When current in a coil changes from 5,A to 2, | vedclass

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(D) According to Faraday's law of electromagnetic induction,the magnitude of induced emf is given by $e = L \left| \frac{di}{dt} \right|$.Given:Initia

Vedclass · Accepted Answer

$1.67$

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(D) According to Faraday's law of electromagnetic induction,the magnitude of induced emf is given by $e = L \left| \frac{di}{dt} ight|$.Given:Initial current $i_1 = 5\,A$Final current $i_2 = 2\,A$Change in current $di = i_1 - i_2 = 5\,A - 2\,A = 3\,A$Time interval $dt = 0.1\,s$Induced emf $e = 50\,V$Substituting the values into the formula:$50 = L \left( \frac{3}{0.1} ight)$$50 = L 	imes 30$$L = \frac{50}{30} = \frac{5}{3} \approx 1.67\,H$.Therefore,the self-inductance of the coil is $1.67\,H$.

When current in a coil changes from $5\,A$ to $2\,A$ in $0.1\,s$,an average voltage of $50\,V$ is produced. The self-inductance of the coil is $............H$.

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