A) 4H
B) 0.4H
C) 0.04H
D) 5H
Correct Answer: C
Solution :
If a current \[i\] is passed in the circuit and it is changed with a rate\[\frac{di}{dt}\], the induced emf\[e\]produced in the circuit is directly proportional to the rate of change of current. Thus, \[e\propto \frac{di}{dt}\] When the proportionality constant is removed, the constant L comes here. \[\therefore \] \[e=-L\frac{di}{dt}\] ... (i) The minus sign here is a reflection of Lenzs law. Here,\[di=(2-10)A=-8A\], \[dt=0.1\,\,s\] \[e=3.28\,\,V\]. \[\therefore \] \[3.28=\frac{L(-8)}{0.1}\] \[\therefore \] \[L=\frac{3.28\times 0.1}{8}=0.04\,\,H\]
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