The current in a coil falls from 5.0 A to 0.0 A in 0.1 s. If average emf of 200 V is induced, the value of self inductance of coil is:
Answer & explanation
Correct answer: option 2
$ |\xi| = L \frac{\Delta I}{\Delta t}$
$ L = \xi \frac{\Delta t}{\Delta I} = 200\times \frac{0.1}{5} = 4H$