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THREE PHASE
CIRCUIT
SINGLE PHASE TWO WIRE
V p 
Objectives
Explain the differences between singlephase, two-phase and three-phase.
 Compute and define the Balanced ThreePhase voltages.
 Determine the phase and line
voltages/currents for Three-Phase
systems.

SINGLE PHASE SYSTEM
A generator connected through a pair of wire
to a load – Single Phase Two Wire.
 Vp is the magnitude of the source voltage, and
 is the phase.

SINLGE PHASE THREE WIRE
V p 
V p 
SINGLE PHASE SYSTEM
Most common in practice: two identical
sources connected to two loads by two outer
wires and the neutral: Single Phase Three Wire.
 Terminal voltages have same magnitude and the
same phase.

POLYPHASE SYSTEM
 Circuit
or system in which AC
sources operate at the same
frequency but different phases are
known as polyphase.
TWO PHASE SYSTEM THREE WIRE
V p 
V p   90
POLYPHASE SYSTEM

Two Phase System:
◦ A generator consists of two coils placed
perpendicular to each other
◦ The voltage generated by one lags the other by 90.
POLYPHASE SYSTEM

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.
THREE PHASE FOUR WIRE
IMPORTANCE OF THREE PHASE SYSTEM

All electric power is generated and
distributed in three phase.
◦ One phase, two phase, or more than three phase
input can be taken from three phase system rather
than generated independently.
◦ Melting purposes need 48 phases supply.
IMPORTANCE OF THREE PHASE
SYSTEM

Uniform power transmission and less vibration
of three phase machines.
◦ The instantaneous power in a 3 system can be
constant (not pulsating).
◦ High power motors prefer a steady torque
especially one created by a rotating magnetic field.
IMPORTANCE OF THREE PHASE
SYSTEM

Three phase system is more economical than
the single phase.
◦ The amount of wire required for a three phase
system is less than required for an equivalent single
phase system.
◦ Conductor: Copper, Aluminum, etc
THREE PHASE
GENERATION
FARADAYS LAW



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.
THREE PHASE GENERATOR
GENERATOR WORK

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.
GENERATION OF THREE-PHASE AC
S
x
x
N
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.
GENERATION OF 3 VOLTAGES
Phase 1Phase 2 Phase 3
S
x
Phase 1 is ready to go positive.
Phase 2 is going more negative.
Phase 3 is going less positive.
x
N
THREE PHASE
QUANTITIES
BALANCED 3 VOLTAGES

Balanced three phase voltages:
◦ same magnitude (VM )
◦ 120 phase shift
v an (t )  VM cos t 
vbn (t )  VM cos t  120
vcn (t )  VM cos t  240  VM cos t  120
BALANCED 3 CURRENTS

Balanced three phase currents:
◦ same magnitude (IM )
◦ 120 phase shift
ia (t )  I M cos t   
ib (t )  I M cos t    120
ic (t )  I M cos t    240
PHASE SEQUENCE
van (t )  VM cos t
vbn (t )  VM cost  120
vcn (t )  VM cost  120
Van  VM 0
Van  VM 0
Vbn  VM   120
Vbn  VM   120
Vcn  VM   120
Vcn  VM   120
POSITIVE
SEQUENCE
NEGATIVE
SEQUENCE
PHASE SEQUENCE
EXAMPLE # 1

Determine the phase sequence of the set
voltages:
van  200 cost  10
vbn  200 cost  230
vcn  200 cost  110
BALANCED VOLTAGE AND
LOAD
Balanced Phase Voltage: all phase voltages are
equal in magnitude and are out of phase with
each other by 120.
 Balanced Load: the phase impedances are
equal in magnitude and in phase.

THREE PHASE CIRCUIT

POWER
◦ The instantaneous power is constant
p(t )  pa (t )  pb (t )  pc (t )
VM I M
3
cos 
2
 3Vrms I rms cos( )
THREE PHASE CIRCUIT

Three Phase Power,
S T  S A  S B  S C  3 S
THREE PHASE QUANTITIES
QUANTITY
SYMBOL
Phase current
I
Line current
IL
Phase voltage
V
Line voltage
VL
PHASE VOLTAGES and LINE
VOLTAGES
Phase voltage is measured between the neutral
and any line: line to neutral voltage
 Line voltage is measured between any two of
the three lines: line to line voltage.

PHASE CURRENTS and LINE
CURRENTS

Line current (IL) is the current in each line
of the source or load.

Phase current (I) is the current in each
phase of the source or load.
THREE PHASE
CONNECTION
SOURCE-LOAD CONNECTION
SOURCE
LOAD
CONNECTION
Wye
Wye
Y-Y
Wye
Delta
Y-
Delta
Delta
- 
Delta
Wye
-Y
SOURCE-LOAD CONNECTION

Common connection of source:WYE
◦ Delta connected sources: the circulating current
may result in the delta mesh if the three phase
voltages are slightly unbalanced.

