6. SELF-INDUCTANCE OF CONTINUOUS-CURRENT CIRCUITS 30. Self-inductance makes itself felt in continuous-current circuits only in starting and stopping or, in general, when the current changes in value. Starting of Current. If r = resistance, L = inductance of circuit, E = continuous e.m.-f. impressed upon circuit, i = current in circuit at time t after impressing e.m.f. E, and di the increase of current during time moment dt, then the increase of magnetic interlinkages during time dt is IM, and the e.m.f. generated thereby is r di ei = -L~di By Lentz's law it is negative, since it is opposite to the im- pressed e.m.f., its cause. Thus the e.m.f. acting in this moment upon the circuit is E + ei = E - L § CONTINUOUS-CURRENT CIRCUITS 25 and the current is or, transposing, ™ _ r dt rdt di L i- r the integral of which is . E rt . /. E\ . - L --= log, [i - -) - log, c, where — log, c = integration constant. This reduces to - • E _i - 7 i = - + ce L at t = 0, i = 0, and thus E -~ = c. Substituting this value, the current is and the e.m.f. of inductance is Att = co , i'o Substituting these values, _ rf = ir — E = — Ee~ L' r» -' «i = 0. i = iQ (l - e and Z. The expression u = j is called the "attenuation constant," and its reciprocal, — , the "time constant of the circuit."1 1 The name time constant dates back to the early days of telegraphy, where it was applied to the ratio : — , that is, the reciprocal of the attenuation con- stant. This quantity which had gradually come into disuse, again became of importance when investigating transient electric phenomena, and in this work it was found more convenient to denote the value Y as attenuation constant, since this value appears as one term of the more gen- eral constant of the electric circuit ( Y + ~r< ) • (Theory and Calculation of Transient Electric Phenomena and Oscillations, Section IV.) 26 ELEMENTS OF ELECTRICAL ENGINEERING Substituted in the foregoing equation this gives and ei = - = - 0.368 E. 31. Stopping of Current. In a circuit of inductance L and E resistance r, let a current IQ = — be produced by the impressed e.m.f. E, and this e.m.f. E be withdrawn and the circuit closed through a resistance r\. Let the current be i at the time t after withdrawal of the e.m.f. E and the change of current during time moment dt be di. di is negative, that is, the current decreases. The decrease of magnetic interlinkages during moment dt is Ldi. Thus the e.m.f. generated thereby is Tdi ei== ~Ldi It is negative since di is negative, and e\ must be positive, that is, in the same direction as E, to maintain the current or oppose the decrease of current, its cause. Then the current is e\ L di _ = r + n r + ri dt or, transposing, the integral of which is t = log, i - log, c, where— log€ c = integration constant. r+n This reduces to i = ce L 77* for t = 0, t CONTINUOUS-CURRENT CIRCUITS 27 Substituting this value, the current is E (r + n) t i = -€~ L , and the generated e.m.f. is r + r, _fc + ei =• t (r + fi) •» Js — — c i r •p Substituting IQ = — , the current is i = lot and the generated e.m.f. is 6l = H(r-f TI)€ JT-', At * = 0, that is, the generated e.m.f. is increased over the previously impressed e.m.f. in the same ratio as the resistance is increased. When TI = 0, that is, when in withdrawing the impressed e.m.f. E the circuit is short circuited, E _L« _';! i = — c L = io€ z, the current, and _ TL _ !l ei = E € L = i0re £ the generated e.m.f. In this case, at t = 0, e\ = E, that is, the e.m.f. does not rise. In the case r\ = <» , that is, if in withdrawing the e.m.f. E the circuit is broken, we have t = 0 and ei = » , that is, the e.m.f. rises infinitely. The greater r\, the higher is the generated e.m.f. e\, the faster, however, do e\ and i decrease. If n = r, we have at t = 0, 611 = 2E, i = i0, and en — i*» W = | z' Jo that is, the energy stored as magnetism in a circuit of current iQ and inductance L is 2 ' which is independent both of the resistance r of the circuit and the resistance n inserted in breaking the circuit. This energy has to be expended in stopping the current. EXAMPLES 32. (1) In the alternator field in Section 1, Example 4, Sec- tion 2, Example 2, and Section 5, Example 1, how long a time after impressing the required e.m.f. E = 230 volts will it take for the field to reach (a) J/£ strength, (b) %Q strength? (2) If 500 volts are impressed upon the field of this alternator, and a non-inductive resistance inserted in series so as to give the required exciting current of 6.95 amp., how long after impressing the e.m.f. E = 500 volts will it take for the field to reach (a) y% strength, (b) %o strength, (c) and what is the resist- ance required in the rheostat? (3) If 500 volts are impressed upon the field of this alter- nator without insertion of resistance, how long will it take for the field to reach full strength? (4) With full field strength, what is the energy stored as magnetism? (1) The resistance of the alternator field is 33.2 ohms (Section 2, Example 2), the inductance 112 h. (Section 5, Example 1), the impressed e.m.f. is E = 230, the final value of current E io = — = 6.95 amp. Thus the current at time t is t = * - 6 = 6.95 (1 - e-°-296<). CONTINUOUS-CURRENT CIRCUITS 29 (a) M strength: i = ~, hence (1 - €-°-29«0 = 0.5. e-o.296 1 = 0.5, - 0.296 Mog e = log 0.5, t = t = 2.34 seconds. (b) %0 strength: i = 0.9 *0j hence (1 - e-°-2960 = 0.9, and t = 7.8 seconds. (2) To get io = 6.95 amp., with E = 500 volts, a resist- 500 ance r = ^-f-= = 72 ohms, and thus a rheostat having a resist- o.9o ance of 72 — 33.2 = 38.8 ohms, is required. We then have i = io (l € 2) = 6.95 (1 - €-°-6430. (a) i = ^, after t = 1.08 seconds. & (b) i = 0.9 i0, after i = 3.6 seconds. (3) Impressing E = 500 volts upon a circuit of r = 33.2, L = 112, gives = 15.1 (1 - €-°-2960. i = 6.95, or full field strength, gives 6.95 = 15.1 (1 - e-°-2960. 1 - €-°-296 « = 0.46 and t = 2.08 seconds. (4) The stored energy is 6.952 X 112 ~ = 2720 watt-seconds or joules _ _ = 2000 foot-pounds. (1 joule = 0.736 foot-pounds.) Thus in case (3), where the field reaches full strength in 2.08 2000 seconds, the average power input is -c -^ = 960 foot pounds Z.Oo per second = 1.75 hp. In breaking the field circuit of this alternator, 2000 foot- pounds of energy have to be dissipated in the spark, etc. 33. (5) A coil of resistance r = 0.002 ohm and inductance L = 0.005 mh., carrying current / = 90 amp., is short circuited. 30 ELEMENTS OF ELECTRICAL ENGINEERING (a) What is the equation of the current after short circuit? (6) In what time has the current decreased to 0.1. its initial value? _ L* (a) i = /e L = 90 e-400'. (6) i = 0.1 7, c-400< = 0.1, after t = 0.00576 second. (6) When short circuiting the coil in Example 5, an e.m.f. E = 1 volt is inserted in the circuit of this coil, in opposite direc- tion to the current. (a) What is equation of the current? (6) After what time does the current become zero? (c) After what time does the current reverse to its initial value in opposite direction? (d) What impressed e.m.f. is required to make the current die out in Hooo second? (e) What impressed e.m.f. E is required to reverse the current in Kooo second? (a) If e.m.f. — E is inserted, and at time t the current is denoted by i, we have di ei = — L -r, the generated e.m.f. ; Thus, - E + 61 = - E - L jt, the total e.m.f.; and -E + ei E L di ^ i = = -r., the current; r r r dt} Transposing, r , di and integrating, - j- = log, (- + i) - log, c, where — log^ c = integration constant. At t = 0, i Substituting, E At t = 0, i = /, thus c = / + -; t E\ -« E h7r "7' Kon ,—400 1 p;nn *J*S\J C