CHAPTER X. MUTUAL INDUCTANCE. 82. In the preceding chapters, circuits have been considered containing resistance, self-inductance, and capacity, but no mutual inductance; that is, the phenomena which take place in the circuit have been assumed as depending upon the impressed e.m.f. and the constants of the circuit, but not upon the phenomena taking place in any other circuit. Of the magnetic flux produced by the current in a circuit and interlinked with this circuit, a part may be interlinked with a second circuit also, and so by its change generate an e.m.f. in the second circuit, and part of the magnetic flux produced by Fig. 38. Mutual inductance between circuits. the current in a second circuit and interlinked with the second circuit may be interlinked also with the first circuit, and a change of current in the second circuit, that is, a change of magnetic flux produced by the current in the second circuit, then generates an e.m.f. in the first circuit. Diagrammatically the mutual inductance between two circuits can be sketched as shown by M in Fig. 38, by two coaxial coils, while the self-inductance is shown by a single coil L, and the resistance by a zigzag line. 141 142 TRANSIENT PHENOMENA The presence of mutual inductance, with a second circuit, introduces into the equation of the circuit a term depending upon the current in the second circuit. If i^ = the current in the circuit and r1 = the resistance of the circuit, then r^\ = the e.m.f. consumed by the resistance of the circuit. If L1 = the inductance of the circuit, that is, total number of interlinkages between the circuit and the number of lines of magnetic force produced by unit current in the circuit, we have di L{j± = e.m.f. consumed by the inductance, ctt where, t = time. If instead of time t an angle 6 = 2 nft is introduced, where / is some standard frequency, as 60 cycles, di x. 3^ = e.m.f. consumed by the inductance, au where x1 = 2 nfL^ = inductive reactance. If now M = mutual inductance between the circuit and another circuit, that is, number of interlinkages of the circuit with the magnetic flux produced by unit current in the second circuit, and i2 = the current in the second circuit, then M—~= e.m.f. consumed by mutual inductance in the first at circuit, M— - = e.m.f. consumed by mutual inductance in the second at circuit. Introducing xm = 2 nfM = mutual reactance between the two circuits, we have di xm-^= e.m.f. consumed by mutual inductance in the first do circuit, di xm—±= e.m.f. consumed by mutual inductance in the second au circuit. MUTUAL INDUCTANCE 143 If now e^ = the e.m.f. impressed upon the first circuit and e2 = the e.m.f. impressed upon the second circuit, the equations of the circuits are dL di7 e, = r^ + x^-± + xm ^ + xci and r J il dO (1) -*2^+^^+*C2/V^, (2) where r1 = the resistance, xl = 2 7r/L1 = the inductive re- actance, and xci = = the condensive reactance of the first circuit; r2 = the resistance, x2 = 2 rfL2 = the inductive reactance, xca = = the condensive reactance of the second circuit, and xm = 2 nfM = mutual inductive reactance between the two circuits. 83. In these equations, xl and x2 are the total inductive reactance, Ll and L2 the total inductance of the circuit, that is, the number of magnetic interlinkages of the circuit with the total flux produced by unit current in the circuit, the self- inductive flux as well as the mutual inductive flux, and not merely the self-inductive reactance and inductance respectively. In induction apparatus, such as transformers and induction machines, it is usually preferable to separate the total reactance z, into the self-inductive reactance, x81 referring to the magnetic flux interlinked with the inducing circuit only, but with no other circuit, and the mutual inductive reactance, xm, usually represented as a susceptance, which refers to the mutual induc- tive component of the total inductance; in which case x = xs + xm. This is not done in the present case. Furthermore it is assumed that the circuits are inductively related to each other symmetrically, or reduced thereto; that is, where the mutual inductance is due to coils enclosed in the first circuit, interlinked magnetically with coils enclosed in the second circuit, as the primary and the secondary coils of a transformer, or a shunt and a series field winding of a generator, 144 TRANSIENT PHENOMENA the two coils are assumed as of the same number of turns, or reduced thereto. ri, No. turns second circuit If a = — = — =rr— — - -- : - r— , the currents in the nA No. turns first circuit second circuit are multiplied, the e.m.fs. divided by a, the resis- tances and reactances divided by a2, to reduce the second circuit to the first circuit, in the manner customary in dealing with transformers and especially induction machines.