CHAPTER IV VECTOR REPRESENTATION 16. While alternating waves can be, and frequently are, rep- resented graphically in rectangular coordinates, with the time as abscissae, and the instantaneous values of the wave as ordinates, the best insight with regard to the mutual relation of different alternating waves is given by their representation as vectors, in the so-called crank diagram. A vector, equal in length to the maximum value of the alternating wave, revolves at uniform speed so as to make a complete revolution per period, and the pro- jections of this revolving vector on the horizontal then denote the instantaneous values of the wave. Obviously, by this diagram only sine waves can be represented or, in general, waves which are so near sine shape that they can be represented by a sine. Let, for instance, 01 represent in length the maximum value / of a sine wave of current. Assuming then a vector, 01, to revolve, left handed or in counter-clockwise direc- tion, so that it makes a complete revolution during each cycle or period U. If then at a certain moment of time, this vector stands in position 01 1 (Fig. 8), the projec- tion, 0A\, of 0I\ on the horizontal line 0.4 represents the instantaneous value of the current at this moment. At a later mo- ment, 01 has moved farther, to 01 1, and the projection, OA-i, of 0/2 on OA is the instantaneous value at this later moment. The diagram so shows the instantaneous condition of the sine wave: each sine wave reaches its maximum at the moment of time where its revolving vector passes the horizontal, and reaches zero at the moment where its revolving vector passes the vertical. If now the time, t, and thus the angle, ^ = 10 A = 2ir — (where to to = time of one complete cycle or period), is counted from the moment of time where the revolving vector 01 in Fig. 8 stands in position Oil, then this sine wave would be represented by i = I cos (?? - i?i), 19 20 ALTERNATING-CURRENT PHENOMENA where ??i — IiOA may be called the phase of the wave, and I = QJi the amplitude or intensity. At the time, «> = ??i, that is, the angle, ??i, after the moment of time represented by position OIi, i = I, and 01 passes through the horizontal OA, that is, has its maximum value. The phase ^1 thus is the angle representing the time, ti, at which the wave reaches its maximum value. If the time, t, and thus the angle, ??, are counted from the moment at which the revolving vector reaches position Oh, the equation of the wave would be i = I cos (?> - ??2), and t?2 = hOA is the phase. 17. When dealing with one wave only, it obviously is imma- terial from which moment of time as zero value the time and thus the angle, t?, is counted. That is, the phase ??] or t?2 may be chosen anything desired. As^oon, however, as several alternating waves enter the diagram, it is obvious that for all the waves of the same diagram the time must be counted from the same moment, and by choosing the phase angle of one of the waves, that of the others is determined. Thus, let / = the maximum value of a current, lagging behind the maximum value of voltage E by time ti, that is, angle of phase difference ??i = 2 tt - to The phase of the voltage, E, then may be chosen as a, and the voltage represented, in Fig. 9, by vector OE ^ E at phase angle EOA = a. As the current lags by phase difference ??i, the phase of the current then must be /3 = « + t?i, and the current is represented, in Fig. 9, by vector 01 = /, under phase angle (3 = 10 A. The equations of voltage and current then are: e — E cos (t? — a) i = / cos (?? - /3) = I cos (t? — a — ??i). The voltage OE = E, as the first vector, may be plotted in any desired direction, for instance, under angle — a' = EOA in Fig. 10. The current then would be represented by 01 = I, under VECTOR REPRESENTATION 21 Fig. 10. phase angle — /3' = — (a' — ??]) = 10 A, and the equations of voltage and current would be: e = E cos (i? 4- a') - i = I cos {t} + /3') = / cos (?? + «'- ??i). Or, the current 01 = I may be chosen as the first vector, in Fig. 9, under phase angle /3 = 10 A ^ and the voltage then would have the phase angle a = ^ — §1, and be represented by vector OE = E, and the equations would be: i = I cos (?? - /3) e = E cos (?? — a.) = E cos (?? - /3 + ??]). In this vector representation, a current lagging behind its voltage makes a greater angle with the horizontal, OA, that is, the current vector, 01, lags behind the voltage vector, OE, in the direction of rotation, thus passes the zero line, OA, of maximum value, at a later time. Inversely, a leading current passes the zero line OA earlier, that is, is ahead in the direction of rotation. Instead of the maximum value of the rotating vector, the effective value is commonly used, especially where the instan- taneous values are not required, but the diagram intended to represent the relations of the dif- ferent alternating waves to each other. With the length of the rotating vector equal to the effect- ive value of the alternating wave, the maximum value obviously is •\/2 times the length of the vector, and the instantaneous values are ■\/2 times the projections of the vectors on the horizontal. 