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(C) When the key is moved to $(2)$,the capacitor discharges through the inductor,forming an $LC$ oscillator.Let $Q$ be the maximum charge on the capac

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$\frac{\pi \sqrt{LC}}{4}$

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(C) When the key is moved to $(2)$,the capacitor discharges through the inductor,forming an $LC$ oscillator.Let $Q$ be the maximum charge on the capacitor. The total energy of the system is $U_{total} = \frac{Q^2}{2C}$.We want the energy in the capacitor $U_C$ to be equal to the energy in the inductor $U_L$. Since $U_C + U_L = U_{total}$,if $U_C = U_L$,then $U_C = \frac{1}{2} U_{total}$.$\frac{q^2}{2C} = \frac{1}{2} \left( \frac{Q^2}{2C} \right) \Rightarrow q^2 = \frac{Q^2}{2} \Rightarrow q = \frac{Q}{\sqrt{2}}$.The charge on the capacitor at time $t$ is given by $q = Q \cos(\omega t)$,where $\omega = \frac{1}{\sqrt{LC}}$.Setting $q = \frac{Q}{\sqrt{2}}$,we get $Q \cos(\omega t) = \frac{Q}{\sqrt{2}} \Rightarrow \cos(\omega t) = \frac{1}{\sqrt{2}}$.This implies $\omega t = \frac{\pi}{4}$.Therefore,$t = \frac{\pi}{4\omega} = \frac{\pi}{4} \sqrt{LC}$.

Initially,the key was placed on $(1)$ until the capacitor got fully charged. Now,the key is placed on $(2)$ at $t = 0$. Find the minimum time when the energy in both the capacitor and the inductor will be the same.

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