Question

A parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to potential difference $2V$. The charging batteries are now 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. The final energy of the configuration is

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

Answer 1

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

Let $q_1$ and $q_2$ be the charge on $C$ and $2C$ respectively when they are connected to the battery $V$ and $2V$ respectively.

$q_1=CV$

$q_2=\left(2C\right)\left(2V\right)$

$=4CV$
The capacitors are now joined in parallel
Since in above condition, positive
terminal of one capacitor is connected to the negative terminal of other , total charge will be $|q_1-q_2|$

Let $Q_f$ be the total charge

$Q_f=|4CV-CV|$

$=3CV$

Also, in parallel, total capacitance

$C_{eq}=C+2C$

$=3C$

Now, let $V_c$ be the common potential difference across capacitors
$V_c=\frac{Q_f}{c_{eq}}=\frac{3CV}{3C}=V$

$\therefore$ Energy of configuration $=\frac{1}{2}{c}_{eq}(V_c{)}^2$ $=\frac{1}{2}\left(3C\right)\left(V^2\right)=\frac{3C{V}^2}{2}$
Hence, option (c) is correct.