18. To combine different sine waves, their graphical representations as vectors, are combined by the parallelogram law. If, for instance, two sine waves, OEi, and OE2 (Fig- H), are superposed — as, for instance, two e.m.fs. acting in the same cir- cuit— their resultant wave is represented by OE, the diagonal of a parallelogram with OEi and OE2 as sides. As the projection of 22 ALTERNATING-CURRENT PHENOMENA the diagonal of a parallelogram equals the sum of the projections of the sides, during the rotation of the parallelogram OE1EE2, the projection of OE on the horizontal OA, that is, the instan- taneous value of the wave represented by vector OE, is equal to the sum of the projection of the two sides OEi and OE2, that is, the sum of the instantaneous values of the component vectors OEi and OE2. From the foregoing considerations we have the conclusions: The sine wave is represented graphically in the crank diagram, by a vector, which by its length, OE, denotes the intensity, and by its amplitude, AOE, the phase, of the sine wave. Sine waves are combined or resolved graphically, in vector representation, by the law of the parallelogram or the polygon of sine waves. Kirchhofif's laws now assume, for alternating sine waves, the form : (a) The resultant of all the e.m.fs. in a closed circuit, as found by the parallelogram of sine waves, is zero if the counter e.m.fs. of resistance and of reactance are included. (6) The resultant of all the currents toward a distributing point, as found by the parallelogram of sine waves, is zero. The power equation expressed graphically is as follows: The power of an alternating-current circuit is represented in vector representation by the product of the current, I, into the projection of the e.m.f., E, upon the current, or by the e.m.f., E, into the projection of the current, I, £3 Eo upon the e.m.f., or by IE cos d, where A ^,--^^'' ^ ~ angle of phase displacement. ^J,.^--^ / 19. Suppose, as an example, that in ^^ *E ^' a line having the resistance, r, and the reactance, x = 2 irfL — where / = fre- quency and L = inductance — there p, ,r> exists a current of / amp., the line being connected to a non-inductive circuit operating at a voltage of E volts. What will be the voltage required at the generator end of the line? In the vector diagram. Fig. 12, let the phase of the current be assumed as the initial or zero line, 01. Since the receiving cir- cuit is non-inductive, the current is in phase with its voltage. Hence the voltage, E, at the end of the line, impressed upon the receiving circuit, is represented by a vector, OE. To overcome VECTOR REPRESENTATION 23 the resistance, r, of the hne, a voltage, Ir, is required in phase with the current, represented by OEi in the diagram. The inductive reactance of the hne generates an e.m.f. which is pro- portional to the current, /, and the reactance, x, and lags a quarter of a period, or 90°, behind the current. To overcome this counter e.m.f. of inductive reactance, a voltage of the value Ix is required, in phase 90° ahead of the current, hence represented by vector 0E2- Thus resistance consumes voltage in phase, and reactance voltage 90° ahead of the current. The voltage of the generator, Eo, has to give the three voltages E, Ei, E^, hence it is determined as their resultant. Combining by the parallelo- gram law, OEi and OE2, give OEz, the voltage required to over- come the impedance of the line, and similarly OEz and OE give OEa, the voltage required at the generator side of the line, to yield the voltage, E, at the receiving end of the line. Algebraic- ally, we get from Fig. 12 E, = V{E + Iry + {IxY or E = VEo' - {Ixy - Ir. In this example we have considered the voltage consumed by the resistance (in phase with the current) and the voltage con- sumed by the reactance (90° ahead of the current) as parts, or components, of the impressed volt- age, ^0, and have derived E^ by combining Er, Ex, and E. 20. We may, however, introduce the effect of the inductive react- ance directly as an e.m.f., E'l, the counter e.m.f. of inductive react- ance = Ix, and lagging 90° behind the current; and the e.m.f. con- ^ , . , . Fig. 13. sumed by the resistance as a counter e.m.f., E'l =Ir, in opposition to the current, as is done in Fig. 13; and combine the three voltages Eq, E\, E'2, to form a resultant voltage E, which is left at the end of the line. E\ and E'2 combine to form E'3, the counter e.m.f. of impedance; and since £"3 and Eo must combine to form E, Eq is found as the side of a parallelogram, OEqEE's, whose other side, OE's, and diagonal QE, are given. Or we may say (Fig. 14), that to overcome