Question

4 a. Consider the following arcuit with +1(0=0, V(o)=0 loomull EV •Find illt) C=400mF L=100 mH R=4000 homA 102 teo 402 Find i (t) 35H lzov 14hE

Answer 1

Given circuit . parallel RLC 7.C) 100mA 1 100ml 400-2 V 40onf 4 ARC wo w = Jug ² Q² SLO - . 2 RC 21400x400x103 320 1 Srad/sec. , wid 15² 100x400x106 (320)2 + loom A since Q< wo The system shows underdamped response. { (P) = forced natural iforced inatural = ē at (a coswejt + A, sľn wont] -t/320 QCH) = 0.1 + ē CA cos 5t + A2 sin st at t=0, 9210)=0 and dilo). Vco) voolis since no current flows in resistor at t=0", "Total current of lows on Capacitor VLO)=0 dillo) со ZIND dt L SO dt

from ©, Substituting t=0 in 82 (+) 10) = 0.1 + cos 5(0) + Ay sin 5(0)] 0.1+ 1 (4,4 o] 320 e - A from ② . dil(a) = dt -AY 320 +5A2 t=o we know A,=-0. dilt) de An =-6.25x10-5 t=0 23206-0.1cOS 57-6.25x16 los sinst] PX1H)= 0.1te (5) at to circuit reached steady state inductor short circuited, capacitor open circuited 10.1- 402 120 (+) C = 44F LESH istys -120 -- 2.40 10 AD Ito vel-2120+10 62.4)] -967 to

40.2 pit) leso is 96V SH 4RF + ich) No = 96v, =-2.4A 4x5 42 -C R2 4x10-6 0124 >o 402 So, underdomped R App 221 W ug ²-d2 JLC dz4, Woo 223.606, d = 223.371 (t)= * [A COSW, t t hy sin wort] dict) dt z-QA, & wg Az 1=0 PO) = -2.4 -20) (A, CI}+A Az[0] A =-2.4 dico) dr Vio L because inductor is short 0 = 0 5 circuited. voltage across it is 0. -da, tWdAz 0 A23-0.0.43 i(t) -40 [-2.40 .4 COS(223.571+)--00435n(22354)