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.