A 16 Ω wire is bent to form a square loop. A 9 V battery with internal resistance 1 Ω is connected across one of its sides. If a 4 capacitor is connected across one of its diagonals, the energy stored by the capacitor will be x/2, where x= _________ Numerical answer type. Work it out, then open the answer.

Answer: 81. Each side = 4 Ω. Across the battery: 4 Ω ∥ 12 Ω = 3 Ω, so I = 9/3+1 = 9/4 A. Voltage across that side = 9/4×3 = 27/4 V; current in the other three sides = (27/4)/(12) = 9/16 A. A diagonal spans two sides of this path: V = 2×4×9/16 = 4.5 V. U = 1/2CV² = 1/2(4)(4.5)² = 81/2, so x = 81.

A 16 Ω wire is bent to form a square loop. A 9 V battery with… | JEE Main 2024 (29 Jan, Shift 1)

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Capacitors in DC Circuits & RC

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

Numerical answer type. Work it out, then open the answer.

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Answer: 81

Each side $= 4\ \Omega$ . Across the battery: $4\ \Omega \parallel 12\ \Omega = 3\ \Omega$ , so $I = \dfrac{9}{3+1} = \dfrac{9}{4}\text{ A}$ .
Voltage across that side $= \dfrac{9}{4}\times3 = \dfrac{27}{4}\text{ V}$ ; current in the other three sides $= \dfrac{27/4}{12} = \dfrac{9}{16}\text{ A}$ .
A diagonal spans two sides of this path: $V = 2\times4\times\dfrac{9}{16} = 4.5\text{ V}$ .
$U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}(4)(4.5)^2 = \dfrac{81}{2}\ \mu\text{J}$ , so $x = 81$ .

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