In the given circuit, when $S_1$ is closed, the capacitor $C$ gets full charged. Then $S_1$ is kept open and…
In the given circuit, when $S_1$ is closed, the capacitor $C$ gets full charged. Then $S_1$ is kept open and $S_2$ is closed. Hence
The current in the circuit is in the same direction.
The instantaneous current in the circuit may be $V \sqrt{\frac{C}{L}}$.
The energy stored in the circuit is purely in the form of magnetic energy.
There is no exchange of energy between inductor $L$ and capacitor $C$.
Solution
When $S_1$ is closed, potential difference across capacitor is $V$.
Direction of current gets reversed when $S_2$ is closed.
When $S_1$ is opened and $S_2$ is closed,
$\begin{aligned}
& \frac{1}{2} C V^2=\frac{1}{2} L I^2 \\
& \Rightarrow I=V \sqrt{\frac{C}{L}}
\end{aligned}$
Total energy oscillates between capacitor and inductor.