Common connection of load: DELTA
◦ Wye connected load: neutral line may not be
accessible, load can not be added or removed
easily.
WYE CONNECTION
WYE CONNECTED
GENERATOR
WYE CONNECTED LOAD
a
a
b
Y
Z
Z
Y
OR
b
Load
ZY
c
n
ZY
c
n
ZY
ZY
Load
BALANCED Y-Y CONNECTION

PHASE CURRENTS AND LINE
CURRENTS
In Y-Y system:
IL  Iφ
PHASE VOLTAGES, V
Phase voltage is
measured
between the
neutral and any
line: line to
neutral voltage

Van
Vbn
Vcn
PHASE VOLTAGES,V
Van  VM 0
volt
Vbn  VM   120 volt
Vcn  VM 120
volt
LINE VOLTAGES, VL
Ia
Line voltage is
measured between
any two of the
three lines: line to
line voltage.
a

Van
Vab
ab
n
Vbn
Ib
b
Vcn
V
V bc
bc
Ic
c
VVcaca
LINE VOLTAGES,VL
Vab  Van  Vbn
Vbc  Vbn  Vcn
Vca  Vcn  Van
Vab  3VM 30
Vbc  3VM   90
Vca  3VM 150
Van  VM 0
volt
Vbn  VM   120 volt
Vcn  VM 120 volt
LINE
VOLTAGE
(VL)
PHASE
VOLTAGE (V)
Vab  3 VM 30 volt
Vbc  3 VM   90 volt
Vca  3 VM 150 volt
PHASE DIAGRAM OF VL AND V
Vca
Vcn
Vab
30°
120°
Vbn
Vbc
-Vbn
Van

PROPERTIES OF PHASE
VOLTAGE
All phase voltages have the same magnitude,
= V
= V
V  Van 

bn
cn

Out of phase with each other by 120
PROPERTIES OF LINE VOLTAGE

All line voltages have the same magnitude,
VL  Vab = Vbc = Vca

Out of phase with each other by 120
RELATIONSHIP BETWEEN V
and VL
1.
Magnitude
VL  3 V
2.
Phase
- VL LEAD their corresponding V by 30
VL  V  30
EXAMPLE 1

Calculate the line currents
DELTA CONNECTION
DELTA CONNECTED
SOURCES
DELTA CONNECTED
LOAD
OR
BALANCED -  CONNECTION

PHASE VOLTAGE AND LINE
VOLTAGE
In - system, line voltages equal to phase
voltages:
VL  Vφ
PHASE VOLTAGE,V

Phase voltages are equal to the voltages across
the load impedances.



PHASE CURRENTS, I

The phase currents are obtained:
VBC
VCA
VAB
I AB 
, I BC 
, I CA 
ZΔ
ZΔ
ZΔ

LINE CURRENTS, IL
The line currents are obtained from the phase
currents by applying KCL at nodes A,B, and C.



LINE CURRENTS, IL
I a  I AB  I CA
I b  I BC  I AB
I c  I CA  I BC
I a  3 I AB  30
I b  I a   120
I c  I a   120
PHASE
CURRENTS (I)
I AB
I BC
I CA
VAB

ZΔ
VBC

ZΔ
VCA

ZΔ
LINE CURRENTS (IL)
I a  3 I AB  30
I b  I a   120
I c  I a   120
PHASE DIAGRAM OF IL AND I

PROPERTIES OF PHASE
CURRENT
All phase currents have the same magnitude,
Iφ  I AB  I BC  ICA 

Vφ
ZΔ
Out of phase with each other by 120

PROPERTIES OF LINE
CURRENT
All line currents have the same magnitude,
I L  Ia  I b  Ic

Out of phase with each other by 120
1.
RELATIONSHIP BETWEEN I and
IL
Magnitude
I L  3 I
2.
Phase
- IL LAG their corresponding I by 30
I L  I  30
EXAMPLE
A balanced delta connected load having an
impedance 20-j15  is connected to a delta
connected, positive sequence generator having
Vab = 3300 V. Calculate the phase currents of
the load and the line currents.
Given Quantities
 ZΔ  20  j15   25  36.87
 Vab  3300
Phase Currents
VAB
3300
I AB 

 13.236.87A
ZΔ
25  36.87
I BC  I AB  120  13.2 - 83.13A
I CA  I AB  120  13.2156.87A
Line Currents
I a  I AB 3  30


 13.236.87 3  30 A
 22.866.87
I b  I a   120  22.86 - 113.13A
I c  I a   120  22.86126.87A
BALANCED WYE-DELTASYSTEM
THREE PHASE POWER
MEASUREMENT
Unbalanced load
In a three-phase four-wire system the line voltage is 400V
and non-inductive loads of 5 kW, 8 kW and 10 kW are
connected between the three conductors and the neutral.
Calculate:
(a) the current in each phase
(b) the current in the neutral conductor.
VL 400

