Consider an unbalanced phasor has shown in (a). The positive, negative, and zero sequence are shown in (b), (c), and (d) respectively.
Let's define the a-operator
a = e j 120 ° = − 1 2 + j 3 2 a 2 = e j 240 ° = − 1 2 − j 3 2 a=e^{j120\degree}=-\dfrac{1}{2}+j\dfrac{\sqrt{3}}{2}\\
a^2=e^{j240\degree}=-\dfrac{1}{2}-j\dfrac{\sqrt{3}}{2}\\ a = e j 120° = − 2 1 + j 2 3 a 2 = e j 240° = − 2 1 − j 2 3
from (b)
V b 1 = a 2 V a 1 V c 1 = a V a 1 V_{b1}=a^2V_{a1}\\
V_{c1}=aV_{a1}\\ V b 1 = a 2 V a 1 V c 1 = a V a 1
from (c)
V b 2 = a V a 2 V c 2 = a 2 V a 2 V_{b2}=aV_{a2}\\
V_{c2}=a^2V_{a2} V b 2 = a V a 2 V c 2 = a 2 V a 2
from (d)
V a 0 = V b 0 = V c 0 V_{a0}=V_{b0}=V_{c0} V a 0 = V b 0 = V c 0
V a = V a 0 + V a 1 + V a 2 V b = V b 0 + V b 1 + V b 2 V b = V a 0 + a 2 V a 1 + a V a 2 V c = V c 0 + V c 1 + V c 2 V c = V a 0 + a V a 1 + a 2 V a 2 V_{a}=V_{a0}+V_{a1}+V_{a2}\\
V_{b}=V_{b0}+V_{b1}+V_{b2}\\
V_{b}=V_{a0}+a^2V_{a1}+aV_{a2}\\
V_{c}=V_{c0}+V_{c1}+V_{c2}\\
V_{c}=V_{a0}+aV_{a1}+a^2V_{a2}\\ V a = V a 0 + V a 1 + V a 2 V b = V b 0 + V b 1 + V b 2 V b = V a 0 + a 2 V a 1 + a V a 2 V c = V c 0 + V c 1 + V c 2 V c = V a 0 + a V a 1 + a 2 V a 2
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