Strange Delta-Wye Transform

Before we begin, we normalize all currents, voltages and elements by the values given below: $\begin{aligned} \text{voltages:}&&&\SI{1}{V}, &&&\text{currents:}&&&\SI{1}{A}&&& \Rightarrow &&&&R,\: Z:&&&\SI{1}{\ohm} \end{aligned}$ The network equations remain the same in the process, but now all currents, voltages and elements represent their normalized dimensionless counterparts.

We use node analysis and loop analysis to directly get the equations for the delta- and wye-network, respectively. Alternatively, we could also use KCL and KVL, respectively, but it would take a bit longer: $\begin{aligned} \Delta: &&&& \vec{I}&:=\begin{pmatrix} I_A\\I_B \end{pmatrix}=\begin{pmatrix} Y_{AB}+Y_{AC} & -Y_{AB}\\ -Y_{AB} & Y_{AB}+Y_{BC} \end{pmatrix}\begin{pmatrix} V_{AC}\\V_{BC} \end{pmatrix}=\frac{1}{10}\begin{pmatrix} 2&-1\\ -1&2 \end{pmatrix}\begin{pmatrix} V_{AC}\\V_{BC} \end{pmatrix}=:Y_{\Delta}\vec{U}\\[1em] Y:&&&&\vec{U}&=\begin{pmatrix} V_{AC}\\V_{BC} \end{pmatrix}=\begin{pmatrix} Z_A+Z^*&Z^*\\ Z^*&Z+Z^* \end{pmatrix}\begin{pmatrix} I_A\\I_B \end{pmatrix}=\begin{pmatrix} 5+Z^*&Z^*\\ Z^*&Z+Z^* \end{pmatrix}\begin{pmatrix} I_A\\I_B \end{pmatrix}=:Z_{Y}\vec{I} \end{aligned}$ The first equation holds for arbitrary voltages $\vec{U}$ , while the second equation holds for arbitrary currents $\vec{I}$ . By definition, if we choose the symmetric 3-phase voltages $\begin{aligned} V_{AB}&=10, &&& V_{CA}&=10 w_3=-V_{AC},&&&V_{BC}&=10w_3^2=10w_3^*, &&&&&\left| w_3:=e^{j\frac{2\pi}{3}}=-\frac{1}{2}+j\frac{\sqrt{3}}{2} \right. \end{aligned}$ both networks should be equivalent, i.e. both equations should hold at the same time! We insert the first into the second to eliminate $\vec{I}$ and notice we get an eigenvalue equation: $\vec{U}=Z_Y\vec{I}=Z_YY_\Delta\vec{U}=\frac{1}{10}\begin{pmatrix} 10+Z^*&Z^*-5\\ Z^*-Z&2Z+Z^* \end{pmatrix}\vec{U}\qquad\left| \vec{U}\text{ is eigenvector of }Z_YY_\Delta\text{ with eigenvalue }\lambda=1 \right.$ We could start calculating eigenvalues now, but it is easier to just expand the first row with matrix-multiplication: $\begin{aligned} &&V_{AC}&=-10w_3=(10+Z^*)(-w_3) + (Z^*-5)(w_3^*) = -5(2w_3+w_3^*) +Z^*(w_3^*-w_3)\\[.5em] \underset{(..)^*}{\Rightarrow} && Z&=\frac{5w_3}{w_3-w_3^*}=\frac{5w_3}{j\sqrt{3}},\qquad |Z|=\frac{5|w_3|}{\sqrt{3}}=\frac{5}{\sqrt{3}}\approx\boxed{2.887} \end{aligned}$ Inserting $Z$ into the second row, we verify our solution does indeed satisfy the entire eigenvalue equation. We notice $Z$ is inductive, because its imaginary value is positive. Accordingly, $Z^*$ is capacitive, because its imaginary value is negative.