INTRODUCTION TO SYMMETRICAL COMPONENTS 1
INTRODUCTION TO SYMMETRICAL COMPONENTS
Edmund O. Schweitzer, III Ph.D.
Schweitzer Engineering Laboratories, Inc.
Pullman, WA USA
Stanley E. Zocholl
Schweitzer Engineering Laboratories, Inc.
Pullman, WA USA
S
YMMETRICAL
C
OMPONENTS

Figure 1 is the diagram of a balanced three-phase circuit. The impedance of the lines and loads
are the same in each phase, and the source voltages are equal in magnitude and are 120 degrees
apart. The balanced condition allows us to treat one phase as an independent single-phase
circuit. Once we have calculated the current in the A-phase, the balanced condition indicates that
the other phases have the same current magnitude, but are displaced from each other by 120
electrical degrees.
V
a1
I
a1
Z
a1
Z
load

Figure 1 Diagram of a Balanced Three-Phase Circuit
However, the single-phase calculation must take into account the voltage drop across the mutual
impedance caused by the other phase currents. Therefore Z
a1
is special impedance in which the
self-impedance of the line and the mutual impedance between lines are contained as lumped
elements for the special condition of balanced current. The special impedance is called the
positive-sequence impedance of the line and is derived as follows.
I
a
I
b
I
c
Z
m
Z
m
Z
m
Z
s
Z
s
Z
s

Figure 2 Self and Mutual Impedance in a Transposed Line
Figure 2 is the diagram of a transposed transmission line showing the self-impedance Z
s
of each
conductor and equal mutual impedance between each pair of conductors. We can write the
voltage drop in the A-phase as:

(
)
c
b
m
a
s
c
m
b
m
a
s
a
I
I
Z
I
Z
I
Z
I
Z
I
Z
V
+
+
=
+
+
=
(1) 2
For the special case of balanced three-phase current (I
b
+ I
c
) = -I
a
. Consequently:

a
m
s
a
I
)
Z
Z
(
V
-
=
(2)
Dividing both sides of equation (2) by I
a
shows that the positive-sequence impedance of the line
is equal to the self-impedance of the line minus the mutual impedance:

Z
V
I
Z
Z
a
a
a
s
m
1
=
=
-
(
) (3)
This is the impedance seen by a system of balanced three-phase currents, which are equal in
magnitude and displaced by 120 degrees.
By convention, phasors are said to rotate in a counter clockwise direction, and their sequence
refers to the order in which three-phase phasors appear passing a stationary point of observation.
Three systems of balanced phasors can be defined as shown in Figure 3. Each system has
phasors of equal magnitude, but each system has its own unique phase sequence.
I
c1
I
a1
I
b1
I
b2
I
a2
I
c2
I
a0
I
b0
I
c0
Positive-Seq.
Negative-Seq.
Zero-Seq.

Figure 3 Symmetrical Component Systems (A-Gnd Fault)
The first is called the positive-sequence system and has the sequence a-b-c generated by the
system voltage. The second is called the negative-sequence system and has the sequence a-c-b
not generated by the system voltage. The third is called the zero-sequence system because the
phasors have the same phase angle and rotate together. It is also a sequence not generated by the
system voltage. Collectively, they are the symmetrical components.
I
c1
I
a1
I
b2
I
b0
I
c0
I
b1
I
a0
I
a2
I
c2
I
a
= 3I
a0
I
b
= 0
I
c
= 0

Figure 4 Adding Symmetrical Components to Obtain Three-Phase Currents
They are significant because, although each is a balanced system, adding them phase-by-phase
produces unbalanced phase quantities. Consequently, by choosing the magnitude and reference
phase angle of each set, we can use them to represent any degree of phasor unbalance. For 3
example, the symmetrical components in Figure 3 result from a line-to-ground fault where there
is current in Phase A and zero current in B and C. Figure 4 shows the components added phase
by phase to reconstitute the phase currents. Since I
a1
, I
a2
, and I
a0
have the same magnitude and
phase angle, the A-phase current equals 3I
a0
. However, the components for B-phase and C-phase
are balanced phasors that add to zero and produce no current.
As a second example, Figure 5 shows the symmetrical components of the phase currents
resulting from a B-to-C line-to-line fault. The phase-by-phase addition to reconstitute the phase
currents is shown in Figure 6. With no zero-sequence current present, the I
a1
and I
a2
components
cancel to produce zero A-phase current. The positive- and negative- sequence B-phase and C-
phase components combine to produce equal and opposite B and C current of the line-to-line
fault.
I
a1
I
b1
I
c1
I
a2
I
c2
I
b2
Positive-Seq.
Negative-Seq.
No Zero-Seq.

