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(B) $1$. Initially,the switch $S_1$ is closed and $S_2$ is open. The capacitor $C$ charges to the potential $V$ of the battery.$2$. The energy stored

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$V \sqrt{\frac{C}{L}}$

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(B) $1$. Initially,the switch $S_1$ is closed and $S_2$ is open. The capacitor $C$ charges to the potential $V$ of the battery.$2$. The energy stored in the capacitor is $U_C = \frac{1}{2} C V^2$.$3$. When $S_1$ is opened and $S_2$ is closed,the capacitor discharges through the inductor $L$,forming an $LC$ oscillation circuit.$4$. According to the law of conservation of energy,the maximum energy stored in the capacitor is transferred to the inductor as magnetic energy when the current is maximum $(i_{max})$.$5$. $\frac{1}{2} C V^2 = \frac{1}{2} L i_{max}^2$$6$. Solving for $i_{max}$,we get $i_{max}^2 = \frac{C V^2}{L}$,which implies $i_{max} = V \sqrt{\frac{C}{L}}$.

In the circuit given below,the capacitor $C$ is charged by closing the switch $S_1$ and opening the switch $S_2$. After charging,the switch $S_1$ is opened and $S_2$ is closed. Find the maximum current in the circuit.

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