Editing Inductance (section)

====T-circuit====
[[File:Mutual inductance equivalent circuit.svg|thumb|''T'' equivalent circuit of mutually coupled inductors]]
Mutually coupled inductors can equivalently be represented by a T-circuit of inductors as shown. If the coupling is strong and the inductors are of unequal values then the series inductor on the step-down side may take on a negative value.<ref>{{Cite book |last=Eslami |first=Mansour |url=https://archive.org/details/circuitanalysisf0000esla/mode/2up |title=Circuit Analysis Fundamentals |date=May 24, 2005 |publisher=Agile Press |isbn=0-9718239-5-2 |location=Chicago, IL, US |publication-date=May 24, 2005 |pages=194 |language=EN}}</ref>

This can be analyzed as a two port network. With the output terminated with some arbitrary impedance {{nowrap|<math>Z</math>,}} the voltage gain {{nowrap|<math>A_v</math>,}} is given by:

<big><math display=block> A_\mathrm{v}
  = \frac{s M Z}{\, s^2 L_1 L_2 - s^2 M^2 + s L_1 Z \,}
  = \frac{k}{\, s \left(1 - k^2\right) \frac{ \sqrt{L_1 L_2} }{Z} + \sqrt{\frac{L_1}{L_2}} \,}
</math></big>

where <math>k</math> is the coupling constant and <math>s</math> is the [[complex frequency]] variable, as above.
For tightly coupled inductors where <math>k = 1</math> this reduces to

<math display=block> A_\mathrm v = \sqrt {L_2 \over L_1} </math>

which is independent of the load impedance. If the inductors are wound on the same core and with the same geometry, then this expression is equal to the turns ratio of the two inductors because inductance is proportional to the square of turns ratio.

The input impedance of the network is given by:

<big><math display=block>Z_\text{in}
  = \frac {s^2 L_1 L_2 - s^2 M^2 + s L_1 Z}{sL_2 + Z}
  = \frac{L_1}{L_2}\, Z\, \left( \frac{ 1 }{ 1 + \frac{Z}{\, s L_2 \,} } \right) \left( 1  + \frac{1 - k^2}{ \frac{Z}{\, s L_2 \,} } \right)
</math></big>

For <math>k = 1</math> this reduces to

<math display=block> Z_\text{in} = \frac{s L_1 Z}{sL_2 + Z} = \frac{L_1}{L_2}\, Z\, \left( \frac{ 1 }{ 1 + \frac{Z}{\, s L_2 \,} } \right)</math>

Thus, current gain <math>A_i</math> is {{em|not}} independent of load unless the further condition

<math display=block>|sL_2| \gg |Z|</math>

is met, in which case,

<math display=block> Z_\text{in} \approx {L_1 \over L_2} Z </math>

and

<math display=block> A_\text{i} \approx \sqrt {L_1 \over L_2} = {1 \over A_\text{v}} </math>