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
