LIBRARY UNIVERSITY OF CALIFORNIA. Class ,- A CAMBRIDGE PHYSICAL SERIES. GENERAL EDITORS: F. H. NEVILLE, M.A., F.R.S. AXD W. C. D. WHETHAM, M.A., F.R.S. A TREATISE ON THE THEORY OF ALTERNATING CURRENTS Son&on: C. J. CLAY AND SONS, CAMBKLDGE UNIVEKSITY PKESS WAKEHOUSE, AYE MAEIA LANE, AND H. K. LEWIS, 136, GOWER STEEET, W.C. lasgofo: 50, WELLINGTON STREET. ILetpjtg: F. A. BROCKHAUS. fo lark: THE MACMILLAN COMPANY. anH (Calcutta : MACMILLAN AND CO., LTD. [All Rights reserved.] A TREATISE ON THE THEORY OF ALTERNATING CURRENTS by ALEXANDER RUSSELL, M.A., M.I.E.E. Late Scholar and Assistant Lecturer of Gonville and Caius College, Cambridge. Lecturer in Applied Mathematics and Superintendent of the Testing Department, Faraday House, London. Volume I UNIVERSITY Of s4LlF( CAMBRIDGE : at the University Press 1904 PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS. PREFACE. great progress that has been made in the application of alternating currents to industrial purposes makes it desirable to collect together and examine the mathematical theorems which electricians use in their everyday work. In this volume the more general theorems are collected, and proofs are given of the more important of them, due stress being laid on the assumptions that it is sometimes necessary to make, and on the consequent limitations in the use of the formulae. In the second volume the theory of alternators, motors, transformers, converters, and of the transmission of power by polyphase currents will be given. The reader is supposed to be familiar with the elementary theory of electricity and magnetism, and to have a working knowledge of trigonometry and of the elements of the calculus. A knowledge of De Moivre's theorem, De Moivre's property of the circle and of hyperbolic sines and cosines will be found essential. Lord Kelvin's bei and ber functions given in Chapter xvi will be understood at once from their definitions. The references, given at the end of nearly every chapter, are to the books or papers I have consulted when writing the chapter. In many cases they contain more detailed discussions of the same or similar problems, and will be helpful to the student. The proofs of several of the problems could have been considerably shortened by deducing them from the general equations of electro- magnetism, but, in my opinion, the proofs given bring out the physical meaning more clearly to the electrical student. 1G5764 vi PREFACE The early chapters deal mainly with inductance and capacity, and it is hoped that the practical formulae in them will be helpful to the working electrician. In Chapter v, illustrations are given of methods by means of which the capacities of polyphase cables and overhead wires can be calculated. It is also shown how the inductances of these combinations of conductors for the case of surface currents can be found. The definition and the theory of the power factor given in Chapter VI are now almost universally adopted by electricians. It was thought necessary to define terms like 'watt current' and 'wattless current' 'le courant watte' and *le courant dewatte' of French writers as they are so widely used. In Chapter IX, the test room methods of measuring power are given, and in Chapter x the method, when discussing practical problems, of replacing an air-core transformer by its equivalent net-work is explained. In Volume II this method will be extended to iron-core transformers. In Chapter XII some problems in two phase theory are discussed graphically and illustrate how the theorems of solid geometry can be applied usefully in this case. In Chapter XIII the main problems in the theory of phase indicators and induction type watt-hour meters are stated, and approximate solutions are obtained. The theory of rotating magnetic fields, given in Chapter xiv, is founded on a paper which I wrote for the Electrical Review in 1893. In Chapter XV the interesting problem of the nature of the magnetic field round parallel wires, carrying polyphase currents, is discussed. Although complete solutions of the problem of the eddy currents in magnetic metals have not yet been obtained, the approximate solutions obtained by J. J. Thomson and Oliver Heaviside, which are given in Chapter xvi, will be found most helpful in practical work. Heaviside's solution is given in terms of bei and ber functions, and so, in given cases, numerical values can be found readily by PREFACE vii means of the tables given in this volume. In Chapter xvn a slight sketch is given of the useful method of duality. I have to thank Mr G. F. C. Searle, M.A., of St Peter's College, Cambridge, University Lecturer in Experimental Physics, for the many suggestions and emendations he has made, during the printing of the work. Of these suggestions and emendations I have freely availed myself. Particularly I have to acknowledge the valuable help he has given me in the problems connected with capacity, eddy currents and inductance, more especially the inductance when the currents are confined to the surface of the conductors. I have to thank him also for his unwearying kindness in reading the proofs and for checking nearly all the formulae given. 1 take this opportunity of thanking Dr Charles Chree, F.R.S., the Superintendent of the Observatory Department of the National Physical Laboratory for his help and encouragement. I am also indebted to Mr W. C. Dampier Whetham, F.R.S., who has edited this work, for many valuable suggestions and criticisms. Finally, I have to thank the Council of the Institution of Electrical Engineers and The Electrician Printing and Publishing Company for their kind permission to make what use I pleased, in the preparation of this volume, of any of my papers which they have published. A. R. 2 BELLEVUE PLACE, RICHMOND, SURREY. October, 1904. CONTENTS. HAPTER PAGE I. Introduction. Electrostatics. Magnetism. Electrodynamics 1 II. Alternating Current in an Inductive Circuit. Self-Induct- ance Formulae 40 III. Effective Values. Choking Coil and Condenser Currents. Resonance 65 IV. Electrostatic Capacity. Maxwell's Equations. Capacity Formulae for Parallel Cylinders. The Capacities of Three Core Cables in Terms of Maxwell's Coefficients . 89 V. Capacity Formulae for Cables. The Inductances of Parallel Wires with Surface Currents 121 VI. The Theory of the Power Factor 145 VII. The Method of the Complex Variable 161 VIII. Vectors in Space 180 IX. The Measurement of Power 189 X. The Air Core Transformer 211 XL The Theory of Three Phase Currents 219 XII. The Theory of Two Phase Currents 243 XIII. The Conversion of Polyphase Systems. Phase Indicators . 261 XIV. Rotating Magnetic Fields. Gliding Magnetic Fields . . 280 X CONTENTS CHAPTEK PAGE XV. The Magnetic Field round Polyphase Cables. Losses in Single Phase and Three Phase Cables .... 304 XVI. Eddy Current Losses. Oliver Heaviside's and J. J. Thomson's Formulae .......... 350 XVII. The Method of Duality 380 ADDITIONS 400 INDEX 403 SYMBOLS xi ABBREVIATIONS xii ERRATA. P. 9, line 21, after B insert ', B completely surrounding A, and.' P. 51, line 14, for ' .ft' write ' 0*.' P. 87, line 3, dele 'ultimately' and write ', since \ and q are zero at the instant of closing the switch in the main, A is zero, and thus.' r D P. 87, line 21, dele 'ultimately' and add, 'since e l and I j e 2 dt are zero at the moment of closing the switch in the main, A is zero, and thus.'... P. 192, line 12, dele 'therefore' and insert 'but, when the potentials are kept constant, the gain dW in the electrical energy (see p. 400) is equal to the work done on the disc, and is therefore given by.' SYMBOLS. A, effective value of an alternating current; a constant. A lt A 2 , ..., effective values of the currents in polyphase mains. B, magnetic induction. B^ , maximum value of the magnetic induction. (7, direct current ; a constant. E, maximum value of the alternating voltage ; direct voltage. E lt _E 2 > effective values of star voltages. F, force. G, average torque. H, magnetic force. I, intensity of magnetisation ; maximum value of an alternating current when it follows the harmonic law. ^mx.j maximum value of an alternating current. JTj, J 2 , ..., effective values of the mesh currents. K, capacity between two conductors; capacity of a conductor, ^i.u ^i.2> Maxwell's coefficients of self and mutual induction for electro- static charges. K q , capacity when the charges are constant. K v , capacity when the potentials are constant. L, self inductance. L lil5 L l 2 , ..., coefficients of induction in electro-magnetism. J/, mutual inductance ; magnetic moment. P, power. Q, quantity of electricity. R, resultant electrostatic force; resistance; resistance of primary circuit ; radial force. R! , resistance of primary circuit. E 2 , resistance of secondary circuit. S, area of cross section; resistance of secondary circuit. T, periodic time; tangential force ; longitudinal tension. F, electrostatic potential ; magnetic potential ; volume ; effective value of potential difference. F 1>2 , F 1<3 , ..., effective values of the mesh voltages. W, average value of the power ; power ; energy. Z, impedance. c, thermal capacity per unit volume. e, instantaneous value of the electromotive force, [e], Ecosut + \/^lEsinwt, /, force; frequency; fault resistance. /i /2 > fault resistances of the mains. 2 ALTERNATING CURRENT THEORY [CH. sketch is given of a method of duality founded on these analogies, and Maxwell has shown how important they are in advanced theory. It is convenient to divide the electricities generated when certain bodies are rubbed together into two kinds, Electrostatics. ,, -, ... j . T . which are called positive and negative. It is found that a body charged with positive electricity will repel a body charged with the same kind of electricity and will attract a body negatively charged. By means of a torsion balance Coulomb proved that the force of repulsion / between two small bodies in air possessing electric charges q and q f of like sign is given by the formula /-*. where r is the distance between the bodies and & is a constant. If we define unit charge to be that charge which if concentrated at a point would repel an equal like charge concentrated at a point one centimetre away with a force of one dyne, then k is unity and the formula becomes If the bodies be not immersed in air we must write where X is a constant depending on the medium in which the bodies are placed. This constant is generally called the " specific inductive capacity" of the medium, but the term "dielectric coefficient " is coming into use. Faraday mapped out the electric field which surrounds a charged body by means of lines drawn so that the direction of the resultant force at any point on these lines is in the direction of the tangent at that point. By the resultant force at a point is meant the force with which a unit positive charge placed at the point would be urged if it could be placed there without disturbing the electrical distribution elsewhere. These lines he called lines I] ELECTROSTATICS 3 of force. We shall see later on that we can map out by means of tubes of force not only the direction of the field, but also its strength. The electric potential at a point P due to any electrified bodies in the neighbourhood is the amount of work Potential. . . , . , ..... in ergs that has to be done on a unit or positive electricity to bring it from the boundary of the field to P, the electric distribution being supposed to be undisturbed during the process. If the electric potentials at two neighbouring points P and P' be V and V+dV respectively, and if PP' equal ds, then the work done by the electric forces while unit of positive electricity is moved from P to P' will be V- (F + dV), and if F be the average electric force in dynes from P to P' the work done will also be represented by Fds. Therefore Fds = -dV, dV and F = j-. ds Hence, if we know the mathematical expression for the potential at a point, this equation completely specifies the electric force in any direction at the point. We can therefore represent the attractions and repulsions by means of a single symbol F instead of having to give the components of the force at the point in three directions mutually at right angles. This is one of the advantages of the potential method of treating problems in attractions and repulsions. We should notice that the potential function itself is an undirected quantity ; it possesses merely magnitude. The electromotive force between two points P and Q is defined as the work done in taking a unit of positive electricity from one to the other. Thus if Fj and F 2 be the potentials at P and Q, then the electromotive force between P and Q = F x F 2 = I* Fds. It will be seen that the dimensions of electromotive force, or as it is generally written E.M.F., are the same as those of work divided by electrical quantity. Suppose that we have q units of electricity concentrated at a 12 ALTERNATING CURRENT THEORY [CH. point and that we wish to find the potential V at a point P distant r from 0. By definition J = ( J r If we have n charges q^ = dS cos a Now, if we take the sum of R cos adS over the whole sur- face, we get the total flux which traverses the surface. In symbols Fig. 1. Flux of force over the surface I] GAUSS'S THEOREM 5 for q is constant and the sum of all the solid angles at a point is the ratio of the surface of a sphere to the square of its radius, i.e. 4-7T. If we had n particles with charges q lt q 2 ... inside the closed surface, then we have by addition %R cos adS = #! cos ^dS + ^R^ cos c^dS + . . . This is true whatever the shape of the surface may be, provided that it embraces all the charges. It follows that, if we know ^R cos adS over a closed surface, then we can find the sum of the charges enclosed by dividing this sum by 4?r. It also follows that if there are no charges within the surface, then If we choose three axes OX, OY and OZ at right angles to one another, and take the surface integral of the Equation. norma l force over a small rectangular parallelepiped dxdydz, we get an important equation due to Poisson. Con- sider the part of the surface integral contributed by the two faces of the parallelepiped parallel to the plane YOZ. One side dV contributes +^dydz, and the other ax dV. d (dV Hence by addition we see that the surface integral over these two faces gives j i j -y dxdydz. Proceeding similarly for the other four faces, and equating the total sum to 7rp dxdydz, where p is the volume density of the distribution, we get Poisson's equation This equation is generally written or 6 ALTERNATING CURRENT THEORY [CH. In the particular case when F is the potential at a point in free space, we get Laplace's equation V 2 7=0. If the electric field is uniform and the lines, of force are all parallel to the axis of #, then this becomes dx* where A arid B are constants. If the field be symmetrical about an axis and if r be the distance from the axis of a point P in air at which the potential is F, then we can easily show that d ( dV\ -r(2wr-j- =0. dr \ dr J .-. V=A + B\bgr. If we imagine a tube in which the end starts from a positively electrified body and the sides of which are formed by lines of force, we get what Faraday called a tube of induction. As we move along a line of force, the potential continually diminishes, and hence a line of force can never be a closed curve. We imagine then a tube of force as starting from a positively electrified surface, and ending at a negatively electrified surface. Consider the portion of a tube of force intercepted between two equipotential surfaces. Let dS and dS' be the intercepts on them, and let F and F' be the intensities of the force at the two surfaces respectively. By the intensity of the force at a point we mean the electric force that would be exerted on a unit quantity of electricity placed there. Applying Gauss's equation to this element, and noting that the sides of the tube add nothing to the integral ZR cos adS, since R is zero over the sides, we see that Hence along a tube of force the product of the intensity of the force at a point and the area of the section of the equipotential surface through the point intercepted by the sides of the tube is constant. If the section of the positive surface from which the tube starts [] TUBES OF FORCE 7 contains unit quantity of electricity, we get a unit tube. There- fore the force at a point can be measured by the number of unit tubes which pass through unit area of the equipotential surface at the point. Hence both the direction and the magnitude of the field can be mapped out by means of these unit tubes. Coulomb found experimentally that the resultant force near an electrified conductor was at right angles to its for^hTinten 1 ;^ of surface and that the magnitude of the force was a force near a proportional to the. surface density. The exact relation between these two quantities can be found from the following considerations. In order that the electricity on a conducting body may be in equilibrium, the E.M.F. between any two points must be zero; hence the potential must be constant throughout and equal to its surface value. Thus the bounding surface of the conductor is an equipotential surface, and the resultant force at points infinitely near it is normal to the surface and the force at all points in the conductor is zero. Consider now the surface integral of the normal force over a small closed surface formed by an element dS of an equipotential surface very close to the conductor, the tube of force through the boundary of dS, and a surface inside the conductor con- tinuous with the tube and closing it. The tube of force is supposed to enclose a quantity adS' of electricity spread over an area dS' on the surface of the conductor. We see that FdS is the value of the surface integral, for the sides of the tube and the surface in the conductor contribute nothing to it. Hence by Gauss's theorem When the equipotential surface is infinitely close to the charged body dS = dS'; hence F=4>7T(7. This numerical relation was proved by Poisson. If V be the potential at a point near the surface, and dn be an element of the normal to dS drawn outwards, then F a = ^r I dV 4-7T dn 8 ALTERNATING CURRENT THEORY [CH. If the body be immersed in a medium of which the dielectric coefficient is X, then X dV <7 = -T -y- . 4?r an Since FdS is constant along a tube of force, we see that the quantities of electricity at each end of the tube are equal in magnitude but opposite in sign. If a conductor be placed in an electric field, it clearly follows from the principles we have developed that those parts of it where tubes of induction stop must be negatively electrified and those parts where they begin must be positively electrified. The potential F due to a body cannot have a maximum or a minimum value in free space, for if it had then ^FcosadS taken round a small sphere enclosing the point would not be zero. Hence, if the potential be constant round a closed surface it will be constant at all points in that space, as otherwise it would have a maximum or a minimum value at some point in it. If we have various charged bodies enclosed in a metallic envelope, the tubes of force starting from them will all terminate on the inside surface of the envelope, and hence the induced charge will be exactly equal and of opposite sign to the sum of the charges on the bodies. Green proved mathematically that if we suppose an equi- potential surface replaced by a conductor having system of charged the same boundary as the surface and if it be bodies by a con- * ductorofwhich electrified so that the surface density is given by the boundary is an equipotential 1 d V surface. CT = --- = 4-7T an then the electricity on this conductor will be in equilibrium, and will produce the same potential at external points as the charges enclosed by the equipotential surface originally did. A physical proof of this theorem can be given as follows. Suppose the equipotential surface replaced by a metal sheet coincident with it. If the metal be connected to earth, it will be at zero potential, and all points external to it will also be at zero potential. Let F be the potential at an external point due to CAPACITY 9 the enclosed charges and V l be that due to the bound charge on the sheet, then V+V l = and therefore V 1 = V. Also at a point on the conductor itself v + v l = and v l = v = constant by hypothesis; hence the electricity on the conductor would be in equilibrium. Therefore, if we suppose this conductor charged with electricity of the opposite sign to what it has in this case, it will produce a potential V at all external points and have itself a constant potential v. Hence the theorem follows. The method of images is due to Lord Kelvin. It is of great practical value in solving problems connected with the form of the lines of force round conductors suspended parallel to the earth, etc. Suppose for example we have a wire parallel to the earth at a distance h above it. Let the wire be charged with q units of electricity. We imagine an equal parallel wire at a depth h in the earth with a charge of q units. The surface of the earth will then be an equipotential surface of this system, and the distribution of the lines of force in the air can be found by solving the problem of two parallel wires at a distance '2h containing equal and opposite charges. Use will be made of this method in Chapter v. ,"3 ocmpJelely *c* grounding A a Consider two conductors A and ^(both at an infinite distance from all other conductors. Suppose that a charge Capacity. r r q be given to A and that B is always maintained at zero potential. When the electricity is in equilibrium on A, T> the surface density will be given at any point P by cr = - where R is the resultant force at the point P which is, by what we have seen, perpendicular to the conductor. All points on the conductor A will be at the same potential v, and we picture tubes of force all starting perpendicular to A and finishing up at right angles to B. The quantity of electricity on B will be q. Now if we were to give to A a further charge q, then it would distribute itself over A so that the density would be now 20-. The mutual actions between the two coincident charges will be perpendicular to the surface, and so doubling the density of the layer does not disturb the equilibrium. Hence also we shall have a charge 10 ALTERNATING CURRENT THEORY [CH. 2q on B. The potential of A will now be 2v, since a, and therefore R, is doubled. Similarly if we gave a charge nq to A, its potential would become nv. We see then that the charge on A when B is maintained at zero potential, is in a constant ratio to its potential. This constant ratio is denned as the capacity of the conductor A. When there are several conductors at different potentials, the relations between them can be expressed by linear equations. We will consider this problem in Chapter IV. Consider the body A in the preceding paragraph. The relation Potential energy of between q and v may be written an electrified body. jr where K is the capacity of A. Let dq be the increment of charge necessary to raise the potential to v 4- dv. The work done in taking dq from the boundary of the field to A is dq.v, therefore the work in ergs done in charging the conductor with a charge Q is F Kvdv When we have an E.M.F. between two points in a conductor, a current is produced. A current is measured Electric current. i i by the rate at which quantity of electricity flows through any cross section of the conductor ; it is therefore ~r and at will be denoted by i. The work done in taking q from a point where the potential is v l to a point where it is v 2 is q Oi - 2 ) = wt ergs, where w is the work done per second in ergs, and t seconds is the time of working, the rate of working being uniform. Hence differentiating i (X v s ) = w. In this equation i is the current in electrostatic units, v l - v 2 is the I] MAGNETISM 11 potential difference (P.D.) in electrostatic units and w is the power in ergs per second. The corresponding equation in electro- magnetic units is i Oi ~ v 2 ) = w, where i is the current in amperes, Vi v a the P.D. in volts and w the power in joules per second or watts. The ampere, which we will define shortly, is 3 x 10 9 times the electrostatic unit of current and the volt is the three-hundredth part of the electrostatic unit of E.M.F. (Chapter IV.). If any magnet be supported in such a way that it is free to turn about its centre of gravity in the earth's magnetic field, it is found that a particular line through the centre of gravity of the magnet always tends to point in the same direction. This line is called the magnetic axis of the magnet. If we have a long thin cylindrical magnet with its magnetic axis coincident with the axis of the cylinder, then the centres of the circular faces are the poles of the magnet. To a first approximation we can suppose that these poles are centres of force, and the action of the magnet can be calculated by supposing attracting matter m concentrated at the north pole of the magnet, i.e. the pole which points to the north when the magnet is suspended, and repelling matter m concentrated at the south pole, the rest of the magnet consisting of inert matter. If 21 be the distance between the poles, 2lm is called the magnetic moment (M) of the magnet. It is found by experiments with the torsion balance and otherwise that like magnetic poles repel one another and unlike attract with a force which is directly proportional to the strengths of the poles and inversely proportional to the square of the distance between them. We define the unit pole to be that pole which repels an equal like pole one centimetre away with a force of one dyne. Hence, in air or other non-magnetic medium, the law of repulsion of magnetic poles is ,._ mm /"' t Following the electrostatic analogy, we define the magnetic potential at a point to be the work done in ergs in taking unit 12 ALTERNATING CURRENT THEORY [CH. north pole from the boundary of the field to the point in question. Hence if there was only one pole of strength m, the potential at a distance r from it would be given by F=-. Let N and 8 (Fig. 2) Potential near a be tne P oles bar magnet. Q f a baf magnet. Let their strengths be m and m respectively and let 2a be the distance between them. Let be the middle point of NS and let OP equal r. Then if F be the potential at P due to this magnet we have N a O a S Fig. 2. Magnetic Potential at P _ M cos 9 M (5 cos 3 6 - 3 cos 6) a 2 ~+- -+ m m Va 2 + r 2 - 2ar cos Va 2 + r 2 + 2ar cos " If r is greater than (V2 + 1) a we have by the binomial theorem, 1 * n /o- a'M-* = -U- 2-cos0 - a ijl +cos ^ + f| cos ^_l^ 2 + ^ r \2 Hence substituting in (1) and simplifying we get Tr 2ma cos 5 cos 3 3 cos + ma 3f(5cos 3 0-3cos0)a 2 I] MAGNETIC POTENTIAL 13 where M is the magnetic moment of the magnet. Hence if r be large compared to a, Mcos0 It follows that the force in the direction OP will be dV_2Mcose dr~ r ' , and in a direction at right angles to OP, in the direction in which increases, it will be dV _Msm0 rdO~ ~r* " Hence we can easily find the direction and the magnitude of the magnetic force at a point P when r is large compared to the length of the magnet. The equipotential surfaces round an infinitely small magnet will be found from the equation or by giving various values to V. We can prove in exactly the same way as in electrostatics (Gauss's Theorem) that Tubes offeree. Tubes of indue- 2.R COS CldS = 4>TTm where the integral is taken over a closed surface enclosing poles the sum of whose strengths is m. R is the intensity of the magnetic force at dS and a is the angle between the direction of the force and the normal to the surface. Applying this theorem to the part of a tube of force intercepted between two equipotential surfaces, we find that FdS = F'dS', where F is the intensity of the force at a point on dS. Hence we can suppose the magnetic field divided up into tubes of force in exactly the same way as we divided up the electrostatic field, and the tubes of force map out the intensity and direction of the magnetic force at any point. The product R cos adS is called the flux of force through the area dS and we see from Gauss's theorem that the total flux of force from a unit magnetic pole is 4?r. It is proposed to call the init tube of force the maxwell. 14 ALTERNATING CURRENT THEORY [CH. Now when magnetic force acts on a medium, it produces in it magnetic induction. In air the number of tubes of induction produced by the magnetising force is the same as the number of tubes of force. In a magnetic medium the number of tubes of induction may be thousands of times greater than the number of tubes of force. The strength of the field at a point in the medium is the number of tubes of induction per square centi- metre. One tube of induction per square centimetre is called a gauss, and this is the unit in which magnetic induction density is measured. If we break into two portions a long thin bar magnet which has been uniformly magnetised and has poles of Intensity of ' . . r Magnetisation, strength m and m respectively, it will be found that each portion is a magnet with poles of practically the same strength as the original magnet. The axes of the magnets are the two portions of the original axis. This is true also when we divide the magnet into many parts, and hence we can regard such a magnet as made up of a large number of little magnets of which the axes coincide with the axis of the original magnet and with ends perpendicular to that axis. Such a magnet is called a solenoidal 7 \agnet, and we see that no great error is made in assuming that its ends are covered with a layer of strength m of attracting and repelling matter respectively, and that the rest of the magnet is inert. The intensity of the magnetisation of a solenoidal magnet is defined as the magnetic moment per unit volume, and is generally denoted by /. T M ml m I= V = N = S where S is the cross-sectional area of the magnet. We see that / may also be defined as the pole strength per square centimetre of the area of the cross section. If the magnet were not uniformly magnetised, 7 would be different at different points, and so we should have to define it as dM dV l] PERMEABILITY 15 Consider now a circular iron ring uniformly magnetised, the cross section of which is circular. If / be the intensity of the magnetisation, then, if this ring be sawn through we should have what may be regarded as a layer of attracting matter on one side of the air gap and a layer of repelling matter on the other, the surface density being in each case /. If a unit magnetic pole be placed in the air gap on the axis of the ring and at a distance a from either circular face, the force of attraction F to one face will be obviously along the axis, and therefore x cos 6 r 2 where 6 is the angle made by the axis with a line drawn from the pole to an element of the ring with a radius r. Now r = a tan B, therefore dr = a sec 2 Odd = - (a- + r 2 ) dO. Therefore F = 2?r/ I ** sin 0d0 I J o where < is the value of 6 when r is the radius of the cross section. Hence, when a is small, .Fis 2?r/, and since the repelling face will repel the pole with an equal force, we see that the intensity of the field in the air gap is -ivr/. Faraday showed that liese tubes or, as they are more commonly called, lines of force are continuations of lines inside the iron which are called lines of magnetisation. If we saw through the ring at some other point, and then imagined it stretched out straight, the ends of the first gap still remaining the same distance apart, we see that our unit magnetic pole will now be subjected to the attractions and repulsions of the poles at the other ends of the bar; hence the field in which it is situated will be weakened. If the bar were very long however we could neglect the demagnetising effects of the ends. If the bar were placed in a magnetising field of intensity H, the unit pole would be subjected to forces H and 4?r/, where / is the new intensity of the magnetisation. If B is the resultant of H and 4-Tr/, then B will be the strength of the field in the air gap. If H and 4?r/ are in the same direction, then 16 ALTERNATING CURRENT THEORY [CH. The ratio of I to H is called the susceptibility of the iron, and the ratio of B to H is called the permeability. It is this latter ratio that is generally wanted in practice ; it is always denoted by u, so that When iron is magnetised to a certain induction density B, it is found that when the force is withdrawn a certain quantity B is left remanent in the iron, and a certain coercive force has to be applied to get rid of B Q . We will return to this point when we discuss magnetic tests. If we have a thin sheet of iron made up of an infinite number of little magnets with their axes perpendicular to tn sheet and their like poles all pointing in the sheiif y0ftWO same direction, we get what is called a magnetic shell. A study of the properties of these shells is most helpful in understanding the theory of electrodynamics. The strength of a magnetic shell is its magnetic moment per unit area, so that, if g be its strength and h its thickness, j- is the polar strength of the face per unit area. We will now find the mutual potential energy of two such shells, A and B. Consider an elementary small magnet of the shell A ; its potential at a point P is - , i.e. g^day where co is the solid angle at P. Integrating over the whole shell j4, we see that g^i is the potential at P due to the shell, where f^ is the solid angle subtended by its boundary at P. If the strength of the shell B were g z , then the polar strength of an element dS 2 at P would be ~ dS 2 , where dn is the thickness of dn the shell, and we have shown that the potential at P is ^ft^ Hence the work done in taking the polar element dS 2 from infinity to its position at P against the repulsion of the shell A would be and this expression gives the mutual potential energy of dS 2 and MUTUAL POTENTIAL ENERGY 17 the shell A. Similarly the energy of the other pole of the elemen- t magnet the end of which is dS 2 will be n. (7 9 d Hence the total potential energy of the elementary magnet is Therefore, integrating over the whole shell B, we find that the utual potential energy is given by Now ^ - is the resultant force measured normally at the dn shell B due to the shell A. Hence the integral gives the flux of force ! through the shell B due to the shell A. Similarly if cf). 2 were the flux of force through A caused by B, gi 2 would be the mutual potential energy. Therefore # 2 i =#i< 2 , which is a remarkable reciprocal relation of great practical im- portance. When the strengths of the shells are equal, we see that the flux through A caused by B equals the flux through B caused by A. Oersted showed that when a wire carrying an electric current Eiectrod namics produced by a battery was brought near a mag- netic needle, the needle was deflected. Hence an electric current produces a magnetic field. We have seen that an infinitely small magnet, the magnetic moment of which is M, produces a potential 2 at points at a distance r from it, where 6 is the angle which r makes with the length of the bar. Weber proved experimentally that a small closed plane circuit carrying a current not only produced the same field but also was acted on by the same forces as a small solenoidal magnet with an axis perpendicular to the plane of the coil, provided that a certain relation held between the magnetic R. i. 2 18 ALTERNATING CURRENT THEORY [CH. moment of the magnet, the area of the circuit, and the current flowing in it. If S be the area of the circuit, and i the strength of the current, then the potential at any point is k 2 ' , where A; is a constant. If the fields produced by the small magnet and the small circuit are the same, we must have M= kSi. Now in order to simplify our formulae as much as possible, we choose our unit of current so that k is unity, and therefore ir- ft*. Jr Si cos 6 Hence also V . ?- The potential V of the small circuit at a point P, distant r from its centre, may be written where co is the small solid angle subtended by S at P. If we now suppose that an infinite number of these small circuits are all crowded together, forming a network, and that they are all carrying equal currents i flowing in the same sense round the meshes, then it is easy to see that F=i2a> = *n where fl is the solid angle subtended by the boundary of all the small circuits. Where the elementary circuits touch one another we have equal currents flowing in opposite directions, and hence they are neutralised. We see then that this arrangement is equivalent to a circuit coinciding with the boundary of the small circuits and carrying a current i. It follows that the potential V at any point due to an electric circuit is always equal to il, where H is the solid angle which the circuit subtends at the point. We have shown above that the potential due to a magnetic shell of strength g at a point P is given by V = gCl where H is the solid angle subtended by the boundary of the shell at the point. Hence Ampere's theorem follows, namely, that a magnetic shell and a closed circuit are equivalent if g=-i. IJ CIRCULAR CURRENT 19 Consider the case of a circle of wire of radius r carrying a current i. The potential F at a point P on its alar current. . ,. , -, -n i -nfii axis perpendicular to its plane will be ico. With centre P (Fig. 3) describe a sphere passing through the circle. The area of the spherical cap intercepted by this circle is s^ h 27rRh, where R is the radius of the sphere and h the height of the cap, hence 7=i = 27^(1- cos 0) = 27ri (l - r 2 where x is the distance of the point P from the plane of the wire. Hence if F be the force Fig. 3. Potential at P=2*i (l - cos e). on unit pole at P along the axis, we have F ^_dV dx When x is zero In this formula F is in dynes, r in centimetres and i in C.G.S. units. We can thus define the unit current as the current which, flowing in a circle of radius r centimetres, produces a force of - dynes at its centre perpendicular to its plane. The relation between the direction of the current and the lines of force can be remembered by the diagrams shown in Figs. 4 and 5. A current flowing in the direction of the arrowheads shown in Fig. 4 produces magnetic lines upwards through the paper. In practice the unit current adopted is the ampere, which is one-tenth of the absolute C.G.S. unit defined above. A current of 22 20 ALTERNATING CURRENT THEORY [CH. electricity in a conductor is the rate at which a quantity q of elec- tricity is flowing through the conductor. In symbols Fig. 4. Fig. 5. Relation between polarity and direction of current. We may therefore define on the electromagnetic system the unit quantity of electricity as the quantity which flows past any cross section of the conductor every second when unit current is flowing. The practical unit of quantity is the coulomb, which is sometimes called the ampere-second. Joule's Law. Ohm's Law. The quantity of electricity conveyed by a steady current of i amperes flowing for t seconds in a conductor between two points with a potential difference e, will be it. Hence the work done is eit. Now the practical electrical unit of work is the Joule or 10 7 ergs. The work done can be measured by the number H of units of heat (calories) generated in the conductor, and if the dynamical equivalent of heat be denoted by J (in joules per calorie), then JH e it .(1). The unit in which e is measured is called the volt. If i were in centimetre-gram-second units and JH in ergs the number obtained for e would be 10 8 times larger. The volt equals 10 8 c.G.s. electro- magnetic units. Equation (1) gives Joule's law in symbols. I] ELECTROMAGNETIC INDUCTION 21 A difference of potential between two points can be maintained by means of a battery ; a current will flow between them, work being done in the process. Now Ohm proved experimentally that the ratio of the difference of potential to the current was constant so long as the conductor between the two points remained in the same physical state. This constant ratio is called the resistance of the conductor between the points, and Ohm's law may be written i = 6 - .............................. (2)- Hence if W be the number of joules done in time t, If P be the rate at which work is being done between the two points, in joules per second, i.e. if P be the power in watts, Whenever the lines (or tubes) of magnetic force passing Electromagnetic through a circuit alter, an E.M.F. is set up in the induction. circuit. We will first consider the case of a circuit carrying a current i in the neighbourhood of a magnet. Suppose that the flux of force through the circuit is <, and that it alters to + dcf) in a time dt. Then the work done by the magnetic forces in moving the magnet is id(f>, and this amount of work must be supplied by the source of E.M.F. in the circuit. If E be the E.M.F. in the circuit, the work done in the time dt will be Eidt, and this work must be equal to the sum of id and the energy expended in heating the circuit, namely R&dt. Therefore Eidt = Rtfdt + id, Hence the induced E.M.F. in a circuit equals the rate at which the flux embraced by the circuit is altering ; it acts in the positive direction (counter-clockwise) when the flux coming towards the spectator is diminishing. 22 ALTERNATING CURRENT THEORY [CH. This law, as modified by Maxwell, can be stated as follows. The E.M.F. generated in a circuit always tends to produce a current which opposes any change in the value of the flux. If we bring a positive magnetic pole near a circuit, we diminish the number of lines of force coming through it towards the spectator, and hence the induced current must flow- in such a direction that the positive face of the equivalent mag- netic shell may face the positive pole. Therefore the current must flow in the opposite direction to the hands of a watch (Fig. 4). If the flux be increased, the induced current must tend to maintain the initial state of affairs, hence the direction of the induced E.M.F. is with the hands of a watch. To get the positive direction of rotation the flux must diminish, and hence the induced E.M.F. must be written 37 . If 6 be iu maxwells, then r- is at at in C.G.s. units. Hence if e be the counter E.M.F. in volts, = _. dt If cf) traverse n turns of wire, then db e n- r 10~ 8 . dt If we have two circuits A and B carrying currents i\ and i 2 Mutual in- respectively, then, since we may replace them by ductance. magnetic shells, we see that their mutual potential energy may be represented by ^ 2 or by - i^fa, where c/> 2 is the flux through A due to B, and fa is the flux through B due to A. ISJow ( 2 is in direct proportion to ^' 2 ; we may therefore write fa = Mi 2 , where M is a constant depending only on the positions of the circuits. Similarly we can write fa = M'i 1 , and, noting that the two expressions for the potential energy must be equal, we see that M = M'. Hence the flux of force through A due to unit current in B equals the flux of force through B due to unit current in A. If the current in B alters, the flux fa through A alters also, and the induced E.M.F. set up is f' 2 or M -p '. It is dt dt I] INDUCTANCE 23 to be noted that M may be positive or negative, depending on the directions taken as positive for the two currents. In practice the circuits will consist of many turns of wire, and the fluxes through the turns will all be different. In this case the mutual potential energy of the two coils will be where fa' is the flux due to i. 2 embraced by one turn, fa" the flux embraced by the next, and so on. The above definition applies only to non-magnetic circuits. The practical unit of mutual in- ductance is called the henry ; it equals 10 9 times the C.G.s. unit. If we double the strength of a current flowing in a non-magnetic circuit, we double the strength of the equivalent magnetic shell, and therefore we double the number of lines of force due to the current itself, embraced by the circuit. It follows that the current strength and the total flux are directly proportional to one another; we may thus write (f> = Li where L is a constant. This constant is called the self inductance of the circuit. We see that as i increases $ increases, and therefore by Lenz's law the induced E.M.F. tends to set up a current which will retard the increase of <. The value of this E.M.F. is given by d(f> j. di ~~dt = dt' Instead of having a single loop, the circuit may consist of many loops ; in this case we have HI 0i + H 2 02+ = Li where fa is the flux embraced by a group of n^ turns of wire, etc. Hence L may be defined as the number of linkages of the lines of force with the circuit when traversed by unit current. The practical unit of self inductance like that of mutual inductance is the henry. Suppose that we have two circuits A and B respectively, and Eiectromag- that L, N are their self inductances and M their mutual inductance. Then if i, and i a be the currents through the circuits, the flux through A is L^ + Mi^, and 24 ALTERNATING CURRENT THEORY [CH, similarly the flux through B is Ni^ + Mi^ If e, and e. 2 be the E.M.F.'s in each circuit, and R l} R 2 be their resistances, = Ri,+L u - + M"jj. dt dt Similarly Thus, Now R&? + R 2 i 2 2 is the rate at which energy is being expended d in heat, and hence we see that -j- (A- Li^ + Mifa + -J- Nif) must be Out the rate at which energy is being stored up in the field. Hence when the currents in the two coils are ^ and i 2} the energy stored up in the field is This expression may be written in the form and since it must be positive for all values of ^ and i, and there- fore when i^ j- i 2 , we see that M 2 cannot be greater than LN. j When there is only one circuit, we see that the energy stored up in the field is \ Li 2 when the current is i. By the last section we see that this may be written ^i^n^> where ^ is the flux embraced by n^ turns, etc. The above formulae and definitions only apply strictly to elementary tubes or filaments of current. The The self energy of an electric electric currents that we have to consider in practice cannot be regarded merely as filaments of current. In the general case we may write %LP 2 for the self energy stored up in the field, where / is the total current. A useful expression for this energy can be found as follows. Consider first a circuit made up of two filaments of current i] KELVIN'S FORMULA 25 t'i and i. 2) so that / equals i + i. 2 . The self energy of the circuit is where lml and <,. are the fluxes due to the currents i^ and i 2 respectively, and we have where ^0 a and S. 2 are the total fluxes linked with ^ and i. 2 respectively. When we have n filaments of current, we get in the same way the summations being effected over the whole system for all the filaments of current and the fluxes they embrace. The magnetic potential at a point due to a closed tube of current i is ico, where &> is the solid angle sub- tended at the point by the tube. If we take a un ^ positive pole once round a line of force embracing the tube, the change of potential energy is 47Ti, since 4?r is the difference between the initial and final values of the solid angle. Hence if H be the magnetic force at any point where ds is an element of the line of force round which the integral has been taken. Xow if d(f> be the flux of force over an element of an equi- potential surface through the point d = HdS. Hence idd> = -- / Eds x HdS. But HdS is constant along a tube of force, and ds x dS = dv = element of volume of the tube of force. Thus idd>= 4n the integration being taken along a tube of force. 26 ALTERNATING CURRENT THEORY [CH. Hence, integrating over all the equipotential surface, we get id> = ~ 2H*dv, 4-7T where the summation is taken throughout all space. Hence the expression \i$ for the self energy of the system may be written The E.M.F. generated in a conductor cutting lines offeree. Suppose that we have a wire sliding with uniform velocity v on two parallel wires which are joined by another wire at one end, and that the moving wire is cutting a uniform magnetic field which makes an angle 6 with the plane of the wires. If the strength of the field be H, and the slider be perpendicular to the parallel wires, then the number of lines of force cut by it in t seconds is H sin dlvt, where I is the length of the slider the ends of which are on the parallel wires. The increase of the flux round the closed circuit is H sin dlvt Therefore the E.M.F. generated will be e = - -T- (H sin Blvt) = -Hsiu Olv. Cit The E.M.F. generated is in the clockwise direction round the circuit. In this formula e is in C.G.s. units, H is in gausses, I in centimetres and v in centimetres per second. If e is in volts, then e = -Hsin0lv x 10~ 8 . In Fig. 6 is shown the relation Motion between the motion of a wire, the magnetic field, and the direc- tion of the induced E.M.F. and current. If we place the fingers of the right hand so that the 4.1 ,^r ' ji i- , f ,1 thumb is in the direction of the cfi ig. 6. Relation between the motion of a wire and the induced current. FORCE ACTING ON A CONDUCTOR 27 motion and the first finger in the direction of the magnetic field, then, if the second finger be put at right angles to the first and the thumb placed so as to coincide with the wire, the second finger will point out the direction of the induced E.M.F. This E.M.F. will vanish when the motion is parallel to the field of force. If the current in the moving wire be i, then the electric power Force on a generated by it will be ei. Hence if / be the moving wire. resultant force in dynes acting on the wire, and v be expressed in centimetres per second, fv = ei = H sin 6lvi, where f is in dynes and i is also in absolute units. If i be in amperes, then H sin 6li , ~YQ d y nes - The negative sign shows that there is a force acting on the wire tending to make it move so as to increase the number of lines of force embraced by the circuit. This is the principle that is utilised in the electric motor. We see then that a wire carrying a current and placed in a magnetic field is subjected to a certain force acting in a certain direction. If we move the wire in the direction opposite to that in which this force is acting, by Lenz's law the induced E.M.F. must act so as to increase the current flowing and thus impede the motion. Therefore we see that the force would be in the opposite direction to the line marked motion in Fig. 6. Fleming's left-hand rule states this in an easily remembered form. If the fore-finger of the left hand point in the direction of the field, the second finger in the direc- tion of the current in the wire, then the thumb will point in the direction of the force on the wire, if the two fingers be held at right angles to one another and also to the thumb. It will be sufficient for our next purpose to suppose that the current is in a plane. We will calculate the 1 he magnetic force due to an magnetic force at a point (Fig. 7) due to a element of current. Laplace's For- current i in a closed circuit. Suppose that we have a positive pole of strength m at and that 28 ALTERNATING CURRENT THEORY [CH. OA is a fixed line. The strength of the magnetic field at P due m to will be - Hence if the arrow indicate the direction of the r 2 Fig. 7. Magnetic force at due to the current i in a plane closed curve is upwards from the plane of the paper and i sin Ods [ id(f> where = the angle POA. current we see by the preceding paragraph that the force dF on ds will be given by m dF = -- sin Oids dynes, and the force, by the left-hand rule, tends to move ds downwards at right angles to the paper. Using a well-known artifice in Statics, we may replace this force by a force dF at acting down- wards, and a couple with a moment rdF round an axis lying in LAPLACE'S FORMULA 29 the plane of the paper and perpendicular to OP. It is easy to see that rdF= sin dids r m rd(f> . , = -- j ids r as midff), and hence for sin 6 = ~- , where the angle POA = . But round a closed curve for every couple of which the; moment is mid there is another having a moment mid (see Fig. 7) ; hence the resultant couple is zero. Thus the resultant force of the pole on the circuit is a force 'm sin dids acting downwards at 0. Since action and reaction are equal and opposite, we see that the resultant force of the circuit on the pole is T mi sin J r- and acts upwards. We see then that, so long as we consider closed curves, we can calculate the intensity of the magnetic force F at any point by the formula r- fid$ J r If i be in amperes, then the magnetic force is given by The force at the centre of a circle of radius R can easily be found, for r R and is constant, and %d = 27r, Formulae for the hpnnA magnetic forces inside circles and _ 2*7rl rectangles. Ju ^ , which agrees with our former result on page 19. 30 ALTERNATING CURRENT THEORY [CH. If the point be not at the centre of the circle (Fig. 8) let OC = a, CP = R and the angle POG=0, then Therefore But Thus OP = a cos 6 + V]R 2 - a 2 sin 2 0. a 2 sin 2 6 1 a cos 6 + r = ~g V '[ 2nd0 f =z I . Jo r i* f^ ^"E^^^JO _ i x circumference of ellipse .(3). Fig. 8. The magnetic force at due to a current t in the circle is perpendicular to the plane of the paper and equals i x circumference of ellipse The ellipse has for a focus. The force acts towards the reader. FORCE NEAR A STRAIGHT CONDUCTOR 31 M N The ellipse referred to in the formula has C for its centre, for a focus, and has its major axis equal to a diameter of the circle. When a is zero this reduces to formula (2). The case of a rectangular wire (Fig. 9) also admits of an easy solution. If the current i be flow- ing round the wire in the direction indicated by the arrowheads, then the force Fig. 9. Force at due to current in rectangle F on unit pole placed at = ix sum of reci P rocals of perpendiculars will be given by the formula from on LM, MX, NR and EL. (4) , where p l} p. 2 , p s and p are the perpendiculars from on the lines LM, My, yR and RL respectively, and ROM and NOL are drawn parallel to the sides of the rectangle. It is an instructive exercise to find graphically by means of this construction the density of the lines of force at various points inside a rectangle. Let three of the sides of the rectangle move to infinity, then the formula becomes Force near a long straight ^ 2l wire carrying a JP = current. T and we obtain the force due to a long straight current, where r is the perpendicular distance from the point con- sidered to the wire. The force F is perpendicular to r, and the directions of force and current are related in the same way as the directions of current and force in Figs. 4 and 5. If the current be upwards through the plane of the paper, then the lines of force are circles, and act in the direction ft. If the current be downwards through the paper, then the lines of force act in the direction rSt By Ampere's theorem, we could have replaced the wire by a plane magnetic shell bounded by it and extending to infinity. 32 ALTERNATING CURRENT THEORY [CH. Suppose that r makes an angle 6 with the plane of this shell. Draw two planes through the point, one passing through the wire and the other parallel to the shell. The area of the lune intercepted on a sphere of radius R by these planes is 2R 2 8. Therefore the solid angle o> subtended by the infinite plane at the point is ^ = 2(9. Hence the potential at the point is dV Thus = . the force in the direction of r, must be zero, and dr dV -JQ , the force perpendicular to r in the direction in which 2* increases, must be . We see again that the work done in taking unit pole round the wire is The magnetic Let PP f (Fig. 10) be a section of the cylindrical shell, and the point at which we wish to find the force. Divide the shell into an infinite number of filaments parallel to its axis. Consider the f rCe at du6 tO the CUI%rents in the filaments passing into the paper at P and P' respectively. From symmetry, their resultant will be per- pendicular to OC, and hence, since one half of the shell is an image of the other, we see that the total force F acts per- curr^nt'parafle 3 ! to its axis. Fig. 10. The force at points outside a cylindrical sheet of current is the same as if the current were concentrated along the axis of the cylinder. I] CYLINDRICAL CONDUCTOR 33 pendicularly to GO. Hence if i be the total current in the conductor, the angles PCO, POC be 0, $ respectively, and <7P, PO and 00 be a, r and d, then 7/3 lad cos + a 2 Now by expanding in a trigonometrical series (page 58) we see that log (d 2 - 2ad cos 6 + a 2 ) dO = 2?r log d, o since d is greater than a. Hence, differentiating with regard to d, d a cos 6 JQ _ TT I- - 2ad cos 6 + a? ~ 5 ' therefore F = , d Hence the cylindrical conductor acts as if its current were concentrated along its axis. As we can suppose that a solid cylindrical conductor is made up of an infinite number of coaxial tubes, we see that a solid cylinder also acts on external points as if its current were concentrated at the axis. Inside an infinite cylindrical tube carrying a current parallel The magnetic * tne ax i g > since d is less than a, force inside an infinite cylindrical tube carrying a current parallel to ^ and hence, when we differentiate with regard to d, the right-hand side vanishes, and we get (log (d 2 - 2ad cos + a 2 ) dO = 2?r log a, J o Therefore the magnetic potential inside a hollow cylindrical conductor is constant, while outside it is given by where A is a constant. R. i. 34 ALTERNATING CURRENT THEORY [CH. Let us wind an iron ring uniformly with insulated wire, and Magnetic tests of P^ ace a further winding on it connected to the iron. Hysteresis, terminals of a ballistic galvanometer. If we start a current through the first winding, the throw of the ballistic galvanometer enables us to calculate the flux of induction caused by the magnetising force of the current. If (f> be the total flux in the core the induced current i in amperes will be given by if there are n turns of wire in series with the galvanometer, and R is the total resistance of this circuit in ohms. Thus we have lidt = ^ M0> n nlO~* and Q = --g (fa - &), where Q is the number of coulombs that traverse the circuit when the flux increases from cf) l to 2 . The throw of the galvanometer needle gives the value of Q. If we divide the flux by the cross section in square centimetres, we get the mean flux density B, and we shall prove in the next chapter that we can calculate the mean value of H by the formula 4?