A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductance, as a result of which 5A current flows through the inductance. Calculate the value of the inductance.

^{-6} F = 10^{-4} F,

V = 50V, i = 5A

The energy stored in the electric field in the capacitor

= \(\frac{1}{2}\)CV^{2}

The energy stored in the magnetic field in the inductor = \(\frac{1}{2}\)Li^{2}

Here, \(\frac{1}{2}\)CV^{2} = \(\frac{1}{2}\)Li^{2}

∴ L = C\(\frac{V^{2}}{i^{2}}\)

∴ L = C\(\left(\frac{V}{i}\right)^{2}=10^{-4}\left(\frac{50}{5}\right)^{2}\) = 10^{-4} × 10^{2}

= 10^{-2}H

This is the value of the inductance.