ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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 \(x=\) ________.

  1. A \(52\)
  2. B \(42\)
  3. C \(81\)
  4. D \(12\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(81\)

Step-by-step Solution

Detailed explanation

\( I=\frac{V}{R_{e q}} I=\frac{V}{R_{e q}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} \) \( I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} \) \( V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V \)…
From JEE Mains
Explore more questions on app