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)?
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.
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.