* If the ratio of the number of turns is introduced in the equa- tions, that is, in the first equation — xm substituted for xm, in the KI 79 second equation — xm for xm, and the equations then are n2 and di~ n. di. Since the solution and further investigation of these equations (3), (4) are the same as in the case of equations (1) and (2), except that nl and n2 appear as factors, it is preferable to eliminate nl and n2 by reducing one circuit to the other by the ratio of turns M a = — , and then use the simpler equations (1), (2). ni 84. (A) Circuits containing resistance, inductance, and mutual inductance but no capacity. In such a circuit, shown diagrammatically in Fig. 38, we have di. di~ ~ * and e2 = r2i2 + x2— 2 + xm ^ • (6) Differentiating (6) gives de2 _ di2 d?i2 d2il ~dd~~T2dd^ x*~dF~VXm^] * See the chapters on induction machines, etc., in " Theory and Calcula- tion of Alternating Current Phenomena." MUTUAL INDUCTANCE 145 from (5) follows /^ / « v* *"~ , ^ ~^r "' and, differentiated, fo * Substituting (8) and (9) in (7) gives del de2 _ . dil + <*,*,-*-•>£, (10) and analogously, de2 de^ + (*,*, - ^2) ^ • (ID Equations (10) and (11) are the two differential equations of second order, of currents i\ and iv If e/, i/ and e/, i2' are the permanent values of impressed e.m.fs. and of currents in the two circuits, and e/', if and e2", if are their transient terms, we have, e2 - c/ + el', Since the permanent terms must fulfill the differential equations (10) and (11), de' de' . di/ W + x*^g-x'»W"= Wl + (r'X2 + T>X*}^9 (Pi ' + (xiX2-Xm*)^ (12) and del */ • , r + "*" ^ " rr' + (13) 146 TRANSIENT PHENOMENA subtracting equations (12) and (13) from (10) and (11) gives the differential equations of the transient terms, de," def x di" Wi" + x*~dd- ~ Xm^dd"'= W" + (ri*3 + Vi) ~^T cPi " + (xA - xm>) -^ (14) and *£ '* dd "m dd f1'2"2 ' V1~2 ' '2^ dd (Pi, 2 rle2 + x1 -— xm ——- — rjT2i2 + (r^ -f T2Xj) — - (15) These differential equations of the transient terms are the same as the general differential equations (10) and (11) and the differential equations of the permanent terms (12) and (13). 85. If, as is usually the case, the impressed e.m.fs. contain no transient term, that is, the transient terms of current do not react upon the sources of supply of the impressed e.m.fs. and affect them, we have e^ = 0 and - ea" = 0; hence, the differential equations of the transient terms are di (Pi 0 = rjj + (r,x2 + r2x,) — + (x,x2 - xm2) — (16) and are the same for both currents i" and i2", that is, the transient terms of currents differ only by their integration constants, or the terminal conditions. Equation (16) is integrated by the function i = Ae-a9. (17) Substituting (17) in (16) gives A£-a'{rir2 - a (r,x2 + r2xj + a2 (x,x2 - xm2)} - 0; hence, A — indefinite, as integration constant, and _ r 2 ' rr _r2 *^W *t/1*x/2 *°W MUTUAL INDUCTANCE 147 The exponent a is given by a quadratic equation (18). This quadratic equation (18) always has two real roots, and in this respect differs from the quadratic equation appearing in a circuit containing capacity, which latter may have two imaginary roots and so give rise to an oscillation. Mutual induction in the absence of capacity thus always gives a logarithmic transient term; thus, a = (r^2 + T^} \ ; (T^ rf + * T^Xm • (19) •>//y«/y» /y» * \ Z/ ^jU/2 .t/fn / As seen, the term under the radical in (19) is always positive, that is, the two roots al and a2 always real and always positive, since the square root is smaller than the term outside of it. Herefrom then follows the integral equation of one of the currents, for instance iv as o -- <>' ' 4_ A ff~aid4- A e-a*9 f90\ *i T »i P -Aj* r ^12£ i \^) and eliminating from the two equations (5) and (6) the term dL pji + (x^x2 - xm2) - + xme2 - x2e^ , (21) Tx 2m leaving the two integration constants Al and A2 to be deter- mined by the terminal conditions, as 0 = 0, and i2 = i2°. 