Q146 / 155 · WBJEE 2019 · Physics
A capacitor of capacitance $C$ is connected in series with a resistance $R$ and $D C$ source of emf $E$ through a key. The capacitor starts charging when the key is closed. By the time the capacitor has been fully charged. what amount of energy is dissipated in the resistance $R ?$ 
- Correct answerA$\frac{1}{2} C E^{2}$
- B0
- C$\mathrm{CE}^{2}$
- D$\frac{E^{2}}{R}$
Solution
We know that, energy stored in the capacitor
$$
=\frac{1}{2} C E^{2}
$$
and energy supplied by the source of emf
$$
=C E^{2}
$$
$\therefore$ Energy dissipated in resistance $R$ = Energy supplied by the source of emf $E$ - Energy stored in the capacitor
$$
\begin{array}{l}
=C E^{2}-\frac{1}{2} C E^{2} \\
=\left(1-\frac{1}{2}\right) C E^{2}=\frac{1}{2} C E^{2}
\end{array}
$$
Hence, the correct answer is option: 'A'.