generator volt- age is 44.38 + j 70.5 kv. The charging current at the generator end is j 0.333 (44.38 + j 70.5) = -23.5 + j 14.79 amp. The generator current is 135.5 + j 98.8 amp. Prob. 5. Explain the physical reason why a dielectric susceptance in- creases with the frequency, while a magnetic susceptance is inversely proportional to it. Prob. 6. Investigate the influence of a condenser in a circuit to which a non-sinusoidal voltage is applied; give a treatment similar to that in Art. 23. Show that the presence of an elastance accentuates higher harmonics in the current, while an inductance tends to diminish them. Make the physical reason for this difference clear to yourself. 67. Current and Voltage Resonance. Let a condenser be connected in parallel with a pure reactance coil, across an alter- nating-current line. Let the current through the condenser be 5 amp., leading, and that through the coil, 3 amp., lagging. Then the total current supplied from the generator is 2 amp., leading. Thus, we have the paradox that the resultant current is smaller than either of its components. It is even possible to adjust the permittance and inductance to such values as to make the leading and lagging components equal, in which case the gen- erator current is zero. This condition is called current resonance. When the line current is reduced, to zero, total or perfect reson- ance takes place; otherwise the resonance is called partial. The condition for perfect resonance is that the lagging current shall be equal to the leading current, or, what is the same, the dielectric susceptance must be numerically equal to the magnetic suscept- ance. Thus, if there is no resistance in either circuit, 24G-1/04L), from which 2irfVCL = l (350) From this equation, any one of the three quantities /, (7, and L can be determined, when the other two are given. Condition (350) may be fulfilled for the frequency of one of the higher bar- 206 THE ELECTRIC CIRCUIT [A KT . 67 monies of an e.m.f. wave, in which case we have partial resonance for the fundamental wave, and perfect resonance for one of the harmonics. If such is the case, the line current does not contain this harmonic, although it may be present to a considerable amount in the two branch currents. From the point of view of energy, current resonance consists in a periodic transformation of the potential energy of the elec- trostatic field into the kinetic energy of the magnetic field, and vice versa. When the current is at its maximum, the energy of the magnetic field of the reactance coil is also a maximum. But at this moment the voltage, and consequently the electrostatic displacement, are equal to zero, so that the whole energy of the circuit is in the magnetic field. One quarter of a cycle later, the displacement and the stored energy in the condenser are at a maximum, but the current and the magnetic field are equal to zero. At intermediate moments, the energy is contained partly in the electrostatic, and partly in the magnetic field. When condition (350) is satisfied, the maxima of the two energies are numerically equal, and the system " oscillates " freely in the electrical sense, in a manner analogous to the swinging of a pendu- lum. The generator merely maintains the necessary frequency, and supplies the i z r loss. Without this loss, it would not be necessary to have the generator at all; the oscillations, once started, would continue indefinitely at the proper frequency. When the two energies are not equal, there must be a cyclic ex- change of energy between the generator and one of the branches; namely, the one whose storage capacity for energy, at the gener- ator frequency, is larger than that of the other branch. We then have partial current resonance. The presence of resistance in either branch obscures the effect of resonance to some extent, leaving, however, its general char- acter unchanged. The best way to see the influence of resistance is to replace each impedance by its equivalent parallel combina- tion. We then have two pure susceptances with reactive currents, and two conductances, the currents through which are in phase with the line voltage. The energy supplied to the conductances is converted into heat, and thus does not enter into the electrical oscillations. Let now a dielectric reactance be connected in series with a magnetic reactance, across an alternating-current line. The cur- CHAP. XVIII] DIELECTRIC REACTANCE 207 rent through the two devices is the same, and may be taken as the reference vector. Let the dielectric reactance be such as to produce across the condenser a drop of 1000 volts, lagging behind the current by 90 electrical degrees. Let the voltage across the reactance coil be equal to 900 volts, leading the current by 90 degrees. With these conditions, the total line voltage is equal to 100 volts, lagging behind the current by 90 degrees. Thus, with a line voltage of only 100, it is possible to produce partial voltages of 1000 and 900 