Introduction to Three-phase Circuits
Balanced 3-phase systems
Unbalanced 3-phase systems
1
Introduction to 3-phase systems
Single-phase two-wire system:
•
Single source connected to a
load using two-wire system
Single-phase three-wire system:
•
•
Two sources connected to two
loads using three-wire system
Sources have EQUAL
magnitude and are IN PHASE
2
Circuit or system in which AC sources operate at the same
frequency but different phases are known as polyphase.
Balanced Two-phase three-wire system:
•
•
Two sources connected to two loads
using three-wire system
Sources have EQUAL frequency but
DIFFFERENT phases
Two Phase System:
• A generator consists of two coils placed perpendicular to each other
• The voltage generated by one lags the other by 90.
3
Balanced Three-phase four-wire system:
•
•
Three sources connected to 3 loads using
four-wire system
Sources have EQUAL frequency but
DIFFFERENT phases
Three Phase System:
• A generator consists of three coils placed 120 apart.
• The voltage generated are equal in magnitude but, out of phase by 120.
• Three phase is the most economical polyphase system.
4
AC Generation
•
•
•
Three things must be present in order to produce
electrical current:
a) Magnetic field
b) Conductor
c) Relative motion
Conductor cuts lines of magnetic flux, a voltage is
induced in the conductor
Direction and Speed are important
GENERATING A SINGLE PHASE
S
N
Motion is parallel to the flux.
No voltage is induced.
GENERATING A SINGLE PHASE
S
N
Motion is 45° to flux.
Induced voltage is 0.707 of maximum.
GENERATING A SINGLE PHASE
S
x
N
Motion is perpendicular to flux.
Induced voltage is maximum.
GENERATING A SINGLE PHASE
S
N
Motion is 45° to flux.
Induced voltage is 0.707 of maximum.
GENERATING A SINGLE PHASE
S
N
Motion is parallel to flux.
No voltage is induced.
GENERATING A SINGLE PHASE
S
N
Notice current in the
conductor has reversed.
Motion is 45° to flux.
Induced voltage is
0.707 of maximum.
GENERATING A SINGLE PHASE
S
N
Motion is perpendicular to flux.
Induced voltage is maximum.
GENERATING A SINGLE PHASE
S
N
Motion is 45° to flux.
Induced voltage is 0.707 of maximum.
GENERATING A SINGLE PHASE
S
N
Motion is parallel to flux.
No voltage is induced.
Ready to produce another cycle.
GENERATION OF THREE-PHASE AC
S
x
x
 Three Voltages will be induced
across the coils with 120 phase
difference
N
Practical THREE PHASE GENERATOR
 The generator consists of a rotating magnet (rotor) surrounded by a
stationary winding (stator).
 Three separate windings or coils with terminals a-a’, b-b’, and c-c’
are physically placed 120 apart around the stator.
 As the rotor rotates, its magnetic field cuts the flux from the three
coils and induces voltages in the coils.
 The induced voltage have equal magnitude but out of phase by
120.
THREE-PHASE WAVEFORM
Phase 1
120°
Phase 2
Phase 3
120°
120°
240°
Phase 2 lags phase 1 by 120°.
Phase 3 lags phase 1 by 240°.
Phase 2 leads phase 3 by 120°.
Phase 1 leads phase 3 by 240°.
WHY WE STUDY 3 PHASE SYSTEM ?
•
ALL electric power system in the world used 3-phase system to
GENERATE, TRANSMIT and DISTRIBUTE
 One phase, two phase, or three phase ican be taken from three phase
system rather than generated independently.
•
Instantaneous power is constant (not pulsating).– thus smoother
rotation of electrical machines
 High power motors prefer a steady torque
•
More economical than single phase – less wire for the same power
transfer
 The amount of wire required for a three phase system is less than required
for an equivalent single phase system.
18
3-phase systems
Generation, Transmission and Distribution
19
3-phase systems
Generation, Transmission and Distribution
20
Y and  connections
Balanced 3-phase systems can be considered as 3 equal single phase
voltage sources connected either as Y or Delta () to 3 single three loads
connected as either Y or 
SOURCE CONNECTIONS
LOAD CONNECTIONS
Y connected source
Y connected load
 connected source
 connected load
Y-Y
Y-
-Y
-
21
Balance Three-Phase Sources
•
Two possible configurations:
Three-phase voltage sources: (a) Y-connected ; (b) Δ-connected
22
Balanced 3-phase systems
LOAD CONNECTIONS
 connection
Y connection
a
a
Z1
Zb
Zc
n
b
Z2
b
Z3
Za
c
c
Balanced load:
Z1= Z2 = Z3 = Z Y
Za= Z b = Zc = Z 
ZY 
Z
3
Unbalanced load: each phase load may not be the same.
23
Phase Sequence
The phase sequence is the time order in which the voltages pass through
their respective maximum values.
a
Van
o
 Van  Vp0
+

