7. INDUCTANCE IN ALTERNATING-CURRENT CIRCUITS 34. An alternating current i = IQ sin 2irft or i — I0 sin 0 can be represented graphically in rectangular coordinates by a curved line as shown in Fig. 10, with the instantaneous values FIG. 10. — Alternating sine wave. i as ordinates and the time t, or the arc of the angle corresponding to the time, 6 = 2irft, as abscissas, counting the time from the zero value of the rising wave as zero point. If the zero value of current is not chosen as zero point of time, the wave is represented by i = /0 sin 2 IT/ (t - t'), or i = /osin (6 — 8'), where tf and 6' are respectively the time and the corresponding angle at which the current reaches its zero value in the ascendant. If such a sine wave of alternating current i = IQ sin 2 irft or i = IQ sin 6 passes through a circuit of resistance r and induc- tance L, the magnetic flux produced by the current and thus its interlinkages with the current, iL = IoL sin 0, vary in a wave 32 ELEMENTS OF ELECTRICAL ENGINEERING line similar also to that of the current, as shown in Fig. 11 as $. The e.m.f. generated hereby is proportional to the change of iL, and is thus a maximum where iL changes most rapidly, or at its zero point, and zero where iL is a maximum, and according to Lentz's law it is positive during falling and negative during rising current. Thus this generated e.m.f. is a wave following the wave of current by the time t = -,> where tQ is time of one 1 complete period, = -v or by the time angle 6 = 90°. FIG. 11. — Self-induction effects produced by an alternating sine wave of current. This e.m.f. is called the counter e.m.f. of inductance. It is .'•'• '•••• e'*=-Ljt = - 2 TT/L/O cos 2 irft. It is shown in dotted line in Fig. 11 as e'2. The quantity 2 irfL is called the inductive reactance of the circuit, and denoted by x. It is of the nature of a resistance, and expressed in ohms. If L is given in 109 absolute units or henrys, x appears in ohms. The counter e.m.f. of inductance of the current, i = /o sin 2 irft = /o sin 0) of effective value , V"2 IS e'2 = — xI0 cos 2 irft = — xIQ cos 6, having a maximum value of X!Q and an effective value of xh T E, = -- = xl; ALTERNATING-CURRENT CIRCUITS 33 that is, the effective value of the counter e.m.f. of inductance equals the reactance, x, times the effective value of the current, /, and lags 90 time degrees, or a quarter period, behind the current. 35. By the counter e.m.f. of inductance, e'z = — xIQ cos 0, which is generated by the change in flux due to the passage of the current i — IQ sin 0 through the circuit of reactance x, an equal but opposite e.m.f. ez = xIQ cos 0 is consumed, and thus has to be impressed upon the circuit. This e.m.f. is called the e.m.f. consumed by inductance. It is 90 time degrees, or a quarter period, ahead of the current, and shown in Fig. 11 as a drawn line e2. Thus we have to distinguish between counter e.m.f. of induc- tance 90 time degrees lagging, and e.m.f. consumed by inductance 90 time degrees leading. These e.m.fs. stand in the same relation as action and reaction in mechanics. They are shown in Fig. 11 as e'2 and as ez. The e.m.f. consumed by the resistance r of the circuit is pro- portional to the current, 61 = ri = r/0 sin 0, and in phase therewith, that is, reaches its maximum and its zero value at the same time as the current i, as shown by drawn line 61 in Fig. 11. Its effective value is EI = ri. The resistance can also be represented by a (fictitious) counter e.m.f., e\ = — r/0 sin 0, opposite in phase to the current, shown as e\ in dotted line in Fig. 11. The counter e.m.f. of resistance and the e.m.f. consumed by resistance have the same relation to each other as the counter e.m.f. of inductance and the e.m.f. consumed by inductance or inductive reactance. 