Question

A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a voltage $V$. Another capacitor of capacitance $2C$ is similarly charged to a voltage $2V$. The batteries are disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other capacitor. The final energy of the configuration is

• A. $0$
• B. $\frac{3C{V}^2}{2}$
• C. $\frac{25C{V}^2}{6}$
• D. $\frac{9C{V}^2}{2}$

Answer 1

The correct option is B $\frac{3C{V}^2}{2}$

Charge on the first capacitor, $Q_1=CV$


Charge on the second capacitor, $Q_2=4CV$


When they are connected together like this


The common voltage of above condition will be

$V_c=\frac{Q_2-Q_1}{C_1+C_2}$

$\Rightarrow V_c=\frac{4CV-CV}{C+2C}=V$

As they are now in parallel connection, so equivalent capacitance will be

$C_{eq}=C+2C=3C$

So, the energy stored in the capacitor is given by

$U=\frac{1}{2}\left(3C\right)V^2=\frac{3}{2}C{V}^2$

Hence, option $\left(b\right)$ is correct.