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

Here, `L = 5 H, I = 2 A, e = 50 v`
To induce this e.m.f. let the current be reduced to zero in time dt
As `e = L (dI)/(dt) :. 50 = (5 ( - 0))/(dt)`
`dt = (5 xx 2)/(50) = 0.2 sec`.