1. Initial charges on the capacitors

For a capacitor, $Q = CV$

So for the two identical capacitors:

• Capacitor 1 at potential $V$ has charge $Q_1 = CV$
• Capacitor 2 at potential $2V$ has charge $Q_2 = C(2V)=2CV$

Since their negative plates are connected together and then the positive plates are joined, they are effectively connected in parallel with same polarity.

2. Initial total energy

Energy stored in a capacitor is $U = \frac{1}{2}CV^2$

Thus,

• Energy of capacitor 1: $U_1 = \frac{1}{2}CV^2$
• Energy of capacitor 2: $U_2 = \frac{1}{2}C(2V)^2 = 2CV^2$

Therefore, total initial energy is $U_i = \frac{1}{2}CV^2 + 2CV^2 = \frac{5}{2}CV^2$

3. Final common potential after connection

When connected in parallel with like terminals together, charge is conserved on the combined positive node.

Initial total positive charge: $Q_{\text{total}} = CV + 2CV = 3CV$

Equivalent capacitance after joining in parallel: $C_{\text{eq}} = C + C = 2C$

Hence final common potential is $V_f = \frac{Q_{\text{total}}}{C_{\text{eq}}} = \frac{3CV}{2C} = \frac{3V}{2}$

4. Final energy

$U_f = \frac{1}{2}(2C)\left(\frac{3V}{2}\right)^2$

$U_f = C \cdot \frac{9V^2}{4} = \frac{9}{4}CV^2$

5. Decrease in energy

$\Delta U = U_i - U_f$

$\Delta U = \frac{5}{2}CV^2 - \frac{9}{4}CV^2$

$\Delta U = \frac{10-9}{4}CV^2 = \frac{1}{4}CV^2$

6. Option check

The decrease in energy is $\boxed{\frac{1}{4}CV^2}$

So the correct option is A.