o
 Vbn  Vp  120
vbn (t)  2Vp cos(t  120o )
o
 Vcn  Vp120
v cn (t)  2Vp cos(t  120o )
n
Vcn
b
Vbn
v an (t)  2Vp cos(t)
RMS phasors !
c
240o
120o
Van
Vbn
Vcn
24
Phase Sequence
Vcn  Vp120o
a
Van
+

120o
n
Vcn
Vbn
Van  Vp0o
120o
120o
b
c
Vbn  Vp  120o
Phase sequence : Van leads Vbn by 120o and Vbn leads Vcn by 120o
This is a known as abc sequence or positive sequence
25
Phase Sequence
Vbn  Vp120o
What if different phase sequence?
o
 Van  Vp0
van (t )  2V p cos(t )
120o
v cn (t)  2Vp cos(t  120 ) Vcn  Vp  120
o
o
Van  Vp0o
120o
vbn (t)  2Vp cos(t  120o ) Vbn  Vp120o
120o
RMS phasors !
Vcn  Vp  120o
Phase sequence : Van leads Vcn by 120o and Vcn leads Vbn by 120o
This is a known as acb sequence or negative sequence
120o
Van
240o
Vcn
Vbn
26
Example
Determine the phase sequence of the set of voltages.
van  200 cos(t  10)
vbn  200 cos(t  230)
vcn  200 cos(t  110)
Solution:
The voltages can be expressed in phasor form as
Van  (200 / 2 )10 V
Vbn  (200 / 2 )  230 V
Vcn  (200 / 2 )  110 V
We notice that Van leads Vcn by 120° and Vcn in turn leads Vbn by 120°.
Hence, we have an acb sequence.
Balanced 3-phase Y-Y
Ia
Line current = phase current
A
a
Van
Vcn
+

