Show that, "Three unbalanced phasors of a 3- system can be resolved into three balanced system of phasors."
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=ej120°=−12+j32a2=ej240°=−12−j32a=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=ej120°=−21+j23a2=ej240°=−21−j23
from (b)
Vb1=a2Va1Vc1=aVa1V_{b1}=a^2V_{a1}\\ V_{c1}=aV_{a1}\\Vb1=a2Va1Vc1=aVa1
from (c)
Vb2=aVa2Vc2=a2Va2V_{b2}=aV_{a2}\\ V_{c2}=a^2V_{a2}Vb2=aVa2Vc2=a2Va2
from (d)
Va0=Vb0=Vc0V_{a0}=V_{b0}=V_{c0}Va0=Vb0=Vc0
Va=Va0+Va1+Va2Vb=Vb0+Vb1+Vb2Vb=Va0+a2Va1+aVa2Vc=Vc0+Vc1+Vc2Vc=Va0+aVa1+a2Va2V_{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}\\Va=Va0+Va1+Va2Vb=Vb0+Vb1+Vb2Vb=Va0+a2Va1+aVa2Vc=Vc0+Vc1+Vc2Vc=Va0+aVa1+a2Va2
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments