1) A three-phase power distribution system uses a balanced three-phase configuration to transmit power from generators to loads. 2) A balanced three-phase circuit can be analyzed as an equivalent single-phase circuit. This allows determining the unknown voltages and currents by solving for a single phase. 3) For a balanced Y-Y connected three-phase circuit, the line currents are equal and differ in phase by 120 degrees, while the line voltages are √3 times the phase voltages and differ in phase by 30 degrees.

1. 1
Chapter 11
Balanced Three-Phase
Circuits
11.1-2 Three-Phase Systems
11.3 Analysis of the Y-Y Circuit
11.4 Analysis of the Y- Circuit
11.5 Power Calculations in Balanced
Three-Phase Circuits
11.6 Measuring Average Power in Three-
Phase Circuits

2. 2
Overview
 An electric power distribution system looks like:
where the power transmission uses “balanced
three-phase” configuration.

3. 3
Why three-phase?
 Three-phase generators can be driven by
constant force or torque (to be discussed).
 Industrial applications, such as high-power
motors, welding equipments, have constant
power output if they are three-phase systems
(to be discussed).

4. 4
Key points
 What is a three-phase circuit (source, line, load)?
 Why a balanced three-phase circuit can be
analyzed by an equivalent one-phase circuit?
 How to get all the unknowns (e.g. line voltage of
the load) by the result of one-phase circuit
analysis?
 Why the total instantaneous power of a
balanced three-phase circuit is a constant?

5. 5
Section 11.1, 11.2
Three-Phase Systems
1. Three-phase sources
2. Three-phase systems

6. 6
One-phase voltage sources
 One-phase ac generator: static magnets, one
rotating coil, single output voltage v(t)=Vmcost.
(www.ac-motors.us)

7. 7
Three-phase voltage sources
 Three static coils,
rotating magnets,
three output voltages
va(t), vb(t), vc(t).

8. 8
Ideal Y- and -connected voltage sources
Neutral

9. 9
Real Y- and -connected voltage sources
 Internal impedance of a generator is usually
inductive (due to the use of coils).

10. 10
Balanced three-phase voltages
 Three sinusoidal voltages of the same
amplitude, frequency, but differing by 120
phase difference with one another.
 There are two possible sequences:
1. abc (positive) sequence: vb(t) lags va(t) by 120.
2. acb (negative) sequence: vb(t) leads va(t) by
120.

11. 11
abc sequence
 vb(t) lags va(t) by 120 or T/3.
.
120
,
120
,
0 









 m
c
m
b
m
a V
V
V V
V
V


12. 12
Three-phase systems
 Source-load can be connected in four
configurations: Y-Y, Y-, -Y, -
 It’s sufficient to analyze Y-Y, while the others
can be treated by -Y and Y- transformations.
(Y or )
(Y or )

13. 13
Section 11.3
Analysis of the Y-Y Circuit
1. Equivalent one-phase circuit for
balanced Y-Y circuit
2. Line currents, phase and line voltages

14. 14
General Y-Y circuit model
Ref.
The only
essential
node.

15. 15
Unknowns to be solved
Phase voltage
 Line (line-to-line)
voltage: voltage
across any pair of
lines.
Line current
Phase
current
 Phase (line-to-
neutral) voltage:
voltage across a
single phase.
 For Y-connected load, line current equals phase
current.
Line voltage

16. 16
Solution to general three-phase circuit
 No matter it’s balanced or imbalanced three-
phase circuit, KCL leads to one equation:
Impedance
of neutral
line.
Total
impedance
along line aA.
Total
impedance
along line bB.
Total
impedance
along line cC.
),
1
(
,
1
1
1
0
0

C
c
gc
N
n
c
B
b
gb
N
n
b
A
a
ga
N
n
a
N
cC
bB
aA
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z 

















 V
V
V
V
V
V
V
I
I
I
I
which is sufficient to solve VN (thus the entire
circuit).

17. 17
.
0
,
0
3
1
0














 


N
n
c
n
b
n
a
N
Z
Z
Z
V
V
V
V
V


Solution to “balanced” three-phase circuit
 For balanced three-phase circuits,
1. {Va'n, Vb'n, Vc'n} have equal magnitude and 120
relative phases;
2. {Zga = Zgb = Zgc}, {Z1a = Z1b = Z1c}, {ZA = ZB = ZC};
 total impedance along any line is the same
Zga + Z1a + ZA =… = Z.
,
0 

 Z
Z
Z
Z
N
n
c
N
n
b
N
n
a
N V
V
V
V
V
V
V 




 


 Eq. (1) becomes:

18. 18
Meaning of the solution
 VN = 0 means no voltage difference between
nodes n and N in the presence of Z0.  Neutral
line is both short (v = 0) and open (i = 0).
 The three-phase circuit can be separated into 3
one-phase circuits (open), while each of them
has a short between nodes n and N.

