The time constant for the given circuit will be

In the figure, transformer $${T_1}$$ has two secondaries, all three windings having the same number of turns and with polarities having the same number of turns and with polarities as indicated. One secondary is shorted by a $$10\,\Omega $$ resistor $$R,$$ and the other by a $$15\mu F$$ capacitor. The switch $$SW$$ is opened $$(t=0)$$ when the capacitor is charged to $$5$$ $$V$$ with the left plate as positive. At $$t=0+$$ the voltage $${V_P}$$ and current $${I_R}$$ are

In the circuit shown in Fig. switch $$S{w_1}$$ is initially CLOSED and $$S{w_2}$$ is OPEN. The inductor $$L$$ carries a current of $$10$$ $$A$$ and the capacitor is charged to $$10$$ $$V$$ with polarities as indicated. $$S{w_2}$$ in initially CLOSED at $$t = {0^ - }$$ and $$S{w_1}$$ is OPENED at $$t=0.$$ The current through $$C$$ and the voltage across $$L$$ at $$t = {0^ + }$$ is

An ideal capacitor is charged to a voltage $${V_0}$$ and connected at $$t=0$$ across an ideal inductor $$L.$$ (The circuit now consists of a capacitor and inductor alone). If we let $${\omega _0} = 1/\sqrt {LC} ,$$ the voltage across the capacitor at time $$t>0$$ is given by