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The current in self inductance L = 40 mH is to be increased uniformly from 1 amp to 11 amp in 4 milliseconds. The e.m.f. induced in the induct or during the process is
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- 100 volt
- 0.4 volt
- 4.0 volt
- 440 volt
Correct Option: A
| e = L | dt |
Given that L = 40 × 10–3 H,
di = 11 A – 1 A = 10 A
and dt = 4 × 10–3 s
| ∴ e = 40 × 10–3 × |  |  | = 100 V | 4 × 10–3 |
