The mutual inductance between a primary and s | vedclass

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(A) The induced electromotive force $(e_2)$ in the secondary circuit is given by the formula: $e_2 = M \frac{di_1}{dt}$.Since the secondary circuit ha

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$4$

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(A) The induced electromotive force $(e_2)$ in the secondary circuit is given by the formula: $e_2 = M \frac{di_1}{dt}$.Since the secondary circuit has a resistance $R_2$,the induced current $i_2$ is given by $i_2 = \frac{e_2}{R_2}$,which implies $e_2 = i_2 R_2$.Equating the two expressions for $e_2$,we get: $i_2 R_2 = M \frac{di_1}{dt}$.Given values are $M = 0.5 \, H$,$R_2 = 5 \, \Omega$,and $i_2 = 0.4 \, A$.Substituting these values into the equation: $0.4 	imes 5 = 0.5 	imes \frac{di_1}{dt}$.$2.0 = 0.5 	imes \frac{di_1}{dt}$.$\frac{di_1}{dt} = \frac{2.0}{0.5} = 4 \, A/s$.

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