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 50% charged is

• 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.
Thus, energy from battery is $W=qE$
Energy stored in capacitor is $U=\frac{1}{2}qE$
Heat produced $H=$ energy lost$=W-U=qE-\frac{1}{2}qE=\frac{1}{2}qE$
as capacitor 50 % charged so $q=CE/2$.
Thus, $H=\frac{1}{2}\left(CE/2\right)E=\frac{C{E}^2}{4}$