Question

Four identical capacitors are connected with a battery of voltage $\text{V}$ 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. $\frac{1}{2}{\text{CV}}^2$
• B. $\frac{2}{3}{\text{CV}}^2$
• C. $\frac{1}{3}{\text{CV}}^2$
• D. $\frac{1}{4}{\text{CV}}^2$

Answer 1

Answer: (C)

When $k_1$ is closed

$Q=CV$
energy $U_{\text{i}}=\frac{1}{2}{\text{CV}}^2$
When $k_2$ is also closed

The equivalent circuit:

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