Theory and Calculation of Transient Electric Phenomena and Oscillations — part 3 of 20

CONTINUOUS-CURRENT CIRCUITS 27 21. Starting of a continuous-current lighting circuit, or non-in- ductive load. Let e, = 125 volts = impressed e.m.f. of the circuit, and t, = 1000 amperes = current in the circuit under stationary condition; then the effective resistance of the circuit is r = © = 0.125 ohm. : y Assuming 10 per cent drop in feeders and mains, or 12.5 volts, _ gives a resistance, r, = 0.0125 ohm of the supply conductors. In such large conductor the inductance may be estimated as 10 mh. per ohm; hence, L = 0.125 mh. = 0.000125 henry. The current at the moment of starting is7, = 0, and the general ’ equation of the current in the circuit therefore is, by substitution , in (3), i = 1000 (1 — 7), (6) The time during which this current reaches half value, or i = 500 amperes, is given by substitution in (6) 500 = 1000 (1 — e710), hence e000 — 0.5, t = 0.00069 seconds. The time during which the current reaches 90 per cent of its full value, or 7 = 900 amperes, is ¢ = 0.0023 seconds, that is, the current is established in the circuit in a practically inappre- ciable time, a fraction of a hundredth of a second. 22. Excitation of a motor field. Let, in a continuous-current shunt motor, e, = 250 volts = . impressed e.m.f., and the number of poles = 8. Assume the magnetic flux per pole, ®, = 12.5 megalines, and the ampere-turns per pole required to produce this magnetic flux as ¢ = 9000. Assume 1000 watts used for the excitation of the motor | field gives an exciting current ; . — 1000 4 = 959 = 4 amperes, and herefrom the resistance of the total motor field circuit is r =“ — 62.5 ohms. uy

28 TRANSIENT PHENOMENA

To produce * = 9000 ampere-turns, with 7, = 4 amperes,

requires > = 2250 turns per field spool, or a total of n= 18,000

1 turns. n = 18,000 turns interlinked with ®, = 12.5 megalines gives

, a total number of interlinkages for 7, = 4 amperes of n®, =

225 x 10°, or 562.5 X 10° interlinkages per unit current, or

10 amperes, that is, an inductance of the motor field circuit

L = 562.5 henrys. .

The constants of the circuit thus are e, = 250 volts; r = 62.5 ohms; L = 562.5 henrys, and 7, = 0 = current at time ¢ = 0.

Hence, substituting in (3) gives the equation of the exciting current of the motor field as

~=4 (ql _ en OelIttty (7)

Half excitation of the field is reached after the time ¢ = 6.23 seconds; -

90 per cent of full excitation, or 7 = 3.6 amperes, after the time t = 20.8 seconds.

That is, such a motor field takes a very appreciable time after closing the circuit before it has reached approximately full value and the armature circuit may safely be closed.

Assume now the motor field redesigned, or reconnected so as to consume only a part, for instance half, of the impressed e.m.f., the rest being consumed in non-inductive resistance. This may be done by connecting the field spools by two in multiple.

In this case the resistance and the inductance of the motor field are reduced to one-quarter, but the same amount of external resistance has to be added to consume the impressed e.m.f., and the constants of the circuit then are: e, = 250 volts; r = 31.25 ohms; L = 140.6 henrys, and 7, = 0.

The equation of the exciting current (3) then is

i= 8 (1 — eo ote), (8) _ that is, the current rises far more rapidly. It reaches 0.5 value after = 3.11 seconds, 0.9 value after ¢ = 10.4 seconds.

An inductive circuit, as a motor field circuit, may be made to respond to circuit changes more rapidly by inserting non- inductive resistance in series with it and increasing the im-

CONTINUOUS-CURRENT CIRCUITS 29 pressed e.m.f., that is, the larger the part of the impressed e.m.f. consumed by non-inductive resistance, the quicker is the change.

Disconnecting the motor field winding from the impressed e.m.f. and short-circuiting it upon itself, as by leaving it con- nected in shunt with the armature (the armature winding resistance and inductance being negligible compared with that of the field winding), causes the field current and thereby the field magnetism to decrease at the same rate as it increased in (7) and (8), provided the armature instantly comes to a stand- still, that is, its e.m.f. of rotation disappears. This, however, is usually not the case, but the motor armature slows down gradually, its momentum being consumed by friction and other losses, and while still revolving an e.m.f. of gradually decreas- ing intensity is generated in the armature winding; this e.m-f. is impressed upon the field.

The discharge of a motor field winding through the armature winding, after shutting off the power, therefore leads to the case of an inductive circuit with a varying impressed e.m.f. ;

23. Discharge of a motor field winding.

Assume that in the continuous-current shunt motor dis- cussed under 22, the armature comes to rest t, = 40 seconds after the energy supply has been shut off by disconnecting the motor from the source of impressed e.m.f., while leaving the motor field winding still in shunt with the motor armature winding. .

The resisting torque, which brings the motor to rest, may be assumed as approximately constant, and therefore the deceler- ation of the motor armature as constant, that is, the motor speed decreasing proportionally to the time.

If then S = full motor speed, S (1 - *) is the speed of the

. 1 motor at the time ¢ after disconnecting the motor from the source of energy.

