JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Capacitance

Effective capacitance of parallel combination of two capacitors

C_{1}

and

C_{2}

10 μ F

. When these capacitor 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)

Solution

The parallel combination gives

C_{1} + C_{2} = 10 μ F

, and since the energy stored is

U = \frac{1}{2} C V^{2}

at the same voltage, the condition that

C_{2}

stores four times as much as

C_{1}

means

4 \cdot \frac{1}{2} C_{1} V^{2} = \frac{1}{2} C_{2} V^{2}

, i.e.

C_{2} = 4 C_{1}

. Then

C_{1} + 4 C_{1} = 10

gives

C_{1} = 2 μ F

and

C_{2} = 8 μ F

. In series,

C_{e q} = \frac{C _{1} C _{2}}{C _{1} + C _{2}} = \frac{2 \times 8}{10} = 1.6 μ F

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option