The delta-wye transformations are helpful to transform delta-connected voltage sources, current sources and impedances to wye-connected voltage sources, current sources and impedances, and conversely. These transformations are useful when solving problems with three-phase systems.

Consider the balanced three-phase systems shown in Fig. 1 and Fig. 2. From the point of view of the rest of the network, the system shown in Fig. 1 is equivalent to the one shown in Fig. 2 if $$\begin{equation}V_{ab}(t) = \sqrt{3}\cdot V_{an}(\omega t + 30^\circ)\end{equation}$$ $$\begin{equation}V_{bc}(t) = \sqrt{3}\cdot V_{bn}(\omega t + 30^\circ)\end{equation}$$ $$\begin{equation}V_{ca}(t) = \sqrt{3}\cdot V_{cn}(\omega t + 30^\circ)\end{equation}$$

Conversely, the system shown in Fig. 2 is equivalent to the system shown in Fig. 1 if $$\begin{equation}V_{an}(t) = \frac{1}{\sqrt{3}} \cdot V_{ab}(\omega t - 30^\circ)\end{equation}$$ $$\begin{equation}V_{bn}(t) = \frac{1}{\sqrt{3}} \cdot V_{bc}(\omega t - 30^\circ)\end{equation}$$ $$\begin{equation}V_{cn}(t) = \frac{1}{\sqrt{3}} \cdot V_{ca}(\omega t - 30^\circ)\end{equation}$$

From the point of view of the rest of the network, the system shown in Fig. 3 is equivalent to the one shown in Fig. 4 if $$\begin{equation}I_{ab}(t) = \frac{1}{\sqrt{3}}\cdot I_{a}(\omega t + 30^\circ)\end{equation}$$ $$\begin{equation}I_{bc}(t) = \frac{1}{\sqrt{3}}\cdot I_{b}(\omega t + 30^\circ)\end{equation}$$ $$\begin{equation}I_{ca}(t) = \frac{1}{\sqrt{3}}\cdot I_{c}(\omega t + 30^\circ)\end{equation}$$

Conversely, the system shown in Fig. 4 is equivalent to the system shown in Fig. 3 if $$\begin{equation}I_{a}(t) = \sqrt{3} \cdot I_{ab}(\omega t - 30^\circ)\end{equation}$$ $$\begin{equation}I_{b}(t) = \sqrt{3} \cdot I_{bc}(\omega t - 30^\circ)\end{equation}$$ $$\begin{equation}I_{c}(t) = \sqrt{3} \cdot I_{ca}(\omega t - 30^\circ)\end{equation}$$

Even more generally, if the current sources are replaced by any other components (e.g. impedances) and if the currents going through these components satisfy the above equations, than the two diagrams are equivalent from the standpoint of the rest of the network.

Finally, let us consider the networks shown in Fig. 5 and Fig. 6. From the point of view of the rest of the network, the system shown in Fig. 5 is equivalent to the one shown in Fig. 6 if $$\begin{equation}Z_{ab} = \frac{Z_a Z_b+Z_b Z_c+Z_c Z_a}{Z_c}\end{equation}$$ $$\begin{equation}Z_{bc} = \frac{Z_a Z_b+Z_b Z_c+Z_c Z_a}{Z_a}\end{equation}$$ $$\begin{equation}Z_{ca} = \frac{Z_a Z_b+Z_b Z_c+Z_c Z_a}{Z_b}\end{equation}$$

Conversely, the system shown in Fig. 6 is equivalent to the system shown in Fig. 5 if $$\begin{equation}Z_a = \frac{Z_{ab} Z_{ca}}{Z_{ab}+Z_{bc}+Z_{ca}}\end{equation}$$ $$\begin{equation}Z_b = \frac{Z_{bc} Z_{ab}}{Z_{ab}+Z_{bc}+Z_{ca}}\end{equation}$$ $$\begin{equation}Z_c = \frac{Z_{ca} Z_{bc}}{Z_{ab}+Z_{bc}+Z_{ca}}\end{equation}$$

If the impedance networks are balanced than $Z_{ab}=Z_{bc}=Z_{ca}=Z_{\Delta}$ and $Z_{a}=Z_{b}=Z_{c}=Z_{Y}$ and the previous equations become $$\begin{equation}Z_{\Delta} = \frac{Z_{Y}}{3}\end{equation}$$ $$\begin{equation}Z_{Y} = 3 Z_{\Delta}\end{equation}$$

• Problems that can be solved easier using delta-wye tranformations
  Three-phase (D-Y) system with given source voltages, line and load impedances (1 question)
  Three-phase (D-D) system with given source voltages, line and load impedances (1 question)