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(A) Consider the circuit where a capacitor of capacitance $C$ is charged by a battery of voltage $V$.When the capacitor is fully charged,the charge on

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$50$

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(A) Consider the circuit where a capacitor of capacitance $C$ is charged by a battery of voltage $V$.When the capacitor is fully charged,the charge on it is $Q = CV$.The energy stored in the capacitor is $E_{	ext{capacitor}} = \frac{1}{2} CV^2$.The total work done by the battery in supplying this charge is $W = QV = (CV)V = CV^2$.The energy lost as heat in the circuit is the difference between the work done by the battery and the energy stored in the capacitor:$E_{	ext{loss}} = W - E_{	ext{capacitor}} = CV^2 - \frac{1}{2} CV^2 = \frac{1}{2} CV^2$.Comparing this to the total energy supplied by the battery $(W = CV^2)$:$	ext{Percentage loss} = \frac{E_{	ext{loss}}}{W} 	imes 100\% = \frac{\frac{1}{2} CV^2}{CV^2} 	imes 100\% = 50\%$.Thus,$50\%$ of the energy supplied by the battery is lost.

An uncharged capacitor is connected to a battery. While charging the capacitor,how much is the energy lost,from the energy supplied by the battery (in $\%$)?

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