Assume the magnetic flux ® of the motor as approximately proportional to the exciting current, at exciting current 7 the magnetic flux of the motor is P= 7%, where ®, = 12.5 mega-

1 lines is the flux corresponding to full excitation 1, = 4 amperes.

. | 380 TRANSIENT PHENOMENA The e.m.f. generated in the motor armature winding and | thereby impressed upon the field winding is proportional to | the magnetic flux of the field, ®, and to the speed S (1 - >): . 1 and since full speed S and full flux ®, generate ane.mf.e,= — 250 volts, the e.m.f. generated by the flux ® and speed S (1 - 7) , Lb that is, at time ¢ is 7 t CH= Cy aU _ 1)! (9) and since @, 7 =T, uy we have ' e= Ur (1 — ); (10) l or for r = 62.5 ohms, and (, = 40 seconds, we have e = 62.51 (1 — 0.0250). (11) Substituting this equation (10) of the impressed e.m.f. into the differential equation (1) gives the equation of current i during the field discharge, wr (1 - *) art at (12) hence, _ td di a9 tbo i’ integrated by re

• 2 tL = log ct, where the integration constant ¢ is found by t=0, i=i, logei,=0, ¢ ai 1 hence . , re L ~ 20h = log i? (14) or, re tate 7%, (15)

_CONTIN UOUS-CURRENT CIRCUITS 31

This is the equation of the field current during the time in which the motor armature gradually comes to rest.

At the moment when the motor armature stops, or for

t=t, it is - _ mt ue t,=7t¢ 7%, (16)

This is the same value which the current would have with

the armature permanently at rest, that is, without the assistance : . t of the e‘m.f. generated by rotation, at the time ¢ = > .

The rotation of the motor armature therefore reduces the decrease of field current so as to require twice the time to reach value 7,, that it would without rotation.

These equations cease to apply for ¢ > ¢,, that is, after the armature has come to rest, since they are based on the speed

. t . . . equation S (1 - -) , and this equation applies only up to 1 t=, but for ¢ > ¢, the speed is zero, and not negative, as : t given by s(t - ): 1

That is, at the moment ¢ = ¢, a break occurs in the field discharge curve, and after this time the current 7 decreases in accordance with equation (3), that is,

-" t-—t t= iC ), (17) or, substituting (16), BEY v= a (18)

Substituting numerical values in these equations gives:

fort < ¢,, . t=4e7 o-ootsast « (19) fort = t, = 40,

. t = 0.436; (20)

fort > ¢,, , t= 4 er out — 20), (21)

32 TRANSIENT PHENOMENA

Hence, the field has decreased to half its initial value after the time ¢ = 22.15 seconds, and to one tenth of its initial value after ¢ = 40.73 seconds.

"4.0 J +—}— ee } os Oe oe SAECENE CCCP S [go aa oT MCCCCBSRCEEEEEEE PR etal 525 Ce EEE ee Eee ; $0 NON N +4 We} is i a 0 i 1.6 7 “it NCA = ae i +—{ MoT REPRE EEE Ca NE! gb casi SUSE Reo eis ae i ics a ai a re eer)

Seconds .

Fig. 5. Field discharge current.

Fig. 5 shows as curve I the field discharge current, by equations (19), (20), (21), and as curve II the current calculated by the equation

t = 4 e7 Orttat that is, the discharge of the field with the armature at rest, or when short-circuited upon itself and so- not assisted by the e.m.f. of rotation of the armature. ;

The same Fig. 5 shows as curve III the beginning of the field discharge current for L = 4200, that is, the case that the field circuit has a much higher inductance, as given by the equation

t= 4 em 00001850? As seen in the last case, the decrease of field current is very slow, __ the field decreasing to half value in 47.5 seconds.

24. Self-excitation of direct-current generator.

In the preceding, the inductance L of the machine has been assumed as constant, that is, the magnetic ux @as proportional to the exciting current7. For higher values of ®, this is not even approximately the case. The self-excitation of the direct- _ current generator, shunt or series wound, that is, the feature |

CONTINUOUS-CURRENT CIRCUITS 33 that the voltage of the machine after the start gradually builds up from the value given by the residual magnetism to its full value, depends upon the disproportionality of the magnetic flux with the magnetizing current. When considering this phenom- enon, the inductance cannot therefore be assumed as constant. When investigating circuits in which the inductance L is not constant but varies with the current, it is preferable not to use the term “‘inductance” at all, but to introduce the magnetic flux ®. The magnetic flux ® varies with the magnetizing current 7 by an empirical curve, the magnetic characteristic or saturation curve of the machine. This can approximately, within the range considered here, be represented by a parabolic curve, as was first shown by Frohlich in 1882: gi , = -—— ¢ yp) ® teh’ (22) where ¢ = magnetic flux per ampere, in megalines, at low | density. t magnetic saturation value, or maximum magnetic flux,

• in megalines, and a l+h _ | a (3) can be considered as the magnetic exciting reluctance of the machine field circuit, which here appears as linear function of the exciting current 7. Considering the same shunt-wound commutating machine as in (12) and (13), having the constants r = 62.5 ohms = field resistance; ®, = 12.5 megalines = magnetic flux per pole at : normal m.m.f.; ¢ = 9000 ampere-turns = normal m.m-f. per pole; 7 = 18,000 turns = total field turns (field turns per pole = a = 2250), and 7, =4 amperes = current for full excitation, or flux, ®, = 12.5 megalines. Assuming that at full excitation, ®,, the magnetic reluctance . has already increased by 50 per cent above its initial value, that | |

' 84 TRANSIENT PHENOMENA . .. ampere-turns a es _

is, that the ratio magnetic flux’ or =, at = ®, = 12.5 mega

lines and 7 = 7, = 4 amperes, is 50 per cent higher than at low

excitation, it follows that

1 + bi, = 1.5, or (24) b= 0.135.)

