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Consider the circuit shown with respective specifications of elements marked in the figure. Capacitor $- 1$ is charged such that charge on it is $Q_{0}$ and it's left plate is positively charged. While capacitor $- 2$ is uncharged. The switch is closed at $t = 0 .$

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Frequency of oscillations of charge on left plate of capacitor $- 1$ is $\frac{1}{π} \sqrt{\frac{2}{L C}}$

Maximum current through the inductor is $\frac{Q_{0}}{\sqrt{L C}}$

Frequency of oscillations of charge on left plate of capacitor $- 1$ is $\frac{1}{π \sqrt{2 L C}}$

Maximum current through the inductor is $\frac{Q_{0}}{\sqrt{2 L C}}$

answer is A, C.

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Detailed Solution

$f = \frac{1}{2 π \sqrt[]{L C_{e q}}} = \frac{1}{π \sqrt[]{2 L C}}$

When $i$ is max $, \frac{d i}{d t} = 0$

Charge on two capacitors is same

$\frac{1}{2} L i_{0}^{2} = 2 \times \frac{{(\frac{Q_{0}}{2})}^{2}}{2 C} = \frac{{Q_{0}}^{2}}{4 C}$

$i_{0} = \frac{Q_{0}}{\sqrt{2 L C}}$

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