Question

When current changes from 4 A to 0 A in 0.1 s in an inductor the induced emf is found to be 100 V, then find the value of self - inductance.

Answer 1

Change in current $|\Delta I|=4-0=4A$
Rate of change in current $\frac{|\Delta I|}{\Delta t}=\frac{4}{0.1}=40A/s$
Induced voltage $V=100$ volts
Self inductance $L=\frac{V}{\frac{|\Delta I|}{\Delta t}}=\frac{100}{40}=2.5H$