Question

RO R1 R2 TIL t = 0 12 mV+ L C: Figure P12.21: Circuit for Problems 12.21 and 12.22.

Determine iL(t) for in the circuit of Fig. P12.21 for t =45 ms , given that the switch was opened at t = 0 after it had been closed for a long time, vs = 12 mV, R0 = 5 Ω , R1 = 10 Ω,R2 = 20 Ω, L = 0.2 H, and C = 6 mF. Give the current in micro Amperes, Hint, the initial conditions are zero and the steady state current in the inductor will be 800 microAmps.

Correct answer is around 717.

Answer 1

lon 2on 5n w IdiL 12mv of to 30.2H GMK Initially, at tcas switch is closed. initial conditions are zero, and steady state current in inductor is 800 MA I Loo = SOOMAL Ct=o) & at toi At to switch is opened. Now, inductor is having the effect of input voltage 12 mv. current through inductor at inductor at time, too is iz (t) = į(0) +((((o) -il (8) e-the Given, steady state current is sooMA, initially, current is o i(t) = 800MA + (0-800M) e the e is the time constant

&=22 in(t)800 x10 6 +(0-500x166) en here, time constant , ? is where, R=5+10+20=35 - Ilt) at any time, too is T= 2x0.2 -0.011428 35 800x106 (retle) oo (pre-t/olls) = 800 x 10 LCD = 800 (1-e otis) / Mi Amps at too at t=45mser, iclasm) = 800 .asxlo dolls e ) M =800(1-e 3.937+) = 800 (0.983 = 784.4MA . at t=45msec, i (t)=784.4 MA

I hope explaination is clear. Comment for any queries.

Thank you.