Since « = 7, = 4 produces @ = ®, = 12.5, it follows, from (22) and (24)

p = 4.69.

That is, the magnetic characteristic (22) of the machine is approximated by

4.692 - a ere eo)

Let now e, = e.m.f. generated by the rotation of the arma- ture per megaline of field flux.

This e.m.f. e, is proportional to the speed, and depends upon the constants of the machine. At the speed assumed in (12) and (13), ®, = 12.5 megalines, e, = 250 volts, that is,

. = & = a 5 . e ®, 20 volts.

Then, in the field circuit of the machine, the impressed e.m.f., or e.m.f. generated in the armature by its rotation through the magnetic field is,

e =e, = 209; the e.m.f. consumed by the field resistance r is . ir = 62.570; the e.m.f. consumed by the field inductance, that is, generated _in the field coils by the rise of magnetic flux ®, is db db (® being given in megalines).

CONTINUOUS-CURRENT CIRCUITS 35 The differential equation of the field circuit therefore is (1) . n d® e® = ir + i100 di (26) Since this equation contains the differential quotient of ®, it is more convenient to make ® and not 7 the dependent variable; then substitute for 7 from equation (22), . ® . T= $ — bb’ (27) which gives or n db 0? = Fb * 100 dl’ (28) or, transposed, 100 dt _ (p — bb) d® n {(ge.—r) — be, P} (29) This equation is integrated by resolving into partial fraction by the identity g— be 4, 8 ®{ Gen) —be, db} d* genr—be,6' 0) resolved, this gives " fp — bb = A (ge, — r) — (Abe, ® — BS); | -_% hence, A= gear’ (31) B = br , Pe —?T . _ and 100 dt god & brd® Oe ge SO SsSsS———F——sSs—7?2 nm ~ Ge—1)d* Ge—1) er bee) This integrates by the logarithmic functions 10: ¢ r n ~ de, = log ® alge = 7) log (pe,—r—be, ®)+C. (33) .

36 TRANSIENT PHENOMENA

The integration constant C is calculated from the residual magnetic flux of the machine, that is, the remanent magnetism of the field poles at the moment of staft.

Assume, at the time,t = 0, ® = ®, = 0.5 megalines = residual magnetism and substituting in (33),

o-—? loge, ———" - — log (fe, — r — be, ®,) + C

ge. — 7 ad €- (pe. — 7) 6 ° a , and herefrom calculate C.

C substituted in (33) gives

100t ® r ge, — 7 — be

ne Pe —?r log &, €¢ (pe. — r) log Ge. -T- be,” (34) or,

n © ge, —- Tr — | i= 100 ee (pe. — r) joe. log ®, _ log Pe. -T- be.®, (35) substituting e=e@ and

• Om = e-P,, where e», = e.m.f. generated in the armature by the rotation in the residual magnetic field, n e pec — r — be

f= 100 e, (pe, — 7) } $e, log T log de — 7 — bem) (36)

This, then, is the relation between e and ¢, or the equation of the building up of a continuous-current generator from its residual magnetism, its speed being constant.

Substituting the numerical values n = 18,000 turns; ¢ = 4.69 megalines; b = 0.125; e, = 20 volts; r = 62.5 ohms; ®, = 0.5 megaline, and e, = 10 volts, we have

t = 26.8 log ® — 17.9 log (31.25 — 2.56) + 79.6 (87) and t = 26.8 loge — 17.9 log (81.25 — 0.125e) — 0.8. (38) 4 ‘

CONTINUOUS-CURRENT CIRCUITS 37

Fig. 6 shows the e.m.f. e as function of the time ¢. As seen, under the conditions assumed here, it takes several minutes before the e.m.f. of the machine builds up to approximately full value. \

P Ltt dee tt TT TT | | Ph ee ga8035;- Gesnee ae ; me eee eee ce lr |= |62.5 ohms | it) | a otc h oft oT Pa Cela mating a if | i a eas | | Cl | | Fe, al iN lO Nt 1 a Bk a BE Coo a 0 a SOC AAeee Ho ae Sone eae +H PEAT | ==— a8 Se Se eS I 0 20 40 60 80 ,100 120 140 160 180 200Sec. Fig. 6. Building-up ctirve of a shunt generator.

The phenomenon of self-excitation of shunt generators there- oe fore is a transient phenomenon which may be of very long duration.

From equations (35) and (36) it follows that

e= aa = 250 volts (39) is the e.m.f. to which the machine builds up at t = ©, that is, in stationary condition.

