Question

A capacitor of capacitance $C$ has charge $Q$. It is joined in parallel across an identical uncharged capacitor through a resistor. The heat produced in the resistor is,

• A. $\frac{Q^2}{2C}$
• B. $\frac{Q^2}{4C}$
• C. $\frac{Q^2}{8C}$
• D. $\frac{Q^2}{C}$

Answer 1

The correct option is B $\frac{Q^2}{4C}$

When the capacitor $C$, having a charge of $Q$, is connected to an identical capacitor through a resistor, the charge $Q$ will distribute equally between the capacitors.

Initial energy stored in the capacitor is,

$U_i=\frac{Q^2}{2C}$

Final energy stored in both capacitors will be

$E_f=\frac{(Q/2{)}^2}{2C}+\frac{(Q/2{)}^2}{2C}=\frac{Q^2}{4C}$

So, energy lost will be,

$\Delta E=E_i-E_f=$ Initial energy - Final energy

$\Rightarrow \Delta E=\frac{Q^2}{2C}-\frac{Q^2}{4C}=\frac{Q^2}{4C}$

Hence, $\left(B\right)$ is the correct answer.

Key concept- When capacitors are connected in parallel, their potential becomes equal and charge redistribution might take place.