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

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

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 $$t = 0.$$

For the value of $$R$$ obtained in the above question, the time taken for $$95\% $$ of the stored energy to be dissipated is close to