The Delta Delta, Delta Y Three Phase Systems

Electrical Circuit Analysis > Polyphase Systems

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The basic equations necessary to analyze either of the two systems ($\Delta-\Delta, \, \Delta-Y$) have been presented at least once in this chapter. We will therefore proceed directly to two descriptive examples, one with a $\Delta-connected$ load and one with a $Y-connected$ load.

Example 1: For the $\Delta-\Delta$ system shown in Fig. 1:
a. Find the phase angles $\theta_2$ and$\theta_3$ for the specified phase sequence.
b. Find the current in each phase of the load.
c. Find the magnitude of the line currents.

Fig. 1: Example 1: $\Delta-\Delta$ system.

Solution: a. For an $ A C B $ phase 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

$$ \begin{split} \mathbf{I}_{a b}&=\frac{\mathbf{V}_{a b}}{\mathbf{Z}_{a b}}=\frac{120 \mathrm{~V} \angle 0^{\circ}}{\frac{\left(5 \Omega \angle 0^{\circ}\right)\left(5 \Omega \angle-90^{\circ}\right)}{5 \Omega-j 5 \Omega}}\\ &=\frac{120 \mathrm{~V} \angle 0^{\circ}}{\frac{25 \Omega \angle-90^{\circ}}{7.071 \angle-45^{\circ}}} \\ &=\frac{120 \mathrm{~V} \angle 0^{\circ}}{3.54 \Omega \angle-45^{\circ}}=\mathbf{3 3 . 9} \mathrm{A} \angle 45^{\circ}\\ \mathbf{I}_{b c}&=\frac{\mathbf{V}_{b c}}{\mathbf{Z}_{b c}}=\frac{120 \mathrm{~V} \angle 120^{\circ}}{3.54 \Omega \angle-45^{\circ}}=\mathbf{3 3 . 9} \mathbf{A} \angle 165^{\circ} \\ \mathbf{I}_{c a}&=\frac{\mathbf{V}_{c a}}{\mathbf{Z}_{c a}}=\frac{120 \mathrm{~V} \angle-120^{\circ}}{3.54 \Omega \angle-45^{\circ}}=\mathbf{3 3 . 9} \mathbf{A} \angle-75^{\circ} \end{split}$$

c. $$ I_{L}=\sqrt{3} I_{\phi}=(1.73)(34 \mathrm{~A})=58.82 \mathrm{~A} $$ Therefore, $$I_{A a}=I_{B b}=I_{C c}=\mathbf{5 8 . 8 2} \mathbf{A}$$

Example 2: For the $\Delta-Y$ system shown in Fig. 2:
a. Find the voltage across each phase of the load.
b. Find the magnitude of the line voltages.

Fig. 2: Example 2: $\Delta-Y$ system.

Solution: a. $ \mathbf{I}_{\phi L}=\mathbf{I}_{L} $. Therefore, $$\begin{aligned}\mathbf{I}_{a n} &=\mathbf{I}_{A a}=2 \mathrm{~A} \angle 0^{\circ} \\\mathbf{I}_{b n} &=\mathbf{I}_{B b}=2 \mathrm{~A} \angle-120^{\circ} \\\mathbf{I}_{c n} &=\mathbf{I}_{C c}=2 \mathrm{~A} \angle 120^{\circ}\end{aligned}$$ The phase voltages are

$$\begin{split} \mathbf{V}_{a n}&=\mathbf{I}_{a n} \mathbf{Z}_{a n}=\left(2 \mathrm{~A} \angle 0^{\circ}\right)\left(10 \Omega \angle-53.13^{\circ}\right)\\ &=\mathbf{2 0} \mathbf{V} \angle-\mathbf{5 3 . 1 3}{ }^{\circ} \\ \mathbf{V}_{b n}&=\mathbf{I}_{b n} \mathbf{Z}_{b n}=\left(2 \mathrm{~A} \angle-120^{\circ}\right)\left(10 \Omega \angle-53.13^{\circ}\right)\\ &=\mathbf{2 0} \mathbf{V} \angle-\mathbf{1 7 3 . 1 3 ^ { \circ }} \\ \mathbf{V}_{c n}&=\mathbf{I}_{c n} \mathbf{Z}_{c n}=\left(2 \mathrm{~A} \angle 120^{\circ}\right)\left(10 \Omega \angle-53.13^{\circ}\right)\\ &=\mathbf{2 0} \mathbf{V} \angle \mathbf{6 6 . 8 7} 7^{\circ} \end{split}$$

b. $$E_{L}=\sqrt{3} V_{\phi}=(1.73)(20 \mathrm{~V})=34.6 \mathrm{~V}$$ Therefore, $$E_{B A}=E_{C B}=E_{A C}=\mathbf{3 4 . 6} \mathbf{V}$$

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