A $16 \Omega$ wire is bend to form a square loop. A 9 V battery with internal resistance $1 \Omega$ is connected across one of its sides. If a $4 \mu \mathrm{~F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\frac{\mathrm{X}}{2} \mu \mathrm{~J}$, where $\mathrm{x}=$ $\_\_\_\_$.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Current Electricity

16Ω

wire is bend to form a square loop. A 9 V battery with internal resistance

1Ω

is connected across one of its sides. If a

4 μ F

capacitor is connected across one of its diagonals, the energy stored by the capacitor will be

\frac{X}{2} μ J

, where

x =

____

Answer: 81

16 Ω

wire bent into a square loop has

4 Ω

per side. A

9 V

battery of internal resistance

1 Ω

is connected across one side (nodes A, B): this

4 Ω

side is in parallel with the other three sides in series (

12 Ω

), both bridging A and B.

R_{e x t} = 4 ∥ 12 = \frac{4 \times 12}{16} = 3 Ω, I_{t o t a l} = \frac{9}{3 + 1} = 2.25 A

Current through the

12 Ω

path (A to D to C to B):

I^{'} = 2.25 \times \frac{4}{16} = 0.5625 A

.

Each

4 Ω

segment of this path drops

0.5625 \times 4 = 2.25 V

, so the potential difference across the diagonal (A to C, or equivalently B to D) is

2 \times 2.25 = 4.5 V

.

The capacitor, connected across this diagonal, stores:

U = \frac{1}{2} C V^{2} = \frac{1}{2} (4 \times 1 0^{- 6}) (4.5)^{2} = 40.5 μ J = \frac{81}{2} μ J

x = 81

.

Key confirmed: 81.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Integer answer