If the potential difference across a capacitor is changed from 12 V to 24 V, the work done is W. How much work will be done if the potential difference is changed from 24 V to 48 V?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 4 W
The work done in charging a capacitor equals the increase in its stored energy, \( U = \frac{1}{2}CV^{2} \). The capacitance stays the same.
\( \displaystyle W = \frac{1}{2}C\left(24^{2} - 12^{2}\right) = \frac{1}{2}C\,(576 - 144) = 216\,C \)
\( \displaystyle W' = \frac{1}{2}C\left(48^{2} - 24^{2}\right) = \frac{1}{2}C\,(2304 - 576) = 864\,C \)
\( \displaystyle \frac{W'}{W} = \frac{864}{216} = 4 \;\Rightarrow\; W' = 4W \)
Both voltages are doubled, and the energy goes as \( V^{2} \), so the work becomes \( 2^{2} = 4 \) times.