^ ^oT' where ?ij is the number of turns in the primary winding which carries a current of i amperes, and I is the mean circumference of the iron ring. If the magnetisation of the iron ring be reversed several times, until it gets into what is called its steady cyclic state, we get curves like those shown in Fig. 11. Suppose that, when the current has its maximum positive value, OA is the value of H, and QA is the corresponding value of B. Then, as the current diminishes, the extremity of the ordinate representing B moves along QR. When the current is zero, the density of the flux left in the ring is represented by OR, and is called the remanence. When the flux is zero, H is OS, and this quantity is called the coercive force. UNIVERSITY HYSTERESIS LOOPS 35 An inspection of the figure will show that, when the current diminishes from its highest value to zero, the change in the flux density is only A Q OR, but in increasing numerically from zero 25 3A H Fig. 11. Curves of Magnetisation of steel strips. OLPQ = Permeability Curve. Hysteresis loops with B^ =3900 and B^, = 7300. to its maximum negative value it changes from OR to AQ or by an amount A Q 4- OR. On diminishing the current again, the extremity of the ordinate representing B moves along Q'R. Finally on increasing the current up to its maximum positive value, the point arrives at the point Q from which it originally started. It will be seen that the magnetic induction lags behind the magnetic force producing it. This phenomenon is called hysteresis. The curve OPQ in the figure gives the positions of the top 32 36 ALTERNATING CURRENT THEORY [CH. points of the cyclic curves corresponding to various maximum values of H. The permeability p of the iron for various values of H is found from this curve. Suppose that the iron is placed in a uniform field of which the intensity is H. If this field be produced by taSnglron m an infinitely long solenoid which has n turns of wire through a mag- i n a i en nrth I and carries a current i, then, replacing netic cycle. the turns of wire by equivalent magnetic shells, we see that The induced E.M.F. in n turns of the solenoid when the flux varies will be given by dd> e = n dt where $ is the area of the cross section. Hence the work W done in taking a volume 18 of the iron through a complete cycle is given by fT W= eidt J o -1$ [ T rr d ? ~4^rJo di =^! HdB ' where V is the volume of the iron in cubic centimetres, and the integral is taken over a whole cycle. From Fig. 11 we see that [HdB gives the area of the hysteresis loop corresponding to a given maximum value of B. If B be measured in c.G.S. units, W will be in ergs. The following empirical formula due to Steinmetz is found of Steinmetz's USG Formula. where W represents the ergs lost in taking V cubic centimetres of iron from a maximum induction density of -6 max to a density PERMEABILITY OF STEEL STRIPS 37 -Bj^ , and then back again to 5 max . The constant tj depends on the kind of iron used. If / be the frequency of the alternating current which mag- netises the iron, the formula may be written where IF now represents the power lost in joules per second, or in watts owing to hysteresis. For values of j6 max between 2000 and 14000, this formula is sufficiently trustworthy for practical work. For good soft iron and steel 77 is usually less than 0'002. The following figures give the result of a test on steel strips for the British Electric Plant Co., the curves obtained for these strips being shown in Fig. 11. Permeability Table. H B p. 1 1580 1580 2 4930 2465 3 7000 2333 10 11,800 1180 20 13,840 692 50 15,800 316 100 16,600 166 160 17,980 112 Hysteresis. Maximum B Ergs per c.c. per cycle Watts per c.c. at 100 ~ per sec. Remanence Coercive force Steinmetz's coefficient V 3900 864 0-00864 2920 0-82 0-00152 7300 2386 0-02386 5470 1-08 0-00157 38 ALTERNATING CURRENT THEORY [CH. Taking 77 = 0*0015 we see that the loss in watts per kilo- gramme of iron of which the specific gravity is 7 '8, when B = 4000 and /= 100, would be 1 000 = 0-0015 x 100 x J: x (4000) r(i x 10~ r 7o = 1*1 nearly. Number l*5th power 1 '55th power l-6tl 1000 31,620 44,670 2000 89,440 130,800 1 3000 164,300 245,200 3 4000 253,000 383,000 5 5000 353,600 541,300 8 6000 464,800 718,000 1,1 7000 585,700 911,800 1,4 8000 715,600 1,122,000 1,7 9000 853,800 1,346,000 2,1 10,000 1,000,000 1,585,000 2,5 11,000 1,154,000 1,837,000 2,9 12,000 1,315,000 2,103,000 3,3 13,000 1,482,000 2,380,000 3,8 14,000 1,657,000 2,670,000 4,3 15,000 1,837,000 : 2,971,000 4,8 16,000 2,024,000 3,284,000 5,3 17,000 2,217,000 3,607,000 5,8 18,000 2,415,000 3,942,000 6,4 19,000 2,619,000 4,286,000 7,0 20,000 2,828,000 4,641,000 7,6 21,000 3,043,000 5,005,000 8,2 22,000 3,263,000 5,380,000 8,8 23,000 3,488,000 5,763,000 9,5 24,000 3,718,000 6,156,000 10,1 25,000 3,953,000 6,559,000 10,8 63,100 191,300 365,900 579,800 828,600 ,109,000 1,420,000 A more accurate way of stating the formula for the hysteresis loss is where n is a number which varies slightly over different parts of the curve of W and B and also varies for different kinds of iron and steel. For example Ewing and Klaassen found experimentally for a particular specimen of iron that n was nearly T55 from B STEINMETZ'S FORMULA 39 equal to 1000 to B equal to 2000 and that it was only 1-475 from B equal to 2000 to B equal to 8000. For values of B between 8000 and 14000 they found that for this specimen n was 1'7. Tables of the 1'oth, 1'5 5th and l*6th powers of the numbers representing the induction densities usually wanted in practice are given above. REFERENCES. SILVAXUS THOMPSON, Elementary Lessons in Electricity and Magnetism. R. T. GLAZEBROOK and W. N. SHAW, Practical Physics. J. J. THOMSON, Elements of the Mathematical Theory of Electricity and Magnetism. J. A. EWING, Magnetic Induction in Iron and other Metals. CLERK MAXWELL, Electricity and Magnetism, Vols. 1 and 2. OLIVER HEAVISIDE, Electrical Papers. Vol. 1. G. CHRYSTAL, 'Electricity,' Encyclopaedia Britannica, Ninth Edition. G. F. C. SEARLE, 'The Determination of Currents in Absolute Electro- magnetic Measure, 5 The Electrician, Vols. 27 and 28, 1891. G. M. MINCHIX, ' The Magnetic Field of a Rectangular Current,' The Elec- trician, Vol. 35, p. 603, 1895. The Electrician, Vol. 31, p. 282, 1893, 'Measuring Magnetic Fields.' The Electrician, Vol. 36, p. 50, 1895, 'The Magnetic Force at any Point in a Plane due to Electric Currents.' C. P. STEINMETZ, J. of the Am. I. of E. E. Vol. 9, p. 3, Jan. 19, 1892. J. A. EWIXG and Miss KLAASSEN, Philosophical Transactions Royal Society, 1893, p. 1017. G. F. C. SEARLE and T. G. BEDFORD, 'On the Measurement of Magnetic Hysteresis,' Philosophical Transactions of Royal Society, Vol. 198 A, pp. 33-104, 1902. G. F. C. SEARLE, ' The Ballistic Measurement of Hysteresis,' The Electrician, May 9 and 30, 1902. CHAPTER II. Alternating curreht in an inductive circuit. The calculation of inductance. The E.M.F. required to produce a current i in the inner conductor of a concentric main. Flux of force inside a circular ring of rectangular section. Flux inside a ring of circular cross section. The magnetic analogy of Ohm's law. Reluctance. Inductance formulae for anchor rings. Self inductance of a concentric main. The self inductance of two parallel cylindrical wires. The self inductance of a circuit formed by three equal parallel cylinders, the axes of which lie along the edges of an equilateral prism. Triple concentric main. Minimum self inductance. The repulsive force between two parallel wires carry- ing equal currents in opposite directions. References. THE rotation of the armature of an alternator causes a rapid periodic change in the flux of force through its coils and hence the potential difference between the collector rings is a periodic function of the time. We may therefore express the difference of potential mathematically by the expression f(t) where t is the time in seconds, and if T be the time required for the P.D. to go through all its values, i.e. if it is the period, then Also, since the north and south poles of the field magnets in actual machines are always made alike, it follows that the wave for the first half period is the same as for the second half with the sign changed, so that /<)= - By Fourier's formula f(t) may be expressed in this case by the series f(t) = A, sin (^ t + cU + A 3 sin f 3 *-^~ t + 3 J + ... , [. II] CURRENT IX AN INDUCTIVE CIRCUIT 41 where A l} A 3 ...etc. are the amplitudes of the first, third,... harmonics which can be determined by the usual formulae when f(t) is known. It is customary to express -^ by o>, and we see that &> is the angular velocity of a line rotating -~, times a second. If /be the frequency, i.e. the number of times per second that the potential difference goes through all its values, then 2-rr ft) = -^ = 27T/. Now suppose that an alternating P.D., of which the instanta- neous value is e, is applied to a circuit of resistance R and self inductance L. Then if i be the instantaneous value of the current in the circuit L -=- will be the E.M.F. due to induction, at and we have by Ohm's law, di .! sn A sa sn 2 {(2n Also D sin {(27i + 1) cot 4- 2 , l+1 = ( '2n 4- 1) to cos {(2?i 4- 1) tot + o.> w+1 }, .*. D 2 sin {(2?i 4- 1) o> 4- 0^+1} = (2?i 4- l) 2 &) 2 sin {(2w 4- 1) < 4- 2 +i}. Therefore, using symbolic methods, we can replace D 2 by -(2n + l) a or. 42 ALTERNATING CURRENT THEORY [CH. Substituting and simplifying we find that where tan p 2w+1 = o Also, since the solution of the equation is where B is a constant, we see that the complete solution of the equation (a) above is A w sin (2n + 1) tot + In this formula .B is a constant which depends on the initial conditions, and e is 2 + a s - &) Suppose that the applied P.D. is a simple harmonic function , and that the switch is closed when t is equal to ^. Then since the circuit can not receive finite energy in an infinitely short time, we see that the initial value of i must be zero, and the general solution becomes in this case 77? = sn cot ~ a ~ sn t ~ f T> \ l sin (cot ~ a) ~ sin ^ tl ~ a) 6 " z< ' i ...... (6)> where tan a = . XL When ^ is very large the initial disturbance caused by closing II] INITIAL DISTURBANCE 43 --- the switch will be negligible as the damping factor e L ( 1; will be practically zero, and hence we can write E _. sin (wt a). We see. in this case, that whenever t equals - + nT, then i is di Eco E sin a L the moment of closing the switch n- T di L sin cot l = Hi -f iv -JT , cfa' 7? sin a^j therefore -^ = f , at L since i is zero at this instant. Hence, comparing the initial value of -j- with its value for t I- nT after the current has become at co purely alternating, we see that if cot l = a, i.e. if t l = -~ T, then there will be no initial disturbance, and the current will at once become purely alternating. This can also be seen from equation (b) above, the coefficient of the exponential term vanishing when t, is - . It follows from this equation that the initial disturbance co - is a maximum when a T a , T It is easy to see also that the maximum value of the current after switching on can never be as great as 2E m If ^ lies in value between T and ^ T + then the maximum ZTT ZTT 2i values of i on the positive side are smaller initially than E 44 ALTERNATING CURRENT THEORY [CH. but the maximum values of i on the negative side are greater than E The number of coulombs that circulate during the first interval T after closing the circuit is LEsm((ot 1 -a) idt = -~ ,. --{1--6 L R Similarly for the second period E 2J2 T R Hence we see that this number rapidly gets smaller and smaller and the current soon becomes purely alternating. When the coil has an iron core, the back E.M.F. is no longer proportional to -= , and the equation to determine the initial ctt disturbance is much more complicated. It is easy to see that the remanence of the iron in the core initially is a principal factor in determining the magnitude of the disturbance on closing the switch. In practice this factor is generally unknown. What happens on closing the switch can be determined experimentally by getting a record of the current with an oscillograph. It is found that, when we have iron in the core, the initial fluctuations of the current are sometimes very large. The solution given above for the growth of current in an The calculation alternating current circuit shows that in making cal- of inductance. culations a knowledge of the inductance of a circuit is quite as important as a knowledge of its resistance. The induct- ance, as a rule, is not an easy thing to measure and formulae for it can only be given in some very simple cases. We will investigate some of these formulae in this chapter. These formulae are arrived at on the assumptions that the current density is constant across the section of the wires, or that we can consider that it is all on the circumference, or that the wires are of negligible cross section. With high frequencies the first assumption is not admissible, as there is a tendency for the current density ] MAXWELL'S EQUATION 45 to be greater near the circumference of a solid wire than near its axis. To see what causes the current density to be variable consider the case of a solid cylindrical wire. We may suppose that it is built up of a system of concentric cylindrical shells, and that the current density is constant in any particular shell. When a current starts in an inner cylinder, it produces no inductive effects inside it, as the magnetic force is always zero there (Chapter I.). It produces however inductive effects outside it in exactly the same way as if all the current were concentrated along the axis. The power for these induced currents must be taken from the inducing circuit, and hence the electromagnetic inertia of currents flowing along the cylindrical shells near the axis must be greater than that of cylindrical currents near the circumference. By solving the general equations of the E.M.F. in an electro- The E.M.F. re- magnetic field, Clerk Maxwell shows that if e be the P D - between the ends of the inner inner conductor of conductor then a concentric main. 1 I d~i 1 I d s i 1 I d*i 12rdi? + 4&7*dt 3 ~18Qr*dP + '"' where r is, in absolute units, the resistance of unit length of the inner conductor, the length of which is I, and A is a constant depending on the return conductor. When the current follows the harmonic law, d z i _ n . , d*i ' _ di where co = 2?r/ and / is the frequency of the alternating current. In this case the above equation becomes L ft) 2 1 ft) 4 a 7| />_!__ I f 12 r 180 r- ^ 1 1 co- 13 a; 4 \di + / I A i I I v I XI n | ^ j t t rt I t~\ r* A f\ ^ I 7 * 46 ALTERNATING CURRENT THEORY [CH. Hence if R l and L^ be the effective resistance and inductance of a length I centimetres of the inner conductor, then R,_ l_^_J^*> 4 , T 12 r 180 r 5 L,_ I 1 13 *_ r = 2 48 r 2 8640 r 4 In these formulae ^ and A are in absolute units. To reduce them to ohms and henrys respectively, we have to multiply by 10~ 9 . If R be the resistance of the whole length I of the conductor in ohms, then Hence, if R l and L^ be expressed in ohms and henrys, the formulae can be written as follows : 1 / lay la V ) ffipr "r 7 f , 1 1 / la> y 13 / Ito V . 1 f ~ + 8640 " For example, suppose that the conductor was a solid copper rod half an inch in diameter and 1000 feet long ; then I = 30480 cms. and R = 0'04 ohm. Suppose also that the frequency was 50, then w = 314-2. Hence L' = 0-0573 and = 0-0033. Therefore R^ = 0'04 (1 + 0'0048 - O'OOOO} = 0-04019 ohm. The formulae for R l and L^ show that the resistance and inductance of the cylindrical conductor depend on the frequency of the alternating current flowing in it. This is due to the smaller inductance of the paths for the current near the circum- ference of the cylinder and the consequent irregular distribution of the current. At very high frequencies the current may be practically confined to a shallow layer near the circumference of the conductor. This phenomenon is generally referred to as the L SK1X EFFECT 47 skin ' effect. It is to be noted that the formulae given above are arrived at on the assumption that the conductor is straight, that the current, and therefore also the applied P.D., follows the har- monic law, and that the lines of force round the conductor are circles. In the case of the outer conductor, the formulae for the effective resistance and inductance are more complicated than for the inner conductor. With high frequencies the current is nearly confined to the inner surface of the conductor. If we suppose the frequency to be infinite then the current would be confined to the outer surface of the inner conductor and the inner surface of the outer conductor so that no magnetic force would be produced in the interior of the conductors. If a constant P.D. were suddenly applied to the conductors at one end of a concentric main, the conductors being short circuited at the far end, then the current would attain its steady value first on the outer layer of the inner conductor and on the inner layer of the outer conductor. It might hence be pictured as soaking inwards and outwards respectively. Similarly, when the circuit is broken, the currents in the outer layer of the inner conductor and in the inner layer of the outer conductor will die away the most rapidly. Heaviside has solved the problem of the propagation of electromagnetic waves along concentric cylinders. He has shown that the energy apparently reaches the inner wire from the outside medium and so the electric current is set up first in the outer layer of the wire and takes time to penetrate into the middle. When an alternating potential difference is applied, the currents in the elementary concentric tubes, of which we may suppose the cylinders to be built up, differ not only in amplitude but also in phase, so that at a certain depth the current may be flowing in an opposite direction to that which it has at the surface. If the conductivity of the conductors were infinite, there could be no current set up in the interior of the wires, and there would be no dissipation of energy. This shows that magnetic induction could not penetrate into an ideally perfect conductor. Some- what analogous problems will be discussed later in the chapter on eddy currents. 48 ALTERNATING CURRENT THEORY [CH. Let R be the mean radius of a circular ring of rectangular Flux of force in- section, a its radial depth, b its breadth, and N o i f d r e ect C anguiar ring tne total number of turns of wire wrapped round section. ^ in such a way that we may consider the turns to be in planes passing through the axis of the ring. We will also suppose that everything is symmetrical, so that the whole flux is inside the ring. Now all parts of the core of the ring at the same distance x from its axis are subjected to the same magnetising force. Suppose that the core is of non-magnetic material, then the element of flux d contained in the rectangle N'PNP' (Fig. 12) of breadth dx is kirmbdx where m is the polar o Fig. 12. Flux inside a ring of rectangular cross section. strength per unit area of the face of the equivalent magnetic shell in this place replacing one turn. Now i = ml, where I is the thickness of the shell. Also IN = ZTTX and hence m = N N Therefore x Thus . ibdx dx FLUX IN IRON RINGS If B m denote the mean flux density over the cross section of this ring, then *Ni ( a 1/JLY+ 1 a \2R + 3(''2R) + '"\ 4>7rNi a- a h where n is the number of turns of wire per centimetre length of the mean circumference. If R were infinite in comparison with. a, B m would equal 4>7rni, and since, in this case, B m is independent of a, it is constant whatever the cross section may be and is equal to B. Since in air B equals H, we see that the magnetising force in an infinite solenoid equals 4?r?n'. If i is in amperes 4:7rni H 10 When the core of the ring is magnetic, since H we see that the magnetising force varies at different points of the cross section and therefore, from the properties of iron, it follows that //, also varies. If however x only vary very little, that is, if -^ be small, then the ratio of B m to = , where I is the length xt I of the mean circumference of the ring, gives us an approximate value of IJL corresponding to the value of H at the centre of the cross section of the ring. Owing to the importance of this problem we will work out the case when the section of the ring is circular. R. i. 4 50 ALTERNATING CURRENT THEORY [CH. Suppose that the radius of the cross section is a circle Flux inside a ring of radius a (Fig. 13) then, making the same assumptions as before, , dx. o Fig. 13. Flux inside a ring of circular cross section. /*-Z2+a A/ /V2 ^7? x Therefore <> = 4>Ni I - dx. J R-a % Writing x = R - a sin 6, we get _ 4>Nia- f * I C os 2 \l-~sme\- d6 H 1 1 o^ 1.3 a 4 .'.H m = x In this case also, if -^ be small and i is in amperes, we can XL assume in practical work that [] MAGNETOMOTIVE FORCE 51 If be the flux of induction embraced by an infinitely long 1 solenoid and if n be the number of turns The magnetic ana- logy of ohm's on a length / ot it, then, by what we nave law. Reluctance. n shown, 4;7rniS which is produced by the magnetising force - ni. The amount of the flux produced depends on the value of ~, which is called the reluctance of the length I of the pS % solenoid. We see that the reluctance 'S of a portion of a tube of induction varies as the length of the portion, and varies inversely as the area of the cross section and as the permeability. Hence we get the magnetic analogy to Ohm's law , _ Magnetomotive Force Reluctance The magnetomotive force ni is the difference of magnetic potential between the two ends of the tube, and it is important to notice that it is independent of the shape of its section. It merely depends on the number of turns and the amperes. Hence electricians often talk about ampere turns per centimetre instead of magnetomotive force. Since H = ~ ni JL\)L we see that one ampere turn per centimetre is 1'257 gausses, and that one gauss is nearly (V8 ampere turn per centimetre. 42 52 ALTERNATING CURRENT THEORY [CH. Let us imagine an anchor ring wound uniformly with two windings of N-, and J\ 7 2 turns respectively. Let Inductance for- . i i " j- muiae for anchor its mean radius be R centimetres and the radius of its cross section a centimetres. Then if we can take a as the radius of the inner coil, since all the flux gene- rated by unit current in the inner coil passes through the outer coil, M = birNiN* {R - ^ 2 - a 2 }. When the section is rectangular, of breadth b and radial depth a, we have In the above formulae M is in centimetres ; to obtain the value in henrys we multiply by 10~ 9 . To get the self inductance of the rings when both are wound with N turns of wire, put JVj = N z = N in the above formulae. For a ring of circular cross section this gives nd for a ring of rectangular cross section The most important practical cases, however, in which formulae are required for self inductance, are for concentric cylinders and for cylindrical wires parallel to one another. We will therefore give complete proofs of the formulae for these cases. The method adopted is to calculate the self energy of the circuit when a current / is flowing in it and then equate this expression to A concentric main consists of two hollow copper cylinders, one self inductance of inside and coaxial with the other, and insulated a concentric main. from it The outer CQpper con d UC tor IS generally protected by a lead sheath which is insulated from it. The Il] INDUCTANCE OF CONCENTRIC MAIN 53 alternating current flows in one cylinder and comes back by the other, and vice versa. Hence in practice the cross sectional areas of the two copper cylinders are made equal to one another as each has to carry the same current. H- \Ve will use Kelvin's formula 2 ^ dv to calculate the self O7T energy of the main. Suppose that the current / flows along the inner copper cylinder and returns by the concentric outer cylinder. Now the magnetic force produced by a cylindrical current sheet at a point outside is the same as if the current were concentrated at its axis, and hence, since the currents in the inner and outer cylinders are equal and flowing in opposite directions, there will be no magnetic force outside the main. Again by page 33 the outside cylindrical current sheet produces no magnetic force inside, and hence, if H be the magnetic force between the two conductors at a distance x from the axis, then X Suppose that the outer and inner radii of the outer and inner cylinders are b. 2 , b lt a 2 and c^ respectively, and that the current density is uniform over their cross sections. Then if H. 2 and H t are the magnetic forces at points in the copper of the outer and inner conductors, then 27 2/^(0*-^) 21 X X TT^-b, 2 ) lf- H JLI ir(x*-a*) = 27 / g*\ x TT (a 2 2 aj 2 ) a 2 2 af \ x )' Hence if L be the self inductance of a length I of the concentric cable, L 2 f^ 2 t 2 f6. H- . 2 [**HS T = 7 rJ -.27rxdx+ T , ^-.torxdaf+jrl - . I 1-J bl STT I-J Qa STT I 2 J ai 8tr Therefore L b 2a 4 a., 1 o 2 - 3^ 54 ALTERNATING CURRENT THEORY [CH. Since in practice the cross sectional areas of the cylinders are equal, we have TT (& 2 2 - b*) = TT (a 2 2 - a^). Therefore If the inner cylinder were solid, a x would be zero and 6. 2 2 would equal b-? + a 2 2 . Using these values in the formula and noting that dj 4 log &! is zero when a-^ is zero we get = 2W + + - 2 +-- 8 a 2 2^3 6i 2 12 6^30 6 X 6 We see that the least possible value of L is when 6j equals a 2 ; in this case I = (4 log. 2 -2) 2 = 0-7726 1. With very high frequencies the current in the inner conductor would be concentrated along its outer circumference, and the current in the outer conductor would be concentrated along its inner circumference. The magnetic force is thus confined to the space between the two cylinders, and hence 21 frH* -J-; - Z 1 " J a OT a a 2* log 2, a 2 where a 2 is the outer radius of the inner cylinder and b l is the inner radius of the outer cylinder. If L be in henrys and I in miles, the general formula is Il] SELF ENERGY FORMULAE 55 where a is a quantity that depends on the frequency. For low frequencies a may be taken equal to unity and for very high frequencies it is zero. For example, suppose that 6j and a are 0*406 and 0"192 inch respectively, then the inductance in henrys per mile is given by L = 0-000241 + 0-000161 (0'5 + 0-075 - 0*004 +...) = 0-000241 + 0-000092a. Maxwell calculates the self inductance of two parallel wires . The self induct- by finding the value of % - dv and equating it ance of two ^^ parallel cyiindri- to \LI~. It is easier however in this case to calculate the self energy of the circuit from the formula \^<$i (page 25) where i is the number of tubes of current embraced by and producing the flux (f>. As the problem is one of considerable importance in electrical engineering it will be useful to give a direct proof of the equivalence of the two formulae for the self energy of a circuit. Suppose that the circuit can be divided up into n parts and that we can calculate ^^(j>i for each of those parts separately. Let H lt Ho, ..., H n be the forces due to the n portions of the circuit, and let H be the resultant force. Then H equals the sum of the components of H lt H, ..., H n , in the direction of H. Let 6 l denote the angle between H l and H, 6. 2 the angle between H z and H, etc. ; then JET = #! cos 0! + ZT 2 cos 0- + ..., therefore Now to find the volume integral of ZT 1 5"cos0 1 ; take the tube of force corresponding to H^ and let dS be a section of any equi- potential surface by this tube. Then H^S is constant along the tube and equal to O l for the single tube is l H cos BidSds = H cos B^ds Summing for all the tubes of force, we find that the complete volume integral of H 1 HcosO l is 47rS^ 1 t 1 . Hence finally we see that z , -f J-2. We know that the lines of force due to the current in this cylinder are circles and that 27 / oA , is zero. The value of (f>i per unit length in the conductor itself is 2I 2 r* tf - a** . - T since i = - /, c? 2 2 - i 2 the current being supposed to be uniform over the cross section. Hence the value of Si due to values of x between d b. 2 and d 4- 6 2 and the current in the first cylinder is d+b, 27 log .(3). To find the linkages with the tubes of current in the second cylinder, divide its section into a series of concentric rings and consider one of them with radius r and thickness d?' (Fig. 14). Let Fig. 14. The self inductance of two parallel hollow cylinders. = x,O l P = r and the angle PO, = 6. Then ^ = r 2 -f d- - 2dr cos d, .'. xdx = rdsiu 6 .dd when r is kept constant. The current ^ in the cylinder of which 58 ALTERNATING CURRENT THEORY [CEL the radius is r and thickness dr is equal to , - . . Hence TT (6 2 - 6j 2 ) the part of 2i. Jt also contributes from d + r up to d + 6 2 I ij cfo = 2/ij log (d + & 2 ) + 2/z'j log (cZ + r). . . J d+r II] FORMULAE FOR INDUCTANCE 59 Hence the total contribution by this elementary cylinder is (a) + (0), i.e. 2/1*! log (d 4- 6 2 ) + 27% log d. Substituting ,- -+- for ^ and integrating from r = 6 X to r = b 2 we get - 2/ 2 log (d + 6 2 ) + 2/ 2 log d ............... (4) as the contribution of the second cylinder. For values of x greater than d + b. 2 , every tube of force will embrace both the currents / and / respectively, and hence will contribute nothing to 2(/>i. By adding (1), (2), (3) and (4) together we find that ^(f>i per unit length, where is due to the current in the first cylinder, is d 2a 1 4 /- a 2 P-af-Sof- We can write down from symmetry the value of 20i for the second cylinder; and hence adding the two quantities together and dividing by 7 2 , we find that L d* 2a/ a 2 1 a 2 2 - So, 2 If we put A 2 = a 2 2 - ar and A; 3 this may be written in the form L d~ I/A 2 , , In calculating this formula we have assumed that the density of the current was constant over the cross sections. With high frequencies the current will be practically confined to a thin layer of metal on the circumference of each of the cylinders, the current flowing in such a way as to produce no magnetic force in the interior of the conductors. The lines of force in this case are the 60 ALTERNATING CURRENT THEORY [CH. same as the lines of equal potential for two parallel cylinders maintained at potentials + V and F respectively. Hence as we will see in Chapter v. the formula for the inductance with very high frequencies can be deduced from the corresponding formula for the electrostatic capacity between the two cylinders. When the currents are uniformly distributed over the cross sections and the cylinders are solid and have equal radii, then noting that a^log^ is zero when c^ is zero we have 3 a where I is the length of either cylinder, a its radius and d is the distance between their axes. If a is small compared to d, then we can write where a is a quantity the value of which depends on the frequency. For low frequencies we can take a equal to unity and for very high frequencies a is zero. In these formulae, if I is in centimetres so also is L. To reduce to henrys we must multiply by 10~ 9 . If I be in statute miles, the following formula will be found useful in practice, L = I JO-00148 Iog 10 - + O'OOOieial henrys, I a J and for low frequencies we can take a equal to unity. For example if -d = 40 and a = 1 then the self inductance would be 0'00253 henrys per mile at low frequencies. If the current be uniformly distributed over the cross sections, then and the minimum value of L is obtained when d = 2a. In this case L = 2-7726 Z + I = 3-7726 I, where Z'and L are in centimetres. [] THREE PHASE MAINS 61 For very high frequencies we shall see in Chapter v. that L = 4J log [ence when d is equal to 2a, L is zero and the currents simply mcentrate themselves along the line of contact. We may now consider three mains connected at their far ends, and suppose that a current ^ flows into No. 1 The self induct- main, and that currents i z and i s come back by ance of a circuit . . formed by three ^s o. 2 and No. 3 mams respectively, so that equal parallel cylinders the axes V _j_ j of which lie along Iqui'ilfeTa^pSm. Let the radius of each cylinder be a, let d be the distance between their axes and suppose that the currents are evenly distributed over their cross sections. Forming 2 contained in a tube of force embrace both the going and return current, then i for this tube will be zero. Hence in order to make the self inductance as small as possible, it is necessary that the going and return currents should be very close together. When we consider also that the tubular filaments of current in the conductors themselves add an appreciable amount to the total sum of $1, we see that the conductors should be flat strips of metal separated from one another by the thinnest possible insulating material. In this case however the circuit possesses considerable electrostatic capacity. LONGITUDINAL TENSION 63 The repulsive force between two parallel wires carrying equal currents in oppo- site directions. From their equivalence to magnetic shells, we see that two wires carrying currents flowing in opposite direc- tions repel one another. Let F be the force of repulsion between portions of length I of two parallel wires carrying currents i and i respec- tively at a distance x apart. If the wires are cylindrical, then we have shown in Chapter I. that the force exerted by the tubes of flow in one of them at external points the same as if the current were concentrated along its axis, imilarly it is easy to show that the resultant force on the other ylinder is the same as if the current in the other cylinder were ncentrated along its axis. Therefore F =- x il x 'his formula was given by Ampere. The same result can be deduced from the formula for the self iductance of two parallel wires forming part of one circuit. For ippose that we keep the current i constant while x becomes -I- dx, then the rate of working of the E.M.F. per unit length = i -j- (Li) Clti . dL the rate of working against the external mechanical forces dx F Tt' and the rate of increase of the magnetic energy of the system -!<*&> ., dL r,dx d therefore i- -r- = F -r- + -r. hence '-** - dx x 64 ALTERNATING CURRENT THEORY [CH. II If the two wires form part of a circuit, there will also be a tension along each of them due to the action of the magnetic field on the conductors which must close the circuit. If \T denote the tension along each conductor, then, proceeding as above, we get where a is the radius of either wire and x the distance between them. For example, suppose that iis 1000 amperes or 100 C.G.S. units, 97 and that - is 10, then a r=100 2 (21og e 10 + 0'5) = 100 2 x 5-105 = 51,050 dynes = 52 grams weight nearly. Hence the pull along each conductor will be only equal to the weight of 26 grams. If the wires are surrounded by a magnetic medium, the forces will be much greater. REFERENCES. CLERK MAXWELL, Electricity and Magnetism, Vol. 2. OLIVER HEAVISIDE, Electrical Papers, Vol. 2. LORD EAYLEIGH, Scientific Papers. ERIC GERARD, Legons sur V Electricity ANDREW GRAY, Absolute Measurements in Electricity and Magnetism, Vol. 2, Part i. CHAPTER III. Effective values of complex currents and pressures. Mean value of an alternating current or pressure. Graphical methods of finding mean square values. Equivolt curves. Choking coil .currents. Condenser currents. Effect of altering the resistance of a circuit on the form of the current wave. A simple sine wave of E.M.F. produces the maximum current in an inductive coil and the minimum current in a condenser. Resonance. Resonance of currents. Numerical examples. Method of neutralising the inductive effect of a choking coil. Method of neutralising the capacity effect of a shunted condenser. Comparison of inductances by means of a voltmeter. References. THE frequency of the alternating currents employed in practice varies between 25 and 100 cycles per second. For lighting purposes it is probable that 50 and for power transmission purposes that 25 will become the standard frequency. Now the instruments used to measure alternating amperes and volts are comparatively speaking sluggish in their action, and so their indications do not give the instantaneous values. Consider for example a Siemens electro-dynamometer. The effect of a current i through the fixed coil acting on the movable coil which is traversed by the same current is to produce a couple of which the instantaneous values are proportional to i 2 . The couple will therefore have the same sign in whichever direction the current passes through the dynamometer, and hence a deflection will be- produced by an alternating current. If the periodic time of the fluctuations be small compared with the periodic time of the moving coil, then the coil is insensible to the fluctuations of i 3 , and the mean torque on the moving coil, when its plane is R. i. 5 66 ALTERNATING CURRENT THEORY [CH. perpendicular to the plane of the fixed coil, can be measured by the torsion of the spiral spring required to bring the coil back to this position. This torsion is measured by the angle through which a pointer has to be turned. Hence if R be the reading, k*R = the mean value of i 2 , where k 2 is a constant, determined by experiments with direct current. When i is constant, k fJR gives us its value in amperes ; when i is variable, k*jR gives us the square root of its mean square in the same units. The instrument therefore tells us the square root of the mean square (which is sometimes called the K.M.S.), or the effective value, of the current. When i varies harmonically with the time, it is easy to find the effective value of the current in terms of its maximum value. For example, suppose that i = I sin at, where o> = 27T/" and / is the frequency, then i 2 = I 2 sin 2 cat /2 /2 = y ~ y cos 2a>t Now the mean value of cos Scot over a whole period is zero, for if we plot out the cosine curve, a glance will show that for every positive ordinate in the first and fourth quarter periods there is an equal negative ordinate in the second and third quarter periods, and hence, if we add them all together and divide by the number of them in order to get their mean value, the result will be zero. /2 Therefore the mean value of i 2 is . If we call the effective value of the current A, then Therefore the effective value in this case is about 71 per cent, of the maximum value. In like manner we can show that those alternating current voltmeters, of which the readings depend on the expansion of a heated wire, or on the electromagnetic attractions and repulsions Ill] EFFECTIVE VALUES 67 of moving coils, or on the electrostatic repulsions and attractions of movable segments, give us the square root of the mean square values of the voltages which we measure with them. If the instantaneous value e of the potential difference between the voltmeter terminals be given by e = E sin cot, then, as before, its effective value V is given by V=~E= 0-707LE. If i do not vary harmonically, then by Fourier's Theorem it mav be written as follows : Effective values 7, sin (o, -<*,) + /s sin (3a>i-a 3 ) + ...... (1). ce squaring and taking mean v that the effective value A of the current is given by Hence squaring and taking mean values, we find If R be the ohmic resistance of the conductor in which i is flowing, then RA* = RA* + RA 3 * + RA* + . . . , where A lt A 3 , A 5 ... are the effective values of the harmonics of the current. We thus see that each harmonic component produces its own heating effect on the conductor in exactly the same way as if all the other harmonics were absent. If the alternating current i have a direct current component C, then i = C + Jj sin (cot ttj) 4- 7 3 sin (Scot - 8 ) + . . . . .-. A*=C* + A? + A*+ .................. (2). Hence, to find the heating effect on a conductor produced by a combined direct and alternating current, we calculate the heating effect produced by each separately and add up the results. For example a complex current consisting of 10 amperes of direct current and 10 effective . amperes of alternating current would only heat a conductor as much as a direct or an alternating current of 14*14 amperes, and an alternating current ammeter placed in the circuit would only read 14'14 amperes. We shall 52 68 ALTERNATING CURRENT THEORY [CH. see later the importance of this result in the theory of rotary converters, and it is utilised in systems of polycyclic distribution. If the potential difference applied to the voltmeter terminals be given by e = P + E l sin (cot - a^ + E 3 sin (Scot 8 ) + . . . , then F 2 = P 2 + Vf+ F 3 2 + ..................... (3), where V* = \E*, F 8 ' = J# 8 .... Hence if we put 100 volts direct current pressure in series with 100 volts alternating current pressure, the reading on a voltmeter would be 100 \/2, i.e. 141*4 volts, no matter what was the shape of the wave of alternating pressure. Although in a few cases the negative half of a wave of alter- nating current or pressure is not similar to the Mean value of an . A alternating cur- positive half, as for example when we have alu- minium electrodes or electric arcs between metals in the circuit, yet in general they are exactly alike, so that their mean value over a whole period is zero. In a few cases it is useful to know their mean value over half a period. The mean value of Esmcot, for example, over half a period, starting from t equal to zero, is T 2 f 2 2 w \ Esmwtdt=-E=0-6366E] -L J o TT this is less than Q'*7Q7lE, which is the effective value. The root mean square value of a variable quantity is always greater than its mean value. This follows at once from the algebraical theorem that is than n except when all the n quantities are equal to one another when the two expressions are equal to each other. In practice the readings on electrostatic, hot wire and moving coil instruments give us at once the square root methodJToffind- f tne mean square of the alternating currents and pressures. We can also by oscillographs, ondographs, etc. find the shape of the current [II] GRAPHICAL METHODS 69 and pressure waves. It is important to test whether the root mean square pressure or current got from these waves agrees with the voltmeter or ammeter reading. There are several graphical methods of doing this ; the following method is particularly con- venient when the curve is drawn on sectional paper. O Ki E X Fig. 15. Graphical construction for finding effective values. F 2 = p (area RPED). In Fig. 15 suppose that OB RE is the positive part of any wave curve and that the negative half is exactly similar. Draw a line through R, the highest point of the curve, parallel to OX, and draw any chord BD parallel to it. Draw BM and DN perpen- dicular to RM and OX respectively and join MN, cutting BD in P. Construct RPE the curve locus of P. Then the volume of the solid generated by the revolution of ORE about OX Also = P J o where I is the breadth and h the height of the curve. rl rh y-dx = Z y.BDdy. Jo Jo 70 ALTERNATING CURRENT THEORY [CH. But y.BD = h.PD. .-. I* fdx= 2 { h.PDdy where A' is the area of the curve RPED. If F 2 be the mean square value of y, then ' fdx Now as the curve RPE can easily be constructed, especially when the given curve is drawn on sectional paper, this is a rapid and accurate graphical method of finding F. When the curve has several maxima and minima values the above method becomes inconvenient for we have to divide the curve into several portions, and this entails drawing several loci curves. In this case it is better to draw n lines from a point making angles - - with one another, and measure along these lines lengths OP ly OP 2 ... equal in length to the value of n equi- distant ordinates of the given curve. Then the area A of the curve drawn through P l} P 2 ... will be given by A = Therefore Y A useful formula for F can be found by equating the two well- known expressions for the volume of the solid of revolution formed by rotating the given curve round its time axis, fi TT I J o and Y*=2y m y (4). F is the R.M.s. value of y, while y m is its mean value, which equals - , and y is the height of the centre of gravity of the plane figure Ill] EQUIVOLT CURVES 71 ORE above OX, which can easily be found by the ordinary methods. When there is only one maximum point the first method of finding y is the most accurate in practice, but formula (4) is a useful one. Equivolt curves. It is instructive to draw curves the mean square values of the ordinates of which are all constant. In practice for example, the shape of the wave producing an effective voltage of 50 might be any of the curves drawn in Figs. 6, 17, 18, 19 and 20. We shall see later on that the shape of the wave has a considerable bearing on the economical working of transformers and motors and the following equations to families of waves all having the same effective voltage will repay study. 110 100 90 80 70 30 20 10 A if \ \V 10 20 30 40 50 60 70 80 90 100 Time. Fig. 16. Equivolt curves. The effective voltage of each curve is 50. 72 ALTERNATING CURRENT THEORY [CH. (1) The equations to the curves shown in Fig. 16 are -*< T from t = to t = 4 XW fT \ n T m and the negative half of the wave is supposed to be similar. In these equations e is the instantaneous value of the E.M.F., t is the time in seconds and T is the period of the alternating current. It follows that where E is the maximum value of e and V is its effective value. Hence for a given V the maximum value E increases with n. Now in Fig. 16 three of the curves are drawn, namely those corresponding to n equal to J, 1 and 2. If n equals zero we get a rectangle. When n lies between and 1 we get curves similar to (a) in Fig. 16, namely two curves meeting at an angle and concave to the time axis. When n equals 1 we get the triangle (b), and when n is greater than 1 we get two curves (c) meeting at an angle and convex to the axis. We see that the mere knowledge of the value of V tells us nothing at all about the value of E. (2) Hyperbolic sine curves. The equation to this family is t T e = B sinh n from t = to and e = B sinh n where B= V In this case E = V . , n 2 sinh - n We get the triangle (a) shown in Fig. 17 when n is and the curve (b) is for n equal to 10. Ill] EQUIVOLT CURVES 73 Tables of sinh 6 and cosh 6 are given at the end of Chapter xvi. 120| 110 100 90 80 70 60 50 40 3C 20 10 7. 7 L V \\ \ 10 20 30 40 ' 50 60 70 80 90 100 Time. Fig. 17. Equivolt hyperbolic sine curves. (3) Ellipse and Parabola. The equation in this case is T (c), where and Note that F (n + 1) = nT (n\ T (1) = 1 and T (J) = V?r. 74 ALTERNATING CURRENT THEORY [CH. The curves in Fig. 18 are (a) w = J an ellipse, (b) n I a parabola, and (c) n = 2 a biquadratic. Tables of log F (n) are given in Williamson's Integral Calculus. 10 50 60 70 80 90 100 Time. 10 20 30 40 Fig. 18. Equivolt curves, (a) Ellipse, (b) Parabola. (4) Sine curves. The equation to the E.M.F. is n + rp = E sin" -=- from t = to -= .(d). , /2n_+_l\ V 2 ; When n = we get a rectangle, n \ we get the curve (a) in Fig. 19, when n=l the sine curve (b), and n = 2 the curve (c). I] EQUIVOLT CURVES 75 It will be noted that all the waves figured above are sym- metrical waves, i.e. those in which (5) Distorted waves. If ef(t) be the equation of a symmetrical wave, then 10 20 30 40 50 60 70 80 90 100 Fig. 19. Equivolt sine curves. represents a distorted wave of E.M.F. which has the same maximum height, the same R.M.S. height, the same breadth, and the same area as the original wave. 76 ALTERNATING CURRENT THEORY [CH. This is easily proved. For example, if V be the R.M.S. of the values of e given by the above equations, then T i / T T 2~ Hence F is independent of the values of r. Similarly we can prove that its area etc. are the same as that of the curve e =f(t). T Hence since r may have any value between and -= there are an infinite number of waves which have the same maximum height, \ 10 20 30 40 5CL.. 60 70 80 90 100 Fig. 20. Equivolt sine curves of equal height. Ill] CHOKING COIL CURRENTS 77 the same area and the same R.M.S. height as the original sym- metrical wave. In Fig. 20 the middle curve shown is the sine curve and the others are distorted members of the same family. We will refer to the family of waves given by the equations (e) above as a family of waves of equal height. It must be borne in mind, however, that there is an infinite number of families of waves of equal height. For example (a) in Fig. 16 and the sine wave (b) in Fig. 19 are the symmetrical members of two families of waves whose maximum heights are equal. We have seen (page 41) that the equation connecting the applied P.D. and the current in an inductive coil is Choking coil -t currents. r> , r ^ 6 = Hi + L -J-. dt Now when R is so small that the effective value of the term Ri is negligible compared with the effective value of L -7- , the coil is \AJV called a choking coil, and di This equation shows that when e is zero, i has either a maximum or a minimum value. If the equation to the P. D. wave is e = E sin cot, then E cos cot Leo when the current assumes its normal shape (page 43), and -4 = 0-1591 J J L where A is the effective value of i and/ is the frequency. Similarly when the wave of P.D. is given by the equations (a) above, we can show that 2(2n + l) U V As n increases from zero, A increases until it attains its maximum value, for = ^1 = 07247; it then diminishes for greater values of n. 78 ALTERNATING CURRENT THEORY [CH. y The maximum value of A = 0*1589 -^ , and E is then 1*565 F. J L y And for a sine curve A = 0*1591 -^ , and E = 1-414 F. /^ It will be seen that the sine wave of P.D. produces a larger choking coil current than any of the first family of waves considered. For the hyperbolic sine curves. 2n 6 sinh -= + n cosh ^ If K be the capacity in farads (see Chapter IV.) of a condenser condenser of which the terminals have a P.D. of v volts, then currents. where q is the number of coulombs in the condenser. Hence if i be the current flowing into the condenser, -*$ If A be the effective value of the condenser current when waves of P.D. similar to those illustrated above are applied to the terminals of the condenser, then Minimum value. (a) A = 4 n ^^-^fKV- 6-661/JTF. fKV; n n |smh (*>) n n (c) 9 ti (d) A=-^ T fKV- 6'283/^F (sine curve). Ill] WAVE SHAPE OF THE CURRENT 79 It will be seen that the sine wave produces the least effective value of the condenser current of any of the waves we have con- sidered. It is also not difficult to show that of all E.M.F. waves of equal height (Fig. 20), applied to choking coils and condensers, the symmetrical wave produces the maximum effective current in. the choking coil, and the minimum effective current in the condenser. We have seen (page 42) that the solution for the current in an inductive coil is Effect of altering . = ff. sin (art + .-.) jr.rinCSarf + a.-A) E 1 sin (arf + a,- When R is very large &, /3 3 ... are all very small, and the curve i is practically the same as the curve e on a diminished scale. The more we diminish R, the smaller do the amplitudes of the higher harmonics become as compared with the amplitude of the fundamental harmonic. Hence by diminishing R in an inductive circuit we make i more like a simple sine curve. If we have resistance in series with a condenser in a circuit, then fidt J - .And . = E l sin (a>t + a x + 71) , E 3 sin (Sat + a s + y s ) Hence, proceeding as before, we deduce the following working rule. Diminishing the resistance in a condenser circuit makes the current wave less like a sine wave. 80 ALTERNATING CURRENT THEORY [CH. The equation of the current in an inductive circuit is D . r di A simple sine 6 = Ml + L -j- . waveofE.M.F. U>t produces the maximum current _,, in an inductive lilUS coil and the minimum current in a condenser. f % ft r> r ^l 7 . -, since 2RLi.dt vanishes. J( o dt Now from page 42, And Vf B + , where a a = Now the numerator of this fraction is greater than the de- nominator except when 7 r =F.= ... = 0. Hence a has its minimum value unity when the 'applied E.M.F. is sine-shaped. and the denominator has its minimum value, and therefore A has its maximum value, when a is unity. Therefore the sine wave produces the maximum effective current in an inductive coil. Similarly for a condenser circuit, and /3 has its minimum value unity when the wave is sine-shaped. Therefore the sine-shaped wave produces the minimum effec- tive current in a condenser circuit. Ill] RESONANCE 81 When we have both inductance and capacity in the circuit the problem becomes of great practical importance, Resonance. . l owing to the high pressures and large- currents that are produced in distributing systems through the effects of reso- nance by comparatively low E.M.F/S. Suppose that we have an - 1 B Fig. 21. inductive coil (Fig. 21) in series with a condenser, and that an alternating P.D. is applied to the terminals A and C. Then in order to find the current we have to solve the equation 1 l # 3 sin (3o>* + a 3 ) + ...... (8). (Lt J\. The solution of this equation consists of two parts. There is first the particular integral =2 sin n + la)t where The other part of the solution the complementary function is found by solving the equation ,. di fi 4>L When R- is greater than -~ the solution is of the form (10), where A and B are constants, and , R R* 1 a m * = 9T "" V ~7a ~" TTr ' B. I. 82 ALTERNATING CURRENT THEORY [CH. When -=.- is greater than R 2 the solution is of the form where A and a are constants. The complete solution of (8) is thus given by adding (9) and (10), or (9) and (11) together. When R is large (10) shows that the initial non -oscillatory disturbance of the current wave rapidly subsides, and when R is small, (11) shows that an oscillatory disturbance of gradually diminishing amplitude is superposed initially on the current. The period of oscillation of this disturbing effect is (12). & LK 4i* If the applied P.D. wave were absolutely constant in shape, then, in practice, after a second or two (9) alone would give us the value of the current. Nowif ZJ2"(2n + l) a fi> a =l .................. (13), we see from (9) that /3 2n +i = 0, and hence that the (2n + l)th harmonic of i is in phase with the (2n + l)th harmonic of the & applied P.D. Its amplitude also is simply -^ , and if R is small JTL compared to Leo, this term practically swamps all the other terms ; thus the frequency of the alternating current is practically W , and its effective value only slightly greater than ZTT Y +1 . Also in this case the effective value of the P.D. across the choking coil will be nearly equal to f ' 271+1 A ' and the effective value of the P.D. across the condenser is approxi mately equal to or It will be seen 'that if R be very small, or if the frequency - - - be very high, then these voltages may attain enormous 2?r Ill] RESONANCE WITH DIRECT CURRENT 83 values. The phase of the choking coil P.D. will be practically 90 degrees in advance, and that of the condenser P.D. 90 degrees behind the current. Hence the two P.D.'s are nearly in opposition to one another, and their resultant or the applied P.D. between A and C (Fig. 21) will be very small compared with either of them. We see from (13) that the lowest value of the frequency at which resonance can occur is ; hence if the highest 27TVLK harmonic in the applied P.D. have a frequency less than this there will be no danger of resonance. It is also to be noted that the above investigation shows that it is quite possible to obtain resonance effects in direct current circuits. The voltage curve of a direct current dynamo is never an exact straight line. The voltage has an alternating component due to the facts that the number of bars round the commutator is finite, and that the slots in the armature give rise to pulsations of the magnetic field. By suitably adjusting the value of the induct- ance of a choking coil put in series with a condenser between the mains, it is possible to get resonance, so that a large alternating current component starts in the circuit, and the pressure across the condenser attains a very high value. Duddell has shown that if we put a resonant circuit across a direct current arc formed between hard carbons, then resonance often ensues, and a large alternating current flows across the arc, causing it to emit a musical note. This happens even when accumulators are used instead of dynamos, and must be due to constantly recurring irregularities in the burning of the arc which continually renew oscillatory waves of the form given by (11). In Fig. 22 we have supposed that the curve (/), which is parabolic, represents the shape of the wave of current in a circuit formed by a choking coil and condenser in series, and we have calculated the applied P.D. (E) necessary to produce this wave. It will be seen that it is a peaky wave very different from a sine curve. Er, is the wave of P.D. at the terminals of the choking coil and is triangular in shape. E l gives the shape of the P.D. at the terminals of the condenser, and is very similar to a sine curve The diagram illustrates the general theorem that except when 62 84 ALTERNATING CURRENT THEORY [CH. the applied P.D. wave is a sine curve, the wave of P.D. across the choking coil terminals is much more distorted from the sine shape than the wave of P.D. across the terminals of the condenser. ^ X Fig. 22. Choking coil (EJ and condenser (EJ P.D.'S when the current wave (I) is part of a parabola. If we have (Fis:. 23) a condenser with capacity K shunted by- Resonance of a choking coil of inductance L, then in certain cases currents. ^e curren t i n the main can be very small compared with that either in the choking coil or in the condenser. Let e, i* r mnnn B Pig. 23. Resonance of currents. and i 3 be the instantaneous values of the P.D., the current in the condenser and the current in the choking coil respectively, then de di 2 are III] RESONANCE OF CURRENTS 85 Now by equations (6) and (7) the effective values of the currents V and A 2 = acoL ' where a and /9 are constants depending on the shape of e ; these constants have their minimum value unity when the wave is sine- shaped. If i be the current in the main, then t.st and hence, noting that we find that COS = - , where is the phase difference (see Chapter VI.) between AI and A 9 . If K and V are fixed, then the minimum value of the main current is A sin , and in this case the choking coil current is A^ cos <. .'. LK(oLa>)* = \ ..................... (14). If L and V are fixed, then the minimum value of the current in the main is A 2 sin $, aud it has this value when = \ ........................ (15). 1. For a sine wave, is 180 degrees, and the main current is Numerical Zero when examples. LKd) 2 = 1. 2. For a parabolic wave < is 173 46', and if the condenser current is constant and equal to A lt the minimum value of the current in the main is Q'WSQA^ and it has this value when I'QQILKu* = 1. Similarly if the current A 2 in the choking coil be kept constant, the minimum value of the main current is 0*1086-A a > an d it has this value when I'OlSLKw* = 1. 3. For a triangular wave is 155 54/. When the condenser current is constant and equal to A lt then the minimum value of the main current is 0'41 is a constant and is negative. Similarly the charge on the nth conductor will be Kn^v^ Now consider another state of equi- librium, when the potentials of the conductors are 0, v a , 0...0. In this case the charge on the first conductor will be Ki,. 2 v z and on the others K 2f2 v. 2 , K s ^v. 2 ... K n ^v z . Similarly we can write down expressions for the charges on the conductors when the pth is at a potential v p and all the others are at zero potential. Now if we superpose all these systems, we get another system in a 90 ALTERNATING CURRENT THEORY [CH. state of equilibrium, in which the charges on the conductors will be given by the linear equations r i r . Again we may write the self energy of a single conductor in* the form J qV. Now by Gauss's theorem 4>7rq = 2,RdS, where we may suppose that dS is an element of the surface of the conductor itself. Also if ds be an element of the axis of the tube of force starting from dS, V = fRds. Therefore the self energy of the conductor But along ds, RdS is constant, therefore RdSfRds = fR*dSds. Now dSds = dv the element of volume of a tube of force. Hence RdSfRds = fR 2 dv, the integration being taken along the tube of force standing on dS. Therefore, integrating for the whole surface of the conductor, we see that is an expression for the self energy of an electrified conductor, the integration being taken throughout all the space occupied by tubes of force. Compare this with the expression for the self energy of an electric current. The definition of capacity given on page 9 is a particular case capacity of a of the following more general definition due to conductor. Maxwell. The capacity of a conductor is its charge when its own potential is unity and that of all the other conductors is zero. In other words, it is the coefficient of self induction of the conductor for electrostatic charges. It is to be noted that any 92 ALTERNATING CURRENT THEORY [CH. alteration in the position of any of the conductors generally alters the capacity of the conductor. In those practical cases, however, where we want to know this coefficient, all the conductors are fixed in position. The capacity of a conductor is found by measuring the charge q which flows into it when it is connected to one terminal of an insulated battery whose electromotive force is v, the other terminal being connected to earth or to any of the surrounding conductors which are all earthed in this case as required by the definition. The ratio of q to v gives us the required capacity. Electricians generally refer to the capacity of a conductor as the capacity between the conductor and all neighbouring con- ductors in parallel with the earth. In order to find the coefficients of mutual induction for electrostatic charges between the various conductors we find relations between the coefficients K lmli K^^, by measuring what is called the capacity between two of the conductors or between two groups of the conductors. The capacity between two conductors may be defined as follows. Let there be any number of conductors 1, 2, 3...n, and let Capacity between 1 an d 2 be insulated. Then, if, when all the two conductors. conductors are initially uncharged a charge q be given to 1 and a charge q to 2, and if the potential of 1 now exceed that of 2 by t^ v z , then the ratio of q to v l v 2 is constant and is called the capacity between the two conductors. It is to be noticed that any alteration in the position of any of the conductors 3, 4, ... n generally alters the capacity between 1 and 2. Also connecting by fine wires any of the insulated conductors to one another or to the earth in general alters the capacity between 1 and 2, and so it is necessary to specify which of the conductors 3, 4, ...n are insulated from earth and whether any are joined together. In practice the equal and opposite charges are given to the two conductors by connecting the terminals of an insulated battery to them, and the capacity between them is measured in exactly the same way as the capacity of an ordinary condenser. Maxwell's equations on page 90 enable us to find in all cases an expression for the capacity between two conductors in terms IV] CAPACITY BETWEEN TWO CONDUCTORS 93 of the coefficients K lfl , K lfZ We shall apply them to find the capacity between two conductors, 1 and 2, when all other con- ductors in the neighbourhood are earthed. Give a charge q to 1 and a charge q to 2 and let their potentials be v l and v 2 respectively, then since v s , v 4 ,... are all zero, .