When a current changes in a coil from 4A to 6 A in 0.05s and it induces an e.m.f of 8 V. The co-efficient of self-inductance is:
Correct answer: B. 0.2H
- A. 0.1H
- B. 0.2H
- C. 0.4H
- D. 0.8H
Explanation
The calculation is as follows: L = E/(∆I÷∆t) = 8/(2÷0.05) = 8/40 = 0.2H
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