Mutual inductive coupling in a nodal admittance matrix

Question

I have this question that has been haunting for quite a while now. To compute the nodal admittance matrix \$Y_N\$ of an electrical network, one can make use of the incidence matrix and the branch (or edge) admittance matrix. However, in the case of a transformer or a winding who have inductively coupled turns, how is this taken into account in the nodal admittance matrix? I can't comprehend it, and if it's not taken into account how can I take it into account?

Adding schematic to ease comprehension: Take for instance this circuit:

Its graph is the following:

From which you can deduce the nodal admittance matrix: $$Y_N = A_f Y A_f^{T} (1)$$ with both \$A_f\$ and \$Y\$ being the incidence matrix and the branch matrix, respectively, defined as follows:

and,

Eventually, we have the following nodal admittance matrix:

The difficulty that I am facing is for mutually coupled inductors (those in the circuit). I do not know how the take the mutual Mij into account in my nodal admittance matrix (well first in my branch matrix then \$Y_N\$ will be deduced from equation(1))

Thanks in advance :)

Answer 1

In general I don't think you can. The method used in most circuit simulators is modified nodal analysis. Here a couple of extra equations are introduced where you solve for the currents through the inductors as well:

from this MNA lecture

In some situations you can eliminate the mutual inductance, for example if the two bottom legs of the transformer are connected there are transformations to circuits with three uncoupled inductors which allows you to use simple nodal analysis.

Answer 2

Start with the impedance matrix Z (including mutual impedances), invert it to an admittance matrix before calculating Yn as in (1) and then invert it again to get the reduced impedance matrix.