Question

A particular inductor is connected to a circuit where it experiences a change in current of 0.8 amps every 0.10 sec. If the inductor has a self-inductance of 2.0 V, what is the inductance

Answer 1

Answer:

0.4

Explanation:

Given that a particular inductor is connected to a circuit where it experiences a change in current of 0.8 amps every 0.10 sec. If the inductor has a self-inductance of 2.0 V, what is the inductance

Using the power formula

P = IV

Substitute all the parameters

P = 0.8 × 2

P = 1.6 W

But P = I^2 R

Substitute power and current

1.6 = 0.8^2 R

R = 1.6 / 0.64

R = 2.5 ohms

Inductance = reciprocal of resistance

Inductance = 1 / 2.5

Inductance = 0.4