The circuit shown in the figure contains an inductor L , a...

(P) When $K_1$ is closed current in $R_0$ is $I_1$ At $\mathrm{t}=0$; circuit will be $$\begin{aligned} & \mathrm{I}_1=0 \\ & \mathrm{P} \rightarrow(1)\end{aligned}$$ (Q) After long time inductor behave as a wire so $\mathrm{I}_2$$\begin{aligned} & \mathrm{I}_2=\frac{20}{5}=4 \mathrm{ A} \\ & \mathrm{Q} \rightarrow(3)\end{aligned}$(R) When $\mathrm{K}_2$ is closed and $\mathrm{K}_1$ open\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \omega_0=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{5 \times 10^{-4}} \\ & \omega_0=2 \times 10^3 \mathrm{rad} / \mathrm{s} \\ & \omega_0=2 \text{ kilo-radian } / \mathrm{s} \\ & \mathrm{R} \rightarrow(2)\end{aligned}(S) Now $\mathrm{K}_2$ is closed and $\mathrm{K}_1$ open\begin{aligned} & \frac{1}{2} \mathrm{LI}_2^2=\frac{1}{2} \mathrm{CV}_0^2 \\ & 25 \times 10^{-3} \times(4)^2=10 \times 10^{-6} \times \mathrm{V}_0^2 \\ & \mathrm{ V}_0^2=2500 \times 16 \\ & \mathrm{ V}_0=50 \times 4=200 \mathrm{ V} \\ & \mathrm{ S} \rightarrow(5)\end{aligned}$$