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When the current in a coil charges from 2A to 4A in 0.05 s, emf of 8 volt is induced in the coil. The coefficient of self induction of the coil is -
A. 0.1 H
B. 0.2 H
C. 0.3 H
D. 0.4 H

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Best answer
Correct Answer - B
`e=L(dI)/(dt)`
`therefore L=(e)/(dI//dt)=(8)/(2//0.05)=0.2`H

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