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(C) When a capacitor of capacitance $C$ is charged through a resistance $R$ by a battery of $EMF$ $V$,the charge $q$ on the capacitor at any time $t$

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$C V^2$

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(C) When a capacitor of capacitance $C$ is charged through a resistance $R$ by a battery of $EMF$ $V$,the charge $q$ on the capacitor at any time $t$ is given by $q = C V (1 - e^{-t/RC})$.As $t 	o \infty$,the final charge on the capacitor becomes $q_{final} = C V$.The total charge supplied by the battery is $Q = C V$.The work done by the battery (or energy supplied by the battery) is given by $W = Q V = (C V) V = C V^2$.Note that the energy stored in the capacitor is $U = \frac{1}{2} C V^2$,which means half of the energy supplied by the battery is dissipated as heat in the resistor.

As shown in the figure,if a capacitor of capacitance $C$ is charged by connecting it with a resistance $R$ and a battery of $EMF$ $V$,then the total energy supplied by the battery is:

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