N
n
Ib
Vbn
Ia 
Ib 
Ic 
ZY
B
b Ic
Vp0o
ZY
Phase
voltages
Vcn  Vp120o
ZY
C
measured between the neutral
and any line
(line to neutral voltage)
Vab  Va  Vb  Va  Vb  Vn  Vn 
 Van  Vnb  V p 0 o  V p 60 o
ZY
Vp  120o
Vbn  Vp  120o
ZY
In
c
Van  Vp0o
line
currents
 3V p 30 o
Vp120o
ZY
Ia  Ib  Ic  In  0
The wire connecting n and N can be removed !
Vbc  Vbn  Vnc
 3 Vp  90o
Vca  Vcn  Vna
 3 Vp150o
line-line
voltages
OR
Line
voltages
28
Balanced 3-phase systems
Balanced Y-Y Connection
Vab  Van  Vnb
 Vp 0o  Vp 60o
 3 Vp 30o
29
Balanced 3-phase systems
Vab  Van  Vnb
 Vp 0o  Vp 60o
Balanced Y-Y Connection
Vcn
Van
 3 Vp 30o
Vbn
30
Balanced 3-phase systems
Balanced Y-Y Connection
Vab  Van  Vnb
 Vp 0o  Vp 60o
 3 Vp 30
o
Van
Vnb
31
Balanced 3-phase systems
Balanced Y-Y Connection
Vab  Van  Vnb
 Vp 0o  Vp 60o
 3 Vp 30
Van
o
Vnb
Vbn
Vbc  Vbn  Vnc
 3 Vp  90o
Vnc
32
Balanced 3-phase systems
Balanced Y-Y Connection
Vna
Vab  Van  Vnb
Vcn
 Vp 0o  Vp 60o
 3 Vp 30
Van
o
Vnb
Vbn
Vbc  Vbn  Vnc
 3 Vp  90o
Vnc
Vca  Vcn  Vna
 3 Vp150o
33
Balanced 3-phase systems
Balanced Y-Y Connection
Vca
Vab  Van  Vnb
 Vp 0o  Vp 60o
Vab
30o Vcn
30o
Van
 3 Vp 30o
Vbn
Vbc  Vbn  Vnc
30o
 3 Vp  90o
Vbc
Vca  Vcn  Vna
 3 Vp150o
VL  3 Vp
where
VL  Vab  Vbc  Vca
and Vp  Van  Vbn  Vcn
Line voltage LEADS phase voltage by 30o
34
Balanced 3-phase systems
Balanced Y-Y Connection
For a balanced Y-Y connection, analysis can be performed using an
equivalent per-phase circuit: e.g. for phase A:
Ia
A
a
Van
Vcn
+

ZY
In=0
N
n
Ib
c
Vbn
b Ic
ZY
B
ZY
C
35
Balanced 3-phase systems
Balanced Y-Y Connection
For a balanced Y-Y connection, analysis can be performed using an
equivalent per-phase circuit: e.g. for phase A:
Ia
A
a
Van
+

ZY
N
n
Ia 
Van
ZY
Based on the sequence, the other line currents can be
obtained from:
Ib  Ia  120o
Ic  Ia120o
36
Balanced 3-phase systems
Balanced Y- Connection
Ia
a
Van
Vcn
A
+

Z
n
Ib
c
Vbn
Vab  3 Vp30o
B
V
 AB
Z
Vbn  Vp  120o
Z
I AB Z
IBC
b Ic
IAB
 VAB
C
ICA
Using KCL,
Vcn  Vp120o
Ia  IAB  ICA
 IAB (1  1120o )
 IAB 3  30o
Vbc  3 Vp  90
o
IBC 
 VBC
Vca  3 Vp150o
 VCA
Van  Vp0o
ICA 
VBC
Z
VCA
Z
Phase
currents
Ib  IBC  IAB
 IBC (1  1120o )
 IBC 3  30o
Ic  ICA 3  30o
37
Balanced 3-phase systems
Balanced Y- Connection
Ic
ICA
30o
30o
IBC
Ib
IL  3Ip
IAB
Ia  IAB  ICA
30o
 IAB (1  1120o )
 IAB 3  30o
Ia
Ib  IBC  IAB
 IBC (1  1120o )
 IBC 3  30o
and Ip  IAB  IBC  ICA o
where IL  Ia  Ib  Ic
Ic  ICA 3  30
38
Phase current LEADS line current by 30o
Balanced 3-phase -
Line-line voltage is the
same as phase voltage in
- 
Vab  Vp0o
Ia
a
A
Vca
Vab
Ib
+

c
Vbc
Vab  VAB
Z
B
Z
I AB Z
IBC
b Ic
IAB
V
 AB
Z
Vbc  Vp  120o
Using KCL,
C
ICA
Vcn  Vp120o
Ia  IAB  ICA
 IAB (1  1120o )
 IAB 3  30o
Vbc  VBC
Vca  VCA
IBC 
ICA 
VBC
Z
VCA
Z
Phase
currents
Ib  IBC  IAB
line
currents
 IBC (1  1120o )
 IBC 3  30o
Ic  ICA 3  30o
39
Balanced 3-phase systems
Balanced - Connection
Ia
a
Vab  Vp0o
A
Vca
Z
Ib
+

