Question

4. Consider the circuit below. Assume the switch has been in position A for a long time. It moves to position Bat time t = 0s, and back to position A at time t = 3s. Determine the values of Rand L such that i(t) = 2 - 2e-2 at time t = 3s and i(t) = (2-2e-2) e-4 at time t =9s. B t=0 22 2A ŽR

rate for correct answer

Answer 1

fortso → No Source is available for suppling Current to Inductor so i Current through i (0-) =OA Inductor Property Inductor Que to i (8t) = oA circait for tro ~ 1+1 212 2A for - supply fully [Steady state] Inductor will get Charged with given and will behave as Short circuit so circuit becomes for t o i 913 ment Due to Short circent, no current flow the 80 (m) = 2 A flow through el

Tow Calculate time it Contern (z) cariching Constant (2) which is Sec Calculating e, all Independent Source should be off. 80 In cirait for to A Rev 2n} so Per = 21) and T= Genral equation for Corrent through Inductor ilt) = .() + (ilot) - ico)et-tas/t_ ci ta-time at which Switch mored 90 - (t-o) iti= 2 + (0-2) e 12/27 ilt A ilt = 2 (1 – é & A for t-3 sec. ilti-2 (1-é & x3) A

Compare with given equation for t= 3 sec. [[= 3 H Answer sec, switch moved to Tosition A. I so at t=3 for tyg i(+) wt we need only this Portion for analysis 7 itt) 3 L=34 Now auctor again no souro available, so stored from (t=0 sec to 3 sec) will decay is Resistor R And in last - Ti ( =OA) and from analysis i(3) = 2-22-2

gain use equation (1) and Hove to pin 3 % 80 ilt- ot(i (3) - o) é et It-3) itt) = 13) -(4-3) A o it) = (2-962)(-+-$t->) for t=9 sec. ittia (-229(e-$-) itu = (2-94-) (6-2) Compare with geven equation in question 2R=4 [R= 28 Arsen