Two identical capacitors have same capacitance $$C$$. One of them is charged to the potential $$V$$ and other to the potential $$2V$$. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

Let the capacitance of each capacitor be $$C$$. Capacitor 1 is charged to potential $$V$$ and capacitor 2 is charged to potential $$2V$$.

Charge on capacitor 1: $$Q_1 = C \times V = CV$$.
Charge on capacitor 2: $$Q_2 = C \times 2V = 2CV$$.

Initial energy of capacitor 1: $$U_1 = \frac12 C V^2$$.
Initial energy of capacitor 2: $$U_2 = \frac12 C (2V)^2 = \frac12 C \times 4V^2 = 2CV^2$$.

Total initial energy, $$U_i = U_1 + U_2 = \frac12 C V^2 + 2CV^2 = \frac{1}{2}CV^2 + 2CV^2 = \frac{5}{2}CV^2$$. $$-(1)$$

After connecting the negative ends together and then the positive ends, the total charge on the combined positive terminal is:
$$Q_{\text{total}} = Q_1 + Q_2 = CV + 2CV = 3CV$$.

Equivalent capacitance of two identical capacitors in parallel is $$C_{\text{eq}} = C + C = 2C$$.

Final common potential $$V_f$$ is given by the charge‐capacitance relation:
$$V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{3CV}{2C} = \frac{3V}{2}$$. $$-(2)$$

Final energy of the system:
$$U_f = \frac12 C_{\text{eq}} V_f^2 = \frac12 \times 2C \times \Bigl(\frac{3V}{2}\Bigr)^2 = C \times \frac{9V^2}{4} = \frac{9}{4}CV^2$$.

Decrease in energy:
$$\Delta U = U_i - U_f = \frac{5}{2}CV^2 - \frac{9}{4}CV^2 = \frac{10}{4}CV^2 - \frac{9}{4}CV^2 = \frac{1}{4}CV^2$$.

Therefore, the decrease in energy of the combined system is $$\frac{1}{4}CV^2$$, which corresponds to Option A.