A wire is bent to form a square loop. A 9V battery with internal resistance is connected across one of its sides. If a capacitor is connected across one of its diagonals, the energy stored by the capacitor will be , where

Question

A $16 Ω$ wire is bent to form a square loop. A 9V 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 $x$ = ______. [2024]

Smart Achievers · Accepted Answer

(81)

At steady state, the current supplied from the battery,

$1 Ω$

$I_{1} = \frac{9}{4} \times \frac{4}{16} = \frac{9}{16} A$

$V_{A} - V_{B} = I_{1} \times 8 = \frac{9}{16} \times 8 = \frac{9}{2} V$

$∴$ $U = \frac{1}{2} \times 4 \times \frac{81}{4} μ J = \frac{81}{2} μ J$

$∴$ $x = 81$

Solution

Q.
A $16 Ω$ wire is bent to form a square loop. A 9V 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 $x$ = ______. [2024]

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Solution

Q. A 16Ω wire is bent to form a square loop. A 9V battery with internal resistance 1Ω is connected across one of its sides. If a 4μF capacitor is connected across one of its diagonals, the energy stored by the capacitor will be x/2 μJ, where x = ______. [2024]

Ans. (81) At steady state, the current supplied from the battery, I=VReq=91+12×412+4=94A I1=94×416=916A VA-VB=I1×8=916×8=92V ∴ U=12×4×814 μJ=812 μJ ∴ x=81

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Q.
A $16 Ω$ wire is bent to form a square loop. A 9V 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 $x$ = ______. [2024]