respectively. This condition is called voltage resonance. When the two reactances in series are equal, we have complete voltage resonance; otherwise the resonance is partial. The student will readily see that the condition for complete volt- age resonance is also expressed by eq. (350). In this case, the presence of resistance has no effect upon the correctness of the equation. By reading again the foregoing discussion of current resonance, and applying it to voltage resonance, the points of similarity and the differences between the two will be easily One has to be on guard against possible resonance and a dangerous rise in potential in the operation of transmission lines and extended cable systems, because there the presence of per- mittance and inductance offers favorable conditions for surges between the dielectric and magnetic energies. These surges either produce large currents which open the circuit-protecting devices and interrupt the service, or the potential is raised to a value at which the insulation of the system is broken down. With a clear understanding of the principle of interchange of energy explained above, the student ought to be able to follow without difficulty special works on the subject. 1 Prob. 1. A magnetic reactance of 65 ohms is connected in parallel with a permittance of 73.6 mf., across a 2200-volt, 25-cycle circuit. De- termine the total current, and the component currents, through the react- ance and through the condenser. Ans. 33.85 25.44 = 8.41 amp. (lagging). This is a case of partial current resonance, the total current being smaller than one of its com- ponents. 1 See W. S. Franklin, Electric Waves; C. P. Steinmetz, Electric Dis- charges, Waves, and Impulses; also his larger work on Transient Electric Phenomena and Oscillations. Some elementary experiments and curves of resonance will be found in V. Karapetoff's Experimental Electrical Engineer- ing, Vol. II, Arts. 440 to 445. 208 THE ELECTRIC CIRCUIT [ART. 68 Prob. 2. The pcrmittor and the reactance coil given in the preceding problem are connected across the same line in series, instead of in parallel. Find the total current and the component voltages. AIMS. 102.3 amp. (leading) ; 2200 = 8850 - 6650 volts. This is a case of partial voltage resonance, the voltage drop across each of the two devices being larger than the applied voltage. Prob. 3. The elastance of a 60-cycle underground cable system is equal to 11.5 kilodarafs; at what value of the inductance in the circuit is resonance of the seventh harmonic to be feared? Ans. 1.65 millihenry. 68. Voltage Regulation of a Transmission Line, Taking Its Distributed Permittance into Account. The voltage regulation of a transmission line, disregarding its permittance, is treated in Art. 33. The value of the permittance of a single-phase line is deduced in Art. 60, while in Chapter 17 the effect of the charging current in a three-phase line is considered, and it is shown how to calculate the permittance of the equivalent single-phase line. The inductance of transmission lines is treated in Chapter 11 of the author's Magnetic Circuit. It remains now to show how to determine, for a given load, the relation between the generator and receiver voltages of an equivalent single-phase line, knowing its constants; viz., the values of the distributed resistance, mag- netic reactance, and dielectric susceptance. Let the total resistance of the equivalent single-phase line be r ohms, and its magnetic reactance x ohms. Then the series impedance of the line is Z = r + jx (351) Let the dielectric susceptance of the line be b mhos, where &, ac- cording to eq. (347), is a negative quantity; and let the leakage conductance to the ground be g mhos. Then the shunted ad- mittance of the line is Y = g-jb (352) The leakage conductance is due to imperfect insulation of the line, and may also be made to take into account the corona loss, if any exists. The value of g can only be estimated, and in most cases may be safely neglected. It is introduced here in order to obtain a more general result, at the same time making the ex- pressions for Z and Y symmetrical. Since Y is uniformly distributed along the line, the current changes as the distance from the generator increases; and there- CHAP. XVIII] DIELECTRIC REACTANCE 209 fore it is necessary to consider the electrical relations in an in- finitesimal length ds, at some intermediate point of the line. Let the line voltage at this point be E, and the line current, /. The series impedance of the element ds is (Z/l) ds, and its shunted admittance is (Y/l) ds, where I is the total length of the line. Let dE be the increment in the voltage in the length ds, and let dl be the corresponding increment in the line current due to the shunted admittance. We have then dE =-I(Z/l)ds, ..... (353) and dl = - E(Y/l) ds ...... (354) The minus sign is needed on the right-hand side of eq. (353), because the drop in voltage / (Z/l) ds causes a decrement in E. Likewise the charging current E(Y/l)ds causes a decrement in the line current. Equations (353) and (354) contain two dependent variables, E and /. To eliminate /, we divide both sides of eq. (353) by ds and take the derivative with respect to s. The result is Substituting the value of dl/ds from eq. (354), we obtain d*E/ds 2 = EZY/l* ....... (355) This is a differential equation of the second order for E. We shall omit the solution of it and give only the result, for two reasons: first, because most students are not familiar with the methods of integration of differential equations; and secondly, because the solution is most conveniently expressed in hyperbolic functions of a complex variable, a form of function unknown to most students of engineering. 