86. If the impressed e.m.fs. e1 and e2 are constant, we have hence, the equations of the permanent terms (12) and (13) give thus: and ^ and tV - - ; (22) T T (23) where, A/ and A2 follow from Al and A2 by equation (21). 148 TRANSIENT PHENOMENA If the mutual inductance between the two circuits is perfect, that is, xm2 = x,xv (24) - r 2 •^m equation (18) becomes, by multiplication with - r^2 -r / ^ a=rXr+\x ; (25) that is, only one transient term exists. As example may be considered a circuit having the following constants: et =" 100 volts; e2 = 0; rt = 5 ohms; r2 = 5 ohms; xl = 100 ohms; x2 = 100 ohms, and xm = 80 ohms. This gives i' = 20 amp. and i2 = 0, and a2 - 0.278 a + 0.00695 = 0; the roots are at = 0.0278 and a2 = 0.251 and By equation (21), ^=-25 + 1.25^ + 9^; hence, For 6 = 0 let if = 18 amp., or the current 10 per cent below the normal, and if = 0; then substituted, gives: 18 = 20 + Al + A2 and 0 = Al - A2, hence, A^ = A2 = — 1; and we have il = 20 — (£-°-0278* + £-°'2510) and i2=- (e-°-0278 -£-°-251'). MUTUAL INDUCTANCE 149 87. An interesting application of the preceding is the inves- tigation of the building up of an overcompounded direct-current generator, with sudden changes of load, or the building up, or down, of a compound wound direct-current booster. While it would be desirable that a generator or booster, under sudden changes of load, should instantly adjust its voltage to the change so as to avoid a temporary fluctuation of voltage, actually an appreciable time must elapse. A 600-kw. 8-pole direct-current generator overcompounds from 500 volts at no load to 600 volts at terminals at full load of 1000 amperes. The circuit constants are: resistance of armature winding, r0 = 0.01 ohm; resistance of series field winding, r2' = 0.003 ohm; number of turns per pole in shunt field winding, n1= 1000, and magnetic flux per pole at 500 volts, 4> = 10 megalines. At 600 volts full load terminal voltage (or voltage from brush to brush) the generated e.m.f. is e + irQ = 610 volts. From the saturation curve or magnetic characteristics of the machine, we have: At no load and 500 volts : 5000 ampere-turns, 10 megalines and 5 amp. in shunt field circuit. At no load and 600 volts : 7000 ampere-turns and 12 megalines. At no load and 610 volts: 7200 ampere-turns and 12.2 megalines. At full load and 600 volts: 8500 ampere-turns, 12.2 megalines and 6 amp. in shunt field. Hence the demagnetizing force of the armature, due to the shift of brushes, is 1300 ampere-turns per pole. At 600 volts and full load the shunt field winding takes 6 amperes, and gives 6000 ampere-turns, so that the series field winding has to supply 2500 ampere-turns per pole, of which 1300 are consumed by the armature reaction and 1200 magnetize. At 1000 amp. full load the series field winding thus has 2.5 turns per pole, of which 1.3 neutralize the armature reaction and ft 2 — 1.2 turns are effective magnetizing turns. 150 TRANSIENT PHENOMENA The ratio of effective turns in series field winding and in shunt field winding is a = ^ = 1.2 X 10"3. This then is the reduc- n, tion factor of the shunt circuit to the series circuit. It is convenient to reduce the phenomena taking place in the shunt field winding to the same number of turns as the series field winding, by the factors a and a2 respectively. If then 6 = terminal voltage of the armature, or voltage impressed upon the main circuit consisting of series field winding and external circuit, the same voltage is impressed upon the shunt field winding and reduced to the main circuit by factor «, gives e, = ae = 1.2 X W~3 e. Since at 500 volts impressed the shunt field current is 5 amperes, the field rheostat must be set so as to give to the shunt 500 field circuit the total resistance of r/ = - - = 100 ohms. 5 Reduced to the main circuit by the square of the ratio of turns, this gives the resistance, fi = av/ = 144 X 10~6 ohms. An increase of ampere-turns from 5000 to 7000, corresponding to an increase of current in the shunt field winding by 2 amperes, increases the generated e.m.f. from 500 to 600 volts, and the magnetic