
334 Appendix A: Symmetrical Components
These two indications lead us to the sequence network shown in Figure A.3. For this
circuit, therefore, the total driving voltage divided by the total impedance gives the
current. The total driving voltage is the sum of three sequence voltages (that is, V
˜
1
+
V
˜
2
+ V
˜
o
), and the total impedance is Z
˜
1
+ Z
˜
2 + 20
+ (3 Z
˜
f
+ 3 Z
˜
g
+ 3 Z
˜
n
).
∴ I
˜
1
= I
˜
2
= I
˜
o
= (V
˜
1
+ V
˜
2 +
V
˜
o
) ÷ (Z
˜
1
+ Z
˜
2
+ Z
˜
o
+ 3 Z
˜
f
+ 3 Z
˜
g
+ 3 Z
˜
n
) (A.10)
a
V
1
Z
1
Positive
sequence
network
V
2
Z
2
Negative
sequence
network
V
0
Z
0
Zero
sequence
network
V
a
3 Z
f
3 Z
n
3 Z
g
FIGURE A.3 Sequence networks connected in series for L-G fault (depen ...