Effective capacitance of parallel combination of two capacitors C1 and C2 is 10 μF. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor C2 is 4 times that of C1. If these capacitors are connected in series, their effective capacitance will be :

Capacitor PYQ — JEE Main 2020 · 8 Jan · Shift 1 · Q53 with Solution

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Correct answer: C

Parallel combination of $C_1$ and $C_2$ has effective capacitance: $C_1 + C_2 = 10\,\mu F$
Each capacitor is individually connected to a $1\,V$ source.
Energy stored in $C_2$ is 4 times that in $C_1$ .

For a capacitor connected to voltage $V$ : $U = \frac{1}{2}CV^2$

Since both are connected to the same voltage $V=1\,V$ , energy is directly proportional to capacitance.

So, $\frac{U_2}{U_1} = \frac{C_2}{C_1} = 4$

Thus, $C_2 = 4C_1$

$C_1 + C_2 = 10$ $C_1 + 4C_1 = 10$ $5C_1 = 10$ $C_1 = 2\,\mu F$

Then, $C_2 = 8\,\mu F$

For two capacitors in series: $C_s = \frac{C_1C_2}{C_1+C_2}$

Substitute values: $C_s = \frac{(2)(8)}{2+8}$ $C_s = \frac{16}{10} = 1.6\,\mu F$

Therefore, the correct answer is Option C.