Q. 84.4( 20 Votes )

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.

Answer :

Given: Initial current I1 = 5.0 A

Final current I2 = 0.0 A


Time = 0.1 seconds


Average e.m.f = 200 V


Change in the current can be calculated as:


dI = I1 - I2 = 5.0A – 0.0A


dI = 5 A


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



or


substituting the values, we get



L = 4 Henry


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Electromagnetic InductionElectromagnetic InductionElectromagnetic Induction50 mins
Source of Electromagnetic waveSource of Electromagnetic waveSource of Electromagnetic wave6 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :