For the RC circuit the switch is closed for a long time and

For the RC circuit, the switch is closed for a long time and then is opened at time t = 0. For the RL circuit, the switch is in position B for a long time and then it is switched to position A at time t = 0.

Solution

at Vc(0-) switch is closed and assume that circuit has reached the steady state.Cacpitor is opened.Therefore Vc(0-) = is obtained by voltage divider rule.Therefore.Vc(0-)= V/2.Switch is opened now.The circuit reduces to Capacitor and Resistor.Applying KVL Vc+RC*dVc/dt =0 Therefore.applying Lapalce transform, RCSVc(s)-RCVc(0-)+Vc(s)= 0. =>

Vc(s)[RCS+1] = RCVc(0-)= RCV/2 => v(t) = V/2*e^(-t/RC) [After applying inverse laplace transform]Therefore Vc(0+) = V/2*e^(-t/RC) and Ic = CdVc/dt =-V/2Re^(-t/RC)

secnd bit,When switch is at B assume current and voltage to be zero ie,IL(0-) = 0 in the inductor as there is no current source or voltage source attached in the position B loop.When switch goes to position A then V=iR + Ldi/dt applying laplace transform, V/S = I(s)[R+LS] => V/S(LS+R) =I(s) => I(s) = V/R[(1/S) - (1/(S+R/L))].Therefore IL(t) = V/R*[1-e^-(tR/L)] =IL0+).VL = LdiL/dt = V/L(e^-(tR/L))