Two capacitors of capacitances $$C$$ and $$2C$$ are charged to potential differences $$V$$ and $$2V$$, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:

We have two capacitors. The first has capacitance $$C$$ and was originally charged to the potential difference $$V$$, while the second has capacitance $$2C$$ and was originally charged to the potential difference $$2V$$.

For every capacitor the charge on its positive plate is related to the potential difference by the basic relation

$$Q = C \, \Delta V,$$

where $$\Delta V$$ is the potential of the positive plate minus that of the negative plate.

Hence the initial charges are

$$Q_1 = C \times V = CV,$$

$$Q_2 = (2C) \times (2V) = 4CV.$$

The plates therefore carry

positive plate of $$C$$ : $$+CV,$$    negative plate of $$C$$ : $$-CV,$$

positive plate of $$2C$$ : $$+4CV,$$    negative plate of $$2C$$ : $$-4CV.$$

Now the capacitors are connected in parallel but with opposite polarity: the positive plate of the first is joined to the negative plate of the second, and simultaneously the remaining two plates are joined together. Let us call the common point of the joined positive-negative plates node X, and the common point of the other two plates node Y.

The total charge that already resides on each node before any redistribution is obtained by simple addition:

$$Q_X^{\text{(initial)}} = (+CV) + (-4CV) = -3CV,$$

$$Q_Y^{\text{(initial)}} = (-CV) + (+4CV) = +3CV.$$

After the connection the charges are free to move inside each node until both plates in the same node come to a common potential. However, no charge can leave a node, so these total charges stay the same in the final state:

$$Q_X^{\text{(final)}} = -3CV, \qquad Q_Y^{\text{(final)}} = +3CV.$$

Let the final potentials of the two nodes be $$V_X$$ and $$V_Y$$, so that the final potential difference across each capacitor is

$$V_f = V_X - V_Y.$$

Using the basic formula again, the charges on the plates after equilibrium are

on node X from capacitor $$C$$ : $$Q_{1X}=C(V_X - V_Y),$$

on node X from capacitor $$2C$$ : the plate here is negative for this capacitor, so its charge is $$Q_{2X}=+2C(V_X - V_Y).$$

Therefore the total charge on node X in the final state is

$$Q_X^{\text{(final)}} = Q_{1X} + Q_{2X} = C(V_X - V_Y) + 2C(V_X - V_Y)= 3C(V_X - V_Y).$$

But this must equal the value already fixed for the node, namely $$-3CV$$. Equating,

$$3C(V_X - V_Y) = -3CV \quad\Longrightarrow\quad V_X - V_Y = -V.$$

The magnitude of the final potential difference is therefore

$$|V_f| = |V_X - V_Y| = V.$$

Thus each capacitor is finally at a potential difference of just $$V$$ in magnitude.

The energy stored in a capacitor is given by the well-known expression

$$U = \frac12 C (\Delta V)^2.$$

Applying this to the two capacitors after equilibrium:

$$U_1 = \frac12\,C\,(V)^2 = \frac12\,CV^2,$$

$$U_2 = \frac12\,(2C)\,(V)^2 = CV^2.$$

The total final energy is therefore

$$U_{\text{final}} = U_1 + U_2 = \frac12\,CV^2 + CV^2 = \frac32\,CV^2.$$

Hence, the correct answer is Option B.