Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

V
V
RN
YN
=
∠ − ° +
∠ − ° +
∠ − ° =
∠ −
°
=
∠
200
10
100
50
25
30
308 8
23 64
200
V rms
.
.
−−
° +
∠ ° +
∠ − ° =
∠ −
°
=
∠
° +
130
100 70
25
30
109 4
131 7
200 110
10
V rms
.
.
V
BN
00
170
25
30
214 66 132 6
∠ −
° +
∠ − ° =
∠
°
V rms
.
.
The phase voltage set is an unbalanced one and we have expressed it as a sum of three components –
one balanced positive sequence three-phase voltage, one balanced negative sequence three-phase 
voltage and one single-phase voltage.
We could have expressed these equations in another format too.
V
V
RN
YN
= ×
∠ − ° + ×
∠ − ° + ×
∠ − °
= ∠ −
° ×
∠ − ° +
1 200
10
1 100
50
1 25
30
1
120
200
10
1
∠
∠
° ×
∠ − ° + × ∠ − °
= ∠
° ×
∠ − ° + ∠ −
° ×
∠ −
120
100
50
1 25
30
1 120
200
10
1
120
100
V
BN
550
1 25
30
° + × ∠ − °
Let us define a constant complex number 
a
as 
a
 
=
 
1
∠
120
°
. When a phasor is multiplied by this 
operator, its magnitude stays unaffected, but it gets rotated in counter clockwise direction by 120
°
. 
Then, 
a
2
=
1
∠
240
°
=
1
∠-
120
°
is the operator that can rotate a phasor by 120
°
in clockwise direction. 
Note that 1
+ 
a
+ 
a
2
=
0, 
a
2 
=
a
*
, 
a
3
=
1
∠
0
°
and 
a
4
=
a
. Now, the above equations can be expressed in 
terms of this operator 
a
as
V
V
a
a
RN
YN
= ×
∠ − ° + ×
∠ − ° + ×
∠ − °
=
×
∠ − ° + ×
∠
1 200
10
1 100
50
1 25
30
200
10
100
2
−− ° + × ∠ − °
= ×
∠ − ° +
×
∠ − ° + × ∠ − °
50
1 25
30
200
10
100
50
1 25
30
2
V
a
a
BN
Now, let us introduce 
X
+ 
, 
X
-
and 
X
0
as the values of positive, negative and zero sequence 
components (note that they are phasors) of some voltage or current instead of the three numbers – 
200
∠-
10
°
, 100
∠-
50
°
and 25
∠-
30
°
– we used till now. Then, the unbalanced quantities are given in 
terms of symmetrical components (i.e., 
X
+ 
, 
X
-
and 
X
0
) by 
X
X
X
X
X
a
X
a X
X
X
a X
a
X
X
R
Y
B
= ×
+ ×
+ ×
=
×
+ ×
+ ×
= ×
+
×
+ ×
+
−
+
−
+
−
1
1
1
1
1
0
2
0
2
0
We can express this equation set in matrix form as below.
X
X
X
a
a
a
a
X
X
X
R
Y
B
0










=




















+
−
1
1
1
1
1
2
2
(8.5-1)
X
R
, 
X
Y
and 
X
B
can represent any phase voltage, phase current, line voltage or line current phasors in 
an unbalanced system. The 3
× 
3 matrix in Eqn. 8.5-1 is called the Symmetric Transformation Matrix. 
The inverse equation required for determining sequence components from phase quantities is as below.
X
X
X
a
a
a
a
X
X
X
0
R
Y
B
+
−










=





















1
3
1
1
1
1
1
2
2
2









(8.5-2)

