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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
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Verified Answer
The correct answer is:
The instantaneous current in the circuit may be $V \sqrt{\frac{C}{L}}$.
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.
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.
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