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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 enter image description here

Which is, according to solutions, this : enter image description here So what the primary sees is this: enter image description here

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: enter image description here

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?

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