Question

Use the Diagrams document. Determine il(t) for in the circuit of Fig. P12.21 fort =13 ms, given that the switch was opened at t = 0 after it had been closed for a long time, vs = 12 mV, RO = 50, R1 = 102, R2 = 2012, 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. RO Ri R2 JIL t=0 12 mV+ L Figure P12.21: Circuit for Problems 12.21 and 12.22.

I got the answer wrong but the correct answer is 348.7539

Answer 1

conditions, taplace By assuming initial equivalent circuit diagram for too 5 10 www is as follows. Ves) 20 N1,05) 0.93 18m 1000 65 applying KOL at Ves) VCS) – 12x103 Vas) VOS) S + 0.25 20 + 1000 15 65 35 VOS) 12 [is I+ 15 o] 0.25 60S500 1000 st AS vis) 35 12 [ 15 +0.95 3 S + ] COS + 500 1000S 15 12X VOS) [ QIS'+1000 S + 7500 3 5(60S+ 500) ]- 1000 S 15 VO) = 36 GOS+500) x 15 COIS +1000 5+7 500) 1000 we Icnow that 105) 36 (60S +500) Vos) x103 0.25 0.95 (215+1000 5+7500) 15 (6%) (605 +500 ) S($+ 1000 21 I(S) - 2500 ) 15 Xi03 S+ IS) (607) (605 +500) S(3+38.3) (5+9.39)

[At B C I(S) = + 5+38.3 S+9.33 AS همراه5.1 12 SO B = (S+38.3) TCS) \s=-38.3 - -13.9 ca(579.33) ICS) 152-9.3 = 1.9 5.-9.3 ICS) - 1113 + 1.9 S49.33 13.9 S+383 ] x 10 atow by inverse laplace toans form ict) = 12 +1.9e ē a 9.33 15 -38.32 - 13.90 mA (13ms) = =(19+ -9.33x13x103 + 19. 38.3x13x103 -13.9 e Jx103 15 (3m): 348.969 pe A