1 Fortunately, even for the longest transmission 1 The simple theory of hyperbolic functions and the solution of eq. (355) may be found, among others, in the following works and articles : McMahon, Hyperbolic Functions; Dr. Kennelly, Applications of Hyperbolic Functions to Electrical Engineering; Dr. Steinmetz, Transient Electric Phenomena; Fender and Thomson, " The Mechanical and Electrical Characteristics of Transmission Lines," Trans. Amer. Inst. Electr. Engrs., Vol. 30 (1911); W. E. Miller, " Hyper- bolic Functions and Their Application to Transmission Line Problems," General Electric Review, Vol. 13 (1910), p. 177; M. W. Franklin, "Transmission Line Calculations," ibid., p. 74. For a proof of expansion (356), see Blondel and Le Roy, "Calcul des Lignes de Transport d'Energie a Courants Alternatifs en tenant compte de la Capacity et de la Perditance Reparties," La Lumiere 210 THE ELECTRIC CIRCUIT [ART. 68 lines built or projected, the solution can be represented with suffi- cient accuracy by a few terms of an infinite series, as follows: E, = E 2 (l + %YZ + & F 2 Z 2 + etc.) + 7,Z(1 + | YZ + T W F 2 Z 2 + etc.). (356) In this equation E\ is the generator voltage, E 2 the receiver volt- age, the same as in Art. 33, and / 2 is the load current. Both F and Z are known complex quantities, and therefore their product and the square of the product .are also known. The terms in- volving F 2 Z 2 are negligibly small in many cases. Instead of eliminating / from eqs. (353) and (354), E may be eliminated by a similar process, giving a differential equation for /, analogous to eq. (355). The solution of this equation is !i = /(! + I YZ + ^ F 2 Z 2 + etc.) + E 2 Y (1 + | YZ + T io F 2 Z 2 + etc.), (357) where /i is the generator current, and 1% the load current. The general form of eqs. (356) and (357) is the same as that of the corresponding equations in Art. 33, and in Chapters 11, 12, and 13, so that the methods of calculation indicated there are appli- cable here, with self-evident modifications. Neglecting the leakage conductance g in eq. (352), and using the value of permittance given in Art. 60 above, also the value of inductance from Art. 61 of the Magnetic Circuit, we find that YZ = (Z/1000) 2 (-v+jw), .... (358) and consequently F 2 Z 2 = (Z/1000) 4 (*> 2 - w* - 2jvw), . . . (359) where v = 0.09514 (0.1 /) 2 [0.46 + 0.05/log (b/d)], . (360) and w = 0.1515 /r'/log(6/a) (361) In these expressions, I is the length of the line in kilometers, and r' is the resistance per kilometer of one conductor, in ohms. With Elcclrique, Vol. 7, 2nd Scries (1909), p. 355; also J. F. H. Douglas, "Trans- mission Line Calculations," Electrical World, Vol. 55 (1910), p. 1066; arid Dr. Stoinmetz, Engineering Mathematics, p. 204. The best tables of hyperbolic functions are those published by the Smithsonian Institution; briefer tables will be found in McMahon's book and in the General Electric Review, Vol. 13, supplement to No. 5. See also Seaver's Mathematical Handbook, pp. 85 and 266. CHAP. XVIII] DIELECTRIC REACTANCE 211 the extreme values of the ratio b/a of say 10 and 1000, the value of log (b/a) varies within the narrow limits of 1 to 3, so that the second term in the brackets in eq. (360) is comparatively small. In practice, the value of the whole expression in the brackets in formula (360) is usually between 0.48 and 0.50. This fact is taken advantage of in numerical calculations which do not require particular accuracy. Prob. 1. Check the numerical coefficients in formulae (360) and (361). Prob. 2. For a given receiver voltage, calculate the generator voltage, at no load and at full load, for some very long transmission line, the dimensions of which are taken from a descriptive article. Prob. 3. Solve problem 2 by the use of tables of hyperbolic functions, following the method indicated in one of the references in the footnote. Compare the results with those obtained in problem 2, and make clear to yourself the relative simplicity, and the limits of accuracy, of the series when one, two, or three terms are used. 69. Approximate Formulae for the Voltage Regulation of a Transmission Line, Considering Its Permittance Concentrated at One or More Points. Instead of treating the permittance of a