the counter e.m.f. 24 ALTERNATING-CURRENT PHENOMENA of impedance, OE'^, of the line, the component, OEz, of the impressed voltage is required which, with the other component, OE, must give the impressed voltage, OEq. As shown, we can represent the voltages produced in a circuit in two ways — either as counter e.m.fs., which combine with the impressed voltage, or as parts, or components, of the impressed voltage, in the latter case being of opposite phase. According to the nature of the problem, either the one or the other way may be preferable. Fio. 14. As an example, the voltage consumed by the resistance is Ir, and in phase with the current; the counter e.m.f. of resistance is in opposition to the current. The voltage consumed by the reactance is Ix, and 90° ahead of the current, while the counter e.m.f. of reactance is 90° behind the current; so that, if, in Fig. 15, 01 is the current. OEi = voltage consumed by resistance, OE'i = counter e.m.f. of resistance, OE2 = voltage consumed l:)y inductive reactance, OE'2 = counter e.m.f. of inductive reactance, OE3 = voltage consumed by impedance, OE'i = counter e.m.f. of impedance. Obviously, these counter e.m.fs. are different from, for instance, the counter e.m.f. of a synchronous motor, in so far as they have no independent existence, but exist only through, and as long as the current exists. In this respect they are analogous to the opposing force of friction in mechanics. 21. Coming back to the equation found for the voltage at the generator end of the line, Eo - V{E + /r)2 + (7rc)2 VECTOR REPRESENTATION 25 we find, as the drop of potential in the line, e ^ Eo- E = V{E + /r)2 + (/x)* E. This is different from, and less than, the e.m.f. of impedance, E^ = Iz = iVr'- + x\ Hence it is wrong to calculate the drop of potential in a circuit by multiplying the current by the impedance; and the drop of potential in the line depends, with a given current fed over the line into a non-inductive circuit, not only upon the constants of the line, r and x, but also upon the voltage, E, at the end of line, as can readily be seen from the diagrams. 22. If the receiver circuit is inductive, that is, if the current, I, lags behind the voltage, £^_by an angle, 8, and we choose again as the zero line, the current 01 (Fig. 16), the voltage, OE, is ahead of Eo E2 Ei^^^y^ / 0 m\ , 1 ^E, Fig. 16. Fig. 17. the current by the angle, Q. The voltage consumed by the resist- ance, //', is in phase with the current, and represented by 0E\ the voltage consumed by the reactance, Ix, is 90° ahead of the current, and represented by OEi. Combining OE^ OEi, and OE2, we get OEo, the voltage required at the generator end of the line. Comparing Fig. 16 with Fig. 12, we see that in the former OEq is larger; or conversely, if £'0 is the same, E will be less with an inductive load. In other words, the drop of potential in an inductive line is greater if the receiving circuit is inductive than if it is non-inductive. From Fig. 16, Ea = V{E COS 6 4- /r)2 + (E sin 6 + Ixy. If, however, the current in the receiving circuit is leading, as 26 ALTERNATING-CURRENT PHENOMENA is the case when feeding condensers or synchronous motors whose counter e.m.f. is larger than the impressed voltage, then the voltage will be. represented, in Fig. 17, by a vector, OE, lagging behind the current, 01, by the angle of lead, d'; and in this case we get, by combining OE with OEi, in phase with the current, and OEi, 90° ahead of the current, the generator voltage, OEq, which in this case is not only less than in Fig. 16 and in Fig. 12, but may be even less than E', that is, the voltage rises in the line. In other words, in a circuit with leading current, the inductive reactance of the line raises the voltage, so that the drop of voltage is less than with a non-inductive load, or may even be negative, and the voltage at the generator lower than at the other end of the line. These diagrams. Figs. 12 to 17, can be considered vector dia- grams of an alternating-current generator of a generated e.m.f., E(i, a resistance voltage. Ex = Ir, a reactance voltage, E2 = Ix, and a difference of potential, E, at the alternator terminals; and we see, in this case, that with the same generated e.m.f., with an inductive load the potential difference at the alternator terminals will be lower than with a non-inductive load, and that with a non-inductive load it will be lower than when feeding into a cir- cuit with leading current, as for instance, a synchronous motor circuit under the circumstances stated above. 