flux from 10 to 12, or by 2 megalines per pole. In the induction range covered by the overcompounding from 500 to 600 volts, 1 ampere increase in the shunt field increases the flux by 1 megaline per pole, and so, with nl = 1000 turns, gives 109' magnetic interlinkages per pole, or 8 X 109 interlinkages with 8 poles, per ampere, hence 80 X 109 interlinkages per unit current or 10 amperes, that is, an inductance of 80 henrys. Reduced to the main circuit this gives an inductance of 1.22 X 10~6 X 80 = 115.2 X 10~6 henrys. This is the inductance due to the magnetic flux in the field poles, which interlinks with shunt and series coil, or the mutual inductance, M = 115.2 X 10~6 henrys. Assuming the total inductance Lt of the shunt field winding as 10 per cent higher than the mutual inductance M, that is, assuming 10 per cent stray flux, we have Lx = 1.1 M = 126.7 X 10~6 henrys. MUTUAL INDUCTANCE 151 In the main circuit, full load is 1000 amp. at 600 volts. This gives the effective resistance of the main circuit as r = 0.6 ohm. The quantities referring to the main circuit may be denoted without index. The total inductance of the main circuit depends upon the character of the load. Assuming an average railway motor load, the inductance may be estimated as about L = 2000 X 10~6 henry s. * In the present problem the impressed e.m.fs. are not constant but depend upon the currents, that is, the sum i + iv where t\ = shunt field current reduced to the main circuit by the ratio of turns. The' impressed e.m.f., e, is approximately proportional to the magnetic flux <£, hence less than proportional to the current, in consequence of magnetic saturation. Thus we have e = 500 volts for 5000 ampere-turns, 5000 or i + il — — — - = 4170 amp. and \.2i e = 600 volts for 7200 ampere-turns, 7200 or i + it = - - = 6000 amp. ; \.2i hence, 1830 amp. produce a 'rise of voltage of 100, or 1 amp. 100 1 raises the voltage by 1830 18.3 6000 At 6000 amp. the voltage is — - = 328 volts higher than at lo.o 0 amp., that is, the voltage in the range of saturation between 500 and 600 volts, when assuming the saturation curve in this range as straight line, is given by the equation The impressed e.m.f. of the shunt field is the same, hence, reduced to the main circuit by the ratio of turns, a = 1.2 X 10~3, is e, = 152 TRANSIENT PHENOMENA Assuming now as standard frequency, / = 60 cycles per sec., the constants of the two mutually inductive circuits shown diagrammatically in Fig. 38 are : Main Circuit. Shunt Field Circuit. Current i amp. il amp Impressed e.m.f... Resistance *= 272 1^ VOHS T = 6 ohms e1 = (272+~Ml.2X10-3volts rt = 0 144X 10~3 ohms Inductance Reactance, 2xfL. . Mutual inductance Mutual reactance . L = 2000X ID"6 henrys x = 755X 10-3 ohms M = 115.2 > xm = 43.5 > L1= 126. 7X10-6 henrys Xl= 47.8X10-3 ohms ( 10-6 henrys ; 10~3 ohms This gives the differential equations of the problem as and (26) (27) 88. Eliminating — from equations (26) and (27) gives do di — l- = 0.695 i - 0.0712 ^ - 338. Equation (28) substituted in (26) gives i = 13.07 ~n + 9.95 i - 4950. ad Equation (29) substituted in (28) gives J± = _ o.93 — - 0.015 i + 15. Equation (29) differentiated, and equated with (30), gives + 0.828 0.00115 i- 1.15 = 0. (28) (29) (30) (31) MUTUAL INDUCTANCE 153 Equation (31) is integrated by i = i'0 + A£-°*. Substituting this in (31) gives Ara0{a? - 0.828 a + 0.00115} + {0.00115 iQ - 1.15} -= 0, hence, i0 = 1000, A is indefinite, as integration constant, and a2 - 0.828 a + 0.00115 - 0; thus a = 0.414 ± 0.4126, and the roots are at = 0.0014 and a2 = 0.827. Therefore i = 1000 + A/-0-0014' + A/"0'827'. (32) Substituting (32) in (29) gives i, - 5000 + 9.932 A/"0-0014'- 0.85 A2r °'827'. (33) Substituting in (32) and (33) the terminal conditions 0 = 0, i = 0, and il = 4170, gives At + A2 = -- 1000 and 9.932 At - 0.85 A2 = - 830, that is, At = - 156 and A2 = - 844. Therefore i = 1000 - 156 e-°'™9 - 844 r0'827' (34) and i\ = 5000 - 1550 r °-00140 + 720 r °'827fl ; (35) or the shunt field current t\ reduced back to the number of turns of the shunt field by the factor a « 1.2 X 10~3 is i/ = 6 - 1.86 r0'™140 + 0.86 r0'827, (36) 154 TRANSIENT PHENOMENA and the terminal voltage of the machine is e 272+^, 18.3 — 0.8270 or> e = 600 - 93.2 e~{ - 6.8 £~l "y. (37) As seen, of the two exponential terms one disappears very quickly, the other very slowly. Introducing now instead of the angle 6 = 2 nft the time, t, gives the main current as i = 1000 - 156 £-°'53' - 844 r*1" the shunt field current as i> = 6- 1.86 £-°-53< + 0.86 and the terminal voltage as e = 600 - 93.2 £-°'53< - 6.8 e- (38) 89. Fig. 39 shows these three quantities, with the time, t, as abscissas. 