A 142 H inductor carries a steady current of 0655 A When the

A 1.42 H inductor carries a steady current of 0.655 A. When the voltage source in the circuit is shorted out, the current disappears in 7.4 ms. What is the average induced emf in the inductor during this time? Answer in units of V.

Solution

Inductance L = 1.42 H

Steady current i = 0.655 A

Time taken to disappears the current, dt = 7.4 ms = 7.4 x10 ^-3 s

Change in current in this time interval di = 0.655A - 0 A = 0.655 A

the average induced emf in the inductor during this time E = L di/ dt

                                                                                   = 1.42 (0.655)/(7.4 x10 ^-3 )

                                                                                   = 125.7 volt