Question

If

Vab=400VV_{ab} = 400 V

in a balanced Y-connected three-phase generator, find the phase voltages, assuming the phase sequence is: (a) abc (b) acb

Solution

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a) We have in abc phase sequence

Vab=3⋅Va∠30∘,\begin{align*} V_{ab}=\sqrt{3}\cdot V_{a}\angle30^{\circ}, \end{align*}

so

Va=231∠−30∘ V.\begin{align*} V_{a}=231\angle-30^{\circ}\,\text{V}.\end{align*}

Then we can write

Vb=Va∠−120∘\begin{align*} V_{b}=V_{a}\angle-120^{\circ} \end{align*}

Vc=Va∠120∘.\begin{align*} V_{c}=V_{a}\angle120^{\circ}. \end{align*}

Thus

Vb=231∠−150∘ V\begin{align*} V_{b}=231\angle-150^{\circ}\,\text{V} \end{align*}

Vc=231∠+90∘ V.\begin{align*} V_{c}=231\angle+90^{\circ}\,\text{V}. \end{align*}

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