(2). Solving these equations for v l and v 2 we find that IT i TT . f) T7- -ft-l. i T -ft-2.2 ' ***!. 8 and *M.l (3), where K is the capacity between the two conductors. .KTi.i == -ST 2 .2> then If (4). An important practical case arises when the conductor 1 completely encloses the conductor 2. In this case when the charge on the conductor 2 is q, the induced charge on the inside of 1 will be q. By definition, the difference of potential between 1 and 2 is the work done against the electric forces when we take a unit of positive electricity from 1 to 2. Hence the difference of potential between 1 and 2 depends only on the charge on the conductor 2, since the space inside 1 is completely screened from electrostatic induction from the outside. Hence, also, Maxwell's equation for q is 9 = ^0.! ^+#2.2 V 2 . If v l equals v 2 , q must be zero, therefore we must have K 2tt equal to ^r hence Since this is true in all cases, it is true when the charge on the outside of the conductor 1 is zero, and hence it follows from our definition that the capacity between the conductors 1 and 2 is 7^2.2 or Jfj.,,. If the conductor 2 be a metal sphere the centre of which 94 ALTERNATING CURRENT THEORY [CH. coincides with the centre of a spherical cavity in the conductor 1, then if K be the capacity between 1 and 2, I? T? V TI r * J-L ^2.2 ^M.2 ~> where r is the radius of the spherical cavity and r 2 is the radius of the metal sphere. When r t is infinitely great K equals r 2 . If the conductors instead of being separated from one another by air were separated by insulating materials like india-rubber, paper, oil, etc., then the capacities and coefficients of induction of the conductors would be altered. If they were embedded in a homogeneous insulating mass whose dielectric coefficient (specific inductive capacity) was X, then the new constants would be XlTi.!, X/Ci.2, etc. In electric lighting cables the conductors are as a rule separated by various materials whose dielectric coefficients are different. This considerably increases the difficulty of calcu- lating the capacities between the various conductors, but as they are generally arranged in a symmetrical manner inside a metal sheath various useful formulae can be found giving all the capacities in terms of two or three constants. We will first however find an expression for the capacity of a condenser and calculate the capacities of concentric mains and of two parallel cylinders. A condenser consists of two equal insulated conductors 1 and 2, capacity of a whose coefficients KI,!, K 2 . 2 , KI.I are very large compared with the mutual coefficients between 1 or 2 and the earth or other conductors. For a condenser we have, therefore, approximately, since K lfl is taken to be equal to K z ^, Now these equations must be true when v a is zero and thus K lml = K lm t approximately. Using this relation, we at once find that the capacity between the two conductors, or briefly the capacity of the condenser, is approximately K lfl or K lf2 . In the theoretical condenser we suppose that jfiT 1<2 is infinitely great compared with the other mutual coefficients, and hence K ltl or .Ki.a is the capacity of the theoretical condenser. IV CONCENTRIC MAIN 95 Let a be the outer radius of the inner cylindrical conductor capacity of a con- an d & the inner radius of the outer conductor. centric main. L et also F x and F 2 be their potentials, + q and - q be their charges per unit length, and let k be the capacity per unit length, then q = k(V 1 -V 2 ) ........................ (5). Now, from symmetry, the equipotential surfaces between the two cylinders are coaxial cylinders no matter how small a may be. 'Also the potential inside due to the charge on the outer cylinder is a constant and hence, if V be the potential at a point P distant dV x from the axis, where x lies between a and b, the force -. - on unit dx of positive electricity placed at P will be the same, by Green's theorem (see page 8), as if the charge q were concentrated along an infinitely thin conductor coincident with the axis. Hence if dz be an element of the axis, dV qxdz This could also have been proved easily from Laplace's equation, which in this case gives us (page 6) V=A + B\ogx ; therefore (page 7) 1 dV_ B 4>7r dx 4>7ra on the inner cylinder. Now q = Zjracr and hence B = 2q. If X be the dielectric coefficient of the insulating medium, = dx Xa? ' Comparing this result with (5), we see that k= 96 ALTERNATING CURRENT THEORY [CH. or if K be the capacity, in electrostatic units, of a length I centi- metres of concentric main, then K=kl .(6). The unit of capacity in either the electrostatic or the electro- Ratio of electro- magnetic system is denned as the capacity of a mfgrU\lc e un*.~ condenser in which unit charge produces unit The microfarad, difference of potential between the conductors. Hence we may write Q e = K e V e and Q m = K m V m , where the suffix e denotes that the quantities are measured in electrostatic units and the suffix m denotes that the quantities are measured in electromagnetic units. Now we can calculate by (6) the capacity K e of a concentric main in electrostatic units. Also by charging it to a known P.D. through a ballistic galvanometer we can find its capacity in absolute electromagnetic units. The ratio of the electromagnetic to the electrostatic unit of capacity is theoretically equal to v 2 where v is the velocity of light. This quantity v has been shown experimentally by several observers to be 3 x 10 10 centimetres per second, a number closely approximating to the observed speed of light. We may therefore write %*-$%' The C.G.S. unit of electromagnetic capacity is too great for prac- tical use. We therefore use two smaller units, the farad and the microfarad, the latter being the millionth part of the farad. In practice the microfarad is always used. The farad is the capacity of a condenser which has a P.D. of one volt between its terminals when charged with one coulomb. Since the coulomb is the tenth of the C.G.S. unit of quantity, and the volt equals 10 8 C.G.S. units, we get 1 farad = - -- - C.G.S. unit, 1 microfarad = - C.G.S. unit. IV] RATIO OF UNITS 97 Hence if K be the capacity of a condenser in microfarads, and K e be its capacity in electrostatic units, TO 15 = 900000 ** For example, the capacity of a concentric main of length I miles (160,900 I centimetres) in microfarads is given by #= 0-0776 ^-r ........................ (7), 21og- 6 a where the logarithm is to the base 10. If we assume that the magnetic permeability and the dielectric coefficient have no dimensions, then the dimensions of the ratio of the electromagnetic to the electrostatic unit of capacity become (velocity) 2 . By Coulomb's Law, assuming that the dielectric coefficient has no dimensions, we may write Q.' Force x (distance) 2 thus Q e x space x Vforce. Also Q e . V e x work x force x space ; therefore V e oc Vforce. Similarly, assuming that the magnetic permeability has no di- mensions, since the repulsion between two wires carrying equal currents i in opposite directions varies as i* (page 63), i oc Vforce ; and Q m oc it oc Vforce x time. Also Q m V m x work x force x space ; therefore V m oc Vforce x velocity ; ,, Q e space thus ^-x -f x velocity, Q m time V e and ,,, x V m velocity ' K V 1 Hence ^x. <- ^e F m Q, (velocity) 2 R. I. 98 ALTERNATING CURRENT THEORY [CH. To reduce the value of a given capacity when measured in electro- static units to its value when measured in electromagnetic units we have to divide the first value by v*. The electromagnetic C.G.S. unit is therefore v 2 times larger than the electrostatic. From Maxwell's equations expressing the propagation of an electro- magnetic disturbance through a uniform medium, it follows that the velocity of propagation would be v. The agreement between v and the velocity of light is strong evidence of the soundness of his theory that light is a series of electromagnetic waves. Let a be the outside radius of the inner main, 6 and c the radii of the middle main, and d the inside radius of the outer main. Then, denoting the capacity between the inner and the middle main by C 1<2 , etc. we get The capacities of a triple concentric main. .(8). In the case of (7 1>3 the middle main is insulated. The presence of the inner cylinder thus increases the capacity between the two outers. IV] INVERSE POINTS 99 Consider first the equipotential lines round two thin cylinders which are placed with their axes A and B (Fig. 25) ^denser formed perpendicular to the plane of the paper. Suppose ral ' that they are so thin that they may be regarded as lines, and that the cylinder A is charged with a quantity of electricity +q per unit length and that the B cylinder has a charge q. Now join AB and bisect it in ; let AO = ^ . Let V A be the potential at any point P (Fig. 25) due to the action of the A wire alone and let V A be the potential at 0. Then if the medium be air, dr r e Thus V A -V A ' = l J T! where AP = r x . Similarly if V B and v s ' be potentials at P and due to the action of the wire B, *\ V B - V B = %q log n~ where BP = r z . Xow by the principle of superposition, if v be the potential at P due to both, then Also by symmetry the potential of will be zero. Therefore = V A 4- V B . Hence v = 2q log . Therefore the equation to the equipotential surface the potential of which is V is V =-<2 l g^ (1)> or - = constant. 72 100 ALTERNATING CURRENT THEORY [CH. Now by a well-known geometrical theorem the locus of a point P which moves so that the ratio of its distances from two fixed points A and B is constant, is a circle, and if C be its centre and r its radius, CA.CB = r* (2). These points are called inverse points with regard to the circle. It will be seen that the equipotential surfaces are a series of cylinders surrounding A and B, all of which have A and B for inverse points. A particular case is the plane bisecting AB at right angles and passing through 0. Now it follows from Green's theorem that, if we distribute electricity over one of the equi- potential cylinders surrounding the wire A and if the surface R density of the distribution at any point be j , where R is the normal force at that point, then this distribution will be in equilibrium, the potential at external points will be unaltered, and the potential at all points inside this cylinder will be constant. Similarly we can replace the wire B by one of the equipotential cylinders surrounding it. Fig. 26. A and B are the inverse points of the circles. CA . CB=CL* ; DA . DB = DM*. CL = a, DM=b, CD = d, AB = c. Suppose then that we have a solid cylinder LL' (Fig. 26) surrounding A and another MM' surrounding B. Let their potentials be V 1 and F 2 respectively, and their charges + q and q per unit length as before, then from (1) IV] TWO PARALLEL CYLINDERS 101 k' where k is the capacity per unit length. '' k 3L AM (2)> 'BM Now from Fig. 26, BL BL' 2.BC BC CL }. (3). AM AM' 2. AD AD Also 'BM'~2.MD MD Let the radii of the two cylinders be a and b respectively, let d be the distance between their centres, and let AB be c. Then from (3), BL.AM = BC.AD AL.BM~ ab _b.CB ~ a.DB' since DA.DB = b*. We also have CB (CB - c) = a 2 , And Also d = CB + BD v< *r an equation from which c can be rapidly found by the aid of logarithmic tables. 102 ALTERNATING CURRENT THEORY [CH If X be the dielectric coefficient of the medium and I the length of the cylinders, we get finally for the capacity K the formula 2 {log. tail (90- 1) -log. tan | where tan X = , and tan # 2 = , and c is given by equation (4). c c This equation may also be easily put in the form where a = Formula (5) is however the most convenient one to use. If d be large compared with a + b , then approximately When a equals b, the exact formula is, /I / ~v" ~ * , y y \ i And finally, if cZ be large compared to a, .(9). 4 lOpr --- - ie In formulae (5) to (9) if I be in miles and the logarithms are to the base 10, then to get K in microfarads we must multiply the right-hand side of the equations by 0'0776. IV] CYLINDRICAL CONDENSER 103 The cross section of the two cylinders is shown in Fig. 27. The capacity of a ^ ne inner one * s supposed to have a charge + q per unit length, and the outer one a charge q. Let A and B be the inverse points common to the two circles, then, using the same notation as in the preceding example, we have condenser formed by two long parallel cylinders one wholly enclosed by the other. Fig. 27r A and B are the inverse points of the two circles. CA . CB = CL* ; DA . DB = CL = a, DM=b, CD = d, Also Hence where tan 6, = - , and tan 2 = , and c is found from (1). c c 104 ALTERNATING CURRENT THEORY [CH. Suppose that we have three cables whose sections are shown in Fig. 28 and suppose that they are I miles long and that X is the (i) (2) Fig. 28. Capacities of (1), (2) and (3) are K, 1-Q5K and dielectric coefficient of the insulating material used. Then the following table gives their constants, where K 0'0644\Z micro- farads. Condenser 6 a d c calc. from (1) V * 2 Capacity in microfarads (1) 4 1 00 K (2) 4 1 1 13-9 30 8-2 1-05 K (3) 4 1 2 5-12 57'4 21-3 1-30 # Formula (2) may be put into the form \l K (3), where a 2 + fr 2 - ~2ab If d* be small compared with b' 2 a 2 , then we have the ap- proximate formula We see that the capacity is a minimum when the axis of the inner cylinder coincides with the axis of the outer one. The O 2 potential energy ~ 7 , is therefore a maximum and the inner ZJ\. cylinder would be in unstable equilibrium if it were free to move in any direction. IV] TWO CORE CABLE 105 If another metal cylinder be placed symmetrically inside the outer cylinder whose section is shown in (3) Fig. 28, then we should have a two core cable. In this case we can see from elementary principles that the capacity between the inner cylinder and the outer one will be increased by the presence of the new cylinder. Hence this capacity will be greater than 1'30-iT, where K = 0-0776 Manufacturers sometimes use formulae of the form al to predetermine the capacity of cables of various sizes, where a is a constant found experimentally, and they find such formulae of practical use. Formulae for the capacity of polyphase cables will be given in the next chapter. In this case if K be the capacity of the cable measured in microfarads, then the effective value of the con- SS.'ES^i. denser current A is given by cables and in two parallel overhead wires. A = afKVlQ-* amperes, where / is the frequency, V the effective voltage between the wires, and a a constant which has its minimum value 27r when the applied wave is sine-shaped. We have assumed that the potential drop due to the resistance of the cable is negligible. The two conductors are embedded in insulating material Two core cable. (Fig. 29) and are en- closed in a metal sheath which is con- nected to earth and is therefore at zero potential. Let K^ . l be the capacity per mile of No. 1 conductor when No. 2 and the sheath are earthed, as defined on p. 91, let v l be its potential at any instant, and let K lfZ be the co- efficient of mutual induction per mile Fig. 29. Two core cable. 106 ALTERNATING CURRENT THEORY [CH. between the two conductors. We will assume that Vi is constant throughout the whole length of the cable at any instant. Then if q l be the quantity of electricity on a mile of the No. 1 conductor, I i where v 2 is the instantaneous value of the potential of No. 2. Similarly q= K 2 <>v<, + K i v, . / i- 1* 21 9 4.11 These equations may be written : ^i = C^i. i + #1.2) (vi - 0) - K lt2 (vj, - O, ?2 = (#2.2 + Kl.9) (V 2 - 0) - Jf 1>2 (V a - Wj). Now if we have two small bodies 1 and 2 (Fig. 30) which are connected with each other and with an earthed con- ductor through condensers of capacities K^ K^ and K a as in the figure, and if the potentials of 1 and 2 are v x and # 2 then the charges on the conductors connected with 1 and 2 will be equal to q^ and q 2 provided that and K 2 = 71 = ^T 2-2 , and thus If, in addition, the alternator be insulated from earth and everything is symmetrical with regard to the middle point of the external load, then where v is the potential difference between the terminals of the machine. Now the condenser current for any main is defined as Fig. 30. Equivalent condensers. IV] CONDENSER CURRENTS 107 the rate of increase of charge upon the main, and hence, in this case, if 4- i and i be the condenser currents for the two mains, Leaving the symmetrical case for the moment, let No. 2 main be earthed at the alternator terminal, so that v 2 is zero. Then since v l equals v, the condenser currents are dv If v is the same in magnitude and wave form in the two cases, then at corresponding instants 77- 9 77- ^i *Ai.i 1 2 ^-"-1.2 The currents have therefore constant ratios to one another, and if A y AI and A 2 be their effective values we have Fig. 31 represents the section of a lead covered twin cable. The capacity between the two con- ductors is 0'345 microfarad per mile, and the capacity per mile between one conductor aud the other in parallel with the sheathing is 0'53. Hence KI.I= 0'53, -#!.,) = 0-345, Fig. 31. Two core cable. Therefore if A be the condenser current when both mains are insu- lated, then when No. 2 main is earthed A l will equal 1'54 A and A 2 will equal 0'46 A and the current in the sheathing will be (1'54 - 0'46) A, i.e. T08 A. 108 ALTERNATING CURRENT THEORY [CH. We will suppose that three conductors are symmetrically Three phase embedded in a dielectric cables. an( j surroun( ied by a metal sheath (Fig. 32). Then if V , v lt v 2 and v s be the potentials of the sheath and of the three conductors respectively we have, with our usual notation, q o = #o. 00 + #o.lVl + #0.202 + #0.303, 2l = #l. Vo + #l.lVr+ #1.202+ #1.308, q z = #2.9^0 + #2.101 + #2.202 + #2.303, q., = #3. + #8.lVl + #3.202 + #3.303- If we make v l = v 2 = v 3 = v , then there will be no charge on any of the internal conductors, since in practice they are completely screened by the sheath from electrostatic induction from the outside. Hence we obtain the three equations, = #!. o + #!.! + #!. 2 + #!. = #2. ~H #2. 1 H~ #2. 2 ~t~ #2. 3 ) ^ = #3.0 + #3.1 + #3.2 + #3.3- Now from symmetry, #1.1 = #2.2 = #3.3. ^1.2 = #2.3 #3.1 > #0. 1 = #0. 2 = #0. 3 Using these values we reduce the three equations to the single equation #0.1 + # 1 . 1 + 2# 1 . 2 = (a). Hence if we know K lnl and # 1-2 we can find # .i, and then we shall be able to calculate the capacities which can be obtained with various combinations of the three conductors and the sheath. For example, suppose that we wish to find the capacity between the conductors 1 and 2 when 3 and the sheath S are insulated. Putting qi = q, q z = q in the general equations, we have q = KI, O V O + K l . l v 1 + # 1>2 v 2 + #!. 2 V B , ~ q = #1.0 ^0 + #1.2 ^1 + #1.1 02 + #1.2 03- Hence 2gr = (# 1 . 1 -# 1 . 8 )(!; 1 -^), and therefore by definition the capacity in question is IV] CAPACITIES OF THREE PHASE CABLE 109 We obviously get the same result when S and 3 are joined together by a fine wire or are put to earth. Again, suppose we require the capacity between S and the conductor formed by joining 1 and 2, when 3 is insulated. Give equal charges ^q to each of the mains 1 and 2. The induced charge on the inside of the sheath will be q, and if there is no charge on the outside of the sheath this will be the total charge on S. Since the conductors 1 and 2 are practically screened by S, the capacity between them and S will be inde- pendent of the absolute value of the potential of S, and hence it will simplify our equations to put V Q zero. Since v t equals v 2 , the first and fourth equations now become and hence q = -2v l K . l l - . \ -M. Employing (a) we obtain for the capacity The following is the complete list of the capacities that can be got from a three core cable. (1) Capacity between 1 and 2 (2) Capacity between 1 and 2, 3 -!(*,.,-*,.,). (3) Capacity between 1 and S (2 and 3 insulated) f*i.i (4) Capacity between 1 and S, 2 (3 insulated) (5) Capacity between 1 and S, 2, 3 (6) Capacity between S and 1, 2 (3 insulated) 110 ALTERNATING CURRENT THEORY [CH. (7) Capacity between 1, S and 2, 3 (8) Capacity between S and 1, 2, 3 If we measure (5) in the ordinary way, by reading the throw on a mirror galvanometer and comparing with the throw given by a standard condenser, we get K lml . A further measurement of (7) or (8) will give us a simple equation to find K ifZ . Let us take as an example the three phase ' clover leaf extra Numerical n ^^ tension cable supplied to the Manchester Cor- exampie. poration by the British Insulated Wire Co. The working pressure between the conductors is 6500 volts. Working pressure between any conductor and the sheathing = 3750 volts. Section of a conductor = 0*15 square inch = 0*97 sq. cm. Minimum distance between conductor and sheathing) ,. , 6 y = 0-86 cm. Minimum distance between any two conductors J Insulating material, specially prepared paper. Mean dielectric coefficient X = 2"8. By measurement, (7) was found to be 0'436 microfarad per mile, and (8) was 0'488 microfarad per mile. Therefore 2(# 1 . 1 + ^ 1 . 2 ) = 0-436) 3A^ 1 . 1 + 6^ 1 . 2 = 0-488) ' Whence K^ = 0-273, ^ 12 = _ 0-0553. We deduce the other capacities by the formulae given above. The results are expressed in microfarads per mile. (1) Capacity between 1 and 2 (2) 1 and 2, 3 (3) S and 1 Fig. 33. 'Clover leaf cable. 0-164. 0-219. 0-245. IV] EQUIVALENT CONDENSERS 111 (4) Capacity between S, 1 and 2 = 0'262. (5) S, 1,2 and 3 = 0'273. (6) Sand 2, 3 =0-391. (7) S, 1 and 2, 3 = 0*436. (8) flandl, 2, 3 = 0-488. We also see that K 0ml , the coefficient of electrostatic in- duction between the sheath and a conductor, is 0'163. In practical work v is zero, and when the load is balanced (see Condenser Chapter XI.) we have currents in three ^ phase working. Vi + V + V 3 = 0. In this case our equations become l . 2 (v 2 Similarly = (K^ . l - K^ . a ) v 2 , Hence since we get dq l - dv^ lt i t and i 3 Hence n where K = ^(K l , l K lf2 ) and are the capacity currents, calculating the capacity currents we can suppose that the conductors have no capacity, and are joined to the sheathing (Fig. 34) by three condensers each of capacity 2K. For example, in the Manchester cable Fig. 34. Equivalent condensers. 112 ALTERNATING CURRENT THEORY [CH. described above, 27T is double the capacity between any two con- ductors, it therefore equals 0'328 microfarad per mile. If the working pressure between a conductor and the sheathing be 3750 volts, and the frequency be 50, then the minimum value of the condenser current in each conductor is 27rf'2KlV10~ 6 , i.e. 0'386 ampere, where I is the length of the cable in miles. The exact calculation of the capacity currents when the potential differences between the conductors and earth are not balanced is difficult, but a minimum limit can be fixed to the sum of the three condenser currents. Let i l} i 2 and i s be the three condenser currents, then dv, dv, Thus i, - i, = (K lml - K^ (v, - v 9 ) Similarly =<=> 7 /// And i s ii = 2K - = a s , at where v' t v" and v'" are the P.D.'s between adjacent conductors, i.e. the mesh voltages of the three phases. In general, when the P.D.'s to earth of the three conductors are out of balance, the waves of P.D. v', v" and v" between the three terminals of the machine are all of different shapes. Hence if A lt A 2 and A 3 be the effective values of a lt a? and a 3) we must write where the values of a, ft or 7 cannot be less than 2?r, and /is the frequency. In these formulae K is in farads. Now we shall show in Chapter vili. that the above equations prove that A l) A z and A 3 IV] FOUR CORE CABLE 113 can be represented by the sides of a triangle ABC. They also prove that J 15 7 2 and J 3 , the effective values of i l> i z and i 3 , can be represented by lines OA, OB and 0(7, where is a point which is not necessarily in the plane of the triangle ABC, but since OA + OB + OC is never less than therefore I l + I 2 + I 3 is never less than \ (A^ + A 2 + A 3 ) and a fortiori it is never less than 2?r (7' + V" + 7"') /, where K is the capacity between any two conductors. Using the same notation as before, we have Two phase cables g 1 = J l 1 V 1 + KI. 2 V 2 + li. 3 V 3 + K\.iUi with four separate conductors. when the sheath is earthed. Similarly we can write down the three other equations. From symmetry (see Fig. 35) Now if the system is balanced (see Chapter XIL), V 2 + V 4 = 0. Fig. 35. Four core cable.. TT ' ^1 Hence ii = ^rr E. I. 114 ALTERNATING CURRENT THEORY [CH. Therefore when we neglect the resistance of the conductors, the effect of capacity can be shown by imagin- ing that the conductors have no capacity, but are joined by four condensers each of capacity K lml Ki, 3 connected star-wise between the conductors (Fig. 36). This capacity is double the capacity between two opposite conductors. If we measure the capacity K l between 1 and 2, 3, 4, $ joined in parallel, then ] 1 = J ST 1 . Similarly, if we measure the capacity K S, then Equivalent condensers. between 1, 3 and 2, 4, ..) t . Hence ^.3 = - (^ ~PT 2 ). For example, in a lead-covered four core cable, #> 0-234 and 7f 2 = Therefore J S r 1 . 1 = O234 and K lmS .-. K^-Ki. 3 = 0-241. If V be the effective pressure between any conductor and earth then the minimum value of the capacity current in a conductor is %TrVKf 10~ 6 , where K is 0'241 microfarad per mile. It is not difficult to find expressions for all the capacities of the various condensers that can be made out of the four con- ductors and the sheath. For example, suppose we wish to find the capacity between 1 and 3 (Fig. 35). We have on giving charges + q and q to 1 and 3 respec- tively, and - q = K 0fl v + K li3 ^ + 7f 1>2 Thus 2q = (Ki.i-K and therefore q = J (K-^ . x K l . 3 ) (^ v s ). Hence the capacity between a pair of opposite conductors is %(K lml KI.S), which is half the capacities of the condensers shown in Fig. 36. IV] TWIN CONCENTRIC CABLE 115 Fig. 37. Twin concentric cable. In the twin concentric cable, a section of which is shown in Fig. Twin concentric 37, 1 and 4 are copper cable - conductors and X is a third cylindrical copper conductor enclosing the other two and itself enclosed by a lead sheath. This cable is used for two phase work- ing ; 1 and 4 are what are ordinarily called the two outside conductors, and 2, 3 or X ia their common return. The copper used in X is T414 times the copper used in either 1 or 4. When the system is balanced, the P.D. between 1 and X is equal to the P.D. between 4 and X as regards effective value but differs in phase from it by 90 degrees. The effective value of the P.D. between 1 and 4 is T414 times the effective value of the P.D. between 1 and X or between 4 and X. We also know that its phase difference from either of them is 135 degrees (Chapter xn.). Let Vi, v x and v 4 be the potentials from earth of 1, X and 4 respectively, then as before and q 4 = -fi^.i Vi + K 4 . x v x + ir 4 . 4 v 4 . From symmetry ^T 1 . 1 = ^T 4 . 4 and K l . x = K 4 . x . Since one conductor surrounds the other two, we must have q l and q 4 equal to zero when i^, v x and v 4 are all equal, therefore ^.1 + ^.1 + ^4.1 = 0. Hence > Kp.q, for its coefficients of mutual induction for electrostatic charges. The number of condensers required to construct the model would be n + \n(n 1), that is, ^n(n + l). IV] MODEL OF A THREE CORE CABLE 119 Three core cable. In the particular case of a three core cable, the capacities can be represented as in Fig. 38. The capacities of the condensers joining the conductors to the sheath will be K lfQ , ^2.0 and K 3 . , and the capacities of the condensers joining the conductors will be - ^1.2, K 2 . 3 a,ud K 9 ^ respectively. Fig. 38. Model of a three core cable. When the three core cable is symmetrical, and ^.2 = ^2.3 = ^3.!. Hence when K lm0 and ^T 1>2 are known, all the capacities of the cable can be found. It is an instructive exercise to find these capacities from Fig. 38. It will be found that they agree with the values which we have found for them earlier in the chapter. The capacities can also be represented as in Fig. 39. 120 ALTERNATING CURRENT THEORY [CH. IV. -K 1.3 - K 2.3 Fig. 39. Reciprocal model of a three core cable. REFERENCES. CLERK MAXWELL, Electricity and Magnetism, Vol. 1. G. CHRYSTAL, ' Electricity,' Encyclopaedia Britannica, Ninth Edition. A. G. WEBSTER, Electricity and Magnetism. Cambridge University and College Examination Papers, 'Capacities of Parallel Cylinders, etc.' St Catharine's College published the solution in a form very similar to the one given in this chapter, in 1880. C. E. GUYE, 'Les Courants de Capacite dans les Lignes Polyphasees Symetriques,' L'Eclairage Mectrique, Jan. 20, 1900. F. A. C. PERRINE and F. G. BAUM, 'Aluminium Wires and Polyphase Transmission,' Journal of the Am. I. of E. E., Trans. 17, p. 391, 1900. J. of the I. E. E., Vol. 30, p. 1022, ' The Capacities of Polyphase Cables,' 1901. A. BELLA RICCIA, ' Capacity of Polyphase Cables,' Soc. Beige JSlect. Bull., 1 9, p. 318, 1902. M. B. FIELD, ' A Study of the Phenomenon of Resonance in Electric Circuits by the aid of Oscillograms,' J. I. E. E., Vol. 33, p. 647, 1903. CHAPTER V. Formulae for a three core cable. Formula for a four core cable. Cable with n cores. The capacity of a cylinder parallel to the earth. The capacity between two horizontal parallel wires when near the earth. The capacity between two parallel horizontal wires one vertically over the other. The capacity of three phase overhead wires. The inductances of parallel wires with surface currents. IN Chapter IV. we have considered the mutual relations between the capacities of the cores and the sheathing in polyphase cables. In this chapter we will investigate formulae for these capacities. These formulae are in some cases only approximate, but the approximations are sufficiently close to be practically useful, and the simple method employed, combined with the method of electrical images due to Lord Kelvin, is so powerful that it is deserving of attentive study. We will suppose that the copper conductors or cores as we shall call them are three parallel cylinders, and we Formulae for a will first consider the case when each has a charge three core cable. ^ per unit length. The equipotential surface o whose potential is v is given by ^ = (7-2|logr 1 -2|logr 2 -2|logr 3 , where C is a constant and r l} ?* 2 and r 3 are the distances of a point on the surface from the axes of the three cores respectively. If A be a constant this equation may be written in the form 122 ALTERNATING CURRENT THEORY [CH. Now, if the axes of the cores are at the angular points of an equilateral triangle whose centre is 0, and if OP equals r, then it is easy to prove directly or by means of De Moivre's property of the circle (Loney's Trigonometry) that the last equation may be written r 6 - 2a 3 r 3 cos SO + a 6 = rfrfr,? = A 6 (1), where 6 is the angle which OP makes with a line passing through and one of the angular points of the equilateral triangle, and a is the radius of the circle circumscribing it. When the constant in (1) is zero, the curves are simply the three points where the axes of the cores cut the plane of the paper. When the constant is small, the equipotential curves are ovals which are nearly circular in shape and enclose the three points. When the constant equals a 6 , the curves are given by r 3 = 3a 3 cos 30, which represents three loops, each enclosing a core. Each of these loops has a double point and two tangents at the origin, the angle between the tangents being 60 degrees. When the constant is greater than a 6 , we get a single curve enclosing the three cores and having three elevations and three depressions on it. For the curve passing through the point r = 2a, = the constant equals (7a 3 ) 2 , and the maximum value of r for this curve is 2a and the minimum value is l'82a. Hence the radii of this curve differ from the radius of the circle r = l'91a by less than five per cent. Now, by Green's theorem, we can replace any conductor by another surrounding it provided that the surface of the outer con- ductor is an equipotential surface of the system of distribution. Suppose then that the three core cable has the section shown in Fig. 40. If b is the minimum distance of a core from the centre of the cable, then the equation to the boundaries of the cross sections of the three cores is r 6 - 2a 3 r 3 cos 30 + a 6 = (a 3 - 6 3 ) 2 (2). This equation has equal roots when , 6 3 (2a 3 -6 3 ) cos 2 30 = ^ '- . V] THREE CORE CABLE 123 If ft is the positive solution of this equation, 2ft is the angular breadth of the core as seen from the centre of the cable, and 120 2ft is the angular breadth of the space between the cores as seen from the centre. Fig. 40. Section of a three core cable, the equipotential surfaces being given by r 6 - 2a 3 r 3 cos 36 + a 6 = constant, when the three cores are at the same potential. If b = J a, as in Fig. 40, then ft = 20*3 nearly, and hence 2ft = 40'6 and 120 - 2ft = 79'4. If c be the maximum distance of a point on a core from the centre of the cable, then from (2) and hence, if b is \a, c will be T23a. This is the case illustrated in Fig. 40, the boundary of the lead sheath being supposed to coincide with the curve r 6 - 2aV cos 3(9 + a 6 = (7a 3 ) 2 , 124 ALTERNATING CURRENT THEORY [CH. so that its maxima and minima radii are 2a and l'82a re- spectively. If v l be the potential of each core, and v 2 be the potential of the lead sheath, then l -a-2|logr,r i f,-0-8|log4 i . Putting 6 = and r = b we have by (1), - 6 3 8 2 If R is a maximum radius of the lead sheath, 2 and Vi - V2 = . Therefore the capacity between the three cores in parallel and the lead sheath is 3X3 where I is the length of the conductor, X the dielectric coefficient, R the maximum inner radius of the sheath and b and c are the minimum and maximum distances of points on the cores from the centre of the cable. The formula may also be written in the form XZ - 01 R 2 1 ^ 21og- + -log p a 3 since 2a 3 = 6 3 + c 3 . METHOD OF IMAGES 125 Hence when - is greater than -p , the capacity is less than that a -ft of a concentric main whose inner radius is a and outer radius R. When - equals ^ the capacity equals that of this concentric main, CL -K 7 and when - is less than ^ it is greater than it. a H With our usual notation (see page 107) \l 2 log .(1). a 3 - 6 3 If the sections of the cores, instead of haviug the shapes shown in Fig. 40, were true circles, then we should expect that the equipotential surfaces would still be very similar to the curves r^r^r^ = constant, and hence that the above formulae could be used as first ap- proximations. The exact shapes of the equipotential curves in this case could be found by the well-known laboratory method of tracing out the equipotential lines on a circular sheet of tinfoil whose boundary was maintained at zero potential whilst three circular copper electrodes were pressed on it at symmetrical points and maintained at constant equal potentials by a suitable battery. If, however, we make the supposition that the circular cross sections of the wires are small compared with the cross section of the sheath we can find by the method of images approximate formulae to give us the values of K lml and K lf2 . Let (Fig. 40*) be the centre of the section of the sheath by a plane perpendicular to its axis. Give charges + ^ to each of o the conductors the sections of which we suppose to be almost coincident with the points A, B and C. Let OA, OB and OC be each equal to a and let the angles between them be each equal to 120. Let A', B' and C' be the inverse points (page 99) of A, B and G 126 ALTERNATING CURRENT THEORY [CH. with respect to the circle formed by the section of the inner surface of the sheath. Then OA.OA^OB.OB'^OC.OC'^R* where R is the inner radius of the sheath. Let charges f be given o to fine wires passing through the points A' t B' and C' and parallel to the three original wires. Then if r iy r z, r a> r i> r 2 and r 3 ' are the distances of A, B, C, A', B' and C' from a point P where the poten- tial is v, we should have if the sheath were removed fl = C-2|log^^,...(2). Now for all points on the circle, we f have (page 101) Fig. 40*. The images of the three wires A, B and C are at A', B' and C" where OA.OA'=OB. OB' = 00 .00' = R 2 . In finding formulae we replace the sheath by these images. 7 = , = ,= -=: = constant. J Hence, substituting these values in (2), we see that the inner surface of the sheath is an equipotentiai surface of the six charged wires. Therefore, by Green's theorem, we can replace the three outside wires by the sheath without disturbing the equipotentiai surfaces inside. Conversely, as we desire in this case, we can replace the sheath by the three outside wires. It is necessary to suppose that the section of the wires inside is very small, other- wise the bounding surfaces of the three wires cannot be considered as equipotentiai surfaces which are determined approximately by (2). When the sheath is at zero potential (2) becomes Now if r be the radius of the circular cross section of a wire and #! be its potential, then, noting that (Fig. 40*) R 2 CJL and AB" = OB"* + OA* + OB' . OA THREE CORE CABLE 127 Hence since the capacity between the three conductors in parallel and the sheath is 3 (#1.1 + 2^.,,) we get 2 log If we put a r for b in (1) and suppose that - and I -^ ) are negli- gibly small, it is easy to see that the two formulae agree. To find the capacity between the wires A and B we give a charge + q to A and a charge q to B. We replace the sheath by wires A' and B' having charges q and -f q respectively. Assuming that there is no charge on C the equipotential surfaces are given by the equation v = C-2q\og^,. r ^. ^1 '*2 Hence when the sheath is at zero potential, If v : be the potential of A and v 2 be the potential of B, then i 4 + R-a 2 + a 4 \* ~tf ) Hence since the capacity between the two wires is we get and therefore -^,,) = - X? - ...... (4). , a yo XL a g ~~ ' 4 128 ALTERNATING CURRENT THEORY [CH. From equations (3) and (4) K ltl and ^T 1>2 can be readily deter- mined. By the help of the formulae given in Chapter IV. we can calculate all the capacities of a three core cable when the cores are of small section and not too close to one another in terms of these approximate values of K ltl and K^ z . Assume that the equation to the boundaries of the sections of Formula for a four the four COreS is r 8 - 2a*r* cos 4(9 + a 8 = (a 4 - 6 4 ) 2 where a is the distance of the axes of the cores from the centre of the cable, and b is the minimum distance of a conductor from the centre. Also assume that the equation to the boundary of the lead sheath is r 8 - 2aV cos 40 + a 8 = (JR 4 - a 4 ) 2 where R is the greatest value of the radius vector. Then proceeding in the same way as for a three core cable, we find that the equation to the equipotential surfaces is v = C- 2 | log r.r^r,. Hence if v t be the potential of the four cores and v 2 be the potential of the sheath t* = 0-2 | log (a*-*), Thus v 1 - v 2 = 2 7 The capacity between the four cores in parallel and the sheath is therefore - a If R = ma and b = , this becomes m 2 log m ' V] POLYCORE CABLE 129 which is the same formula as for a concentric main whose outer radius is m times its inner one. If c be the maximum distance of a point on the boundary of the cross section of a core from the centre of the cable, then c 4 = 2 4 - 6 4 . And hence the formula becomes When 6 and c are nearly equal to a, we can find approximate formulae for K lml , K lt2 and K lt3 by methods similar to that employed for the three core cable. Suppose that the n cores are all equal and parallel, and that they are arranged symmetrically in the sheath. Cable with n cores. Q . , . If ~ be the charge per unit length on each wire, then the potential at any point P inside the cylinder will be given by v = C-2% log n - 2 ^ log ?' 2 - ...... -2^1ogr n If the axes of the cores are arranged on a circle of radius a, this becomes i v = C - q log [r~ n - 2a n r n cos nO + a- n } n where 6 is the angle which OP makes with OA, where A is one of the points of intersection of the axes of the conductors with a perpendicular plane. The equation to the equipotential curve passing through the point r = d, 6 = is r* 1 - 2a n r n cos n6 + a 2n = (d n - a 11 ) 2 . This only meets the line = when d is greater than 2a. When i d is very little greater than 2 n a, the curve is rippled symmetri- cally and encloses the n cores. The maximum values of the radius R. i. 9 130 ALTERNATING CURRENT THEORY [CH. vector are when 6 = 0, ...... , and the minimum values when n n 0= - ....... The maximum values are each equal to d n n n i and the minimum values to (d n 2a n ) n . For example, when n is 20 and d is 1*2 a, the curve would differ from the circle r = T199 a by less than one part in a thousand. Hence no great error is introduced by the assumption that the equipotential lines near the outer cylinder are circles. i When d = 2 n a the equipotential lines are n loops, each en- closing a core and each having a double point at the origin. The angle between the tangents at the origin to one of these loops is -. Hence the length of the loops is much greater than then- breadth. :i When d is less than 2 w a, there are n oval curves, one round each wire, and the smaller d is, the rounder these curves become. When d equals a, cos nB must equal unity, and therefore 6 is 0, -- , - , ...... that is, the curves are reduced to points which n n coincide with the axes of the cores. Hence, when d is nearly equal to a, the equipotential curves are n small rounded curves, and we may suppose the sections of the n cores to coincide with them. Let V-L be the potential of each core whose minimum distance from the centre of the cable is b, then = C - 2 Jog (a n - b n ). Similarly, if v 2 be the potential of the sheath whose maximum inner radius is R, then v 2 = C - 2 2 log (R n - a 11 ), q . R n - a n and ^- V2 = 2log - V] CYLINDER PARALLEL TO THE EARTH 131 If c be the maximum distance of any point on a conductor from the centre of the cable, we have c n = 2a n b n . Hence the capacity between the n wires in parallel and the sheath is This ma also be written in the form Now - will in general be greater than ^, and hence the a H capacity will be less than that of a concentric main whose inner and outer radii are a and R respectively. If bR = a 2 , or if n is infinite, the capacity will be the same as that of this concentric main. The general appearance of the equipotential surfaces for the case of eight cores can be understood from Fig. 117, Chapter XV. When b is greater than 1'5 times a, the equipotential surfaces are practically circular cylinders having the same axis as the cable. In the case of overhead wires which, unlike the conductors in cables, are not screened from outside influences by an enclosing metallic screen, we have to take into account the effect of the earth. We will first give the exact solution of the capacity of a single cylindrical wire parallel to the earth, and then give various practical approximate solutions for the case of several wires in parallel. Suppose that we have two equal cylinders parallel to one another, one charged with a quantity q of elec- The capacity of a J * cylinder parallel to tncity per unit of length, and the other charged with a quantity q. Then, by pages 9 and 99, the equipotential surfaces are given by T* v = 2q log -^ , 92 132 ALTERNATING CURRENT THEORY [CH. where r and r 2 are the distances of a point P from the inverse points A and B of the two circular sections made by a plane cutting the cylinders at right angles. Now, at every point on the plane bisecting AB at right angles, r a equals r z , and therefore v equals zero, so that this plane is an equipotential surface. Since the earth is at zero potential, we see by Green's theorem, that the equation 7* v = 2q log *i gives the equipotential surfaces for a single cylinder and the earth. If a be the radius of the cylinder, v 1 its potential, and h be the height of its axis above the earth, then where is the centre of the cylinder. But OA.OB = a 2 , and OA + OB = 2A, thus OA=h-Sh*-a* t and therefore h + */h*-a* 2 log a We suppose that the wires are at the same height, and we shall first find the capacity between the two wires The capacity be- . . _ tween two hori- in parallel and the earth. Let the charge on each zontal parallel the* earth en wire be -4- ^ per unit length. It is easy to see by the method of images that the equipotential surfaces will be the same as those due to four parallel wires, the new wires being the images of the old wires and charged with negative elec- tricity. Let h be the height of the wires above the earth, then CAPACITIES OF TWO HORIZONTAL WIRES 133 the distance between a wire and its image will be 2h, and the equipotential surfaces will be given by It is easy to see that C is zero ; hence ?*o /', v = g log . >V- If d be the horizontal distance between the wires, and if a, the radius of each wire, be small compared with d and h, we get the following approximate equation to give i\ the potential of the wires, > - ad - Now if K lfl be the coefficient of self induction for electro- static charges of each wire and K lm the coefficient of mutual induction, the capacity of the two wires in parallel is 2 (K lm and thus We shall now find the capacity between the two wires. Suppose that the charge on one wire is 4- q per unit length and that the charge on the other wire is q ; then, since the capacity between the two wires is ^(K ltl KI. S ), we find by the method of images that t r I r V In deducing these equations we have supposed that d and h are large compared with a. Solving the equations we find a 134 ALTERNATING CURRENT THEORY [CH. and Comparing this with the formula obtained for a single cylinder, we see that the presence of a neighbouring cylinder increases the value of K lfl . If we make h infinite in the above formulae, then I and a The capacity between the wires in this case is approximately Let d be the distance between the wires, and h the height of the lower wire above the ground. Let the radius between a two of each wire be a, and suppose that it is small " tal compared with either d or h. Then, if the charge on ^ e l wer w * re ^ e 3 P er um ^ length and that on the upper wire q per unit length, the equi- potential surfaces are given by where 1\ and r-[ are the distances of a point on the surface from the axis of the lower wire and its image respectively, and r. 2 and n' are its distances from the upper wire and its image. If v t and v 2 be the potentials of the lower and upper wires, then 2/i . d + 2/i v, = 2g log -- - 2q log d v] CAPACITIES OF THREE PHASE WIRES 135 Similarly , 20 log Thus (d + h) a 3 (d + 2/i) 2 Therefore the capacity between the two wires is approximately I If d be small compared with 2/i, this may be written _ I _ d_ 2eP ' g ~ If the wires had been in the same horizontal plane at a height h above the ground, then the capacity would be I Now since is less than d' 2 the capacity between the wires for a given distance between them is a little smaller when they are arranged one over the other than when they are placed side by side, provided that the height of the lower wire in the one case is the same as the height of the two wires in the other. If however the mean height is the same in both cases we see, by writing h \d instead of h, that the capacities for the two arrangements are practically equal when d is small compared with 2h. Suppose that the three wires are parallel to, and equidistant from, one another, the plane through the axes of The capacity of three phase over- the lower two being parallel to the earth. We will find the capacity between the lower two and the top wire. Let the charge on each of the lower two (1) and 136 ALTERNATING CURRENT THEORY [CH. (3) be + 1 per unit length, and on the top wire (2) be q, then, taking images (Fig. 41), we find for the equation of the equipotential surfaces I r. 2 = ? lo g^ ? ?- Let d be the distance be- tween the axes of the wires and h the height of the two lower wires above the ground. Let a be the radius of each wire, which is supposed to be small compared with d and h. Then if v : be the potential of the lower wires, and v 2 the potential of the upper one, we have approximately %hd(4h* = 2 log Fig. 41. Image of overhead three phase mains. Thus Vi - v 2 = q log - 2h (4/i d \/3) 2 Therefore the capacity is 2hd \/3) 2 3 The expressions for the capacity between one of the lower wires and the other two, or between the three in parallel and the e.arth, can be written down similarly without much difficulty. They are however much more complicated. V] CAPACITIES IX PRACTICE 137 In applying the above formulae it has to be remembered that we have made the assumption that the earth in the neighbourhood of the wires is a perfectly level plane of perfectly conducting matter. This assumption is in many cases not permissible. The presence of trees, rocks, buildings, etc. con- siderably complicates the problem ; moreover the heights of the wires above the earth are generally not constant. In addition the wires are often ' spiralled ' or ' barrelled ' relatively to one another. For example if a, b and c be the insulators on the first pole and ~j~^f-> Cut dt" and thus there is in this case no definite value of L. The problem becomes a little simpler when / is a simple harmonic function of the time, for then we can speak of the effective value of the self inductance. We can also calculate it, as Maxwell has done, for the simple case considered on page 46. In two cases only can we obtain definite values of L. The first case, in which the currents are constant and the conductivity is finite, has been sufficiently considered in pp. 52 62. The second case is that in which either the conductivity is infinite or the frequency is infinite. It is simpler to consider that the conductivity is infinite and we shall proceed on this assumption. Now the E.M.F. between the ends of all the conductors of one set is the same and in the same direction, and hence, since their con- ductivity is infinite, the total flux of force passing between any two conductors of the same set is constant. If initially, before the E.M.F. was applied between A and B, there were no currents in the conductors, this flux of force is always zero. Hence, by what we have shown above, all the conductors of one set in the electro- static problem are at the same potential. Since the total flux between any two conductors of the same set is zero, the total flux between any two conductors of opposite sets is the same, and hence also the total flux of induction which is linked with any conductor of either set is the same for all the conductors of that set. The total flux between two conductors of opposite sets is easily seen to be simply LI. For, by page 56, we have where i is the total current embraced by the tube of force in which the flux is (j>, and the summation takes in all the tubes of force. Since there is no magnetic force inside a conductor, i is constant Y] SELF INDUCTANCE 141 for every tube linked with that conductor, and thus we may write LI* = 3> A $i A + $>,2i B , since 2< has a constant value, say <& A) for every conductor of one set and a constant value <& B for every conductor of the other set. But the total flux between two conductors of opposite sets is *&A + &B' an d since 2i' 4 equals / and 2i B also equals /, it follows that the total flux between two conductors of opposite sets equals LI, which proves our assertion. We thus see that when the conductivity is infinite the total flux of force between two conductors of opposite sets, when the total current is unity, is equal to the self inductance of the circuit ; this result is only true in the special case of infinite conductivity, though language is often used which implies that it is true when the conductivity is finite. The result is not true when the current is distributed over the sections of the conductors, because in this case some tubes of current are em- braced by more tubes of magnetic force than others. The calculation given on pp. 55 to 59 definitely takes account of this variation in the number of tubes of magnetic force embraced by the various current tubes. We can now easily express L in terms of K, the capacity between the two sets of cylinders, the cylinders of each set being connected ' in parallel.' For the flux of force <> A -f 4> B between two conductors of opposite sets per unit length is LI /I, where I is the length of either conductor. Hence in the electrostatic problem if v be the difference of potential between two conductors of opposite sets, then since v must equal <& A -f- <#, we have and therefore, since / equals q, We can now write down the value of L for a pair of equal and parallel circular cylinders by the aid of the formula (8) of page 102. Thus 'a 142 ALTERNATING CURRENT THEORY [CH. or when d is large compared with a f d a N The inductance L of a circuit formed by two parallel circular cylinders, one wholly enclosed by the other, is given by L = '21 log e a 2 2 where The radii of the cylinders are a and b respectively, and d is the distance between their axes (page 104). When d 2 is small compared with 6 2 a 2 this becomes b d 2 . T - a b*-a 2 and when d is zero (see also page 54), We can write down in a similar manner approximate formulae for the inductances of the circuits formed by connecting in various ways the cores and the sheath of a polyphase cable by the aid of the formulae for the capacities given at the beginning of this chapter. For example, when the frequency of the alternating current is very high, the inductance of the circuit formed by the three cores in parallel and the sheath of the three phase cable illustrated in Fig. 40 can be found by the formula We can find in a similar way the formulae for the inductances of parallel wire circuits suspended above the earth and carrying currents of very high frequency by the help of the formula, For example, if part of a circuit be formed by a wire of length I and radius a at a height h above the earth, and if the earth form POWER TRAXSMISSK the remainder of the circuit, the earth currents will be concen- trated on the surface of the earth, and we shall have by page 132 a or, when a is small compared with h, . Cb The logarithms in this chapter are all to the base e. In proving the above formulae it has been assumed that the total current in a conductor of either set has the same strength at all points on that conductor. This implies that the rates of variation of the charges which appear on the surfaces of the conductors, due to the E.M.F. between the two sets, are negligible, that is to say that the condenser currents are negligible. Now this is true when we are dealing with direct currents which have arrived at their steady values. It is also true with alternating currents when the wires are infinitely thin and so have zero capacity. In the general case it is not true. For example, on a single phase line for the transmission of electric power at high pressure, the effective values of the currents in the two conductors at the generating station may be as high as fifty amperes, although the ends of the two conductors are not joined at the distributing station. As we proceed along the line from the generating station to the distributing station, the effective value of the current con- tinually diminishes from fifty amperes to zero. The calculation of the electromagnetic energy at a given instant in this case is difficult even although we make the assumption that the conductivity of the wires is infinite. If v be the instantaneous value of the potential difference between the mains the electrostatic energy \Ktf can however be found at once. Let the current at the generating end of the line- when the line is loaded be i + i, where i is the capacity component of the current and i is the current taken by the power station. Then the electro- magnetic energy stored in the field round the transmission lines lies in value between \L (i + i ) 2 and ^Li*, where L is the value of the inductance calculated by the formulae given above. If the load 144 ALTERNATING CURRENT THEORY [CH. V current be large compared with the charging current, then %Li- will be approximately equal to the total electromagnetic energy. When the line is very long, or when the frequency is so high that the distance travelled by light and therefore also the distance travelled by the electric disturbance during one alternation of the applied potential difference is comparable to the length of the conductors, other and more complex phenomena arise. The formulae given in this chapter should not be used in these cases without investigating whether the assumptions on which they are founded are permissible. We have also neglected the brush discharge from the wires. When the voltage is 20,000 or upwards, the brush discharges from the wires are similar to those got from ordinary frictional electrical machines. If a thin wire be connected with a main at a high potential, the air immediately around it appears to glow, and, if the end of the wire is pointed, a current of air comes from it. Pieces of cotton hung from the mains repel one another, and all the ordinary electrostatic phenomena are produced. It is found in practice that the current and power taken to maintain this brush discharge is appreciable, and for this reason very thin wires must not be used for the transmission of electric power at very high pressures. REFERENCES. OLIVER HEAVISIDE, Electrical Papers, Vol. 1, p. 42, 'On the Electrostatic Capacity of Suspended Wires,' and also p. 101, 'The Inductances of Suspended Wires.' H. A. ROWLAND, 'Electromagnetic Waves and Oscillations at the Surface of Conductors.' American Journal of Mathematics, Vol. 11, p. 373, 1889. Sir W. THOMSON [Lord KELVIN], 'On Alternate Currents in Parallel Conductors of Homogeneous or Heterogeneous Substance.' B.A. Report, 189CL p. 732. For diaJKms of equipotential surfaces in special cases see the figures in Chapter xv. and the references given there to papers discussing these curves. CHAPTER VI. The power factor. When the power factor is unity, the volt and ampere waves are similar. The maximum value of the power factor is unity. Geometrical interpretation of the power factor. Definition of phase difference. Time lag. Numerical examples. Zero power factor. Watt E. M. F. and wattless E. M. F. Impedance. Reactance. Watt current and wattless current. IF e represent the instantaneous value of the P.D. across a circuit and i the instantaneous value of the current in The power factor. it, then ei gives the instantaneous value of the watts expended in the circuit. The mean value of ei over a whole period gives us the rate at which work is being done in the circuit, or the power ( W) being expended in it. The value of ei is sometimes negative for a fraction of a period and hence its mean value can be very small. Now the reading V of the volt- meter gives us the R.M.S. value of e, and A the reading of the ammeter gives us the R.M.S. value of i, but VA will not in general give us the mean value of ei. It is found convenient in practice to call the ratio of W to VA the power factor of the circuit, and we shall show that the maximum possible value of the power factor is unity. It may therefore be denoted by cos , where is a certain auxiliary angle of great use in graphical calculations. In mathematical symbols the power factor is defined by the equation 1 rT rT I eidt eidt *- ' o Jo (1 [ T , 1 [ T . I* (T 7 7^\ p 7 *"^ j I e-dt.