transmission line in the correct manner described in the preceding article, it is sometimes assumed to be concentrated at one or more points along the line. The calculation of voltage regulation then becomes similar to the treatment in Chapters 9 to 13. There is no particular advantage in this approximate treatment as far as the simplicity of numerical computations is concerned, because the formulse obtained are similar to eq. (356). It is advisable, however, for the student to deduce such formulae in order to see for himself that the form of eq. (356) is a rational one; more- over, this gives him one more exercise in the use of complex quantities. (a) The simplest assumption is to consider one half of the line permittance (and leakage, if any) concentrated at the gener- ator end of the line, the other half at the receiver end. The load current is in this case apparently increased by the current E 2 % Y through the permittance \ Y connected in parallel with the load, so that the total receiver current is equal to / 2 + \ E 2 Y. Hence, the generator voltage is #1 = E 2 (l + i FZ) + 7 2 Z. . (362) 212 THE ELECTRIC CIRCUIT [ART. 69 Comparing this formula with eq. (356), we see that the principal terms are identical, the difference being in the additional terms containing higher powers of YZ. If the influence of the line per- mittance is small, for instance in short lines, the results calculated by means of both formulae differ from each other but very little. There is no reason, however, why the accurate expansion (356) should not be used in all cases, taking as many terms as are re- quired in a given problem. The generator current, with the capacity concentrated at both ends, is 7, = 7, + i YE, + i YE lt or, substituting the value of E\ from eq. (362), !i = JU + *YZ) + E 2 Y(1 + J YZ). . . (363) This formula is similar to eq. (357), and differs from it only in the values of the coefficients of the minor terms. (b) The line permittance and leakage may also be concen- trated at the middle point of the line, in which case a diagram of connections is obtained similar to Fig. 42, except that the suscept- ance is dielectric and not magnetic. Introducing the voltage at the center of the line as an auxiliary unknown quantity, and eliminating it from the result, we obtain Ei = E 2 (l + i YZ) + / a Z(l + I YZ), . . (364) and Ii = | 2 (1 + YZ) + E,Y (365) (c) A closer approximation is obtained by assuming a part of the line permittance concentrated at the middle point, and the rest at both ends of the line. The fractions of the total permit- tance to be assigned to these three points are determined from Simpson's Rule for approximate integration; namely, according to this "parabolic " rule, y ave = [l/(3 n)] [y Q + 4 ( yi + y s + etc. + y n ^) + 2 (y t + y, + etc. + y n - 2 ) + y n ], (366) where y ave is the average ordinate of a given curve, n is the num- ber of equal parts into which the total width of the curve is sub- divided, and t/o, yi, etc., are the actual ordinates at the points of division. In the above formula, n must be an even number. Let the given curve represent some arbitrary distribution of the CHAP. XVIII] DIELECTRIC REACTANCE 213 permittance along the line, and let n = 2. The foregoing formula gives <7 2 '), .... (367) where the C's are marked with the prime sign to indicate that they refer to unit length of the line. But in reality the permit- tance is uniformly distributed over the length of the line, so that Ca,e' = Co' = Ci = C 2 f . Multiplying both sides of eq. (367) by the length I of the line, we obtain C = iC+fC + |C ...... (368) This means that two thirds of the total permittance must be con- centrated at the middle of the line, and one sixth at each end. 1 With this distribution of permittance it is again convenient to introduce the voltage at the center of the line as an auxiliary quantity. The relation between the load voltage and the gener- ator voltage is calculated in the well-known manner, by adding the voltage drop in the line to the load voltage. The result is Ei = E. 2 (l + YZ + & Y*Z 2 ) + I 2 Z(1 -H FZ); . (369) !i = /i(l + * YZ + A F 2 Z 2 ) + E 2 Y(l + & YZ + *}- ir r 2 Z 2 ) ............ (370) These formulae come closer to eqs. (356) and (357) than those obtained in the preceding two approximations. Prob. 1. Check formulae (364) and (365) by actually performing the algebraic transformations. Prob. 2. Check formula (369) and (370) by actually performing the algebraic transformations. Prob. 3. If it be desired to have the permittance concentrated at five equidistant points along the line, show that according to Simpson's Rule one sixth of the total permittance must be placed in the middle, one twelfth at each end, and the rest at one quarter and three quarters of the length of the line. 1 This result has been first indicated by Dr. Steinmetz, in his Alternating- Current Phenomena, in the chapter on " Distributed Capacity." APPENDIX. THE AMPERE-OHM SYSTEM OF UNITS. THE ampere and