Question

Capacity of a capacitor is $ 48\, \mu F $ . When it is charged from $ 0.1 \,C $ to $ 0.5\, C $ , change in the energy stored is

• A. $ 2500\, J $
• B. $ 2.5 \times 10^{-3}\,J $
• C. $ 2.5 \times 10^{6}\,J $
• D. $ 2.42 \times 10^{-2}\,J $

Answer 1

Answer: (A)

Change in energy $\Delta U=\frac{1}{2}\left[\frac{q_{1}^{2}-q_{2}^{2}}{C}\right]$
$=\frac{1}{2}\left[\frac{(0.5)^{2}-(0.1)^{2}}{48 \times 10^{-6}}\right] $
$=\frac{1}{2}\left[\frac{0.25-0.01}{48 \times 10^{-6}}\right] $
$=\frac{1}{2}\left[\frac{0.24}{48 \times 10^{-6}}\right] $
$=\frac{1}{2}\left[\frac{10^{4}}{2}\right] $
$=2500 \,J$