Figure 5 Symmetrical Components Resulting for a Line-to-Line Fault
I
a1
I
b1
I
c1
I
a2
I
c2
I
b2
I
a
= 0
I
b
I
c
I
b
= 1.732|I
a1
|
I
c
= 1.732|I
a1
|

Figure 6 Adding the Symmetrical Components for a B-to-C Fault
We can now observe that three-phase currents are composed of symmetrical components that
exist simultaneously and independently in the network. Furthermore, each of the components
encounters a unique impedance, which incorporates the self, and mutual impedance in a manner
determined by its sequence. We have already seen the derivation of the positive-sequence
impedance. Because the relation (I
b
+ I
c
) = -I
a
, negative-sequence current encounters a negative-
sequence impedance equal to the positive-sequence impedance: 4

)
Z
Z
(
I
V
Z
m
s
2
a
2
a
2
a
-
=
=
(4)
However, since the zero-sequence currents I
a0
, I
b0
, and I
c0
are all in phase, the zero-sequence in-
corporates the self and mutual impedance as follows. The zero-sequence current flowing in the
transposed line shown in Figure 7 causes the voltage drop:

(
)
0
c
0
b
m
0
a
s
0
c
m
0
b
m
0
a
s
0
a
I
I
Z
I
Z
I
Z
I
Z
I
Z
V
+
+
=
+
+
=
(5)
For the special balanced case of zero-sequence current I
b0
+ I
c0
= 2I
a0
. Consequently, the voltage
drop can be written as:

0
a
m
s
0
a
I
)
Z
2
Z
(
V
+
=
(6)
Dividing both sides of Equation 6 by I
a0
gives the zero-sequence impedance:

)
Z
2
Z
(
I
V
Z
m
s
0
a
0
a
0
a
+
=
=
(7)
I
a0
I
b0
I
c0
Z
m
Z
m
Z
m
Z
s
Z
s
Z
s

Figure 7 Zero-Sequence Current in a Transposed Line
By superposition, we can consider the symmetrical components as flowing in three separate
networks called the positive-, negative-, and zero-sequence. Figure 8 shows the one-line diagram
of a generator G feeding a motor M through a two-section transmission line. The generator is
resistance grounded through resistor R
0
, while the motor remains ungrounded. Figure 9 shows
the positive-, negative-, and zero-sequence networks of the system.
G
M
L1
L2
R
0

Figure 8 System One-Line Diagram 5
VG
a1
VM
a1
ZG
1
ZL1
1
ZL2
1
ZL1
2
ZL2
2
ZG
2
ZM
1
ZM
2
3R
0
ZG
0
ZL1
0
ZL2
0
ZM
0
Positive-Seq.
Negative-Seq.
Zero-Seq.

Figure 9 Symmetrical Component Networks
The positive-sequence network contains the positive-sequence voltage of the generator, the
internal voltage of the motor, and the positive-sequence impedance of the generator, motor, and
lines. The negative- and positive-sequence networks contain only sequence impedance.
In Figure 9, there is no generated negative- or zero-sequence voltage and no connection between
the networks. Therefore, the diagram shows the normal balanced operating state of the system
where only positive-sequence current flows, and there is no negative- or zero-sequence current.
The presence of negative- and zero-sequence current indicates an unbalanced fault condition and
a connection between networks. We have already seen in Figure 3 that equal current in the A-
phase of each network characterizes the Phase A-to-ground fault. Connecting the networks in
series at the point of fault produces this condition. The location of the fault is shown in Figure
10, and the connection of the sequence networks is shown in Figure 11. Note that because the
motor is ungrounded, all the zero-sequence current flows in the grounding resistor. A resistance
of 3R
0
is used to produce the same voltage drop as the total ground current 3I
0
flowing in the
ground resistor.
G
M
L1
L2
R
0

Figure 10 A-Phase Fault Location 6
VG
a1
VM
a1
ZG
1
ZL1
1
ZL2
1
ZL1
2
ZL2
2
ZG
2
ZM
1
ZM
2
3R
0
ZG
0
ZL1
0
ZL2
0
ZM
0
Positive-Seq.
Negative-Seq.
Zero-Seq.
I
a0
I
a1
I
a2

Figure 11 Sequence Network Connection for Bus 2 A-to-Ground Fault
We have seen in Figure 5 that only positive- and negative-sequence current are present in a line-
to-line fault. Furthermore, we observe that a B-to-C fault is charact