X and Y are two circuits having a coefficient | vedclass

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(B) Given: Mutual inductance $M = 3 	ext{ mH} = 3 	imes 10^{-3} 	ext{ H}$,Resistance of circuit $Y$ $R_2 = 4 	ext{ } \Omega$,Induced current in $Y

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$0.16$

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(B) Given: Mutual inductance $M = 3 	ext{ mH} = 3 	imes 10^{-3} 	ext{ H}$,Resistance of circuit $Y$ $R_2 = 4 	ext{ } \Omega$,Induced current in $Y$ $I_2 = 60 	imes 10^{-4} 	ext{ A}$,Time interval $\Delta t = 0.02 	ext{ s}$.The induced electromotive force $(e_2)$ in circuit $Y$ due to change in current in circuit $X$ is given by $e_2 = M \frac{\Delta I_1}{\Delta t}$.The induced current in circuit $Y$ is $I_2 = \frac{e_2}{R_2}$.Substituting the expression for $e_2$,we get $I_2 = \frac{M \cdot \Delta I_1}{R_2 \cdot \Delta t}$.Rearranging to solve for the change in current $\Delta I_1$ in circuit $X$:$\Delta I_1 = \frac{I_2 \cdot R_2 \cdot \Delta t}{M}$.Substituting the values:$\Delta I_1 = \frac{60 	imes 10^{-4} 	imes 4 	imes 0.02}{3 	imes 10^{-3}}$.$\Delta I_1 = \frac{60 	imes 10^{-4} 	imes 0.08}{3 	imes 10^{-3}} = \frac{4.8 	imes 10^{-4}}{3 	imes 10^{-3}} = 1.6 	imes 10^{-1} = 0.16 	ext{ A}$.

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