A) \[0.2\] H
B) \[0.4\] H
C) \[0.8\] H
D) \[0.1\] H
Correct Answer: D
Solution :
| [d] The induced emf, \[e=-L\frac{di}{dt}=-L\frac{(-2-2)}{0.05}\] |
| \[8=L\frac{(4)}{0.05}\] |
| \[\therefore \] \[L=\frac{8\times 0.05}{4}0.1\,H\] |
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