A capacitor is charged to a potential of $V_{0} .$ It is connected with an inductor through a switch S. The switch is closed at time t=0. Which of the following statement(s) is/are correct?

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Energy stored in the inductor at $t = \frac{π}{2} \sqrt{L C}$ is $\frac{1}{4} C V_{0}^{2}$

The maximum current in the circuit is $V_{0} \sqrt{\frac{C}{L}} .$

Potential difference across the capacitor become zero for the first time at $t = π \sqrt{L C}$

Maximum energy of stored in the inductor is $\frac{1}{2} C V_{0}^{2}$

answer is A, D.

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Detailed Solution

This is a LC-circuit Therefore
$q = q_{0} \cos ω t$ and $V = V_{0} \cos ω t$ , where $ω = \frac{1}{\sqrt{L C}}$ (or) $T = 2 π \sqrt{L C}$
$i = \frac{d q}{d t} = - q_{0} ω \sin ω t$
A) Maximum current in the circuit is $i_{\max} = q_{0} ω = V_{0} \sqrt{\frac{C}{L}}$
B) P.D. across the capacitor becomes zero after time $t = \frac{T}{4} = \frac{π}{2} \sqrt{L C}$
C) at time $t = \frac{π}{2} \sqrt{L C} = \frac{T}{4}$ energy stored in the capacitor is zero and the total initial energy $\frac{1}{2} C V_{0}^{2}$ now gets stored in the inductor.
D) The maximum energy stored in the inductor will be $\frac{1}{2} C V_{0}^{2}$