19. 19
Equivalent one-phase circuit
 Directly giving the line current & phase voltages:
 Unknowns of phases b, c can be determined by
the fixed (abc or acb) sequence relation.
Inn=0IaA
Line current
Phase
voltage
of load
  ,
1 
Z
Z
Z
Z A
a
ga
N
n
a
aA




  V
V
I
Phase
voltage
of source
,
A
aA
AN Z
I
V   .
1 A
a
aA
an Z
Z 
 I
V

20. 20
,
120



 
aA
n
b
bB
Z
I
V
I

The 3 line and phase currents in abc sequence
aA
I
bB
I
cC
I
 Given the other 2 line currents are:
,

Z
n
a
aA 
 V
I
which still
follow the abc
sequence
relation.
,
120


 
aA
n
c
cC
Z
I
V
I


21. 21
The phase & line voltages of the load in abc seq.
Phase
voltage
Line
voltage
 
,
30
3
120











AN
AN
AN
BN
AN
AB
V
V
V
V
V
V
.
120
,
120
, 







 
 AN
CN
AN
B
n
b
BN
A
n
a
AN
Z
Z
Z
Z
V
V
V
V
V
V
V


(abc sequence)
   
 
.
150
3
120
,
90
3
120
120





















AN
AN
AN
CA
AN
AN
AN
BC
V
V
V
V
V
V
V
V

22. 22
The phase & line voltages of the load in acb seq.
Phase
voltage
Line
voltage
 
   
 
.
150
3
120
,
90
3
120
120
,
30
3
120
































AN
AN
AN
CA
AN
AN
AN
BC
AN
AN
AN
BN
AN
AB
V
V
V
V
V
V
V
V
V
V
V
V
V
V
 Line voltages are 3 times bigger, leading (abc)
or lagging (acb) the phase voltages by 30.
(acb
sequence)

23. 23
Example 11.1 (1)
Phase voltages
Zga Z1a
ZA
 Q: What are the line currents, phase and line
voltages of the load and source, respectively?
Z = Zga + Z1a + ZA = 40 + j30 .
(abc sequence)

24. 24
Example 11.1 (2)
 The 3 line currents (of both load & source) are:
 The 3 phase voltages of the load are:
 
 
  .
A
13
.
83
4
.
2
120
,
A
87
.
156
4
.
2
120
,
A
87
.
36
4
.
2
30
40
0
120
1


























 
aA
cC
aA
bB
A
a
ga
n
a
aA
j
Z
Z
Z
I
I
I
I
V
I
    
 
  .
V
81
.
118
22
.
115
120
,
V
19
.
121
22
.
115
120
.
V
19
.
1
22
.
115
28
39
87
.
36
4
.
2


























AN
CN
AN
BN
A
aA
AN j
Z
V
V
V
V
I
V

25. 25
Example 11.1 (3)
 The 3 line voltages of the load are:
 
  
 
 
  .
V
81
.
148
58
.
199
120
,
V
19
.
91
58
.
199
120
,
V
81
.
28
58
.
199
19
.
1
22
.
115
30
3
30
3





























AB
CA
AB
BC
AN
AB
V
V
V
V
V
V

26. 26
Example 11.1 (4)
 The three line voltages of the source are:
 The 3 phase voltages of the source are:
  
 
 
  .
V
68
.
119
9
.
118
120
,
V
32
.
120
9
.
118
120
,
V
32
.
0
9
.
118
5
0
2
0
87
.
36
4
.
2
120



























 
an
cn
an
bn
ga
aA
n
a
an .
j
.
Z
V
V
V
V
I
V
V
    
 
 
  .
V
68
.
149
94
.
205
120
,
V
32
.
90
94
.
205
120
,
V
68
.
29
94
.
205
32
.
0
9
.
118
30
3
30
3





























ab
ca
ab
bc
an
ab
V
V
V
V
V
V

27. 27
Section 11.4
Analysis of the Y- Circuit

28. 28
Load in  configuration
Phase current
Line current
Line voltage =
Phase voltage

