A 100 μF 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:

0.01 H

The correct answer is Option (1) → 0.01 H

Given:

Capacitance $C = 100 \mu F = 100 \times 10^{-6} F$

Voltage $V = 50 \, V$

Current $I = 5 \, A$

Energy stored in capacitor:

$E_C = \frac{1}{2} C V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times (50)^2$

$E_C = \frac{1}{2} \times 100 \times 10^{-6} \times 2500$

$E_C = 0.125 \, J$

Energy stored in inductor:

$E_L = \frac{1}{2} L I^2$

Equating: $E_C = E_L$

$0.125 = \frac{1}{2} L (5^2)$

$0.125 = \frac{1}{2} L (25)$

$0.125 = 12.5 L$

$L = \frac{0.125}{12.5} = 0.01 \, H$

Final Answer: $L = 0.01 \, H$