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When the current changes from +2 A to −2 A in 0.05 second, an e.m.f. of 8 V is induced in a coil. The coefficient of self-induction of the coil is
- 0.2 H
- 0.4 H
- 0.8 H
- 0.1 H
Correct answer: 0.1 H
Solution
The coefficient of self-induction (L) can be calculated using the formula L = - (e.m.f. / (di/dt)). Here, the change in current (di) is 4 A (from +2 A to -2 A) and the time interval (dt) is 0.05 seconds, resulting in di/dt = 80 A/s. Substituting the values gives L = 8 V / 80 A/s = 0.1 H.
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