# 3 phase transformers and equivalent circuits

There is an exercise in my textbook that goes like this:

3 single phase transformers are connected in wye-delta configuration. Each transformer has the following parameters : R1, X1, Rc = $\infty$, Xm = $\infty$, R2, X2. (R1, X1, R2, X2 being the copper resistance/reactance of the primary and secondary). The primary is connected with a 3-phase line with line-line voltage Vs. Secondary is connected to a balanced load composed of 3 impedances Z=R+jX connected in delta fashion.

So we're asked to draw the equivalent single phase circuit of

Which is, according to solutions, this : So what the primary sees is this:

Now, I understand why $$a' = \frac{V_{AN1}}{V_{AN2}} = \frac{N_1}{N_2} \sqrt{3} = a \sqrt{3}$$ and why how Z is divided by 3 to convert the impedance to the wye configuration.

What confuses me is: why are $R_2$ and $X_2$ divided by 3 on the secondary side? I would think those impedances are in series with the secondary voltage, thus I doubt we could apply $Z_Y = Z_\Delta / 3$ here.

Secondly, what would happen with $R_2$ and $X_2$ if we had this circuit made of 3 single phase transformers like this when we derive the equivalent single phase circuit:

Would $R_2$ and $X_2$ be multiplied by 3 or would it be $R_1$ and $X_1$ that would be divided by 3 in the equivalent single phase circuit?