1=0.010.02 i 6 900 4.8 f 800 4.6 700 4.4 600 4.2] 500 6. 400 5.8 1 300 5.6 200 5.4 100 5.2] 1000 5. 900 4.81 800 4.6 700 4.4 600 L. 4.o = Termina i --MUncn Seconds 0.04 0.05 0.06 0.07 0.08 voltiige currei 3 ohms at. ohms ^ -96-mh Mr t =-12345678 Seconds Fig. 39. Building-up of over-compounded direct-current generator from 600 volts no load to 600 volts load. The upper part of Fig. 39 shows the first part of the curve with 100 times the scale of abscissas as the lower part. As seen, the transient phenomenon consists of two distinctly different MUTUAL INDUCTANCE 155 periods: first a very rapid change covering a part of the range of current or e.m.f., and then a very gradual adjustment to the final condition. So the main current rises from zero to 800 amp. in 0.01 sec., but requires for the next 100 amp., or to rise to a total of 900 amp., about a second, reaching 95 per cent of full value in 2.25 sec. During this time the shunt field current first falls very rapidly, from 5 amp. at start to 4.2 amp. in 0.01 sec., and then, after a minimum of 4.16 amp., at t = 0.015, gradually and very slowly rises, reaching 5 amp., or its starting point, again after somewhat more than a second. After 2.5 sec. the shunt field current has completed half of its change, and after 5.5 sec. 90 per cent of its change. The terminal voltage first rises quickly by a few volts, and then rises slowly, completing 50 per cent of its change in 1.2 sec., 90 per cent in 4.5 sec., and 95 per cent in 5.5 sec. Physically, this means that the terminal voltage of the machine rises very slowly, requiring several seconds to approach station- ary conditions. First, the main current rises very rapidly, at a rate depending upon the inductance of the external circuit, to the value corresponding to the resistance of the external circuit and the initial or no load terminal voltage, and during this period of about 0.01 sec. the magnetizing action of the main current is neutralized by a rapid drop of the shunt field current. Then gradually the terminal voltage of the machine builds up, and the shunt field current recovers to its initial value in 1.15 sec., and then rises, together with the main current, in corre- spondence with the rising terminal voltage of the machine. It is interesting to note, however, that a very appreciable time elapses before approximately constant conditions are reached. 90. In the preceding example, as well as in the discussion of the building up of shunt or series generators in Chapter II, the e.m.fs. and thus currents produced in the iron of the magnetic field by the change of the field magnetization have not been considered. The results therefore directly apply to a machine with laminated field, but only approximately to one with solid iron poles. In machines with solid iron in the magnetic circuit, currents produced in the iron act as a second electric circuit in inductive 156 TRANSIENT PHENOMENA relation to the field exciting circuit, and the transition period thus is slower. As example may be considered the excitation of a series booster with solid and with laminated poles; that is, a machine with series field winding, inserted in the main circuit of a feeder, for the purpose of introducing into the circuit a voltage propor- tional to the load, and thus to compensate for the increasing drop of voltage with increase of load. Due to the production of eddy currents in the solid iron of the field magnetic circuit, the magnetic flux density is not uniform throughout the whole field section during a change of the mag- netic field, since the outer shell of the field iron is magnetized by the field coil only, while the central part of the iron is acted upon by the impressed m.m.f. of the field coil and the m.m.f. of the eddy currents in the outer part of the iron, and the change of magnetic flux density in the