A capacitor of capacitance \( C = 10\,\mu\mathrm{F} \) 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
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.88\,\mathrm{mJ} \)
Use conservation of energy: the work done by the battery = change in the energy stored in the capacitor + heat in the wires.
At first the plate connected to the positive terminal carries \( +CV \). After reversal the same plate must carry \( -CV \), so a charge \( 2CV \) passes through the battery.
\( \displaystyle q = CV = 10\times10^{-6}\,\mathrm{F} \times 12\,\mathrm{V} = 1.2\times10^{-4}\,\mathrm{C} \)
\( \displaystyle W_{\mathrm{battery}} = (2CV)\,V = 2CV^{2} \)
The stored energy is \( \frac{1}{2}CV^{2} \) both before and after (only the polarity changes), so \( \Delta U = 0 \) and all the work done by the battery becomes heat.
\( \displaystyle H = 2CV^{2} = 2 \times 10\times10^{-6} \times 12^{2} = 2.88\times10^{-3}\,\mathrm{J} = 2.88\,\mathrm{mJ} \)