29. 29
-Y transformation for balanced 3-phase load
 The impedance of each leg in Y-configuration
(ZY) is one-third of that in -configuration (Z):
.
,
,
3
2
1
c
b
a
b
a
c
b
a
a
c
c
b
a
c
b
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z









.
3
3







Z
Z
Z
Z
ZY

30. 30
Equivalent one-phase circuit
 The 1-phase equivalent circuit in Y-Y config.
continues to work if ZA is replaced by Z/3:
Line current
directly giving the line current: ,
1 A
a
ga
n
a
aA
Z
Z
Z 

 
V
I
and line-to-neutral voltage: .
A
aA
AN Z
I
V 
Line-to-neutral
voltage 
Phase voltage
 Line voltage

31. 31
The 3 phase currents of the load in abc seq.
 Can be solved by 3 node equations once the 3
line currents IaA, IbB, IcC are known:
Line current
Phase
current
.
,
, BC
CA
cC
AB
BC
bB
CA
AB
aA I
I
I
I
I
I
I
I
I 





Line current
Phase
current
(abc
sequence)

32. 32
Section 11.5
Power Calculations in
Balanced Three-Phase
Circuits
1. Complex powers of one-phase and
the entire Y-Load
2. The total instantaneous power

33. 33


















.
,
,
3
,
cos
A
L
aA
L
AN
A
Z
I
V
I
I
V
V
I
V
P










I
V
Average power of balanced Y-Load
 The average power delivered to ZA is:
(rms value)
 The total power delivered to the Y-Load is:
.
cos
3
cos
3
3 


 
 L
L
A
tot I
V
I
V
P
P 



34. 34
Complex power of a balanced Y-Load
 The reactive powers of one phase and the
entire Y-Load are:








.
sin
3
sin
3
,
sin











L
L
tot I
V
I
V
Q
I
V
Q
 The complex powers of one phase and the
entire Y-Load are:












.
3
3
3
;
*
















j
L
L
j
tot
j
e
I
V
e
I
V
S
S
e
I
V
jQ
P
S I
V

35. 35
One-phase instantaneous powers
 The instantaneous power of load ZA is:
).
cos(
cos
)
(
)
(
)
( 


 

 t
t
I
V
t
i
t
v
t
p m
m
aA
AN
A
 The instantaneous
powers of ZA, ZC are:
 
 
 
 .
120
cos
120
cos
)
(
,
120
cos
120
cos
)
(
)
(
)
(





















t
t
I
V
t
p
t
t
I
V
t
i
t
v
t
p
m
m
C
m
m
bB
BN
B
(abc sequence)

36. 36
Total instantaneous power
   .
cos
3
cos
2
2
5
.
1
cos
5
.
1
)
(
)
(
)
(
)
(










I
V
I
V
I
V
t
p
t
p
t
p
t
p m
m
C
B
A
tot






 The instantaneous power of the entire Y-Load
is a constant independent of time!
 The torque developed at the shaft of a 3-phase
motor is constant,  less vibration in
machinery powered by 3-phase motors.
 The torque required to empower a 3-phase
generator is constant,  need steady input.

37. 37
Example 11.5 (1)
 Q: What are the complex powers provided by
the source and dissipated by the line of a-phase?
 The equivalent one-phase circuit in Y-Y
configuration is:
Z1a
S
(rms value)

38. 38
Example 11.5 (2)
 The line current of a-phase can be calculated by
the complex power is:
 
 A.
87
.
36
35
.
577
,
3
600
10
120
160
, *
3
*








aA
aA
j
S
I
I
I
V 


 The a-phase voltage of the source is:
  
 V.
57
.
1
51
.
357
025
.
0
005
.
0
87
.
36
35
.
577
3
600
1











j
Z a
aA
AN
an I
V
V

39. 39
Example 11.5 (3)
 The complex power provided by the source of a-
phase is:
 The complex power dissipated by the line of a-
phase is:
  
 kVA.
44
.
38
41
.
206
87
.
36
35
.
577
57
.
1
51
.
357
*








 aA
an
an
S I
V
   
 kVA.
66
.
78
50
.
8
025
.
0
005
.
0
35
.
577
2
1
2





 j
Z
S a
aA
aA I

40. 40
Key points
 What is a three-phase circuit (source, line, load)?
 Why a balanced three-phase circuit can be
analyzed by an equivalent one-phase circuit?
 How to get all the unknowns (e.g. line voltage of
the load) by the result of one-phase circuit
analysis?
 Why the total instantaneous power of a
balanced three-phase circuit is a constant?