---

8.26


SinusoidalSteady-StateinThree-PhaseCircuits
The reader is urged to verify the matrix inversion involved. 
The circuit interpretation of symmetrical components is given in Fig. 8.5-1. Here, an unbalanced 
star-connected voltage source is resolved into its symmetrical components. Symmetrical Components 
theorem assures us that the source in Fig. 8.5-1(a) can be viewed as the composite source in Fig. 
8.5-1(b) in which each source limb contains three sources in series – one each from each sequence 
component set. Further, the superposition principle assures us that applying the source in Fig. 8.5-1(b) 
is the same as applying the three source sets shown in Fig. 8.5-1(c) individually.
Y
B
R
V
0
V
–
V
+
V
+
V
+
V
0
V
0
V
RN
V
BN
V
YN
–
–
–
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
(a)
(b)
(c)
+
+
+
+
N
N
R
V
–
a
2
V
–
a
2
a
2
V
–
a
V
–
a
a
B
Y
Y
V
0
V
0
V
0
–
–
–
+
+
+
N
R
B
Y
V
–
–
–
–
+
+
+
N
R
B
Y
V
–
a
2
V
–
a
2
V
+
–
–
–
+
+
+
N
R
B
Y
Fig. 8.5-1 
Interpretationofsymmetricalcomponents
8.5.2 
the Zero sequence component
This component needs special attention. We get an expression for zero sequence component from
Eqn. 8.5-2 as 
X
X
X
X
0
1
3
=
+
+
(
).
R
Y
B
Zerosequencecomponentistheaverageofthethreethree-phasequantities.
Line voltages in any three-phase circuit, balanced or unbalanced, will add up to zero by KVL. 
Therefore, line voltages in a three-phase system cannot have a zero sequence content anywhere in the 
system.

---

SymmetricalComponents

8.27
Line currents in a three-phase three-wire system will have to add up to zero by KCL. Therefore, 
line currents in a three-phase three-wire system cannot have a zero sequence content anywhere in the 
system. 
Phase voltages in a three-phase system need not add up to zero. Hence, phase voltages can have 
zero sequence content.
Line currents in a four-wire system need not add up to zero. The sum 
I
R
+ 
I
Y
+ 
I
B
can flow through 
the fourth wire (neutral wire) in the return direction. Therefore, three-phase four-wire systems can 
have zero sequence content in their line currents.
8.5.3 
active Power in sequence components
Let the phase voltages across a balanced three-phase load circuit be 
V
RN
, 
V
YN
and 
V
BN
where N is the 
neutral point in the load itself or in its Y-equivalent. Let 
I
R
,
 I
Y
and
 I
B
be the line currents flowing into 
the load. Further, let 
V
0
, 
V
+ 
and 
V
-
be the sequence components of phase voltages and 
I
0
, 
I
+ 
and 
I
-
be 
the sequence components of line currents.
Then, the total active power that flows into the load,
P
V I
V I
V I
V I
V I
V I
=
+
+
=
+
+
Re[
] Re[
] Re[
]
Re[
*
*
*
*
*
*
RN R
YN Y
BN B
RN R
YN Y
BN B
]]
Let us define four column vectors as below.
V
V
V
V
I
I
I
I
V
V
V
V
ryb
RN
YN
BN
ryb
R
Y
B
0+
0
=










=










=
−
+
−
;
;










=










−
+
−
;
I
I
I
I
0+
0
Then, the power equation can be expressed as 
P
V I
ryb
t
ryb
=
Re[
]
*
where the superscripts 
t
indicates 
matrix transpose operation and 
*
indicates complex conjugation operation.
Eqn. 8.5-1 expresses the three-phase quantities in terms of its sequence components. It is reproduced 
below.
X
X
X
a
a
a
a
X
X
X
R
Y
B
0










=




















+
−
1
1
1
1
1
2
2
Using this equation, we write the column vector 
V
ryb
in terms of the column vector 
V
0
+ -
and write 
the column vector 
I
ryb
in terms of the column vector 
I
0
+ -
as below.
V
AV
I
AI
A
a
a
a
a
ryb
ryb
=
=
=










+−
+−
0
0
2
2
1
1
1
1
1
and 
, where 
Then,
P
t
t
=
=
=
+−
+−
+−
+−
Re[
] Re[(
) (
) ] Re[
]
*
*
*
V I
AV
AI
V
A A I
A
ryb ryb
t
t
0
0
0
0
*
t
==










=








1
1
1
1
1
1
1
1
1
1
2
2
2
2
a
a
a
a