Question

1. Response of the RL circuit - After having been in position 1 for a long time, the switch in the circuit below is moved to position 2 at the time t = 0. Given that Vo = 12 V, R1 = 3092, R2 = 12092, R3 = 6022 and L = 0.2 H, determine; a) iz(0-) and vi(0-) b) iz(0) and vi(0) c) il(o) and vi(0) d) il(t) for t20 e) vi(t) for t20 RU AN

Answer 1

m2 to X tro H ¢ & ve R2 1. R2 & 1 Vo = 12v, RL=3002 Ra=1208, R3=600 L=0.2H

Lay For tco &R2 {R3 & VL since there is no power supply for tão so No current will flow across Inductor i_105) = 0 A since in (0)=0 Now At t=00, Inductor behaves as short circut wire. so original circuit diagam at to is

ILLO-) {R2 R3 YIL V.10-) so VL (0)=on ( For short circut No potential drop) <by Now Inductol follow current continuity and capacitor follow Voltage continuity for all power source except Impulse (8 (t)) Source ie. ilco-) = Il 10tJ=I2(0) And Nelo) 'Velo-) = Vc (ot) = Velo) IL10-) = 0 A do Il 10+1 =IL10) = OA

Now ciruel For ty, Žd when t oo Inductor goes in saturation state and Act as short aircuit wire so iruil for t o mRz VIL { Rz It Now wire will Across R3, there is short circut parallel to it. So No current flow through R3.

so Equivalent ciruint at toto MILO a Velo) Illos) = Vo-o I RI = 12 A 30 | Il los) = 04A VL (0)= ov l short circuit across Inductor ) Now We will use transient formula For Inductor current i. (H) = 4(0)-hi (0)-in lotge i where t is time constant of circond.

Il 100) = 0.4 A - in (ot) = OA t = 1 where across t Rth us Thevenin for tyo Equivalent - {R39 RTH parallel Both R3 and Ro are in so RH = RillR3 = 30x60 - 30460 Rw = 2012 I= 0.2 H T= = b 2 = 0.01 sec RTH 20

+7,0 in (t) = Izcal-Lid Cos) - incoryf et = 04 - 604-05 e o za = ouble this as = 0.411-2 Tot ja t20 From This Equation we can find all values of current for t710 Example t=0 -OXO iLLO) = 0.4 li-e = 0.4 11-1) 50A Already calculated in part b i1100) = 0:4 (1-ēto = 0.4(1-dot) = 0.4 (1-0) = 0.4A

VL (2) = L di -10t -0.2 d 04 (1-és It = 0.08 / d()-de 1 lot O ldt dt -lot v = | V.It) = 0.08X(-11X45) e 0.4 e lot , Here t=0 V210) = 0.4 elox VL(0) = 0.4v - Part b V. (0) VL (0) = 0:4 - OV 2000 I etloxo > Already calculated in part c