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CUET · PHYSICS · PYQ PAPER 2025

A \( 100 \mu F \) capacitor is charged with a 50 V source supply. Then the source supply is removed, and the capacitor is connected across a pure inductor coil. As a result of which 5 A current flows through the inductor. The value of the self-inductance of the coil is:

  1. A 0.01 H
  2. B 0.02 H
  3. C 0.1 H
  4. D 0.17 H
Verified Solution

Answer & Solution

Correct Answer

(A) 0.01 H

Step-by-step Solution

Detailed explanation

\( \frac{1}{2} C V^2 = \frac{1}{2} L I^2 \) \( L = \frac{C V^2}{I^2} = \frac{(100 \times 10^{-6} F) (50 V)^2}{(5 A)^2} \) \( L = 0.01 H \)
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