Question

Current in a circuit falls from 5a to 1a in 0.1 second. If an average emf of 200 volts is induced, find the self - inductance of the coil.

Answer 1

Initial current I1= 5.0 A

Final current i2 = 0.0 A

Time,t = 0.1 seconds

Average e.m.f = 200 V

Change in the current can be calculated as ,di = 5.0A – 1.0A

di = 4 A

The self-conductance of the circuit is calculated using the formula,

e=L{di}/{dt}. e×dt=Ldi

L=e*dt/di

Now, L = (200V ×0.1)/4 =5H