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?

  • \$\begingroup\$ It will change the total inductance. If you place in phase inductance will increase. If you place opposite direction of magnetic direction mutual inductance will reduce the self inductance \$\endgroup\$
    – Photon001
    Commented Jan 9, 2016 at 16:18

3 Answers 3


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.

  • \$\begingroup\$ Great answer. The circuit that is confusing me has a capacitor in parallel with each winding to form two tank circuits. I have seen the argument that you can use the primary inductance of each respective winding to determine the resonance of each tank circuit. This seemed funny because it is being treated as an isolated inductor instead of a transformer. \$\endgroup\$ Commented Jan 9, 2016 at 17:25
  • \$\begingroup\$ That's a slightly different problem - and a classic - a double-tuned transformer, often used as an IF stage. All the above applies - but sometimes the coupling is relatively low (mutual inductance is high) so that you can, to a first approximation, treat each L-C as independent. (Or even adjustable to control the flatness of the passband) \$\endgroup\$
    – user16324
    Commented Jan 9, 2016 at 17:39

Let me shed some light on the physics background here.

Inductance of each coil can be calculated as \$ L = \frac{N^2}{R} \$.

The reluctance term is calculates as: \$ R = \frac{core-length}{\mu A_c} \$.

Therefore, the inductance is proportional to the permeability \$ \mu \$, which is a product of absolute permeability \$ \mu_o \$ and the relative permeability \$ \mu_r \$, which is 1 for air and ~thousands for magnetic materials.

Therefore, if you have two inductors with closed-loop magnetic path close to each other, the leakage fluxes will affect each other very little, well below 1%, since all the flux is contained in a low reluctance magnetic circuit .

However, if you have two air coils or somehow connect the magnetic circuits of the two individual coils, you will get much stronger mutual coupling since the magnetic flux is not constrained to the magnetic core.

Air cores inductors will have their individual inductances unchanged because the reluctance will not change.

Magnetic core inductors will see some change or reluctance (decrease) and therefore the individual inductances will decrease as well. This is similar to adding inductors in parallel but the coupling is not done electrically but magnetically.


Are their individual inductances still L1 and L2?

Think about measuring the inductance of L1 with L2 open circuit - you will measure exactly the same inductance for L1 if it were a million miles away from L2 because L2 has no current flowing. Sure it is now producing an open circuit voltage (just like a transformer would) but there are no amps in the secondary that can affect the magnetic field produced by L1 when measuring it.


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