A 16 Ω wire is bend to form a square loop. A 9V battery with internal resistance 1 Ω is connected across one of its sides. If a 4 μF capacitor is connected across one of its diagonals, the energy stored by the capacitor will be x/2 μJ. where x = _________.
Physics · Capacitor · RC circuit
JEE Mains 2024 29 Jan Morning Shift

Question

A 16 Ω wire is bend to form a square loop. A 9V battery with internal resistance 1 Ω is connected across one of its sides. If a 4 μF capacitor is connected across one of its diagonals, the energy stored by the capacitor will be x/2 μJ. where x = _________.

Physics · Capacitor · RC circuit

JEE Mains 2024 29 Jan Morning Shift

G1 Physics · Accepted Answer

The 16 Ω wire forms a square, so each side has resistance: 16/4 = 4 Ω Between the battery terminals, one path is 4 Ω and the other path is 12 Ω. Their parallel equivalent is: R(p) = (4 × 12)/(4 + 12) = 3 Ω Including internal resistance 1 Ω: R(total) = 4 Ω Battery current: I = 9/4 A Current through the 12 Ω branch: I₁ = I × 4/(12 + 4) = (9/4) × (4/16) = 9/16 A Potential difference across the capacitor diagonal: V(A) - V(B) = I₁ × 8 = (9/16) × 8 = 9/2 V Energy stored: U = ½CV² = ½ × 4 μF × (9/2)²…