\[\left[\begin{array}{c} V_{a 0} \\ V_{a 1} \\ V_{a 2} \end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & a & a^{2} \\ 1 & a^{2} & a \end{array}\right]\left[\begin{array}{l} V_{a} \\ V_{b} \\ V_{c} \end{array}\right]\]
\[\boxed{a=e^{j 120^{0}}=-\frac{1}{2}+j \frac{\sqrt{3}}{2}}\]
\[\left[\begin{array}{c} \mathrm{V}_{\mathrm{a}} \\ \mathrm{V}_{\mathrm{b}} \\ \mathrm{V}_{\mathrm{c}} \end{array}\right]=\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & \mathrm{a}^{2} & \mathrm{a} \\ 1 & \mathrm{a} & \mathrm{a}^{2} \end{array}\right]\left[\begin{array}{l} \mathrm{V}_{\mathrm{a} 0} \\ \mathrm{~V}_{\mathrm{a} 1} \\ \mathrm{~V}_{\mathrm{a} 2} \end{array}\right]\]
The voltages across a 3-phase unbalanced load are \[V_a = 300 \angle 20^{\circ}~V, ~V_b = 360\angle 90^{\circ}~V, ~\text{and}~V_c = 500 \angle -140^{\circ}~V\]
Determine the symmetrical components of voltages
Phase sequence is abc
Symmetrical component voltages:
\[\left[\begin{array}{c} V_{a 0} \\ V_{a 1} \\ V_{a 2} \end{array}\right]=\frac{1}{3}\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & a & a^{2} \\ 1 & a^{2} & a \end{array}\right]\left[\begin{array}{l} V_{a} \\ V_{b} \\ V_{c} \end{array}\right]\]
On expansion: \[\begin{aligned} \mathrm{V}_{\mathrm{a} 0}&=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{V}_{\mathrm{b}}+\mathrm{V}_{\mathrm{c}}\right] \\ \mathrm{V}_{\mathrm{a} 1}&=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{aV}_{\mathrm{b}}+\mathrm{a}^{2} \mathrm{~V}_{\mathrm{c}}\right] \\ \mathrm{V}_{\mathrm{a} 2}&=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{a}^{2} \mathrm{~V}_{\mathrm{b}}+\mathrm{a} \mathrm{V}_{\mathrm{c}}\right] \end{aligned}\]
Given data:
\[\begin{aligned} \mathrm{V}_{\mathrm{a}} &=300 \angle 20^{\circ} \mathrm{V}=281.91+\mathrm{j} 102.61 \mathrm{~V} \\ \mathrm{~V}_{\mathrm{b}} &=360 \angle 90^{\circ} \mathrm{V}=0+\mathrm{j} 360 \mathrm{~V} \\ \mathrm{~V}_{\mathrm{c}} &=500 \angle-140^{\circ} \mathrm{V}=-383.02-\mathrm{j} 321.39 \mathrm{~V} \end{aligned}\]
Find values:
\[ \begin{aligned} a V_{b}&=1 \angle 120^{\circ} \times 360 \angle 90^{\circ}=360 \angle 210^{\circ}=-311.77-j 180 \mathrm{~V} \\ a^{2} V_{b}&=1 \angle 240^{\circ} \times 360 \angle 90^{\circ}=360 \angle 330^{\circ}=311.77-j 180 \mathrm{~V} \\ a V_{c}&=1 \angle 120^{\circ} \times 500 \angle-140^{\circ}=500 \angle-20^{\circ}=469.85-j 171.01 \mathrm{~V} \\ a^{2} \mathrm{~V}_{c}&=1 \angle 240^{\circ} \times 500 \angle-140^{\circ}=500 \angle 100^{\circ}=-86.82+\mathrm{j} 492.40 \mathrm{~V} \end{aligned} \]
Symmetrical voltages of phase-a: \[ \begin{aligned} \mathrm{V}_{\mathrm{a} 0} &=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{V}_{\mathrm{b}}+\mathrm{V}_{\mathrm{c}}\right]\\ &=\frac{1}{3}(281.91+\mathrm{j} 102.61+0+\mathrm{j} 360-383.02-\mathrm{j} 321.39) \\ &=\frac{1}{3}(-101.11+\mathrm{j} 141.22)=-33.70+\mathrm{j} 47.07=57.89 \angle 126^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{a} 1} &=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{a} \mathrm{V}_{\mathrm{b}}+\mathrm{a}^{2} \mathrm{~V}_{\mathrm{c}}\right]\\ &=\frac{1}{3}(281.91+\mathrm{j} 102.61-311.77-\mathrm{j} 180-86.82+\mathrm{j} 492.40) \\ &=\frac{1}{3}(-116.68+\mathrm{j} 415.01)=-38.89+\mathrm{j} 138.34=143.70 \angle 106^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{a} 2} &=\frac{1}{3}\left[\mathrm{~V}_{\mathrm{a}}+\mathrm{a}^{2} \mathrm{~V}_{\mathrm{b}}+\mathrm{a} \mathrm{V}_{\mathrm{c}}\right]\\ &=\frac{1}{3}(281.91+\mathrm{j} 102.61+311.77-\mathrm{j} 180+469.85-\mathrm{j} 171.01) \\ &=\frac{1}{3}(1063.53-\mathrm{j} 248.40)=354.51-\mathrm{j} 82.80=364.05 \angle-13^{\circ} \mathrm{V} \end{aligned} \]
We know that \(V_{a 0}=V_{b 0}=V_{c 0}\)
The zero sequence components are \[\begin{array}{l} \mathrm{V}_{\mathrm{a} 0}=57.89 \angle 126^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{b} 0}=57.89 \angle 126^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{co}}=57.89 \angle 126^{\circ} \mathrm{V} \end{array}\]
