Chapter: AC Circuits· Physics

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

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.

Radhika Asked 1 Nov 2022 Student doubt Save WhatsApp Report

1 answer

Best answer

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.

Radhika · 1 Nov 2022 Report

Was this answer helpful?

Practise this chapter

AC Circuits

Take a 10-question test new questions each time · marked instantly · free

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

1 answer

Your answer