^l vdt\ -( e 2 ^. i'cw}- I- 1 ' ^ .'o j (Jo Jo ) 10 146 ALTERNATING CURRENT THEORY [CH. To prove this we notice, from the meaning of the integral When the power si g n > that factor is unity the -, fT 2 2 2 2 volt and ampere L I ^T, _ y K \ K 2 t7 3 ~r . . . ~t t? n waves are similar. Hi I * <** ~ " > -L JO n=<*> n where e l} 2 , ..., e n , are equidistant ordinates of the P.D. wave. Dividing up the current wave into the same number of ordinates, we see that, if the power factor equals unity, then from (1), thus (e^ + e^ + ...) 2 - (ef + ej + ...) (# + ^' 2 2 + ...) = 0, and (e } i 2 - e^ + (e^ - e^ + . . . = 0. Now since the square of a number is alwa}^s positive, every term on the left-hand side is positive ; and since the sum of them is zero, every term must be zero. Hence e^ e^ = ; e^i* e^ = ; etc. Thus 5 = 5 = 5 = ...=^. *i *2 ^3 *n Therefore, at every instant, the ratio of the volts to the amperes is constant, which proves the theorem. Again since (e^ e^)- + (e^i s e z i-tf + ... is always greater than zero except when each term equals zero, it The maximum g - 1 A value of the power easily follows by Sfoing 1 through the above proof factor is unity. J ' . . backwards, that the power factor is less than unity except in the very particular case when the current and P.D. waves are the same curve drawn on different scales. It is convenient to call these waves similar waves. Hence if the power factor of a circuit is unity, the volt and ampere waves are similar, vanishing at the same instant and attaining all their maximum and minimum values at the same instants. We may show in a similar manner that the power factor can not be less than 1. When the power factor is 1, e is negative when i is positive and vice versa, but, as before, the ratio e to i is constant. Hence in this case also the waves are similar waves. They are drawn however on opposite sides of the axis. Vl] POWER FACTOR 147 It will be seen that cos < can only be equal to its limiting values, +1 and 1, in very special cases. As the angle is mainly useful in the application of graphical methods to alter- nating current problems, it is convenient to make the limitation that < lies between and 180. If a simple circuit is absolutely non-inductive and has no capacity, then e = Ei, where R is the resistance of the circuit. Substituting this value of e in (I) we see that the power factor equals unity. This could be proved directly as follows. We have e- = R 2 i- } thus V 2 and V=RA. Also ei=Ri 2 , therefore the mean value of ei = mean value of Ri*, hence W = RA- W and the power factor ~~V~A ~ 1- By defining the power factor as the cosine of a certain angle , Geometrical in we are a ^^ e ^ n man y cases to give a geometrical terpretation of the interpretation to the quantities involved, and can easily prove many algebraical relations between them by the help of known theorems in algebra and trigonometry. We will consider a simple case. B 'VVVV Bi VVV B UvAAAy c Fig. 42. Inductive resistance in series with non-inductive resistances. Suppose that B. 2 C (Fig. 42) is part of an alternating current circuit. Let the resistances between BB l and B^B^ be R and R respectively, and suppose them non-inductive. Let e l) e' be the instantaneous values of the P.D. between B^ and C and between 102 148 ALTERNATING CURRENT THEORY [CH. B and C respectively, and let the instantaneous value of the current be i. Then evidently we have always thus e'' 2 = e^ + Rfi* - ZR^i. Hence by taking mean values for a whole period, V'^Vf + RfAt-ZRiW .................. (2) where V , V 1 and A are the effective values of the volts and amperes, and W is the mean value of e^i, that is, the power being expended in the circuit B 1 C. Denoting the power factor of this circuit by cos fa , we have W= V,A cos fa, and substituting in (2) we get F' 2 = V* + RfA* - 2R, V,A cos fa . If we now construct a triangle CBB l (Fig. 43) whose sides GB, BB l and J B 1 (7 are V, R^A, and V l respectively, we see by trigonometry that the angle CB 1 B is fa. The cosine of CB^B is the power factor of the circuit OB l (Fig. 42). C 3, B Fig. 43. This diagram shows graphically the magnitudes and phases of the potential differences in the circuit shown in Fig. 42. Similarly producing BB l to B 2 (Fig. 43) and making B^., equal to R^A, it is easy to show that the cosine of the angle GBc,B is the power factor of the circuit CB 2 (Fig. 42) and that CB 2 (Fig. 43) is the effective value of the P.D. between C and B. 2 . Vl] PHASE DIFFERENCE 149 If F> denote this P.D. and (f> 2 denote the angle CB. 2 B, then by trigonometry, Fo sin . 2 = F] sin fa . If we denote the angle CB K by $', then F! cos ! - V cos ' = R^A, thus V,A cos (/>! - V'A cos (/>' = .ft^ 2 . Since Fj A cos j is the work done between S l and (7, it follows from this equation that V'A cos ' is the work done in the circuit BC (Fig. 42). Hence cos <' is the power factor of the circuit BC. Fig. 43, then, can be used to prove many relations between the volts, amperes and watts in the circuit. Thus to define the power factor as the cosine of an angle is a real help in understanding the relations of the various quantities involved. It is convenient to call the angle whose cosine is the power Definition of phase factor the phase difference between the waves of difference. P D an( ^ curre nt. More generally, if e l and e 2 be two periodic functions of the same frequency, and < be their phase difference, then T r \ J0 \ e*dt. (JQ JO (3), being an angle between deg. and 180 deg. The principal advantage of determining the value of this angle is to enable us to employ graphical methods. For example, if e be the resultant of two P.D.'S e 1 and e 2 > then at every instant and e- e-c + e. 2 ~ -f Hence, taking mean values over a whole period, F^F l 2 +F/ + 2F 1 F 2 cos<#> ............... (4), where < is given by equation (3). We see then that F, the effective P. D. of the resultant, is the diagonal of the parallelogram constructed on Fj and F 2 as adjacent sides when the angle between them is <. 150 ALTERNATING CURRENT THEORY [CH. By the time lag of two periodic functions of the same frequency we mean the interval that elapses between the instants when they pass through their zero values in the positive direction. The angle of time lag may be defined as the angle described in an interval equal to the time lag by a uniformly rotating radius which makes one revolution in a time equal to the period of the alternating current. If ti t t 2 be the epochs at which e l and e 2 pass through zero in the positive direction, we can write o)^ = QL-I and cot 2 = 2 . The time lag between ^ and e 2 is ^ 1 2 , and the angle of time lag is ft> (ti tz), that is j 2 . If e l = E l sin (at a^ and e 2 = E 2 sin (cot 2 ), by substituting in (3) we find that cos = cos (! a 2 ), therefore = i 2 . In this case, then, the phase difference is numerically equal to the angle of time lag between e t and e. 2) and, when ^ and e 2 vanish at the same instant, then < is zero or 180. When e l and e 2 are not similar curves, then for no value of the time lag is the phase difference zero or 180. This will be best understood by solving a few numerical examples. In order to simplify the calculations we will suppose that one Numerical f the curves is a sine curve, and will find the phase examples, difference between it and the curves of which the positive halves are shown in Fig. 44. All the curves drawn give an effective voltage of 50, but the absolute value of the voltage has nothing to do with the phase difference, which depends only on the class of curve and its position relatively to the sine curve. If one of the curves represents a P.D. curve and the other the current curve to which it gives rise, then the cosine of the angle of phase difference between them will give the power factor. VI] NUMERICAL EXAMPLES 151 The equations to the first halves of the curves shown in Fig. 44 are (a) Rectangle, e = V, (b) Parabola, e (c) Sine Curve, e _ 2 (d) Triangle, e = (e) Inverted Parabolas, e = (/) Inverted Cubics, e = All these curves have the same effective voltage V. We shall find their phase differences with the curves 2<7r T - * A T / 2<7r 4 \ I sin -~- 1 and i Isiu(-^-t CM. Curve Maximum value of e Height of C.G. cos with in degrees cos with (a) V 0-5 F 0-9003 25-8 0-9003 cos a 1-370 F 0-5476 F 0-9995 2-2 0-9995 cos a i 1-414 F 1-732 F 0-5552 F 0-5773 F 1-0000 0-9928 6-75 cos a 0-9928 cos a w 2-236F 0-6708 F 0-9322 21-2 0-9322 cos a 2-646 F 0-7560 F 0-8628 30-4 0-8628 cos a We might also have calculated the phase difference between any two of the curves shown in the figure. For example, the phase difference (f> between the rectangle (a) and the very peaky curve (/) is 48'6 degrees, and the power factor (cos) is 0*6613. We have proved at the beginning of this chapter that the power factor can only be unity, and consequently the phase difference can only be zero, when the ratio of e to i is constant throughout the whole wave. A first essential condition for zero 152 ALTERNATING CURRENT THEORY [CH. phase difference is, thus, that e and i should both vanish at the same instant. This makes possible the second essential condition, namely, that the ratio of the ordinates of the two waves should be constant. 10 Fig. 44. Voltage curves, each of which has an effective value of 50. (a) Rectangle. (b) Parabola, (c) Sine curve, (d) Triangle, (e) Inverted parabolas. (/) Cubic curves. Each of the curves shown in Fig. 44 is the first half of a symmetrical alternating curve. In such a curve the curve from = to \T is symmetrical about the ordinate corresponding to VI] TIME LAG AND PHASE DIFFERENCE 153 t = i T, and the curve from t = J T to T is the exact counterpart of the curve from = to \T, except that it lies on the opposite side of the axis. We may express these conditions by the equations f(t)=f(^T-t) = -f^T+t) ................ (c). Since the curve is a continuous one it follows that f(t) = when t = and when tT and also when t = J T. We can now show that, if < be the phase difference when the time lag is zero, and if a be the angle of lag between a symmetrical alternating curve and a sine curve, then cos = cos cosa ....................... (5). To prove this, we have * f (t) sin ^ -a)dt= j* f(t) S ~ dt cos a T I f (t) cos -=- dt sin a. Jo 1 o T Putting t = - - x, we get, from (c), r (T N f ( "g "~ X I COS 2 T The last integral therefore vanishes. If the limits in the last integral had been \T to T, the integral would have vanished on account of the symmetry of f(t) about the ordinate corresponding to t = T. We thus obtain J - a] dt = eosa fftOnt^* ...... (d). sn O \ 4 / J Q Hence from (d) with the aid of (1) or (3) cos = cos < cos a. It is instructive to give a graphical interpretation to this formula. Let OAB and OBC (Fig. 45) be two planes at right angles to one another, and let the angle AOB equal , and the angle BOG equal a, then the angle AOC will be <. 154 ALTERNATING CURRENT THEORY [CH. Draw EC (Fig. 45) perpendicular to OC and join AC. Then OA* = OB 2 + AB* = OC* + CB* + AB* Therefore the angle OCA is a right angle. Fig. 45. Time lag and phase difference. For all positions of OC, the angle BOG gives the angle of time lag and the angle AOC gives the angle of phase difference between the two periodic quantities. The minimum value of the angle of phase difference is the angle AOB. Now OC = OA cos A OC, and OC = OB cos a = OA cos ( cos a. Thus cos A OC = cos < cos a ; therefore by (5) the angle AOC is the phase difference. This gives us a graphical construction for the phase difference between a symmetrical wave and any sine wave. It is to be noted that this construction is in three dimensions. We shall return to this method of representing phase differences in Chapter vm. In practice it is the exception to have both curves symmetrical. Suppose that the current is a sine curve, and that there is no time lag between it and the P.D. curve. Then if we consider a Vl] WAVES OF EQUAL HEIGHT 155 family of P.D. curves all of the same height (see Fig. 20, Chap, ill.), a little consideration will show that the power factor is a maximum when the peak of the wave occurs at the quarter period. Suppose now that the current is lagging, and that the P.D. wave has its maximum value before the quarter period. Then it is evident that the power factor will be less than if the P.D. wave were sym- metrical. The maximum value of the power factor in this case occurs when the peak of the P.D. wave is in the second quarter. The curves considered above have only one maximum ordinate during the half wave ; they might however have several maxima and minima ordinates. Suppose for example the P.D. and current were given by the curves in Fig. 46. In this case the power factor would be 075 and the phase difference 41 '4 degrees. It is fairly obvious that if the P.D. have a minimum value when the current has its maximum value, then the power factor will be low and the phase difference large even if there be no time lag between the current and the P.D. Volts Amperes Fig. 46. Phase difference 41-4. Power factor 0-75. We shall now consider what happens when the power factor vanishes. We see from equation (1) that we have Zero power factor. in this case r 156 ALTERNATING CURRENT THEORY [CH. Since in practice when the time is increased by \T the values e l and i become e\ and ^ respectively, we see that r2 rT I eidt = \ eidt, Jo JT /T r-2. eidt = 2 I eidt. j Jo Let us suppose that the curve e only cuts the axis at points which are at distances \T apart, and let us also make the same supposi- tion about i. When the time lag between e and i is zero, then ei is positive rT over the half period, and therefore the integral I eidt is positive. Jo When the time lag between e and i is \T, ei is negative over the half period and hence the integral is negative. When the time lag r is less than \ T, then from to T, ei is negative and from T to \ T, ei is positive. We see then that, as r increases from zero [ T to \T, I eidt gradually changes from positive to negative. J o [ T Hence eidt is a continuous function of T, and therefore for some Jo value of r the integral will vanish. It is therefore always possible to get a power factor of zero with voltage and current waves of any shape by giving to the time lag between them a proper value. In the particular case when both waves are symmetrical, the value of this time lag is a quarter of a period. This may be proved graphically. An analytical proof can be given as follows. Let f(t) and JPU + ^-j be the functions which represent the voltage and current waves respectively. Then, since they are symmetrical, alternating curves, both/() and F(t) satisfy (c), and thus Vl] ZERO POWER FACTOR 157 T putting *= T- x, this = - j"f (*-f ) F (x -%T)dx T -//<*)*()*< T 2 Thus This proves that, when both the current and voltage waves are symmetrical, the power factor vanishes when the time lag is a quarter of a period. We have seen (page 78) that the charging current of a condenser is r-de where e is the potential difference at the condenser terminals. In this case T dt \T rT C eidt = K Jo Jo [H = 0, since e has the same value after an interval equal to the period. d& We see also that when i is zero - - is zero, and therefore e has a dt maximum or a minimum value. Assuming that e has only one maximum value in a period, we see that the time lag between e and i equals the time e takes to increase from zero to its maximum value. Now e can have its maximum value at any time during the half period when it is positive. Therefore the time lag between e and i can have any value between and ^T. In all cases however the power factor is zero. This proves that we can infer nothing concerning the value of the time lag from a mere knowledge that the power factor is zero. 158 ALTERNATING CURRENT THEORY [CH. If W be the wattmeter reading in an alternating current watt E.M.F. and circuit, V and A the voltmeter and ammeter wattless E.M.F. readings respectively, and

........................ (6). This follows from the definitions (1) and (3) above. Now if we suppose that V is resolved into two components whose values are Fcos and Fsin respectively, then these components are called the watt E.M.F. and the wattless E.M.F. respectively. We see from (6) that Fcos $ or the watt E.M.F. multiplied by the effective value of the current gives us the true mean power expended in the circuit. If we have a simple inductive coil subjected to an alternating P.D., then therefore VA cos = RA 1 TT" T~* A A *-* J / W/t/ \ 7 and F 2 = ./t 2 ^. 2 + -~ f -y- a. ^ j o \*/ Thus the effective value of Ri is Fcos(, and the effective value di of Z j, is Fsin<. Hence we can suppose the applied P.D. split di up into two components Ri and L -=- , which have a phase diffe- rence of 90 degrees, and it is convenient to give names to the effective values of the two components. In the general case, however, when iron is present, we need to be careful when reasoning about the watt and wattless components of the E.M.F. as they do not seem to have much physical significance. The impedance of a circuit is the ratio of the applied effective voltage to the effective value of the current pro- duced. If V be the reading of a voltmeter placed Vl] IMPEDANCE 159 across the circuit and A the reading of an ammeter in the circuit, then where Z denotes the impedance. When direct current is used, Z is simply the resistance R of the circuit. With alternating currents, Z may be a very com- plicated function, as it depends on the shape and frequency of the applied potential difference wave, on eddy currents, the position of neighbouring circuits, capacity, inductance and magnetic per- meability. In the case of a simple coil whose self inductance is constant, we have (page 43), when the potential difference wave is sine shaped, Z- = R 2 + orL 2 . If the wave be not sine shaped we can write (page 80) Z- = R* + a.-arL-, where a has its minimum value unity when the applied wave is sine shaped. If V be the applied P.D., A the current and cos < the power Reactance, factor of a circuit, then is called the reactance A of the circuit. If there is no iron near the circuit and there are ( F 2 ) * no eddy currents, then the reactance equals \ ^ R 2 y or awL, and if in addition the applied P.D. be sine shaped, then the reactance is simply coL. The reactance may also be defined as the ratio of the wattless P.D. to the current. Instead of supposing that the E.M.F. is resolved into two watt current and components, we may suppose that the current is wattless current. SQ reso l ve d. I n fafe case A COS IS the Watt current and A sin is the wattless current. For an inductive coil e = Ri + L^. at 160 ALTERNATING CURRENT THEORY [CH. VI When e is of the form E sin cot, then R wL ^ = W^L* E sm mt ~ & + <*D E c s wt ' The first term on the right-hand side, being in phase with e, may be considered as the watt component of the current, and the other term may be considered as the wattless component of the current. In this case we have . RV RA* the watt current = A cos < = ^ 2 + ^2 = -pr-> coLV and the wattless current = A sin 6 = -^- 2 If e were the parabolic wave whose equation (6) is given above, then we can show that This proves that in general A cos < is a very complicated function of the quantities involved, and reasoning based on it has to be closely examined. REFERENCES. The Electrician, Vol. 44, p. 49, ' The Theory of the Power Factor.' 1899. The Electrical Review, Vol. 45, p. 744, 'The Current produced in an Inductive Coil by a Parabolic Wave of E. M. F.' 1899. CHAPTER VII. Argand's method of representing a complex variable. Vectors. Addition of vectors. Polygon law for compounding vectors. Multiplication of a vector by a complex number. Division of a vector by a complex number. Application to the theory of alternating currents. The currents in a divided circuit. Inductive coil in series with choking coil shunted by a non-inductive resistance. The apparent resistance and inductance of branched circuits. The currents in a branched circuit when mutual inductance is taken into account. Condition that the energy expended should have a stationary value. Important consequences in this case. Graphical solution. References. IT is convenient in Algebra and Trigonometry to introduce the conception of the square root of negative unity, and to make the convention that it must obey all the ordinary algebraical laws. An expression of the form x + y V 1 is called a complex number, and has many properties which make it a great aid in calculation. The expression x y*J 1 is said to be conjugate to x + y V 1, and V& 2 4- y' 2 , the square root of the product of the two, is called the modulus of either. The following three fundamental theorems are proved in text-books on algebra. 1. If a, b, c and d are real quantities and a + b\/- 1 = c + d\/ 1, then a = c and b = d. 1 a b o c + d V 1 _ ac + bd ad be / =- a + 6V ^1 ~ a 2 + b 2 a 2 + b 2 R. i. 11 162 ALTERNATING CURRENT THEORY [CH. If we agree that the abscissa represents the real part of the complex vari- Argand's method of representing a complex variable. able and the ordinate the imaginary part, then a line OP (Fig. 47) can be repre- sented by x + y*J 1. The length of OP, V# 2 + 2/ 2 , is the modulus of the complex vari- able and its inclination to the axis of x is tan" 1 - . If r and x o Fig. 47. OP is the graphical representa- be the polar coordinates of tion of the complex variable P, then and tan 6 = If OP rotate about the point P with uniform angular velocity vectors, ft) and if OX be the initial position of OP, then OM = T COS tot, PM = r sin cat, where t is the time in seconds since OP coincided with OX. The line OP is called a vector and may be represented by r cos cot + V 1 r sin cot. By trigonometry this expression may be written in the form re 2 i. If * is greater than p- -, then x is greater than a 2 > and a' ./l! -t^2 is greater than o. 2 , but less than a l . Hence the current ^ lags behind i, but the phase of the current i z is in advance of i. Again, since [e] = [pj [ij = [/> 2 ] [i' 2 ], we have and Thus and Lt = ' 2 {cos fa 4- a, - a') + V- 1 sin fa + a 2 a')]-. A R = --V,- 2 cos fa + a., - a'), ^w = -^7 2 sin fa + o 2 a 7 ), and where ^K is the equivalent resistance, L the equivalent inductance, and Z the equivalent impedance of the two branches in parallel. We see that i lags behind e by an angle a l + Oo a 7 . The above formulae for the effective resistance and inductance of the branched circuit may be written in the form Z' 2 (L.R, - LI ^y ' 2 j' where ^ y = {(^ + R^ + o> 2 (A + Hence, as the frequency increases, the apparent resistance ^o . . . increases from ^ ^ , its minimum value, to L. 2 ~R l ., , its maximum value, and the apparent self inductance diminishes from , its maximum value, to , 1 ' , its minimum value. It is also worth noting that when the time constants of the circuits are equal, R and L are independent of the frequency. 168 ALTERNATING CURRENT THEORY [CH. Inductive coil in series with chok- ing coil (0, Z,) shunted by a non- inductive resist- ance x. Suppose (Fig. 50) that we maintain the effective p. D. between A and B con- stant. Suppose also that the resistance of the inductive coil is r, and its self induct- ance I, then by the preceding formulae, AAAA/Wi (o,L) Pig. 50. Current is a maximum for a par- ticular value of x when the effective value of the applied p. D. is constant. where Z is the impedance of the circuit AB. Now by the differential calculus this is a maximum or a minimum when H(L+W) i.e. when x - afD = 0, ( ^ Since x must be positive, we have prefixed the positive sign to the radical. It is easy to see that this value of x makes Z a maximum, and therefore the current through the inductive coil a minimum. Hence shunting a choking coil with a non-inductive resistance sometimes increases the apparent resistance of a circuit, a result which has been noticed in practical work. Let The apparent re- sistance and inductance of branched circuits. L 2 )... be the resistances and self inductances of the n branches, then, if we neglect mutual in- ductance, 169 If i be the current in the main, then = O'i] + H + + FJ -i^- -1-. lw IW [p*\ 1 1 ) " 1 J w where ^ is the equivalent resistance and L the equivalent induct- ance of the n branches in parallel. Therefore J? ft And Thus Hence and L(o 1 !_ a 2 + P' a/ va b Va 2 We also see that the current in the main lags behind the applied P.D. by an angle 6 where Let (^j, Zj), (ft, Z 2 ) be the resistances and inductances of the two branches, and let M be their mutual branched circuit inductance, then with the usual notation when mutual in- ductance is taken e = (JR + L^)^ + MDlo,, . into account. _ _ '? .~ > (a). 170 ALTERNATING CURRENT THEORY [CH. Assuming that the functions are simple harmonic, we may write toV 1 for D, and solving the equations we find i, {R.R, + (M* - AZ.K + (L 2 R, + LiRJa V^l} = [R 2 + (L, - M) wV^T} e, it [R,R 2 4- (M 2 - LMco 2 + (L 2 R, + Aiy 6>V"^~i} = {^ + (L, - M)co\/^l}e. Hence since i the current in the main equals i\ + i a , we get e = a+W -I ac + bd ad be -nrV-lM a 2 + 6 2 a 2 + 6 2 where a = R l + R 2 , c = R 1 R 2 + (M 2 ^ L 2 ) &> 2 , b = (L l + L 2 - 2M) w, d = (L,R 2 Hence ac + bd (R, + R*f + a> 2 (L, - 2 M ) 2 Hence, as the frequency increases, the effective resistance increases from (L 2 - It is to be noted that in the above equations M may be either positive or negative, depending on how the coils are connected with the mains. When the frequency is zero L-L I -* /] If, for example, we measured the self inductance of the two coils in parallel by Maxwell's method, the self inductance L found would be the L given by this formula. A strict proof of this VII] HIGH FREQUENCY CURRENTS 171 particular formula can be given by other methods. As the frequency increases the effective inductance L diminishes, and it has its least possible value when the frequency is infinite. We have shown on page 24 that L^Lo A/ 2 is zero or positive. The case when it is zero need not be considered, as it is impossible in practice to arrange two circuits so as to satisfy this condition. In practice, L^ M* is positive, and therefore, since M is less than I^L^, Li + L^ ZM is positive. The equations (a) may be written in the form e = (R l + ADK + MDit = (R, + L,D)L + MDi,. If i denote the current in the main, we must have, at every instant, i equal to the sum of ^ and i 2 . Substituting i ^ for i. 2 in the above equation we get ( R, + L,D) i, + MD (i - i,) = (R, + L,D) (i - i,) + MDi, and thus (R l + R. 2 )i l + (L l + L,-2M)Di l = R,i-{-(L 2 -j\I)Di ............ (6). If i-L and i. 2 are harmonic functions we get , 2= , (R, + Rtf + o> 2 (A + L, - ' ' When the frequency is very high L,-M - '- ' We choose the signs in these formulae so that A^ and A 2 are positive. Suppose that M is less than L^ and Z 2 , then the positive signs must be chosen and we have AI + -4 2 = " Thus the phase difference between the vectors representing A 1 and A 2 is zero. 172 ALTERNATING CURRENT THEORY [CH. Now the average value of the electromagnetic energy stored in the field is s cos <. Since cos < is unity and A 2 equals A A l} this becomes \L,A? + \L, (A - AJ + MA, (A - A,). If we suppose that A is a constant, then this equation is a function of A! and attains its minimum value when L 2 -M Hence, with very high frequencies, we see that in this case the values of the currents are such that the average value of the energy stored in the field is a minimum. When M is greater than L 2 but less than L lt then Thus the phase difference < between the vectors of A l and A 2 is 180. Proceeding as before we find that the average value of the energy stored in the field is again a minimum. Therefore since M cannot be greater than L and L 2 we see that this theorem is true in all cases. The above theorems may also be deduced from the following more general theorem. In the particular case of high frequency currents the terms (jR x + R 2 ) ^ and R 2 i in equation (6) can be neglected in comparison with the other terms, we thus get (A + L 2 - 2M) Di, = (L 2 - M) Di and integrating (Zj + L 2 - 2M) i : = (L 2 -M)i + constant. The constant must vanish, for ^ and i are periodic functions of the time the mean values of which are zero, and thus (A + L 2 - 2M) i, = (L 2 - M) i and similarly (L-^ +L 2 2M) i 2 = (L 2 M) i. It is to be noted that in proving these equations we have made no assumption as to the shape of the applied wave of electromotive force. VII] ENERGY THEOREM 173 Now the value of the energy stored in the field, at the instant when the currents are i t and i. 2 respectively, is L^ i^ + i A i. Hence PR is co (L, - M) A, and PS is o> (L. 2 M)A Z . Since the angles at R and S are right angles, the 176 ALTERNATING CURRENT THEORY [CH. circle described on OP as diameter passes through R and S. Hence = OP 2 . To construct the diagram when the values of R 1} R 2 , L^L^ M y f, and F are given, we proceed as follows. Draw any line OR and, choosing a convenient scale, make its length equal to R t . Draw RL at right angles to OR and equal to coLi where co equals 27T/1 Make LP equal to coM so that HP equals co (L 1 M). Join OP and describe a circle on OP as diameter. Make the angle POS equal to the angle whose tangent co(L,-M) is p -- ^2 Produce SP to JV so that SN equals Through iV and Z draw lines NE and Z^ parallel to RL and respectively. Then OE will be the effective value of the applied potential difference which produces unit current in the branch (jRj , Z t ). Now the phase differences and the relative magnitudes of the various vectors are independent of the absolute value of the applied potential difference provided that the frequency does not alter. Hence if we choose the scale of the diagram so that OE represents F, then, on this scale, OR will represent jRj. A^ arid OS will represent R 2 A% in magnitude and phase, and thus, by dividing these lengths by R } and R. 2 respectively, we get A l and A z . If be the phase difference between A^ and A 2 and tana and tan fi be equal to L,-M L 9 -M a) -= and a) -J2 /L! ^ 2 respectively, we have tan (f> = tan (a /3) (c). VII] GRAPHICAL SOLUTION 177 We can see at once from equation (c) that when the frequency is very low the angle of phase difference between A l and A 2 is very small. When the frequency is very high, the phase difference is again very small. It is easy to show that it is a maximum when If we divide the lengths of the lines NE, LE and PE by coM we get the magnitudes of the currents A l} A z and A. If we draw EV (Fig. 51) perpendicular to OE, then the angles NE, LE and PE make with EV are the respective phase differences between the currents and the applied P.D. Fig. 52. Currents in a divided circuit. OE = applied p. D. ; OR = R 1 A 1 -, is greater than L l and less than L 2 . Fig. 52 gives the graphical construction when M is greater than L and less than L 2 . In this case A l and A 2 are in quadrature when -(A and the phase difference increases as the frequency increases. When the frequency is high, the currents are nearly in opposition in phase, and hence the current in the main is nearly equal to their difference. If L^ be approximately equal to L z then the current in R. i. 12 178 ALTERNATING CURRENT THEORY [CH. the main may be very much smaller than the branch currents. The inductive effects produce as it were a whirlpool of current in the two branches. Large circulating currents produced in this way are often useful in alternating current work. The diagram in Fig. 52 is drawn to scale for the case of two coils (100, 0'5), (100, 1) the mutual inductance being 0'6 henry and the frequency being 16, so that co is nearly 100. On the given scale, OR is equal to 100, RL equals 50 and RP equals 10. A circle is described on OP as diameter. The angle POS equals the angle whose tangent is 0*4 and SN equals f SP. The parallelogram PNEL is then completed and OE is joined. If we wish to find the currents and the phase differences when the applied potential difference is one thousand volts, we choose a new scale so that OE is 1000 volts. On this scale OR is found to be 625 and 0$ is 583. Thus A^ equals 6'25 amperes and A 2 equals 5'83 amperes. The angle of lag of A l is the angle EOR and it equals 38'5. The angle EOS equals the lag of A 2 and measured by a protractor it is found to equal 66 degrees. If the frequency of the applied potential difference be increased to 80, then it is easy to show graphically that A-L = 1*41 amperes, a x = 45, J. 2 = 2'83 amperes, a 2 =135. Thus the vectors A^ and A z are at right angles to one another and so A = \A* + A = 316 amperes. In Fig. 53 M is negative. It follows from (c) that is zero when o> is zero and again when CD is infinite. The angle $ has its maximum value when the frequency is determined by (d). The diagram in Fig. 53 is worked out for the case of two coils (53, 0'29), (117, 018) and M equal to 014. The frequency has been taken equal to 80 so that o> is nearly equal to 500. We find by measuring the lines and angles that A l = 9 '3 amperes, a x = 40, A z = 10'2 amperes, 2 18, A = 19'2 amperes, Impedance = 52 ohms. VII] NEGATIVE MUTUAL INDUCTANCE 179 It is to be noted that all the theorems and formulae in this Chapter, deduced by the method of the complex variable, are proved on the assump- tion of a sine wave of E. M. F. Moreover they only give the particular integrals of the differential equations in- volved, and hence it is best to regard them merely as giving a first approximate solution of the general problem. As a rule, it is better for engineers to solve problems graphically, as the graphical method is quite ac- curate enough and far more instructive than the mere algebraical manipulation of complex quantities in which the physical laws are not SO apparent. In the next Fig. 53. Currents in a divided circuit. Chapter we will consider = applied p. D. ; the graphical method and 08=ILJU its limitations. M is negative. REFERENCES. G. CHRYSTAL, Text Book on Algebra, Chapter xn, ' Complex Numbers.' S. L. LONEY, Plane Trigonometry -, Chapter xu, ' Geometrical Representation of Complex Quantities.' LORD RAYLEIGH, Phil. Mag., p. 379, ' On Forced Harmonic Oscillations of Various Periods.' May, 1886. J. J. THOMSON, Recent Researches in Electricity and Magnetism, Chapter vi. H. E. HARRISON, The Electrician, Vol. 31, p. 119, 'Effects Produced by using an Inductionless Shunt with a Choking Coil.' 1893. The Electrician, Vol. 33, p. 595, ' Alternating Currents in a Divided Circuit.' C. P. STEINMETZ, Alternating Current Phenomena. W. G. RHODES, An Elementary Treatise on Alternating Currents. 122 CHAPTER VIII. Parallelogram of vectors. Extension of theorem. Condition that three vectors lie in a plane. Vector of a constant quantity. Vectors in space. Resultant of three vectors. Condition that four vectors can be represented graphically. Failure of graphical methods. Numerical example. References. WHEN we are dealing with harmonically varying quantities, we Parallelogram of can represent them graphically by lines drawn in vectors. a plane. For example, if we have two E.M.F.'S E l sin27rft and E 2 siu(27rft + a.) we see that their values are the projections of two lines OP l and OP 2 , which are inclined to one another at an angle a and rotate round 'f times per second, upon a fixed line OA drawn at right angles to the initial position of OP^ The resultant got by adding the two E.M.F.'s together, ^ sin 2?i/ + .#2 sin (27T/J5 + a), is easily shown to be the projection upon OA of the diagonal OR of the parallelogram constructed with OPj and OP 2 for adjacent sides. Since in practice we are generally only concerned with the R.M.S. values of the E.M.F.'S and their phase differences, and since the R.M.s. value of a simple harmonic quantity bears a constant ratio to its maximum value, it is usual to represent the E.M.F.'S by lines OP 15 OP 2 and OR equal to their R.M.S. values and inclined to one another at angles equal to their phase differences. These lines are called vectors and the above theorem is the parallelogram of vectors. It can obviously be generalised into the polygon of vectors. In alternating current work we seldom have to do with Extension of h arm m c waves of P. D. and current, but still graphi- theorem. ca j methods are found to be convenient in many cases. We have therefore to investigate the limits of accuracy of [. VJIl] VECTORS IN ONE PLANE 181 }hese methods. Let us first consider the case of two periodic quantities, for example alternating electromotive forces. Let e and e. 2 be the instantaneous values of the two E.M.F.'S and let e be their resultant, theii Now e-L and e. 2 may be dissimilar curves and e may be dissimilar to both of them, hence their R.M.s. values do not bear a constant ratio to their maximum values. In all cases however we have and thus V- = F x 2 + F 2 2 + 2 V l F 2 cos a x . where [ I COS gl . 2 = , - rr ............... (1). Hence the parallelogram construction will still hold good for compounding lines representing the effective values of two periodic quantities if we define phase difference by equation (1). This definition we have already given in Chapter VI. It is customary to call the lines representing V lt V. 2 and V vectors. The two vectors and their resultant are lines in one plane and the angles between the various vectors will give their phase differences in accordance with definition (1). In general if a linear relation hold between the instantaneous condition that values of three periodic alternating functions e l} in e ' 2 anc ^ &3 ' ea k f wn i c h nas t ne same period, then their R.M.S. values can be represented by lines drawn in one plane, the angles between the lines being the phase differences. For example, suppose that at every instant le^ + me 2 + ne 3 = 0, where I, m and n are constants and e l} e 2 and e 3 are periodic functions of the time, then thus ^Fj 2 = m 2 F 2 2 + ri*V 3 z + 2mnF 2 F 3 cos 0^3. 182 ALTERNATING CURRENT THEORY [CH. Hence IV, is equal in magnitude to the resultant got by com- pounding two lines inclined to one another at an angle a 2 . 3 and the lengths of which are mF 2 and nV s respectively, by the parallelo- gram construction. The line representing IV, must however be drawn in the opposite direction to their resultant, since le, = (me 2 + ne s ). We also have i. 2 + 2.s+ a s.i= 2?r and, just as in statics, IV, rnV^ jiF^ sin L 2 sin 2 . 3 sin 3 . x ' These lines IV,, mF 2 and nV s are called vectors in engineering practice, but the instantaneous values of e,, e* and e 3 can no longer be represented by the projections of lines rotating with constant angular velocity. It follows from the definition of phase difference given in (1) vector of aeon- that the cosine of the phase difference between a stant quantity. constant and a periodic quantity is always zero, and hence the phase difference is ninety degrees. Hence the R.M.s. value of the sum of an alternating and direct current, for example, is represented in magnitude and phase by the diagonal of the rectangle constructed with the R.M.s. values of the direct and alternating components as adjacent sides. If how- ever we have two alternating current components which are not in the same phase, then the line representing the direct current component must be drawn at right angles to the plane containing the two lines representing the alternating current components so as to be at right angles to the three lines representing the two alternating components and their resultant. We must therefore, if we are going to use graphical methods in this case, have recourse to solid geometry. In general, when we have three periodic functions and there is no linear relation connecting" them, we can repre- Vectors in space. . f sent their R.M.s. values graphically by three lines drawn in space, the angles between the lines being the phase differences as determined by equation (1). In order to prove this VECTORS IN SPACE 183 we have to show that a, /3 and 7, the angles of phase difference, can always form a solid angle. We have to prove therefore that a -f ft + 7 can never be greater than 4?r, and also that no two of the angles can be greater than the third. Let x, y and z be the three periodic functions and let a be the phase difference between y and z, /3 between z and x and 7 between x and y respectively. From the definition of phase difference we have cos a rT yzdt Jo * Divide the period T into a large number (n) of equal intervals, and let x l} x Z) ... x n \ y lt i/ 2 , ... y n ; 1} z%, ... n be the values of the functions at the eod of successive intervals. Then from the meaning of integration we have, when n is infinitely large, = with corresponding values for cos 2 /3 and cos 2 7. Then 1 cos 2 a cos 2 /3 cos 2 7 + 2 cos a cos /3 cos 7 XYZ-XA*- Y&-Z& + 2ABC XYZ Now X(XYZ-XA*- YB*- = (XY- &) (XZ - B*) - (XA - CB)\ Also XY -C* = (x.y, - x. 2 y^f + (^y, - x t ytf + ........... (6), XZ-R- = (x^ - x^ztf + (x^z, - x,z^ + ............ (c), XA -CB = (x,y 2 - x*yi) (x,z 2 - x^) + (x,y 3 - x^) (x^ - x^) + . . .(d). Now it is easy to show that 184 ALTERNATING CURRENT THEORY [CH. is not less than (Pp + Qq + ...) 2 . It therefore follows from (b), (c) and (d) that (XY- C 2 ) (XZ- B 2 ) - (XA - CB)* is never negative. Hence from (a) 1 cos 2 a cos 2 fi cos 2 7 + 2 cos a cos fi cos 7 is never negative. Now the trigonometrical expression can be written in the form {cos 7 cos (a + /3)} {cos (a /3) cos 7). Hence we see that if it vanishes, the sum of two of the angles equals the third or the sum of the three equals four right angles. The three vectors in this case are therefore in one plane. It may also be written in the form . a + yS+7 . ft + ry-oc. . 7 + a-/3 . a + /3 - 7 4sm--|-^sm^-^ sm Z _____ ^ sm _ _|_ I _(<>). By definition, a, /3 and 7 lie between deg. and 180 deg. As the expression (e) is positive, all the terms may be positive, or two of them may be negative, or all four of them may be negative. If they are all positive, then a -f ft + 7 is less than four right angles, and also any two of the angles are together greater than the third. Suppose, now, that the first two terms of (e) are negative. Then a + ft + 7 is greater than 2w, and a yS 7 is greater than zero. Therefore a is greater than TT, which is contrary to the definition. Similarly, no other two terms can be negative and a fortiori the whole four cannot be negative. In all cases, therefore, we see that the sum of the three angles is not greater than four right angles, and that the sum of two of them is never greater than the third. Hence the three angles can always form a solid angle. We shall apply the above theorem to find graphically the Resultant of effective value of the sum of three alternating three vectors, periodic functions. Let e 2 , e s and e 4 be the three functions, and let e l be their resultant, then and 6i 2 = BI e. 2 + #1 #3 + ^i #4 > thus F! S = F 1 F 2 cosa 1 . 2 + P_Ficosai. s + F 1 F 4 cosa 1>4 . VIII] RESULTANT OF THREE VECTORS 185 IS Hence F x = F, cos a 1-2 + F 3 cos a 1>3 + F 4 cos a 1-4 similarly F x cos a x . 2 = F, + F 3 cos o^. 3 + F 4 cos c^. 4 Fj cos i . 3 = F 2 cos 2 . 3 + F 3 + F 4 cos a 3 . 4 F! cos ati.4 = Fo cos a 2-4 + F 3 cos a 3 . 4 -4- F 4 And F X 2 =F 2 2 +F 3 2 + + 2 F 2 F 3 cos a 2 . 3 In these equations V n is the effective value of e n and the phase difference between e n and e m . Construct the solid angle at (see Fig. 54), so that the angles POQ, QOR and ROP equal 02.3, s.4 and 4 . 2 respectively. Complete the parallelepiped OPQRS and let OS be the diagonal. Then from (b) we see that OS equals V l . Hence from the equations (a) we see that the angles POS, QOS and ROS equal i. 2 , cti.s and a^ 4 respectively. We can thus extend graphical methods to solid geometry for the case of three periodic quantities. V, Fig. 54. OS is the resultant of OR, OP and OQ. In this case we can show from equations (a) that F, F 2 _ __ sin (02. 3 , 02. 4 , 03.4) sin(a 1 . 8 , 0^.4, a 3 . 4 ) sin(a 1 . 2 , tti.4, 2 . 4 ) 8^(^.2, ctj.3, a 2 . 3 ) where sin (a, ^8, 7) = {1 cos 2 a cos 2 /3 cos 2 7 + 2 cos a cos /& cos In general when four periodic quantities of the same frequency are connected by a linear relation, le 4- me 2 + ne s + pe 4 = Q: then their R.M.S. values can be represented in magnitude and phase by lines drawn from a point in space. Condition that four vectors can be represented graphically. 186 ALTERNATING CURRENT THEORY [CH. In statics these lines would represent a system of four forces in equilibrium, and we shall see later on that many statical theorems have their counterpart in electrical theory. If no linear relation connects the four periodic quantities then, Failure of graphi- as & rule, we cannot represent them graphically by lines drawn in space. For example, suppose that one of them is constant, then it would have to be represented by a line drawn at right angles to the other three, which is impossi- ble when they form a solid angle. If however we replace two of the vectors by their resultant, we can represent this resultant and the other two vectors graphically. 1-6 Fig. 55. Resonance of currents. Numerical example. A variable condenser (Fig. 55) was used to shunt a choking coil on a 2000 volt circuit ; the power factor of the choking coil was 0*041 and of the condenser 0*124. The current in the choking coil was 6 amperes, and the current in the main was 1*6 amperes when the condenser was adjusted so that the current in its circuit was 6 amperes. From these data let us find the minimum possible power factor of the shunted choking coil. Let 0V (Fig. 56) represent the P.D. across the terminals of the choking coil and the condenser. Now if a and /3 be the phase differences between the choking coil current and VIII] NUMERICAL EXAMPLE 187 le P.D. and between the condenser current and the P.D., then (Chapter vi) cos a = 0-041, cos /3 = 0-124. Thus a is 87 39' and j3 is 82 53'. In Fig. 56 OF represents the ap- plied voltage, 00 the choking coil current and OK the condenser current. The angle FOC' is a and the angle VOK is /3. Now it every instant Fig. 56. where i is the current in the main and i l} ^ are the currents in the branches. Hence a linear relation connects the three currents and the three current vectors will be in one plane. Let 7 be the phase difference between ^ and i. 2) then, since their effective values are each equal to 6, therefore 7 = 164 26'. Since a + /3 equals 170 32' it follows that OF, 00 and OK are not in one plane. But OR, which represents the current in the main, is always in the plane COK. The diagram (Fig. 56) shows us exactly what happens when we increase or diminish OK, since R always lies on CR which is a line drawn parallel to OK. We suppose that the phase differences between the applied potential difference, the condenser current and the choking coil current remain constant for all values of the condenser current. Hence the angles forming the solid angle at remain constant. The power factor of the combined circuit is cos < where c/> is the angle FOE. Now the minimum value of c/> is got when the planes VOR and COK are at right angles to one another. In this case, by the formulae of spherical trigono- metry, we get sin w sin 7 where sin &> = 2 Vsin a sin (cr a) sin (cr ft) sin (cr 7) 188 ALTERNATING CURRENT THEORY [CH. VIII and a is half the sum of the angles a, ft and 7. Substituting the above values of a, /3 and 7 in this equation, we find that < is 52 24' and therefore the maximum possible value of the power factor is 0'61. We have assumed that the shape of the wave of the applied P.D. does not alter as the capacity of the condenser is altered. An alteration in its shape would alter the power factors and phase differences, and hence the angles a, ft and 7 would vary with the shape of the wave. The diagram (Fig. 56) shows why it is in general impossible to make the power factor of a choking coil unity by shunting it with a condenser. If it were unity the angle VOR would be zero, and the vectors representing the two currents and the applied P.D. would lie in one plane. REFERENCES. W. E. SUMPNER, Proceedings of the Royal Society, Vol. 61, p. 465, ' The Vector Properties of Alternating Currents and other Periodic Quantities.' 1897. The Electrician, Vol. 44, p. 49. 1897. J. of the I. E. E., Vol. 31, p. 440, ' The Limitations of Graphical Methods in Electrical Theory.' 1901. W. M. MORDEY, J. of the I. E. E., Vol. 30, p. 364, ' Capacity in Alternate Current Working.' 1901. CHAPTER IX. The measurement of power. The quadrant electrometer. The electrostatic wattmeter. Electrostatic wattmeter shunted. The electromagnetic wattmeter. Electromagnetic wattmeter, with mutual inductance. Watt- hour meters. Eeisz's method of power measurement. The three volt- meter method. The three ammeter method. Transformer methods. Resonance methods. References. WE have seen in Chapter vi that if W be the mean power in The measurement watts expended in an alternating current circuit, of power. Y and A the effective volts and amperes, and cos a the power factor, then W= VAcosa. The maximum possible value of the power is VA, and it has this value only when the phase difference a is zero, and this can only occur when the ratio of the instantaneous volts to the instantaneous amperes 1-1 is always constant (Chapter vi). If, as is generally the case, a is not zero, then in order to find W we should need to know the shapes of the volt and ampere curves and their time lag relatively to one another. This could be done by means of an oscillograph, as it is not difficult to find the mean value of ei, that is W, from the curves. In practice, however, this is best done by means of some form of wattmeter, or by some of the methods described below. Before describing the electrostatic wattmeter, we will give the theory of the quadrant electrometer as the principle of the two instruments is the same. 190 ALTERNATING CURRENT THEORY [CH. In the quadrant electrometer, invented by Lord Kelvin, use is The quadrant made of electrostatic attractions and repulsions in electrometer. order to measure potential differences. In the ordinary form of this instrument there is an aluminium disc shaped like the figure 8 placed inside an insulated cylindrical metal box which is completely divided into four quadrants (Fig. 57). This flat disc 3 is suspended by a torsion fibre per- pendicular to its plane, and its position of equilibrium when the quadrants are at the same potential is shown in Fig. 57. Fig. 57. The Kelvin Quadrant Electrometer. The opposite quadrants 1 and 1' are permanently connected by wires, so that they are always at the same potential, and so also are the quadrants 2 and 2'. Let the potential of the quadrants 1 and V be V lt of the quadrants 2 and 2' be F 2 and of the disc be F 3 . There will evidently be repulsion between 3 and 1 and attraction between 3 and 2, hence there will be a torque in the direction of the arrowhead. Now if Q lt Q 2 , Q 3 and Q be respectively the quantities IX] THE QUADRANT ELECTROMETER 191 of electricity on the two pairs of quadrants, on the disc and on the inside of the earthed screen, which in the case of an electro- static wattmeter is a metal cover practically enclosing the quadrants, then ft + Q, + & + Q,=o ..................... (i). By Chapter iv, we have, since F is zero Ql = Kl.lVl + -^1.3^*2 Now (1) must be satisfied for all values of V lt F 2 and F 3 . Therefore K^ + K^ + K^-}- R tml = ............... (2)j 1* and K ., + K,., + K,., + K,., = Q ............... (3) Now let the disc turn through an angle 6, and let K' lfl ,K' lm , ... be the new values of the coefficients. These coefficients must still satisfy equations (2) and (3). Now, owing to the shape of the disc, we see, since the gaps between the quadrants are very narrow, that so long as the edges of 1 and 3 or 1' and 3' are not brought too close together K' z , 3 = K 3 . 3 = constant, for the motion merely brings a different part of 3 opposite the gaps between 1 and 2 and between 1' and 2'. Also since the motion of the disc does not appreciably alter the coefficients of mutual induction for electrostatic charges between 1 and 2, and 1, and and 2, we have K'l.2 = -^1.2. K'o.l ^0.1 > -^'o.2= ^o.2' Hence from (2) = constant = *Ti.i + *i.. ..................... (4). Similarly A" 2 . a + K'^ = K 2 ., + K 2 . 3 ..................... (5). As 6 increases, the parts of the quadrants 1 and I' opposed to the disc 3 diminish uniformly, and hence we may write ^'1.1 = #1.1 -X0.. ....... (6) 192 ALTERNATING CURRENT THEORY [CH. where X is a constant. For the same reason K'^ = K^ + \0 ........................ (7). From (4) and (6) K'^ z = K l ,.+ \0, and from (5) and (7) " 2 . 8 = K 2 , 3 - \6. The electrical energy W of the system in the new position is given by (see p. 90) W = %K\ tl V? + iJT 2 . 2 F a a + JJr,. 8 F 8 a + K'^V.V, + K'.^V,V, + K\.,V,V 2 . Now the work done by the electrical forces on the disc when it turns through an angle d6 = The moment of the forces about its axis x dO, t>+ t ftre kepr constant ,f tie .gain et**f 'tKe *l?t\f\t*V >< *<' ^'i^^re^^Tf = Torque x cZ(9. m Hence, Torque = If therefore Fj, T 7 ^ and T 7 ^ are kept constant, we see that the torque equals 1 TT^a- 1 - 1 i i F 2 2 - 2 _j_ F F 2 - 3 -i- F F 2 Kl ~~ f 2 Ka T" K ^ 3 " KsKl = X(F 1 -F 2 ){F 3 -J(F 1 +F 2 )}. If the needle be put in metallic connection with the quadrants, 1 and 1', then F 3 equals Fj and the formula becomes Torque = | (F,-F 2 ) 2 . Used in this manner the electrometer becomes a voltmeter. Equilibrium is attained when the torsional couple is equal to the electrical couple. With direct currents the reading is propor- tional to the square of the P.D. between its terminals, and with alternating currents it is proportional to the mean value of the square of the P.D. The most satisfactory way of reading the deflection is by means of a ray of light reflected on to a scale from a light mirror fixed to the axis of the disc. If the scale is direct reading, then the divisions on the upper part of the scale will be much larger than those lower down as the de- flections increase as the square of the applied voltage. IX] ELECTROSTATIC WATTMETER 193 This instrument, shown diagrammatically in Fig. 58, is practically a slight modification of the quadrant electrometer described above. In the figure, Q l represents one pair of the quadrants, Q. 2 another pair, and N the needle or disc. The disc is generally The electrostatic wattmeter. A' B' Fig. 58. Connections of the Electrostatic Wattmeter. connected to an insulated terminal on the case of the instrument by means of a phosphor bronze suspension strip which replaces the torsion fibre in the electrometer. Suppose that we wish to measure the electric power being expended in the coil AB. Place a small resistance R (DC) in the main BB r and connect the disc N of the instrument to A. Let Vi be the instantaneous value of the potential of A, v 2 the potential of B, and v s the potential of D. Then, by what we have shown above, the torque g acting on the disc will be given by - v 3 ) fa - J (v 2 + v 3 )}, where X is a constant depending on the instrument. If DC be non-inductive, then v. 2 v 3 = Ri, where i is the instantaneous value of the current. Hence g = XRi fa -%(v, + t;,)}. Now, if E be the middle point of DC, its potential is ^ (v 2 + v 3 ) R. I. 13 194 ALTERNATING CURRENT THEORY [CH. and v l ^(v 2 -{-v 3 ) is the P - D - between A and E. Therefore, if G be the mean value of the torque g, G = \R x watts expended in AE where W is the true value of the watts expended in AB and A is the effective value of the current. Now, if 6 be the deflection of the disc, we can write G = \kO, where & is a constant, hence therefore W = ~ -\A*R ........................ (1). This equation will give us the true value of the watts. If we replace R by another resistance R f , we have Thus we can increase the range of the loads that the wattmeter can measure by making a suitable set of resistances to put in series with the mains. In practice it is not desirable that the potential drop at full load across the series resistance should be more than one per cent, of the potential difference between the mains. The constant k can be found by using the instrument to measure the power expended in a non-inductive circuit. For example, if V be the effective voltage across AB and 6 l be the deflection in this case, then In practice it is not convenient to have a P.D. of more than about 200 volts between the disc and the quadrants. Hence for measuring power in high tension circuits the instrument has to be used with a shunt. IX] WATTMETER WITH SHUNT 195 The connections for this case are shown in Fig. 59. A'E' A' A vww Fig. 59. Electrostatic Wattmeter with shunt, represents a non-inductive shunt which is put across Electrostatic r wattmeter the mains and the disc is connected to a point L in the shunt. Let the resistance of A' L be R^ and of LB' be R 2 . Let v x and v/ be the potentials of 2!' and Z respectively, then since the shunt is non-inductive, v - v - v thus where As formerly, Thus and W + therefore = the multiplying power of the shunt. * fa v 3 %(v. 2 v 3 )} \Ri L ( Vl - v s ) - g (v s - = ( F + RA*) - RA n - (2)- 132 196 ALTERNATING CURRENT THEORY [CH. In this formula for W, k has the same value as when the wattmeter is unshunted. An interesting case arises when we use the wattmeter with a shunt multiplier of 2. The formula then becomes (3) and no correction has to be applied for the series resistance. The instrument used in this way is a very accurate one for measuring power. Even if the shunt be inductive, formula (3) is correct provided that the two divisions of the shunt are symmetrical. In practice, when using a shunt for which N is greater than 2, it is possible to get a zero or even a negative deflection of the instrument with low power factors. Mr Addenbrooke, who first called the author's attention to this interesting fact, showed him that an electrostatic wattmeter with a shunt for which N is equal to 10 gave a large negative deflection when measuring the power absorbed by a condenser load. It is easy to see that when W is N 2 less than RA 2 we must get a negative reading on the in- strument. When the deflection is zero, then (4). Hence the instrument could be used to measure power by means of a null method. R z could easily be made variable so that it could be adjusted until the deflection were zero. Then ~D \ ~p> since N = --. 