23. As a further example, we may consider the diagram of an alternating-current transformer, feeding through its secondary circuit an inductive load. For simplicity, we may neglect here the magnetic hysteresis, the effect of which will be fully treated in a separate chapter on this subject. Let the time be counted from the moment when the magnetic flux is zero and rising. The magnetic flux then passes its maxi- mum at the time ?? = 90°, and the phase of the magnetic flux thus is ?? = 90°, the flux thus represented by the vector 0* in Fig. 18, vertically downward. The e.m.f. generated by this mag- netic flux in the secondary circuit, Ei, lags 90° behind the flux; thus its vector, OEi, passes the zero line, OA 90°, later than the magnetic flux vector, or at the time t? = 180°; that is, the e.m.f. generated in the secondary by the magnetic flux, OEi, has the phase t? = 180°. The secondary current, /i, lags behind the e.m.f., El, by an angle, 61, which is determined by the resistance and inductive reactance of the secondary circuit; that is, by the VECTOR REPRESENTATION 27 load in the secondary circuit, and is represented in the diagram by the vector, OFi, of phase 180 + ^i. Instead of the secondary current, /i, we plot, however, the secondary m.m.f., Fi = nili, where ni is the number of secondary turns, and i^i is given in ampere-turns. This makes us inde- pendent of the ratio of transformation. Ei-^ From the secondary e.m.f., Ei, we get the flux, , required to induce this e.m.f., from the equation El = ^/2 7^/^l/$ 10-8; where El = secondary e.m.f., in effective volts, / = frequency, in cycles per second, Wi = number of secondary turns, $ = maximum value of magnetic flux, in lines of magnetic force. The derivation of this equation has been given in a preceding chapter. This magnetic flux, $, is represented by a vector, 0^, QO'' in phase, and to produce it a m.m.f., F, is required, which is de- termined by the magnetic characteristic of the iron and the section and length of the magnetic circuit of the transformer; this m.m.f. is in phase with the flux, $, and is represented by the vector, OF, in effective ampere-turns. The effect of hysteresis, neglected at present, is to shift OF ahead of 0, by an angle, a, the angle of hysteretic lead. (See Chapter on Hysteresis.) This m.m.f., F, is the resultant of the secondary m.m.f., Fi, 28 ALTERNATING-CURRENT PHENOMENA and the primary m.m.f., Fp; or graphically, OF is the diagonal of a parallelogram with OFi and OFq as sides, OFi and OF being known, we find OFq, the primary ampere-turns, and therefrom and the number of primary turns, no, the primary current, 7o = Fo — , which corresponds to the secondary, 7i. Wo To overcome the resistance, ro, of the primary coil, a voltage, Er — loVo, is required, jn phase with the current, h, and repre- sented by the vector, OEr. To overcome the reactance, Xo = 2x/Io, of the primary coil, a voltage, Ex = IqXq, is required, 90° ahead of the current, 7o, and represented by vector, OEx. The resultant magnetic flux, , which generates in the second- ary coil the e.m.f., Ei, generates in the primary coil an e.m.f. pro- portional to El by the ratio of turns — and in phase with Ei, or, ill E'i no 71 1 Er, which is represented by the vector, OE'i. To overcome this counter e.m.f., E'i^ a primary voltage, Ei, is required, equal but in phase opposition to E'l, and represented by the vector, OEi. Ei-< «; Fig. 19. The primary impressed e.m.f., Ep, must thus consist of the three components OEi, OEr, and OE^, and is, therefore, their resultant OEo, while the difference of phase in the primary cir- cuit is found to be 00 = EoOFo. 24. Thus, in Figs. 18 to 20, the diagram of a transformer is drawn for the same secondary e.m.f., E^, secondary current, /i, and therefore secondary m.m.f., Fi, but with different conditions of secondary phase displacement: VECTOR REPRESENTATION 29 In Fig. 18 the secondary current, /i, lags 60° behind the sec- ondary e.m.f., El. In Fig. 19, the secondary current, /i, is in phase with the sec- ondary e.m.f., El. In Fig. 20 the secondary current, /i, leads by 60° the secondary e.m.f.. El. These diagrams show that lag of the current in the secondary circuit increases and lead decreases the primary current and pri- mary impressed e.m.f. required to produce in the secondary circuit the same e.m.f. and current; or conversely, at a given primary impressed e.m.f., Eo, the secondary e.m.f., Ei, will be smaller with an inductive, and larger with a condensive (leading current), load than with a non-inductive load. Ei-* At the same time we see that a difference of phase existing in the secondary circuit of a transformer reappears in the primary circuit, somewhat decreased, if the current is leading, and slightly increased if lagging in phase. Later we shall see that hysteresis reduces the displacement in the primary circuit, so that, with an excessive lag in the secondary circuit, the lag in the primary circuit may be less than in the secondary. A conclusion from the foregoing is that the transformer is not suitable for producing currents of displaced phase, since primary and secondary current are, except at very light loads, very nearly in phase, or rather in opposition, to each other.