Question

Two capacitors of capacitances $C$ and $2C$ are charged to potential differences $V$ and $2V$ respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:

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

Answer 1

Answer: (D)

$Q_1=CV$ $Q_2=2C\times 2V=4CV$

$\Rightarrow$ By conservation of charge
$q_i=q_f$
$Q_1+Q_2=q_1+q_2$
$4CV-CV=(C+2C)V_C$
$V_C=\frac{3CV}{3C}\Rightarrow V$

$\Rightarrow \frac{1}{2}\times (3C)\times V_c^2$
$=\frac{1}{2}\times 3C\times V^2=\frac{3}{2}C{V}^2$