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

The correct option is A 2500 J

Energy stored in a capacitor of capacitance $C$ having a charge of $Q=\frac{Q^2}{2C}$

Thus the change in energy stored when the charge changes from $Q_1$ to $Q_2=\frac{Q_2^2}{2C}-\frac{Q_1^2}{2C}$

$=\frac{1}{2\times 48\times {10}^{-6}}\times ({0.5}^2-{0.1}^2)=2500J$