Question

13. In the given circuit, the switch has been closed for a long time prior to t=0, then it opens at t = 0 and closes again at t = 1s. Find iz(t) for all t. 232 w 212 WW

Answer 1

case a fortzo switch was alosed from therefore, Indenter behave long as time short circuit steady state, - 16) = {co) = {(ot) = 0 for t o Indentor switch behave was open Again as short circuit lb 24 o open 2A TO) Rt

In (0) = 2A* 2 = LA [corrent devider] (ln() = JA * calculation of time constant, (E) Rth = 20 ton (becaupe worrent open circuit to find Rth 1 Rth=4u Tz Rita= It ad pec = 1 sec. * Now corrent across Inductor is written ap- in4 = ['m (s) + (16+) - Inco ēEIT in 4 = 1+ Lord] ēt inct = trēta for 77,0 but t<25 Cope ® Now At tzt's in (2-) = to ēd z o. 632A 22(25) = 16+) – įQ) = 0.632A * Again t=ds closed ewitch then After long time , steady state Indeutor behaves as short circuit,

in ( 90A Do2A UL@ RI Rth - an due to and wrrent switch closed open cheniz Rth time constant = = 2 = = 2 Bel. row corrent written ap- in 4) = lncolt ['Gt) – I.(osette in4 = ot [0.632-o] et Inq) = 0.632 ē 0.6367-); 77 t der those fare, it = = o ; z 20. ln et A.; ost s 1 sec. azonF-b) A ; taber. 0.632 e