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An inductor of 5H inductance carries a steady current of 2A. How can a 50V self induced emf be made to appear in the inductor?

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An Intiative by Sri Chaitanya

t = 0.5 sec

t = 0.4 sec

t = 0.6 sec

t = 0.2 sec

answer is A.

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Detailed Solution

$e = L \frac{d i}{d t}$ (or)

$d t = \frac{L d i}{e} = \frac{5 \times 2}{50} = \frac{1}{5} = 0.2 \sec$

So, the desired emf can be produced by reducing the given current to zero in 0.2second.

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