Question

An initially uncharged capacitor $C$ is being charged by a battery of emf $E$ through a resistance $R$ upto the instant when the capacitor is charged to the potential $E/2$, the ratio of the work done by the battery to the heat dissipated by the resistor is given by :-

• A. 2 : 1
• B. 3 : 1
• C. 4 : 3
• D. 4 : 1

Answer 1

Answer: (C)

$i=\frac{E}{R}{e}^{-t/RC},Q=CE\left(1-e^{-t/RC}\right)$
Capacitor is charged to $\frac{E}{2}$,
So $Q=\frac{CE}{2}$
$\therefore \frac{CE}{2}=CE\left(1-e^{-t/RC}\right)$
$\frac{1}{2}=e^{-t/RC}$
$t=RC\ln 2$
Work done by battery $=\left(Q_{\text{flown }}\right)(\Delta V)$
$=\left(\frac{CE}{2}\right)(E)=\frac{C{E}^2}{2}$
Heat dissipated $=\underset{0}{\overset{RC\ell n2}{\int }}{i}^{2}Rdt$
$=\frac{E^2}{R}\underset{0}{\overset{RC\ell n2}{\int }}{e}^{-2t/RC}\cdot dt$
$=\frac{3}{4}\left(\frac{C{E}^2}{2}\right)$
$\frac{\text{ Work done }}{\text{ Heat dissipated }}=\frac{C{E}^2/2}{\frac{3}{4}\left(\frac{C{E}^2}{2}\right)}=\frac{4}{3}$