In the transformer equivalent circuit,why the primary and secondary windings are not coupled with magnetizing inductance?

Question

^{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.

Answer 1

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: -

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.

Answer 2

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.