2 , W could be found from (4) and the reading of an ammeter in the circuit. In the general form of electromagnetic wattmeter we have two The eiectromag- coils, of which one is fixed and carries the main netic wattmeter, current while the other, which is movable and in series with a high nearly non-inductive resistance, is placed as a shunt across the circuit. The coils are placed with their axes at right angles, so that when currents are passing through both there IX] ELECTROMAGNETIC WATTMETER 197 is a force tending to turn them so as to make their axes coincide. The torque on the movable coil is proportional to the product of the two currents so long as that coil is kept in the same position. The movable coil is brought back to its initial position by means of a torsion head connected to it by a spiral spring so that the angle turned round by the head to which the pointer is attached is proportional to the torque acting on the coil. If g be the torque, we may write g = \H 1} where i and ^ are the currents in the series and shunt coils respectively. With direct currents, if 6 be the angle turned through by the torsion head, then g equals k'\0, where k' is a constant, hence Mi Fig. 60. Electromagnetic Wattmeter. Now if 8 be the total resistance of the shunt circuit (Fig. 60) and e the P.D. applied at its terminals, k'\0 \i . o Therefore and thus W=k0 (1) where k equals Sk r and is constant. The watts then will be proportional to the reading of the instrument, and, since in this 198 ALTERNATING CURRENT THEORY [CH. case they are equal to the product of the volts and the amperes, k can easily be determined. With alternating currents we still have 9 = ^w'i > where g, i and ^ are the instantaneous values of the torque and amperes respectively. Therefore the mean value of g (G) is given by G X x the mean value of ^'^ 1 = \AA l cosa ................................. (2) where a is the phase difference between the periodic functions i and ^ and A and A^ are their R.M.S. values. Now if there is no iron in the shunt coil, no mutual inductance between the coils, and the eddy currents in the instrument itself are negligible, we have SAi>= VA.cosj, and thus 8A l = Fcos7 ........................... (3) where V is the effective value of the applied P.D. and 7 is the angle of phase difference between V and A lt Also if W be the true power in the load, together with the power RA 2 expended in heating the series coil the resistance of which is R, W=VAcos/3 ........................... (4) where /3 is the phase difference between Fand A. Substituting from (3) and (4) in (2), we get X COB a cosy S cos/3 Now G equals \k'6, where 6 is the mean value of the deflection. With the frequencies used in practice, 6 is constant, hence finally W=k6x-^- ...(5). cos a cos 7 Now the wattmeter is calibrated with direct currents, and therefore indicates kd watts. We see that, when the wattmeter is used on an alternating current circuit, the readings of the instrument must be multiplied by cos 3 cos a cos 7 IX] CORRECTING FACTOR 199 In Fig. 61, where V, A and A l are represented graphically, it must be remembered that we cannot assume that their vectors are in one plane. In general therefore ft will be less than a 4- 7. If there were no phase difference between the applied P.D. and the current in the shunt coil, then 7 would be zero, and ft would be equal to a. In this case the instrument would indi- cate alternating current power correctly. In practice however, 7, although small, cannot be neglected, as it may introduce a large error when measuring the electric power in circuits with a low power factor. Let a = ft 7 -f x, then from the geometry of Fig. 61 we see that ft + 7 cannot be less than a, and therefore the maximum possible value of x is 7. Hence if 7 is small x also is small. Therefore, if these angles be expressed in circular measure, we may write cos 7 = cos (7 x) = 1 , sin 7 = 7 and sin (7 x) = 7 x very approximately. The correcting factor becomes cos ft cos ft Fig. 61. Currents in shunt and series coils of wattmeter. Vectors are in dif- ferent planes. cos a cos 7 cos (ft - (7 - x)} cos ft .(6). cos ft + (7 x) sin ft For very low power factors, when ft is nearly ninety degrees, (6) may have almost any value, as it depends chiefly on the value of 7 x. If the potential difference and current waves are sine shaped, V, OA and OA l will be in the same plane, and if the inductance of the shunt coil be greater than its capacity, as it generally is in practice, OA and OA l will lie on the same side of 0V for inductive loads and x will be zero. If the waves are 200 ALTERNATING CURRENT THEORY [CH. approximately sine shaped, the vectors will be nearly in one plane aad x will be small. It will be seen from (6) that if 7 is appreci- able, then in this case the wattmeter generally reads too high in alternating current measurements. The electromagnetic wattmeter as ordinarily constructed cannot be used when the power factor of the circuit is very low. On a condenser load the phase difference a between A and A t is large. With sine waves, OA and OA : would lie in one plane and be on opposite sides of OF. In this case we may therefore write a = /3 + 7 - x, and proceeding as before we find that the correcting factor is approximately ^l a} cos ft (7 - x) sin {3 When 7 - x = cot ft we see that the wattmeter reads zero, and when 7 x is greater than cot j3 we get a negative deflection. Now when cos/3 is very small so also is cot/9. The wattmeter therefore does not give trustworthy readings on a condenser load when the power factor is low. The errors of this instrument are as a rule greater the higher the frequency, as 7 is practically proportional to the frequency. In the theory of the electromagnetic wattmeter discussed Electromagnetic above we have assumed that the shunt and series mTtui e ?ndt h coils are so situated that the mutual inductance ance - between them is zero. We shall now consider the case when the mutual inductance can not be neglected. Let the constants of the shunt coil be (S, Zj), of the series coil (R, L) and let M be the mutual inductance between the coils. Then if e and v are the instantaneous values of the potential differences across the shunt coil and across the load respectively, our equations are .- + . + /;* + ** (8 ) e = Si 1 + L + M (9). IX] WATT-HOUR METER 201 Multiplying both sides of equations (8) and (9) by i and taking mean values we get T and VA cos /3 = SAAj. cos a + -^ i-j-dt J. j o at Thus eliminating the integral we have When M is zero this reduces to the formula (4) which we have discussed above. In the particular case when M equals Z x the formula (10) becomes W = k6, and therefore, when the self inductance of the shunt coil equals the mutual inductance between the coils, the instrument reads correctly. The above theorem suggests the following method of con- structing an electromagnetic wattmeter. Suspend the shunt coil so that its self inductance equals the mutual inductance between the coils, and adjust the controlling spring till the pointer reads zero in this position. When it is used to measure power, and currents are passing through both coils, the shunt coil is brought back to its initial position by means of a torsion head and the pointer reads the power. The scale is evenly divided, and one reading with direct current is sufficient to find the constant of the instrument. In order that M may be equal to L^ it is necessary that the self inductance of the series coil should be greater than the self inductance of the shunt coil. Various kinds of watt-hour meters are used in practice to watt-hour measure the total energy expended in a given time in meters. an installation. If the meter be theoretically correct, the number of revolutions of the spindle in a given time must 202 ALTERNATING CURRENT THEORY [CH. be proportional to the watt-seconds consumed. If r be the rate of the meter, then Revolutions Watt-Seconds n = Wt' Hence W= U . rt Therefore if r be known W can be found. Most watt-hour meters cannot, however, be trusted to read accurately when the power factor of the circuit is low unless very special precautions are taken in their manufacture. Consider for example the well-known Elihu Thomson watt-hour meter. In this meter the shunt circuit consists of a drum wound armature, the axis of which is the spindle. The armature is in series with a high nearly non-inductive resistance. This is placed in the field due to the series coils, and rotates for the same reason that the armature of a motor rotates. Now, if A 1 be the current in the shunt circuit, then, since the field due to the series coils is pro- portional to the current in them, that is to the main current A, we see that the mean value of the driving torque is proportional to AA l cosoi. The retarding torque when the spindle rotates is due to the eddy currents induced by permanent magnets in a disc of copper fixed to the spindle. Owing to the induct-' ance of the circuits in which the eddy currents flow, they do not come instantaneously into existence, neither do they die away instantaneously. By Lenz's law they retard the motion of the disc and, since the field due to the permanent magnets is constant, their values are proportional to the angular velocity (2irn) of the spindle. Therefore, neglecting friction, the retarding torque is proportional to n, and the driving torque to A A l cos a. Proceeding as in the case of the electromagnetic wattmeter, and assuming that the mutual inductance between the shunt and series circuits is negligible, we find that w= n cos/3 rt cos a cos 7 where r is the rate found by direct currents. Hence we see that IX] REISZ'S METHOD 203 . unless 7 is zero, i.e. unless the shunt circuit is absolutely non- inductive, the reading on an inductive load will in general be too high. Also on a condenser load if 7 x is greater than cot/3, the meter will run backwards. In actual meters a few extra windings of the shunt circuit are arranged so as to help the field due to the main current and thus enable the meter to start on a small load. It is found that the friction of the spindle in its bearings is very nearly constant, and, if the extra windings be arranged to compensate for this, then AA l cosa will be very approximately proportional to n. For this method we require an electrostatic voltmeter, an ammeter and two high non-inductive resistances of known values R^ and R 2 . Let BAG (Fig. 62) be the circuit in which we wish to measure the power. Reisz's method of power measurement. Fig. 62. Eeisz's method. Place an ammeter A in the main circuit, and place R^ as a shunt across the circuit BAG. Then if ^ be the main current, e the P.D. across BC and i the current in BAG, we have Hence e 2 Sf where w = ei = the instantaneous watts expended in BAG. 204 ALTERNATING CURRENT THEORY [CH. F 2 '2W By summation A^ = A 2 + ^ -f ^ Similarly 4 a * = 4 2 + - + jtt 2 _fi 2 Therefore JF = 6 (A? - Af) - a V* ........................ (1) where a= Ki = If R. 2 is infinite, then (1) becomes W-%<4f--AJ)- ..................... (2). If the power factor were zero, W would be zero, and we see from (1) that b(A l 2 A 2 2 ) would then have its minimum value aF 2 . For low power factors, therefore. W is the difference of two large quantities and a very small error in the ammeter or voltmeter readings may introduce a large error in the value of W calculated from (1). It is also to be noted that we make the assumption that the shape of the wave of the applied potential difference is not altered by shunting the circuit by the high non-inductive resistances. In this method a non-inductive resistance R (Fig. 63) is put The three volt- m series with the load EC. Let e, e^ and e 2 be the meter method, instantaneous values of the P.D.'s across AC, AB and BC respectively, then if i be the current in the circuit, we shall have at every instant e = ei + e. 2 . Therefore e 2 = ei* + e? + 2e l e 2 Fig. 63. Three Voltmeter method. IX] THKEE VOLTMETER METHOD 205 If, therefore, W be the average value of the power ie expended in BC we have by summation and thus W = ~ ( F 2 - F, 2 - F 2 2 ) (i) where A is the effective value of the current measured by an am- meter and V, Fj and F 2 are the readings of voltmeters placed across AC, A B and BC respectively. If the supply voltage is sufficiently steady, one voltmeter can be used to read F, V 1 and F 2 . Suppose that three voltmeters are used and that + a, b, c are the percentage errors of the readings F, Fj and F 2 and that the ammeter reads correctly, then it can be shown that the maximum possible error in the value of W given by (1) is a minimum when F-F A/ 1 V a + b' and in this case the maximum percentage error in W is where cos < is the power factor of the circuit. Suppose that the three voltmeters are all equally trustworthy, then the maximum percentage error is 4a 3o H 7 . COS (f) If the voltmeters could be trusted to read correctly to within 0'2 per cent., then for a power factor of unity the maximum error would be 1'4 per cent, and for a power factor of 0*01 it would be 80' 6 per cent. Owing to the unavoidable errors of observation, with ordinary commercial instruments and on circuits where the voltage is unsteady, formula (1) often gives negative values for W when cos < is small. The practical objection to this method is that the testing pressure applied between A and C has to be nearly double the working pressure applied between B and G, and this increased 206 ALTERNATING CURRENT THEORY [CH. pressure is not always available. Another difficulty is that the shape of the wave of the applied potential difference is generally altered when we put a non-inductive resistance in series with the load. The first of these difficulties can be met by using the three ammeter method. In this method a non-inductive resistance R (Fig. 64) is placed The three ammeter as a shunt across the load, and ammeters are placed in the branch circuits and in the main. If i be the instantaneous value of the current in the main, then Therefore i 2 = i^ + i 2 2 + 2{ 1 t* 2 = i 2 I ^2 I _ g^' where e is the applied P.D. Hence by summation A* = A* + As + ^ ^ M , Fig. 64. Three Ammeter method. IX] TRANSFORMER METHODS 207 Thus W=^(A*-A 1 *-Af) = (A*-A?-Af) .................. (1) where Fis the applied P.D. and A t the current in the non-inductive resistance. When the power factor is very small, W is small and hence, from (1), A 2 is nearly equal to A-f + Af. Therefore in this case a small error in measuring any of the currents may introduce a large error into the value of W calculated from (1). There are several methods of measuring power by means of Transformer transformers. These are in general based on the fact methods. ^^ fa e phase difference between the primary and secondary voltage of a transformer on a light load is almost exactly 180 degrees. Let e and e z be the primary and secondary voltages, \ and i' 2 , r x and ?- 2 , 7i l and 7i 2 be the currents, resistances and turns respectively ; then if be the flux in the core and there is no magnetic leakage, 2 2 = _ 2 --_, therefore e l + e. 2 = n^ -- -r z i z ........................ (1). Yc-fj 7to INow, in a modern closed iron circuit commercial transformer on open secondary, the maximum value of & is about ten thousand times greater than the maximum value of r^ and therefore, in this case, we can write This equation shows that &> and e^ are similar curves and differ in phase by 180 degrees. The method shown in Fig. 65 is due to Albert Campbell. A small non-inductive resistance R is placed in series with the load, and the primary of a suitable transformer is placed across it. By means of a reversing switch 8 the secondary voltage &> of this transformer can be added or subtracted from the P.D. e applied 208 ALTERNATING CURRENT THEORY [CH. across the load, and the resultant measured by an electrostatic voltmeter. Let v l and v 2 be the two resultant voltages, then v l = e + 2 n. 2 where i' is the current in R. Fig. 65. Campbell's method. Now, if the magnetising current of the transformer be very small compared with the current i in the load, then i' is approxi- mately equal to i and V-L = e 2 Ri. Thus IX] RESONANCE METHODS 209 Similarly F 2 9 = F 2 + ^ R*A* + 2 ^ R W . 71^ n^ r> V-V* Hence TF = ^ r 1 ................................. (1) F 1 ) ..................... (2) where A; is a constant. This method, like the three voltmeter method, fails when W is very small ; but as a series of values of F 2 and V l can rapidly be taken it is practically convenient in other cases. Elihu Thomson, Aron and other watt-hour meters for high tension circuits are often provided with a transformer for the volt coil and make use of the fact that for light loads the primary and secondary volt waves are in opposition in phase. The series coils of meters also are sometimes connected across the secondaries of transformers the primaries of which are in series with high tension mains. In this case the magnetising current of the transformer must be small, as it is assumed that the primary and secondary currents are in a constant ratio to one another and that they are in opposition in phase. For measuring the dielectric losses in condensers, Rosa and Resonance Smith make use of the principle of resonance. A methods. drum of fl ex ible cable AD (Fig. 66) is put in series with the condenser K, which, for example, may be a concentric or a polyphase cable on open circuit. A and B are connected to low pressure supply mains, and a wattmeter W is placed in the circuit so that it measures the power taken by both the inductive coil and the condenser. If the core of the cable be made of many strands of very fine wire and there be no iron or metal near and the frequency be not too high, then the power consumed by the coil will be very nearly A 2 R where A is the current in the circuit and R is the resistance of the coil. Now if the inductance of the coil be adjusted by unwinding some of the cable, then, for a certain inductance, resonance of the first harmonic will ensue and the voltage V across the condenser may be 20 or 30 times greater than the P.D. applied across AB. In this case the power factor of the resonant circuit may be nearly unity and the wattmeter will read the load W accurately. Therefore, subtracting the power R. i. 14 210 ALTERNATING CURRENT THEORY [CH. IX expended in the inductive coil, we find W A^R&s the condenser loss. In this case, although the power factor may be as small as 0*01, the loss could be measured with fair accuracy. Fig. 66. Resonance method. REFERENCES. J. HOPKINSON, Proceedings of the Physical Society of London, Vol. 7, p. 7, 'On the Quadrant Electrometer.' March 14, 1885. G. L. ADDENBROOKE, Electrician, Vol. 45, p. 901, ' Alternate Current Measure- ment by Electrometers.' 1900. C. V. DRYSDALE, Electrician, Vol. 46, p. 774, ' Alternate Current Wattmeter.' 1901. W. E. AYRTON and W. E. SUMPNER, Proc. Roy. Soc., Vol. 49, p. 434, ' On the Measurement of the Power given by any Electric Current to*auy Circuit/ 9th April, 1891. Electrician, Vol. 30, p. 241, ' Errors of Observation in the Three Voltmeter Method.' 1892. ALBERT CAMPBELL, J. I. E. E., Vol. 30, p. 889, ' Alternate Current Measure- ment.' 1901. W. M. MORDEY, J. /. E. E., Vol. 30, p. 864, ' Capacity in Alternate Current Working.' 1901. E. B. ROSA arid A. W. SMITH, Phys. Rev., Vol. 8, p. 1, 'Condenser Losses.' 1899. E. REISZ, Elektrotechn. Zeitschr., Vol. 21, p. 713, 'Measurement of Power in Inductive Circuits.' 1900. CHAPTER X. The air core transformer. Equivalent net-work. Variable secondary load. Secondary load inductive. Condenser in secondary load. Circle diagram for a leaky transformer. Transformer with no leakage. Maxwell's formula. Bimington's theorem. Leading primary current with con- denser load. Transformer with constant primary current. IF we have two neighbouring electric circuits and an alternating The air core current be flowing in one of them, then, owing to transformer. ^^ mu t ua l induction, an alternating E.M.F. will be induced in the other, and if this be a closed circuit an alternating current will be set up in it. The magnitude of this induced current will depend on the relative position of the two circuits and the permeability of the medium in which they are placed. We shall for the present consider that there are no magnetic materials near the circuits, so that we may take the permeability of the medium to be unity ; the two circuits will then have a constant mutual inductance. This is the problem of the air-core transformer and, as its solution is of fundamental importance in alternating current theory, we will first of all attempt to solve it without making any assumption as to the shape of the wave of the applied P.D. Let e be the instantaneous value of the P.D. applied to the Equivalent primary circuit, whose resistance and inductance are net-work. ft ohms and L henrys respectively, and let S and N be the resistance and inductance of the secondary circuit, and M the mutual inductance of the two circuits. If ^ and i z be the U 2 212 ALTERNATING CURRENT THEORY [CH. instantaneous values of the currents in the primary and secondary circuits we have S (2). Ui(j UUU These equations may be written in the forms e = Ri l + L^ (3), .(4), M . where / = i t + -y i 2 Ju M* and " =l The quantity a is called the leakage factor of the transformer. We shall see in Volume II that a knowledge of its value is of fundamental importance in the theory of transformers. Eliminating -^- between (3) and (4) we get -^(e-Ri^^Si^Na^ (5), Ju at -*h-jp& + &#*-& (6), i' = Hence finally where i' = T i. 2 (7). Li T &J*-*'+JP^ (9) > -RH=J w These three equations (8) (9) and (10) show us that the problem of finding ^ is the same as that of finding the current in the coil AB in Fig. 67, where the resistance of AB is R, the EQUIVALENT NET-WORK 213 resistance and inductance of BGE are -^ 8 and ^ No- respectively, and the inductance of the choking coil BDE is L. The solution of the transformer problem is therefore identical with the solution of the problem of finding the currents in the comparatively simple (seeing that there is no mutual induction) net-work shown in Fig. 67, where the primary P.D. (e) is supposed to be applied between A and E. By (7) we see that the current in the secondary circuit is in exact opposition in phase to the , vwwww Tnnr-i Fig. 67. Equivalent net-work of a transformer. Eesistance of AB is R ; resist- ance and inductance of BCE are Z/ 2 L 2 and ^ No- which is equal to ; iVfl-y-jyj . Inductance of BDE is L. Current in the primary is the same as the current in AB. current in BCE and its magnitude is -^ times this current. Again, if there is a non-inductive resistance across the secondary terminals, the P. D. between the terminals will be in phase with the secondary current and therefore in opposition to the phase of the current in BCE. Variable secondary load. This construction still applies when 5 is a variable resistance, as for example a spark gap. Hence we should expect that when a transformer is supplying an electric arc, the peculiar shape of the P.D. wave at the terminals of the arc would be transmitted through the transformer into the primary circuit. This has been proved experimentally by Duddell and Marchant. 214 ALTERNATING CURRENT THEORY [CH. Again, if we put an inductive load on the secondary, then, from Secondary load equation (2), we alter the value of N and therefore inductive. also the value of a S u pp 0se that the inductance of the secondary load is N'. Then the inductance of the imagi- nary coil BCE (Fig. 67) is 7-2 r -" / -m-r . -T/\ / JL* N' Therefore the leakage factor becomes cr + -== . Thus the effect of adding inductance to the secondary is exactly the same as the effect of increasing the magnetic leakage. If y then A = ^ , whatever be the value of 8. If $ be the angle of phase difference between the primary current and the applied P.D., then we have in this case tan 41 = 360 (2). It also follows by the ' rule of sines ' that T7I 77 77T flj - JTj ,-j J2j o T) sin Un 3 Q sin u 3 ! T sin v-\ a From any point (Fig. 74) draw OL "equal to E l angle LON equal to 3>1 and make ON equal to E 3 . Make the angle NOM equal to 2-3 and OM equal to E. 2 . Then by (2) the angle LOM equals 6^., Now, Make the (4) Fig. 74. Fig. 74. is the c. of G. of - , - and - at L, M and p q r N respectively. 224 ALTERNATING CURRENT THEORY [CH. Therefore LM equals F 1>2 . Similarly, MN and NL equal F 2 . 3 and F 8>1 respectively. Hence the triangle LMN is the voltage triangle already found (Fig. 72). Produce LO, MO and NO to cut the sides of the triangle in P } Q and R respectively. MPOMsmMOP Sim oa r , y _ PN~ON sin NOP _E 2 sm 1>2 ~~ E 3 sin 8 . j = 2 from (3). It follows from these relations that is the centre of gravity of three particles of masses 1/p, 1./q and l/r placed at L, M and N respectively. Construct a triangle LMN the lengths of whose sides represent Rule for finding the voltages between the mains. Find the centre the voltages across o f gravity of masses 1/p, l/q and l/r placed at a star load when J the resistances of L, M and N. Then OL equals E l} the voltage across the terminals of the resistance p (Fig. 73), OM equals E 2 , and ON equals E.,. The angles LOM, MON and NOL give the angles of phase difference between the various voltages and also, since the arms are non-inductive, between the currents in the arms. The supplements of the angles of the triangle LMN give the angles of phase difference between the voltages represented by the sides of the triangle. In practice the voltages between the mains are nearly equal. Algebraical for- In this case, if the voltages be V, V + xV and Seang^^th! V + yV, we can find the angles of the triangle in voltage triangle, degrees by the approximate formulae 60-33(ar + y); 60 + 33 (2y - x) ; 60 + 33 (2# - y). Xl] STAR AND MESH VOLTAGES 225 Suppose that the P.D.'s between the mains are 2000, 2060 and 2140 volts respectively. By the formulae the angles of the voltage triangle are 56*7, 597 and 63'6 degrees. Their true values are 56'6, 59'6 and 63'8 degrees. The supple- ments of these angles will give the phase differences between the three voltages. Since (Fig. 74) is the centre of gravity of masses 1/p, l/q and 1/r placed at L, M and N, it follows that star ^a^fd/us 1 ! * the moment of inertia of these masses about an the pe" that axis through perpendicular to the plane of the rSnhrfum * is a paper is less than their moment of inertia about any other parallel axis, hence OD OM* ON* -- 1 --- h - - p q r is a minimum. Since p, q and r are non-inductive, this expression represents the power being expended in the star load. Since -l + + J = Q ) p q r Algebraical rela- .-. -, -, tions between the i / -- \ _ Vi. 2 _ ^3. l various voltages. \p q T J ~ ~q~ ~T~ ' Hence, squaring and taking mean values, q* But 2K,. 2 F 3 . 1 cos /3 = F= 2 . 3 - F 2 ,,, - F 2 ,.,. /I 1 1\ 2 /F 2 F2 \ /I 1\ "Ps Hence E{ ( - + - + - ) = [ ^ + - 3 " l } (- + -}- - - -(5)- \p q r) \ q r J \q r J qr We can write down two similar equations by symmetry. Similarly . / pq \ P q \p q r and two other similar equations. Hence, also, 2 172 *- 2 l K 2 -3 | p q r pq qr rp R. i. 15 226 ALTERNATING CURRENT THEORY [CH. It follows that if p, q and r are all equal, then n. 2 +F^ 3 +n. 1 =3(# 1 2 +# 2 2 +#3 2 ) ......... (8). If in addition V 1% 2 = F 2 . 3 = F 3 . lt then Fi.aWatfj ................. .......... (9). This can also easily be proved from Fig. 74, since in this case LMN is an equilateral triangle. It is to be noted that no assumption is made as to the shape of the P.D. waves. Let v lt v. 2 and v 3 be the potentials of the three mains, and let /!, / 2 and / 3 be their fault resistances to earth. The potentials to * " J * J * earth of the By the fault resistance of a main we mean the resistance of all those leakage paths from it to earth which do not pass through the other mains. Then since if the capacity currents in the sheath can be neglected, the sum of the leakage currents to earth must be zero, Proceeding as before, we see that if is the centre of gravity of masses l//i, l// 2 and l// 3 placed at the angles L, M, N of the voltage triangle, then OL, OM and ON represent V lt F 2 and V s respectively in magnitude and phase. It follows also that, if the fault resistances of the mains vary, then their potentials adjust themselves so that the power ex- pended in leakage currents is a minimum. An analogous theorem holds true for the case of three wire direct current systems. Let p, q and r be the resistances of the arms of the star load, tid the potential of the and let x be the potential of the centre. Then To find the centre of a star V l " X , * ~ X , V * ~ X = Q load which is f) Q T insulated from /I 1 1\ v, v, v, therefore x{ - -\ [--=-+- + -. \p q rj p q r Now as the P.D. waves have different shapes, a little consideration will show that the frequency of the alternating potential is an unknown quantity. XI] CAPACITY CURRENTS 227 If V denote the effective value of oc, then V cos + r 3 i 3 = 0. Hence as formerly n^i r 2 / 2 r 3 I 3 , sn sn sin l . where # 2 .3 is the phase difference between I 2 and I 3 . If rj/j, n/o and 7*3/3 represent forces acting at a point 0, equations (3) show that they will be in equilibrium. Now it is known from statics that is the centre of gravity of three equal masses placed at the extremities of r l l l , r 2 / 2 and r 3 / 3 respectively. Hence will also be the centre of gravity of masses proportional to r l} n and r 3 placed at the extremities of lines equal in length to /j , 7 2 and I 3 respectively. Now from (1) A? = 7 3 2 + / 2 2 - 2/ 3 / 2 cos 2 . 8 . If we draw therefore from (Fig. 77) lines proportional to I lt 7 2 and / 3 and inclined to one another at angles # 2 .3> #3.1 and 1>2 , we see that the lines joining the extremities of these lines give us A lt A z and A 3 respectively. Hence the triangle is the same triangle as in Fig. 76. Knowing the currents in the mains and the resistances of the arms of the . mesh, we can find the currents in the arms by the following construction. M Fig. 77. The currents in the mains and in the arms of the mesh load. 230 ALTERNATING CURRENT THEORY [CH. Make a triangle LMN (see Fig. 77) whose sides are equal to AI, A 2 and A s respectively. Find the centre of gravity of masses r l5 r. 2 and r s placed at 7, M and N. Then OL will give /u OM will give 7 2 , and (W will give 7 3 . Again, if in Fig. 77 we produce OL to L', etc. so that OL'=r l .OL ) OM' = r 2 .OM, and ON' = r 3 .ON, then if the re- sistances are non-inductive, L'M'N' is the triangle giving the voltages between the mains and OL, OM' and ON' give the voltages between the mains and the centre of the star load. We can prove the following algebraical relations between the currents in the same way as we proved the relations Algebraical formulae for the between the P. D.'s. currents in a mesh load. r^A* = (r. 2 L* + rjf) (r 2 + r,) - rflf (4) and two similar equations. Similarly, and two similar equations. Hence, also (Ifri + 7 2 2 r 2 + 7 3 2 r 3 ) (n + r 2 + r a ) The case when r 1} r 2 and r s are all equal is important. (Fig. 78) is now the centre of gravity of the triangle LMN. Hence, from geometry or from equa- tions (4) and (5) above, and four similar equations. Also from (6) M A, N Fig. 78. The relations be- tween the currents. and the curious theorem is also true that AS + AJ + AJ = (37?) 2 + (3/ 2 2 ) 2 + (3/ 3 When the currents in the mains are all equal, (10). (11). It is to be noted that these formulae are perfectly general, no assumptions having been made as to the shape of the wave. XI] EQUATION TO THE WAVE 231 When we have a star winding, the currents in the branches are of course equal to the currents in the mains. Their phase differences can be got by constructing a triangle whose sides are proportional to the currents. The exterior angles of this triangle are the phase differences required. If we have both a star and mesh winding, the graphical representation of all the currents by a figure drawn in one plane is rarely possible. We will first suppose that the effective P.D.'S between the mains are all equal to one another. Let Vf(t) represent the instantaneous value of one of these P.D.'S, where F is the effective voltage and t is the time in seconds from the era of reckoning. Then the other two will (T\ / 2T\ t + - Q and Vf ( t 4- - - - ) where T is the o / V o / period of the alternating current. Since the sum of the instan- taneous values of the P.D.'S must always be zero, we must have a). This functional equation limits the possible forms of f(t). Solving it by Laplace's method we get where X and Y are constants or functions of t whose values do T 2T not alter when t + -= or t+ - - is substituted for t. o o Since, however, / (t) is an alternating periodic function, we have another equation to satisfy, namely This adds the condition that X and Y do not change when T . t + is written for t in them. Hence X and Y are functions T T 2T of t that do not alter when + -5 , + > or t 4- -^ is written for t. o 2 o 232 ALTERNATING CURRENT THEORY Yr \ Fig. 79. y = sin f -^ + ~ sin 6 -~ J . Possible form of P. D. wave in balanced three phase system. Positive half only shown. [CH. 1 r^ \ I \ \ \ \ \ r \ I \ 1 \ 1 \ / \ 1 \ / \ I \ / \ / ^ \ 1 ^^ V / \ J V / \ / \ Fig. 80. y = l + - sin 6 -J sin - . Another possible form. XI] WAVE FORM 233 In Fig. 79, X has been taken equal to 1, and Y equal to Examples. - sin 6 -^- . so that the equation to the curve is . a = sm - + sin 6 -- . In Fig. 80 the equation to the curve is 1 The equation (1) can also be solved by Fourier's method. The Fourier series in this case may be written ......... (2). As a rule it will be found that the expression X sin f-^- - + TJ is more convenient to work with than the series given by (2). If it is possible for the potential difference waves between the mains to be similar curves when their The P. D. waves between the mains effective values are different, let the P.D.'s can only be similar curves when their between the mams be represented by Vif(t), effective values are / T\ / 9T\ V,f ( t + 4 ) and V s f(t + 51 J respectively, where \ o/ \ o / Fj, F 2 and F 3 are positive. Then, at time t, VJ(t} + V,f(t + ) + V 3 f(t + ^) = 0, 7' / T\ / 9T\ at time t + | , F 3 /(() + F,/ (< + ^ J + F 2 / (< + i J = 0, 9T 7 at time t + -* , V t f(t) + F,/ + + i + 9'7 7 \ y ) = = 0. Eliminating the functions by determinants or otherwise, we get and therefore V l = F 2 = F 3 , since V 1 + F 2 + F 3 must be positive. Hence, if we make the assumption that the P.D. waves between the terminals of a three phase alternator are all similar, we also make the assumption that their values are all equal. 234 ALTERNATING CURRENT THEORY [CH. By an almost identical proof we can show that the P.D. waves between the mains and earth can only be similar curves when the leakage currents to earth are all equal to one another. In Fig. 81 let A, B and C be the three terminals of a three phase alternator, the armature of which is star wound. Let also AD, BE and CF be the mains. The points ^ anc * ^ are somet i mes P ut to earth or connected by a wire. We will investigate the frequency of J the current in this wire when the loads on the of a balanced three phase system. three arms are equal. Let Cf(t), Gf\t r r -= and C / 27 7 \ f(t + --\ \ o / Fig. 81. Frequency of the current in the fourth wire equals 3 (2n + l)f. be the instantaneous values of the currents in the mains and let (7 (t) be the current in the fourth wire. Then and t + -^ and thus Hence (t + ^ = - (t + ) = <#> (t). XI] THREE PHASE LOAD 235 It follows that, if the frequency of the currents in the mains be/", the frequency of the current in the fourth wire is of the form 3(2n + l)y, where n is a positive integer. Its lowest value is therefore 3y. It is easy to show that, if the current waves in the mains be triangular or parabolic in shape, then the current wave in the fourth wire is also triangular or parabolic, and, since its frequency is three times as great as that of the wave in any main, its effective value is one-third that of the current in each main. If the currents in the mains were sine curves, its value would be zero. Since a sine curve and a parabola differ very little in shape from one another (see Fig. 44), this shows how important a small modification in the shape of the wave may be. In the general case the shape of the resultant wave is quite different from the shape of the component waves. The above results prove that in practical three phase working slight causes may considerably alter the shape of the E.M.F. and current waves. For example, a slight variation in the resistance or, a fortiori, in the inductance of the fourth wire, will alter the shapes of the current waves in the other three wires. We have shown that, when everything is symmetrical in a three phase system, the potentials of the mains Value of load on r J ' can be expressed by functions of the form a three phase alternator at every instant. X sin ( -~- + Y] , where X and Y are periodic V T functions of t whose frequency is ^ or 6f. If the mesh load between the mains be three non-inductive resistances each equal to R, and the star load arms be each equal to r, then the power at any instant is ^Xs 1 n( 2 J + F)-Xsn 1 ( 2 f + F + |)} 2 + ... + ... 236 ALTERNATING CURRENT THEORY [CH. where X is a function of t whose frequency is 6/. Hence the instantaneous value of the power is only constant in special cases. Fig. 82. The measurement of power. Suppose that there is both a mesh and a star load as in The measurement of power in three phase circuits. Fig. 82. Leto^c be the instantaneous values of the currents in the mains, in the arms of the mesh load and in the arms of the star load respectively. Then a 1 = i' 3 i z + i x , Thus aj + a 2 + a 3 = i x + i y + i = 0, if the three phase machine is insulated from the earth and the leakage currents to the mains are inappreciable. Let w be the instantaneous value of the watts, then Now and Thus by symmetry and Similarly = v l . 2 (i s - i* 2 + 4) + v 3 , 2 (i a - ^ + i z ) (1). + V 2 a 2 + v s a 3 (2). Xl] THE TWO WATTMETER METHOD 237 Let us next consider the case when the point is not main- tained at zero potential and let v x be its potential. Then the power being expended in all the paths of the current a x from the point 1 where the potential is v-^ to the point or points where the potential is v x is (^ v x ) a lf Similarly (v 2 v x ) a 2 and (v s v x ) a 3 will be the values of the power in the paths of the two other main currents respectively. Thus the total power w in the load is given by w = (v^- v x ) tt! + (v 2 -v x )a 2 + (v 3 -v x )a 3 ......... (3). If we make the supposition that is insulated from the earth, then, ! + a. 2 + a s must be zero and we get w = v 1 a- L + v 2 a 2 + v 3 a 3 . Similarly w = v lm3 a^ -f v. 2 . 3 a 2 and two similar equations. Therefore equations (1) and (2) still hold in this case. When is connected with the earth by a wire and the leakage currents from the mains are appreciable, or when the centre 0' of the star winding of the armature is also connected with the earth, or when and 0' are joined by a fourth wire, then e^ + 2 + #3 is not necessarily zero. Let i' be the current in the wire joining to the earth, then we must always have ! + a 2 + a 3 + i = 0, and thus by (3) w = v l a l + v 2 a 2 + v 3 a 3 + v x i . Therefore v^ + v. 2 a, 2 + v 3 a 3 is not equal to the power in the load in this case. We also have w = i . 3 i + v 2 . 3 a, - (v 3 - v x ) i . When i' is zero these formulae simplify to (2) and (1) given above. The formulae (1) and (3) give the methods of measuring power in three phase circuits. The first method is to use two wattmeters. The ampere coil of one of them is put in No. 1 main and the volt coil is connected across 1 and 2. The ampere coil of the other is put in No. 2 main and the volt coil is connected across 3 and 2. Suppose that w 1 is the reading on one meter and that w 2 is the reading on the other, and suppose that W-L is greater than w. 2 . Then the power given to the circuit is w-^ w. 2 . If the 238 ALTERNATING CURRENT THEORY [CH. phase difference between a 3 and # 3 .. 2 is less than 90 degrees, w 2 is positive, but if greater, w. 2 is negative. We must be careful, when measuring, to note whether we have to reverse the shunt con- nections or not in order that the needle may deflect the right way. If we have to reverse, then w 2 is negative. The second method is to use three wattmeters, their ampere coils being put in the main circuits and their volt coils across and 1, and 2 and and 3 respectively. It will be seen that the three wattmeter method is applicable to three phase systems which use a fourth wire. In this case the two wattmeter method cannot be used. These methods also apply when the load is arranged in any manner between the mains. For example, it may be arranged as in Fig. 83, where 1, 2 and 3 are the terminals for the mains. Fig. 83. Possible connections. An algebraical proof may be given as follows. Let 6j, e 2 and e 3 denote the P.D.'s between 1 and L, 2 and M and 3 and N respectively, and let be insulated from earth. XI] POWER MEASUREMENT 239 Then w = e^ 4- e 2 2 4- e s a s = (ei -e* + V LM ) aj 4- (e B -e 2 + V NM ) a, = VL 2^1 + ^3. 23 as before. Similarly w = e l ^ 4- e 2 2 + e s a s 4- v 3 a 4- v 3f 2 + VNO a s It is to be noted that, if all the mains are equally loaded and the loads are non-inductive, one meter is sufficient. If the volt coil be connected across two of the mains we multiply the reading by 2. If it be connected from one main to the centre of the system, then the multiplying factor is 3. This latter method is also true for balanced inductive loads. An important case arises when the volt coil of the meter forms one of the arms of a star load, made up of two equal high resist- ances connected with the volt coil. An arrangement of this kind is generally called a star-box. If the three arms are of equal resistance, the multiplying factor is 3. If, however, as is often the case in practice, the resistance of the volt coil be different from that of the other two arms, then the multiplying factor for balanced loads is 24--^, where JR, is the resistance of the volt coil K and r that of either of the other arms of the box. Let (Fig. 84) be the centre of gravity of masses l/R, l/r arid l/r placed at A, B and C re- spectively. Then OA, OB and OC will A be the three P.D.'S to the centre of the star-box. Also \ r r ) \r r! r B Thus OA* 2-f 4j =3. \ -ti/ where G is the centre of gravity of the triangle ABC; Fig. 84. Voltages in star-box. therefore 240 ALTERNATING CURRENT THEORY [CH. Now if the arms had been equal, GA would have been the P.D. 3 GA across the volt coil. Hence the required multiplying factor is ~r\-r- > T and this we have shown equals 2 + -^ . In Fig. 85, G is the generator and M is the motor. If we watt-hour assume that the three arms are equally loaded, then meters. fa Q ener gy expended can be measured by means of an ordinary single phase watt-hour meter W and a star-box (S in w Fig. 85. Connections of watt-hour meter to measure balanced loads. the figure). The ampere coil A 1 A 2 (see Chapter ix) is put in one of the mains 22' and the volt coil A^V^ is connected to 22' and the centre of the star-box. The units registered by the meter multiplied by three will give the total energy expended on the motor. If the arms are unequally loaded, a watt-hour meter has to be put in each arm and the volt coils connected to 0. The sum of the three readings will then give the energy consumed in the given time. It will be seen that the three wattmeters can easily be made into a single one containing a star-box and having six ampere terminals A lt A 2 , etc., three being connected with the supply mains and three being connected with the mains for the load, the volt connections being permanently made inside the meter. If we connect two single phase watt-hour meters, as in Fig. 86, then by the formulae given above the sum of their readings will XI] WATT-HOUR METERS 241 give the energy expended. If one of them runs backwards, then its reading has to be subtracted from the reading of the other, Fig. 86. Connections of Watt-hour Meters to measure all loads. The ampere coils of the two meters are put in two of the mains and the volt coils are connected with the third main. which will always be the greater. A true three phase meter can be made by combining the two meters (see Chapter xin). It will have four ampere terminals and one volt terminal. If the load be balanced and non-inductive, the reading on either instrument shown in Fig. 86 multiplied by two will give the true units expended. It can be shown mathematically that if we take a point Minimum value within a triangle LMN of the sum of the (Fig 1 . 87) then three voltages in a star load. OL + OM + ON is a minimum when the angles LOM, MON and NOL are each equal to 120 degrees, provided, of course, that no M angle of the triangle is equal to or Fig. 87. In a triangle greater than 120 degrees. If one angle OL + OM+ON of the triangle is equal to 120 degrees, i* a minimum when the angles ,, , f^ T r\*/r r\-*r at are all equal. then, to make OL + OM -f N a mim- R. i. 16 242 ALTERNATING CURRENT THEORY [CH. XI mum, must coincide with this angle. It follows therefore from Fig. 74 that the sum of the three voltages across the arms of a star-box is a minimum when their phase differences are each equal to 120 degrees. In this case the currents in the arms are all equal to one another. Describe the voltage triangle LMN. On each side of it describe a segment of a circle which will contain sistancesofthe an angle of 120 degrees. These will intersect in a sin g le P oint 0. Produce LO to cut MN in P, symmetrical three tnen fa e Y ^tio Q f J/p to p]\f ^lll be the ratio of phase currents. the resistances q and r in the desired load. In this case E l = E 2 = E 3 and is therefore the centre of the circumscribing circle. Let A, B and C be the To find the ratio of , , . . , the resistances in angles of the voltage triangle, then a star load in , order that the C7. 2 . 3 = ZA, QIC. voltages to the centre may be . , -C/i JL% -C/ 3 equal. AlSO - : - TJ - = -- "^ --- = - . - ^ - . 2>8UL0 c .s q sin u~ , rsin0 1<2 Hence p sin 2A = q sin 2B = r sin 2C. It will be seen that three phase problems, considered graphi- cally, generally resolve themselves into problems connected with the trigonometry of a triangle. When the arms of the load are inductive, then, as a rule, only approximate solutions can be got graphically, as the vectors can no longer be accurately represented geometrically (see Chapter vin). REFERENCES. SILVANUS THOMPSON, 'Polyphase Electric Currents and Alternate Current Motors.' A. E. KENNELLY, Electrical World and Engineer, 34, p. 413, : On the Equi- valence of Triangles and Stars in Conducting Networks.' 1899. The Electrician, Vol. 47, p. 639, ' The Elements of Three Phase Theory.' 1901. The Electrician, Vol. 48, p. 487, 'P.D. Wave Forms in Three Phase Systems.' 1902. For Laplace's Method of Solving Functional Equations, see GEORGE BOOLE, ' The Calculus of Finite Differences.' CHAPTER XII. Two phase systems. The magnitudes and the phase differences of the P. D.'S between the mains. The voltage tetrahedron. The graphical representa- tion of the voltages in the four arms of a star load. Kule for finding the voltages across the arms of a star load when the resistances of the arms are given. The voltages to the centre of the load so adjust themselves that the power expended on it is a minimum. The potentials to earth of the mains. The currents in a mesh load. The conditions under which it is possible for the P.D. waves between the four mains and earth to be similar curves. The p. D. waves between the mains are only similar in a special case. To find the phase differ- ences between the opposite voltages and between the diagonal voltages in a two phase system. The measurement of power in two phase circuits. Two phase meters. When the currents in a star load are equal the sum of the voltages to the centre is a minimum. DIAGRAMS 70 and 71 in the last chapter illustrate the funda- TWO phase mental principle of three phase machines with mesh and star wound armatures respectively. The prin- ciple of two phase machines is the same, but instead of the armature being wound in three sections it is wound in four. There is, however, in this case a third method of winding, which is illustrated in Fig. 88. A, B, C and D are the four terminals of the machine. A and C are the terminals of one winding of the armature and B and D are the terminals of the other. From symmetry the P.D.'s generated between A and C and between B and D when the field magnet revolves will differ in phase by ninety degrees. The currents flowing in circuits connected across the mains to A and C and in similar circuits connected across the mains to B and D will also differ in phase by a right angle and we shall see later on that it is easy to produce a rotary magnetic 162 244 ALTERNATING CURRENT THEORY [CH. field by means of these currents, arid hence two phase motors are simple to construct. In practice instead of having four mains connected to A, B, C and D respectively it is customary to have only three, one of which is connected to two adjacent terminals, as for example A Fig. 88. Two Phase Machine with two separate windings on armature. and B. In this case the P.D.'S between A and C and between B or A and D will still differ in phase by a right angle, and they will from symmetry be equal to one another. Now with our usual notation Thus and = -V AC -V DA . V CD + V DA + V AC = 0, V CD + V DB + V CA = 0. TWO PHASE SYSTEMS XII] Therefore V- CD = V- since V DB and V AC are in quad- rature. They are also equal, and thus 245 AC> The main connected to A and B is generally referred to as the 'common return' and the other mains are sometimes referred to as the ' outers.' In Fig. 89, OX and OY represent the P.D.'s be- tween the outers and the common return and OZ represents the P.D. between the two outers in mag- nitude and phase. The magnitudes and the phase differences of the P.D.'s between the mains. Fig. 89. Potential Differences in two phase three wire system with balanced load. In the general case, that is, when four mains are used, it is im- possible to repre- sent the P.D.'S by vectors drawn in one plane. They can, however, be represented by vectors drawn in space. Let i\, t'o, v 3 and v 4 be the potentials of the four mains. Let v 1-2 , t' 2>3 , v 3 . 4 and v 4-1 be the P.D.'S between the mains 1 and 2, 2 and 3, 3 and 4, and 4 and 1, respectively. Then v l . 2 = v l v.>, Hence v lm2 + v 2 , 3 + y 3 . 4 + v 4tl = 0. A linear relation therefore connects the four instantaneous values, and hence by Chapter vin their effective values can be represented by lines drawn in space. In Fig. 54, page 185, if we produce SO backwards to T making OT equal to OS, then OT will represent V l in this case, and the relations between the 246 ALTERNATING CURRENT THEORY [CH. voltages and phase differences can be written down by the formulae given in that chapter. If N be tjie fourth corner of the parallelogram ORQ (Fig. 54, The voltage page 185) and we join tetrahedron. Q N an( J ^ we get the voltage tetrahedron 1, 2, 3, 4 (Fig. 90). The angles between the lines drawn in this figure are the supplements of the phase differences between the P. D.'s these lines represent. In Fig. 90 12, 23, etc. represent the P. D.'s between the mains 1 and 2, be- tween the mains 2 and 3, etc. Let 1, 2, 3 and 4 (Fig. 91) be the four The graphical representation of and let be the the voltages in the four arms of a centre of the star star load. , -, T . load. Let r lt r 2 , r s and r 4 be the resistances (non- Fi * 90 - The Voltage Tetrahedron. inductive) of the four arms, and let e lt e*, e s and e+ be the P. D.'s be- tween the mains and the centre of the load. Then, since the Ol algebraical sum of the currents at must be zero, we have Hence - , , and can be represented by vectors in the ways shown in Fig. 90 page 185. We in the ways shown in Fig. 90 and in Fig. 54, page 185. may write (1) in the form r Fig. 91. Star Load. 01.3 ,01.4 n r. XII] where v 1> THE VOLTAGE EQUATIONS e P.D. between the mains 1 and 2 247 Now by the voltage tetrahedron (Fig. 90) we see that F 1<2 , F 2 . 3 and F 3 .;i form a triangle. Hence if a be the angle of phase difference between V lt and F 1<3 we find by trigonometry, that .3 cos a, and thus nr. Hence, squaring (2), taking mean values and substituting for the cosines of the phase differences, we get 11 1 1\ 2 FV 2 . V\. s . F 2 , 4 , F^ + F^-F 2 ^ 2 (! + 1 + 1 + 1 V = JLJJ + y l - s + v l - 4 + J-J F- 1 _1_ 172 T72 172 I TT2 1^2 1.3 "I K 1.4 ' 3.4 , K 1.4 "t" V 1.2 K 2.4 ^ III r z r 3 r 4 F 2 , F-J T7"2 3.4 ^ 4. .(3). The other three equations giving ^ 2 , ^ and ^ in terms of the six P.D.'s between the mains can easily be written down by symmetry. By adding up these four equations and cancelling out the common factor we deduce (E? E.-? E 3 2 E 4 2 \ /I + - + - + ---) - V r, r. 2 r 3 r 4 J \n 1 1 1 - + - + - r, r r 2 TT2 TT2 il + 21^ + JljJ (4) . If the resistances are all equal (5). If, in addition, the voltages between adjacent mains are all equal and the diagonal voltages, namely F 1>3 and F 2t4 , are also equal, then from (3) And from (5) S# 1 2 =2F 2 1 .,+ F 2 1- (6). 248 ALTERNATING CURRENT THEORY [CH. If, finally, F lt and F 2 . 3 are in quadrature y-2 _ y-2 I y-2 ' 1.3 K 1.2'' 2.3 Hence VE^F^ .............................. (7). In getting the equation (7) we have made no assumption as to the shapes of the waves of P.D. but the shapes are restricted by equation (1). Looking back at Fig. 54, page 185, we see that if OP, OQ, OH and OT (drawn equal and opposite to OS) represent forces acting at 0, they will be in equilibrium. Now, if G be the centre of gravity of equal particles placed at P, Q, R and T, the resultant of the forces OP, OQ, OR and OT will be 4 . OG ; but since they are in equilibrium, this resultant must be zero, and therefore G must coincide with 0. Hence is the centre of gravity of equal masses placed at P, Q, R and T. -& Let OP represent - , and let OQ, OR and OT represent TI T? T? V , - and - Then, since the resistances are non-inductive, r,' r, r, n.OP, r t .OQ, r s .OR and r,.OT represent E L , E z> E 3 and E, respectively. Let OP' = r, . OP, OQ = r 2 . OQ, etc., then P'Q'R'T' will be the tetrahedron that gives the P.D.'S between the mains. To prove this it is sufficient to notice that l/^ 9 == \j\ ~~ @o Therefore V\ . 2 = E? 4- E? - 2E, E, cos ^ . , , and hence the rest follows by trigonometry. Construct the tetrahedron that gives the voltages between the mains. In order to do this we need to take six Rule for finding voltage readings, namely, the four readings be- the voltages across . . . .. the arms of a star tween adjacent mams and the two diagonal re'fstances^f the readings. Find the centre of gravity G of masses arms are given. Ill 1 , , - and placed at the four angular points i\ r z r s r* 4 of the tetrahedron. Then the lines joining G to the four angular points will give the phase differences and the magnitudes of the required voltages. XII] THE POTENTIALS OF THE MAINS 249 Let L, M, N and R be the angular points of the tetrahedron representing the voltages between the mains, and The voltages to we suppose that the dynamo maintains these load"" adjust * voltages constant. Then, by a well-known theorem ex hat in statics, if be any point in space OL* + OW + ON* + _ n r, r, ' r 4 is a minimum when is the centre of gravity of masses , , - and - placed at L, M, N and R respectively. But since the resistances are non-inductive, this expression represents the power expended on the load, and hence the theorem follows. Let /!, f 2 , / 3 and / 4 be the fault resistances of the four mains : then, by Kirchhoff's law when the condenser currents in the sheath are negligible, we have . /i /> /a /4 Proceeding as above, we see that, if G be the centre of gravity of masses -? , 7 , 7 and -. placed at the vertices of the voltage /I ft /I /4 tetrahedron LMNR, then GL, GM, GN and GR represent V lf 7 2 , V 3 and F" 4 respectively in magnitude and phase. It follows that if the fault resistances of the mains vary, then their potentials adjust themselves so that the power lost in leakage currents is a minimum. Let !, a 2 , a 3 and a 4 (Fig. 92) be the currents in the mains, The currents in an( l ^ et ^ *s, *o an( i *4 be the currents in the a mesh load. mesn Then i = *i *4, a. 2 = i. 2 Therefore a^ + a. 2 + a s + a = 0. This also follows at once since there is no accumulation of current in the load. 250 ALTERNATING CURRENT THEORY [CH. Hence, like the voltages, A l} A. 2) A s and A^ can be represented by lines drawn from a point in space, and this point is the centre Fig. 92. The Currents in a Mesh Load. of gravity of four equal particles placed at their extremities. The currents can also be represented by a skew quadrilateral, the angles of which are the supplements of the angles of phase difference between the four currents. Again, if r lt r 2 , r 3 and r 4 be the resistances of the arms LM, MN, NR and RL in Fig. 92, then Therefore r^ + r^ + r s i' 3 + r 4 i' 4 = (2). Hence, if 7*1/1, r 2/2 ^3/3 an d ^4/4 be represented by lines drawn from a point 0, this point will be the centre of gravity of equal masses placed at the extremities of these lines. The lines joining these extremities give the voltage tetrahedron of the P.D.'s between the mains. XII] THE CURRENT TETRAHEDRON 251 Again, if we draw I lt L, I 3 and 7 4 (Fig. 93) from the point 0, it can easily be proved that is ^ the centre of gravity of particles, whose masses are proportional to ?*!, r. 2 , r 3 and r 4 , placed at the extremities of 7 l5 / 2 , 7 3 and 7 4 . Equations (1) show us that the lines joining the extremities of 7j and 7 4 , 7 2 and 7 1} 7 3 and 7 2 and 7 4 and 7 3 represent A 1} A, A 3 and ^i 4 . Let (Fig. 93) A 5 be equal to the length of the line joining 1 to 3 Fig " 93 ' The CuiTent Tetrahed n - and let A 9 be the length of the line joining 2 to 4, then A,- = A. 2 - + A 3 - + 2^3^.3 cos L 2 . An inspection of Fig. 93 will show that there are several other expressions for A 5 and A 6 . From (2) where i a . 2 = i l i, = a 2 , etc. Proceeding in exactly the same way as we did with the corresponding voltage equation, we find and three similar equations. Hence also + r 3 r. 2 ?' 4 If the resistances are all equal, then 4 (Jj 2 + 7 2 - + 7 3 2 + / 4 2 ) = A? + AS + ^ 3 2 + A* + A* + A 6 2 . . .(5). If, in addition, the currents in the mains are all equal and the phases are such that A 5 equals A$, then from (3) Hence (6). 252 ALTERNATING CURRENT THEORY [CH. If the currents in the mains are in quadrature, then (Fig. 93) Hence (7). In order to get the factor \/2, which is used in practical work, we have therefore to make important assumptions. The ] skew quadrilateral reduces simply to a square in this case (Fig. 94) and if be the intersection of the diagonals the magnitudes and phases A, of the currents are as shown in this figure. Let the equation to the wave of P.D. be- tween a main and earth be of the form The conditions under which it is possible for the P.D. waves be- tween the four mains and earth to be similar Fig. 94. The currents in a balanced load. Let r l} r. 2 , r s and r 4 be the fault resistances of the mains and let their potentials be T\ / T\ / 37\ ij- r '/(' + j) and M* + T) respectively, then V\ r, T T Now this is true for all values of t, hence writing t + -r , t + -= , and 4 ~ t+ . for t in (1) successively, we get four equations from which the functions can easily be eliminated by determinants and we get that rv 1* V I I i* v v v 1 '2 '3 4 ' \ * I '2 '3 '4 u_u + ^-- 4 =o. V F\- v,-^) XII] Hence either CONDITION FOR SIMILAR WAVES V V V V E' + ll'.i+IJ or and V. 253 -(2), .(3). In what precedes the only assumption we have made is that f(t) is a periodic function. If we make the additional assumption that it is an alternating function, i.e. that /(<). we get from (1) that As this has to be true for all values of t, we must hav6 .(3). Now r V V V V ' \ *3 i ' i '4 = and - SB ^ r s ?