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.
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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
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