In the circuit shows in Fig. switch `k_(2)` is open and switch `k_(1)` is closed at `t = 0`. At time `t = t_(0)`, switch `k_(1)` is opened and switch `k_(2)` is simultaneosuly closed. The variation of inductor current with time is

In the given circuit, the switch K_2 is opened and the switch K_1 is closed at time t=0. At time t=t_0 , the switch K_1 is opened and the switch K_2 is simultaneously closed. Sketch the variation of the inductor current I with time.

In the circuit shown in figure the switch S is closed at t = 0. A long time after closing the switch

In the circuit as shown in figure the switch is closed at t = 0 . At the instant of closing the switch

In the circuit as shown in figure the switch is closed at t = 0 . A long time after closing the switch

The circuit shown in figure is in the steady state with switch S_(1) closed.At t=0 , S_(1) is opened and switch S_(2) is closed. The first instant t when energy in inductor becomes one third of that in capacitor is equal to (pi xx 10^(-5))/x sec . Then finout value of x .

The circuit shown in figure is in the steady state with switch S_(1) closed.At t=0 , S_(1) is opened and switch S_(2) is closed. The first instant t when energy in inductor becomes one third of that in capacitor is equal to (pi xx 10^(-5))/x sec . Then finout value of x .

The circuit shows in Fig. is in the steady state with switch S_(1) closed. At t = 0,S_(1) is opened and switch S_(2) is closed. a. Derive expression for the charge on capacitor C_(2) as a function of time. b. Determine the first instant t , when the energy in the inductor becomes one-third of that in the capacitor.