- 2 r 4 and F 4 are determined by finding the centre of gravity of masses , , - and placed at the angular points i\ ? % 2 r 3 r* of the voltage tetrahedron and joining this point to the angular points, hence (3) can only be true in very special cases. We are therefore not justified in assuming that the potential waves between the mains and earth are similar curves. Suppose that the P.D. waves between the mains are hf),F s /(i + !)and The P. D. waves between the mains are only similar in a special case. Then since the sum of them must be zero and /<)-.-/(*+i we get (y\~ Hence V l must equal V.> and V. 2 must equal F 4 . OFTHE UNIVERSITY OF 254 ALTERNATING CURRENT THEORY [CH. Let LMNR (Fig. 95) be the voltage tetrahedron. Let A, B, C, D, E and F be the middle points of its edges. Then from geometry we see that ABCD, etc. are parallelograms, and that AC, BD and EF intersect in a point which is the centre of gravity of the tetrahedron. We will first find the angle of phase difference between the diagonal voltages LN and MR. To find the phase differences between the opposite voltages and between the diagonal voltages in a two phase system. Fig. 95. Finding the Phase Differences in a Two Phase System. XII] PHASE DIFFERENCES 255 We have 2LN . MR cos < = 4 . 2AB . BC cos ABC = 2 [AC 2 + EF 2 - (BD* + EF*)} = 4 {AE* + E&- (EB 2 + ED*)} * + LR* - (LM* + NR*), and therefore cosc/)= ' 3 11 ' 2 ............ (1). Similarly if <' be the angle of phase difference between the opposite voltages F li8 and F 3 . 4 , then The phase differences between the other voltages can easily be got from the triangular faces of the voltage tetrahedron (Fig. 95). If the diagonal voltages be in quadrature FY.+ FYi=FY,+jrV< .................. (3). Conversely, if (3) be true, then from (1), the phase difference of the diagonal voltages is 90 degrees. If F 1<2 and F 3 . 4 (LM and NR in the figure) be in opposition in phase, then from (2) F i ,.,+ FV4-FV,+ n., + 8Fi. 1 F,. 4 ............ (4). In this case, since LM and NR are parallel, the four points L, M, N and R are in one plane, and (4) could easily be proved otherwise. If F 1>2 and F 3 . 4 and also F 2 . 3 and F 4<1 be in opposite phases, then F 1>2 = F 3 . 4 and F 2 . 3 = F 4il . If, in addition, F,. 3 and F 2 . 4 be in quadrature, then F F V V V V 1 - 3 2 - 4 v "~ "- Let 1, 2, 3 and 4 (Fig. 96) be the mains and suppose that there is both a mesh winding and a star winding, The measurement . . ' of power in two the centre of the star winding being 0. phase circuits. T 11.1 .1 Let !, tt 2 , 3 and a 4 be the currents in the mains, i lf . 2 the current in the mesh winding joining 1 and 2, 256 ALTERNATING CURRENT THEORY [CH. and let 4 be the current in the star winding from 1 to 0. Then 1 = 4 + 4. -2 -4. . 3-.2 1 Therefore a 4 = + .i -.4' (^ + a., + a s + a 4 = 0. a 3 /3 Fig. 96. The Measurement of Power. Let 0! be the P.D. from 1 to 0, v lt2 the P.D. from 1 to 2, and the power expended in the windings. Then w = V& + V 2 i 2 + v 3 i s + v 4 *4 + (^1 ~ 2 )*1.2 + (02 - 3 ) 4.3 + (03 - 04> 4.4 + (l>4 ~ 0l) 4.1 = 0!^ 4- 2 a 2 4- 3 a 3 + 4 a4 (1)> = 0i . 4 ai + 2 . 4 a. 2 + 3 . 4 a 3 (2). XII] TWO PHASE METERS 257 Formulae (1) and (2) give the two methods of measuring power in two phase circuits. For the first method we require four wattmeters, the ampere coils being put in series with the mains, and the volt coils being connected from the mains to the centre of the star system. For the second method we require three wattmeters. The ampere coils are put in any three of the mains and the volt coils of the three wattmeters are connected from these mains with the other main. Care must be taken to find out whether any of the wattmeters are reading negatively. If the system is symmetrical, then 10 = 4^! (3), and w = 2v 1 . 4 a 1 + v. 2 . 4 a 2 (4). Hence one wattmeter is required for the first method, the multiplying factor being 4, and for the second two wattmeters are required. Their ampere coils are put in adjacent mains, No. 1 and No. 2 for example, and their volt coils are connected between No. 1 and No. 4 and No. 2 and No. 4 respectively. Twice the .reading of the first wattmeter added to the reading of the second will give us the watts expended in the circuit. The first of these methods is obviously the preferable one. When part of the load is in series with the mains as in Fig. 83 Chapter xi, the formulae still apply. The proof is practically the same as for the three phase case. Just as in the case of three phase circuits, if we have an TWO phase ordinary watt-hour meter and the arms are equally loaded, we can measure the energy consumed by connecting up as in Fig. 97, where S is a star-box connected with the four mains. The energy recorded multiplied by four will give the total energy expended in the given time. If the arms are unequally loaded, we should require a watt-hour meter in each arm and the sum of the meter readings would give the required energy. If we connect up as in Fig. 98, then the sum of the three readings will measure the units consumed. Thus we can make a two phase meter with six ampere terminals and one volt terminal. If the arms are equally loaded and the load be non-inductive, then R. i. 17 258 ALTERNATING CURRENT THEORY [CH. the reading of either of the meters AV or A"V" (Fig. 98) multi- plied by four or the reading of the meter A'V multiplied by two will give the required energy. Fig. 97. Connections of watt-hour meter for balanced working. A l and ^ 2 are the ampere terminals of the meter and V is the volt terminal. G is the generator, M the motor and the centre of the star-box S. Fig. 98. Connections of three watt-hour meters for measuring the energy expended in any two phase circuit. Fig. 99 shows the connections for the case of a common return. The sum of the two watt-hour meter readings will obviously give the units consumed. The two meters could be combined into a single instrument having four ampere terminals and one volt terminal which has to be connected to the common return. By comparison with Fig. 86, Chapter xi, it will be seen that this form of meter XII] GEOMETRICAL APPLICATIONS 259 measures the units consumed either on three phase circuits or on two phase circuits with a common return. It would also measure the units consumed in a three wire direct current system. Fig. 99. Two meters sufficient when there is a common return. Let LMNR be the voltage tetrahedron, the magnitude of which we suppose fixed, and any point, then OL + OM + ON + OR is a minimum when any two of the opposite edges subtend equal angles at 0. In this case the solid angles (see Chapter vin) formed by any three of the vectors OL, OM, ON and OR are equal, and hence When the currents in a star load are equal the sum of the voltages to the centre is a mini- mum. . i.e. the currents in the four arms of the star load are equal. It is interesting to note that the equations for the electrical methods of measuring power in two and three phase circuits, when interpreted geometrically, give rise to numerous theorems connected with the trigonometry of triangles and tetrahedrons. 172 260 ALTERNATING CURRENT THEORY [CH. XII For example, if ABC be the voltage triangle for a three phase system and D, E and F are the middle points of the sides, then if R be the resistance (non-inductive) of each arm of a mesh load, the watts expended in it are -^ (V\ 2 + F a 2 .3 + F 2 3tl ). Hence by the two wattmeter formula V^A, cos DAB + V 3 . 2 A 3 cosBCF= i (F 2 lt2 + F 2 2 . 8 + F 2 3>1 ) or interpreting trigonometrically 2(c.AD cos D4-B + a . CFcos BCF) = a 2 + 6 2 4 c 2 , a known theorem. Similarly, by taking the three wattmeter formula and noting that the star load can be negligible, we get OAcoa(OAG) + OBcos( OBG) + OCcos( OCG) = 3.AGcos, where ABC is an equilateral triangle, G its centre, any point within it and any angle. The proper signs to be prefixed to the angles OAG, etc. depend on the position of 0. For example, if be inside the triangle AGE where E is the middle point of AC, then the positive sign must be taken in the first two terms and the negative in the third. CHAPTER XIII. Conversion of a two phase system to a three phase system. Conversion of a three phase system to a two phase system. The power factor of a three phase system. Wattmeter method of finding cos 7 * and ^ be the effective currents in 266 ALTERNATING CURRENT THEORY [CH. them. Then, if cos l5 cos c 2 and cos $ 3 be the power factors of the three arms, we have W, W 2 W cos <= > OOB-.-r-,, COS ( = =-. * 1.2-*3 The power factor (cos ) of the load may be defined as the ratio of the true watts to the sum of the apparent watts, and hence WI + TT.+TT, ' 2.sA~t~ ^3.1-^2+ ' 1.2-*3 It therefore lies in value between the greatest and the least values of the power factors of the arms. When the load is balanced, T + TFo + TT W cos we must therefore measure the power expended in the load by the two wattmeter method; if a star- box is available one wattmeter will be sufficient. We must also measure the pressure between two mains and the current in either of them. For a star connected load we find in a similar manner ** s *E 1 When the load is balanced we have T^ + Fa + TTs W cos d> = = - - - = VSF^^ JZVA When only one wattmeter is available, the following approxi- mate method of finding the power factor is sometimes used. Let the P. D.'s and currents of the balanced load be represented by the lines drawn in Figf. 101. We have made Wattmeter / method of finding the assumption that the current and P.O. vectors can be represented by lines drawn in one plane. This assumption is true when the currents and potential differ- ences follow the harmonic law. XIII] BALANCED LOAD The P.D. vectors are always in one plane, for 267 and the current vectors are also in one plane since but all the vectors would only be in one plane when a linear relation connected every three of them (Chapter vm). Let < be the angle be- tween 7 3 and F 1>2 and also between 7 2 and F 3-1 . Since the load is balanced, the angle between 7 2 and 7 3 is 120 de- grees and that between 7 3 and A l is 30 degrees, for A l is equal to the resultant of 7 3 and To. Fig. 101. P.IX'S and currents on a three phase inductive mesh load when balanced. Put the ampere coil of the wattmeter in the main 1 and connect the volt coil first between 1 and 2 and then between 1 and 3 ; let W l and W 2 be the readings in the two cases, then where fa is the angle between F 1>2 and A l in Fig. 101. Since the angle between 7 3 and F 1<2 is and between 7 3 and A! is 30 degrees, therefore fa = 30 - fa Hence W l = V l .^A 1 cos (30 - <) .................. (1). If we now disconnect the terminal of the volt coil from No. 2 main and connect it with No. 3, the reading W on the watt- meter will be where fa is the angle between F 1>3 and A lt Now fa + fa is the angle between F 1>2 and F 1<3 and is therefore equal to 60 degrees. Thus F 2 = F 1 . 3 ud 1 cos(30 + (^) (2). 268 ALTERNATING CURRENT THEORY [CH. By adding (1) and (2) and noting that F 1>2 and F 2 . 3 both equal "Pi. 3 , we get = the power expended on the balanced load. This method of measuring the power expended on a balanced load is sometimes used in practice and it will be seen from formula (b) given below that it is applicable when the current and the electromotive force waves do not follow the harmonic law. Dividing (1) by (2) we get W, _ cos (30- 0) TF 2 ~cos(30+4>) _ V3 + tan (/> ~ V3 - tan (f) ' and therefore tan = V3 ^ W i -f- rr 2 w, + w, s = This method gives the value of cos (/> by means of two watt- meter readings when the waves follow the harmonic law. In other cases the formula is only approximate. We shall now consider the theory of the phase indicator, which is an instrument to measure the power factor of Phase indicator. a balanced load on a polyphase system. The scale of this instrument may be graduated by making its readings equal the phase differences calculated from the readings of a wattmeter, an ammeter and a voltmeter when they are all connected in the proper manner with a balanced load the power factor of which can be varied ; and this is the method adopted in practice. It is instructive, however, to consider the theoretical basis for the design of instruments of this class, as this enables us to find out their limitations and shows how some of their defects may be remedied. Let us take the case of a three phase instrument. We saw on page 236 that the instantaneous value w of the total power taken by a three phase load is given by w = v^i 4- v^a. 2 + v s a s . I XIII] PHASE INDICATOR 269 If W be the mean value of w, then when the load is balanced, we have, from symmetry, = V3 V lfZ A l cos(j) (a), where cos is the power factor of the balanced load and < is the phase difference between v l and a^. We get in a similar manner, from formula (1) page 236, that W = FLS^-I cos fa + V SmZ A 3 cos fa', where fa is the phase difference between v 1-2 and a x and fa' is the phase difference between v Si2 and 3 . Similarly we have W = F, 3 .4<> cos < 2 + F! 3 ^-! cos fa. If the load be balanced, we have, by symmetry, F le s-4.1 cos fa' = F 3> 2^3 cos0 3 ', and thus we have By equating (a) and (b) we get, since F 1>2 and F 1>8 are equal on a balanced load, cos fa + cos fa' = V3 cos (c). Since we always have "Pi.2 ^2.3 and F 3 .! form a triangle and when the sides are all equal F 1-2 and F 3il are inclined to one another at an angle of 120. We shall now make the assumption that v 1>3 , v s<1 and a x follow the harmonic law so that the three vectors F 1-2 , F 3-1 and A t are in one plane. Since fa is the angle between F 1<2 and A-^ and 180 fa is the angle between F 3-1 and A l and the angle between F 1-2 and F 3tl is 120, we must have 180 -fa- fa = I 20, and therefore fa + fa' = 60. Substituting 60 - fa for fa' in (c) we get cos fa + cos (60 fa) = V3 cos 0. Therefore 2 cos 30 cos (30 - fa) = V3 cos fa and thus fa = 30 - and fa = 30 + (/>] We thus see that if we make the assumption that the currents and the potential differences follow the harmonic law, the angle 270 ALTERNATING CURRENT THEORY [CH. between F 1-2 and A^ is 30 2 and the current in the other fixed coil is in phase with v 3tl . Hence, if the time be reckoned from the instant when the potential differ- ence between the mains 1 and 2 is a maximum, the current in the fixed coil between 1 and 2, and therefore the strength of the mag- netic field in the direction OA due to this current, is proportional to cos o)t and we shall denote it by h^ cos cot. Similarly the strength of the field in the direction OB due to the current in the other fixed coil may be denoted by Ti^ cos (cot i/r), where ty is the phase difference between v lmZ and v Stl . Let the angle AOB in Fig. 102 be denoted by a, and let OM bisect the angle A OB. Then, if h x and h y be the components of the resultant magnetic field due to the currents in the two fixed coils perpendicular to and along OM respectively, we have cos cot sin ~ + - ^ cos (cot sin - ~ 2/^j sin - cos -|- cos ( cot -^ XIIl] THEORY OF INDICATOR 271 and h y = h^ cos cot cos 9 ^1 cos (cot ty) cos - a . ">r . cos ; sm -- sin \|r\ - -|- j . We see that when a equals A/T, /^ and h y are the projections of a line of length 2Aj sin ^ cos 5 or /^ sin a, which revolves with a con- stant angular velocity co. If, therefore, the angle AOB is made equal to i/r, the resultant magnetic field due to the currents in the fixed coils is of constant strength and rotates with uniform angular velocity. Since the indicator is only used to find the power factor on balanced loads, i/r is constant and equals 120 degrees. Let us suppose that the angle AOB is made equal to this, so that we always have a pure rotating field. The strength h of the magnetic field resolved along OP is given by h = h x sin <' + h y cos ' = A! sin a sin f (' + -= cot] , and since a equals 120 we get .(e). Now the moveable coil carries an alternating current which is in phase with a lf We can therefore suppose that it is replaced by a magnet the polarity of which alternates with the current. If the coil be small, the action of this magnet may be imitated exactly by two small permanent magnets which rotate in opposite directions with equal angular velocities o> and the moments of which are half the maximum moment of the alternating current magnet. To prove this, consider two small magnets each of moment \H rotating side by side in opposite directions with equal angular velocities o>, and let the time be reckoned from the instant when the magnetic fields due to each are pointing in the same direction. Then at the time t, the component field in this direction at unit distance is H cos cot + H cos ( wt), that is, 2H cos cot, and the component field perpendicular to this direction is \ H sin at + %H sin (- wt) 272 ALTERNATING CURRENT THEORY [CH. or zero (see page 13). We see therefore that they are equivalent to the single alternating current magnet which, at unit distance, produces the field 2H cos cot in the direction of its axis. Now the action of the rotating field on the equivalent per- manent magnet which is rotating in the same direction as itself is to produce a couple which acts so as to diminish the angle between this magnet and the direction of the rotating field. The mean couple produced by the rotating field on the equivalent magnet which rotates in the other direction is zero. Thus the moveable coil is in a position of equilibrium when the direction of the rota- ting field due to the fixed coils coincides with the direction of the equivalent magnet which rotates in the same direction. Let OP (Fig. 102) be the position of the axis of the moveable coil when in equilibrium. At the instant when OP is the direc- tion of the rotating field, h, which acts along this direction, has its maximum value and both the equivalent permanent magnets are also pointing in this direction. Thus h is in phase with a l5 and therefore from (e) the phase difference between a x and v ltZ is 30 <'. But, by the equation (d), this angle is equal to 30 when the load is balanced, and thus ' must be equal to . If therefore the scale of the instrument be divided into degrees, the cosine of the reading will give the power factor, when the potential differences and the currents follow the harmonic law. If the axes of the fixed coils be not placed so as to include an angle of 120 degrees, then in general, both the angular velocity and the magnitude of the rotating field due to the fixed coils vary at different instants even when the applied waves are sine shaped. It is found however that the pointer gives a definite reading for a load of given power factor, and so, with the aid of a wattmeter, an ammeter and a voltmeter the scale can be marked. In Fig. 103 are shown the connections of the Heap phase indicator which is constructed on the above principle. K and R are resistances in series with the fixed coils, and the moveable coil carries the whole current. In the Heap phase indicators for high pressure working the fixed coils are the secondaries of transformers the primaries of which are connected between 2 and 1 and between 1 and 3. The moveable coil sometimes carries only a fraction of the current in the main 1, and sometimes it is the secondary of a XIII] PHASE INDICATOR 273 transformer whose primary consists of one or two turns placed in series with the main 1. It is thus possible to arrange the indicator in the circuit so that only low pressure wires come to the instrument. 3 1 2 Fig. 103. Connections of Heap Phase Indicator for three phase working. Let us suppose that the phase indicator is connected in the usual manner with the three mains supplying power to the windings of the armature of a three phase synchronous motor (see Vol. II, Chap. v). Then, it will be proved in Vol. II that, when the potential differences and the currents follow the har- monic law, we can alter the power factor by varying the direct current excitation required by the machine, although the me- chanical power given out by the motor is maintained at a constant value. For a particular excitation we can show that the power factor is unity. For excitations less than this the current in a winding lags behind the potential difference applied at the terminals of the winding, and for excitations greater than this it leads the potential difference, the winding of one phase acting like a condenser with a non-inductive resistance in series with it. Thus, as we increase the excitation of the motor, the pointer of jhe phase indicator will move from one side of the scale to the R. i. 18 274 ALTERNATING CURRENT THEORY [CH. other, passing through the position where it reads zero, and where consequently the power factor is unity. In practice the waves of potential difference and current are not sine shaped, but the pointer moves in a similar manner from one side of the scale to the other as the excitation varies from a low to a high value, and thus for one particular excitation the instrument apparently indicates zero phase difference and therefore a power factor of unity. Now we have seen on p. 146 that when the power factor is unity the volt and ampere waves are similar, and if the instrument reads correctly in this case, it would follow that it is always possible to adjust the excitation of a synchronous motor, which is simply an ordinary alternator running as a motor, until the shape of the waves of current supplying the motor were exactly the same as the shape of the potential difference waves between the mains. As a matter of fact, however, a zero reading of a phase indicator only corresponds to a power factor of unity when the waves are sine shaped. In other cases there is no simple relation between the reading of the instrument and the power factor, although when the instrument is always used in the same supply circuit its readings are of value to the engineer. If we make the shunt circuit of an electromagnetic watt- Meter for the meter, which is constructed on the dynamometer wattless current. p r j nc ipi e arK j nas no mutual inductance between its coils in the zero position, very inductive, then the current A l in it will lag in phase by nearly ninety degrees behind the applied potential difference V. Such an instrument may be used to measure the wattless current (see page 159). The reading of the instru- ment will be proportional to AA l cos a, where A is the main current and a is the phase difference between A and A^. If the power factor of the load be cos <, A will lag behind V by (/> degrees and A l lags behind V by ninety degrees. When the applied potential difference and the currents follow the harmonic law, the three vectors representing F", A and A lie in one plane (see Chapter Vin), and therefore a is 90 <. Since AI is proportional to F, the reading of the instrument is proportional to VA cos (90 ), that is, to VA sin . Thus when V is known we can find the wattless current A sin . XIIl] INDUCTION METER 275 If one or more of the waves do not follow the harmonic law, the three vectors form a solid angle, and + a is therefore greater than ninety degrees. Let us suppose that $ + a = 90 + x and that x is small. Then the reading of the instrument divided by V will be proportional to A cos {90 -(-x)} } that is, to A sin c/> x . A cos t - a), / 2 ' = / 2 cos (at - ), ij = n^I-L cos (at tti), it = ^2^/2 cos (cot ft), *V = n \ k I i cos (cot a?), i 2 = n 2 l 2 I 2 cos (cot {3. 2 ), XIII] THEORY OF METER 277 The values of l l} 1%, ..., a lt ft, ..., depend on the values of the inductance and resistance of the paths of the eddy currents. Their values also depend on the frequency. Let the phase difference between the currents in the magnet windings be 7, then we shall have y = p-a = P 1 -a 1 = .... Let also ^ be the phase difference between ^ and // ; then S x will also be equal to the phase difference between i z and / 2 '- Thus we have Let also Substituting in (1) we get g = ?i 2 ?z 1 / 2 / 1 cos (cot ft) {&!?! cos (cot o^) + & 2 2 cos (cot 2 ) + '. n-^n^Ii / 2 cos (cot a) [^ ^ cos (cot ft) + & 2 ^ 2 cos (w^ ft) + . . . If G denote the mean torque, we have G = 2 * - [&i ^ cos (! 0) -f & 2 2 cos ( 2 ft) + . . . - k-h cos (ft - a) - k 2 l z cos (ft - a) + . . .} ilJ^sfo g^ ^ sin ^ - + ...h = n l n^ l2 sn 7 sn The values of 8 lt S 2 ... depend on the conductivity of the metal disc. If the temperature and the frequency remain constant, the expression k^ sin 8 l + k 2 l 2 siu S 2 + ... will be a constant. We thus see that, when we make the sine curve assumption, G will be proportional to I^zSmy. Now, since we can neglect the actions of the small eddy currents, /j is proportional to the voltage V applied to the load and 7 2 is proportional to the effective value A of the load current. The accelerating torque G is therefore proportional to VA sin 7. The angle 7 is the phase difference between the currents in the windings of the shunt and series magnets. On a non-inductive load it is nearly ninety degrees; and since we are assuming that all 278 ALTERNATING CURRENT THEORY [CH. the currents follow the harmonic law, y will equal 90 when cos is the power factor of the load. It therefore follows that the resultant accelerating torque is proportional to VA cos c/>, that is, to the power being expended on the load. The retarding torque is produced by the eddy currents gene- rated in the rotating disc by the field due to permanent magnets. These magnets are generally (7-shaped, and the circumference of the disc rotates in the air-gap. The retarding torque will therefore be proportional to the angular velocity of the spindle. It is to be noted that, since the eddy currents due to the series and shunt magnets are alternating, the mean effect of the permanent magnets on them is zero. The shunt and series magnets also have on the average no effect on the eddy currents due to the permanent mag- nets. When the motion is steady, the accelerating torque is equal to the retarding torque, and thus the power being expended on the load is proportional to the angular velocity of the spindle. The spindle is connected with the counting mechanism by worm gearing and so a record is made of the energy expended. Instead of a disc, a light hollow cylinder of aluminium is some- times used, the shunt and series magnets being placed near to one another and facing the circumference of the cylinder. As before the retarding torque is produced by the eddy currents due to per- manent magnets. The theory of the action of this instrument is practically identical with that of the disc meter. In meters which have an aluminium disc, a compound magnet made up of m-shaped iron stampings is often used instead of two separate magnets for the shunt and series windings. The shunt coil is wound on the middle limb and the series coils on one or both of the outer limbs. The compound magnet is placed near the circumference of the aluminium disc, and strips of iron are fixed underneath the disc so as to reduce the reluctance of the magnetic circuits of the compound magnet. When there is a current in the series coil, the mean values of the magnetic fluxes in the two outer limbs are unequal, as the magnetising force of the shunt coil increases the resultant magnetising force in one limb and diminishes it in the other. This effect alters the ratio of the angular velocity of the disc to the effective current in the series coils, and tends to make the meter read inaccurately. It can POLYPHASE METERS 279 be neutralised by placing a few turns of series winding on the middle limb. This type of meter can easily be adapted to register the energy expended in a two or three phase load. All that we need do is to apply the two wattmeter method (see Chapters XI and xil) and combine the two watt-hour meters into one instrument. Two re- shaped magnets are arranged in this case to act on the same aluminium disc. They are placed facing the top surface on opposite sides of the centre of the disc. If they are suitably wound and connected with the mains as in the two wattmeter method, the sum of the accelerating torques, that is, the resultant torque, will be proportional to the total power being expended on the load. In other meters of the induction type, we have a metal disc placed in a rotating magnetic field. The principle of these meters will easily be understood from the theory of the induction motor developed in Volume II. A serious objection to the use of meters, the action of which depends on the generation of eddy currents in masses of metal, is their large temperature coefficient. The change of the resistivity of copper and aluminium, the metals usually employed, is about O'-i per cent, per degree centigrade. In the expression given above for the accelerating torque it will be seen that 1 ,/ 2 -- all depend on the resistance of the paths of the eddy currents. They therefore vary with the temperature. Unless some compensating device be employed, the meter will only read correctly at a particular tem- perature. It will also only read correctly at a given frequency. REFERENCES. C. F. SCOTT, The Electrician, Vol. 32, p. 640, ' Polyphase Transmission,' 1894. "W. H. BROWNE, Trans, of the Amer. Inst. El. Eng., Vol. 18, p. 475, 'Power- Factor Indicators,' 1901. The Electrician, Vol. 51, p. 168, 'A Direct-reading Power-factor Indicator,' 1903. K. EDGCUMBE and F. PUNGA, J. of the I. E. E. Vol. 33, ' Direct-reading Measur- ing Instruments for Switchboard Use.' March 1904. CHAUMAT, Bulletin de la Societe Internationale des Electriciens, Vol. 4, p. 173, ' filectrogoniometre de MM. Grammont et Routin.' (Phase indicating device.) March 1904. CHAPTER XIV. Rotating magnetic fields. Resultant field due to n equal and symmetrically placed poles supplied with n phase currents. Equal poles unevenly spaced; phase differences of magnetic vectors equal to their angular distances apart. General case. Properties of rotating and alternating magnetic fields. Magnetic field in the air-gap of polyphase machines. Rotating field in the air-gap of a polyphase induction motor. Gliding magnetic fields. Rotating magnetic fields when the currents are not sine shaped. Rotating magnetic field producing a constant effective E.M.F. in a search coil placed with its plane perpendicular to the field. Extension to three phase theory. Arn6's phase indicator. References. MAGNETIC forces are compounded by the parallelogram law, and hence we may apply statical constructions in order to find their resultant. For example, suppose that the magnetic forces at the point (Fig. 104) are represented in magnitude and direction by the lines OA l} OA 2) OA 3 and OA 4 . Then, if G be the centre Fig. 104. Resultant of OA lt OA Z , OA 3 and CH. XIV] ROTATING MAGNETIC FIELDS 281 of gravity of equal masses placed at A 1} A 2 , A & and A+, the resultant of the forces will be represented in magnitude by 4 . OG, and in direction by OG. If there had been n forces, then the resultant would have been equal to n . OG. To find G, we bisect A 1 A 2 in g l} and then make 9i92 = ^g^A 3 , g 2 g 3 = ig 2 A 4 , etc. This construction is very simple in practice. It is easy to see that it is true whether the lines are in one plane or not. The necessary and sufficient condition that the forces are in equili- brium is that the centre of gravity of equal masses placed at the extremities of the lines representing the forces coincides with 0. We have seen that this theorem also holds for alternating current vectors, and we have already made use of it in the chapters on two and three phase theory. In a rotating magnetic field the direction of the magnetic Rotating magnetic force is continually revolving. If we move a strong permanent magnet round a small compass needle, keeping the same end of the magnet always pointing to the needle, we produce a rotating magnetic field at the centre of the small compass. The needle at any instant points out the direction of the magnetic field, and its angular velocity measures the velocity of rotation of the field. If the core of an electromagnet be made up of thin strips of soft iron insulated from one another by means of shellac varnish, or by paper pasted on one side of each strip, we get an alternating current magnet. When the windings of such a magnet are supplied with alternating currents, the polarity of the ends of the magnet alter- nates with the same frequency as the alternating currents and a varying magnetic field is produced in the neighbourhood. The magnet is usually made with a straight core. The iron in the core needs to be laminated, for otherwise its temperature would rise excessively owing to the heat developed by the eddy currents that would be induced in it. A simple way of making such a magnet is to take for the core a cylindrical bundle of insulated iron wires, the lengths of which are parallel to the axis of the cylinder, and then to wind insulated copper wire round the cylinder to carry the alternating current required to magnetise the core. 282 ALTERNATING CURRENT THEORY [CH. Now place two alternating current magnets (Fig. 105) at right angles to one another, with axes pointing in the direc- tions AOA' and BOB' respec- tively. If the magnetic force at due to the first be Hcoscot and that due to the second be (7T\ mt ~ 2 J ' and if (Fig, 105) OQ = Hcos cot, and OP = Hcos tot - ?) = then, if OR be the resultant of OP and OQ, we have and therefore B' Fig. 105. Pure Rotating Field produced by two alternating fields in quadrature. OR = H a constant, and the angle OR makes with OA will be cot, so that OR will rotate with constant angular velocity to. In this case, then, we obtain a pure rotating magnetic field, that is, one which is not only constant in magnitude but rotates with constant angular velocity. If we used two solenoids without iron which were supplied with sine shaped current waves from a two phase machine, we could produce this field. The rotating fields produced in practice, although often constant in magnitude, rarely rotate with constant angular velocity. For the present we will consider the case of pure rotating fields, and consequently we make the supposition that the alternating currents inducing the fields are sine shaped. Consider the case of the field produced by n equal poles arranged evenly round a circle, the currents pro- O?A ducing the poles differing in phase by - - degrees. Resultant field due to n equal and symmetrically placed poles sup- plied with n phase currents. n Let the strength of the field at 0, due to the pole Pj, be NCOS cot in the direction Op lt xiv] ELLIPTIC FIELD imilarly let the field at 0, due to P 2 , be H cos t&t direction Op. 2 , and so on for the remaining poles. With centre and radius equal to H de- scribe a circle, and let a circle TT of radius - roll round this circle times a second, the 2-7T centre of the moving circle lying on OP l when t is zero. Then the intercepts Op,, Op 2 , ...... will represent H cos cot, 360 283 - ) in the n J 360\ H cos ( cot -- 1 ...... , Fig. 106. Kesultant Field of n phase currents equals that is the values of the component strengths of the fields. It is easily shown that p lt p 2 ,... are fixed points on the moving circle. Since the angles^O/Jo, p. 2 0p 3 , ... are all equal, p^p 2 ... p n is a regular polygon. If C be the centre of the rolling circle, C is always the centre of gravity of equal masses placed at p l} p 2 , .... Therefore the resultant magne- tic force is represented in mag- nitude and direction by n . 00. Hence the resulting field is a pure rotating one of magnitude As in the last case, suppose that a circle rolls inside another of double its Equal poles un- diameter. Then evenly spaced; phase differences , -(!*) H Thus G lies on an ellipse whose centre is and axes H + 2c and H 2c respectively. Since the field is represented in direction and magnitude by n . OG, we see that we do not get a pure rotating field in this case. If G^ be the angular velocity of OG about 0, then OG _dy x dx y ~dt'OG~dt'OG' and thus OG 2 . co, = (~ - d*\ &>. It is proved in Thomson and Tait's Natural Philosophy, Part I, 66, that a point, whose motion is the resultant of any number of simple harmonic motions of the same period in any directions and with any phases, moves on the circumference of an ellipse, and that its radius vector sweeps out equal areas in equal times. Hence, if H be the strength of the field at a point due to any number of magnetic poles of any strengths with any directions, and with any phase differences between them, provided they follow the sine law, then H*! is the angular velocity with which the resultant field revolves. The important case in practice is when the vectors are all in one plane. Let us suppose that the direction of the alternating XIV] GENERAL THEOREM 285 field ITjCos (at oO makes an angle ^ with the axis of x. Then, fx and y be the coordinates of the extremity of the vector repre- nting the resultant field, we have x = 2 H cos (cot a.) cos ty = a cos cot + b sin cot, and y = ' H cos (o> a) sin ty = ccos cot+ d sin cot, where a, b, c and d are quantities which do not vary with the time. B Solving these equations for cos cot and sin cot we get dx- by COS t= da^Vc> ex ay and sm cot = -7 --, . co ad Hence, we get (dx by)- + (ex ay)' 2 = (ad be) 2 , which shows that the locus of the extremity of the vector repre- senting the resultant field is an ellipse. In the special case when a = b and c = d the ellipse becomes a straight line, and thus the resultant field is purely alternating. Again, when a = d and 6 = + c the ellipse becomes a circle and the strength of the resultant field is constant at every instant. Also, since d6 dy dx r ^'di = x di~ y dt' therefore r 2 .co l = co (ad be), where o^ is the angular velocity of the vector of the resultant field; and thus, if r is constant, co^ is also constant. If the rotating field be produced by currents of different frequencies, we get all manner of varying fields. The curves that would be described by the extremity of the vector representing the instantaneous value of the resultant magnetic force in some of these cases are given in treatises on the theory of sound under 286 ALTERNATING CURRENT THEORY [CH. the head of Lissajous's figures, and various mechanical devices are described for drawing them. We shall now consider a few of the properties of rotating and alternating magnetic fields. By a pure rotating Properties of ro- . , , tating and alter- held we mean one whose strength and angular fiefds g magnetic velocity are both constant, and by an alternating field we mean one whose direction is constant but whose strength is a periodic function of the time obeying the sine law. When we consider more than one field, all the fields will be supposed to be parallel to one plane and the frequency both of the rotations of the rotating fields and of the alternations of the alternating fields will be supposed to have the same value. We will suppose the fields represented by rotating and alternating vectors, which may be drawn in one plane. The following theorems will be found useful in practice. (1) Two vectors rotating in the same direction in one plane are equivalent to a single rotating vector. Let Op and Oq be the two vectors. Since they rotate with the same angular velocity, the angle pOq remains constant. Construct a parallelogram on Op and Oq as adjacent sides, and let OR be the diagonal. This represents the resultant field, which will obviously rotate with the same angular velocity as Op and Oq. Hence two vectors rotating in the same direction compound into a single rotating vector. (2) Two equal vectors rotating in opposite directions in one plane are equivalent to a single alternating vector, and conversely a single alternating vector is equivalent to two equal vectors rotating in opposite directions. Let Op and Oq be the two equal vectors. At any instant their resultant is 2 . Or, where r is the middle point of pq. Also since Or is perpendicular to pq, and Op and Oq are rotating with equal angular velocities in opposite directions, Or is fixed in direction. Thus two equal vectors rotating in opposite directions compound into a pure alternating vector the amplitude of which equals 2 . Op. Similarly an alternating vector whose amplitude is Or may be replaced by two equal rotating vectors whose magnitudes are each . equal to \ . Or. MAGNETIC VECTORS 287 (3) Two unequal vectors rotating in opposite directions are equivalent to an alternating vector and a rotating vector, and, con- versely, an alternating vector and a rotating vector are equivalent to two unequal vectors rotating in opposite directions. Let Op and Oq be the mag- nitudes of the two vectors. Make Oq equal to Op. Then, since Oq is equivalent to the sum of two rotating fields, in phase with one another, whose magnitudes are Oq and q'q respectively, and since by (2) Op and Oq' com- pound into a vector alternating along the line Oa (Fig. 108), which bisects the angle pOq, and having the amplitude 2 . Op, Fig ' 108 ' Two unequal vectors ro " 1 . tating in opposite directions are equi- we see that the given vectors valent to an alternating and a rotating compound into an alternating vector. vector of amplitude 2 . Op and a rotating vector whose magnitude is Oq Op. (4) Two unequal alternating vectors can, in general, be replaced by a simple alternating vector and a rotating vector. By (2) we can replace the alternating vectors whose directions are Or and Or' (Fig. 109) by four rotating vectors Op, Oq and Op, Oq of which Op and Op' rotate in one direction, and Oq and Oq' in the other, and where Op is half the maximum magnitude of Or and Op' is half the maximum magnitude of Or'. Hence, by (1), we can replace Op and Op by OP, and Oq and Oq' by OQ where OP and OQ are vectors rotating in opposite directions. If OP and OQ are equal to one another, the resultant field is, by (2), a purely alternating one. If either OP or OQ be zero, the field is a purely rotating one. In all other cases we see, by (3), that it can be represented by an alternating vector and a rotating vector. \Yhen OP equals OQ the resultant field is purely alternating, 288 ALTERNATING CURRENT THEORY [CH. and it is easy to show that in this case the two component fields must be in phase with one another. To get a purely rotating Fig. 109. Two alternating vectors can be replaced by an alternating and a rotating vector. field we must have either OP or OQ equal to zero. Hence either Op = Op' or Oq= Oq' '; in either case the amplitudes of the two alternating fields must be equal in magnitude. In the first case (Fig. 110) Op and Op' must be in the same straight line but Fig. 110. Op and Op' are in the same line, when the alternating vectors produce a pure rotating field. XIV] PURE ROTATING FIELD 289 pointing in opposite ways, so that the angle between Op and Op' is 7T. Hence cf> + a = TT, where is the angle between the directions of the two alternating vectors and a is their phase difference. For if (Fig. 110) Op be in the same line as Op', then + = rOr' - rOp + r'Op' = r'6p + r'Op' = -jr. We can prove this important theorem analytically as follows. Let the strengths of the fields in the directions sr and s'r' (Fig. Ill) be given by H l cos cot and H. 2 cos (cot a) respectively, and let A B Fig. 111. When the angle rOr is the supplement of the phase difference between the fields alternating in the directions sr and s'r' we get a pure rotating field. be the angle rOr. If sr make an angle with AB (Fig. Ill) the magnetic force h along OA at the time t is given by h = H l cos cot cos 6 + H 2 cos (cot a.) cos (c/> 0) = {If} cos 6 + H. 2 cos a cos (c/> - 6)} cos a> + ^T 2 sin a cos ( - 0) sin a> = #" sin (cot + 7), where H 2 = H, 2 cos 2 + H 2 2 cos 2 ((/> - 0) + 2# a # 2 cos a cos <9 cos (0 - 6\ and tan 7 = HI C S ^ + H * CQS a CQS (( ^ ~ ^ ^/osin acos( 0) R. i. 19 290 ALTERNATING CURRENT THEORY [CH. Now if h be independent of 6, that is, if the amplitude of the alternating field in every direction at (Fig. Ill) be the same, the value of H must be the same when 6 is and when it is - . Hence we must have H^ equal to H 2 and H 2 = H? {cos 2 9 + cos 2 ( - 9) + 2 cos a cos 9 cos (< - 9)}. Now cos 2 9 = sin 2 + cos 2 9 sin 2 < = sin 2 + cos (< + 0) cos ( - 0). Therefore # 2 = H* [sin 2 + cos (0 - 0) {cos (< + (9) + cos ( - 0) + 2 cos a cos 9\] = #! 2 {sin 2 (f> + 2 cos (< - 0) cos (cos < -I- cos a)). Thus, when cos + cos a is zero, the value of H is ^ sin <, and is therefore the same for all values of 9. In this case we must have c + a a A 2 cos ^ cos r -^ = 0, and therefore, since a is not greater than TT, + a must be equal to TT. In this case h = H! sin < sin (a)t + $ 6) and 7 = $ 9. (5) ^.m/ number of alternating vectors are in general equivalent to a simple alternating vector and a rotating vector. We can replace every alternating vector of amplitude Or by two equal vectors Op and Oq rotating in opposite directions. All the component vectors like Op rotating in the same direction can be replaced by a single rotating vector OP. Similarly the Oq components compound into OQ. Hence the theorem follows from (3). When OP and OQ are equal, the rssultant field is a purely alternating one. When either OP or OQ is zero, the resultant field is a purely rotating one. If OP is zero, the resultant of all the Op components is zero, and thus a closed polygon can be constructed whose sides are equal and parallel to all the Op components, and a similar theorem holds when OQ is zero. XIV] PROPERTIES OF VECTORS 291 (6) If p i} p 2 , ... and q l ,q. 2) ... be any vectors and r ^ 8 be the angle between p r and q s , then 22jo r ? cos r ,g = PQ cos 4>, where P is the resultant of all the p vectors, Q is the resultant of all the q vectors, and O is the angle between P and Q. Resolve all the p vectors along q lf then by projections p l cos 2 . ! + . . . + p n cos <. j = P cos ^j. Hence q 1 2p r cos r . l = Pq l cos ^ . Similarly q^p r cos r . 2 = Pq^ cos ^ 2 Therefore ^p r q* cos < r . g But Sgcos^ is the sum of the projections of all the q components upon P, and it therefore equals Qcos. Therefore 22/) r . (7) //' jo be the amplitude of an alternating vector and q be a rotating one, the mean value of pqcoscf) is ^pqcosS, where 8 is the angle between the directions of p and q when p has its maximum value. We can replace the alternating vector p by two vectors, rotating in opposite directions, the magnitudes of which are each equal to ^p. The mean value of the product of q and the com- ponent vector ^p rotating in the same direction and the cosine of the angle between them is ^pq cos &, because the angle between them is always equal to B. The mean value of the corresponding product for the other vector is zero, since, for a complete revolution, the mean value of cos (cot + S) is zero. (8) If p and q be two alternating vectors whose directions are inclined to one another at an angle , then the mean value of pqcos is 22 {cos(0 + /3 - a) + cos(< + a- )}, where a (B is the phase difference between the two vectors. The angle between p and q is always <, so that we have only to find the mean value of the instantaneous value of the product of the two vectors. This can be done by (7) for we can replace p >y two vectors each equal to ^p rotating in opposite directions. 192 292 ALTERNATING CURRENT THEORY [CH. The mean value of one of these components multiplied by q and the cosine of the angle between them is by the preceding theorem and 8 equals c/> + /3 a or + /3 depending on which component we take. The sum of the two will obviously give the mean value of pqcoscf). This mean value, by adding the cosines, may be written in the shape ^ cos (/> cos (a yS). (9) If p and q be vectors which rotate with different angular velocities, the mean value of pq cos (f> is zero. Suppose that they are rotating in the same direction with angular velocities ^ and a> 2 ; then in the time t where (a) l w 2 )t equals 2?r, the angle will have increased from to 2?r, and therefore its mean value taken over the time t will be zero. Hence, if the mean value be taken over a time ^ which is large compared with t, it will be zero if ^ be a multiple of t and will be very small in other cases. If the vectors rotate in opposite directions, then the time t which the angle (/> takes to increase from to 2?r is given by (! + o) 2 ) t == 2?r and the mean value over a time t l} large compared with t, will be zero or very nearly zero. When we are dealing with the magnetic fields in the air-gaps Magnetic field in between the rotating and the stationary parts (the p^pha^ rotor and the stator) of several kinds of polyphase machines. machines, a modification of the above method be- comes necessary. The following way of treating the problem is due to A. Potier. In the case considered, we have a hollow laminated cylinder of soft iron with slots along the interior, parallel to the axis, which carry evenly distributed windings for the polyphase currents. When a current flows in one of these windings, it divides the interior into p segments of North and p segments of South polarity. If I be the breadth of one of these segments, then XIV] POLYPHASE MACHINES 293 is the inner circumference of the cylinder. We shall suppose that the next winding is displaced from the first by a distance - on the circumference of the cylinder, where q is the number of phases. The rotor consists of a laminated iron cylinder axial with the stator and suitably wound with copper conductors. Like the stator it will have 2p poles, but the phases of the currents in it may have any values. The air-gap between the two cylinders may be as small as half a millimetre. Let us consider the case of a polyphase induction motor Rotating field in (Vol. II, Chap. XII). We shall suppose that the p h o!y a pha g se p indu a c- windings of the stator are connected with the. poly- tion motor. phase supply mains, and that the windings of the rotor consist of closed coils so that the currents in them are due to induction only. When this kind of motor is running, the fre- quency of the alternating currents in the rotor is small when compared with the frequency of the currents in the stator. We will for the present neglect the magnetising effects of the rotor currents. Fig. 112. Field in the air-gap of a polyphase induction motor. Take some point (Fig. 112) on the inner surface of the stator and make where the points A lt A. 2 , A 3 ... are on the circumference and the distances between them are measured along the circumference. 294 ALTERNATING CURRENT THEORY [CH. If the radial magnetic force at be zero at the instant when t is zero, the magnitude of the first harmonic of the radial com- ponent of the field may be indicated by the wavy line in the figure, the force being positive from to A lf negative from A l to A z , etc. Consider the magnetic force perpendicular to the surface of the stator at a point P, where OP equals x, due to a single alternating current in one of the windings. We may evidently write this in the form f (t) $ (x) where f(t) is some function of the time, and (x) = $ (x + 1) = (x + 21) = etc. Suppose that the second winding is similar to the first and that it is displaced to the right to a distance - , and that the T current in it lags by an interval behind the current in the first winding. Then the radial component at P due to this winding will be Hence, when all the windings are excited by polyphase currents, we find that the radial force H at P is given by _( 9 _i)a-( ? -i) ...... a). T It is easily seen that, when t is' increased by and x by - , the value of H is not altered. If, therefore, we only examine T the field at intervals of time ^- , it will appear to glide round the XIV] FIELD IN THE AIR-GAP 295 air-gap, without altering in shape, with a linear velocity -^ . If 2p be the number of poles and r the radius of the rotor, 2?rr = 2pl, and hence the angular velocity of the field is - or - , where co is JL 2?r times the frequency of the alternating current. If f (t) = B sin cot and $ (#) = C sin TT y , then, substituting in (1) and summing by the ordinary trigono- metrical formula, we find Jf = BC cos I cot TT j ) (2). Hence, when the magnetic force due to the current follows the harmonic law and when the distribution of the flux round the air-gap is a sine function of the space, the resultant radial field is sine shaped and glides round the air-gap with constant speed. In this case also, if A be the effective current in each phase, the maximum value of the resultant magnetic field would equal that produced by a current | A in one phase, provided that the permeability of the iron were constant. When f(t) and (x) are not sine functions, they may be developed in two series of sine functions by Fourier's theorem. The field will thus be decomposed into a series of magnetic fields rotating with angular velocities - - , where m and n are integers. Some of these fields, also, may rotate in the reverse direction. We shall consider this case more fully in Vol. II, Chapters XII and xiv, where also the shape of the magnetic fields produced by various simple windings is considered. It is shown that the in- duced currents in the rotor windings produce a magnetic field which rotates in space with the same angular velocity as the magnetic field due to the polyphase currents in the stator windings. The magnetic field which is the resultant of the fields due to the currents in the stator and rotor windings respectively must rotate in space with the same angular velocity as its two components. Thus when we take into account the rotor currents, 296 ALTERNATING CURRENT THEORY [CH. we still have in the air-gap a magnetic field in which the distribution of the magnetic flux is always exactly the same at intervals of time T which differ by multiples of ^-. At any instant also we have p Zq segments of North and p segments of South polarity. This kind of magnetic field we shall call a gliding magnetic field. If one phase only of the stator windings be excited and the rotor be stationary, then we have as before p segments of North and p segments of South polarity. The polarity of these segments alternates with the frequency of the supply current, but the field is a stationary one, the lines dividing the segments being fixed in position and the magnetic force along these lines being always zero. When we are discussing the magnetic field in the air-gap of an alternating current machine, we shall refer to a field of this kind as an alternating field. Unlike the alternating fields con- sidered earlier in the chapter, the amplitude of the magnetic force is different at different points of the field. We will now consider what happens when gliding fields which Gliding magnetic follow the harmonic law are superposed on one fields - another. (1) Two fields gliding in the same direction. Time lag but no space lag. In this case TT TT ( ^ X \ , TT ( ^ X \ H = //! sin ( cot j- J + H 2 sin f cot a j- J R sin io)t P TT-J where R 2 = H, 2 + H* + 2H.H, cos COOL H 2 sin coa and tan coB = TT I2 l + JH 2 COS COOL Hence the resultant field glides in the same direction with the same velocity, and its magnitude and phase are given by the parallelogram construction. (2) Two fields, gliding in the same direction. Space lag but no time lag. GLIDING MAGNETIC FIELDS 297 We have H H l sin f tot j- } + H 2 sin I cot TT j J . H sin . / x-b\ in I cot TT j J Tra where R- = H-,- + H.? + 277, H 9 cos -? 2 Tra Trb and tan -=- = I TT -TT 7T<^ H^ + HoCos The resultant field therefore glides in the same direction with the same velocity, and its magnitude and phase can be got graphically by the parallelogram construction. (3) Two equal fields gliding in opposite directions are equi- valent to an alternating field. In this case, tx \ / x \ at- TT-J + a l j+H l sin (&>Z + TT 7 + 02) = 2fl, sin (a,t + "' + ^ cos ( J + ^) . The resultant field is therefore a purely alternating one, jET always being zero at the points given by where n is an integer. The greatest value of H is obviously (4) An alternating magnetic field whose maximum value is H is equivalent to two equal fields gliding in opposite directions whose maximum values are each ^H. This follows from the theorem that TT . x ti . f TTX\ i . f nrx\ R sin tot cos TT y = -- sin [ant + j- } + TT sm ( ^ J~ ) I 2i \ I J 2i \ I } The angular velocities of the fields are - and , where 2p is the number of poles. 298 ALTERNATING CURRENT THEORY [CH. (5) The resultant of any number of alternating and gliding fields of the same period is in general two fields gliding in opposite directions. This follows from (1), (2) and (4). It is to be noted however that we make the assumption that they are all sine shaped. If the two systems of currents represented by ( ^ \ (at TT j + ! ) V If J and i z = / 2 sin cot + TT + 2 x are superposed in the distributed conductors, then the heat gene- rated in the conductors will be that due to each system of currents /2 J 2 separately, for the mean value of (i L + i 2 ) 2 is -^-+-^-. We shall call ij a system of currents turning to the right. (6) The mean torque produced by a magnetic field turning to the left on a system of currents turning to the right is zero. The currents may be flowing, for example, in the windings of the rotor of an induction motor. The torque on any conductor will be proportional to the product . / , X \ / X \ sm I cat TT -j + a x 1 sin (cot + TT -= + a. 2 . \ I / \ I ) The mean value of this expression from t equal to to t equal to Tis l cos (ZTTJ + ,-,) ..................... (1). \ * / And the mean value of (1) from x equal to to x equal to 2pl is obviously zero, and thus the mean torque between the field and the system of currents is zero. This theorem is analogous to the theorems in the undulatory theory of optics regarding the non-interference of circular vibrations in opposite directions. Several other theorems might be adapted from optics, and it will be found that these theorems are useful when we come to the theory of asynchronous motors. The artifice of replacing a fixed alternating field by two fields rotating in opposite directions, used first by Fresnel in optics arid adapted by Ferraris to alter- nating current theory, is invaluable in this connection. XI Vl EFFECTIVE VALUES 299 Consider the case of two coils the axes of which intersect at some point 0. We will consider the field at in the plane determined by the two axes. Let h l} h, n0t ke t ^ ie va ^ ues at f tne magnetic forces along the axes Op and Oq respectively, and let the angle pOq equal a. Now if Op and Oq be equal to h-^ and h 2 and r be the middle point of pq, 2 Or will be the resultant force, and it will be seen that, as h^ and h 2 alter, the locus of r may be a complicated curve and the angular velocity of Or may vary in an erratic manner. Draw any line Oa in the plane pOq, and let the angle pOa equal 0. Then, if h be the resultant magnetic force resolved along this line, h = h l cos 6 + h. 2 cos (a 0). Therefore h- = hf cos 2 + /* 2 2 cos- (a - 0) -f 2A 1 /< 2 cos cos (a - 0). Now if the currents in the coils be adjusted until the effective or root mean square value of h^ equals the R.M.S. value of h z , and if capital letters denote the R.M.S. values, H 2 = H? {cos 2 + cos 2 (a - 0) -f 2 cos cos (a 0) cos }, where is the phase difference between A x and A. 2 (see Chap. vi). Noting that cos 2 = sin 2 a. + cos (a + 0) cos (a - 0), we can easily prove that the above equation may be written in the form H- = H* {(cos a + cos a -20) (cos a + cos c/>) + sin 2 a} . . .(1). Now if a = TT 2 . 3 . T2n4- TT IT TJ -Dili 42 i I22 = *l3' Thus cos 2.3 = ^, and hence 2 . 3 =120. If h be the magnetic force resolved along a line drawn through the centre making an angle with h lt then f* 2?rN \ //i 4 ^ N h = h l cos + A 2 cos -- + h s cos ,;- \ 6 / \ 6 {/' 9 \ / 4< COS 2 + COS 2 (0 - *5-j + COS 2 ( - - \ o / \ 3 t a 27T\ 27T\ 4>7T COS COS [ ^ r COS ;- COS - o / V 3 / \ o cos I s~ 1 cos [ , V 3 / J and hence H = ^H l . Therefore H is independent of and is the same for every line in the plane of the circle. In practical work, when the coils have iron cores, h^ + h 2 + h s is not zero at every instant. We can, however, represent H^ H. 2 and XIV THREE PHASE THEORY 301 HZ by lines forming a solid angle (Chap, vm) and if the coils be equal and symmetrical and the currents magnetising them be equal, then, from symmetry, the phase differences between any two will be equal and will be less than 120 degrees. Hence we may put . m COS (,.3=- , where m is less than 1, and thus H- = cos ; m -{cos -C-T)*--}] and hence H = ^H^ \/6 + 3m. Therefore, in this case also, H is independent of 6. If A be the strength of the field perpendicular to the plane of the coil, the E.M.F. induced in it will obviously be proportional to Rotating magnetic - . When we have two coils, we find in the same fields producing a at constant effective , r , E.M.F. in a search manner as before that coil placed with its plane perpen- E~ dicular to the field. j 2 [(cos a + cos 2 - 20) (cos a + cos ^Jr) -f sin 2 a} . . .