Chapter 11
Balanced Three-Phase
Circuits
11.1-2
11.3
11.4
11.5
11.6
Three-Phase Systems
Analysis of the Y-Y Circuit
Analysis of the Y- Circuit
Power Calculations in Balanced
Three-Phase Circuits
Measuring Average Power in ThreePhase Circuits
1
Overview

An electric power distribution system looks like:
where the power transmission uses “balanced
three-phase” configuration.
2
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).
3
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?
4
Section 11.1, 11.2
Three-Phase Systems
1.
2.
Three-phase sources
Three-phase systems
5
One-phase voltage sources

One-phase ac generator: static magnets, one
rotating coil, single output voltage v(t)=Vmcost.
(www.ac-motors.us)
6
Three-phase voltage sources

Three static coils,
rotating magnets,
three output voltages
va(t), vb(t), vc(t).
7
Ideal Y- and -connected voltage sources
Neutral
8
Real Y- and -connected voltage sources

Internal impedance of a generator is usually
inductive (due to the use of coils).
9
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:
abc (positive) sequence: vb(t) lags va(t) by 120.
acb (negative) sequence: vb(t) leads va(t) by
120.
1.
2.
10
abc sequence

vb(t) lags va(t) by 120 or T/3.

Va  Vm0 , Vb  Vm  120 , Vc  Vm  120.
11
Three-phase systems
(Y or )
(Y or )

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.
12
Section 11.3
Analysis of the Y-Y Circuit
1.
2.
Equivalent one-phase circuit for
balanced Y-Y circuit
Line currents, phase and line voltages
13
General Y-Y circuit model
Ref.
The only
essential
node.
14
Unknowns to be solved



Line (line-to-line)
voltage: voltage
across any pair of
lines.
Phase (line-toneutral) voltage:
voltage across a
single phase.
Line current
Line voltage
Phase
current
Phase voltage
For Y-connected load, line current equals phase
current.
15
Solution to general three-phase circuit

No matter it’s balanced or imbalanced threephase circuit, KCL leads to one equation:
I 0  I aA  I bB  I cC , 
VN
Van  VN
Vbn  VN
Vcn  VN



 (1),
Z 0 Z ga  Z1a  Z A Z gb  Z1b  Z B Z gc  Z1c  Z C
Impedance
of neutral
line.
Total
impedance
along line aA.
Total
impedance
along line bB.
Total
impedance
along line cC.
which is sufficient to solve VN (thus the entire
circuit).
16
Solution to “balanced” three-phase circuit

1.
2.
For balanced three-phase circuits,
{Va'n, Vb'n, Vc'n} have equal magnitude and 120
relative phases;
{Zga = Zgb = Zgc}, {Z1a = Z1b = Z1c}, {ZA = ZB = ZC};
 total impedance along any line is the same
Zga + Z1a + ZA =… = Z.
VN Van  VN Vbn  VN Vcn  VN



,
 Eq. (1) becomes:
Z0
Z
Z
Z
 1
 Van  Vbn  Vcn
3
 VN    
 0, VN  0.
Z

Z
Z
0



17
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.
18
Equivalent one-phase circuit
Phase
voltage
of source
Line current
Phase
voltage
of load
Inn = 0  IaA

Directly giving the line current & phase voltages:
Van  VN
I aA 
, VAN  I aA Z A , Van  I aA Z1a  Z A .
Z ga  Z1a  Z A   Z

Unknowns of phases b, c can be determined by
the fixed (abc or acb) sequence relation.
19
The 3 line and phase currents in abc sequence

Given I aA  Van Z , the other 2 line currents are:
I bB
Vbn

 I aA  120 ,
Z
I cC
Vcn

 I aA120 ,
Z
which still
follow the abc
sequence
relation.
I cC
I aA
I bB
20
The phase & line voltages of the load in abc seq.
VAN
ZA
ZB
 Van
, VBN  Vbn
 VAN   120 , VCN  VAN 120.
Z
Z
VAB  VAN  VBN
 VAN  VAN   120 
(abc sequence)
Line
voltage
 3VAN   30 ,
VBC  VAN   120   VAN   120 
 3VAN   90 ,
Phase
voltage
VCA  VAN   120   VAN
 3VAN   150.
21
The phase & line voltages of the load in acb seq.
VAB  VAN  VBN
 VAN  VAN   120 
 3VAN   30 ,
(acb
sequence)
VBC  VAN   120   VAN   120 
Phase
voltage
 3VAN   90 ,
VCA  VAN   120   VAN
 3VAN   150.

