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When modelling an ideal transformer, why do we not consider the inductance of the primary and secondary windings of the transformer. Wouldn't this inductance affect the current pulled from the source?

(By ideal, I mean no leakage reactance, winding resistance, iron/copper losses, no magnetizing current.)

Why do we model the ideal transformer with no magnetic properties (i.e. inductance)?

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  • \$\begingroup\$ Look at your own definition of "ideal transformer". Where does inductance fit into that? \$\endgroup\$ – Dave Tweed Apr 26 '15 at 15:52
  • \$\begingroup\$ An ideal transformer, by definition is ideal and there are no losses and no magnetization current and 100% coupling. If you modelled any of those just mentioned it is NOT and ideal transformer. \$\endgroup\$ – Andy aka Apr 26 '15 at 16:02
  • \$\begingroup\$ @Andyaka Thank you for your reply. A transformer is nothing but wound wire around a core, and this wound wire will posses an inductance of its own. Even though there is 100% coupling (no leakage inductance) why do we not consider the inherent inductance of the wound coils of the transformer. \$\endgroup\$ – Sada93 Apr 26 '15 at 16:18
  • \$\begingroup\$ @Akash IDEAL means we don't; NON-IDEAL means we do. \$\endgroup\$ – Andy aka Apr 26 '15 at 17:38
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When modeling an ideal transformer, why do we not consider the inductance of the primary and secondary winding of the transformer.

For an ideal transformer, the primary and secondary inductances are arbitrarily large ('infinite'). This must be so since, for an ideal transformer, there is no frequency dependence.

To see this, consider the equations (in the phasor domain) for ideally coupled ideal inductors:

$$V_1 = j\omega L_1I_1 - j\omega M I_2$$

$$V_2 = j \omega M I_1 - j \omega L_2 I_2$$

where

$$M = \sqrt{L_1L_2}$$

Solving for \$V_2\$ yields

$$V_2 = \left(\sqrt{\frac{L_2}{L_1}}\right)V_1 = \frac{N_2}{N_1}V_1$$

Now, assume the primary is driven by a voltage source and that there is an impedance \$Z_2\$ connected to the secondary such that

$$V_2 = I_2 Z_2$$

It follows that

$$I_2 = \frac{j \omega M}{Z_2 + j \omega L_2}I_1 = \left(\sqrt{\frac{L_1}{L_2}}\cdot\frac{1}{1 + \frac{Z_2}{j\omega L_2}}\right)I_1 = \left(\frac{N_1}{N_2}\cdot\frac{1}{1 + \frac{Z_2}{j\omega L_2}}\right)I_1$$

This is certainly not the behaviour of an ideal transformer where we expect

$$I_2 = \frac{N_1}{N_2}I_1$$

But notice that in the case that \$j\omega L_2 \gg Z_2\$ we have

$$I_2 \approx \frac{N_1}{N_2}I_1$$

which is exact in the limit that \$\frac{Z_2}{j \omega L_2} \rightarrow 0\$

Thus, we recover the ideal transformer equations from the ideally coupled ideal inductors in the limit that \$L_1, L_2\$ go to infinity (keeping their ratio constant).

In summary, we don't consider the inductances for the ideal transformer since, as shown above, the ideal transformer equations hold only in the limit of arbitrarily large primary and secondary inductances.

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A a real transformer acts more like an ideal transformer when there is a substantial load on the secondary. The impedance looking into the primary looks pretty much like the impedance looking out from the secondary, scaled by the square of the turns ratio. The resistance effects in the primary and secondary are usually noticeable at substantial loads. T

Also, there is a (usually) slight difference due to the magnetizing inductance and the leakage inductance, core losses, as well as capacitances. Looking into the primary, the magnetizing inductance is effectively in parallel with a much lower impedance from the reflected impedance connected to the secondary, so the latter dominates.

Remove the load entirely, and all you have left are those slight effects, which are now 100% of what you can measure.

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  • \$\begingroup\$ I agree with the resistance looking at the primary, this is essentially why we can use transformers for impedance matching. I also agree with the magnetizing inductance being small hence the approximate model of a transformer. But why do we not consider the inductance of the actual transformer winding? \$\endgroup\$ – Sada93 Apr 26 '15 at 15:23
  • \$\begingroup\$ The magnetizing inductance is relatively large, so the current due to it is small in relation to the full load current, and it's reactive anyway, so no real power is involved, other than losses in winding resistance due to magnetizing current. \$\endgroup\$ – Spehro Pefhany Apr 26 '15 at 16:57

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