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(D) At $t = 0$, the switch $S$ is closed. The circuit consists of an inductor $L$ in parallel with two capacitors $C_1$ and $C_2$ which are also in pa

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maximum charge on $C_1 = I_0 C_1 \sqrt{\frac{L}{C_1 + C_2}}$

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(D) At $t = 0$, the switch $S$ is closed. The circuit consists of an inductor $L$ in parallel with two capacitors $C_1$ and $C_2$ which are also in parallel with each other. The equivalent capacitance is $C_{eq} = C_1 + C_2$.At $t = 0$, the energy stored in the inductor is $U_L = \frac{1}{2} L I_0^2$ and the energy in the capacitors is $U_C = 0$.As $t$ increases, the energy oscillates between the inductor and the equivalent capacitor. By conservation of energy, the maximum energy in the capacitor equals the initial energy in the inductor:$\frac{1}{2} C_{eq} V_{max}^2 = \frac{1}{2} L I_0^2$$V_{max}^2 = \frac{L I_0^2}{C_1 + C_2}$$V_{max} = I_0 \sqrt{\frac{L}{C_1 + C_2}}$The maximum charge on capacitor $C_1$ is $Q_{1,max} = C_1 V_{max} = C_1 I_0 \sqrt{\frac{L}{C_1 + C_2}}$.Thus, option $(D)$ is correct.

At a moment $t = 0$ when the charge on capacitor $C_1$ is zero, the switch $S$ is closed. If $I_0$ is the current through the inductor at that instant, then for $t > 0$,

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