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(B) When a capacitor of capacitance $C$ is charged by a battery of $EMF$ $V$,the total work done by the battery is $W = QV = (CV)V = CV^2$.However,the

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Half the energy supplied is stored in the capacitor

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(B) When a capacitor of capacitance $C$ is charged by a battery of $EMF$ $V$,the total work done by the battery is $W = QV = (CV)V = CV^2$.However,the energy stored in the capacitor is given by $U = \frac{1}{2}CV^2$.Comparing these,we see that the energy stored in the capacitor is exactly half of the total work done (energy supplied) by the battery.The remaining half of the energy is dissipated as heat in the connecting wires or internal resistance of the circuit.

An uncharged capacitor is connected to a battery. On charging the capacitor,

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