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(B) Initial state (switch at $A$): The capacitor $C$ is charged by the battery of $EMF$ $E$. The charge on the capacitor is $Q = CE$. The initial ener

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$\frac{3}{8}\frac{Q^2}{C}$

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(B) Initial state (switch at $A$): The capacitor $C$ is charged by the battery of $EMF$ $E$. The charge on the capacitor is $Q = CE$. The initial energy stored is $U_i = \frac{1}{2}CE^2 = \frac{1}{2}\frac{Q^2}{C}$.Final state (switch at $B$): The capacitor $C$ is connected in parallel with the capacitor $3C$. The total charge $Q$ is redistributed between the two capacitors. Since they are in parallel,the final potential difference $V$ across both is the same. The total capacitance is $C_{eq} = C + 3C = 4C$. The final potential difference is $V = \frac{Q}{C_{eq}} = \frac{Q}{4C} = \frac{E}{4}$.The final energy stored is $U_f = \frac{1}{2}C_{eq}V^2 = \frac{1}{2}(4C)\left(\frac{E}{4}\right)^2 = \frac{1}{2}(4C)\frac{E^2}{16} = \frac{1}{8}CE^2 = \frac{1}{8}\frac{Q^2}{C}$.The energy dissipated in the circuit is $\Delta U = U_i - U_f = \frac{1}{2}\frac{Q^2}{C} - \frac{1}{8}\frac{Q^2}{C} = \left(\frac{4-1}{8}\right)\frac{Q^2}{C} = \frac{3}{8}\frac{Q^2}{C}$.

In the figure shown,after the switch $S$ is turned from position $A$ to position $B$,the energy dissipated in the circuit in terms of capacitance $C$ and total charge $Q$ is

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