Question

A capacitor $C$ is fully charged with voltage $V_0$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C/2. The energy loss in the process after the charge is distributed between the two capacitors is:

• A. $\frac{1}{2}C{V_0}^2$
• B. $\frac{1}{4}C{V_0}^2$
• C. $\frac{1}{3}C{V_0}^2$
• D. $\frac{1}{6}C{V_0}^2$

Answer 1

The correct option is D

$\frac{1}{6}C{V_0}^2$

Step 1. Given data:

Capacitance of fully charged capacitor is $C$ and voltage is $V_0$.

Capacitance of uncharged capacitor is $\frac{C}{2}$

Step 2. Solving for energy loss:

Since the two capacitors are connected in parallel, using $C_p=C_1+C_2$

The capacity of new combination is $C_f=\left(C+\frac{C}{2}\right)=\frac{3C}{2}$

Since the total charge $(Q=CV)$ is constant, we have

$C_f{V}_f=C{V}_0$

$\frac{3C}{2}{V}_f=C{V}_0$

$\Rightarrow V_f=\frac{2}{3}{V}_0$

The initial energy is $U_0=\frac{1}{2}C{V_0}^2$

The energy of new configuration is, $U_f=\frac{1}{2}\frac{3C}{2}{\left(\frac{2V_0}{3}\right)}^2$

$\Rightarrow U_f=\frac{1}{2}\frac{2}{3}C{V_0}^2$

The Energy loss is, $U_0-U_f=\left(1-\frac{2}{3}\right)\frac{1}{2}C{V_0}^2$

$\Rightarrow U_0-U_f=\frac{1}{3}.\frac{1}{2}C{V_0}^2=\frac{1}{6}C{V_0}^2$

Hence, option D is correct