Question

Four identical capacitors are connected with a battery of voltage $\textit{V}$ and two switches $K_{1}$ and $K_{2}$ as shown in the figure below. Initially, $K_{1}$ is closed, now if $K_{2}$ is also closed, find the heat loss.

• A. $\frac{1}{2}\textit{CV}^{2}$
• B. $\frac{2}{3}\textit{CV}^{2}$
• C. $\frac{1}{3}\textit{CV}^{2}$
• D. $\frac{1}{4}\textit{CV}^{2}$

Answer 1

Answer: (C)

When $k_{1}$ is closed

$Q=CV$
energy $U_{\text{i}}=\frac{1}{2}\textit{CV}^{2}$
When $k_{2}$ is also closed

The equivalent circuit:

$C_{\text{eq}}=C+\frac{2 C}{3}=\frac{5 C}{3}$
Energy $U_{\text{f}}=\frac{1}{2}\times \frac{5 C}{3}V^{2}=\frac{5}{6}\textit{CV}^{2}$
charge supplied by battery after closing $k_2$
$=\frac{5}{3}\textit{CV}-\textit{CV}=\frac{2}{3}\textit{CV}$
the energy supplied by the battery $=U_{\text{f}}-U_{\text{i}}+\Delta H$
$\Rightarrow \frac{2}{3}CV^{2}=\frac{5}{6}\textit{CV}^{2}-\frac{1}{2}\textit{CV}^{2}+\Delta \textit{H}$
$\therefore \Delta H=\frac{1}{3}CV^{2}$