simulate this circuit – Schematic created using CircuitLab

The attached diagram is the equivalent circuit of a practical transformer. Why is the coupling shown only between primary winding and secondary winding of ideal transformer?

Suppose secondary is open. So 'i_s' is zero.Hence 'i_p' should also be zero (ideal transformer). If both primary winding and secondary winding currents are zero, then as per the model,no mutual flux is produced by ideal primary and secondary windings, and hence no EMF should be induced in both. (Note that in the model, coupling is only between primary and secondary of ideal transformer.) But 'v_p' is in parallel with 'Lm' and hence 'v_p' will not be zero. Similarly we know that 'v_s' is also not zero under open circuit conditions.

To explain this behavior, shouldn't both primary and secondary windings of ideal transformer have magnetic coupling with magnetizing inductance'Lm'? (Since model should have the same behavior as the actual device.)

Please note that I understand 'Lm' is not a physical inductor but it is only for representing the behavior of actual transformer.

  • 1
    \$\begingroup\$ I'm struggling to see what your point is. An ideal transformer is an ideal transformer and you shouldn't attribute any magnetic properties to it at all. \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2014 at 13:22
  • \$\begingroup\$ oops..missed the whole point that ideal transformer itself is a model just to take care of the transformation ratio. ie,You cannot explain the operation of a perfect transformer(no leakage, no losses) using ideal transformer model.The moment u say about flux linkages, u have automatically included magnetising inductance into picture.Hence ideal transformer is just a box which tells that output volt is turns ratio times input voltage, without really telling about how this is done. \$\endgroup\$ Commented Mar 21, 2014 at 13:48
  • \$\begingroup\$ Do you want me to leave a formal answer? \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2014 at 13:58
  • \$\begingroup\$ Could you please clarify this doubt as well:By introducing a coupling between ideal transformer windings and magnetising winding,what changes does it cause in the model behavior?OR Is it like you cant even talk about any form of coupling with ideal transformer windings?Then what is the point of even assosiating a coupling between primary and secondary of ideal transformers,when we cant even explain the coupling in terms of flux linkages? \$\endgroup\$ Commented Mar 21, 2014 at 14:06
  • \$\begingroup\$ You have to forget thinking about doing anything within the ideal part of the model - this is untouchable - it is a device that allows you to model the components around it. Don't even start thinking that the ideal part has windings - it's a black box. \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2014 at 14:22

2 Answers 2


What you may be getting confused about is the "ideal transformer" in the equivalent circuit. You should not regard it as having any magnetic qualities at all. Try and see it like this: -

enter image description here

Whatever voltage you have on the input to the ideal transformer, \$V_P\$ is converted to \$V_S\$ on the output of this "theoretical" and perfect device. It converts power in to power out without loss or degradation such that: -

\$V_P\cdot I_P = V_S\cdot I_S\$

The ratio of \$V_P\$ to \$V_S\$ happens to be also called the turns ratio and almost quite literally it is on an unloaded transformer because there will be no volt-drop across R3 and L3 and only the tiniest of volt-drops on R1 and L1.

This means you now have a relatively easy way of constructing scenarios of load effects and recognizing the volt drops across the leakage components that are present.

The equivalent circuit of the transformer is really quite good once you accept that the ideal power converter is "untouchable" and should just be regarded as a black box. For instance you can measure both R1 and R3 and, by shorting the secondary you can get a pretty good idea what L3 and L1 are. With open circuit secondary you can measure the current into the primary and get a pretty good idea what \$L_M\$ is too.

  • \$\begingroup\$ thanks. Another point i forgot was ideal transformer can be considered to have infinite inductance and hence it can develop a voltage across it even if rate of change of flux linkage is almost zero.This can explain v_p and v_s as induced voltage even if you consider almost zero rate of change of flux linkage. \$\endgroup\$ Commented Mar 21, 2014 at 15:32
  • \$\begingroup\$ It should be regarded as having infinite input impedance (in either direction) with nothing connected to the other two terminals. Also it is an impedance transformer - put 1kohm on one side and for a 1:1 ratio 1kohm is on the other side. If the ratio is 10:1 the impedance magnifier is ratio-squared. \$\endgroup\$
    – Andy aka
    Commented Mar 21, 2014 at 15:51
  • \$\begingroup\$ I understand the black box concept and how its untouchable but if this black box has been defined to have infinite inductance (infinite impedance) then how can this black box draw any current at all? \$\endgroup\$
    – user56467
    Commented Oct 19, 2014 at 1:02
  • \$\begingroup\$ @xPTPCREWx the load is in parallel with the infinite magnetizing inductance - in my previous comment I did say "with nothing connected to the other two terminals" and in the diagram the load (via the "ideal power converter" is seen to be in parallel. \$\endgroup\$
    – Andy aka
    Commented Oct 19, 2014 at 9:38
  • \$\begingroup\$ Andy aka, thanks for the clarification. So can you explain what the purpose of the infinite magnetizing reactance is in the ideal transformer model? And why is it there or even mentioned to have an infinite inductance if it hides inside this black box of the "power converter"? It seems that it really acts as an open in terms of analysis, but like you said reflects the secondary load, so why have it? \$\endgroup\$ Commented Oct 20, 2014 at 8:08

As this is a model the magnetizing inductance could be represented by inductance on either side of the "ideal transformer", and the resulting behavior would be the same (assuming that you take the ratio into account). For that reason there is no need for a separate primary Lm and secondary Lm.


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