Electronic – Does mutual inductance change individual inductances

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If I have two inductors, with inductances \$L_1\$ and \$L_2\$. I put them really close to each other to make a transformer.

Are their individual inductances still \$L_1\$ and \$L_2\$?

Or has the mutual inductance changed their individual inductance?

Best Answer

It depends what is connected to the other winding, which is why "mutual inductance" is also called "coupling factor" (They are not identical, but closely related terms).

The classic way of characterising a transformer's performance (after establishing n, the turns ratio) is to first measure the inductance of the primary - with the secondary open circuit. This measurement is the "primary inductance" - effectively unaffected by the other winding since no current flows in it.

And the primary inductance is an impedance connected across the power source - effectively wasted power, and as it is a low impedance at low frequency it determines the low frequency performance of the transformer.

Then re-measure the primary, but with the secondary short circuited. This is the "leakage inductance" (technically it's the parallel combination of primary and leakage inductances, but the primary inductance is usually a large enough impedance that it can be regarded as infinite, and ignored). Anyway the "leakage inductance" is essentially the coupling factor of the transformer into a short circuit - so in a good transformer it will be a very low impedance.

(The same pair of measurements can be made on the secondary, with the primary open/short circuit. It should give you the same result, scaled by n^2).

So the leakage inductance doesn't change the winding inductances - it couples one winding to the other, allowing the load impedance (scaled by 1/n^2) to appear in parallel with the winding inductance.

And the series combination of source impedance and primary inductance determine the LF response, while the series combination of leakage inductance and (load impedance/n^2) determine the HF response.

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