An inductor of 5 H inductance carries a steady current of 2A. How can a 50 V self-induced emf 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)/(5)s= 0.2s`
So, the desired emf can be produced by reducing the given current to zero in 0.2 second.