A 16 Ω wire is bend to form a square loop. A 9 V battery with internal resistance 1 Ω 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 x/2 μ J . where x =

Physics · Capacitors and R-C Circuits

JEE Main 2024 — 29 January, Shift 1 — Question 55

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 $\frac{x}{2} \mu \mathrm{~J}$ . where $\mathrm{x}=$ $\qquad$

Answer: 81

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}} \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4}$ $\mathrm{I}_{1}=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16}$ $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\mathrm{I}_{1} \times 8=\frac{9}{16} \times 8=\frac{9}{2} \mathbf{V}$ $\therefore \mathrm{U}=\frac{1}{2} \times 4 \times \frac{81}{4} \mu \mathrm{~J}$

$\therefore \mathrm{U}=\frac{81}{2} \mu \mathrm{~J}$

$\therefore \mathrm{x}=81$

Answer key and solution verified before publishing.

Practise Capacitors and R-C Circuits

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Charging and Discharging of R-C Circuits