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:
- A 0.01 H
- B 0.02 H
- C 0.1 H
- D 0.17 H
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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