An inductor of 5 H inductance carries a steady current of 2 A. How can a 50 V self induced e.m.f. be made to appear in the inductor ?

Explanation

`L = 5H, |e| = 50V`, Let us produce the required emf by reducing current to zero
Now, `|e| = L(dI)/(dt)` or `dt = (LdI)/(|e|) = (5 xx 2)/(50) s`
`(10)/(50)s = (1)/(8)s = 0.2 s`
So, the desireed emf can be produced by reducing the given current to zero in `0.2` second.