To make the machine self-exciting, the condition

: ge. -—r>0 (40) must obtain, that is, the field winding resistance must be r< ¢e,, or, (41)

• 1 < 93.8 ohms, or, inversely, e,, which is proportional to the speed, must be €. > 7 c ¢ or, (42) €.> 13.3 volts. Y

88 TRANSIENT PHENOMENA

The time required by the machine to build up decreases with -; increasing e,, that is, increasing speed; and increases with . increasing 7, that is, increasing field resistance. :

25. Self-excitation of direct-current series machine.

Of interest is the phenomenon of self-excitation in a series. machine, as a railway motor, since when using the railway motor as brake, by closing its circuit upon a resistance, its usefulness depends upon the rapidity of building up as generator. ;

Assuming a 4-polar railway motor, designed for e,= 600 volts — and 2,= 200 amperes, let, at current 7 = 7,= 200 amperes, the magnetic flux per pole of the motor be ®,= 10 megalines, and ~ 8000 ampere-turns per field pole be required to produce this ~ flux. This gives 40 exciting turns per pole, or a total of m= __ 160 turns.

Estimating 8 per cent loss in the conductors of field and armature at 200 amperes, this gives a resistance of the motor

; circuit r= 0.24 ohms.

To limit the current to the full load value of 7, = 200 amperes, with the machine generating e,= 600 volts, requires a total | resistance of the circuit, internal plus external, of

r = 3 ohms, or an external resistance of 2.76 ohms.

600 volts generated by 10 megalines gives |

e-= 60 volts per megaline per field pole.

Since in railway motors at heavy load the magnetic flux is carried up to high values of saturation, at 7,= 200 amperes the magnetic reluctance of the motor field may be assumed as three times the value which it has at low density, that is, in equation (22), 1 + bi, = 3, on, b= 0.01, and since for 7 = 200, ® = 10, we have in (22)

¢ = 0.15, 0.157 @ = ——___ hence, 1+ 0.01% (43) represents the magnetic characteristic of the machine.

CONTINUOUS-CURRENT CIRCUITS 89 Ci«

Assuming a residual magnetism of 10 per cent, or ®, =

1 megaline, hence e,, = e, ®,= 60 volts, and substituting in equation (36) gives n = 160 turns;, ¢ = 0.15 megaline; 6 = ‘ 0.01; e,= 60 volts; r = 3 ohms; ®,= 1 megaline, and-en = 60 volts, t = 0.04 log e — 0.01333 log (600 — e) — 0.08. (44) This gives for e = 300, or 0.5 excitation, t = 0.072 seconds; and for e = 540, or 0.9 excitation, ¢ = 0.117 seconds; that is, such a motor excites itself as series generator practically instantly, or in a small fraction of a second. . The lowest value of e, at which self-excitation still takes place is given by equation (42) as r & === 20, ¢ that is, at one-third of full speed.

If this series motor, with field and armature windings connected in generator position, —that is, reverse position, —short-circuits upon itself,

• r = 0.24 ohms,

we have .

t = 0.0274 log e — 0.00073 log (876 — e) — 0.1075, (45) that is, self-excitation is practically instantaneous:

e = 300 volts is reached after ¢ = 0.044 seconds.

Since for e = 300 volts, the current 7 = < = 1250 amperes, the power is p = ev = 375 kw., that is, a series motor short- circuited in generator position instantly stops.

Short-circuited upon itself, r = 0.24, this series motor still builds up at é, =3 = 1.6, and since at full load speed e,= 60, é.= 1.6 is 2.67 per cent of full load speed, that is, the motor acts as brake down to 2.67 per cent of full speed.

It must be considered, however, that the parabolic equation (22) is only an approximation of the magnetic characteristic,

| 40 TRANSIENT PHENOMENA and the results based on this equation therefore are approximate only. | One of the most important transient phenomena of direct- | current circuits is the reversal of current in the armature coil short-circuited by the commutator brush in the commutating machine. Regarding this, see “ Theoretical Elements of Elec- __ trical Engineering,” Part II, Section B. | | ; .
{ ‘ h 7 . 1 . , yo wo 0 Vv - ‘ . . e a % . . , a t. ;

! CHAPTER IV. INDUCTANCE AND RESISTANCE IN ALTERNATING- CURRENT CIRCUITS.

26. In alternating-current circuits, the inductance L, or, as it is usually employed, the reactance x = 2 xfL, where f = fre- quency, enters the expression of the transient as well as the permanent term.

At the moment @ = 0, let the emf. e = Ecos (@ — 6,) be impressed upon a circuit of resistance r and inductance L, thus inductive reactance x = 2 zfL; let the time 0 = 2 zft be counted , from the moment of closing the circuit, and 6, be the phase of the impressed e.m.f. at this moment.

In this case the e.m.f. consumed by the. resistance = 7r, where 7 = instantaneous value of current.