c
Vab
Vbc
b Ic
B
Vbc  Vp  120o
Z
I AB Z
IBC
C
ICA
Vcn  Vp120o
Alternatively, by transforming the  connections to the equivalent Y
connections per phase equivalent circuit analysis can be performed.
40
Balanced -Y Connection
Balanced 3-phase systems
Ia
A
a
Vca
N
Ib
Vbc
ZY
 Ia  Ib 
Vab
ZY
B
b Ic
Loop1 - Vab  Z YIa  Z YIb  0
Since circuit is balanced, Ib = Ia-120o
Ia 
Vp
ZY
3
Vca  Vp120o
ZY
How to find Ia ?
Therefore
Vbc  Vp  120o
ZY
Loop1
+

c
Vab
Vab  Vp0o
C
 Ia  Ib  Ia (1 1(120o ))
 Ia 330o
  30o
41
Balanced -Y Connection
Balanced 3-phase systems
Ia
A
a
Vca
N
Ib
Vbc
Vbc  Vp  120o
ZY
Vab
+

c
Vab  Vp0o
b Ic
ZY
B
Vca  Vp120o
ZY
C
How to find Ia ? (Alternative)
Transform the delta source connection to an equivalent Y and then
perform the per phase circuit analysis
42
 A balanced Y-Y system, showing the source, line and load impedances.
Line Impedance
Source Impedance
Load Impedance
Equivalent Circuit
43
Three-phase Circuits
Unbalanced 3-phase systems
Power in 3-phase system
44
UNBALANCED DELTA-CONNECTED LOAD
The line currents will not be equal nor will they have a 120° phase
difference as was the case with balanced loads.
45
UNBALANCED FOUR-WIRE, WYE-CONNECTED LOAD
 On a four-wire system the neutral conductor will carry a current when the load
is unbalanced
 The voltage across each of the load impedances remains fixed with the same
magnitude as the line to neutral voltage.
 The line currents are unequal and do not have a 120° phase difference.
46
UNBALANCED THREE-WIRE, WYE-CONNECTED LOAD
 The common point of the three load impedances is not at the
potential of the neutral and is marked "O" instead of N.
 The voltages across the three impedances can vary considerably
from line to neutral magnitude, as shown by the voltage triangle
which relates all of the voltages in the circuit.
 Draw the circuit diagram and select mesh currents as shown in Fig.
 Write the corresponding matrix equations (Crammer Rule)
47
UNBALANCED THREE-WIRE, WYE-CONNECTED LOAD
Now the voltages across the three impedances are given by the products
of the line currents and the corresponding impedances.
48
POWER IN BALANCED THREE-PHASE LOADS
 Since the phase impedances of balanced wye or delta loads contain
equal currents, the phase power is one-third of the total power.
• The voltage across is line voltage
• The current is phase current.
• The angle between V & I is the angle
on the impedance.
• The voltage across is phase voltage
• The current is line current.
• The angle between V & I is the angle
on the impedance.
Phase power
Phase power
Total power
Total power
For a balanced Δ-connected loads:
For a balanced Y-connected loads:
49
POWER IN BALANCED THREE-PHASE LOADS
50
INSTANTANEOUS THREE-PHASE POWER
 Remember:
The instantaneous Single-phase power
51
INSTANTANEOUS THREE-PHASE POWER
The instantaneous 3-phase power
p 𝑡 = 𝑉𝐴𝑁 𝐼𝑎 + 𝑉𝐵𝑁 𝐼𝑏 + 𝑉𝐶𝑁 𝐼𝐶
52
𝑉3𝑝
𝑉1𝑝
55
=
3 ∗(𝑙 ∗𝑆3𝑝)
2∗(𝑙∗𝑆1𝑝)
=
3 ∗(𝑆1𝑝/2)
2∗(𝑆1𝑝)
=
3
4