Starting from "The Effective Resistance of a Load Resistance attached to a Transformer" at page 8 of this document there is an example where it demonstrates impedance matching using an ideal transformer. There are sections as the paragraph below: enter image description here

The document says the winding resistance of the ideal transformer is zero. So far so good. I don't have any problem with that.

But then it writes the equation:

Vo = Ip*Ri + Vp

I have two problems/confusions with this equation:

1-) Vp is across the primary winding. The primary winding is a pure inductive reactance. So shouldn't correct the formula use phasors or simply be?:

Vo^2 = (Ip*Ri)^2 + Vp^2

(I mean why does the doemunet treat Vp and Ip*Ri as they are in phase?)

2-) This is a bit more confusing. According to the definition of ideal transformer it is written:

An ideal transformer is a theoretical linear transformer that is lossless and perfectly coupled. Perfect coupling implies infinitely high core magnetic permeability and winding inductances and zero net magnetomotive force (i.e. ipnp - isns = 0).

This means ideal transformer has infinite inductance. And if the inductance is infinite, the voltage across the winding at any non zero frequency should become infinite from s*L equation.

But then back to the document, Ip must be zero since inductive reactance of the primary would be infinite. But the document treats as if it not zero. Where am I thinking wrong here?

  • 1
    \$\begingroup\$ Why infinite inductance, why not zero? When a real transformer is loaded, voltage and current are almost in phase, in ideal transformer they are always in phase. \$\endgroup\$ Commented May 7, 2019 at 11:20
  • \$\begingroup\$ Isn't primary side like this electrical4u.com/electrical/wp-content/uploads/2013/05/… There must be 90 degree phase shift between the voltage across Ri and Vp. Also are u sure isnt the definition of ideal transformer involves infinite inductance? \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 11:36

3 Answers 3


For a normal power transformer, it will have an equivalent circuit like this: -

enter image description here

To make it ideal we turn the series components Lp, Rp, Rs and Ls into short circuits and we open circuit the components Rc and Lm.

This leaves us with a perfect ideal transformer and that will do two things: -

  1. Take only the power needed by the load from the driving (primary voltage)
  2. Reduce or increase the secondary voltage by a factor called the "turns ratio".

The impact of this is that the secondary load appears at the primary terminals multiplied by the turns ratio squared.

Picture taken from this site. That site might also help you understand how transformers work.

For a more practical transformer, we can choose to introduce Lm and that introduction causes a primary current in addition to the primary referred load current AND, that additional current through Lm will be 90 degree lagging. Maybe that is where you are getting confused. In an ideal transformer, Lm is assumed to be infinite.

  • \$\begingroup\$ Interesting can you take this detailed transformer model and reduce it to ideal transformer step by step? So it would be easier to see the difference between the ideal and non-ideal- That would be GREAT help! \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 12:03
  • \$\begingroup\$ Replace Lp, Rp, Ls and Rs with short circuits and remove Rc and Lm. It's that simple. \$\endgroup\$
    – Andy aka
    Commented May 7, 2019 at 12:04
  • \$\begingroup\$ I see from this model things make more sense. One last question regarding Lm. So if I want to mimic an ideal transformer in SPICE simulator. I can take Lp Rp Rs Ls as short Rc as open and coupled inductor inductances very high? Something like here ltwiki.org/LTspiceHelp/image/xformer.gif if we take L1 and L2 very high and the rest is short and Rc open, can we use this as an ideal transformer? So L1 and L2 represents Lm in your transformer circuit model correct? \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 12:29
  • \$\begingroup\$ Or have you tried to simulate yours i.sstatic.net/Ugdq5.png in spice? The ideal part is a bit tricky to model I guess. \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 12:31
  • \$\begingroup\$ The ideal part can be modeled as two inductors with a coupling coefficient of 1. The inductances can be set "unfeasibly" high so as to not cause a significant primary current to flow when the secondary is unloaded. You still need to set the correct ratio of inductances to properly represent the turns ratio i.e. if a step down ratio of 5:1 is required and you choose a primary inductance of (say) 100 henries, the secondary inductance will be \$(5:1)^2\$ lower at 4 henries. This is because on a transformer with perfect coupling, inductance is proportional to turns squared. \$\endgroup\$
    – Andy aka
    Commented May 7, 2019 at 13:09

The document assumes that you have a resistive load on the secondary side. For an ideal transformer, the combination of transformer plus resistor just looks like another resistor. so everything is "real" valued.

I think, the purpose of the exercise to calculate the value of this "transformed resistor" as seen on the primary side.

  • \$\begingroup\$ Just focus on the primary side. Transformer primary winding is nothing but a pure inductor. So the primary side is Ri in series with sL. So to me the voltage across Ri and Vp must have 90 degree phase shift. One cannot just add them algebraically. Or? \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 11:18
  • \$\begingroup\$ The ideal transformier is NOT a pure inductor. It's a transformed impedance on whatever is on the secondary side. You can't just look at the primary side alone. \$\endgroup\$
    – Hilmar
    Commented May 7, 2019 at 15:41

The Ri is the internal resistance of the signal generator, not the transformer winding. The transformer in the exercise is an ideal transformer, so Np/Ns=Vp/Vs and Np/Ns=Is/Ip, none inductivity and none resistance in the transformer!

  • \$\begingroup\$ I know what Ri is. But to me the equation should follow this electrical4u.com/electrical/wp-content/uploads/2013/05/… but how come they are in phase in the example? \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 11:37
  • \$\begingroup\$ Because it is an ideal transformer that doesn't exist, for the simplicity of your exercise. \$\endgroup\$ Commented May 7, 2019 at 11:39
  • \$\begingroup\$ But ideal means what in the model? Infinite inductance or zero inductance? \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 11:40
  • \$\begingroup\$ Which model are you referring at? The ideal transformer model is, again: Np/Ns=Vp/Vs and Np/Ns=Is/Ip. No inductances and no resistances. \$\endgroup\$ Commented May 7, 2019 at 11:46
  • \$\begingroup\$ I see because they draw the inductor shapes in ideal transformer I was not treating them as if they dont exist as an inductor. \$\endgroup\$
    – user1245
    Commented May 7, 2019 at 11:53

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