Electrical Power Systems

Hence,
rom eqn. (i),
V b = V c
Unbalanced ault Analysis 267
...(i)
Ib = – Ic ...(ii)
Va = 0 ...(iii)
b 2 V a1 + bV a2 + V a0 = bV a1 + b 2 V a2 + V a0
\ V a1 = V a2 ...(iv)
rom eqn. (iii),
V a1 + V a2 + V a0 = 0 ...(v)
rom eqns. (iv) and (v), we get
rom eqn. (ii), we get
V a1 = V a2 = – V a0
b 2 I a1 + b I a2 + I a0 = – (b I a1 + b 2 I a2 +I ao )
2
...(vi)
\ I a1 + I a2 = 2 I a0 ...(vii)
Sequence network connection is shown in ig. 10.16.
ig. 10.16: Sequence networks of Example 10.6.
As V a1 = V a2 , the positive and negative sequence networks are connected in parallel. As
I a1 + I a2 = 2I a0, the zero sequence network is connected in series with the parallel combination
of positive and negative sequence networks.
Example 10.7: ig. 10.17 shows a three phase generator in which phases b and c are sort
circuited and connected through an impedance Z f to phase a. Draw the equivalent sequence
network.