Question

A 500 pF capacitor is charged up so that it has 10μC of charge on its plates. The capacitor is then quicklyconnected to a 10 H inductor. Calculate themaximum energystored in the magnetic field of the inductoras the circuit oscillates. What is the current through the inductor when the maximum energy stored in theinductor is reached?

Answer 1

Answer:

Explanation:

Given that,

A capacitor of capacitance

C = 500pF

Charge on capacitor is

Q = 10μC

Capacitor is then connected to an inductor of inductance 10H

L = 10H

Since we want to calculate the maximum energy stored by the inductor, then, we will assume all the energy from the capacitor is transfer to the inductor

So energy stored in capacitor can be determined by using

U = ½CV²

Then, Q = CV

Therefore V = Q/C

U = ½ C • (Q/C)² = ½ C × Q²/C²

U = ½Q² / C

Then,

U = ½ × (10 × 10^-6)² / (500 × 10^-9)

U = 1 × 10^-4 J

U = 0.1 mJ

So the energy stored in this capacitor is transfers to the inductor.

So, energy stored in the inductor is 0.1mJ

B. Current through the inductor

Energy in the inductor is given as

U = ½Li²

1 × 10^-4 = ½ × 10 × i²

1 × 10^-4 = 5× i²

i² = 1 × 10^-4 / 5

i² = 2 × 10^-5

I = √(2×10^-5)

I = 4.47 × 10^-3 Amps

Then,

I = 4.47 mA