Question

An ac generator has a frequency of 5.0 kHz and a voltage of 45 V. When an inductor is connected between the terminals of this generator, the current in the inductor is 65 mA. What is the inductance of the inductor?

Answer 1

Answer:

0.022 H

Explanation:

From Alternating current,

V = (XL)I..................... Equation 1

Where V = Voltage of the generator, I = current in the inductor, XL = Inductive reactance of the generator.

Make XL the subject of the equation

XL = V/I................. Equation 2

Given: V = 45 V, I = 65 mA = 0.065 A

Substitute into equation 2

XL = 45/0.065

XL = 692.31 Ω

But,

XL = 2πFL........................ Equation 3

Where F = frequency of the generator, L = Inductance of the inductor

Make L the subject of the equation

L = XL/(2πF).................. Equation 4

Given: F = 5.0 kHz = 5000 Hz, π = 3.14, XL =  692.31 Ω

L = 692.31/(2×3.14×5000)

L = 0.022 H

Answer 2

Answer:

Explanation:

Given that,

Frequency f = 5 kHz = 5000Hz

Voltage V=45V

Current In inductor I = 65mA

I = 65 × 10^-3 = 0.065A

We want to find inductance L

We know that the reactive inductance cam be given as

XL = 2πFL

Where

XL is reactive inductance

F is frequency

L is inductance

Then,

L = XL/2πF

From ohms law

V = IR

We can calculate the receive reactance of the inductor

V = I•XL

Then, XL = V/I

XL = 45/0.065

XL = 692.31 ohms

Then,

L = XL/2πF

L = 692.31/(2π×5000)

L = 0.02204 H

Then, L = 22.04 mH

The inductance of the inductor is 22.04mH