36. If an alternating current i = 70 sin 0 of effective value I = —^ exists in a circuit of resistance r and inductance L, that is, of reactance x = 2 irfL, we have to distinguish ; 3 34 ELEMENTS OF ELECTRICAL ENGINEERING E.m.f. consumed by resistance, e\ = r!0 sin 6, of effective value EI = rl, and in phase with the current. Counter e.m.f. of resistance, e'\ = — r/o sin 6, of effective value EI = rl, and in opposition or 180 time degrees displaced from the current. E.m.f. consumed by reactance, ez = X!Q cos 6, of effective value E2 = xl, and leading the current by 90 time degrees or a quarter period. Counter e.m.f. of reactance, e'z = xlo cos 0, of effective value E'z = xl, and lagging 90 time degrees or a quarter period behind the current. The e.m.fs. consumed by resistance and by reactance are the e.m.fs. which have to be impressed upon the circuit to overcome the counter e.m.fs. of resistance and of reactance. Thus, the total counter e.m.f. of the circuit is e' = e'i + e'z = — IQ (r sin 6 + x cos-0), and the total impressed e.m.f., or e.m.f. consumed by the circuit, is e = ei + ez = /o (r sin 6 + x cos 0). Substituting 9S - = tan 00 and VV2 + r2 = z, it follows that x = z sin 00, r = z cos 00, and we have as the total impressed e.m.f. e = Z!Q sin (0 + 00), shown by heavy drawn line e in Fig. 11, and total counter e.m.f. e' = - zI0 sin (0 + 00), shown by heavy dotted line e' in Fig. 11, both of effective value e = zl. For 0 = — 00, e — 0, that is, the zero value of e is ahead of the zero value of current by the time angle 0o, or the current lags behind the impressed e.m.f. by the angle 00. 0o is called the angle of lag of the current, and z = \A"2 + x2 the impedance of the circuit, e is called the e.m.f. consumed by impedance, e' the counter e.m.f. of impedance. ALTERNATING-CURRENT CIRCUITS 35 Since Ei = rl is the e.m.f. consumed by resistance, Ez = xl is the e.m.f. consumed by reactance, and E = zl = \/r2 + x2 1 is the e.m.f. consumed by impe- dance, we have E = VES + #22, the total e.m.f. and Ei = E cos 00, E% = E sin 0o, its components. The tangent of the angle of lag is x 27T/L tan 00 = - = — — > and the time constant of the circuit is L _ tan &o r = ~27f" The total e.m.f., e, impressed upon the circuit consists of two components, one, e\t in phase with the current, the other one, e2, in quadrature with the current. Their effective values are E, E cos 00) E sin 00. EXAMPLES 37. (1) What is the reactance per wire of a transmission line of length Z, if ld = diameter of the wire, 18 = spacing of the wires, and/ = frequency? If / = current, in absolute units, in one wire of the trans- mission line, the m.m.f. is I; thus the magnetizing force in a zone dlx at distance lx from center of wire (Fig. 12) is / = 0 7 Z TTlx and the field intensity in this zone is H = 4 irf = 2 y— Thus Lx the magnetic flux in this zone is d* . H ldli m hence, the total magnetic flux between the wire and the return wire is L XI* d* = $ — | CfcSF = ^.f6| -y— = 2 1 1 IQge -j — > LX I'd 2 2 neglecting the flux inside the transmission wire. 36 ELEMENTS OF ELECTRICAL ENGINEERING The inductance is L = -y = 2 I loge -y^ absolute units -/ 'd 2 Z loge s 1(T9 h., I'd 21 s and the reactance x = 2 irfL = 4 irfl loge -y-, in absolute units; or x = 4 7T/7 loge -y^ 10~9, in ohms. 