A \( 100\,\mu\mathrm{F} \) 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:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.01\,\mathrm{H} \)
In an ideal LC circuit there is no loss, so by conservation of energy the energy stored in the capacitor, \( CV^{2}/2 \), is converted into the energy in the inductor, \( LI^{2}/2 \), at the instant when the current is maximum and the capacitor is discharged.
\( \displaystyle \frac{1}{2}\,C V^{2} = \frac{1}{2}\,L I^{2} \)
\( \displaystyle L = \frac{C V^{2}}{I^{2}} = \frac{100\times10^{-6} \times (50)^{2}}{(5)^{2}} = \frac{0.25}{25} = 0.01\,\mathrm{H} \)