Q 12-02-153JEE MainJEE Main 2020 (8 Jan, Shift 1)Easy
The effective capacitance of a parallel combination of two capacitors $C_1$ and $C_2$ is $10\ \mu\text{F}$. When these capacitors are individually connected to a voltage source of $1$ V, 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:
Answer: (C) $1.6\ \mu\text{F}$
At the same voltage $U \propto C$, so $C_2 = 4C_1$. With $C_1 + C_2 = 10\ \mu$F: $C_1 = 2\ \mu$F, $C_2 = 8\ \mu$F.
$$C_s = \frac{2\times8}{2 + 8} = 1.6\ \mu\text{F}$$
Solution by Sreeraj P, M.Sc Physics