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:
Answer & explanation
Correct answer: option 3
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 = $Ei-Ef$
$=\frac{1}{2}{C_1V_1}^2+\frac{1}{2}{C_2V_2}^2-\frac{1}{2}(C_1C_2)V^2$
$=\frac{C_1 C_2\left(V_1-V_2\right)^2}{2\left(C_1+C_2\right)}$