Question

A capacitor, of capacity $C$, is fully charged with voltage $V_0$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}$. 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. $\frac{1}{3}C{V}_0^2$
• C. $\frac{1}{4}C{V}_0^2$
• D. $\frac{1}{6}C{V}_0^2$

Answer 1

The correct option is D $\frac{1}{6}C{V}_0^2$

When two capacitors with capacitance $C_1\text{and}{C}_2$, at potential $V_1\text{and}{V}_2$ respectively, are connected to each other by wire, charge begins to flow from higher to lower potential, till they acquire common potential.

Initially, $U_i=\frac{1}{2}C{V}_0^2$

Let $V$ is the final potential across both the capacitors.

$\Rightarrow U_f=\frac{1}{2}(C+\frac{C}{2})V^2=\frac{3C}{4}{V}^2$

Here, $V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}=\frac{C{V}_0}{\left(\frac{3C}{2}\right)}=\frac{2V_0}{3}$
$\therefore U_f=\frac{C{V}_0^2}{3}$

So, loss of energy will be,

$\Delta U=\frac{C{V}_0^2}{2}-\frac{C{V}_0^2}{3}=\frac{C{V}_0^2}{6}$

Hence, $\left(D\right)$ is the correct answer.