the ohm can be now considered as two arbitrary fundamental units established by an international agree- ment. Their values can be reproduced to a fraction of a per cent according to detailed specifications adopted by practically all civilized nations. These two units, together with the centi- meter and the second, permit the determination of the values of all other electric and magnetic quantities. The units of mass and of temperature do not enter explicitly into the formulae, but are contained in the legal definition of the ampere and the ohm. The dimension of resistance can be expressed through those of power and current, according to the equation P = PR, but it is more convenient to consider the dimension of R as fundamental, in order to avoid the explicit use of the dimension of mass [M]. Besides, there is no direct proof that the physical dimensions of electric power are the same as those of mechanical power. All we know is that the two kinds of power are equivalent one to the other. For the engineer there is no need of using the electrostatic or the electromagnetic units; for him there is but one ampere-ohm system, which is neither electrostatic nor electromagnetic. The ampere has not only a magnitude, but a physical dimension as well, a dimension which, with our present knowledge, is fun- damental; that is, it cannot be reduced to a combination of the dimensions of length, time, and mass (or energy). Let the dimen- sion of current be denoted by [I] and that of resistance by [R]; let the dimensions of length and time be denoted respectively by the commonly recognized symbols [L] and [T]. The magnitudes and dimensions of all other electric units can be expressed through these foui, as shown in the following table. For the correspond- ing expressions of the magnetic units in the ampere-ohm system, see Appendix I to the author's Magnetic Circuit. 215 216 THE ELECTRIC CIRCUIT TABLE OF ELECTRIC UNITS, AND THEIR DIMENSIONS IN THE AMPERE-OHM SYSTEM Symbol and Formula Quantity Dimensions Name of the Unit j Current [I] Ampere U=I/A Current density [IL- 2 J Ampere per square centimeter Q=IT Quantity of electricity and dielectric flux [IT]* Coulomb (ampere- second) D = Q/A Dielectric flux density [ITL- 2 ] Coulomb per square centimeter E=IR Voltage, difference of [IR] Volt potential, or e.m.f. G=E/l Voltage gradient, elec- [IRL-i] Volt per centimeter tric intensity, or di- electric stress T Resistance j X = 2irfL Reactance [R] Ohm 2=(r 2 +x 2 ) i Impedance ) g =l/r p Conductance j b = l/x p Susceptance [ [R- 1 ] Mho y= 1/2= (0 2 -|-> 2 )* Admittance ) P = G/U Resistivity [RL] Ohm per centimeter cube y=U/G=l/ P Conductivity [R-^- 1 ] Mho per centimeter cube S=E/Q Elastance [RT- 1 ] Daraf C = Q/E=l/S Permittance (capacity) [R-iT] Farad a = G/D Elastivity [RT-iL] Daraf per centime- ter cube K =D/G=l/a Permittivity [R-iTL- 1 ] Farad per centime- ter cube P=EI Power [PR] Watt P' = P/V Density of power [PRL- 3 ] Watt per cubic cen- timeter W'-\GD Stored electric energy Density of electric en- [PRT] [PRTL- 3 ] Joule (watt-second) Joule per cubic cen- ergy timeter F=W/l Force [PRTL- 1 ] Joulecen L " WI " Inductance [RT] Henry * These are also the dimensions of the electric pole strength. The con- cept of pole strength is of no use in electrical engineering, and, in the author's opinion, its usefulness in physics is more than doubtful. The whole elemen- tary theory of electrostatics can and ought to be built up on the idea of stresses and displacements in the dielectric, as is done in this work. THE AMPERE-OHM SYSTEM 217 Other units of more convenient magnitude are easily created by multiplying the tabulated units by powers of 10, or by adding the prefixes milli-, micro-, kilo-, mega-, etc. A study of the physical dimensions of the electric and magnetic quantities is interesting in itself, and gives a better insight into the nature of these quantities. Moreover, for- mute can be checked and errors detected by comparing phys- ical dimensions on both sides of the equation. Let, for instance, a formula for energy be given, W = aQDl/K, where a is a numerical coefficient. Substituting the physical di- mensions of all the quantities on the right-hand side of the equa- tion from the table below, the result will be found to be of the dimensions of energy. This fact adds to one's assurance that the given formula is theoretically correct. A slight irregularity in the system as outlined above is caused by the use of the kilogram as the unit force, because it leads to two units for energy and torque, viz., the kilogram-meter and the joule; 1 kg.