35. What is the self-inductance of a coil which produces $5 V$ when the current changes from 3 ampere to 2 ampere in one millisecond? (1993)
Solution:
Induced emf is given by $e = L \left|\frac{\Delta I}{\Delta t}\right|$.
Here $e = 5 V$, $|\Delta I| = |2 - 3| = 1 A$, and $\Delta t = 1 ms = 10^{-3} s$.
$L = \frac{e}{|\Delta I / \Delta t|} = \frac{5}{1 / 10^{-3}} = 5 \times 10^{-3} H = 5 mili-henry$.
Leave a Reply