(2), where the effective values of - - and - are each equal to E lt at at and T/T is the angle of phase difference between them. Hence when -^ + a equals TT, the search coil indicates the same effective E.M.F. E l sin a, no matter what the angle 6 may be, and since a can easily be measured $* can be found. Now in a choking coil the applied potential difference e is proportional to -7- . If, therefore, at/ we have two choking coils and we adjust them until the search coil connected to the terminals of an electrostatic voltmeter indicates the same effective E.M.F. for all positions of the search coil such that its plane is perpendicular to the plane of the axes of the choking coils, then the phase difference between the applied p.D.'s is the supplement of the angle between the axes. 302 ALTERNATING CURRENT THEORY [CH. Similarly, if three choking coils with their axes at angles of 120 degrees apart be magnetised by a system of three phase currents, we can show that the effective E.M.F. E induced in the search coil, no matter what the value of 6 may be, is equal to \E^ V6 + 3m, where m is less than unity and E l is the E.M.F. that would be induced in the search coil by one of the coils acting alone when the plane of the search coil is perpendicular to the axis of the other coil. Thus it is possible, since E and EI can be found, to find m, and thus the phase difference between the rates at which the magnetisations in the coils are altering. This is equal to the phase difference between the electromotive forces. In the form of phase indicator invented by Riccardo Amo's phase Arno, two circular coils have a common diameter, indicator. about which they can rotate, arid the currents, whose phase differences are to be measured, are passed through them. A search coil at the centre of their common axis and entirely enclosed by the two coils can also rotate about this axis which is in its plane. The search coil is connected to an electro-dynamo- meter or a hot wire ammeter, of negligible resistance, which measures the current in it. Now, in general, for different positions of the search coil, we get different readings in the ammeter. If, however, the current in one of the given coils be varied until the effective values of the magnetic forces produced at the common centre of the two coils be equal, it is possible, by adjusting the angle a between the planes of the two coils carrying the currents, to find a position in which the reading on the ammeter is always the same whatever the position of the search coil. If f be the resolved magnetic force perpendicular to the plane of the search coil, then z. d f- 7?y , T di k ~dt- Hl + L dt> where R is the resistance and L the inductance of the search coil circuit and k is a constant. If R be negligible, then idf_ , di te dt" L Jt' Therefore, integrating, kf = Li + const., XIV] PHASE INDICATOR 303 and, since/ and i are alternating functions, the constant must be zero ; thus taking effective values kF=LA. Now, since A is constant in all positions of the search coil, F must also be constant, and therefore, by the converse of the theorem proved on p. 299 = 180 -a, where < is the angle of phase difference between /i and / 2 , i.e. the angle of phase difference between the currents, since there is no iron in the circuit, and a is the angle between the planes of the two coils. REFERENCES. G. FERRARIS, Turin Acad., 'Rotazioni Elettro-dynamiche.' March 1888. The Electrical Revieic, 'Rotary Magnetic Fields.' June 1893. ANDRE BLONDEL, VEclairage Electrique, Vol. 4, p. 241, ' Quelques Proprietes Generales des Champs Magnetiques Tournants.' 1895. G. FERRARIS, translated in the Electrician, Vol. 33, p. 110, 'A Method for the Treatment of Rotating or Alternating Vectors.' 1894. A. POTIER, International Physical Congress at Paris, Report 3, p. 197, 'Sur les Courants Polyphases.' 1900. A. CAMPBELL, J.I.E.E., Vol. 30, p. 889, 'Test Room Methods of Alternate Current Measurement.' 1901. G. B. AIRY, 'On the Undulatory Theory of Optics.' CHAPTER XV. The magnetic fields ; round polyphase cables. The magnetic field round two long parallel wires carrying equal currents flowing in opposite directions. Bipolar circles. Currents equal and flowing in the same direction. Cassinian ovals. Lines of force when the currents are unequal. How the magnetic field alters round wires carrying single phase currents. Lines of force round three phase cables. Two phase cables. Twin concentric cables. Field of force round n parallel wires symmetrically arranged with their axes on a circle. Concentric cable. The strength of the magnetic field round n parallel wires. The strength of the field round a concentric main. The losses in cables. Three core cables. Duality. References. THE magnetic fields round polyphase cables when carrying currents are very complex, but the equations to the fie h ids m round tic lines of force can easily be found and it will be PJyp s hase instructive to study them. In practice, we have to investigate whether the field will affect neighbouring telephone or telegraph wires, and whether it will produce appreci- able eddy current losses in the lead sheath or in the copper shield, which is sometimes placed immediately inside the lead sheath to insure tbat in the event of one of the copper conductors getting accidentally into contact with it, the main fuse may act promptly. If there were no earth shield and one of the conductors made contact with the sheath, then, if the resistance of the earth in the neighbourhood of the place of contact were high, the sheath might be maintained at a high potential and be dangerous. In armoured cables we have also to investigate the hysteresis and eddy current losses in the steel strip or galvanised iron wires used to protect the cables. CH. XV] MAGNETIC FORCES 305 We will first of all consider the magnetic field round two parallel wires. This would be the case of the The magnetic field ma i ns O f a t?wo wire direct current system or of round two long parallel wires a single phase alternating current system. Let carrying equal , , . , ,-, , r , currents flowing the wires be perpendicular to the plane ot the t?ons P S paper, and let A and B (Fig. 113) be the points where their axes cut this plane, the current in A flowing towards the observer and in B away from him. We suppose that the wires are circular in section. So far therefore as the magnetic force at points external to them is concerned we may suppose that the currents are concentrated along the axes of the wires (p. 33). We shall only consider the magnetic forces at points external to the conductors. The magnetic force at any point P will be the resultant of two forces and ^7^ which are Hr A B C Fig. 113. Currents in opposite directions. F=PR= 2 ^i. r,r, perpendicular to AP and BP respectively and in the plane of the paper. Let PQ and PS represent these forces. Let AP = r lt BP r^ AB=2a, i = the value of the current in each wire in C.G.S. measure, and let F=PR = ihe resultant magnetic force at P. Now the angles APB and PSR are each the supplement of the angle QPS and are therefore equal to one lother. R. I. 20 306 ALTERNATING CURRENT THEORY [CH. 2* PS_r 2 _r 1 _AP AIS RS~2i~r~BP' Hence by Euclid (vi. 6) the triangles RPS and BAP are similar. Therefore the angle SPR equals the angle PAB and the angle QPR equals the angle PBA. PR AB Again, PS=AP and therefore .P = . 7*1^2 If the angle ABP were a right angle, then PS would be parallel to AB and hence APR would be a straight line. Thus at any point P f on the line through B perpendicular to AB, the direction of the resultant magnetic force will be AP f . In Fig. 113 if we draw PC perpendicular to PR to meet AB produced in C, then the angles CPB and SPR are each the complement of SPC and are therefore equal to one another. Hence the angle CPB is equal to the angle PAB and therefore PC is the tangent and PR the normal to the circle passing through A, B and P. Therefore every circle passing through A and B is a line of equal magnetic potential as the direction of the resultant magnetic force at any point P on it is aormal to the curve. Also, since the angle PC A is common to the triangles CPB, CAP, they are similar triangles, and thus <7P 2 = CA . CB. If, then, with centre C and radius tfCA . CB we describe a circle any radius CP of this circle will be a tangent to the circle through A, B and P\ the tangent therefore at every point on the circle, whose centre is (7, will be in the direction of the resultant mag- netic force at that point. Hence this circle will be a line of force. The points A and B are inverse points with respect to the circle. XV] BIPOLAR CIRCLES 307 It follows that all circles which have A and B for inverse points (Fig. 114) will be lines of force round the wires. The polar of A with regard to all the circles sur- rounding B will pass through B and vice versa. In other words Bipolar circles. Fig. 114. The lines of force round wires carrying equal currents flowing in opposite directions are circles. the chord of contact of the two tangents from A, to any circle surrounding B, passes through B. These circles and the circles passing through A and B are called bipolar circles. The bipolar equation to the circles round B (Fig. 113) is r, AP CA constant = m, or mr. The equation to any line of equal magnetic potential will obviously be 2 _ 1 = a) where a is a constant and 2 , 6 l are the angles PBC, PAG respectively. We have shown above that the magnitude of the resultant agnetic force is given by . 202 308 ALTERNATING CURRENT THEORY [CH. Thus the bipolar equation of any line of equal magnetic force is r-jr-a = constant. These curves are Cassinian ovals and are shown in Fig. 116. Since with single phase alternating currents the currents in the mains are at every instant equal in magnitude and opposite in direction, the lines of force due to the currents will always be the circles which have A and B for inverse points. Hence at any point P the magnetic field will be fixed in direction, and the vector representing it will simply oscillate backwards and forwards through P. It will therefore be a purely oscillatory field, and its maximum value will be - , where / is the nrj, maximum value of the current in either main. This is also true for concentric mains when the axis of the outer conducting cylinder is not coincident with the axis of the inner conductor. In this case the lines of force external to the outer conductor are bipolar circles, and induction effects are produced on neighbouring wires. Eddy currents will also be set up in the lead sheath. Let F be the resultant magnetic force at P (Fig. 115) and let the angle APB equal a, then, with the same Currents equal . . _ and flowing in the notation as before, same direction. 4i* 4i* 4^ 2 F ' = + -? + 2 - - cos a r,. r 2 7-^2 -< Therefore Also, since in Fig. 115 PB ~ PQ ' XV] CASSINIAX OVALS 309 and the angle APB equals the angle QPS, it follows that the parallelogram described on AP and PB as adjacent sides will be Fig. 115. Currents flowing in the same direction. , =pa= 4.0P.i similar to the parallelogram PQRS. Thus the diagonals of the two parallelograms will be inclined equally to their respective sides. But the diagonal of the parallelogram described on AP and PB as adjacent sides is represented by 2 . PO in magnitude and direction. Therefore SPR=OPA and RPQ=OPB. Also and Hence, for example, in the particular case when APR is a straight line, OPB is a right angle, and therefore at every point P f on the circle described on OB as diameter, AP' gives the direction of the resultant force. The bipolar equation to the lines of force can be found as follows. Cassinian ovals. If NI be the number of lines of force, per unit length of the wires, crossing OP (Fig. 115) due to the current i in the wire whose axis cuts the plane of the paper perpendicularly at A, we have 310 ALTERNATING CURRENT THEORY [CH. where A is the origin and the limits of r are AO and AP. Thus we get If N be the total number of lines of force, we obviously have = 2i log r a r 2 2i log a 2 . If P trace out a line of force the number of lines crossing Of* must remain constant. Hence along a line of force r^a = constant = m\ and this is the required equation. If we transform this equation into polar coordinates with as pole and OB as initial line (Fig. 115), we get r 4 - 2aV 2 cos 20 + a 4 - ra 4 = 0. Fig. 116. The lines of force when equal currents are flowing in the same direction are Cassinian ovals. If m be less than a, we get two ovals (Fig. 116), when m equals a we get the lemniscate, the lines of force apparently crossing at XV LINES OF FORCE 11 the origin where the magnetic force is zero, and when ra is greater than a we get only one oval. If m be greater than \/2a there is no point of inflexion on the oval, and if m be very large the lines of force are approximately circular. Again, since the sum of the magnetic potentials due to the two currents is a constant at every point on a line of equal magnetic potential, we have by p. 32 2ift + 2ift = a constant where a is a constant. Thus the lines of equal magnetic potential are given by the equation 6, + ft = a, which represents a series of equiaxial hyperbolas passing through A and B and cutting the Cassinian ovals at right angles. When the currents in the long parallel wires are unequal in magnitude, the equations to the lines of force can Lines of force when the currents be written down at once in bipolar coordinates by the above method. We shall first, however, use Cartesian coordinates as this method is instructive and will be found convenient in solving problems when the earth's field has to be taken into account. Let us suppose that the instantaneous value of the current in the wire A (Fig. 115) is ^ and in the wire B, i 2 , then if (x, y) be the coordinates of the point P, we have by resolving the magnetic forces at P horizontally and vertically, 9V 2i Fcos6 = sin ft + 2 sin ft, r i r 2 and F sin d = - - l cos ft - 2 cos ft, r \ r. 2 where 6 is the angle which F makes with OX, ft is the angle PA B and ft is the angle PBX. Also n a x and TV 2 = v 2 + (a x)~ ; sin ^o = - ; cos 6., = r z r* 312 ALTERNATING CURRENT THEORY [CH. Hence if (x, y) be a point on the line of force through P, dii -f- tan 6 dx - i> 2 2 (a + x) 4- t>i 2 (a x) i^y + i^y x a I Q 2 + y 2 + a 2 ) + 2aa? . i "' 1 7 where I = ^ cZ , . and thus -j- (IM?) = v . ^ 2 + *i + 1 a; 2 + ?/ 2 + a 2 where ^ = du Hence x -=- = r ca? u + 7 and therefore = -- - du. x w 2 1 Hence integrating and simplifying we get (ux x) l+l = m 2 (ux + a?)*" 1 , where m is a constant. Substituting for u and I their values we get finally r-firj* = constant ........................ (1), as the equation which gives all the lines of force. From the equation (a) given above we see that the tangents to the lines of force at points where they cut the curve a? + y* -f a 2 + Zlax = 0, or x a . are at right angles to the line joining the axes of the two wires. If the currents are flowing in the same direction, the curve is imaginary, but if they are flowing in opposite directions it is a circle with its centre at the point XV] EQUIPOTENTIAL LINES / j ^ \i and the radius of this circle equals 2a ^ . If C be its centre 4C=a^-and BC=a^- t and therefore The centre of this circle is therefore the point where the magnetic forces due to the currents in the two wires balance. In the differential equation (a) if we write on the left-hand fj /Y* (ill side T- instead of -f- we sret the differential equation of the dy dx lines of equal magnetic potential round the wires. The solution in this case may be written in the form = constant (2). The curves given by this equation are the same as the stream lines of current in an infinite metal plate, two points A and B of which are maintained at potentials which are proportional to ^ and i. 2 respectively. The above equations (1) and (2) can also be proved as follows by using bipolar coordinates. Since there is no magnetic force at right angles to a line of force, we have COS o COS (3), where fa and fa are the angles between the radii AP and BP and the resultant magnetic force PR (Fig. 115) which is a tangent to the required curve. Now dr<> -t , <> cos &! = -^- , and cos -ur. 2 = - 1 - . as Hence substituting we get *i dr, i. 2 dr _ ~T~ ' ~T~ U> rj ds r. 2 ds and therefore i^ log ^ + i' 2 log r 2 = constant. Thus r 1 *r 2 *'* = constant. 314 ALTERNATING CURRENT THEORY [CH. If fa', fa be the angles which the normal to the line of force at P makes with r x and r 2 , then fa = fa , and fa = fa y hence substituting in (3) we get 2ij . , 2^2 . , A sin ^T! H sin T/r 2 = 0. r i ^2 Now if ds' be an element of the equipotential curve through P, dO^ dd z sm fa = 7*! -jg , and sin fa = r 2 ^-, , and therefore i-^dd^ + i. 2 d9 2 = 0. Hence ^ ^! -H t' 2 ^ 2 = constant. In Fig. 117 the lines of force are shown when the currents are flowing in opposite directions and ^ is four times i z . Fig. 117. Currents flowing in opposite directions, one being four times greater than the other. The neutral point, that is, the point where the magnetic force is zero, ought to be specially noted. The bipolar equation to the looped line of force through this point is 27r 1 4 =2048r 2 a 3 . In general, if N be the neutral point and A and B the points where the axes of the wires cut the plane of the paper, then, THREE PHASE CABLE If the currents are flowing in the same direction, N is between the wires, if in opposite directions N is on the side next the weaker current. When N is at an infinite distance, i l = i. 2 and the lines of force are the circles shown in Fig. 114. When ij= 4^ the lines of force are as in Fig. 117. When i. 2 equals zero the lines of force are circles round A. When ^=41*0, .ZV divides AB in the ratio of 4 to 1, and a crossed line of force through this point loops A with B. Finally, when ^ = i 2 the lines of force are the Cassinian ovals shown in Fig. 116. If the ratio of ^ to i 2 is always constant, that is, if the phase difference between i, and L is either zero or 180 How the magnetic field alters when degrees, then the lines of force are fixed. In this case the field at any point is a purely oscillatory ?', and i-2 are alternating currents. one. If however the phase difference between the currents is neither zero nor 180 degrees, the magnetic field is continually changing in direction. It is not difficult to form a mental picture of what happens by considering how the neutral point N oscillates in any given case and by studying the Figures 114, 116 and 117. In general the force at any point is continually changing in magnitude and direction and hence is partly oscillatory and partly rotary. Suppose that the three copper conductors are parallel cylinders Ma netic field an< ^ ^at their axes cut the paper at the angular round a three points A, B and C of an equilateral triangle whose sides are of length d. Then if i lt i, 2 and i 3 be the instantaneous values of the currents, the magnetic force (f) at any instant will be the resultant of the three magnetic forces 2t 2i 2i* , and -- acting at P and perpendicular to AP, BP and CP 1*1 7*2 r s respectively. By a well-known theorem in Statics, we have But and thus cos & . a = r 2 2 + r s 2 - d* 2r. 2 r 3 316 ALTERNATING CURRENT THEORY [CH. Now, in practice, we generally have ii + a + i = 0, and therefore 24t' 3 = i-f if i. On substituting this value of 2i 2 i 3 in the above equation and simplifying, it is easy to see that Let be the centre of the circle through A, B and C, a its radius and let the angle POA be 9, then we have r* = r 2 - 2ar cos + a\ r* = r *- 2ar cos (6 - |TT) + a 2 , and r 3 2 = r 2 2ar cos (0 |TT) + a 2 . Substituting these values of r lf r 2 and r* 3 in the formula for/ 2 and noting that d? is 3a 2 and that ri 2 r 2 2 r 3 2 = r 6 - Ztfr 3 cos 3^ + a 6 , we get ^ = r- 2aVcos 3(9 + a 6 ( * 12 + * 8 * + * 32) 24/7V - }i 1 2 c cos 30 + a 6 1 At the centre of the circle through A, B and C, r is zero and thus 6(^ +4 *) a 2 In order that the field at the centre of the circle may be of constant strength we must have i a -1- V 2 + 4 2 = constant. This is true, for example, when the currents follow the harmonic law. Let us now suppose that the currents follow the harmonic law so that we can write i 1 = Icoscot, i 2 = I cos(c0t |TT), i s = Icos(a)t |TT). Then we have i? + i? + if = f/ 2 , and i' 1 2 cos#+...-f ... =/ 2 {(1 +cos 2cot)cos 6 + ... + ...} XV] TANGENTIAL FORCE ZERO 317 Substituting these values in the formula for/ 2 , we get a * ^ + r2 + 2ar cos ^ wt + ^ ' At points on the circumference of the circle passing through ABC, r equals a, and thus + f= J a ' 3(9 Fig. 118. The currents in the wires are supposed to be flowing in the same direction. Let us now consider the magnetic force at a point P on the circumference of this circle when the currents in the cores are i l} L and i 3 respectively (Fig. 118). 318 ALTERNATING CURRENT THEORY [CH. Since ABC is an equilateral triangle the angle CPE is 120 and AP bisects it. Let be the centre of the circle and let the angle POD be 6 and let OA be a. The angle OP A is obviously so that the angles CPO and BPO are 60 - and 60 + \Q. We have also n = AP = 2a cos 40, r 2 = 2a cos (60 + 40) and r 3 = 2a cos (60 - 0). Let PT be the tangent at the point P and let OP be produced to R. The tangential force I 7 along PT will be given by T = cos 6 + 2 cos (60 + 40) + ^ cos (60 - 0) T! r 2 r 3 Hence, at any instant, the tangential force has the same value at all points on this circle which are external to the cores, whatever the value of the currents. In practice we generally have ^ + i z + i s equal to zero. In this case the tangential force is zero at all points which lie on the circle passing through the axes of the cores of a three phase cable but are external to the cores. The resultant force at these points is therefore represented in magnitude and direction by the radial component R. Resolving the forces at P radially (Fig. 118) we get - E = - l sin J0 + 2 sin (60 + 0) - 2 * 3 sin (60 - J0) TI r 2 r s = - {^ tan 10 + 4 tan (60 + 0) - i 3 tan (60 - 0)}. a Writing i 2 i s for ^ and simplifying we get _ = V3 tg cos (60 - ^0) - 4 cos (60 + 0) 2a ' cos J0 cos (60 - J0) cos (60 We see that R vanishes and therefore also the resultant force vanishes when t a cos (60 - J0) - i s cos (60 + 0) = 0, i 2 cos (60 + ^0) and therefore when i 3 cos (60 -40) XV] ROTATION OF THE NEUTRAL POINT 319 This theorem gives us a simple construction for finding the neutral point. Join the axes B and C of the two cores in which the currents are flowing in the same directions. Divide this line at M so that BM is to CM in the ratio of i' 2 to z' 3 . Produce AM to meet the circle at P ; then, since PM bisects the angle CPB, P is the neutral point. If we write 7 cos (cot |TT) and I cos (cot |TT) for i 2 and i 3 in the above formula, we get _ 37 sin (cot - J0) " ~a ' cos (2 cos 01) __ 37 sin (cot ^B) a ' cos J# Writing 180 6 for in this formula, we see, by comparing it with the formula we have found for f (p. 317) that R must be equal to f. This also follows since, in this case, T is zero. We see that the neutral point is determined by 7/1 6 = 2cot, so that -E = 2ew, at and thus as long as it is outside the cores of the conductors it moves round the circumference of the circle with an angular velocity 2&). The amplitude of the magnetic force at a point P on the circumference of the circle is 37/a cos f 0. Thus the magnetic force has the minimum amplitude 37/a where 6 is zero. The amplitude of the magnetic force increases as 6 increases. When 6 is 50 it is nearly four times as large as the minimum value and for higher values of 6 it increases very rapidly. The formula is not applicable to points inside the cores. Let us now consider the magnetic force at any point P, not on the circle ABC, when the currents follow the harmonic law. This force is the resultant of three forces fixed in direction which oscillate according to the same law. The extremity of the line representing the resultant force at P therefore (Chapter xiv) traces out an ellipse, and if / be the value of the resultant force and co l its angular velocity at any instant we have f*a>i = constant. 320 ALTERNATING CURRENT THEORY [CH. This is true even for points inside the metal cores. It follows that if f vanish at any instant it must either be zero always or it must oscillate in a straight line. Since there is no neutral point outside two of the cores when the current in the third vanishes, it follows that there is no point outside the cores where the magnetic force is always zero. Similarly there is no point inside the cores where the magnetic force is always zero. If the force, therefore, ever vanish at a point the field at that point must be purely oscillatory. Along the axis of a three phase cable carrying currents which follow the sine law, the field is purely rotary, that is, the extremity of the line representing the resultant force describes a circle. As we approach the circle ABC the fields become elliptical and their eccentricity increases as we approach the circumference. At points on this circle, outside the cores, the magaetic forces are purely oscillatory, the oscillations taking place in the radial direction. Outside this circle the field at any point is elliptical but the eccentricity is practically zero when the distance r of the point from the cable is large compared with the radius a of the circle ABC. We shall see later on that, when a 2 /r* can be neglected compared with a/r, we get a rotating field the strength of w r hich is 3a//r 2 where / is the maximum value of the current in a core. The locus of the positions of the neutral points inside the cores lies outside of the circle ABC. At all points on this curve the field is purely oscillatory. The equations to the lines of force can easily be found in terms of tripolar coordinates. Since the resultant magnetic force at any point of a line of force is by definition a tangent to the curve, it follows that the sum of the component forces resolved at right angles to a line of force vanishes. At every point on a line of force we therefore have /! COS ^ +/ 2 COS T/r 2 -f / 3 COS ^ 3 = 0, where fa is the angle between AP and the tangent to the line of force through P, etc., and thus we get 2ij dr^ 2i 2 dr 2 2i 3 dr s ~7~ ~l -- ~T~ ~i -- ~ ~T = " r 1 ! ds r 2 ds r s ds XV] THREE PHASE CABLE 321 Therefore ^ log n + i z log r 2 + i 3 log r 3 = constant, or r^rj^rj 3 = constant. For given values of i 1} i 2 and i' 8 we get the equations to all the lines of force by giving different values to the constant. Since, in practice, ii + 1' 2 + i* = 0, the equation to the lines of force may be written r l i >rj* = mr 3 i i +i *. To illustrate the curves represented by this equation, we shall draw the lines of force in particular cases. Suppose, for example, that h^'H 8 " tii then, at this instant, all the lines of force are given by Fig. 119. Lines of force in a three phase cable when R. I. 21 322 ALTERNATING CURRENT THEORY [CH. If we take as origin (Fig. 119) a point on the circle circum- scribing A, B and C, so that is equidistant from A and J3, and if we take 00 as the axis of x, the equation to the lines of force i3 Vs \ 2 ) ( / A/3 \ 2 y--* R ) <-t*y+i+- 8 -* N where R is the radius of the circumscribing circle. Putting this equation into polar coordinates we get (r 2 - rR cos 6 + R 2 ) 2 - m 2 (r 2 - 4rft cos + 4.R 2 ) 2 = 3r 2 JB* sin 2 6. Some of these curves are shown in Fig. 119. When m is greater than unity the curves are ovals round C. When m equals unity we get (2r 2 - 5Rr cos + 5J? 2 ) (r cos - R) = Rr* sin 2 6 as the equation of the line of force through the centre of the circumscribing circle. When r is large this curve almost coincides with 2r cos = 3R, which is the equation to the straight line perpendicular to the axis of x and midway between the centre and C. When m is less than unity but greater than J we get ovals surrounding both A and B and having points of inflexion. When m equals J the equation is or 2 = SRr cos - 8R* cos 2(9, which represents the looped curve round A and J9, the origin therefore being a point of zero magnetic force. At the origin cos 20 equals zero so that the tangents there make angles of 45 with the axes; the tangents are therefore perpendicular to one another. When m is less than J the curves arc ovals surrounding either A or B. We can now form a picture of the magnetic field round a three core cable. Suppose that i z is zero, then ^ equals i 3 and the lines of force are circles having A and C for inverse points (Fig. 114). The twelfth of a period later i z and ^ are each equal to Jig, and the lines of force are as in Fig. 119. The twelfth of a period after this, they will be circles round B and C as inverse points and so on. The neutral point, outside the cores, makes two XV] TWO PHASE CABLE 323 complete revolutions round the circle passing through A, B and C during the period of the alternating current. It must be remembered that in practice the cores, instead of being cylinders, are stranded cables made up of 1 + 6 + 12 + 18 + ... wires, the usual numbers being nineteen and thirty-seven. The outside layers generally have a ' lay ' of about twenty times the diameter, that is, the wires make a complete twist round the axis of a core in a length equal to twenty times the core's diameter. In addition, the three stranded conductors or cores are spiralled relatively to one another, each making a complete turn round the axis of the cable in about eight feet. The section of a conductor is sometimes shaped like the sector of a circle, the vertical angle of the sector being 120 degrees (Fig. 33). The main features of the field of force round the cable are, however, similar to the fields we have described above. In this cable (see Fig. 35) we have four cores and the sections of the axes by the plane of the paper are at the Magnetic field . J TO round a two phase angular points of a square. If r lt r 2) r z and r 4 be the distances of these four points from P and if i lt i z , i 3 and i 4 be the values of the currents at any instant, the lines of force will be given by r 1 t 'r 2 *r 8 'r 4 * = constant. During the normal working of the system, we have i 1 + i 3 = Q and i 2 + i' 4 = 0, and thus the equation becomes i = constant. When i z is zero the lines of force are circles having 1 and 3 as inverse points. An eighth of a period later all the currents are equal in magnitude and therefore gives the equations to the lines of force. When ra equals unity we get a straight line passing through the centre and dividing the field into two symmetrical portions. There are in general two neutral points on the circle circumscribing 1, 2, 3 and 4, and the resultant magnetic force at all other points on this circle, which 212 324 ALTERNATING CURRENT THEORY [CH. are external to the cores, is normal to the circle. When ij equals t' 2 , the equation to the looped lines of force passing through the neutral points which are external to the cores is Hence it is easy to draw a rough diagram of the lines of force in this case. An eighth of a period later, the lines of force are circles having 2 and 4 as inverse points, and so on. When the currents follow the harmonic law, the neutral points, when outside the cores, travel with uniform speed round the circle which passes through the axes of the four cores. At the instants when ii and i 3 or i z and i 4 vanish, the lines of force are circles (Fig. 114) and there are neutral points only inside the cores. In the case of a twin concentric cable (Fig. 37), the phase Twin concentric difference between the currents in the two inner cables - conductors is ninety degrees, and the current in the outer cylindrical conductor enveloping them is equal, at every instant, to the sum of the two inner currents. The current in the outer return conductor produces no magnetic field inside its inner radius. The magnetic field inside the outer conductor will there- fore be due to the currents in the two inner conductors. The neutral point, due to these two currents, will oscillate on the line joining their axes and it is easy to draw the lines of force inside at any instant. If be the centre of the cable and A and B the points where the axes of the inner conductors cut the plane of the paper, will be the middle point of AB. Let OA = a and ON = x, where N is the neutral point outside the cable, then 2i, | 24 = 2(^+1,) x + a x a x ii -\- i Therefore x = a r - *. *i- ^ Hence the position of the neutral point is determined and the field outside the cable can be drawn. When t' 2 equals i lf the equation to the lines of force outside the cable is ^r^raOP 2 , or (r 2 + a 2 ) 2 - 4a 2 r 2 cos 2 6 = raV 4 . When m equals unity this gives an equiaxial hyperbola. XV] PARALLEL WIRES 325 The equation to the lines of force in a plane perpendicular to Field of force tne axes ^ tne w i res > which we may suppose to round n parallel b e the plane of the paper, is wires symmetri- A A *** ... r> = constant, where r l5 r 2 ... r n are the distances of a point P in this plane from the axes of the wires, and ^, i z ... i n are the currents in the wires. If the currents are all equal and flowing in the same direction this equation becomes r 1 r^...r n = constant. By De Moivre's property of the circle, we can write {r 211 - 2a n r n cos nd + a 271 }* = r : r 2 ...r n = constant, where a is the radius of the circle, r the distance of P from its centre 0, and 6 the angle between OP and a line joining to one of the points of intersection of an axis of the conductor with the plane of the paper. At the centre of the circle, r l = r 2 =...=r n = a, The equation to the lines of force through the centre is therefore r n = 2a n cos n6. This gives us n loops having a multiple point at the centre, where the force is zero. The equation to the line of force passing through the point (-> <-=) is r 271 - 2a n r n cos n0 + a 271 = 4a 271 . This is a curve with n ripples, the minimum value of r being a and the maximum value being 3 n a. If n were 40, the maximum deviation of this line of force from the circle passing through the axes of the wires would be only 2*8 per cent, of the radius of that circle. The equation to the line of force passing through the point (r=I'oa, 6=- \ n is r 271 - 2a n r n cos nO + a in = {(l'5) n + I} 2 a 271 . 826 ALTERNATING CURRENT THEORY [CH. It is easy to see that l'5a is the minimum value of r and that i {(l*5) n 4- 2} n a is its maximum value. When n is 8 the maximum value of r is l*514a, and hence this line of force never deviates from the circle whose radius is ToOTa by as much as the half of one per cent, of that radius. Fig. 120. Lines of force round eight wires carrying currents flowing in the same direction. The lines of force when there are eight wires are shown in Fig. 120, the equation to the curves being r 16 - 2aV cos 80 + a 16 = m 16 , where m is a constant which cannot be less than a. When m is large the lines of force are practically indistinguishable from circles. XV] CONCENTRIC CABLE 327 Let the inner conductor be a circular cylinder and let the outer consist of n wires each of which carries an equal share of the current. Let the distance of a point P from the axis of the inner conductor be r and from the axes of the wires be r lt r 2 , ... respectively. The equation to the lines of force will be mr n = r 1 r 2 ...r n , where m is a constant. Hence, by De Moivre's property of the circle, m*r m = r 271 - 2a n r n cos nd + a. When m equals unity, we get the curves ct n r n cos nO = . It is easy to see that this represents an open symmetrical curve round each wire, the vertex of which is directed towards the centre, and the equation to the asymptotes is where p is an integer. When n is even, r is imaginary when 6 lies between - a - and \ - and -=, etc. When n is odd, it 2n 2n 2n 2n is negative between these values. When m is less than unity, cos nd cannot be less than Vl m 2 and hence, in this case, for each value of m we get a loop round each wire, the loops getting narrower and shorter as rn diminishes. When m is greater than unity we get rippled lines of force which only embrace the central wire. The equation to one of those passing through the point is \ ( 7 - 1 [ r~ n + 2a n r n cos nd - a = 0. (V o n I } It follows from this equation that the minimum value of r for a given value of b is ab T* 328 ALTERNATING CURRENT THEORY [CH. Since b is less than a this is approximately equal to 2 Hence when n is large and b is small we see that the amplitude of the ripples is very small. i When b equals a, the minimum value of r is a/3 n . If there were eight wires (Fig. 121) then the maximum and minimum Fig. 121. Concentric cable in which the outer conductor consists of eight wires. values of r for this line of force would be a and 0'87a respectively. Hence it would differ from the circle r = 0*935a by less than 7 per cent. When b is greater than a then the minimum value of b is a approximately ^ 2 XV] MAXWELL'S VECTOR POTENTIAL 329 which is always less than a/2 n . Hence, the lines of force which are close to the point i r = a/2, = 0, but pass inside it, have very large ripples, which extend in some cases to great distances beyond the circle r = a. Some of the lines of force for a concentric cable having eight wires symmetrically arranged for its outer conductor are shown in Fig. 121. If it were carrying direct current, the needle of a little compass would change its direction 2n times on being taken round the cable. Since, with alternating currents, the current in the inner conductor is always in exact opposition in phase to the currents in any of the outer wires, the magnetic field at any point in the neighbourhood is in this case a purely oscillatory one. In general, if we have a series of parallel wires carrying alternating currents of any magnitudes which are always in step with one another, that is, which are either in phase or in exact opposition in phase, then the magnetic field in their neighbour- hood is purely oscillatory. The strength of the magnetic field in the two preceding The strength of problems is easily found by the following method, nTn parallel due to <> F. C. Searle, which furnishes an example of the use of Maxwell's Vector Potential. Fig. 122. A, B, C, D... are the points where the axes of n long parallel wires cut at right angles the plane of the paper. 330 ALTERNATING CURRENT THEORY [CH. Let us first consider the case when the magnetic force due to the current in the return conductor is negligible. Let the current in each of the n wires be i/n and let A, B, G... (Fig. 122) be the points where the axes of the wires cut the plane of the paper. Consider two lines through P and through a fixed point K, parallel to the wires, and let N be the number of lines of force which pass between these two lines per unit length. Then, if PA=r^ P = r 2 ,..., KA=p l} KB = p 2 ,..., OP = r and OA=a, we have IV where X is a constant. Hence, by De Moivre's property of the circle, a) ......... (a). Now let R (Fig. 123) denote the radial magnetic force at the do O A Fig. 123. R is the radial magnetic force. point P. If the field were uniform it would be measured by the number of lines of force, per unit area, parallel to OP. When it is not uniform, R multiplied by an element of area at P at right angles to OP, will give the number of lines of force through that area. Thus, we have R.rdO = ~.dd, da IdN and therefore R = ^ . r dd XV] RADIAL AND TANGENTIAL FORCES 331 2z r n a n sinn0 Hence, by (), R = - - . ^_ 2r n a n cosne + a > ' -O> Similarly, from Fig. 124, we see that the tangential force T is given by T=--. O A Fig. 124. T is the tangential magnetic force. Thus, from (a), we get >2n r n a n cos n9 _ = ' 7 ' r 2 " - 2r n a n cos w0 + a 2 * 1 If F be the resultant force at any point in the field, since R and T are at right angles. Therefore 9V r n -r-v ^ * r ' jr 271 - 2r 7l a n cos ?i0 + ei 2n j* 2i r n Now at all points on a line of force we have r^ r 2 . . . r n = constant, and therefore, along a line of force, F will vary as r n ~ l . For example, along any of the lines of force shown in Fig. 120 F varies as r 7 . Again, from (7), we see that the tangential force T, that is, the force at right angles to the radius vector, is zero at all points on the curves r n = a 11 cos nd. These are curves exactly similar to the looped curves shown in 332 ALTERNATING CURRENT THEORY [CH. Fig. 120, the extremities of the loops passing through the axes of the wires. If P be a point, therefore, on one of these loops, OP is the direction of the magnetic force at that point. For example, when n equals 2, we see that at any point P on the lemniscate r 3 = a 2 cos 26, the direction of the resultant magnetic force is OP and the magnitude of the force is proportional to OP. In the preceding problem, if we suppose that a current i Thestren th of fl ws in a return wire through 0, parallel to the the field round a n wires, we get the case of a concentric main. concentric main. . Ine effect of this return current is to add to the tangential force T, due to the currents in the n wires, the amount 2i and to leave the radial component R unaltered. If T' be the new tangential force, we have r n a n cos nO a? n cos nO + a 2n ' This vanishes at all points on the curves r n cos n6 = a n . These curves are similar to the open curves shown in Fig. 121 and they pass through the axes of the wires. At any point P on these curves OP gives the direction of the magnetic force. If F' be the resultant magnetic force at any point P, and thus F' = - r The magnetic force at any point on a given line of force (Fig. 121) is therefore inversely proportional to r. Again, from (a), when r is greater than a, we get + log (l - XV] FIELD ROUND CONCENTRIC MAIN 333 where N\i refers to the n wires only. Thus we have T" + " O/l* /y^Tl "I /7^H = 2i log r -I f cos nd + - . cos 2n6 + ... n \r n 2 r 371 Similarly when r is less than a, we may write 2V, /**n ^ 2n \ N \i = 2i log a -\ ( cos nO -f ^ . - cos 2n# + . . . ) ?i \a z a / Therefore, when r is greater than a, E-- 7* du oy //yH ^7^** \ = f ^r sin n6 + - sin 2nd +...), r \r n r J' T , dN 2i and T = -^ dr r Similarly when r is less than a, 2?* 2?* /r n T 271 \ = -- -sin?<<9+ -sin2n<9+... , r \a n a 271 / ' and r r \a The above series show clearly the degree of approximation of the field to that due to a longitudinal current uniformly distributed over the surface of a hollow cylinder and returning by a thin coaxial solid conductor. If we have an alternating current, of effective value A, flowing The losses in i Q a conductor of resistance R, then RA 2 is the cables - least possible value of the mean power lost in the conductor. Even when the conductor is the inner conductor of a concentric main so that it is shielded from the inductive effects of the return current, yet owing to the skin effect (p. 47), the power 334 ALTERNATING CURRENT THEORY [CH. expended in it is greater than RA' 2 . In polyphase cables the wires that form the cores are not shielded from the inductive effects of the currents in neighbouring wires, or of the currents in neighbouring cores and so, except in the special case when the cores are made up of fine wires not too tightly pressed together, the eddy current losses in them are appreciable. Let us first consider the simple problem of single phase mains. When the consumer's terminals are connected with the mains of an alternating current supply company by two conductors run through two separate iron pipes, it is found that, at times of heavy load, there is a great diminution of the voltage between the consumer's terminals. The presence of the iron magnifies very considerably the back electromotive force due to the in- ductance of the conductors, and thus the potential difference V 1 between the consumer's terminals is much smaller than the voltage V between the mains. In this case the potential difference F 2 between the ends of a conductor joining a main to a house terminal, that is, the ' voltage drop ' along the conductor, will be large. In general the phase difference between the voltage drop and the current in a conductor is large. It will only be small in the exceptional case when the eddy current and hysteresis loss is large. If the load in the house circuit consist of incandescent lamps so that its power factor is practically unity and if, as is generally the case in practice, the connecting conductors be equal and the losses in them are the same and take place in the same manner, we can easily measure W, the power lost in the conductors and in the pipes surrounding them. Since the voltage drop along one conductor is in phase with that along the other their resultant value is 2F 2 and so the power W is given by the three voltmeter formula (p. 205). We have, therefore, where A is the reading of an alternating current ammeter in the circuit. If we subtract 2A Z R from W, where R is the resistance of either of the connecting conductors, we get the power expended in hysteresis and eddy currents in the pipes. XV] LOSSES IX IRON CONDUITS 335 As V, V l and F 2 differ in phase, VV^ is generally much smaller than 2F 2 . For example, we could have F=260, F 1 = 240, F 2 = 40, .1 = 100. In this case the value of W, in watts, is given by = 750. Since W must be positive we see that when F is 260 and V l is 240, F 2 must be less than 50. This follows since 260 2 - 240 2 = (2 x 50) 2 . It has been noticed that when the iron pipes are insulated from the earth the voltage drop F 2 for a given current A is less than when the pipes are laid in the earth. This shows that the eddy currents are increased when the pipe is earthed throughout its length. To avoid this excessive drop in the pressure, the two conductors are put into the same pipe. This is found, in practice, to be a satisfactory solution. Let us suppose that the axes of the con- ductors and the axis of the pipe are in one plane and that these axes are parallel. Let us also suppose that the pipe is of very thin metal so that the distortion of the field produced by the eddy currents in it can be neglected. The lines of force will then practically coincide with the circles shown in Fig. 114, and an inspection of this figure will show that the plane containing the straight lines of force bisects the pipe. If the pipe be long, then at any instant the induced current will be flowing down one half of the pipe and back along the other half. The magnetic force at any point P can be easily found. Let N be the number of lines of force per unit length of the system between P and a point on the sheath equidistant from the two wires. Then, as on p. 310, we can easily show that T = - i log (r 2 - 2ar cos 6 + a 2 ) 4- i log (r 2 -f 2ar cos 9 + a 2 ), where i is the current in either conductor, 2a the distance between the axes of the conductors and (r, 0) are the polar coordinates of the point P, the origin being on the axis of the pipe and the initial line passing through the axis of a conductor. Thus, if R 336 ALTERNATING CURRENT THEORY [CH. and T be the radial and tangential components of the magnetic force at P, we get 1 dN R== r'~de 2i 2ar sin 6 (r 2 + a 2 ) r V - 2a 2 r 2 cos 20 + a 4 ' dZV and T = - -=- dr _2i 2ar cos 6 (r 2 - a 2 ) = 7 ' r 4 - 2a 2 r 2 cos 26 + a 4 ' Hence also the resultant force F is given by F 2 = R 2 + T 2 , and thus F = fif, a result which agrees with the formula given on p. 306. For the following method of considering the eddy current losses in a pipe, when it is insulated from earth, the author is indebted to G. F. C. Searle. We have seen that the number N of the lines of force between a point P (r, 6) on the pipe and a point equidistant from the axes of the two conductors is given by Ni log (r 2 + 2ar cos 6 + a 2 ) - i log (r 2 - 2ar cos 6 + a 2 ), when the walls of the pipe are very thin and the metal of which it is made is non-magnetic. Using the method employed on p. 332 and noting that r is greater than a, we have - cos + cos 20 + . r 2 r 2 - cos + i ^ cos 3(9 + \ -^ cos 50 + . . r 3 r 3 5 r 5 Thus since, by symmetry, the current vanishes at the points on the pipe equidistant from the two wires, if follows that if u be the current density at P and cr the resistivity of the pipe, we have, making the assumption that every tube of eddy current is in quadrature with the flux it embraces, dN cos + cos dt r 3 r $ XV] LOSSES IN IRON CONDUITS 337 If i equal /cos wt, the mean value of (di/dt) 2 is |o> 2 / 2 and thus the mean value of o- 2 w 2 is 3 O w T" Hence, if the thickness of the sheath be h, we find that the mean power W lost, per unit length of the sheath, is given by w ^j^j-L ose + ^ cos2d+ J> rde i +l*4+4 ro- where A is the effective value of the current. Thus, if (o^/Sr 2 ) 2 be negligible compared with unity, the eddy current loss W in the pipe, in ergs per second, is given by W=(Aa<*y l, where Z is the length of the pipe, and all the quantities are measured in absolute units. We see that, when the reactance of every eddy current path is negligible compared with its resistance, the eddy current loss, per unit length, varies as the square of the current, the square of the distance between the axes of the wires and the square of the frequency. It also varies directly as the thickness of the pipe and inversely as its radius and its resistivity. Let us now consider the hysteresis loss in a very thin iron pipe. If r be the radius of the pipe, the magnetic force F at a point P on its circumference is given by and thus by p. 12, It can be shown that this series is convergent when r is greater than a. F has its maximum value when 9 is zero and its minimum value when 6 is -= . Thus the amplitudes of F in c.G.s. 2 R. i. 22 338 ALTERNATING CURRENT THEORY [CH. measure at points in the pipe lie between 4/ max.<* d 4/ max .g where / max . denotes the maximum value of the current i. If we know Steinmetz's coefficient for the iron (p. 37) and the perme- ability curve we can easily find a superior limit to the hysteresis loss. The following data concerning three core cables and the method of their manufacture will show the nature Three core cables. . of the problems that arise m practical work. The cores consist of stranded conductors; a 19/14 core, for example, would be one made up of 19 strands of wire of No. 14 Standard Wire Gauge, that is, wire with a diameter of O'OSO of an inch. Strips of paper are wound round the cores, and the cores are spiralled relatively to one another making a complete turn round the axis in from four to eight feet, the space between the three cores being filled either with jute or with a paper core. The cores are now wrapped together with more strips of paper, and after immersion in a bath of a special compound, are passed to the lead covering press. In the low temperature process, the lead covering is put on under very considerable pressure and at a temperature well below the melting point of lead. In most cases a covering of compounded tape or jute yarn is laid on (served) over the lead by a special machine. The armouring comes over the compounded tape and generally consists of galvanised iron wires laid on longitudinally with sufficient ' spiral ' to keep them together. In some cases this armouring is served with compounded jute. The cross section of a 19/14 core is nearly 0*1 of a square inch, so that if we assume that a current density of 800 amperes per square inch is permissible, then each core can carry 80 amperes. In some cases the armouring consists of steel ribbon or strip. This strip is put on in two layers, the first layer being spiralled in one direction and the next in the opposite direction, thus 'breaking joint' arid affording efficient protection for the cable. This is the standard armouring for ordinary direct current cables, but reasons will be given later on why it should not be used for XV] THREE CORE CABLES 339 three core cables, unless the hysteresis losses in it when subjected to appreciable magnetising forces are negligible. The minimum distance of the cores from one another and from the sheath depends on the voltage for which the cable is designed. For voltages up to 5000 it is customary to allow a thickness of 0'05 of an inch of dielectric for every thousand volts. For higher pressures the thicknesses used in practice are less than that given by this rule. If the cable be for 11,000 volts mesh working, the minimum distance between the cores is generally about 0'4 of an inch. The inner diameter of the lead sheath would be slightly less than 25 inches if the cores were 0'2 of a square inch in section and would be about 3 inches if the cores were 0'4 of a square inch in section. The thickness of the lead is generally about 0*16 of an inch (160 mils). The jute between the lead and the armouring is about 50 mils in thickness but in some cables the armouring is placed directly on the lead. When steel strip is used it is generally about 40 mils thick and as there are two strips one over the other this gives a steel cylinder 80 mils thick surrounding the cable and having a diameter somewhat greater than three inches for large cables. If this be covered with jute the thickness of the jute will be about 100 mils (0'254 cm.). The section of the cores, in the smallest of the three core cables used by the Underground Electric Railway Company of London, is 015 of a square inch and, in the largest, 0*25 of a square inch. The effective voltage between any pair of cores is 11,000 and the minimum distance between the cores is 0*44 of an inch. The coefficient of self induction for electrostatic charges of the cores in the smallest size of cable is 26 of a microfarad per mile, and in the cable which has cores 0'25 of a square inch in section the coefficients of the cores are about 0'3 of a microfarad per mile. The cables are plain lead covered. On the power station circuit of the Metropolitan Electric Tramways Company of London, the three core cables are sheathed under the lead with copper tape as an earthing shield for protective purposes, and are drawn into earthenware ducts. The cores of most of the cables are 01 of a square inch in section, and the minimum distance between the cores is 0'38 of an inch. The 222 340 ALTERNATING CURRENT THEORY [CH. coefficients of the cores in this class of cable are about 0'27 of a microfarad per mile. The following are the principal data for a three core cable, for 11,000 volt mesh working, made by the British Insulated Wire Company. The section of each core is roughly similar to the segment of a circle which has a vertical angle of 120 degrees. There are 37 strands of wire in a core, the diameter of each wire being 0'082 of an inch. Each core is wrapped round with paper to a radial depth of O'lS of an inch and the minimum distance between the cores is about 0'37 of an inch. Over this another layer of paper 018 inch thick is wrapped. The radial thickness of the lead sheath is - 16 of an inch and its outer diameter is 2'5 inches. The lead in this case is served with compounded jute and over this sixty galvanised iron wires are laid, each 0'104 of an inch in diameter. Only sufficient ' spiral ' is given to the wires to keep them together. Wire armouring is never put on by coiling the wire round the cable. The weight of each of the copper cores in the above cable is about 4000 Ibs. per mile. The weight of the lead sheath is over thirteen tons per mile and the weight of the wire armouring is about 10,000 Ibs. per mile. The coefficient of self induction for electrostatic charges of each core is approximately 0'37 of a microfarad per mile. In many situations, as for example in mines, mechanical protection is necessary for three core cables. They are therefore either drawn into iron pipes or heavily armoured. When this is done a copper shield under the lead is in general unnecessary. In addition to the eddy current losses in the cores themselves and in the lead sheath, there may be considerable hysteresis and eddy current losses in the armouring. It is important therefore to know the value of the magnetic forces due to the currents at points near a three phase cable. We shall now find the direction and the magnitude of the Formulae for the magnetic force produced at any point by the ?u?t n o e th C re f e r ph S ase currents in the cores of a three phase cable. We cables, shall assume that the load is balanced, and that the cores can be regarded as cylinders whose axes are parallel. XV] THE VECTOR POTENTIAL 341 Our formulae will be approximately true when the conductors are spiralled. Let i^ i 2 , and i 3 be the currents in the cores. With the notation of p. 330, if we choose the centre of the cable as K, JV= 2ij log a + 2^ log a + 2i 3 log a - 2i, log n - 2{ 2 log r a - 2i 3 log r 3 = - 2^ log n - 2i' 2 log r 2 - 2* 8 log r 3 , since ^ + z 2 + i z 0. Thus, we get by p. 58 = 2^ {- cos 6> + i ^ cos 2(9 + i ^ cos 3(9+ ...1 IT* 2t T o / J From this equation R and I 7 can be found at once by the formulae IdN dN R = --JA and 2 f =-- 3 -. T* cZ^ rfr In order to simplify the formulae for the radial and tangential forces we shall suppose that the currents follow the harmonic law, so that we can. write I*! = / cos o)t, i. 2 = / cos (o)t | TT) and i 3 = I cos (cot f TT). Now, we have 2 cos cot cos pO = cos (cot+pd) + cos (cot pB\ 2 cos (cot - | TT) cos p (0 - f TT) = cos [cot +p0 - | (p + 1) TT} + cos {o> -jp^ + | (^ - 1) TT}, 342 ALTERNATING CURRENT THEORY [CH. and 2 cos (cot - | TT) cosp (6 f TT) = cos{+;)<9-f (p + l)7r) -f cos [ft>-j9(9 + J^-l)^}. If $ be the sum of these quantities, we have o = cos (o) +pd - | (JP + 1) TT) sin (j? + 1) TT sin J(p + l)7r cos [cot - p6 + | (ff 1) TT) sin ( ff - 1) TT Now, the value of sin #7r/sin J #TT is zero, when x is neither zero nor an integral multiple of 3. As x approaches any integral multiple of 3 the quantity approaches the limit 3. Using these results we obtain = 37 - cos (art - 6) + cos (tat + 2(9) + j ^ cos (erf - 4(9) + i ^ cos (w* + 5<9) Hence we find the following value for R 7? _ zzzi ~ r d6 = j- sin (tot -0)~ sin (*> + 2(9) r [r r 2 a 4 . , . m a 5 . + sin (&j^ 40) sin ^ / r 5 Putting their exponential values for sin 6, cos 6, sin 20, etc. we get simple geometrical progressions for the coefficients of sin wt and cos wt. Summing the series and replacing the expo- nential values by sines and cosines we find that jj = 3/ a 2 {(r 4 + a 4 } sin (tot - (9) - ar (r 2 + a 2 ) sin (qrf + 2(9)} a r 6 2aV 3 cos 30 + a 6 XV THE TANGENTIAL FORCE 343 It can be shown also, that this formula is true when r is equal to or less than a. Again, we have dN T=- dr p cos (tot - 0) + ^ cos (cot + 2(9) cos (eo 40) + -j cos (cot + 50) _37 a 2 {(r 4 - a 4 ) cos Q - 6) + ar (r 2 - a 2 ) cos M + 2fl)} ~~a'~ r 6 - 2a 3 ?^ cos 3d + a 6 This formula is also true when r is less than or equal to a. We see at once that T vanishes when r equals a. The amplitude of T equals 3/ a? (r 2 - a 2 ) {(r 2 + a 2 ) 2 + 2ar (?- 2 + a 2 ) cos 30 + aV 2 }* a^ ' r 6 - 2a 3 r 3 cos 3d + a 6 For a given value of r this has obviously its greatest value T max when cos 30 is 1 and its least value T min when cos 30 is 1. Thus we have _37 a 2 (r+a) -^ mav * a ' r'-a 3 and a r 3 + a 3 It is easy to show that T min has a maximum value when r is 1-679 a and r min . is then equal to x O1184. An inspection of the following table will show how the coefficients of in the formulae for T max and T min vary as r increases. CL r a 1 1-5 2 2-5 3 4 10 100 a 2 (r + a) CO 1-053 0-429 0-239 0-154 0-079 0-011 o-ooo ?