A capacitor of capacitance C = 10 μF is charged by connecting it to a battery of emf 12 V. The capacitor is now disconnected and reconnected to the battery with the polarity reversed. The heat developed in the connecting wires will be

2.88 mJ

The correct answer is Option (1) → 2.88 mJ

Given:

Capacitance: $C = 10\ \mu\text{F} = 10 \times 10^{-6}\ \text{F}$

Battery voltage: $V = 12\ \text{V}$

Initially, energy stored: $U_1 = \frac{1}{2} C V^2 = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot 12^2 = 0.00072\ \text{J} = 0.72\ \text{mJ}$

When reconnected with reversed polarity, capacitor voltage changes from +12 V to −12 V. The total change in voltage = 24 V.

Heat developed: $H = \frac{1}{2} C (V_\text{final} - V_\text{initial})^2 = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot (24)^2 = 2.88 \times 10^{-3}\ \text{J} = 2.88\ \text{mJ}$

∴ Heat developed in the connecting wires = 2.88 mJ