Question

A capacitor of capacity $C$ is charged to a potential difference $V$ and another capacitor of capacity $2C$ is charged to a potential difference $2V.$ The charging batteries are disconnected and the two capacitors are connected to each other in parallel with reverse polarity. The energy lost in this process will be:

• A. $\frac{1}{2}C{V}^2$
• B. $\frac{2C{V}^2}{3}$
• C. $\frac{3}{2}C{V}^2$
• D. $3C{V}^2$

Answer 1

The correct option is D $3C{V}^2$

The net charge shared between the capacitors;

$Q^{\prime }=Q_1-Q_2=C_2{V}_2-C_1{V}_1=2C.2V-CV=3CV$

Potential will be same $V^{\prime }$

Net capacitance $=3C$

Combination

Net $V^{\prime }=\frac{3CV}{3C}=V$

$E_i=\frac{1}{2}C{V}^2+\frac{1}{2}(2C.4V^2)$

$E_i=\frac{9}{2}C{V}^2$

Energy of capacitors = $\frac{3}{2}C{V}^2$

$\Delta E=E_i-E_f=\frac{9}{2}C{V}^2-\frac{3}{2}C{V}^2=3C{V}^2$