Consider a current in a circuit falls from 6. | vedclass

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(B) Given: Initial current $I_1 = 6.0 \,A$,final current $I_2 = 1.0 \,A$,time interval $\Delta t = 0.2 \,s$,and average induced emf $e = 150 \,V$.The

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$6$

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(B) Given: Initial current $I_1 = 6.0 \,A$,final current $I_2 = 1.0 \,A$,time interval $\Delta t = 0.2 \,s$,and average induced emf $e = 150 \,V$.The formula for the average emf induced in an inductor is given by $e = L \frac{|\Delta I|}{\Delta t}$,where $\Delta I = I_1 - I_2$.Substituting the given values:$150 = L \frac{(6.0 - 1.0)}{0.2}$$150 = L \frac{5.0}{0.2}$$150 = L 	imes 25$$L = \frac{150}{25} = 6 \,H$.Therefore,the self-inductance of the circuit is $6 \,H$. The correct option is $B$.

Consider a current in a circuit falls from $6.0 \,A$ to $1.0 \,A$ in $0.2 \,s$. If an average emf of $150 \,V$ is induced by the circuit,then the self inductance of the circuit is (in $\,H$)

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