Question

Capacity of a capacitor is 48 µF. When it is charged from 0.1 C to 0.5 C, change in energy stored is

• A. 2500 J
• B. 2.5 x 10⁻³ J
• C. 2.5 × 10⁶ J
• D. 2.42 x 10⁻² J

Answer 1

Answer: (A)

2500 J

Answer 2

The energy stored in a capacitor is given by

$$U = \frac{q^2}{2C}$$

For a charge change from $q_1 = 0.1$ C to $q_2 = 0.5$ C, the change in energy is

$$\Delta U = \frac{q_2^2 - q_1^2}{2C}$$

Given $C = 48 \ \mu\text{F} = 48 \times 10^{-6} \text{F}$, we calculate:

$$q_2^2 = (0.5)^2 = 0.25, \quad q_1^2 = (0.1)^2 = 0.01$$ $$q_2^2 - q_1^2 = 0.25 - 0.01 = 0.24$$ $$2C = 2 \times 48 \times 10^{-6} = 96 \times 10^{-6} = 9.6 \times 10^{-5} \text{F}$$ $$\Delta U = \frac{0.24}{9.6 \times 10^{-5}} = 2500 \text{ J}$$

Change in energy is calculated using $\Delta U = \frac{q_2^2 - q_1^2}{2C}$. Substitute values and simplify to get 2500 J.