interior thus lags behind that of the outside of the iron. As result hereof the eddy currents in the different layers of the structure differ in intensity and in phase. A complete investigation of the distribution of magnetism in this case leads to a transient phenom- enon in space, and is discussed 'in Section III. For the present purpose, where the total m.m.f. of the eddy currents is small compared with that of the main field, we can approxi- mate the effect of eddy currents in the iron by a closed circuit second- ary conductor, that is, can assume uniform intensity and phase of secondary currents in an outer layer Fig' 40' tSection °f a mag' ,, , , . J , , , . ., ,, netic circuit. of the iron, that is, consider the outer layer of the iron, up to a certain depth, as a closed circuit secondary. Let Fig. 40 represent a section of the magnetic circuit of the machine, and assume uniform flux density. If <£ = the total magnetic flux, lr= the radius of the field section, then at a distance I from the center, the magnetic flux enclosed by a / l\2 circle with radius I is f — J 4>, and the e.m.f. generated in the zone MUTUAL IXDUCTANCE 157 / l\2 at distance I from the center is proportional to f— J 4>,that /l\2 is,e = a (— j 4>. The current density of the eddy currents in this zone, which has the length 2 Til, is therefore proportional to e bl — - , or is i = — <£. This current density acts as a m.m.f . upon 2i Til lr /l\2 the space enclosed by it, that is, upon f— j of the total field section, and the magnetic reaction of the secondary current at /l\2 distance I from the center therefore is proportional to i f — j , or vr/ x is & = — $, and therefore the total magnetic reaction of the eddy currents is At the outer periphery of the field iron, the generated e.m.f. is et = a<£, the current density therefore il = y, and therefore lr that is, the magnetic reaction of the eddy currents, assuming uniform flux density in the field poles, is the same as that of the currents produced in a closed circuit of a thickness -j, or one- fourth the depth of the pole iron, of the material of the field pole and surrounch'ng the field pole, that is, fully induced and fully magnetizing. The eddy currents in the solid material of the field poles thus can be represented by a closed secondary circuit of depth -^ surrounding the field poles. The magnitude of the depth of the field copper on the spools 158 TRANS I EN T PHENOMENA is probably about one-fourth the depth of the field poles. Assum- ing then the width of the band of iron which represents the eddy current circuit as about twice the width of the field coils — since eddy currents are produced also in the yoke of the machine, etc. — and the conductivity of the iron as about 0.1 that of the field copper, the effective resistance of the eddy current circuit, reduced to the field circuit, approximates five times that of the field circuit. Hence, if r2 — resistance of main field winding, rl = 5 r2 = resistance of the secondary short circuit which represents the eddy currents. Since the eddy currents extend beyond the space covered by the field poles, and considerably down into the iron, the self- inductance of the eddy current circuit is considerably greater than its mutual inductance with the main field circuit, and thus may be assumed as twice the latter. 91. As example, consider a 20Okw. series booster covering the range of voltage from 0 to 200, that is, giving a full load value of 1000 amperes at 200 volts. Making the assumptions set forth in the preceding paragraph, the following constants are taken: the armature resistance = 0.008 ohms and the series field 'winding resistance^ 0.004 ohm; hence, the short circuit — or eddy current resistance — rl = 0.02 ohm. Further- more let M = 900 X 10~6 henry = mutual inductance between main field and short-circuited secondary; hence, xm = 0.34 ohm = mutual reactance, and therefore, assuming a leakage flux of the secondary equal to the main flux, Ll = 1800 X 10~6 henry and xl = 0.68 ohm. The booster is inserted into a constant potential circuit of 550 volts, so as to raise the voltage from 550 volts no load to 750 volts at 1000 amperes. The total resistance of the circuit at full load, including main circuit and booster, therefore is