In the circuit shown, the capacitor is initially charged to a potential V. At t=0 .switch S is closed. When magnetic energy stored to the inductor is equal to the electric energy stored in the capacitor, current flowing in the circuit is.

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In the circuit shown, the capacitor is initially charged to a potential V. At t=0 .switch S is closed. When magnetic energy stored to the inductor is equal to the electric energy stored in the capacitor, current flowing in the circuit is.

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$V \sqrt{\frac{C}{2 L}}$

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

$2 V \sqrt{\frac{C}{L}}$

$\frac{V}{2} \sqrt{\frac{L}{C}}$

answer is A.

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Detailed Solution

$U_{B} = \frac{1}{2} {Li}^{2}, U_{E} = U_{B} = \frac{1}{2} {Li}^{2}$
Conserving energy , $\frac{1}{2} {Li}^{2} + \frac{1}{2} {Li}^{2} = \frac{1}{2} {CV}^{2} \Rightarrow i = V \sqrt{\frac{C}{2 L}}$