Potential, $E^{\prime}=E\left(1-e^{-t / R C}\right)$
$\therefore$ Charge, $Q=C E\left(1-e^{\frac{-t}{R C}}\right) \quad\left(\because Q=C E^{\prime}\right)$
As capacitor is charged to $\frac{E}{2}$.
$\Rightarrow Q =\frac{C E}{2}$
$\therefore \frac{C E}{2}=C E\left(1-e^{\frac{-t}{R C}}\right)$
$\Rightarrow \frac{1}{2}=e^{\frac{-t}{R C}} \text { or } t=\frac{R C}{\ln 2}$
Work done by battery,
Work done by battery,
$W=Q \times \Delta V=\frac{C E}{2} \times E=\frac{C E^{2}}{2}$
Heat dissipated $=\int \limits_{0}^{R C / ln 2} i R d t$
$=\frac{E^{2}}{R} \int \limits_{0}^{R C / \ln 2} e^{\frac{-2 t}{R C}} d t=\frac{3}{4}\left(\frac{C E^{2}}{2}\right)$
$\therefore$ Work done $\frac{\left(\frac{C E^{2}}{2}\right)}{\frac{3}{8} C E^{2}}$
$=\frac{4}{3}$ or $4: 3$
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