The Y-Delta System

There is no neutral connection for the $Y-\Delta$ system of Fig. 1. Any variation in the impedance of a phase that produces an unbalanced system will simply vary the line and phase currents of the system.
Fig. 1: Y-connected generator with a $\Delta$-connected load
For a balanced load,
$$\mathbf{Z}_{1}=\mathbf{Z}_{2}=\mathbf{Z}_{3}$$
The voltage across each phase of the load is equal to the line voltage of the generator for a balanced or an unbalanced load:
$$\mathbf{V}_{\phi}=\mathbf{E}_{L}$$
The relationship between the line currents and phase currents of a balanced $ \Delta $ load can be found using an approach very similar to that used in Section $ 22.3 $ to find the relationship between the line voltages and phase voltages of a Y-connected generator. For this case, however, Kirchhoff's current law is employed instead of Kirchhoff's voltage law.The results obtained are
$$I_{L}=\sqrt{3} I_{\phi}$$
and the phase angle between a line current and the nearest phase current is $ 30^{\circ} $. A more detailed discussion of this relationship between the line and phase currents of a $ \Delta $-connected system can be found in Section 22.7.For a balanced load, the line currents will be equal in magnitude, as will the phase currents.
Example 1: For the three-phase system of Fig. 2
a. Find the phase angles $v_2$ and $v_3$.
b. Find the current in each phase of the load.
c. Find the magnitude of the line currents.
Fig. 2: Example 1.
Solution: a. For an $ A B C $ sequence,$$\theta_{2}=-120^{\circ} \text { and } \theta_{3}=+120^{\circ}$$b. $ \mathbf{V}_{\phi}=\mathbf{E}_{L} $. Therefore,$$\mathbf{V}_{a b}=\mathbf{E}_{A B} \quad \mathbf{V}_{c a}=\mathbf{E}_{C A} \quad \mathbf{V}_{b c}=\mathbf{E}_{B C}$$The phase currents are$$\begin{array}{l}\mathbf{I}_{a b}=\frac{\mathbf{V}_{a b}}{\mathbf{Z}_{a b}}=\frac{150 \mathrm{~V} \angle 0^{\circ}}{6 \Omega+j 8 \Omega}=\frac{150 \mathrm{~V} \angle 0^{\circ}}{10 \Omega \angle 53.13^{\circ}}=\mathbf{1 5} \mathbf{A} \angle-\mathbf{5 3 . 1 3}{ }^{\circ} \\\mathbf{I}_{b c}=\frac{\mathbf{V}_{b c}}{\mathbf{Z}_{b c}}=\frac{150 \mathrm{~V} \angle-120^{\circ}}{10 \Omega \angle 53.13^{\circ}}=\mathbf{1 5} \mathbf{A} \angle-173.13^{\circ} \\\mathbf{I}_{c a}=\frac{\mathbf{V}_{c a}}{\mathbf{Z}_{c a}}=\frac{150 \mathrm{~V} \angle+120^{\circ}}{10 \Omega \angle 53.13^{\circ}}=\mathbf{1 5} \mathbf{A} \angle \mathbf{6 6 . 8 7 ^ { \circ }}\end{array}$$c. $ I_{L}=\sqrt{3} I_{\phi}=(1.73)(15 \mathrm{~A})=25.95 \mathrm{~A} $. Therefore,$$I_{A a}=I_{B b}=I_{C c}=25.95 \mathrm{~A}$$