The e.m.f. consumed by the inductance L is proportional to L and to the rate of change of the current, e, thus, is Z oo or, by substituting 0 = 22ft, x = 2xfL, the e.m.f. consumed

. . dt by inductance is th

Since e = Ecos (0 — 0,) = impressed e.m_f.,

Ecos (9 -0,)'-irt2” (1) do is the differential equation of the problem. This equation is integrated by the function buce t= 1cos(0— 9) + Ae®, (2) where « = basis of natural logarithms = 2.7183. Substituting (2) in (1), Ecos (6 — 6) = Ir cos (06 — 6) + Are~™ — Ix sin (0 — 8) — Aaze~™, or, rearranged : . (E cos 6, — Ir cos é — Ixsin 6) cos 8 + (E sin 6, — Irsind

• Iz cos 6) sin@ — Ae~® (ax — r) = 0. ; 41

42 TRANSIENT PHENOMENA . Since this equation must be fulfilled for any value of 6, if (2) . is the integral of (1), the coefficients of cos 0, sin 0, e~% must | vanish separately. . That is, , E cos 6, — Ir cos é — Ixsinéd = 0, E sin 0, — Irsinéd + Iz cosd = 0, (3) and ax—r=0. Herefrom it follows that ~~ r a= t . (4) Substituting in (8), tan 0, = . and- (6) z=Vr4 2, where 6, = lag angle and z = impedance of circuit, we have E cos 6, — Iz cos (8 — 4,) = 0 and | E sin 6, — Iz sin (6 — 0,) = 0, and herefrom 1! and 6) d=06,4+ 4, Thus, by substituting (4) and (6) in (2), the integral equation becomes E -"e t= Fc0s (6 — 0,— 6,)+ Ae -, (7) where A is still indefinite, and is determined by the initial con- ditions of the circuit, as follows: for 6 =0, t=0; hence, substituting in (7), . E 0= 5 008 (0, + 9,) + A,

ALTERNATING-CURRENT CIRCUITS 43 or, , E A=- 3s (4, + 6,), (8) and, substituted in (7), ; . ££ ~te i == }c0s (0 — 0, 0)-—e * cos (0, + 8,)} (9) is the general expression of the current in the circuit.

If at the starting moment # = 0 the current is not zero but = 2,, we have, substituted in (7),

1, = 5 cos (0, + 9,) + A,

. ££

A=1a4- 508 (6, + 4,),

. £E @\ -58

i= 2 Seos (0 — 0, — 0,)—(c0s (0, + 04) id), = t. (10) ’

27. The equation of current (9) contains a permanent term E cos (9 — 4 — 6,), which usually is the only term considered, 2 r

. E -3 and a transient term z e 7 COS (A) + 4,). The greater the resistance r and smaller the reactance z, the . E -*e . more rapidly the term 3 * cos (4 + 0,) disappears.

This transient term is a maximum if the circuit is closed at the moment @, = — 6,, that is, at the moment when the permanent value of current, = cos (9 — 4, — 0,), should be a maximum, and is then

EO 6 . —e . 2

The transient term disappears if the circuit is closed at the moment 0, = 90° — 6,, or when the stationary term of current . passes the zero value.

44 TRANSIENT PHENOMENA As example is shown, in Fig. 7, the starting of the current under the conditions of maximum transient term, or 0, = — 9,, in a circuit of the following constants: - = 0.1, corresponding . approximately to a lighting circuit, where the permanent value : LEE SGRESEG ECC aes a FO al T_| Er SO eh NCEE tT la Degrees (RAE RaR SRE SR oS =} a tt ee ghee elt eh eS aes ARARMEPRBERRES: S ReGeoen TT tT | Ly Fig. 7. Starting current of an inductive circuit. . . x of current is reached in a small fraction of a half wave; >= 0.5, corresponding to the starting of an induction motor with rheo- stat in the secondary circuit; = = 1.5, corresponding to an unloaded transformer, or to the starting of an induction motor with short-circuited secondary, and ~ = 10, corresponding to a reactive coil. : SMBe 4 FX NANEEE/A\Cea aan a's

• 0. Na 360) 360 (540, 64) [720] Sip 9) 1080 1140 SENN “TN “Webel eRe INCE: Tt (
     CRYO SEC [eae 2 ae By Se {ttt Fig. 8. Starting current of an inductive circuit. , . Of the last case, = = 10, a series of successive waves are plotted in Fig. 8, showing the very gradual approach to perma- nent condition.

ALTERNATING-CURRENT CIRCUITS 45 Fig. 9 shows, for the circuit ~ = 1.5, the current when closing the circuit 0°, 30°, 60°, 90°, 120°, 150° respectively behind the zero value of permanent current. The permanent value of current is usually shown in these diagrams in dotted line. EIVIN TTT TTT 7A ee CN JAAJIEN ET ET TTT A tt WAAAINE TT TET TAT TTA KLV AW TT NN See eee att TT AW TT TZ TT TT att TEM EET TT Tt aQeEET TEI TTA att TTT IAL TET TTT PT TTT TT] SY TT Et tt 0 oo 10 190 20 800 300 C7) Co) 50 . Degrees ‘ Fig. 9. Starting current of an inductive circuit. 28. Instead of considering, in Fig. 9, the current wave as consisting of the superposition of the permanent term _"6@ I cos (9—9,) and the transient term — Je * cos 0, the current wave can directly be represented by the permanent term (LLL ASN EEE ttt YET INA TET TTT TT TT at IAT TT PN TT tA EAL RERAL TET TTT op ZT | PRE Ee aAtti tit? tt Att tT? Tt tty atti ttt TT IN TT TT TT YT att te EEE TN, TT TT eee oLLLi TT tt ttt | tN TT | | Fig. 10. Current wave represented directly. 1 cos (@ — 9,) by considering the zero line of the diagram as -—76@ deflected exponentially to the curve Je * cos 0, in Fig. 10. That is, the instantaneous values of current are the vertical ‘

46 TRANSIENT PHENOMENA

distances of the sine wave I cos (9 — @,) from the exponential a |

curve Ie * cos 9,, starting at the initial value of perma-

nent current.