-meter = 9.806 joules. Force ought to be measured in joules per centimeter length, to avoid the odd multiplier. Such a unit is equal to about 10.2 kg., and could be properly called the joulecen (= 10 7 dynes). There is not much prospect in sight of introducing this unit of force into practice, because the kilo- gram is too well established in common use. The next best thing to do is to derive formulae and perform calculations, when- ever convenient, in joulecens, and to convert the results into kilograms by multiplying them by g = 9.806. Thus, leaving aside all historical precedents and justifica- tions, the whole system of electric and magnetic units is re- duced to this simple scheme: In addition to the centimeter, the gram, the second, and the degree Centigrade, two other funda- mental units are recognized, the ohm and the ampere. All other electric and magnetic units have dimensions and values which are connected with those of the fundamental six in a simple and almost self-evident manner, as shown in the table above. To appreciate fully the advantages of the practical ampere- ohm system over the C.G.S. electrostatic and electromagnetic systems, one has only to compare the dimensions, for instance, of 218 THE ELECTRIC CIRCUIT current density and voltage gradient in these three systems, as shown below. The Am- pere-ohm System C.G.S. Electro- magnetic System C.G.S. Electro- static System Dimension of current density. . . Dimension of voltage gradient. . IL- 2 IRL" 1 L-MT-V* LM*T-V L-M*T-'* L-*M*T-\-* BIBLIOGRAPHY. ALTERNATING-CURRENT PHENOMENA, by Chas. P. Steinmetz. THEORETICAL ELEMENTS OF ELECTRICAL ENGINEERING, by Chas. P. Stein- metz. ALTERNATING CURRENTS, by Bedell and Crehore. VECTORS AND VECTOR DIAGRAMS, by Cramp and Smith. REVOLVING VECTORS, by G. W. Patterson. ELECTRICAL ENGINEERING, by Thomalen. DlE WlSSENSCHAFTLICHEN GRUNDLAGEN DER ELEKTROTECHNIK, by Benischke. DIE WECHSELSTROMTECHNIK, by E. Arnold. PROBLEMS IN ELECTRICAL ENGINEERING, by Waldo V. Lyon. ELECTRICAL PROBLEMS, by Hooper and Wells. THE ELEMENTS OF ELECTRICAL ENGINEERING, by Franklin and Eaty. PRINCIPLES OF ELECTRICAL ENGINEERING, by H. Pender. MODERN VIEWS OF ELECTRICITY, by Oliver Lodge. ELECTRIC WAVES, by W. S. Franklin. ELEMENTS OF ELECTROMAGNETIC THEORY, by S. J. Barnett. KAPAZITAT UNO INDUKTIVITAT, by Ernst Orlich. 219 INDEX PAGE Addition and subtraction of sinusoidal currents and voltages 40 Admittance, definition of 76 expressed as a complex quantity or operator 89 Admittances in parallel 76 in series 80 Air, dielectric strength of 165 permittivity of 151 Alternating-current power when current and voltage are in phase 45 currents, advantages of 31 currents, polyphase 99 Ampere-ohm system of units 3, 215 Amplitude factor, definition of 51 Analogy, hydraulic, to an inductive circuit 65 hydraulic, to Ohm's law 2 hydraulic, to the dielectric circuit 145 hydraulic, to the flow of electricity 25 mechanical, to a charged dielectric 159 thermal, to Ohm's law 2 thermal, to the flow of electricity 24 Apparent power, definition of 56 Average value of alternating current or voltage 50 Cable, elastance of single-core 171 insulation resistance of 26 Cables, grading insulation of 174 Capacity, see Elastance and Permittance. electrostatic, definition of 147 specific inductive 151 Charging current of condenser 203 current of transmission line, see also Elastance 193 currents of three-phase line with symmetrical spacing 196 currents of three-phase line with unsymmetrical spacing 201 Circle coefficient of induction motor 138 diagram of induction motor or transformer 136 Circuit, alternating-current 31 dielectric 143 dielectric, hydraulic analogue 145 direct-current 1 polyphase 99 221 222 INDEX Coefficient, leakage, of induction motor 138 of self-induction see Inductance. temperature of electric resistivity 5 Complex expression for admittance expression for impedance quantity, definition of 85 Component, energy, of current or voltage 56 reactive, of current or voltage 56 Condenser, charging current of 203 definition of 143 Conductance and resistance, how related, in an A.C. circuit 79 definition of 2 dielectric 109 Conductances, addition of 8 Conductivity, definition of 14 Conductor, definition of 1 of variable cross-section 22 unit, definition of 13 Continuous current, see Current, direct. Core loss of transformer Ill Corona, electrostatic 167 Current, alternating 31 density, definition of 15 direct 1 due to non-sinusoidal voltage 71 effective value of alternating 48 energy and reactive components of 56 primary, of induction motor 126 radial flow of 26 refraction, law of 28 resonance 205 transient, in opening and closing a circuit 71 Currents, polyphase alternating 99 Cycle of alternating wave, definition of 33 Cylinders, elastance between two large parallel 188 Daraf, definition of 148 Delta-connected three-phase system 105 Dielectric circuit 143 conductance 169 elastivity of 152 energy stored in 158 flux density 154 flux, refraction of 164 hysteresis 169 nature of 144 permittance of 147 INDEX 223 PAGE Dielectric, permittivity of 151 reactance 204 strength 164 stress 156 susceptance 203 Dimensions of units, table of 216 Dispersion factor of induction motor 138 Displacement, electric, illustrated 144 Disruptive voltage, see Dielectric strength. Effective value in terms of harmonics 54 value of variable current 49 values of alternating