A 5 $$\mu$$F capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 $$\mu$$F capacitor. If the energy change during the charge redistribution is $$\frac{X}{100}$$ J then value of X to the nearest integer is:

Initial energy stored in the first capacitor:

$$U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times (5 \times 10^{-6}) \times 220^2 = \frac{1}{2} \times 5 \times 10^{-6} \times 48400 = 0.121\text{ J}$$

Total initial charge on the first capacitor:

$$Q = C_1 V_1 = 5 \times 10^{-6} \times 220 = 1.1 \times 10^{-3}\text{ C}$$

Equivalent capacitance when connected in parallel loop redistribution:

$$C_{\text{eq}} = C_1 + C_2 = 5 + 2.5 = 7.5\text{ }\mu\text{F}$$

Final energy stored in the system:

$$U_f = \frac{Q^2}{2 C_{\text{eq}}} = \frac{(1.1 \times 10^{-3})^2}{2 \times 7.5 \times 10^{-6}} = \frac{1.21 \times 10^{-6}}{15 \times 10^{-6}} = \frac{1.21}{15} $$

Energy change during charge redistribution:

$$\Delta U = U_i - U_f = 0.121 - \frac{1.21}{15} = \frac{1.815 - 1.21}{15} = \frac{0.605}{15} = \frac{1.21}{30} \approx 0.040\text{ J}$$

$$\frac{X}{100} = 0.040 \implies X = 4.0$$