CHAPTER 6:
THREE-PHASE CIRCUIT
Contents
•
•
•
Balanced Three-Phase Voltages
Balanced Connections: Wye-Wye,
Wye-Delta, Delta-Delta, Delta-Wye
Power in a Balanced System
Introduction
• The earliest application of a.c was electric lamps 
single phase
• Then a.c motor was developed  single phase not
suitable
• Need 3-phase system
• In general, the generation, transmission and distribution
of electrical power are more efficient using 3-phase
system.
• For economic reasons, 3-phase systems are usually
designed to operate in the balanced state.
Introduction
Single phase systems
a) two-wire type
b) three-wire type
Introduction
The basic structure of a 3-phase systems consist of a
voltage source connected to loads by means transformers
and transmission lines. To analyze such circuit, we can
reduce it to a voltage source connected to a load via a
transmission line as shown in this figure
3-phase voltage
source
3-phase load
3-phase transmission line
Three Phase System
Balanced Three Phase
Voltages
Three-phase voltage sources
a) wye-connected source
b) delta-connected source
Balanced Three Phase
Voltages
Phase sequences
Van  Vp 00
Vbn  Vp   1200
Vcn  Vp   2400  Vp   1200
a) abc or positive sequence
* Vp is the effective or rms value
Balanced Three Phase
Voltages
Phase sequences
Van  Vp 00
Vcn  Vp   1200
Vbn  Vp   2400  Vp   1200
b) acb or negative sequence
* Vp is the effective or rms value
Balanced Three Phase
Voltages
If the voltage source have the same amplitude
and frequency ω and are out of phase with each
other by 120o, the voltage are said to be
balanced.
Van  Vbn  Vcn  0
Van  Vbn  Vcn
Balanced phase voltages are equal in
magnitude and out of phase with each other by
120o
Balanced Three Phase
Voltages
Two possible three-phase load configurations:
a) a wye-connected load
b) a delta-connected load
Balanced Three Phase
Voltages
A balanced load is one in which the phase impedances
are equal in magnitude and in phase.
For a balanced wye connected load:
Z1  Z2  Z3  ZY
For a balanced delta connected load:
Za  Z b  Zc  Z 
Z  3ZY
1
ZY  Z
3
Source and load connection
•
•
•
•
Y – Y (Wye – Wye connection)
Y - ∆ (Wye – Delta connection)
∆ - ∆ (Delta – Delta connection)
∆ - Y (Delta – Wye connection)
Example
Determine the phase sequence of the set of voltages
van = 200 cos(ωt + 10◦)
vbn = 200 cos(ωt − 230◦), vcn = 200 cos(ωt − 110◦)
Solution:
The voltages can be expressed in phasor form as
Van  20010V
Vbn  200  230V
Vcn  200  110V
We notice that Van leads Vcn by 120◦ and Vcn in turn leads
Vbn by 120◦.
acb sequence.
Exercise
Given that Vbn  11030V
, find Van and Vcn, assuming
positive (abc) sequence.
Answer:
Van  110150V
Vcn  110  90V
Balanced Wye-Wye
Connection
How to draw Y-Y connection?
Balanced Wye-Wye
Connection
A balanced Y-Y system is a three phase system with a
balanced Y connected source and balanced Y connected load.
Zs 
Source impedance
Z 
Line impedance
ZL 
ZY 
Load
impedance
Total impedance per phase
Z Y  Zs  Z  Z L
If Zs and Zl very small then,
ZY  Z L
Balanced Wye-Wye
Connection
Phase voltages:
Van  V p 00
Vbn  V p   1200
Vcn  V p   2400  V p   1200
Balanced Wye-Wye
Connection
Line to line voltages or line voltages:
Vab  3Vp 300
Vbc  3Vp   900
Vca  3Vp   2100
Magnitude of line voltages:
VL  3Vp
Vp  Van  Vbn  Vcn
VL  Vab  Vbc  Vca
Balanced Wye-Wye
Connection
Phasor diagram illustrating the relationship between line voltages
and phase voltages.
Vab
Vcn
o
120
Vbn
30o
Van
Magnitude of the line voltages is √3
times of the magnitude of the phase
voltages
Balanced Wye-Wye
Connection
Balanced Y-Y
connection
a
A
Ia
+
Van
Zy
-
N
-
-
n
Z
+
+
Vcn
Vbn
b
Ib
Ic
C
ZY
c
Y
In
B
Balanced Wye-Wye
Connection
From the figure, the line currents are:
Van
Ia 
ZY
Vbn Van   1200
Ib 

 I a   1200
ZY
ZY
Vcn Van   2400
Ic 

 I a   2400
ZY
ZY
I a  I b  I c  I n  0
VnN  Zn I n  0
Example
Calculate the line currents in the three wire Y-Y system of figure
below.
Exercise
A Y-connected balanced three-phase generator
with an impedance of 0.4+j0.3 Ω per phase is
connected to a Y-connected balanced load with an
impedance of 24 + j19 Ω per phase. The line
joining the generator and the load has an
impedance of 0.6 + j0.7 Ω per phase.
Assuming a positive sequence for the source
voltages and that Van  12030 0 V
(a) Draw the Y-Y connection
(b) Calculate the line voltages
(c) Calculate the line currents
Balanced Wye-Delta
Connection
A balanced Y- Δ system consists of balanced Y connected
source feeding a balanced Δ connected load.
Balanced Wye-Delta
Connection
Phase voltages:
Van  V p 0
0
Line voltages:
Vab  3Vp 300  VAB
Vbn  V p   1200
Vbc  3Vp   900  VBC
Vcn  V p   2400  V p   1200
Vca  3Vp   2100  VCA
Balanced Wye-Delta
Connection
Line currents:
Phase currents:
I a  I AB  I CA  3I AB   30
I b  I BC  I AB  3I AB   150
I c  I CA  I BC  3I AB 90
I L  I a  Ib  Ic
I AB
V AB