currents and voltages, definition of 48 Elastance between concentric spheres 175 between small spheres 179 between two large parallel cylinders 188 definition of 148 of a single-core cable 171 of a single-phase line 176 of a three-phase line with symmetrical spacing 196 Elastances, addition of 149 Elastivity, definition of 152 Electric displacement illustrated 144 intensity, definition of 16 intensity in the dielectric circuit 155 power 10 Electromotive force, see Voltage. Electrostatic, see also Dielectric. capacity, definition of 147 corona 167 field, nature of 143 Energy component of current or voltage 56 converted into heat 10 density of 158 stored in dielectric 158 stored magnetic 62 unit of electrical 11 Equipotential surfaces denned 22 Equivalent resistance, definition of 7 series and parallel circuits 78 sine-wave, definition of 53 Exciting admittance of transformer Ill Exponential expressions for vectors and operators 97 Farad, definition of 147 Field, electrostatic, see Electrostatic field. Fleming's method for calculating the effective value of an irregular curve. 52 224 INDEX PAGE Flux density, dielectric 154 dielectric, see Dielectric flux. Form factor, definition of 51 Fourier series 43 Frequency of alternating current or voltage, definition of 33 Gradient, voltage, in the dielectric circuit 155 voltage, definition of 16 Ground, influence upon the charging currents in a three-phase line .... 202 influence upon the elastance of a single-phase line 180 Harmonics, definition of 41 effects of elastance and inductance on 205 Heaviside, Oliver, nomenclature of 152 Henry, definition of 62 Heyland diagram of induction motor or transformer 136 Homopolar machine 31 Horse-power, English, defined 10 metric, defined 10 Hydraulic analogue of inductive circuit 65 analogue of the dielectric circuit 145 Hysteresis, dielectric : . . . 169 Images, Kelvin's method of electric 180 Impedance, definition of 67 equivalent, of transformer 116 expressed as a complex quantity or operator 88 Impedances in parallel 80 in series 68 Inductance, definition of 60 influence of, with non-sinusoidal voltage 69 Induct ion motor, approximate analytical treatment 125 characteristics with locked rotor 123 circle coefficient or dispersion factor 138 circle or Heyland diagram of 136 equivalence to a polyphase transformer 122 equivalent electrical diagram of 122 exact analytical treatment 139 input per phase 127 magnetomotive forces in 124 maximum output of 131 primary current and power-factor 126 pull-out torque of 130 secondary resistance and reactance reduced to primary. 133 slip, calculation of 126 slip, defined 123 squirrel-cage rotor 134 INDEX 225 PAGE Induction motor, starting torque of 129 torque of 127 Inductive reactance, see Reactance. Inertia as an analogue to inductance 60 Insulation, see also Dielectric. condenser type 175 grading of 174 Intensity, electric, definition of 16 electric 155 factor, illustrated 10 Irregular paths, resistance and conductance of 27 Joule, definition of 11 relation of, to thermal units 11 Joulecen, definition of 217 Joule's law 10 Kelvin's law of economy 15 Kelvin's method of electric images 180 Kirchhoff's laws 17 Law, Joule's 10 Kirchhoff's first 18 Kirchhoff's second 19 of current refraction 28 of flux refraction 163 of economy, Kelvin's 15 of minimum resistance 27, 160 Ohm's, synopsis of 1 Leakage conductance of transmission line 208 factor of the induction motor 138 Lehmann, Dr., method of finding resistance of irregular conductor 28 Lehmann, Dr., method of mapping irregular field 162 Line, see Transmission line. Magnetizing current of transformer Ill Mean, see Average. Mesh connection of polyphase system 101 Mho, definition of 2 Minimum resistance, law of 160 Motor, induction, see Induction motor. Neutral points of polyphase system 104 Nomenclature xii Notation xiii Ohm's law, for an infinitesimal conductor 25 hydraulic analogy to 2 226 INDEX PAGE Ohm's law, synopsis of 1 thermal analogy to 2 Operator, admittance 89 impedance 88 Operators expressed as exponential functions 97 polar expressions for 93 Output, maximum, of induction motor 131 Parallel connection of conductors 7 connection of impedances 80 connection of susceptances and conductances 77 Performance characteristics of the induction motor 122 characteristics of the transformer 108 characteristics of the transmission line 94, 208 Permittance, see also Elastance. definition of 147 distributed, of transmission lines 208 Permittances, addition of 149 Permittivity, definition of 151 relative . 