We know that, \(\mathrm{V}_{\mathrm{b} 1}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{al}} \quad ; \quad \mathrm{V}_{\mathrm{cl}}=\mathrm{a} \mathrm{V}_{\mathrm{al}}\)
The positive sequence components are \[\begin{array}{l} \mathrm{V}_{\mathrm{a} 1}=143.70 \angle 106^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{b} 1}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 1}=1 \angle 240^{\circ} \times 143.70 \angle 106^{\circ}=143.70 \angle 346^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{c} 1}=\mathrm{aV}_{\mathrm{al}}=1 \angle 120^{\circ} \times 143.70 \angle 106^{\circ}=143.70 \angle 226^{\circ} \mathrm{V} \end{array}\]
We know that \(, \mathrm{V}_{\mathrm{b} 2}=\mathrm{a} \mathrm{V}_{\mathrm{a}} ; \quad \mathrm{V}_{\mathrm{c} 2}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 2}\)
\(\therefore\) The negative sequence components are \[\begin{array}{l} \mathrm{V}_{a2}=364.05 \angle-13^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{b} 2}=\mathrm{aV}_{\mathrm{a} 2}=1 \angle 120^{\circ} \times 364.05 \angle-13^{\circ}=364.05 \angle 107^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{c} 2}=\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 2}=1 \angle 240^{\circ} \times 364.05 \angle-13^{\circ}=364.05 \angle 227^{\circ} \mathrm{V} \end{array}\]
The symmetrical components of phase-a voltage in a 3-phase unbalanced system are \[\mathrm{V}_{\mathrm{a} 0}=10 \angle 180^{\circ} \mathrm{V}, \mathrm{V}_{\mathrm{al}}=50 \angle 0^{\circ} \mathrm{V} \text { and } \mathrm{V}_{\mathrm{a} 2}=20 \angle 90^{\circ} \mathrm{V}\]
Determine the phase voltages \(V_a, V_b,~ \text{and}~V_c\)
The phase voltages of \(\mathrm{V}_{a}, \mathrm{V}_{\mathrm{b}}\) and \(\mathrm{V}_{\mathrm{c}}\) are given by
\[\left[\begin{array}{c} \mathrm{V}_{\mathrm{a}} \\ \mathrm{V}_{\mathrm{b}} \\ \mathrm{V}_{\mathrm{c}} \end{array}\right] = \left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & \mathrm{a}^{2} & \mathrm{a} \\ 1 & \mathrm{a} & \mathrm{a}^{2} \end{array}\right]\left[\begin{array}{l} \mathrm{V}_{\mathrm{a0}} \\ \mathrm{~V}_{\mathrm{a1}} \\ \mathrm{~V}_{\mathrm{a2}} \end{array}\right]\]
On expansion
\[\begin{array}{l} V_{a}=V_{a0}+V_{a1}+V_{a2} \\ V_{b}=V_{a0}+a^{2} V_{a1}+a V_{a2} \\ V_{c}=V_{a0}+a V_{a1}+a^{2} V_{a2} \end{array}\]
Given data
\[\begin{aligned} \mathrm{V}_{\mathrm{a} 0} &=10 \angle 180^{\circ} \mathrm{V}=-10+\mathrm{j} 0 \\ \mathrm{V}_{\mathrm{a} 1} &=50 \angle 0^{\circ} \mathrm{V}=50+\mathrm{j} 0 \\ \mathrm{V}_{22} &=20 \angle 90^{\circ} \mathrm{V}=0+\mathrm{j} 20 \end{aligned}\]
Find values
\[\begin{array}{l} \mathrm{aV}_{\mathrm{a} 1}=1 \angle 120^{\circ} \times 50 \angle 0^{\circ}=50 \angle 120^{\circ}=-25+\mathrm{j} 43.30 \\ \mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 1}=1 \angle 240^{\circ} \times 50 \angle 0^{\circ}=50 \angle 240^{\circ}=-25 .-\mathrm{j} 43.30 \\ \mathrm{aV}_{\mathrm{a} 2}=1 \angle 120^{\circ} \times 20 \angle 90^{\circ}=20 \angle 210^{\circ}=-17.32-\mathrm{j} 10 \\ \mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 2}=1 \angle 240^{\circ} \times 20 \angle 90^{\circ}=20 \angle 330^{\circ}=17.32-\mathrm{j} 10 \end{array}\]
Phase voltages are:
\[\begin{aligned} \mathrm{V}_{\mathrm{a}} &=\mathrm{V}_{\mathrm{a} 0}+\mathrm{V}_{\mathrm{a} 1}+\mathrm{V}_{\mathrm{a} 2}=40+\mathrm{j} 20=44.72 \angle 27^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{b}} &=\mathrm{V}_{\mathrm{a} 0}+\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 1}+\mathrm{a} \mathrm{V}_{\mathrm{a} 2}=-10-25-\mathrm{j} 43.30-17.32-\mathrm{j} 10 \\ &=-52.32-\mathrm{j} 53.30=74.69 \angle-134^{\circ} \mathrm{V} \\ \mathrm{V}_{\mathrm{c}} &=\mathrm{V}_{\mathrm{a} 0}+\mathrm{a} \mathrm{V}_{\mathrm{a} 1}+\mathrm{a}^{2} \mathrm{~V}_{\mathrm{a} 2}=-10-25+\mathrm{j} 43.30+17.32-\mathrm{j} 10 \\ &=-17.68+\mathrm{j} 33.3=37.70 \angle 118^{\circ} \mathrm{V} \end{aligned}\]