r = 0.75 ohm. The inductance of the external circuit may be assumed as L = 4500 X 10~6 henrys; hence, the reactance at/ = 60 cycles per sec. is x = 1.7 ohms. The impressed e.m.f. of the circuit is e = 550 + e', ef being the e.m.f. generated in the booster. Since at no load, for i = 0, e7 = 0, and at full load, for i = 1000, e' = 200, assuming a straight line magnetic characteristic or saturation curve, that is, assuming the effect of magnetic satura- MUTUAL INDUCTANCE 159 tion as negligible within the working range of the booster, we have e = 550 + 0.2 (i + ij. This gives the following constants : Main Circuit. Eddy Current Circuit. Current Impressed e.m.f Resistance Inductance Reactance i amp. e= 550+0.2(i+t,) volts. r=0.75 ohm. L=4500X10~flhenrys. x— I 7 ohms il amp. 0 volts. r, = 0.02 ohm. L1=1800X10-«henrys. xl — 0 68 ohm Mutual inductance. . . Mutual reactance .... M= 900 : xm = 0.34 ( < 10""6 henrys. mm. This gives the differential equations of the problem as and 550 - 0.55 i + 0.2 L - 1.7 - 0.34 = 0 da do 0.02^ + 0.34^- + 0.68^ = 0* ad ad Adding 2 times (39) to (40) gives di 1100 - 1.1 i + 0.42 i. - 3.06 — = 0, dv or t\ = 7.28 ^ + 2.62 i - 2620, herefrom: 0.02 L = 0.1456 4 + 0.0524 i - 52.4, ay and ^ = 4.95^1.781, substituting the last two equations into (40), + 0.458 + 0.0106 i - 10.6 = 0. ad du If then i = in + Ae -aff (39) (40) (41) (42) (43) (44) (45) (46) At-<*(c? - 0.45S a + 0.0106) + 0.0106 it - 10.6 = 0. 160 TRANSIENT PHENOMENA As transient and permanent terms must each equal zero, i0 = 1000 and a2 - 0.458 a + 0.0106 = 0, wherefrom a = 0.229 ± 0.205; the roots are o4 = 0.024 and a2 = 0.434; then we have i = 1000 + A (57) de dc~ If the impressed e.m.fs., et and e2, are constant, -r- and — 162 TRANSIENT PHENOMENA equal zero, and equations (52) and (53) assume the form (56) and (57); that is, equations (56) and (57) are the differential equations of the transient terms, for -the general case of any e.m.fs., e^ and e2, which have no transient terms, and are the general differential equations of the case of constant impressed e.m.fs., el and er From (56) it follows that d?i2 . dil (Pi^ Xm~d(P ~~ ~ Xcih ~ TI ~dd ~ XlW2 ' Differentiating equation (57) twice, and substituting therein (58), gives 2X d4i , N (Pi , - xm2) — + (r,x2 + r2xj — + (xcix2 ,2 2 ci2 C2, ta + ( Vi + V J ^ + xcxC2 i = 0. (59) This is a differential equation of fourth order, symmetrical in r^x^ and r2x2xC2, which therefore applies to both currents, il and ir The expressions of the two currents il and i2 therefore differ only by their integration constants, as determined by the ter- minal conditions. Equation (59) is integrated by i = As"* (60) and substituting (60) in (59) gives for the determination of the exponent a the quartic equation (x& - xm2) a4 - (r^2 + r2xj a3 + (xcix2 - (xcir2 + x^) a + xClxC2 = 0, or 2^1 ~3 i XCt a -\- • ,x2 - xm2 The solution of this quartic equation gives four values of a, and thus gives *-**9. (62) MUTUAL INDUCTANCE 163 The roots, a, may be real, or two real and two imaginary, or all imaginary, and the solution of the equation by approxima- tion therefore is difficult. In the most important case, where the resistance, r, is small compared with the reactances x and xc — and which is the only case where the transient terms are prominent in intensity and duration, and therefore of interest — as in the transformer and the induction coil or Ruhmkorff coil, the equation (61) can be solved by a simple approximation. In this case, the roots, a, are two pairs of conjugate imaginary numbers, and the phenomenon oscillatory. The real components of the roots, a, must be positive, since the exponential s~a9 must decrease with increasing 0. The four roots thus can be written : (63) where a and /? are positive numbers. In the equation (61), the coefficients of a3 and a are small, since they contain the resistances as factor, and this equation thus can be approximated by a4 + e <*a2 + c 2 = 0; (64) x hence, ' xcix2 _ that is, a2 is negative, having two roots, 6t = - /V and 6, - - /?