Line
voltage
Line voltages are 3 times bigger, leading (abc)
or lagging (acb) the phase voltages by 30.
22
Example 11.1 (1)

Q: What are the line currents, phase and line
voltages of the load and source, respectively?
Zga
Z1a
Phase voltages
ZA
(abc sequence)
Z = Zga + Z1a + ZA = 40 + j30 .
23
Example 11.1 (2)

The 3 line currents (of both load & source) are:
1200
Van

 2.4  36.87  A,
I aA 
Z ga  Z1a  Z A 40  j 30
I bB  I aA  120  2.4  156.87  A,
I cC  I aA  120  2.4  83.13  A.

The 3 phase voltages of the load are:
VAN  I aA Z A  2.4  36.87 39  j 28  115.22  1.19  V.
VBN  VAN   120  115.22  121.19  V,
VCN  VAN   120  115.22  118.81  V.
24
Example 11.1 (3)

The 3 line voltages of the load are:


VAB 

330 115.22  1.19 
330 VAN


 199.58  28.81  V,
VBC  VAB  120
 199.58  91.19  V,
VCA  VAB  120
 199.58  148.81  V.
25
Example 11.1 (4)

The 3 phase voltages of the source are:
Van  Va n  I aA Z ga  120  2.4  36.87  0.2  j 0.5
 118 .9  0.32   V,
Vbn  Van   120   118 .9  120 .32   V,
Vcn  Van   120   118 .9  119 .68  V.

The three line voltages of the source are:
Vab 


330 Van 


330 118.9  0.32 
 205.94  29.68  V,
Vbc  Vab  120  205.94  90.32  V,
Vca  Vab  120  205.94  149.68  V.
26
Section 11.4
Analysis of the Y- Circuit
27
Load in  configuration
Line current
Phase current
Line voltage =
Phase voltage
28
-Y transformation for balanced 3-phase load

The impedance of each leg in Y-configuration
(ZY) is one-third of that in -configuration (Z):
ZbZc
Z1 
,
Za  Zb  Zc
ZcZa
Z2 
,
Za  Zb  Zc
Za Zb
Z3 
.
Za  Zb  Zc
ZZ Z
 ZY 

.
3Z 
3
29
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
Line-to-neutral
voltage 
Phase voltage
 Line voltage
Van
,
directly giving the line current: I aA 
Z ga  Z1a  Z A
and line-to-neutral voltage: VAN  I aA Z A .
30
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:
I aA  I AB  ICA , IbB  I BC  I AB , I cC  ICA  I BC .
Line current
Phase
current
(abc
sequence)
Phase
current
Line current
31
Section 11.5
Power Calculations in
Balanced Three-Phase
Circuits
1.
2.
Complex powers of one-phase and
the entire Y-Load
The total instantaneous power
32
Average power of balanced Y-Load

The average power delivered to ZA is:
PA  V I cos  ,
V  VAN  VL 3 ,

(rms value)
 I  I aA  I L ,
  V  I  Z .


A
 

The total power delivered to the Y-Load is:
Ptot  3PA  3V I cos   3VL I L cos  .
33
Complex power of a balanced Y-Load

The reactive powers of one phase and the
entire Y-Load are:
Q  V I sin  ,

Qtot  3V I sin   3VL I L sin  .

The complex powers of one phase and the
entire Y-Load are:
 S  P  jQ  V I e j  V I* ;

j
j
 Stot  3S  3V I e  3VL I L e .
34
One-phase instantaneous powers

The instantaneous power of load ZA is:
p A (t )  v AN (t )iaA (t )  Vm I m cos t cos(t   ).

(abc sequence)
The instantaneous
powers of ZA, ZC are:
pB (t )  vBN (t )ibB (t )
 Vm I m cost  120 
cost    120 ,
pC (t )  Vm I m cost  120 
cost    120 .
35
Total instantaneous power

The instantaneous power of the entire Y-Load
is a constant independent of time!
ptot (t )  p A (t )  pB (t )  pC (t )  1.5Vm I m cos 

 1.5 2V


2 I cos   3V I cos  .

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.
36
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)
37
Example 11.5 (2)

The line current of a-phase can be calculated by
the complex power is:
600 *
S  V I , 160  j12010 
I aA ,
3

 I aA  577.35  36.87  A.
*

3
The a-phase voltage of the source is:
Van  VAN  I aA Z1a
 600
3  577.35  36.87  0.005  j 0.025
 357.511.57  V.
38
Example 11.5 (3)

The complex power provided by the source of aphase is:
San  Van I*aA  357.511.57 577.3536.87 
 206.4138.44  kVA.

The complex power dissipated by the line of aphase is:
SaA  I aA Z1a  577.35 0.005  j 0.025
2
2
 8.5078.66  kVA.
39
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?
40