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

A coil of inductance $$10$$ $$H$$ resistance $$40\,\,\Omega $$ is connected as shown in Fig. After the switch $$S$$ has been in connection with point $$1$$ for a very long time, it is moved to point $$2$$ at $$t = 0.$$

If, at $$t = {0^ + }$$, the voltage across the coil is $$120$$ $$V,$$ the value of resistance $$R$$ is

The circuit shown in the Fig. is in steady state, when the switch is closed at $$t=0.$$ Assuming that the inductance is ideal, the current through the inductor at $$t = {0^ + }$$ equals