Question

A 5000-pF capacitor is charged to 100 V and then quickly connected to an 80-mH inductor.

Determine

(a) the maximum energy stored in the magnetic field of the inductor,

(b) the peak value of the current, and

(c) the frequency of oscillation of the circuit.

Answer 1

Answer:

a)U= 25 x 10⁻⁶ J

b)I = 25 m A

c)f= 7957.74 Hz

Explanation:

Given that

C= 5000 pF

V= 100 V

L= 80 m H

The maximum charge on the capacitor Q = C V

Q= 5000 x 10⁻¹² x 100 C

Q= 5 x 10⁻⁷ C

The frequency ω  given as

$\omega =\dfrac{1}{\sqrt{LC}}$

$\omega =\dfrac{1}{\sqrt{80\times 10^{-3}\times 5000\times 10^{-12}}}$

ω = 50000 rad/s

The frequency in Hz

ω = 2π f

50000=  2π f

f= 7957.74 Hz

The maximum  current given as

I = Q ω

I=  5 x 10⁻⁷  x  50000

I = 25 m A

The maximum stored energy in the inductor U

$U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 80\times 10^{-3}\times (0.025)^2$

U= 25 x 10⁻⁶ J