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(D) When two inductors with self-inductances $L_1$ and $L_2$ are connected in series,the total induced electromotive force $(EMF)$ is the sum of the i

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$L_1 + L_2 + 2M$

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(D) When two inductors with self-inductances $L_1$ and $L_2$ are connected in series,the total induced electromotive force $(EMF)$ is the sum of the individual induced EMFs and the mutually induced EMFs.If the current flows through the coils such that the magnetic flux produced by one coil aids the flux produced by the other (as indicated by the direction of current in the figure),the mutual inductance contributes positively.The total effective inductance $L_{eq}$ is given by the formula:$L_{eq} = L_1 + L_2 + 2M$Thus,the correct option is $D$.

$A$ circuit contains two inductors of self-inductance $L_1$ and $L_2$ connected in series as shown in the figure. If $M$ is the mutual inductance between them,then the effective inductance of the circuit will be

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