The capacitor C is initially without charge. | vedclass

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(A) Let the $EMF$ of the battery be $E$. When $X$ is joined to $Y$,the capacitor $C$ is charged to a potential $E$. The work done by the battery is $W

Vedclass · Accepted Answer

$H_1 = H_2$

Vedclass · Answer

(A) Let the $EMF$ of the battery be $E$. When $X$ is joined to $Y$,the capacitor $C$ is charged to a potential $E$. The work done by the battery is $W = qE = (CE)E = CE^2$. The energy stored in the capacitor is $U = \frac{1}{2}CE^2$. The heat produced in the resistance $R$ is $H_1 = W - U = CE^2 - \frac{1}{2}CE^2 = \frac{1}{2}CE^2$. When $X$ is joined to $Z$,the capacitor discharges completely through the resistance $R$. The heat produced in the resistance $R$ is $H_2 = U = \frac{1}{2}CE^2$. Comparing the two,we get $H_1 = H_2$.

The capacitor $C$ is initially without charge. $X$ is now joined to $Y$ for a long time,during which $H_1$ heat is produced in the resistance $R$. The $X-Y$ connection is removed and $X$ is now joined to $Z$ for a long time,during which heat $H_2$ is produced in $R$.

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