In polar coordinates, in this case J cos (6 — 6,) is the circle,

26 Ie * cos @, the exponential or loxodromic spiral.

As a rule, the transient term in alternating-current circuits containing resistance and inductance is of importance only in circuits containing iron, where hysteresis and magnetic saturation complicate the phenomenon, or in circuits where unidirectional or periodically recurring changes take place, as in rectifiers, and some such cases are considered in the following chapters.

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‘CHAPTER V. RESISTANCE, INDUCTANCE, AND CAPACITY IN SERIES. CONDENSER CHARGE AND DISCHARGE. 29. If a continuous e.m.f. e is impressed upon a circuit contain- ing resistance, inductance, and capacity in series, the stationary condition of the circuit is zero current, = 0, and the poten- tial difference at the condenser equals the impressed e.mf., e, =e, no permanent current exists, but only the transient current of charge or discharge of the condenser. . The capacity C of a condenser is defined by the equation , nde ° t=C a that is, the current into a condenser is proportional to its increase of e.m.f. and to the capacity. It is therefore de = 7 iat, and e= L f tdt (1) Cc is the potential difference at the terminals of a condenser of capacity C with current 7 in the circuit to the condenser. Let then, in a circuit containing resistance, inductance, and capacity in series, e = impressed e.m.f., whether continuous, alternating, pulsating, etc.; 7 = current in the circuit at time ¢; r= resistance; L = inductance, and C = capacity: then the e.m.f. consumed by resistance r is rt; the e.m.f. consumed by inductance L is L ‘ ’ 47

48 TRANSIENT PHENOMENA and the e.m.f. consumed by capacity C is 1 . e, = Cc f dt; hence, the impressed e.m.f. is . . a dl f¢. e=ri + LE + @ fide (2) and herefrom the potential difference at the condenser terminals is 1 . . di Qag fidt-e-n- LS (3) ‘ Equation (2) differentiated and rearranged gives a dol. de Let ate "a (4) as the general differential equation of a circuit containing resist- ance, inductance, and capacity in series. 30. If the impressed e.m.f. is constant,

e = constant, | then S =0, | and equation (4) assumes the form, for continuous-current circuits,

Pi dk 1. |

Let ata = (5) |

This equation is a linear relation between the dependent vari- |

able, 7, and its differential quotients, and as such is integrated | by an exponential function of the general form

t= Ae~%, (6)

(This exponential function also includes the trigonometric functions sine and cosine, which are exponential functions with imaginary exponent a.)

CONDENSER CHARGE AND DISCHARGE 49 . Substituting (6) in (5) gives m 27 2 —at_ . ; ; (a L ar + C Ae 0; this must be an identity, irrespective of the value of t, to make (6) the integral of (5). That is, 1 2 _ - = @L — art+ CG 0. (7) A is still indefinite, and therefore determined by the terminal conditions of the problem. : | From (7) follows | , 4L . a= rev" ¢ vr —£ (8) | = OL ’ : | | hence the two roots, . Ts 1 OL | and (9) . . aot ts 2’ ray : where sar —*e. (10) . Since there are two roots, a, and a,, either of the two expres- ions (6), «~ and e~°%, and therefore also any combination of these two expressions, satisfies the differential equation (5). That is, the general integral equation, or solution of differential equation (5), is r—-a" r+s i=Ae 7 + Ae 2’, (11) Substituting (11) and (9) in equation (3) gives the potential difference at the condenser terminals as 78 _ rts ge fit tae T+ Say mu} (12)

50 TRANSIENT PHENOMENA

31. Equations (11) and (12) contain two indeterminate con- _ stants, A, and A,, which are the integration constants of the — differential equation of second order, (5), and determined by — the terminal conditions, the current and the potential differ- _ ence at the condenser at the moment t = 0.

Inversely, since in a circuit containing inductance and capac- — ity two electric quantities must be given at the moment of | start of the phenomenon, the current and the condenser poten- | tial — representing the values of energy stored at the moment — t = 0 as electromagnetic and as electrostatic energy, respec- — tively — the equations must lead to two integration constants,

° that is, to a differential equation of second order.

Let 7 =1, = current and e, = e, = potential difference at condenser terminals at the moment ¢ = 0; substituting in (11) _ and (12), .

