The capacities of two capacitors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2$. If they are connected with a wire, then the loss of energy is given by:

$\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}$

The correct answer is Option (3) → $\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}$

On connecting, 'V' be the common potential. According to law of conservation of charge -

$Q_{initial}=Q_{final}$

$C_1V_1+C_2V_2=C_1V+C_2V$

$⇒V=\frac{C_1V_1+C_2V_2}{C_1+C_2}$

Energy loss = $Ei-Ef$

$=\frac{1}{2}{C_1V_1}^2+\frac{1}{2}{C_2V_2}^2-\frac{1}{2}(C_1C_2)V^2$

$=\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}$