JEE Mains · Physics · STD 12 - Electric potential and capacitance
- JEE Mains 2024
- Hard
- Numerical
A \(16\ \Omega\) wire is bend to form a square loop. A \(9 \mathrm{~V}\) 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 \(x=\) ________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(81\)
Step-by-step Solution
Detailed explanation
\( I=\frac{V}{R_{e q}} I=\frac{V}{R_{e q}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} \)
\( I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} \)
\( V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V \)
\( \therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu \mathrm{J} \)
\( \therefore U=\frac{81}{2} \mu \mathrm{J} \)
\( \therefore \mathrm{X}=81\)
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