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