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)

$Q_0=Cϵ$
$\frac{Q_1}{C}=\frac{Q_2}{3C}{Q}_1+Q_2=Q_0$
$\Rightarrow Q_1=\frac{Q_0}{4};Q_2=\frac{3Q_0}{4}$
Energy dissipated,
$E=\frac{1}{2}\frac{Q_0^2}{C}-\frac{1}{2}\frac{Q_1^2}{C}-\frac{1}{2}\frac{Q_2^2}{3C}$
$=\frac{1}{2C}\left(Q_0^2-\frac{Q_0^2}{16}-\frac{9Q_0^2}{3\times 16}\right)$
$=\frac{Q_0^2}{32C}\left(16-1-3\right)=\frac{3}{8}\frac{Q_0^2}{C}$