Why is the generator voltage in phase with current?

Question

The figure shows a toy model for power transmission using ideal transfomers.

In this model, the phase angle between the voltage across the load and the current is zero.

Let's take the load voltage as the reference, i.e., \$ \tilde{V}_{L}=30V \angle 0°\$. The load is resistive, so the current is in phase with the load, i.e., \$ \tilde{I}_{L}=1A\angle 0°\$.

Therefore, the current in the transmission wires is going to be \$ \tilde{I}_{T}=0.5A\angle0°\$ and the voltage across the primary of the second step down transformer is \$\tilde{V}_{p2}=100V\angle0°\$.

Applying KVL in the loop between the two transformers, one concludes that \$\tilde{V}_{s1} =100V\angle0°+0.5A(20\Omega)\angle0°=110V\angle0°\$.

Therefore, \$\tilde{V}_{p1}=220V\angle0°\$ and \$\tilde{I}_{G}=0.25A\angle0°\$

How could this be the case? Assuming everything is ideal, shouldn't the magnetizing current, \$\tilde{I}_{G}\$ in this case, lag the voltage \$\tilde{V}_{p1}\$ by a \$\frac{\pi}{2}\$ angle?

Reference of the figure:

I made it myself

Answer 1

An ideal transformer, as the term 'ideal' is used here, has infinite magnetizing inductance and zero leakage inductance. As such, there is no phase shift between the primary and secondary, so if the load on the secondary is purely resistive, the load presented by the primary will also be purely resistive.

Even in the case of a non-ideal transformer, the current wouldn't lag by π/2. There would be an inductive term to it from finite leakage and magnetizing inductances, but (for reasonably good transformers) the impedance seen by the source would still be primarily real. After all, the source has to provide real power for the load to consume real power!