1=A,+A4, and @ = e- "2A, - "Ay | hence, | @—et+ — ty Ay == ———+— | s and (18) r+s., &—e+——t, | A, =+ ————— » s§ and therefore, substituting in (11) and (12), the current is +s. r—s.. @&—e +4, rts, —€@ +i, rts, j= ——— “—-¢ ** —~ —__-"__g % , (14) © 8 8 the condenser potential is . eet Si, r+s e+ ty r-s e,=e 1 (r—s) 2 «2! (r+s) 2 “aE! =e—— —3s)—_cx“—“ _ OE ’ 2 8 8 (15) : |

CONDENSER CHARGE AND DISCHARGE 51 For no condenser charge, or 1, = 0, e, = 0, we have e A, = ; and e A, =— 3 =— A;: substituting in (11) and (12), we get the charging current as © ~ m8 _rts iat}. mL mutt (16) s The condenser potential as emeyl- slo tsye —(r—s)e ~ . (17) For a condenser discharge or i, = 0, € = &, we have 1 & A, ; and ; ey A,=+7=—-A,; 8 ‘ hence, the discharging current is e _TS, atts, in “fe aL , 2h . - (18) The condenser potential is r-s r+s — £0 § ~an' 4. .7 BL! e, aap tse (r—s)e , (19) that is, in condenser discharge and in condenser charge the currents are the same, but opposite in direction, and the con- denser potential rises in one case in the same way as it falls in the other. $2. As example is shown, in Fig. 11, the charge of a con- denser of C = 10 mf. capacity by an impressed e.m.f. of . | | | | . |

52 TRANSIENT PHENOMENA e = 1000 volts through a circuit of r = 250 ohms _ resistance and LZ = 100 mh. inductance; hence, s = 150 ohms, and the charging current is t = 6.667 {27 50! — ¢~20!} amperes. The condenser potential is €, = 1000 {1 — 1.333 e + 0.333 «-7!! volts. . a AO Gk = . pa) EH Ss oe Pe TTT Te Sool fT raed Cremer th S s ake a | da | | | —+— Bee ae —— al | eo eee 8 HH too raat EL Chee OO a) be haf alee | || 4 8 Py 16 20 24 2 32 36 40 Fig. 11. Charging a condenser through a circuit having resistance and induc- tance. Constant potential. Logarithmic charge. 33. The equations (14) to (19) contain the square root, / 4L s = r— Cc’ hence, they apply in their present form only when 4L p>. ace 4L . . . 0

Ifr = wai these equations become indeterminate, or = 7 and if 7? < 7 s is imaginary, and the equations assume a complex imaginary form. In either case they have to be rearranged to assume a form suitable for application.

Three cases have thus to be distinguished:

(a) rr > ae in which the equations of the circuit can be used in their present form. Since the functions are exponen- tial or logarithmic, this is called the logarithmic case.

\

CONDENSER CHARGE AND DISCHARGE 58 4L. we . .

(b) Pr = oc 3s called the critical case, marking the transi- tion between (a) and (c), but belonging to neither.

() r< te . In this case trigonometric functions appear; it is called the trigonometric case, or oscillation.

34. In the logarithmic case,

4L Pr >

C . or, 4L < Cr, that is, with high resistance, or high capacity, or low induc- tance, equations (14) to (19) apply. r—s rts

The terme 22‘ is always greater than « 22‘, since the former has a lower coefficient in the exponent, and the differ- ence of these terms, in the equations of condenser charge and discharge, is always positive. That is, the current rises from zero at ¢ = 0, reaches a maximum and then falls again to zero at ¢ = ©, but it never reverses. The maximum of the

. . e current is less than 7 = -

The exponential term in equations (17) and (19) also never reverses. That is, the condenser potential gradually changes, without ever reversing or exceeding the impressed e.m.f. in the charge or the starting potential in the discharge.

. 4L .

Hence, in the case 7’ > Ge no abnormal voltage is pro- duced in the circuit, and the transient term is of short duration, so that a condenser charge or discharge under these conditions is relatively harmless.

In charging or discharging a condenser, or in general a circuit containing capacity, the insertion of a resistance in series in the circuit of such value that 7° >Z therefore eliminates the danger from abnormal electrostatic or electromagnetic stresses. |

In general, the higher the resistance of a circuit, compared | with inductance and capacity, the more the transient term is suppressed. .

54 TRANSIENT PHENOMENA 35. In a circuit containing resistance and capacity but no inductance, L = 0, we have, substituting in (5), a1. r a + C ‘= 0, (20) or, transposing, di dt tr’ ‘ | which is integrated by ~ _L i= ™, (21) where c = integration constant.

Equation (21) gives for ¢ = 0, 7 = c; that is, the current at - the moment of closing the circuit must have a finite value, or must jump instantly from zero to c. This is not possible, but so also it is not possible to produce a circuit without any induc- tance whatever.

Therefore equation (21) does not apply for very small values of time, ¢t, but for very small ¢ the inductance, L, of the circuit, however small, determines the current.

The potential difference at the condenser terminals from (3) is

é,=e-71 hence it e,=e-—Te (22)

The integration constant c cannot be determined from equation (21) at ¢ = 0, since the current 7 makes a jump at this moment.

But from (22) it follows that if at the moment ¢ = 0, e, = e,, €, =e— 7, . hence, c=,

Tr and herefrom the equations of the non-inductive condenser circuit, t . re ie (e — e,)e (23) Tr and -4, e,=e—(e—e, Je ™. (24) As seen, these equations do not depend upon the current 7, in the circuit at the moment before ¢ = 0.

