Two coils of self-inductance L_{1} and L_{2} | vedclass

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(D) When two coils are connected in series,the total induced electromotive force $(EMF)$ is the sum of the self-induced EMFs and the mutually induced

Vedclass · Accepted Answer

$L_{1} + L_{2} - 2M$

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(D) When two coils are connected in series,the total induced electromotive force $(EMF)$ is the sum of the self-induced EMFs and the mutually induced EMFs.The total $EMF$ is given by:$V = V_{L1} + V_{L2} + V_{M1} + V_{M2}$$V = L_{1} \frac{dI}{dt} + L_{2} \frac{dI}{dt} + M \frac{dI}{dt} + M \frac{dI}{dt}$Looking at the provided figure,the current $I$ enters the first coil and leaves it,then enters the second coil in a way that the magnetic flux produced by the current in the second coil opposes the flux produced by the first coil (or vice versa).Since the currents in the two coils are in opposite directions relative to the winding,the mutual inductance effect is subtractive.Therefore,the equivalent self-inductance $L_{eq}$ is:$L_{eq} = L_{1} + L_{2} - 2M$

Two coils of self-inductance $L_{1}$ and $L_{2}$ are connected in series. If the mutual inductance of the coils is $M$,the equivalent self-inductance of the combination shown in the figure will be:

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