151 Phase angle, definition of 34 Phase displacement expressed by projections of vectors 91 Polar coordinates, vectors and operators in 93 Polyphase system, definition of 99 system, neutral points of 104 Power, alternating-current, when current and voltage are in phase 45 apparent 56 as double-frequency function 59 average, of non-sinusoidal waves 57 electric 10 expressed by projections of vectors 91 expression for average value in alternating-current circuit 56 practical unit of 10 real 56 Power-factor, definition of 56 of induction motor 126 with non-sinusoidal waves 58 Projections of vectors, addition and subtraction of 82 Quadratic mean value 50 Quantity factor, illustrated 10 Quarter-phase system, star- and mesh-connected 101 Radial flow of current 26 Rayleigh, Lord, method of finding permittance of dielectrics of irregular shape 161 method of finding resistance of irregular conductor 28 Reactance and susceptance in an A.C. circuit 79 INDEX 227 PAGE Reactance, definition of 64 dielectric 204 equivalent, of transformer 118 inductive 63 leakage, of transformer 110 secondary, of induction motor reduced to primary 133 Reactive component of current or voltage 56 Refraction of current 28 of dielectric flux .\ 164 Regulation, see also Voltage regulation. speed, of induction motor 123 voltage, of the transformer 108 voltage, of the transmission line 94 Resistance and conductance in A.C. circuits 79 and temperature, relation between 5 definition of 1 equivalent, definition of 7 equivalent, of transformer 118 law of minimum 160 secondary, of induction motor reduced to primary 133 Resistances, addition of 8 Resistivity, definition of 13 Resonance, current 205 voltage 207 Series connection of admittances 80 connection of conductors 7 connection of impedances 68 Series-parallel combination of permittances and elastances 150 combination of resistances 9 Sine-wave, definition of 32 definition of equivalent 53 of current or voltage 31 represented by a vector 36 Single-phase line, effect of the ground upon the elastance of 180 elastance of 176 equations of lines of force and equipotential surfaces 184 Sinusoidal currents and voltages, rule for addition and subtraction of ... 40 Slip, calculation of of induction motor defined 123 Specific capacity, see Permittivity, relative, resistance, see Resistivity. Spheres, elastance between concentric 175 elastance between small 179 equations of lines of force and equipotential surfaces between . . 187 Square root of mean square value, defined 49 Star connection of polyphase system 101 228 INDEX PAGB Steinmetz, Dr. C. P., symbolic notation of 83 Stream lines, definition of 22 Superposition, principle of 177 Susceptance and reactance, how related, in an A.C. circuit 79 definition of 75 dielectric 203 Susceptances in parallel 75 Symbols, list of xiii System, four-wire, two-phase 99 three-wire, two-phase 100 polyphase, definition of . 99 quarter-phase, star- and mesh-connected 101 three-phase, delta-connected 105 three-phase, V- and T-connected 107 three-phase, Y-connected 103 T-connected three-phase system 107 Temperature coefficient 5 Thermal resistance, definition of 2 Three-phase line, influence of the ground upon the charging currents. . . . 202 with symmetrical spacing, elastance and charging current of. ... 196 with unsymmetrical spacing, charging currents of 201 Three-phase system, delta-connected 105 V- and T-connected 107 Y-connected 103 Time constant of electric circuit 72 Torque of induction motor 127 pull-out, of induction motor 130 starting, of induction motor 129 Transformer, constant-potential, definition of 108 core loss of Ill equivalent impedance of 116 equivalent resistance and reactance of 118 exciting admittance of Ill leakage reactance of 110 magnetizing current of Ill ohmic drop in 110 reactive drop in 110 vector diagram of 113 voltage ratio of 109 voltage regulation 108, 115, 120 Transient current in opening and closing a circuit 71 Transmission line, see also Three-phase line and Single-phase line. leakage conductance of 208 voltage regulation of 94 voltage regulation, taking account of distributed per- mittance . 208 INDEX 229 PAGE Tube of current, meaning of 23 Two-phase, four-wire system 99 three-wire system ; 100 Unit conductor, definition of 13 of conductance 2 of electrical energy 11 of resistance 1 Units, C. G. S. and practical systems 3 international electrical 3 table of names and dimensions of 216 the ampere-ohm system 215 V-connected, three-phase system 107 Vector, definition of 36 diagrams, examples of 100, 102, 113 used to represent a sine-wave . 36 Vectors, addition and subtraction of 37 addition and subtraction of projections of 82 expressed as exponential functions 97 in polar coordinates 93 Voltage, effective value of alternating 48 energy and reactive components of 57 gradient in the dielectric circuit 155 gradient, definition of 16 gradient, rupturing values of 165 regulation of the transformer 108, 115, 120 regulation of the transmission line 94 regulation of transmission line, with permittance concentrated. 211 regulation of transmission line, with permittance distributed. . . 208 resonance 207 Watt, definition of 10 Wave form of alternating current or voltage 51 representation of irregular 41 Y-connected, three-phase system 103 Yrneh, definition of 148 UNIVERSITY OF CALIFORNIA, LOS ANGELES THE UNIVERSITY LIBRARY This book is DUE on the last date stamped belo^ aw 000343367 9