Question

The figure shows a capacitor of capacitance $C$ connected to a battery via a switch, having 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

• A. $\frac{1}{8}\frac{Q^{2}}{C}$
• B. $\frac{3}{8}\frac{Q^{2}}{C}$
• C. $\frac{3}{4}\frac{Q^{2}}{C}$
• D. $\frac{5}{8}\frac{Q^{2}}{C}$

Answer 1

Answer: (B)

$Q_{0}=Cε$
$\frac{Q_{1}}{C}=\frac{Q_{2}}{3 C}Q_{1}+Q_{2}=Q_{0}$
$\Rightarrow Q_{1}=\frac{Q_{0}}{4};Q_{2}=\frac{3 Q_{0}}{4}$
Energy dissipated,
$E=\frac{1}{2}\frac{Q_{0}^{ \, 2}}{C}-\frac{1}{2}\frac{Q_{1}^{ \, 2}}{C}-\frac{1}{2}\frac{Q_{2}^{ \, 2}}{3 C}$
$=\frac{1}{2 C}\left(Q_{0}^{2} - \frac{Q_{0}^{2}}{16} - \frac{9 Q_{0}^{2}}{3 \times 16}\right)$
$=\frac{Q_{0}^{2}}{32 C}\left(16 - 1 - 3\right)=\frac{3}{8}\frac{Q_{0}^{2}}{C}$