Editing Inductance (section)

====π-circuit====
[[File:Mutual inductance pi equivalent circuit.svg|thumb|''π'' equivalent circuit of coupled inductors]]
Alternatively, two coupled inductors can be modelled using a ''π'' equivalent circuit with optional ideal transformers at each port. While the circuit is more complicated than a T-circuit, it can be generalized<ref>{{Cite journal |doi = 10.1109/JSSC.2012.2204545|title = Simultaneous 6-Gb/s Data and 10-mW Power Transmission Using Nested Clover Coils for Noncontact Memory Card|journal = IEEE Journal of Solid-State Circuits|volume = 47|issue = 10|pages = 2484–2495|year = 2012|last1 = Radecki|first1 = Andrzej|last2 = Yuan|first2 = Yuxiang|last3 = Miura|first3 = Noriyuki|last4 = Aikawa|first4 = Iori|last5 = Take|first5 = Yasuhiro|last6 = Ishikuro|first6 = Hiroki|last7 = Kuroda|first7 = Tadahiro|bibcode = 2012IJSSC..47.2484R|s2cid = 29266328}}</ref> to circuits consisting of more than two coupled inductors. Equivalent circuit elements {{nowrap|<math>L_\text{s}</math>,}} <math>L_\text{p}</math> have physical meaning, modelling respectively [[magnetic reluctance]]s of coupling paths and [[magnetic reluctance]]s of [[leakage inductance|leakage paths]]. For example, electric currents flowing through these elements correspond to coupling and leakage [[magnetic flux]]es. Ideal transformers normalize all self-inductances to 1&nbsp;Henry to simplify mathematical formulas.

Equivalent circuit element values can be calculated from coupling coefficients with
<math display=block>\begin{align}
L_{S_{ij}} &= \frac{\det(\mathbf{K})}{-\mathbf{C}_{ij}} \\[3pt]
   L_{P_i} &= \frac{\det(\mathbf{K})}{\sum_{j=1}^N\mathbf{C}_{ij}}
\end{align}</math>

where coupling coefficient matrix and its cofactors are defined as
: <math>\mathbf{K} =
  \begin{bmatrix}
         1 & k_{12} & \cdots & k_{1N} \\
    k_{12} &      1 & \cdots & k_{2N} \\
    \vdots & \vdots & \ddots & \vdots \\
    k_{1N} & k_{2N} & \cdots &      1
  \end{bmatrix}\quad
</math> and <math>\quad \mathbf{C}_{ij} = (-1)^{i+j}\,\mathbf{M}_{ij}.</math>

For two coupled inductors, these formulas simplify to
: <math>L_{S_{12}} = \frac{-k_{12}^2 + 1}{k_{12}}\quad</math> and <math>\quad L_{P_1} = L_{P_2} \!=\! k_{12} + 1,</math>

and for three coupled inductors (for brevity shown only for <math>L_\text{s12}</math> and <math>L_\text{p1}</math>)

: <math>
  L_{S_{12}} = \frac{2\,k_{12}\,k_{13}\,k_{23} - k_{12}^2 - k_{13}^2 - k_{23}^2 + 1} {k_{13}\,k_{23} - k_{12}}
\quad</math> and <math>\quad
  L_{P_1} = \frac{2\,k_{12}\,k_{13}\,k_{23} - k_{12}^2 - k_{13}^2 - k_{23}^2 + 1} {k_{12}\,k_{23} + k_{13}\,k_{23} - k_{23}^2 - k_{12}-k_{13} + 1}.
</math>