Question

An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction to the same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor.

Answer 1

Self-inductance is 8 mH

Explanation:

The induced emf is related to the change on current via the coefficient of self-inductance as-

e = $-L\frac {dl}{dt}$

now, as  

e = 0.4 V

dl = l = - 10 A - (10A) = - 20 A (Current)

dt = t = 0.4 s (Time)

0.4 = $-L \frac {(-20)}{0.4}$

hence coefficient of self inductance will be-

L = 0.008 H = 8 mH

Answer 2

Self induction of the inductor is 0.4 henry

Explanation:

Given,

V = 20 V

I_2 = 2.5 A , I_1 = -2.5 A

dI = $I_2 - I_1$

dI = 2.5 - (-2.5)

dI = 5 A

dt = 0.1 s

We know that ,

V =  $L\times \frac{dI}{dt}$

L = $\frac{dt \times V}{dI}$

L = $\frac{0.1\times 20}{5}$

L = 0.4 henry

Self induction of the inductor is 0.4 henry