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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

1. 5 H
2. 6 H
3. 11 H
4. 12 H

Correct answer: 11 H

Solution

The induced emf is given by Faraday's law: \( e = -L \frac{dI}{dt} \). Rearranging, \( L = \frac{e \cdot \Delta t}{\Delta I} \). Substituting \( e = 220 \text{ V}, \Delta t = 0.5 \text{ s}, \Delta I = 10 \text{ A} \), we get \( L = \frac{220 \cdot 0.5}{10} = 11 \text{ H} \).

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