Suppose I have a threephase circuit with a mutual inductor on the first and second phase (V2 and V3). One way of solving the circuit (ie finding the phase currents) would be the following: convert the mutual inductor in the corresponding Y inductor and then solve the threephase circuit as you normally would do. Suppose you are given the following parameters for the mutual inductor:


simulate this circuit – Schematic created using CircuitLab

$$L11 = 10mH$$ $$L22 = 20mH$$ $$Lm = 12mH$$

where Lm is the mutual inductance. By appling the suggested transform method:


simulate this circuit

$$\begin{pmatrix} L1+L3 & L3\\ L3 & L2+L3 \end{pmatrix} = \begin{pmatrix} L11&Lm\\ Lm&L22 \end{pmatrix}$$

I obtain the following. (Note the negative inductance, which in turn gives a negative reactance, not ok in my opinion).


simulate this circuit

Now, by solving the circuit everything looks fine, the results are EXACTLY as the solution states (that would suggest that my steps are correct), however it does not seems ok to me that the inductance L1 is negative! Why is this happening, what am I missing?

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    \$\begingroup\$ Negative inductive reactance is capacitive reactance. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 26 '16 at 23:21
  • \$\begingroup\$ @IgnacioVazquez-Abrams I know that but: 1) how can it be since there are no capacitor in my circuit? 2) My perplexity is however mainly due because our teacher said this is not possible because of the convenctions of current and tension in the Y \$\endgroup\$ – mickkk Jan 26 '16 at 23:24
  • \$\begingroup\$ The mutual inductor you originally had has converted to a capacitor and two other inductors - try simulating it and see that it is the same. The act of conversion can produce odd-ball components but it doesn't mean they are incorrect. \$\endgroup\$ – Andy aka Jan 27 '16 at 9:00

A negative inductance would imply that the current is dependent on voltage varying with time but also negated. (i = -Ldv/dt instead of i = Ldv/dt ). No one can build a physical inductor that will automatically flip the voltage coming into it, but it makes for an easy analysis. A circuit representation is a way to model the physical world.

With models you can get results that are not physical, that model the system of interest just fine. An important part of electrical engineering is being able to model systems, but also realize the differences between the model and the real world. There are also no ideal circuit elements, there are no capacitors, inductors or resistors that have don't have parasitics.

For most things the parasitics don't matter (do you really care if your resistor has a few nanoHenries of inductance when your creating a voltage divider? No, but you will if your trying to run a GHz signal through it).

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  • \$\begingroup\$ When modelling a transformer in class we encountered the same problem and the professor confirmed what you have just stated. Model vs Reality, the model is just a representation of the physical phoenomena. \$\endgroup\$ – mickkk Nov 3 '16 at 13:09

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