Question

A capacitor of capacitance $C$ is charged by a battery of emf $E$ and internal resistance $r$. A resistance $2r$ is also connected in series with the capacitor. The amount of heat liberated inside the battery by the time capacitor is charged to $50\%$ of its steady state value is given as

• A. $\frac{3}{8}{E}^{2}C$
• B. $\frac{E^{2}C}{6}$
• C. $\frac{E^{2}C}{12}$
• D. $\frac{E^{2}C}{24}$

Answer 1

The correct option is D $\frac{E^{2}C}{24}$

During charging of a capacitor, $50\%$ of the energy supplied by the battery is lost and only $50\%$ is stored.
Total energy lost is given as
$U=\frac{1}{2}\frac{q^2}{C}=\frac{1}{2}\frac{{\left(\frac{EC}{2}\right)}^2}{C}=\frac{E^{2}C}{8}$
This energy is divided in the two resistances in same ratio of resistances so, heat dissipated inside the battery is given as
$H_{\text{battery}}=\frac{1}{8}{E}^{2}C\times \frac{r}{r+2r}=\frac{1}{24}{E}^{2}C$