Z
I BC
V BC

Z
I CA
VCA

Z
Ip  IAB  IBC  ICA
Balanced Wye-Delta
Connection
A single-phase equivalent circuit of a balanced Y- circuit
Magnitude of the line currents is √3
times of the magnitude of the phase
currents
Balanced Wye-Delta
Connection
Change Delta load to Wye load, then the system
become Wye – Wye connection
Z
ZY 
3
Van
Van
Ia 

ZY Z  / 3
This only true for the line currents. To obtain
phase current must use this formula
Example
A balanced abc sequence Y-connected source
with Van  10010 0 V
is connected to a Δ-connected balanced load
(8+j4)Ω per phase. Calculate the phase and line
currents.
Exercise
One line voltage of a balanced Y-connected
source is V AB  180  20 0 V
If the source is connected to a Δ -connected
load of 2040 0  , find the phase and line
currents. Assume the abc sequence.
Balanced Delta-Delta
Connection
A balanced Δ - Δ system is one in which both balanced
source and balanced load are Δ connected.
Balanced Delta-Delta
Connection
Line voltages:
Phase currents:
Line currents:
Vab  VAB
I a  I AB  I CA  3I AB   30
Vbc  VBC
I b  I BC  I AB  3I AB   150
Vca  VCA
I c  I CA  I BC  3I AB 90
Magnitude line currents:
IL  Ip 3
Total impedance:
Z
ZY 
3
I AB
V AB

Z
I BC
V BC

Z
I CA
VCA

Z
Example
A balanced Δ connected load having an impedance
20-j15 Ω is connected to a Δ connected, positive
sequence
0
generator having Vab  3300 V
Calculate the phase currents of the load and the line
currents.
Exercise
A positive-sequence, balanced-connected
source supplies a balanced Δ-connected
load. If the impedance per phase of the load
is 18+j12 Ω and I a  22.535 0 A , find IAB and
VAB.
Balanced Delta-Wye
Connection
A balanced Δ -Y system consists of balanced Δ connected
source feeding a balanced Y connected load.
Balanced Delta-Wye
Connection
Applying KVL to loop aANBba:
Ia  Ib 
From:
V p 0 0
ZY
I b  I a   120 0
Ia  Ib  Ia 3300
Line currents:
Ia 
Vp
3
  30 0
ZY
Balanced Delta-Wye
Connection
Replace Δ connected source to
equivalent Y connected source.
Phase voltages:
Van 
Vbn 
Vcn 
Vp
3
Vp
3
Vp
3
  30 0
  150 0
  90 0
Balanced Delta-Wye
Connection
A single phase equivalent circuit
Vp
Van
Ia 

ZY
3
  30 0
ZY
Example
A balanced Y connected load with a phase
resistance of 40 Ω and a reactance of 25 Ω is
supplies by a balanced, positive sequence Δ
connected source with a line voltage of 210 V.
Calculate the phase currents. Use Vab as
reference.
Exercise
0
V

240
15
V
In a balanced -Y circuit, ab
and ZY = (12 + j15) Ω.
Calculate the line currents.
POWER IN A BALANCED
SYSTEM
For Y connected load, the phase
voltage:
v AN 
2V p cos t
vBN 
2V p cos(t  120 0 )
vCN 
2V p cos(t  120 0 )
POWER IN A BALANCED
SYSTEM
Phase current lag phase voltage by θ.
If
ZY  Z
The phase current:
ia 
2I p cos(t  )
ib 
2I p cos(t    120 0 )
ic 
2I p cos(t    120 0 )
POWER IN A BALANCED
SYSTEM
Total instantaneous power:
p  p a  p b  p c  v ANi a  v BNi b  v CN i c
p  3Vp I p cos
Average power per phase:
Pp  Vp I p cos
Apparent power per phase:
Sp  Vp I p
Reactive power per phase:
Qp  Vp I p sin 
Complex power per phase:
Sp  Pp  jQ p  Vp I*p
POWER IN A BALANCED
SYSTEM
Total average power:
P  3Pp  3Vp I p cos  3VL I L cos
Total reactive power:
Q  3Qp  3Vp I p sin   3VL I L sin 
Total complex power:
2
3
V
S  3Sp  3Vp I*p  3I 2p Zp  *p
Zp
S  P  jQ 
3VL I L 
POWER IN A BALANCED
SYSTEM
Power loss in two wires:
2
L
2
L
P
Ploss  2I R  2R
V
2
L
PL : power absorbed by the load
IL : magnitude of line current
VL : line voltage
R : line resistance
POWER IN A BALANCED
SYSTEM
Power loss in three wires:
Ploss  3I L2 R  3R
2
L
2
L
P
3V
R
2
L
2
L
P
V
PL : power absorbed by the load
IL : magnitude of line current
VL : line voltage
R : line resistance
Example
Determine the total real power, reactive power
and complex power at the source and at the load
Exercise
Exercise
END