Effective capacitance of parallel combination of two capacitors $$C_1$$ and $$C_2$$ is 10$$\mu$$F. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor $$C_2$$ is 4 times that of $$C_1$$. If these capacitors are connected in series, their effective capacitance will be:

We are told that two capacitors $$C_1$$ and $$C_2$$ are first connected in parallel and their equivalent (effective) capacitance is measured to be $$10\;\mu\text{F}$$.

For capacitors joined in parallel, the equivalent capacitance is simply the sum of the individual capacitances. Therefore we write the relation first:

$$C_{\text{parallel}} \;=\; C_1 + C_2$$

Substituting the given value $$C_{\text{parallel}} = 10\;\mu\text{F}$$, we have

$$C_1 + C_2 = 10\;\mu\text{F} \quad -(1)$$

Next, each capacitor is separately connected to a voltage source of $$1\;\text{V}$$. The energy stored in a capacitor is given by the basic formula

$$U = \dfrac{1}{2}\,C\,V^{2}$$

Applying this formula to the first capacitor, which has capacitance $$C_1$$ and voltage $$V = 1\;\text{V}$$, we get

$$U_1 = \dfrac{1}{2}\,C_1\,(1)^2 = \dfrac{1}{2}C_1$$

In the same way, for the second capacitor $$C_2$$ we obtain

$$U_2 = \dfrac{1}{2}\,C_2\,(1)^2 = \dfrac{1}{2}C_2$$

The problem states that the energy stored in $$C_2$$ is four times the energy stored in $$C_1$$. Hence we set up the proportionality condition

$$U_2 = 4\,U_1$$

Substituting the expressions for $$U_1$$ and $$U_2$$ found above, we write

$$\dfrac{1}{2}C_2 = 4 \times \left(\dfrac{1}{2}C_1\right)$$

The factor $$\dfrac{1}{2}$$ is common on both sides, allowing it to be cancelled, giving us the direct relation between the capacitances:

$$C_2 = 4C_1 \quad -(2)$$

Now we possess two linear equations: equation (1) $$C_1 + C_2 = 10$$ and equation (2) $$C_2 = 4C_1$$. We substitute equation (2) into equation (1) to solve for $$C_1$$:

$$C_1 + 4C_1 = 10$$

Simplifying the left-hand side, we find

$$5C_1 = 10$$

Dividing both sides by 5, we obtain

$$C_1 = 2\;\mu\text{F}$$

To find $$C_2$$ we return to equation (2):

$$C_2 = 4C_1 = 4 \times 2\;\mu\text{F} = 8\;\mu\text{F}$$

Having identified the individual capacitances $$C_1 = 2\;\mu\text{F}$$ and $$C_2 = 8\;\mu\text{F}$$, we now connect them in series and determine their new equivalent capacitance. For capacitors in series, the standard formula is

$$\dfrac{1}{C_{\text{series}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$$

Substituting the numerical values, we write

$$\dfrac{1}{C_{\text{series}}} = \dfrac{1}{2\;\mu\text{F}} + \dfrac{1}{8\;\mu\text{F}}$$

We express both fractions with a common denominator of $$8\;\mu\text{F}$$ to combine them easily:

$$\dfrac{1}{2\;\mu\text{F}} = \dfrac{4}{8\;\mu\text{F}}, \quad\text{and}\quad \dfrac{1}{8\;\mu\text{F}} = \dfrac{1}{8\;\mu\text{F}}$$

Adding these two gives

$$\dfrac{1}{C_{\text{series}}} = \dfrac{4}{8\;\mu\text{F}} + \dfrac{1}{8\;\mu\text{F}} = \dfrac{5}{8\;\mu\text{F}}$$

To solve for $$C_{\text{series}}$$ we now invert the fraction:

$$C_{\text{series}} = \dfrac{8\;\mu\text{F}}{5}$$

Carrying out the division, we find

$$C_{\text{series}} = 1.6\;\mu\text{F}$$

Hence, the correct answer is Option C.