In the given circuit,when S_1 is closed,the c | vedclass

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(B) When $S_1$ is closed,the capacitor $C$ charges to a potential difference $V$. When $S_1$ is opened and $S_2$ is closed,the capacitor $C$ and induc

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The instantaneous current in the circuit may be $V \sqrt{\frac{C}{L}}$.

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(B) When $S_1$ is closed,the capacitor $C$ charges to a potential difference $V$. When $S_1$ is opened and $S_2$ is closed,the capacitor $C$ and inductor $L$ form an $LC$ oscillating circuit. The energy initially stored in the capacitor is $U_E = \frac{1}{2} C V^2$. As the capacitor discharges,the energy is transferred to the inductor as magnetic energy $U_B = \frac{1}{2} L I^2$. By the law of conservation of energy,the maximum current $I_{max}$ occurs when all electrostatic energy is converted to magnetic energy: $\frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2$ $I_{max}^2 = \frac{C V^2}{L}$ $I_{max} = V \sqrt{\frac{C}{L}}$. Thus,the instantaneous current in the circuit can reach this maximum value.

In the given circuit,when $S_1$ is closed,the capacitor $C$ gets fully charged. Then $S_1$ is kept open and $S_2$ is closed. Hence

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