Q: An Inductor of 5 H inductance carries a steady current of 2 A. How can a 50 V self induced emf be made to appear in the inductor .

Sol: L = 5H ;|e| = 50V ; let us produce the required emf by reducing current to zero

$\large |e| = L \frac{\Delta I}{\Delta t} $

$\large \Delta t = \frac{L \Delta I}{|e|} = \frac{5 \times 2}{50} $

= 0.2 sec