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Question 6.8: Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-230

Answer»

Given,
INITIAL current i_i = 5.0 A
Final current i_2 = 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 – 0.0A
di = 5 A

The self-conductance of the CIRCUIT is calculated USING the formula,
e=L\frac{di}{dt}\\L=\frac{e}{\frac{di}{dt}}
Now, L = (200V ×0.1)/5 =4H



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