; | CONDENSER CHARGE AND DISCHARGE 55 : 86. These equations do not apply for very small values of t, but in this case the inductance, L, has to be considered; that is, equations (14) to (19) weet For L = 0 the first term in (14) becomes indefinite, as it o contains « ° 7 and therefore has to be evaluated as follows: For L = 0, we have Ss =f, rts 2 —™ "9 and r—s = 7 0 and, developed by the binomial theorem, dropping all but the first term, \ / 4L roearji- 1 - 42 2b, rc and r—s _ 1 , ,

• 8E r+e_r 2L L Substituting these values in equations (14) and (15) gives the current as _ -t ~e—rn, -2 1-2 %, re €7 ly Te i‘ (25) r r and the potential difference at the condenser as +t e,=e—(—ae)e ~; (26) that is, in the equation of the current, the term. _& = % ~ Ty ~ i Tr . .

. 56 TRANSIENT PHENOMENA has to be added to equation (23). This term makes thetransition =~ from the circuit conditions before 4 = 0 to those after! =0, . and is of extremely short duration.

For instance, choosing the same constants as in § 32, namely:

e = 1000 volts; r = 250 ohms; C = 10 mf., but choosing the inductance as low as possible, L = 5mh., gives the equations of condenser charge, i.e., for 7, = 0 and e, = 0,

17=4 {en 100" — e~ 80.0001} : and

e, = 1000 {1 — e-™}.

The second term in the equation of the current, -*"', has ~ decreased already to 1 per cent after t = 17.3 X 10-° seconds, while the first term, «*‘, has during this time decreased only by 0.7 per cent, that is, it has not yet appreciably decreased.

37. In the critical case,

4L P= C and s=0, r a, = 4, = 21’ e—e r a ~ %o _) 0 A,=-A,= - yoo Hence, substituting in equation (14) and rearranging, +) BE! BE!

• r. - a7! € —é i=(e-e—Z4,)e 2b (——). (27) The last term of this equation, ty _ fy Ne?! ~¢ 74 Fe— =—__ =|; ~D s 0 y .

CONDENSER CHARGE AND DISCHARGE 57 that is, becomes indeterminate for s = 0, and therefore is evaluated by differentiation, .

aN ds t F-= DL (28) ds Substituting (28) in (27) gives the equation of current, . i= tle —Fig)e FH (29) The condenser potential is found, by substituting in (15), to be r + $31! a r.\ -soe e =e —(e-e,- rig)e yt = 2 __. 0) The last term of this equation (30) is r+8 opt _T-8 -3rt p+! 2 2 r( se - se 1c Se +h a) TE Lg BT =) ,2L Pa ; Qs \ € + 5) € +e | (31) For s = 0, the first term of this equation (31), by substituting (28), becomes.” , the second term = 1, and substituting in (30), this gives the condenser potential as rt r.\ -soe . ame fit Flee, — File aL. (32) Herefrom it follows that for the condenser charge, 7, = 0 and e, = 0, t= £ ee PE! L and rt! -+ : a nefi-( eter},

| | 58 TRANSIENT PHENOMENA . for the condenser discharge, 1, = 0 and e = 0, . ; t -t | += L eof 27! : and | rt ~~ e= (1+ Fee the, : . 38. As an example are shown, in Fig. 12, the charging current . and the potential difference at the terminals of the condenser, os ee ete any Poe SeSeSReSeSSRe== im See H in a 85000 ao = a0) mh. ere 7 aun EM= 200 ohius] rE Per Pa CL ae ine cee sanaeeannaeee= D000 Wes’ Pee ECR EEE HEHE EHF 4 8 Rn 16 20 24 28 82 36 4u Fig. 12. Charging a condenser through a circuit having resistance and induc- tance. Constant potential. Critical charge. in a circuit having the constants, e = 1000 volts; C = 10 mf.: L = 100 mh., and such resistance as to give the critical start, that is, _ r= Vit = 200 ohms. . C In this case, t = 10,000 te and | e, = 1000 {1 — (1 + 2000) ec ™'}. 39. In the trigonometric or oscillating case,

• 4L re< ron : The term under the square root (10) is negative, that is, the ; square root, s, is imaginary, and a, and a, are complex imaginary quantities, so that the equations (11) and (12) appear in imagi- nary form. They obviously can be reduced to real terms, f

CONDENSER CHARGE AND DISCHARGE 59 ; since the phenomenon is real. Since an exponential function with imaginary exponents is a trigonometric function, and inversely, the solution of the equation thus leads to trigono- metric functions, that is, the phenomenon is periodid¢ or oscil- lating. Substituting s = jg, we have =\ AL q C r (33) and . ri a= oer t ia, 64) | 2 2 Substituting (34) in (11) and (12), and rearranging, . -— + Jd, . ~ 8, iae 2 ) Ae 2L° 4 Axe 22 { (35) a : ig _3 _ ia é, =ete oy Ad jr tidy HE yt st Ag vt (36) Between the exponential function and the trigonometric functions exist the relations et = cosy + jsinv and (87) e~” = cosv — j sin v. Substituting (37) in (35), and rearranging, gives ty , jae 22 JA, + 4) cos 54 t+ j (A, — A,) sin fi. Substituting the two new integration constants, B, =A,+ A, and | (38) B,=7(A, — A,), | gives ar tae 7h $B, cos 0+ Bysin St}. (39)