The figure shows a capacitor of capacitance $C$ connected to a battery via a switch has a total charge $Q$ on it in steady state. When the switch $S$ is turned from position $A$ to position $B$ , the energy dissipated in the circuit is

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$\frac{1}{8} \frac{Q^{2}}{C}$

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

$\frac{3}{4} \frac{Q^{2}}{C}$

$\frac{5}{8} \frac{Q^{2}}{C}$

answer is B.

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Detailed Solution

$Δ U = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2}$
$\begin{array}{l} = \frac{1}{2} \frac{C \times 3 C}{4 C} {(E)}^{2} \\ = \frac{3}{8} C E^{2} = \frac{3}{8} (\frac{Q^{2}}{C}) \end{array}$

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