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A varying current in a coil changes from 10A to zero in 0.5 sec. If the average e.m.f induced in the coil is 220V, the self-inductance of the coil is
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- 5 H
- 6 H
- 11 H
- 12 H
Correct Option: C
Initial current (I1) = 10 A; Final current(I2)= 0; Time (t) = 0.5 sec and induced e.m.f. (ε) = 220 V. Induced e.m.f. (ε)
= - L | = - L | |||
dt | t |
= - L | = 20L | 0.5 |
or, L = | = 11 H | 20 |
[where L = Self inductance of coil]