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

The maximum energy stored in the inductor is 25 mJ, the peak current is 0.707 A, and the frequency of oscillation of the circuit is approximately 447 Hz. These values are derived from the conservation of energy in an LC circuit and the given values for the inductor and capacitor.

Step-by-step explanation:

When a 5000-pF capacitor is charged to 100 V and then connected to an 80-mH inductor, we have a classic example of an LC (inductor-capacitor) circuit. The answers to the student's questions involve several physics concepts, including energy conservation and oscillations in an LC circuit.

(a) Maximum Energy Stored in the Magnetic Field of the Inductor

The maximum energy stored in the magnetic field of the inductor is equal to the energy initially stored in the capacitor, since energy is conserved in an ideal LC circuit, where there is no resistance. The formula for the energy (E) stored in a capacitor is:

E = (1/2) C V2

Plugging in the values, E = (1/2) × 5000 x 10-12F × (100 V)2 yields a maximum energy of 25 mJ.

(b) Peak Value of the Current

The peak current (Imax) can be determined using the relationship between the maximum current and the maximum energy stored in the inductor:

Imax = √(2E/L)

Using the maximum energy calculated previously and the given inductance L, we find Imax to be 0.707 A.

(c) Frequency of Oscillation of the Circuit

The frequency of oscillation (f) for an LC circuit is given by:

f = 1 / (2π√(LC))

We calculate f using L = 80 x 10-3 H and C = 5000 x 10-12 F to get a frequency of approximately 447 Hz.

Answer 2

a)U= 25 x 10⁻⁶ J

b)I = 25 m A

c)f= 7957.74 Hz

Step-by-step 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 =(1)/(√(LC))`

`\omega =\frac{1}{\sqrt{80* 10^(-3)* 5000* 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=(1)/(2)LI^2`

`U=(1)/(2)* 80* 10^(-3)* (0.025)^2`

U= 25 x 10⁻⁶ J