/. This gives the four imaginary roots of a as first approximation : ° 164 TRANSIENT PHENOMENA If av a2, ag, a4 are the four roots of equation (61), this equation can be written /(a) = ( a- oj (a - aa) (a - a3) (a - a4) - 0; or, substituting (63), /(a) = {(a - a,Y + ft'} {(a - «2)2 + ft2} = 0, (67) and comparing (67) with (61) gives as coefficients of a3 and of a, 2 (a, 4- «2) = ? and (68) and since /?j2 and /?22 are given by (65) and (66) as roots of equa- tion (64), av a 2, fv /?2, and hereby the four roots av a2, a3, a4 of equation (61) are approximated by (64), (65), (66), (68). The integration constants Av A2, A3, A4 now follow from the terminal conditions. 93. As an example may be considered the operation of an inductorium, or Ruhmkorff coil, by make and break of a direct- current battery circuit, with a condenser shunting the break, in the usual manner. Let el = 10 volts = impressed e.m.f.; r1 = 0.4 ohm = resistance of primary circuit, giving a current, at closed circuit and in stationary condition, of i0= 25 amp.; r2= 0.2 ohm = resistance of secondary circuit, reduced to the primary by the square of the ratio of primary -=- secondary turns ; xl = 10 ohms = primary inductive reactance; x2 = 10 ohms = secondary inductive reactance, reduced to primary; xm = 8 ohms = mutual inductive reactance; xCi = 4000 ohms = primary condensive reactance of the condenser shunting the break of the interrupter in the battery circuit, and xC2 = 6000 ohms = secondary condensive reactance, due to the capacity of the terminals and the high tension winding. Substituting these values, we have BI = 10 volts i0 = 25 amp. rt = 0.4 ohm xl = 10 ohms xCi = 4000 ohms r2 = 0.2 ohm x2 = 10 ohms xC2 = 6000 ohms xm = 8 ohms. (69) MUTUAL INDUCTANCE 165 These values in equation (61) give / (a) = a4- 0.167 a3 + 2780 a2 - 89 a + 667,000 = 0, (70) and in equation (64) they give ft (a) = a4 + 2780 a2 + 667,000 = 0 and or hence, and =-(1390 ±1125) = -2515, = -265; ft = 50.15 = 16.28. From (68) it follows that ai + a2 = 0.0833 and 265 a^ + 2515 a2 = 44.5; hence, al - 0.073, a2 = 0.010. Introducing for the exponentials with imaginary exponents the trigonometric functions give .! cos 50.15 d + A2 sin 50.15 d } sin 16.280 (71) ^ cos 50.150 + C2 sin 50.150} >! cos 16.28 6 + D2 sin 16.28 0 where the constants C and D depend upon A and B by equations (56), (57), or (58), thus: Substituting (71) into (58), =0 (58) gives an identity, from which, by equating the coefficients of £-°* cos bd and e.'00 sin bd to zero, result four equations, in the coefficients A, B, C, D, A, B, C, D2, 166 TRANSIENT PHENOMENA from which follows, with sufficient approximation, A! = -0.95^ A2 = -0.98(72 Bl = + 1.57 A £2= + 1.57 A; hence, \ = - 0.96 £-0-073^ ci cos 50.15 0 -f C2 sin 50.15 + 1.57 £-™io0 j A cos 16.28 0 + D2 sin 16.28 and substituting (71) and (73) in the equations of the condenser potential, (54) and (55), gives e,f = 10 + 79 s~OJm9{C2 cos 50.15 0 - Ct sin 50.15 0 (72) - 385 £-°-010'{ Z)2 cos 16.28 0 - Dl sin 16.28 0 (74) e2' = 118 £-°-™e{C2 cos 50.15 0 - (7X sin 50.15 0} + 367£-°-010'{A cos 16.280 - A sin 16.280} 94. Substituting now the terminal conditions of the circuit : At the moment where the interrupter opens the primary circuit the current in this circuit is \ = -- - = 25 amp. The condenser in the primary circuit, which is shunted across the break, was short-circuited before the break, hence of zero poten- tial difference. The secondary circuit was dead. This then gives the conditions 0 = 0;^ = 25, ?'2 = 0, e^ = 0, and e/ = 0. Substituting these values in equations (71), (73), (74) gives 25= -0.95 C, + 1.58 A 0 = ' Ci + A 0 = 10 + 79 Ca - 385 D2 = 118(7, +367ZX hence C,= - 10 (72 = - 0.05 =* 0 A = + 10 0.016 =* 0, D2 = MUTUAL INDUCTANCE 167 and i, = 9.6 s-0™9 cos 50.15 0 + 15.7 £-°Moe cos 16.28 6 (75) i2 = -10 £-°073' cos 50.15 (9 + 10 £-°-010' cos 16.28 0 e/ = 10 + 790 £-°-073' sin 50.15 6 + 3850 £-°-010' -sin 16.28 6 Approximately therefore we have i\ = 9.6 £-°-073e cos 50.15 0 + 15.7 £-°-010' cos 16.28 0 ia = - 10 { £-°-0730 cos 50.15 0 - £-°mo0 cos 16.28 0 } e/ = 3850 fi-OJMO« sin 16.28 0 < - -3670 £-°010' sin 16.28 6. The two frequencies of oscillation are 3009 and 977 cycles per sec., hence rather low. The secondary terminal voltage has a maximum of nearly 4000, reduced to the primary, or 400 times as large as corre- sponds to the ratio of turns. In this particular instance, the frequency 3009 is nearly suppressed, and the main oscillation is of the frequency 977.