'd 38. (2) The voltage at the receiving end of a 33.3-cycle three-phase transmission line 14 miles in length shall be 5500 FIG. 12. — Diagram for calculation of inductance between two parallel conductors. between the lines. The line consists of three wires, No. 0 B. & S. (ld = 0.82 cm.), 18 in. (45 cm.) apart, of resistivity p = 1.8 X 10-6. (a) What is the resistance, the reactance, and the impedance per line, and the voltage consumed thereby at 44 amp. ? (6) What is the generator voltage between lines at 44 amp. to a non-inductive load? (c) What is the generator voltage between lines at 44 amp. to a load circuit of 45 degrees lag? (d) What is the generator voltage between lines at 44 amp. to a load circuit of 45 degrees lead? Here I = 14 miles = 14 X 1.6 X 105 = 2.23 X 106 cm. ld = 0.82 cm. Hence the cross section, A = 0.528 sq. cm. ALTERNATING-CURRENT CIRCUITS 37 Z 1.8 X 10-6 X 2.23 X 106 (a) Resistance per line, r = p - = = 7.60 ohms. 2L Reactance per line, x = 4 irfl log, j- X 10~9 = ^ X 33.3 X 2.23 X 106 X log, 110 X 10~9 = 4.35 ohms. The impedance per line, z = \/r2 -f- x'2 = 8.76 ohms. Thus if I = 44 amp. per line, the e.m.f. consumed by resistance is EI = rl = 334 volts, the e.m.f. consumed by reactance is Ez = xl = 192 volts, and the e.m.f. consumed by impedance is E3 = zl = 385 volts. (b) 5500 volts between lines at receiving circuit give — -j= = v 3 3170 volts between line and neutral or zero point (Fig. 13), or per line, corresponding to a maxi- mum voltage of 3170 A/2 = 4500 volts. 44 amp. effective per line gives a maxi- mum value of 44 -\/2 = 62 amp. Denoting the current by i = 62 sin 0, the voltage per line at the receiv- ing end with non-inductive load is e = 4500 sin 0. The e.m.f. consumed by resistance, in phase with the current, of effective JTIG 13 _ Voltage diagram for value 334, and maximum value 334 a three-phase circuit. \/2 = 472, is ei = 472 sin 0. The e.m.f. consumed by reactance, 90 time degrees ahead of the current, of effective value 192, and maximum value 192 -\/2 = 272, is e2 = 272 cos 0. Thus the total voltage required per line at the generator end of the line is eQ = e + el + e2= (4500 + 472) sin 0 + 272 cos 0 = 4972 sin 0 + 272 cos 0. 272 Denoting .n_0 = tan 00, we have tan 0o 272 C°S "° = 4980 1 _ 4972 ~ 4980- 38 ELEMENTS OF ELECTRICAL ENGINEERING Hence, BQ = 4980 (sin 0 cos 0o + cos 6 sin 00) = 4980 sin (0 + 00). Thus 0o is the lag of the current behind the e.m.f. at the generator end of the line, = 3.2 time degrees, and 4980 the 4980 maximum voltage per line at the generator end; thus EQ = = 3520, the effective voltage per line, and 3520 \/3 = 6100, the effective voltage between the lines at the generator. (c) If the current i = 62 sin 0 lags in time 45 degrees behind the e.m.f. at the receiving end of the line, this e.m.f. is expressed by e = 4500 sin (0 + 45) = 3170 (sin 0 + cos 0); that is, it leads the current by 45 time degrees, or is zero at 0 = — 45 time degrees. The e.m.f. consumed by resistance and by reactance being the same as in (6), the generator voltage per line is €o = e -f 6l -f e2 = 3642 sin 0 + 3442 cos 0. 3442 Denoting QA/fo = tan 00, we have OO4Z e0 = 5011 sin (0 -f- 00). Thus 00, the angle of lag of the current- behind the gen- erator e.m.f., is 43 degrees, and 5011 the maximum voltage; hence 3550 the effective voltage per line, and 3550 -\/3 = 6160 the effective voltage between lines at the generator. (d) If the current i = 62 sin 0 leads the e.m.f. by 45 degrees, the e.m.f. at the receiving end is e = 4500 sin (0 - 45) = 3180 (sin 0 - cos 0). Thus at the generator end €Q = e -f 6l + e2 = 3652 sin 0 - 2908 cos 0. 2908 Denoting = tan ^o, it is e0 = 4670 sin (0 - 00). Thus 00, the time angle of lead at the generator, is 39 degrees, and 4654 the maximum voltage; hence 3290 the effective vol- tage per line and 5710 the effective voltage between lines at the generator. POWER IN ALTERNATING-CURRENT CIRCUITS 39