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Consider the following : enter image description here

If we need to figure out the input impedance seen by the source, it will be equal to --> Zin = jw(30u) -jw(27u)+1k+jw(30u)-jw(27u). The first two terms are due to the drops across the first inductor, where the first is the drop due to its self-inductance and the second term is the drop due to the mutual inductance. The same is for the second equivalent inductor. My question is why the drop due to the mutual inductance is negative, how do I determine this voltage drop polarity due to the mutual inductance in general ??

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The drop due to the mutual inductance is negative because the magnetic field generated by the current flowing in the mutually coupled inductor is in the opposite direction. Pay careful attention to the direction the wire wraps around the core. Trace the path of the current around that wire and notice how it is travelling in a direction that will impart a magnetic field in the opposite direction in the core.

If the mutual inductance was equal to the self inductance (perfect coupling) then the two effects would cancel and you would have zero magnetic field. Zero magnetic field means zero inductance.

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I'm not sure the context of your formula. It's not quite accurate to me.

The differential Zin has 30uH cancelling because the windings are CW and CCW on the shared core so if balanced, they cancel. While the Mutual inductance raises both the source and load impedance relative to the common mode stray impedance yet to be defined. If one side of the signal is grounded, then if ωLm is >>1k they become more balanced line and return. Hence the name Balun aka CM choke can balance unbalanced grounded lines.

schematic

For Common Mode \$Z_{cm}(ω){_{IN}}=\dfrac{V_{cm(ω)}}{I_{cm(ω)}}=ω(L_1+M+L_2)\$

For Differential \$Z_{dm}(ω)_{{IN}} = ωL1-ωL2+R=R\$ where the windings on the toroid are perfectly even and the core mu is constant throughout the circumference.

What is neglected there is the stray common mode impedance for the CM return path.

When \$L1=L2=L,~~M = k L\$ for some constant, k the coupling factor for an ideal choke with no RC values.

Pls suggest improvements if I made a mistake.

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It's a common mode choke. Signal current I in the upper line nearly compensates the magnetic effect of I in the lower line. That's intentional.The purpose is to allow the signal to go from the source to the load as is. The attempt is not 100% succesful because the mutual inductance is 10% less than the inductances.

If both wires between the source and the choke happen to capture same noise signal from the air, that signal meets some reactance. If the magnetic material is lossy, that common mode signal can practically vanish in the choke.

Common mode noise signals are harmful because the load often is more complex than a resistor, it's a differential amp circuit which is internally connected to the ground. Common mode noise signal appears as a voltage between the signal line and the ground, both lines have same voltage. That can saturate the load because the voltage can be too high or it's out of the frequency range that the load can handle properly.

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