Two coils have self-inductance L_1 = 4 , mH a | vedclass

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(C) The induced electromotive force $(EMF)$ in a coil is given by $e = L \frac{di}{dt}$.The power supplied to the coil is $P = e \cdot I$,where $I$ is

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$\frac{1}{4}$

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(C) The induced electromotive force $(EMF)$ in a coil is given by $e = L \frac{di}{dt}$.The power supplied to the coil is $P = e \cdot I$,where $I$ is the current.Substituting the expression for $e$,we get $P = L \cdot I \cdot \frac{di}{dt}$.Given that the power $P$ and the rate of change of current $\frac{di}{dt}$ are the same for both coils,we can write:$P = L_1 I_1 \left( \frac{di}{dt} \right) = L_2 I_2 \left( \frac{di}{dt} \right)$.Since $\frac{di}{dt}$ is the same for both,we have $L_1 I_1 = L_2 I_2$.Therefore,the ratio of the currents is $\frac{I_1}{I_2} = \frac{L_2}{L_1}$.Substituting the given values $L_1 = 4 \, mH$ and $L_2 = 1 \, mH$:$\frac{I_1}{I_2} = \frac{1 \, mH}{4 \, mH} = \frac{1}{4}$.

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