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

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A 100 &micro;F 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.

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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.