A 100 µF capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor

Question

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.

Answer 1

Data: C = 100 µF = 100 × 10^-6 F = 10^-4 F,
V = 50V, i = 5A
The energy stored in the electric field in the capacitor
= $\frac{1}{2}$CV²
The energy stored in the magnetic field in the inductor = $\frac{1}{2}$Li²
Here, $\frac{1}{2}$CV² = $\frac{1}{2}$Li²
∴ 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 × 10²
= 10^-2H
This is the value of the inductance.