* - a 3 a 2 (r-a) 0-114 0-111 0-090 0-071 0-046 0-009 o-ooo r< + a 3 344 ALTERNATING CURRENT THEORY [CH. The formulae for R and T given above show us that at all points which are outside the three cores the field is elliptical. Along the axis of the cable the ellipse is a circle, at points on the circle passing through the axes of the three cores T vanishes and thus the ellipse becomes a straight line pointing to the centre of the circle, and when a 2 /r 2 can be neglected in comparison with a/r we again get a pure rotating field the strength of which is 3a//r 2 . If be the centre of the section of the cable, A the point where the axis of the core cuts this section and P a point on OA or OA produced, the values of R and T at P are given by 37 a 2 (r-a) and R a T^I a r 3 a 3 sin cot, cos cot r* a* = 2'max. COS cot. Thus at all points on OA or OA produced the elliptical field has its major axis perpendicular to OA and equal to 2T m&x . Similarly, if OP' bisect the angle BOA where B is a point on the axis of another core, the field at a point P on OP', at a distance r from 0, is given by P_3J a 2 (r + a) a i* + a s sin [cot - and a 2 (r - a) COS [cot -7; - T *n min. (**', The minor axis of the ellipse representing the field at P equals ^min. and its direction is perpendicular to OP. The following table shows how the coefficients of 3//a in the expression for the radial force vary as we move in the directions OA or OP. r a 1 1-5 2 2-5 3 4 10 100 a 2 (r - a) r 3 - a 3 0-333 0-211 0-143 0-103 0-077 0-048 0-009 o-ooo a 2 (r + a) 1 0-571 0-333 0-211 0-143 0-077 0-011 o-ooo 7-3 + a 3 XV] THE RESULTANT FORCE 345 If / be the value of the resultant force at the point (r, 6) we have /- = R- -f T- _ /3/\ 2 a 4 {a 2 + r 2 + 2ar cos " ~ which agrees with the result obtained on p. 317. Since the maximum value of cos 30 is 1 and its minimum value is 1, we find that the maximum and minimum values of / at points at a distance r from the axis of the cable are given by _ 37 a 2 (r+a)_ /max.- a ^_ a , ^max., 37 a 2 (r - a) T /^ = ^- ^ + 0,3 =^nlo, Let us now consider the hysteresis loss in the steel armouring of a three core cable. We shall suppose that the armour consists of very thin steel strip, and we shall neglect the shielding effect of the eddy currents induced in the lead sheath. The integral of the tangential force at any instant taken round the steel cylinder is zero, the lines of induction therefore leave the cylinder and demagnetising effects are produced. If we calculate the hysteresis loss on the assumption that there are no demagnetising effects and that T max acts on every point of the armouring we get a superior limit to the hysteresis losses. If A be the effective value of the alternating current in a core, expressed in amperes, T max . will be given in c.G.s. units by the formula A/2 a * lOa ' As an example we shall take the case of a three core cable designed for 10,000 volt working and having a normal current of about 150 amperes in each core. The area of the cross section of each core is 1*25 square centimetres, the radius a of the circle circumscribing the axes of the cores T8 cms. and the inner radius of the steel armouring 4'5 cms. If the effective current in each core be A amperes, the maximum value of the magnetic force in C.G.S. units is 0*056 A. If, therefore, A were 150 amperes, the maximum value of the magnetic force would be 8'4 C.G.S. units. Hence, if we know the permeability curve and Steinmetz's 346 ALTERNATING CURRENT THEORY [CH. coefficient for the steel (p. 37), we can assign a superior limit to the hysteresis loss. We have seen that the minimum value of the tangential force at a distance r from the axis is given by the formula T = s \ ^ a2 ( r ~ a) 10a ' r 3 + a 3 ' For the cable considered above we get zu.- 0-021 4 = 3*2 c.G.s. units, when A is 150 amperes. This, however, does not enable us to fix a minimum value to the tangential force acting on the steel as we cannot neglect the considerable demagnetising effect produced by the lines of induction leaving the steel. If we make the assumption that the mean value of the maximum flux density is 1000 and take t) = 0*002, we see, by p. 38, that the loss W, in watts per kilogramme of the steel, when the frequency is 50, is given by = 0-002 x 50 x -^ - x (1000) 1 ' 6 x 10~ 7 7*o = 0'08 nearly, assuming that the specific gravity of steel is 7*8. If the steel armouring therefore have a mass of 2000 kilogrammes per mile this would give a loss of 0'16 of a kilowatt per mile. When iron wire is used, with its length parallel to the axis of the cable, the magnetic flux produced will be very much smaller than with steel strip owing to the very large demagnetising effects produced by the flux leaving the wires. The hysteresis loss with wire armouring will therefore be quite negligible. In addition to the hysteresis losses, there are eddy current losses in the cores themselves, in the lead sheath, in the copper earth shield and in the armouring. We shall see in the next chapter that the currents generated near the surface of the sheath screen off the magnetic force from the interior, and thus in the simplest cases even with non- magnetic metals, the calculation of the eddy current losses is difficult. It has to be remembered, in v ^ tt J^^w XV -ITY OF XV] THE ELECTROSTATIC FORCES 347 V estimating possible eddy current losses in sheaths, that these losses do not vary in any simple manner with the resistivity of the metal of which the sheath is made. In addition to the losses that take place in the metallic parts of the cable, we may have losses in the dielectric itself due to the electrostatic forces. When the cores are very thin it is not difficult to find formulae for the resultant electrostatic force at points in the dielectric by the method of images (Chapter v). Let us consider, for example, the electrostatic force /at the axis of the cable. We suppose that all the electrical quantities are given in electrostatic units. The component /i of the force in the direction OA, due to the wire whose axis is at A and the image of this wire at A', is given by (Chapter iv), where OA a, and OA' a and X is the dielectric coefficient. Xow when the load is balanced (Chapter iv), so that v l + v. 2 + v 3 = 0, then #1 = (^1.1 ^1.2) 0i, and thus, since f\ t f$ and f z are inclined to each other at angles of 120, we get by Statics / = /? +/ f +/ s2 -/ 2 / 3 -/,/, -//,, and hence, on substituting, we get 1 fi /'l aV/^ ^ N 2 / ?;2 = b ~~^K A i.i~" A i.2) v0i since aa! = r 2 , where r is the inner radius of the sheath. When the potential differences follow the harmonic law, we get where E is the maximum value of ^ and therefore of the star pressure. Let us take the case of a cable working with an effective pressure of 11,000 volts between the cores. Let a = 1'5, r=4'5 348 ALTERNATING CURRENT THEORY [CH. and let the capacity between two of the cores be O15 of a micro- farad per mile. Then, by p. 92, ^ (K lfl Ki. 2 ) = 0*15 of a microfarad per mile, and thus (K^-K^) E=0'3x 10- 15 x 11,000 Vf x 10 8 C.G.S. electromagnetic units per mile = 0-3 x 10- x 11,000 = 50'2 c.G.s. electrostatic units per centimetre, and thus X/=89'3. If we assume that X is 2'8 (see p. 108) we get /=31*9 dynes per unit charge nearly. This corresponds to a rate of variation of the potential, at the axis of the cable, in the direction of/, equal to 31/9 x 300x2*8, that is, 26,800 volts per centimetre. If the frequency of the alternating current be 25 then /will make 25 revolutions per second. As the force is appreciable, experiments are being made to determine whether the magnitudes of the electrostatic losses are large enough to be measurable. It follows from the formulae we have found for the magnetic force near a three phase cable with straight cores, that there will be an appreciable disturbance set up in telephone circuits if the wires are near the cable and are not twisted together. When the cable is armoured the disturbance will be less. As the armouring is always thin compared with the diameter of the cable and the radial magnetic forces are appreciable, the diminution of the disturbance due to the armouring may, however, only be slight. The intensity of the disturbance at distances greater than a metre from the cable is approximately inversely proportional to the square of the distance provided that the going and return wires of the telephone circuit be close together. The formulae we have found for the magnetic force round three core cables also apply to the forces round overhead wires carrying three phase currents, provided that the wires are equi- distant from one another. In this case, if the telephone wires be carried on the same posts as the wires for the three phase currents and be riot twisted together, a continual hum will be heard in the XV] TELEPHONE DISTURBANCES 349 telephone. This will be slight if the mains are spiralled, and can be made negligible by ' crossing ' the telephone wires at regular intervals, that is, making the higher wire the lower wire for the next section and so on. In the event of a short circuit occurring at an insulator and earthing one of the mains, the currents in them will no longer be balanced, and a loud hum will be set up in the telephone. J. Behn-Eschenburg has used the telephone as a fault signaller for several years on the 30,000 volt transmission line at the Oerlikon Machine Works in Switzerland. The equations and the diagrams for the lines of force and equipotential lines given in this chapter are similar to the equations and the diagrams for the equipotential lines and the lines of force in electrostatics. This is an example of the principle of duality (Chapter xvn). The equipotential line in electrostatics corresponds to the magnetic line of force, and the electrical line of force corresponds to the line of equal magnetic potential (see p. 138). The lines of force in statical electricity are also the same curves as the lines of flow of an electric current through a conducting medium. The diagrams of the lines of force given above can be obtained experimentally by maintaining suitably chosen spots on a sheet of tinfoil at given potentials and mapping out the equipotential lines on the sheets by Carey Foster's method. REFERENCES. G. R. KIRCHHOFF, Poggendorff's Annakn, Vol. 64, p. 497. W. ROBERTSON SMITH, Proc. Roy. Soc. Edin. 1869 70, p. 79. G. CAREY FOSTER and OLIVER J. LODGE, Proc. Phys. Soc. London, Vol. 1, p. 113. For Cassinian ovals and dipolar (bipolar) circles see A. G. GREENHILL, Differential and Integral Calculus. For De Moivre's and Cotes' properties of the circle see S. L. LONEY, Trigonometry. For numerous diagrams of equipotential curves got experimentally see JAHIN and BOUTY, Cours de Physique, Vol. 4, Part I. p. 172, 4th Edition. M. B. FIELD, J. I. E. E. Vol. 33, A Theoretical Consideration of the Currents Induced in Cable Sheaths, and the Losses occasioned thereby. 5 April, 1904. CHAPTER XVI. Eddy currents. Short circuited coil in a uniform alternating magnetic field. Currents in the closed secondary of a constant voltage transformer. Eddy currents in the iron plates of the core of a transformer. Eddy currents produced in an infinite iron plate, placed in a uniform har- monically varying magnetic field, with its sides parallel to the lines of force. Analogy with the theory of heat. Approximate formulae. Eddy currents induced in a metallic cylinder forming the core of a long solenoid. Tables of cosh 0, sinh 0, cos $, sin 6, n , ber (x\ bei (x\ ber' (#), and bei' (#). References. THE calculation of the losses due to the eddy currents, which are venerated in conductors when placed in variable Eddy currents. magnetic fields, is a problem of great practical importance. In most cases only roughly approximate solutions can be obtained, owing to the difficulty of solving the equations even when the conductors are of non-magnetic materials and the strength of the field varies harmonically. Oliver Heaviside has found a solution for the case of a solid cylinder of infinite length, whose axis is parallel to the lines of force of a uniform alternating magnetic field, and J. J. Thomson has solved the corresponding problem for the case of an infinite plate when the surface of the plate is parallel to the lines of force of the field. Before giving an elementary discussion of these problems we shall consider the problem of the currents induced in a closed coil of insulated wire, when placed in an alternating magnetic field. Let us suppose that the coil has n turns of wire, that the mean short circuited area enc l se d by the turns is S and that the induc- coii in a uniform fa ou density parallel to the axis of the coil is alternating mag- netic field. CH. XVI] INDUCED CURRENTS 351 Let R be the resistance of the coil, L its self inductance and i the induced current, then = nSBco cos cot. The effective value A of the current is therefore given by nSBco Hence for a given maximum induction density B, the effective value of the current will continually increase as co increases and therefore as the frequency increases, but it can never be greater than -f T^-. The power expended in heating the coil is RA 2 and -6 V ^ this equals (nSBco)*R This expression continually increases as the frequency increases / Of D \ 2 and its limiting value is (-= ] R. If we suppose that R is the V.L/ Y LI variable, then the heating for a given frequency will be a maximum when R equals Leo or 2?r/Z. Hence, the increase of the resistivity of the metal of the conductor, due to the heating, will increase the power expended in heating the circuit if R is less than ZTT/L but will diminish it if R is greater than 2irfL. From the equations of the air core transformer it is easy to Currents in the sh W (Chapter X), that closed secondary JUT j of a constant 1U ( - E> \ _ D , T _ ^ voltage trans- ~ T ^ l ~ ll) * * * ~dt ' former. where a = 1 = the leakage factor. Thus, squaring and taking mean values, we find that 7lf2 y-i ( V? - 2R, W + ^ 1 2 ^ 1 2 ) = R*A* + ofLfrWAj, L>\ where a is a constant which has its minimum value unity (p. 80) when e l follows the harmonic law. Now W is the mean value of 352 ALTERNATING CURRENT THEORY [CH. the power taken by the primary coil. It must therefore be equal to the mean rate of heat production in both circuits and thus TT7" T> A 2 I Z? A 2 VV == Xl/i-ili T Xl/2-^-2 Hence, on substituting this value of W in the above equation, we get A* (Ri 2 The power expended on the secondary circuit is therefore a. M! ( Fi , _ j^.^^ Rf + 2 ^ E,E 2 + a 2 i 2 V 2 ft) 2 ' Assuming that .^1 is negligible compared with V 1} we see that, when a is not zero, the power expended in heating the secondary circuit continually diminishes as the frequency increases and vanishes when the frequency is infinite. If the secondary resistance vary, then, if we neglect R^j. com- pared with Fj, the power transmitted to the secondary circuit at a given frequency is a maximum when R 2 equals and it then equals 9 1/2 7? +97 2 j^ -Ki + 2 provided that the shape of the current wave in the secondary circuit remain constant. When R 2 is greater than 27r/a 2 o-, an increase in the resistance of the circuit diminishes the loss due to the heating in the circuit, but if R 2 is less than 27r/ajL 2 cr then an increase of R. 2 increases the loss. When RA^ is small compared with F x and the leakage factor is zero, the heating of the secondary circuit is simply equal to and it is therefore independent of the frequency and of the shape of the applied potential difference wave. When the leakage factor is not zero the heating of the secondary circuit for a given effective voltage Fj is a maximum when the applied potential difference wave is sine-shaped. XVI] EDDY CURRENTS 353 The above theorems for insulated fine wire coils, although instructive, do not give us much help in calculating the eddy current losses in masses of metal subjected to an alternating magnetising force. To do this we must find the paths and the magnitudes of the induced currents in the metal. It is found in practice that, unless the iron plates which form Eddy currents in the core of a transformer are less than about half the core ofT* of a millimetre in thickness, the heating due to eddy transformer. currents is appreciable. The cores are therefore built up of finely laminated iron. The lamination appears to have little, if any, effect on the losses due to hysteresis. In the core of a transformer the plates are practically thin strips of iron, the length of the strips being parallel to the direction of the alternating magnetic field. The induced electromotive force will therefore be parallel to the cross section of the strips, and if their thickness be small compared with their breadth, we can suppose that the lines of flow are at right angles to the length of the strips and parallel to the faces of the strips. These currents will alter the value of the magnetising force H, and hence H will have different values at different distances from the surface of the plate. We shall show that there is a diminution in the amplitude of H as we go into the plate. To compensate for this diminution in the value of the magnetising force, we must increase the magnetising current of the transformer, and hence increase the losses in the copper conductors. Again, since by Steinmetz's law the hysteresis loss varies as J5 1 ' 6 , the hysteresis loss will be greater than if the magnetic induction were uniform and had its density equal to the mean value of B. The iron at the centre of the plates is screened, to a certain extent, by the eddy currents, and we shall see that this screening effect is greater the thicker the plates and the higher the frequency of the magnetic force. Let A A' (Fig. 125) be a line perpendicular to the two faces of Eddy currents the plate and let be the middle point of AA'. p'late, Let OZ be parallel to the direction of the applied magnetic force and let OF be at right angles to ' caiiy varying QZ and OX. Suppose that the field is produced magnetic field, . . . . with its sides by harmonic alternating currents flowing in a 1 6 c il f insulated wire wrapped round the plate, R. i. 23 354 ALTERNATING CURRENT THEORY [CH. A' PP f the windings lying in planes parallel to the plane XOY, the coil forming an infinitely long and infinitely broad sole- noid with OZ for its axis, so that the lines of force are everywhere parallel to OZ. By making the breadth of the strip great enough in propor- tion to its thickness, we can consider that the eddy cur- rents, except near the edges, are all flowing in directions parallel to OF. We suppose Fig. 125. AA' the thickness of the plate therefore that every tube of equals 2a, and YOZ is the median plane. current which cutg the plane XOZ at a distance x from OZ also cuts the same plane at a distance x from it. Let, i_ be the current density at a point P (Fig. 125) in the iron, where OP equals x, and let a be the resistivity of iron. Consider a small circuit in the plane XOY of breadth dx and unit length. The electromotive force round this circuit is cr[i+-T-dx} t, when x is + a or a. Let us suppose that H = h cos o)t + k sin cot is a solution of (3) where h and k are functions of x. Substituting this value for H in (3), we get (0(h sin cot -f k cos cot) = .- - ( -y-r cos wt + -j sin wt } , 4<7r/jL \dx* dx* J a relation which must be true for all values of t. Hence we must have a d 2 k cr d z h ^ . and k = T j> - 47r//,co da? 4}7r/jLO) dx? (a and therefore h = -= _ 4m* where m 2 7 dec* where /is the frequency of the magnetising current. Thus m = 27r/ ........................... (5). Now assume that h is Ae nx and substitute this value in (4). We find that 7i 4 = 4m 4 , therefore n 2 = 2m 2 (- V - 1), and n = m(\ V 1). The complete solution of (4) is therefore h = Ae mx cos mx 4- e ma; sin mx + Ce~ mx cos m^ + De~ mx sin m^ 1 , where A, B, C and Z) are constants. Again, we have 1 d*h ~ 2m 2 da? ' On substituting for h and performing the differentiations, we get k = - Ae mx sin ra# + Be mx cos ma? + Ce~^ nx sin m# - De- wa; cos mx. XVl] INFINITE PLATE 357 Now H = h cos cot + k sin cot, and therefore H = u4e ?n * cos we see that, as a? increases from zero, cosh 2mx + cos 2mx increases continually from its least value 2. Thus, we see from (8) that the amplitude of H diminishes as we approach the centre of the plate, where it has its minimum value which is given by (11) below. We see also from (9) that the phase of H is different for different values of x. If the thickness of the plate were - , that is, if m ' ma equals mr, then sin mx would be a factor of both terms in the numerator and 7 would be zero when x is 0, , Hence mm m at depths -- , , ... the magnetising force, in this case, would be 777* < YfL in phase with its surface value. If y c be the phase difference between the phase of the surface value of H and the value of H at the centre, then by (9) tan y c = tan ma tanh ma .................. (10). XVl] SCREENING EFFECT Also, if H c be the amplitude of H at the centre, then 359 .(11). (cosh 2ma + cos 2 In the sheet-iron used in the cores of transformers ordinary values of a and //, are 10,000 C.G.s. units and 2500 respectively. If the frequency were 100, then by (5) m /2500 x V" io,o( 10,000 = 107T. Therefore if ma equal TT, that is, if the thickness of the plates were two millimetres, we see from (10) that y c would be 180 degrees and from (11) by the aid of the table on p. 376 that H c would be #0/11*6. If the plates were four millimetres thick 7T dr The total electromotive force round the circle of radius r is 27m'o- and is also equal to the rate at which the total flux of induction through it is diminishing. Hence, assuming a constant permeability, we get /*>' /J J-T 27m'cr= IJL I ' ZTrrdr .................. (6). J o at Hence from (6) and (a) ra dH r dH Id/ dH\ ^TT/jidH ana theretore -^- [r = = ^ ..................... (c). r dr \ dr ) a dt This equation is identical in form with the corresponding problem of the diffusion of heat into a cylinder. We will assume that the magnetising current follows the harmonic law so that the magnetic force at the surface of the core is H Q cos wt. Since H will be a periodic function of the time, we may assume the solution of (c) to be H = H^ cos wt + H z sin o>, XVI] EDDY CURRENTS IN CYLINDER 369 where H l and H 2 are functions of r, but not of t. Substituting this value for H in (c) we get d f dH 47ru,ft> , 1 d z , and --- r - = r dr\ dr J a r dr\ dr J cr Therefore _ /_cr_Y 1 d_ d^ 1 d_ \ z \ LQ) J r j \4f7TfjLQ) J r dr dr r dr dr where m 2 = -^ . We see also that H 2 satisfies an exactly <7 similar equation. If H l = a Q + a l r + a 2 then, by substituting in (e), we find that a 4 . =. + a 5 .3 2 .5 2 r + a 6 .4 2 .6V+...|. Hence i = 0, a 3 =0, a 5 = 0, ..., , w 4 m 4 anOl QJA ^ -r^ Qn, OJR ~n ~ZT~ &*>, . Therefore IT _ 4a 2 /m 2 m 2 V 2 2 2 2 .4 2 .6 2 "V Kelvin writes this in the form H l a c ber (mr) + --, bei (mr\ where ber (x) = l- .4 2 2 2 .4 2 .6 2 .8 2 and bei ^) = i-2 2 ^ 2 T6 2 + -" If J (a?) be the Bessel's function of zero order, we have ^o(^) = l-2 2 +2T4 2 ~2 2 .4 2 .6 2 + "" R. i. 24 370 ALTERNATING CURRENT THEORY [CH. Hence Therefore ber (tc) - V^l bei (a;) = J. \x (- 1)*}. Similarly ber (a } I ber (mr) a bei (mr) > sin wt. m } Now, when r equals R, the radius of the cylinder, we must have H equal to H cos wt. Hence the boundary conditions give us 4*7 H, = a ber (mR) + ^ bei (mR), = ^1 ber (mR) - a bei (mR). m mu r -^o ber (mR) Therefore a = , -^-V^T^ ^ ber 2 (mR) + bei 2 (mR) , 4a 2 -^o bei (mE) m 2 ber 2 (mJ?) + bei 2 (mR) ' The complete solution is therefore H = G {ber (mR) ber (mr) + bei (mR) bei (mr)} cos wt + C {bei (mJ2) ber (m?-) ber (m,ft) bei (mr)} sin wt . . .(/), XVI] THE MAGNETISING FOKCE 371 TT where C = T ^ ir^7 \ her 2 (mR) + bei 2 (mR) The amplitude of H, at a point at a distance r from the axis of the cylinder, is . j ber 2 (mr) + bei 2 (mr) H .................. ( ' (ber 2 (m) + bei 2 and we shall show later on that this increases as r increases. At the axis of the cylinder, where r equals zero, the amplitude becomes {ber 2 (mR) + bei 2 ( If 7 be the phase difference between the value of IT at a distance r from the axis and its value on the circumference of the cylinder, then _ bei (mR) ber (mr) ber (mR) bei (mr) ,. n 7 = ber (mR) ber (mr) + bei (mR) bei (mr) ...... () ' Tables for ber (x) and bei (x) are given at the end of the chapter. When r is zero, ber (mr) is 1 and bei (mr) is 0, and then bei(mR) . Now Hence, if the frequency of the applied magnetising force be 100 and if the cylinder be made of copper, for which JM equals unity and a equals 1600, we have _87T 2 .100 1600 and therefore m = 2'22 nearly. Now, from the tables, we see that bei (mR) vanishes when mR is zero, it then increases as mR increases and attains a maximum value for some value of mR between 3'5 and 4. It then diminishes and vanishes when mR is a little greater than 5. When mR is 5, 242 372 ALTERNATING CURRENT THEORY [CH. R is 2*2 centimetres nearly, tan y c is approximately zero and the maximum value of H along the axis of the cylinder is H /^2. Now, from (j) and the tables for bei (x) and ber(V) given at the end of the chapter, we see that when mli is zero, 7ri Cm {ber (mR) ber' (mr) + bei (mR) bei' (mr)} cos cot + Cm {bei (mR) ber' (mr) ber (mR) bei' (mr)} sin cot, where ber 7 (x) = -r- {ber (#)}, cLx and " bei' (x) -r- {bei (x)}. The effective value A of the current is therefore given by and since m 2 equals 8?r 2 , we get A 2 Pf TT ber ' 2 (mr) + bei /2 (mr) 4o- ber 2 (mR) + bei 2 (mR) ' XVI] EDDY CURRENT LOSSES 373 Now, if W watts be the mean power expended in a length I of the core, and thus hrif*. J fU {r ber' 2 (mr) + r bei' 2 (mr)} dr . J o -WIH.-I eiBL+e . (m Again, integrating by parts, I ber' (mr) . r ber'(mr) dr = ber (mr) . r ber' (mr) J f fYif -- I ber (mr) -^- {r ber' (mr)} dr, m> J cir and I bei' (mr) . r bei 7 (mr) dr = bei (mr) . r bei' (mr) -- I bei (mr) -=- {r bei' (mr)} dr. m/ J dr By differentiating the series for ber (mr) and bei (mr), it is easy to show that d d , . . , . -j- r -j- ber (mr) = m 2 r bei (mr), dr dr and d~ r d~ ^ ei ( mr ) = m * r k er ( wr )- The sum of the integrals on the right-hand sides of the two equations therefore vanishes when they are taken between the same limits, and thus, integrating between the limits, we have w - 1 7? fi TI 21 A-? ber ( m -K) ber' (mR) 4- bei (mR) bei' (mR) ~ "- - TT-lt /Lt / 6J3 Q J.U . - - i -- 7 - ;J^T - - . - ;prr - . '2m ber 2 (mR) + bei 2 (m#) Since W must always be positive we see that ber (mR) ber' (mR) + bei (wU) bei' (mR), or must always be positive. Thus ber 2 (mR) + bei 2 (mR) increases as mR increases, and thus also, ber 2 (x) + bei 2 (x) increases as x increases. This result is useful in studying some of the equations given above. 374 ALTERNATING CURRENT THEORY [CH. In practice we want to know the average power expended per cubic centimetre of the conductor on account of eddy currents. Hence, dividing W by TrlR 2 and noting that m 2 equals 87r 2 yu//R. (Stokes's Theorem.) /3. If a resistance x (Fig. 1286) be placed in parallel with a coil of resistance R and if the total current through a. Fig. 128 a. The power expended on x is a maximum when x = R (E const.). the two coils in parallel have the constant value C, then the power expended in x is a maximum when x equals the resistance R of the coil, and the maximum power is C 2 ^/4. The preceding theorems may be written symbolically as follows : (a) If c E = ce + c*R (E, R constant), then ce is a maximum when c equals E/2R. KECIPKOCAL THEOREMS XVII] 08) If then ec is a maximum when e equals CR/2. (C, l/R constant), H 383 Fig. 128 6. The power expended on a; is a maximum when x = R (G const.). a. If the potential difference between A and B (Fig. 129 a) be constant, the current in x is a minimum when 2# equals R. R-.T A B Fig. 129 a. The current in is a minimum when 2x equals R (P.D. between A and B constant). /3. If the current in the main be constant the voltage across is a maximum when 2# equals R (Fig. 129 6). R-x B Fig. 129 b. The voltage across x is a maximum when 2x equals R (total current constant). 384 ALTERNATING CURRENT THEORY [CH. The method of duality is particularly suggestive when applied to alternating current theory. We remind the reader that a choking coil is an ideal coil having inductance but no resistance. The equations for a condenser and a choking coil are Capacity and inductance. i = K -v- and ,- di dt' We may thus regard K as the reciprocal of L and that a condenser is the reciprocal of a choking coil. By integrating the preceding equations we get q = Ke and < = Li. Flux and quantity. Hence < and q are reciprocal quantities. We shall now illustrate the method by giving a few reciprocal theorems in which use is made of the following table of reciprocal quantities and connections. (*) e or V r K series (0) i or A 1/r L q parallel a. The current in a choking coil is determined by j di dt' For a given value of the effective voltage V, and for a given frequency, the effective value of the choking coil current A is a maximum when e is sine shaped. ft. The voltage across a condenser is determined by ._ ^de dt' For a given value of A and for a given frequency the condenser potential difference V is a maximum when i is sine shaped. a. The equation to the current in an inductive coil is T di e = Rl + L dt' Inductive coil and leaky condenser. XVIl] RECIPROCAL THEOREMS 385 13. This reciprocates into e An inductive coil therefore reciprocates into a condenser shunted by a non-inductive resistance. a. In sine curve theory, the impedance Z of an inductive coil is given by the formula ft. In sine curve theory, the impedance Z of a leaky con- denser is given by the formula Z V R^ a. In sine curve theory, when the applied potential difference is constant, the mean power expended at a given frequency, in a variable resistance in series with a choking coil, is a maximum when R is La). (J. Hopkinson.) ft. In sine curve theory, when the main current is constant the mean power expended at a given frequency, in a variable resistance shunting a condenser, is a maximum when \JR is Kw. a.. In sine curve theory, a condenser K may be replaced by a choking coil whose resistance is zero and inductance -. 2 . (Rayleigh.) ft. In sine curve theory, a choking coil L may be replaced by a condenser whose resistance is infinite and capacity -j - . a. In sine curve theory, when the current in a resistance coil and choking coil in series lags behind the applied potential difference by an angle 6, then wLi R ~ W.L ft. In sine curve theory, when the potential difference between the terminals of a resistance coil and a condenser in parallel lags behind the main current by an angle 0, then sin 0= l wK j; cosfl= ^ R } ; tau0 = coKR. \& + * J \R 2 + W ~ ) R. i. 25 386 ALTERNATING CURRENT THEORY [CH. a. In sine curve theory, if 6 be the angle of phase difference between the current and the potential difference applied to a resistance coil, a choking coil and a condenser in series, then , , (*-*>> <><> i /3. In sine curve theory, if 6 be the angle of phase difference between the potential difference and the current supplying a resistance coil, a condenser and a choking coil in parallel, then ~. La*) OL. The power expended in n leaky condensers in series is TV TV . The power expended in n coils in parallel is whether the coils are inductive or not. a. The power factor of an inductive coil is given by RA /3. The power factor of a leaky condenser is given by F OL. The resonance of electromotive forces. When a condenser and a choking coil are in series and the effective value of the applied potential difference is constant, the effective potential difference across the terminals of either attains maximum values when LK{(2n+I)a>}*=I. /3. The resonance of currents. When a choking coil and a condenser are in parallel and the main current is constant, the currents in either of them attain maximum values when OL. In sine curve theory, when the effective value of the main XVII] RECIPROCAL THEOREMS 387 current is constant, the effective current in an inductive coil (Fig. 130 a) shunted by a condenser is a maximum, when _L ~ R?~+ Fig. 130 a. The current in the coil L is a maximum when K= The main current is constant. /3. In sine curve theory, when the applied potential difference is constant, the potential difference across a leaky condenser, which is put in series with a choking coil, is a maximum when the self inductance L is given by Fig. 1306. The P.D. across Kis a maximum when it is put in series with a coil whose K self inductance L=. The applied P.D. is constant. The above theorems, which are of importance in practical work, can easily be proved graphically. The main steps in the proof, by the method of Chap, vn, may be written as follows : 252 388 ALTERNATING CURRENT THEORY [CH. a. 0. Inductive coil shunted by a Leaky condenser in series with a condenser. choking coil. i M-/ where where Pi = and ana 1 P2= K J~T\ Thus A 1 Thus V 1 1 T 7 :a) 2 +if Hence, if A" be the only variable, Hence, if L be the only variable, [j is a maximum, when V l is a maximum, when .// -r A COR. I. Ct Q If L be the only variable, A 1 is a If A' be the only variable, Fj is a maximum, when maximum, when 1 1 COR. II. 0. If the frequency be the only If the frequency be the only variable and if 2Z be greater than variable and if 2K be greater than J5T/2 2 , A l is a maximum, when L/R 2 , V l is a maximum, when 2L-KR? _ " = 2AV, 2 ' a. If we have w. coils in parallel and if their time constants are all equal and their mutual inductances zero, so that then XVIl] RECIPROCAL THEOREMS 389 where i\ is the instantaneous value of the current in the coil (R lt A) and i is the instantaneous value of the current in the main. /3. If we have n leaky condensers in series and if their time constants are all equal, so that then Vl where ^ is the instantaneous value of the potential difference across the condenser K lt and v is the instantaneous value of the applied potential difference. a. In sine curve theory, when a constant effective potential difference is applied at the terminals of a non-inductive resistance R in series with a shunted choking coil, the effective current in the main is a minimum when * = |i + i>F-27J' where x is the non-inductive resistance shunting the choking coil L. This is a particular case of a theorem given on page 168. /3. In sine curve theory, when the effective value of the current in the main is constant and we have two branch circuits, one being a resistance R and the other a condenser K in series with a variable resistance x, then the effective potential difference between the terminals of the fixed resistance is a maximum, when (B? 1 )* R -{*+*$- The general theorem given on page 168 may be reciprocated in the same manner (see Figs. 131 a and 131 b). a.. In sine curve theory, if we have a condenser K in series with a resistance R, and if the combination be shunted by a choking coil L in series with a resistance R, then if LKar = 1, the 390 ALTERNATING CURRENT THEORY [CH. combination is equivalent to a non-inductive coil whose resistance R' is given by 7?'- L R M ~ n/WWWVWVn R V Fig. 131 a. The main current is a minimum when R has a certain value. The applied P.D. is constant. Fig. 131 6. The P.D. is a maximum when R has a certain value. The main current is constant. /3. In sine curve theory, if we have a choking coil L shunted by a resistance R, and if the combination be in series with a condenser K shunted by a resistance R, then if KLw 2 = 1, the combination is equivalent to a non-inductive coil whose resistance R' is given by R' 2L^2R' The following is an outline of the analytical proof of the above theorems arranged so as to show that from the mathematical point of view the two problems are identical. /s. L . _ e l jj, dei 1 "T" A 3j7~ .K at a. XVII] RECIPROCAL THEOREMS 391 If the functions follow the harmonic law and if we differentiate twice the equations containing the integral sign and divide by ft) 2 , we obtain i di, we get and where Assuming e = E sin cot, E sin (cot a) Esm(cot + a) tan a = Leo JR. Hence E sin cot h + ^ 2 cos a and R = R Assuming i I sin &)^, / sin (cot a) we get 0! and where _ 7 sin (cot + a) " and a. When a condenser K (Fig. 132 a) shunted by a resistance R is placed in series with a choking coil L shunted by a resistance R, the combination will act like a non-inductive resistance R at all frequencies and whatever the shape of the wave of the applied potential difference, provided that L equals KR 2 (see page 87). L -vwv J Fig. 132 a. When L equals KR" 2 the combination acts like a non-inductive resistance R. 392 ALTERNATING CURRENT THEORY [CH. /3. When a choking coil L (Fig. 132 b) in series with a resistance R is placed in parallel with a condenser K in series with a resistance R, the combination will act like a non-inductive resistance R at all frequencies and whatever the shape of the wave of the main current, provided that K equals L/R- (see page 86). Fig. 132 b. When K equals L/.R 2 the combination acts like a non-inductive resistance JR. Let us now consider whether the coefficients of self and mutual induction for electrostatic charges have the cor- responding electromagnetic coefficients for their reciprocals. Maxwell's equations for the electro- static charges in terms of the potentials of n conductors (page 90) are Electrostatic and electromagnetic coefficients. # If (j) p be the flux through a circuit p, which has L p . p and L p . q for its self and mutual inductances respectively, and the current i p entirely surrounds the flux p , as it does when the circuit has infinite conductivity, then the electromagnetic equations to n circuits carrying currents ^, i' 2 , ... respectively are 01 = 1.1*1 + A. 2*2 + ... + i. n *, 02 = -2.1*1 + ^2.2*2 + + L. 2 . n i n , We see, then, that an electrostatic system of n conductors at given potentials reciprocates into an electromagnetic system of n circuits carrying given currents, if we assume that L p . p is the XVIl] POTENTIAL COEFFICIENTS 393 reciprocal of K p , p and that L p . q is the reciprocal of K p , q . In the above equations it is to be noted that It is also to be noticed that when p and q are different K p . q is always negative but L p . q is not necessarily negative. We have supposed that the conductors and circuits or coils have perfect conductivity. By means of determinants, Maxwell's equations can be written in the form where p L in is given by the equation p Lm A = 3/j. 7n , A being the symmetrical determinant SJT l . 1 lT > .j....JP. J and M L m being the coefficient of X L m in A. Maxwell calls p lfl , p lf . 2 , . . . coefficients of potential. Similarly in the electromagnetic problem we have and (n 1) similar equations, where ^..A'-jri.., A' being the symmetrical determinant + L lfl L. 2 . 2 ... Z n> n and J/'z. w being the coefficient of Z?. m in A x . We may call X K1 , XL 2 , ... coefficients of current. The above equations prove that \ Lm is the reciprocal of pi. m . a. The capacity K v for equal potentials of n conductors is given by for when they are all at the same potential v, t-(ii.i-+^i.i+ +^i.^X? ft = (^'2.1 + /lo.o + . .. + K*. n ) V, and hence K v = (2q)/v = 2 since K 394 ALTERNATING CURRENT THEORY [CH. The electrostatic energy of the system in this case is ^v^q or JjfiTeW 2 . It follows from a theorem proved in a note at the end of this chapter that, if v be maintained constant, the mutual electrical actions of the conductors will tend to move them so that ^K v v* increases. Thus as the conductors separate under the action of the electric forces, K v continually increases. /3. The self inductance L s of n coils in series is given by for when they are all carrying the same current i, = A. 1*1 + i.i*ii and = I/a. 1*1 + -^2. 2V Hence (f>(L 2 . 2 - A. 2 ) = (L^L^- L\^)i^ and Therefore But i\ + 4 is the current in the main and therefore -/2L p . 13. The capacity K q for equal charges of n conductors is given by where p lfl , p lf2 , ... are the potential coefficients. The electro- static energy of the conductors is q 2 /2K q . It is easy to see that the mutual electric forces acting on the conductors tend to increase K q . a. When we have a system of n circuits in parallel, the ratio of (j) to h is constant and is called the effective inductance of the first circuit. /3. When we have n conductors each of which has a charge q, the ratio of q to v l is constant and is called the effective capacity of the first conductor. It is worth noting that, when all the charges on a system of conductors are equal and the charges on all neighbouring con- ductors are zero, the electrostatic energy is q*/2K qt and thus depends only on the charge, the geometrical configuration of the system and its position. Similarly, when all the potentials are main- tained at a known value, the electrostatic energy can be written down at once when K v is known. Now, in many cases, K v can be measured easily experimentally, and in some cases K p can be found. A knowledge of the value of these quantities will be helpful, therefore, in studying the electrical properties of a system of fixed conductors. Similarly a knowledge of L s and L p will be a help in studying a system of coils. We have shown that the following quantities are reciprocals : *,* Kp.q The capacity between two conductors V. Lp.q The self-inductance of two coils in cross parallel XVII] STAR AND MESH 397 Denoting the coefficients of current and potential by \ lmlj tj 2 >--- and PI. i> PI.*)*" we a ^ so have: Pi. i Pi- -2 a. Formula for the electromagnetic wattmeter : The electro- P = kl^. magnetic and Q ne practically non-inductive coil shunts the load, the electrostatic r * m wattmeter. an 1 r-sin0 1>2 ' ft. In a non-inductive mesh load, if p, q and r be the resistances of the arms, sin # 2 . 3 sin #3.! sin # 1>2 ' a. If the load consist of three equal non-inductive resistances connected mesh fashion, then AS + Af + AJ = 9 (// + / 2 4 + / 3 4 ), where A l , A 2 and A 3 are the currents in the three mains and /!, 7 2 and / 3 are the mesh currents. ft. If the load consist of three equal non-inductive resistances connected star fashion, then FV + Fi + F a = 9 + a. The formula for the potential differences across one of the arms of a non-inductive star load is 1 \ 2 / IT" 2 V 2 \ / 1 1 + + 1 = (LJ* + LSI I + 1 - (1 1 I + i n r 3 z r 3 r 2 r ft. The formula for the current 'in one side of a non-inductive mesh load is 2 V 3 ) (r 2 + r,) - ^iV^g. a. The mesh voltages being constant, the sum of the voltages between the centre of a star load and the three mains is a minimum when the currents in the arms are all equal. ft. The currents in the mains being constant the sum of the mesh currents is a minimum when the potential differences between the mains are equal. XVIl] TWO PHASE EXAMPLES 399 An inspection of the formulae and constructions, given in TWO phase Chapter XII for two phase theory, will show how easy examples. ^ j g O reciprocate many of them. a. Rule for finding the voltages across the arms of a non- inductive star load when the voltage tetrahedron and the re- sistances of the arms are given. Find the centre of gravity G of masses - , - , and placed 7*i 7*2 7*3 r 4 at the four angular points of the voltage tetrahedron ; then, the lines joining G to the four angular points will give the magnitudes nd the phase differences of the required voltages. /3. Rule for finding the currents in a non-inductive mesh load when the current tetrahedron and the resistances of the arms are given. Find the centre of gravity G of masses i\, n, r s and r 4 placed at the four angular points of the current tetrahedron ; then, the lines joining G to the four angular points will give the magnitudes and the phase differences of the required currents. (a) The power w expended on a star load (p. 256) is given by w = vfa + V 2 a 2 + v s a s + v 4 4 . (y8) The power w expended on a mesh load is given by W = 12.3^.3 + 4.4^3.4 + *4.lV 4 .l + l'l. 2 Vi. 2 . The above examples could easily be multiplied, but sufficient have been given to prove that the method of duality is as useful in electrical theory as it is in geometry. REFERENCES. LORD RAYLEIGH, Theory of Sound, 'The Reciprocal Theorem,' Vol. 1, 109. (Second edition.) W. E. AYRTOX and W. E. SUMPXER, Proc. Phys. Soc. London, 'Alternate Current and Potential Difference Analogies in the Methods of Measuring Power.' June 1891. P. A. MACMAHOX, The Electrician, Vol. 28, p. 601, 'The Combinations of Resistances.' 1892. H. SIRE DE VILAR, L'Eclairage Electrique, Vol. 27, p. 252, 'La Dualite en Electrotechnique.' 1901. G. F. C. SEARLE, Proc. of the Camb. Phil. Soc., Vol. 12, p. 378, 'On the Calculation of Capacities in Terms of the Coefficients of Electrostatic Induction.' 1904. ADDITIONS. NOTE ON A THEOREM IN ELECTROSTATICS. ON p. 192 it is assumed that when the potentials of a system of conductors, one of which is movable, are maintained constant, the work done on the movable one, during an infinitely small displacement, is equal to the gain in the electrostatic energy of the system. This is a particular case of a well-known theorem in electrostatics which can be proved as follows. Let us consider the case of n charged conductors which form charges a self-contained system screened from outside electric constant. i n fl uence s. To simplify the problem let us suppose that only one of them is movable. Let us also suppose, in the first place, that all the conductors are insulated, so that the charges on them remain constant. By the Conservation of Energy the movable conductor X cannot, unless acted on by external forces, move under the action of the electric forces into a position where the electro- static energy of the system is greater, otherwise the total energy of the system would be increased. The electric forces acting on X move it in the direction along which the electrostatic energy diminishes most rapidly, as the force acting on X will be greatest in this direction. We see therefore that the motion which ensues* due to the action of the electric forces, must diminish the electro- static energy of the system and increase the mechanical energy. Let dw denote the work done on X by the electric forces. during an infinitely small displacement from P to P' t then, with the notation of p. 90, if v l -\-dv lt v z + dv 2) ... be the new values of v l , # 2 , we have ,) - 4^2 (> 2 + dv 2 ) - ... ELECTROSTATIC THEOREM Now q is constant, and since (p. 90) 401 .. . e have = d therefore 4- v a Hence, by substituting in (1), we get . 2 v, + . . .) dv 1 - $ ^ -h ... + v^o.o + .... When the potentials of the TI conductors are maintained constant Potentials aij( ^ have the same initial values as in the last case, the constant. difference between the values of the electrostatic energy when X is in the positions P' and P equals .x + dK lml ) + ... By the preceding paragraph this is equal to c?w and is therefore positive. Hence the electrostatic energy in the position P' is greater than in the position P. Also, since the force acting on X in the position P is exactly the same in the two cases, and P r is infinitely near to P, so that, the electrostatic field being only infinitesimally disturbed by the motion of X, the force at P' is practically the same in the two cases. The mechanical work done on X, therefore, during an infinitesimal displacement is the same whether the charges or the potentials are maintained constant. Thus the work done on X, when the potentials are constant, is also dw, and this equals the gain in the electrostatic energy of the system. It follows, by integration, that the total work done on X during a finite displacement, when the potentials are constant, equals the gain in the electrostatic energy of the system. In an electrostatic voltmeter, for instance, the energy taken from the mains equals twice the mechanical energy required to displace the moving part. R. I. 26 402 ALTERNATING CURRENT THEORY The above theorems can be stated more generally as follows : (1) When, in a system of insulated conductors, the relative positions of the conductors alter owing to their mutual electric actions, the conductors move in such a way that the electrostatic energy is diminished, the diminution being equal to the work done on the conductors by the electric forces. (2) When, in a system of conductors, the potentials of which are maintained constant by means of external sources, the relative positions of the conductors alter owing to their mutual electric actions, the conductors move in such a way that the electrostatic energy is increased by an amount exactly equal to the work done on the conductors by the electric forces. It is instructive to reciprocate these general theorems by the Reciprocal method of Chapter xvn. Reciprocating the second, theorems. we see t na fc w hen the currents in a system of n coils are maintained constant, the motion due to their attractions or repulsions increases the electromagnetic energy of the system. We see also,, that the mechanical work done on the coils is exactly equal to the increase of the electromagnetic energy by the electro- magnetic forces. This theorem can be proved at once by means of Lagrange's general equations of motion (see Maxwell's Electricity and Magnetism, Vol. n. 580). Maxwell notes the resemblance w T ith the corresponding electrostatic problem. The theorem was given by Sir William Thomson (Lord Kelvin) in the second edition of Nichol's Cyclopaedia of Physical Science (Article, " Magnetism, Dynamical Relations of") published in 1860. See also Papers on Electrostatics and Magnetism by Sir William Thomson, p. 446, Second Edition. INDEX. Addenbrooke, G. L., 196 Alternating currents, 40 in inductive circuits, 41 Alternating magnetic fields, 286 Ampere's theorem, 18 Arc, musical, 83 Argand's method, 162 Armagnat, 366 Arno's phase indicator, 302 B, 15 Behn-Eschenburg, J., 349 Bipolar circles, 307 Branched circuits, 167 formulae for, 169, 170 graphical solution, 175 minimum energy in, 172 Cable, two core, 103, 336 three core, 105, 121, 339 three phase, 107, 119, 121, 321, 340, 345, 347 four core, 113, 128, 323 twin concentric, 115, 324 Campbell's, A., method of measuring power, 207 Capacity, 9 and inductance, 384 and wave shape, 80 definitions, 91 et seq. for equal charges, 396 for equal potentials, 393 two parallel cylinders, 101 concentric main, 93 triple concentric main, 98 cable with three cores, 125 cable with four cores, 127 cable with n cores, 129 Capacity (continued) cylindrical condenser, 103 cylinder parallel to the earth, 131 in practice, 137 three phase overhead wires, 135 two wires parallel to the earth, 132, 135 Carey Foster, 349 Cassinian ovals, 309 Choking coil currents, 77 Circle diagram of transformer, 215 Complex, currents, 67 number, 161 variable, 162 Concentric main, inductance, 53 magnetic field round, 327 triple, 61 Condenser, currents, 78 definition, 92 energy stored in, 10 equivalent, 109, 111, 114, 119 in secondary load of a trans- former, 214, 218 leaky, and inductive coil, 384 Conductors connected in cross paral- lel, 395 Conduits, losses in, 337 Constant charges, 400 Conversion of polyphase systems, 265 Coulomb's law, 7 Cross parallel, 395 Current, electric, 10 circular, 19 coefficients, 393 in circle, 30 in rectangle, 31 tetrahedron, two phase, 251 triangle, three phase, 330 404 INDEX Dielectric coefficient, 2 Duality, 380 et seq. Duddell, the direct current arc, 83 transformer waves, 213 Eddy currents, 350 et seq. analogy with heat, 355 and hysteresis losses, 375 in an iron plate, 353 in a copper plate, 367 in a cylinder, 368 in thin sheets, 361 in secondary of transformer, 351 in short-circuited coil, 351 Effective values, 65 graphical methods of finding, 69 Electrodynamics, 17 Electromagnetic, energy, 23 induction, 21 wattmeter, 197 with mutual inductance, 201 Electromotive force, 3 in a conductor cutting lines of force, 26 induced, 22 Electrostatics, 2 Electrostatic, theorem, 400 forces in three phase cable, 347 and electromagnetic coefficients, 392 Electrostatic wattmeter, 193 Elliptic field, 283 Energy in branched circuits, 173 Equipotential lines round parallel wires, 307 et seq. Equivalent net-work of a trans- former, 213 Equivolt curves, 71 of equal height, 76 Ewing, eddy currents and hysteresis, 363 Farad, 94 Ferraris, 298 Fleming's rule, 27 Flux of force, 4 magnetic, 51 Flux and quantity, 384 in circular rings, 48 et seq. Force, magnetic, inside cylindrical cur- rent sheet, 33 near a straight conductor, 31 Force, magnetic (continued) on moving wire, 27 outside cylindrical current sheet, 32 Frequency of the current in the fourth wire of a three phase system, 234 Fresnel, 298 Gauss, the, 14 Gauss's theorem, 4 Geometrical applications of power for- mulae, 259 Gliding magnetic fields, 296 Green's, a theorem of, 8 H, 15 Harmonics, effect of, on eddy currents, 365 effect of, on resonance, 81 in magnetic induction, 366 Heap's phase indicator, 273 Heat and electrical equations identical, 355 Heaviside, Oliver, eddy currents, 350 et seq. electromagnetic waves, 47 formula, 374 Hopkinson's, J., a theorem of, 385 Hysteresis, 35, 37 losses, 375 in armouring of cables, 346 I, 15 Images, electric, 9 Imaginary quantities, 161 currents and E.M.F.S, 165 Impedance, 158 with parabolic wave, 160 Indicator, phase, 268, 273, 302 Induced E.M.F., 22 current and motion, 26 Inductance, 22, 140 and capacity, 384 and wave shape, 80 anchor rings, 52 calculation of, 44 concentric cylinder, 53 minimum, 62 mutual, 22 of surface currents, 142 self, 23 three parallel cylinders, 61 INDEX 405 Inductance (continued) triple concentric mains, 61 two parallel wires, 56 Inductances, comparison of, by a volt- meter, 87 Induction, electromagnetic, 21 electrostatic, 8 Inductive coil and leaky condenser, 384 Intensity of magnetisation, 14 Inverse points, 99 Iron armouring, losses in, 346 Iron conduits, losses in, 335 Iron plate, eddy currents in, 353 et seq. J. J. Thomson's solution, 355 Joule's law, 20 Kelvin, bei and ber functions, 369 formula, 25 self-energy, 25 theory of images, 9, 125, 347 theorem, 402 Thomson and Tait, 284 Kirchhoff's laws, 381 Laplace's formula, 29 Lay of wires, 323 Leaky condenser and induction coil, 384 Lenz's law, 22 Lines of force, 3 round parallel wires, 304 et seq. Load on a three phase alternator, 235 Longitudinal tension, 63 Magnetic analogy with Ohm's law, 51 Magnetic field round polyphase cables, 304 et seq. round n parallel wires, 325 round three phase cables, 315 round concentric main, 328 round twin concentric cables, 325 round two parallel wires, 307, 309, 311 round two phase cables, 323 Magnetic, shell, 16 steel strips, 35 tests, 34 Magnetism, 11 Magnetomotive force, 51 Mathematical tables, 1-5, 1-55 and l'6th powers, 38 6, sine, cos8, sinhfl, cosh0, 376 e n and 6~ n , 377 ber (a), bei (or), ber'(z), bei'(:c), 378 ber 2 (z)+bei 2 0r); etc., 378 Maxwell, the, 13 Maxwell's, electrostatic equations, 90 equation, 45 potential coefficients, 393 self-inductance formula, 54 theory of light, 95 transformer formula, 217 vector potential, 329 Mean value of an alternating function, 68 Mesh and star. 220, 397 Mesh, load, 220 voltages and phase differences, 222, 246 currents and phase differences, 228, 251 Meter, induction type, 275 polyphase, 279 watt-hour, 201 wattless current, 274 Microfarad, 94 Model, polyphase cable, 117 three phase cable, 119 Musical arc, 83 Mutual inductance, 22 Mutual potential energy of two shells, 16 Neutral point, rotation of, 319, 324 Neutralising capacity, 87 inductance, 86 Ohm's law, 20 Parabolic wave impedance, 160 Parallel and series, 381 Permeability, 14 of sheet steel, 37 Phase difference, 149 in two phase system, 255 and time lag, 153 Phase indicator, 268 Heap's, 273 Arno's, 302 Poisson's equations, 5, 7 406 INDEX Polarity, 20 Polycore cable, capacity, 129 inductance, 142 Polyphase cable, magnetic field, 304 et seq. cable, model of, 117 transformer, 261 Potential, electrostatic, 2 magnetic, 11 of a bar magnet, 12 of centre of star load, 226 of three phase mains, 226 of two phase mains, 249 Potentials, constant, 401 Potier, A., 292 Power factor, 145 geometrical interpretation, 154 numerical examples, 151 of a three phase system, 265 unity, 146 wattmeter method of finding, 266 zero, 155 Power, measurement, 189 et seq. three phase loads, 256 two phase loads, 236 Quadrant electrometer, 190 B.M.S., root mean square, 66 Kadial magnetic force, 330 round concentric main, 333 round three phase main, 342 round two parallel wires, 336 round n parallel wires, 331 Katio of units, 97 Kayleigh, 179 Eeactance, 159 Reciprocal quantities and theorems, 380 et seq. Eeferences, 39, 64, 88, 122, 144, 160, 179, 188, 210, 218, 242, 279, 303, 349, 379, 399 Eeisz's method, 203 Eeluctance, 51 Eepulsion of wires, 63 Eesistance, 21 and wave shape, 79 Eesonance, 81 method of measuring power, 209 of E.M.F.S, 81, 386 of currents, 84, 386 Eesonance (continued) with direct current, 83 Eimington's theorem, 217 Eings, iron, 48 Eotating magnetic fields, 281 et seq. due to three phase currents, 300 when the inducing forces are not sine shaped, 299 producing a constant effective E. M.F., 301 properties of, 286 pure, 289 Eotation of neutral point, three phase cable, 319 two phase cable, 324 Scott's, C.F., polyphase transformer, 261 Screening effect of eddy currents, 359 Searle, G. F. C., formulae for magnetic force, 329 formula for eddy currents, 336 Self-energy, of an electric circuit, 24 formulae, 55 Series and parallel, 381 Similar waves, 146 three phase, 233, 397 two phase, 253, 399 Sine waves, effective value of, 66 mean value of, 68 hyperbolic, 72 family of, 74 Skin effect, 47 Specific inductive capacity, 2 Spiral of cable cores, 117, 323 Star and mesh, 220, 397 Star-box, 239 Star, load, 220 theorems, 241, 242 Steinmetz's formula, 36 Stokes, G. G., a theorem of, 382 Surface currents, 139 inductance of, 141 formulae, 142 Symmetrical alternating curve, 152 System of conductors, charges main- tained constant, 396, 400 potentials maintained constant, 393, 401 Tables of, 1 '5,1-55 and l-6th powers, 38 mathematical functions, 376 et seq. INDEX 407 Tangential magnetic force, 331 round concentric main, 333 round three phase cable, 343 round two parallel wires, 336 round n parallel wires, 331 Telephone disturbance, 349 Tetrahedron, current, 251 voltage, 246 Thomson, Elihu, watt-hour meter, 202 Thomson, J. J., 137, 350 eddy currents in infinite plate, 353 Three phase, alternators, 219 cables, 339, et seq. capacity currents, 227 current formulae, 229 magnetic field round, 315 measurement of power, 236 et seq. meters, 240 reciprocal theorems, 397 voltages, 222 voltage rule, 224 wave form on balanced load, 232 wave form, restriction on, 233 Three ammeter method, 206, 397 Three voltmeter method, 205, 334, 397 Time lag, 150 Transformer, air core, 211 et seq. circle diagram, 215 condenser in secondary load, 214 constant current, 218 equivalent net-work, 213 leading primary current, 217 Maxwell's formula, 217 Eimington's theorem, 217 Tubes of force, electrostatic, 6 electromagnetic, 13 Twin concentric cable, 115 magnetic field round, 325 Two phase, alternator, 244 . cable, 111 current tetrahedron, 251 formulae, 251 measurement of power, 255 meters, 258, 259 P.D. waves, 253 systems, 243 et seq. Two phase (continued) voltage tetrahedron, 246 voltage formulae, 247 Unity, power factor, 146 Vector potential, 329 of three phase mains, 341 Vectors, 162 et seq. addition of, 182 condition that they lie in one plane, 181 condition that four can be represented graphically, 185 division of, by a complex number, 164 extension of definition, 182 failure of, 186 in space, 182 multiplication of, 163 of a constant quantity, 182 parallelogram, 180 polygon, 163 resultant of three, 184 Velocity of light, 96 Voltage, tetrahedron, 246 triangle, 222 three phase equations, 225 two phase equations, 247 Watt current, 159 E.M.F., 158 Wattless current, 159 E.M.F., 158 Watt-hour meter, 201 induction type, 275 three phase, 240 two phase, 258, 259 polyphase, 279 Wattmeter, electrostatic, 193 shunted, 195 electromagnetic, 197 method of finding cos 0, 266 with mutual induction, 200 Zero power factor, 155 CAMBRIDGE I PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS. CAMBRIDGE PHYSICAL SERIES. 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