Question

a) A sensitive electronic device of 150 Ω resistance is to be disconnected from an emf source by opening switch S1 (see Fig. 6). An initial current of 30mA runs through the circuit when S2 is closed. The current in the circuit after 48µs is 25mA. What is the R-L time constant ?

b) Find the device current after 40µs if the emf is 5V and S1 is closed. Take the values of resistance and inductance from above (part a).

E S a с w 0000 L R-> i S2 Fig. 6. The R-L circuit

Answer 1

* given circuit ix E ga + Si 150 tecee L R . S2 Sz is closed. initial current of 3oMA seuns through ckt when Si is opend & that means, current through inductor in previous steady state when s, is closed is 30 MA circuit before Sy is closed & s, open E inductor acts as shost ckt lson 30 mo -3 20 mA = 150 X 30 X10 е 150 Ea - 4.5 volts Yoo , when sz is closed the circuit is & s, is open. 30 m 15on m it rea frast order RL circuit Current through inductor is given by I = To é -the e time constant To an initial current lie BOMA through inductor F = 30x10 -3-tle So, Cardent

then given that, when t = 45 usec F = 95 ma d' Current in cht so, substitute these values in es-o to obtain e time constant RL circuit -t/e .45x10 x100/ - 45x15% 30 ln (25) 3 30 Xlo e 25x16 30 xx;} e 25 e -6 - 45 x 10 て 745x16" = 70.182 て 45% 10 = 247-252810 6 now, T: 0.182 80 = I finally Home Constant of RL circuit 243. 252 X 106 we snow that , t = L=8OR (247.252) x 16 x 150 37.083 X10-3 37.087 MH b mou - SV s, is closed. emf the circuit is SV Isor M 37.082mH m it is a first ooder RL eircuit

for a where first order RL circuit with input voltage, current through inductor is given by ficts = ca) + [aco) – Decodjetle f (o) initial current through inductor {(0) final steady state Current through inductor e time constant 247.252 X10 H already determined for same R&L tinitial current through inductor = for (pooblem-2] 6 O UA » in final steady state, circuit & inductor acts as short A COD) 5 150 0.0333 AMPA 33. 33 m По» equation of current through inductor is fct) = f (a) + [16) – f.CoJette as 9 -0. So, {{(t) = f( (o) [i- etle FCH) = 33.33 x103 é t/0247.252xión now Current after 4o. user in Fact) 33:33 x16} [ice 17.282 Xighter 4.938x1013 4.978 MA X103 1- e -40x1041 чо Дяс Ampa So, current through inductor 4.978 mA after is t= 4o lugec