Discussion
A capacitor of capacitance C farads is charged from a battery of emf V volts. Out of the work done by the battery an amount \(\frac{1}{2}CV^{2}\) is stored in the capacitor and the rest is released as heat. The released heat is :
Answer: D
The heat released is \(CV^{2}/2\).
The battery moves a charge \(Q = CV\) through a constant emf V, so it does work \(W = QV = CV^{2}\). The energy stored in the capacitor is \(\frac{1}{2}CV^{2}\), so the rest is dissipated in the resistance of the circuit:
\[ H = CV^{2} - \frac{1}{2}CV^{2} = \frac{1}{2}CV^{2} \]
Remarkably, this half is lost whatever the resistance. The option \(CV^{2}\) is the battery's total work, not the heat, and zero would require charging with no loss, which is impossible from a fixed-voltage source.
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