 230V
3
3
Voltage to neutral
VP 
Current in 10kW resistor
PR 10 4
IR 

 43.5 A
VP 230
Current in 8kW resistor
PY 8 103
IY 

 34.8 A
VP
230
Current in 5kW resistor
PB 5 103
IB 

 21.7 A
VP
230
IR
INV
IN
IYH
IBH
IYV
IBV
IB
INH
IY
Resolve the current components into horizontal and vertical
components.
I H  IY cos 30  I B cos 30o  0.86634.8  21.7  11.3 A
IV  I R  I Y cos 60  I B cos 60o  43.5  0.5(34.8  21.7)  13.0 A
I N  I NH  I NV  11.32  13.02  17.2 A
2
2
A delta –connected load is arranged as in Figure below.
The supply voltage is 400V at 50Hz. Calculate:
(a)The phase currents;
(b)The line currents.
IR
R
400V 400V
Y
400V
R1=100
I1
IY
R2=20 I2
IB
X2=60
C=30F
I3
B
(a)
I1 
VRY 400

 4A
R1 100
I1 is in phase with VRY since there is only resistor in the branch
In branch between YB , there are two components , R2 and X2
I2 
VYB
400

 6.32 A
2
2
ZY
20  60
ZY  R 2  X 2
2
2
X
 Y  tan 1  2
 R2
 20  60
2
IR

400
1 /(2  50  30 10  6 )  90 
o

90
 3.77 A
I1
-I3

 60 
  tan 1    71 34'
 20 

In the branch RB , only capacitor in
it , so the XC is -90 out of phase.
V
I 3  BR
XC
VRY
2
30o
90o
71o34'
VBR
30o
I3
I2
VYB
(b)
I R  I1  I 3
I R2  I12  2 I1 I 3 cos   I 32
=30o
I1
I R2  4.0  24.03.77  cos 30o  3.77   56.3
2
2
I R  7.5 A
120o
 = 71o 34’ -60o= 11o 34’

I Y  I 2  I1
2
2
-I1
I2
I  I  2I1I 2 cos  I
2
Y
2
1
I Y2  6.32  24.06.32 cos11o34'4.0  105.5
2
IY  10.3 A
60o
71o 34'
2
IY
 = 180-30o-11o 34’ = 138o 34’
-I2
I B  I3  I 2
I B2  I 32  2 I 3 I 2 cos   I 22
I2
o
I  6.32  23.77 6.32 cos 138 26'3.77   18.5
2
B
I B  4.3 A
2
90o
11o 34'
2
71o 34'
30o
I3
I2
Power in three phase
Active power per phase = IPVP x power factor
Total active power= 3VPIP x power factor
P  3VP I P cos 
If IL and VL are rms values for line current and line voltage
respectively. Then for delta () connection: VP = VL and IP
= IL/3. therefore:
P  3VL I L cos 
For star connection () : VP = VL/3 and IP = IL. therefore:
P  3VL I L cos 
A three-phase motor operating off a 400V system is developing
20kW at an efficiency of 0.87 p.u and a power factor of 0.82.
Calculate:
(a)The line current;
(b)The phase current if the windings are delta-connected.
(a) Since Efficiency  output power in watts
input power in watts

output power in watts
3 I LVL  p. f
0.87 
20 1000
3  I L  400  0.82
And line current =IL=40.0A
(b) For a delta-connected winding
line current 40.0
Phase current 

 23.1A
3
3
Three identical coils, each having a resistance of 20 and
an inductance of 0.5 H connected in (a) star and (b) delta
to a three phase supply of 400 V; 50 Hz. Calculate the
current and the total power absorbed by both method of
connections.
First of all calculating the impedance of the coils
X P  2  50  0.5  157 
R P  20
Z P  RP  jX P  RP  X P 
2
2
XP
where   tan 
 RP
 157 

 20  157  tan 
  15883
 20 
2
2
1
cos   cos 83  0.1264
1 



Star connection
400V
20
400V
400V
VP  VL  400 V
0.5H
20
0.5H
20
0.5H
VP 400
IP 

 4.38A
Z P 158
P  3VL I L cos   3  400  4.38  0.1264  383W
A balanced three phase load connected in star, each phase consists
of resistance of 100  paralleled with a capacitance of 31.8 F.
The load is connected to a three phase supply of 415 V; 50 Hz.
Calculate:
(a) the line current;
(b) the power absorbed;
(c) total kVA;
(d) power factor .
415
VP 
VL
3

415
3
 240 V
Admittance of the load
1
1
YP 

R P XP

where
XP 
1
jC
1
1
 jC 
 j2  50  31.8  10 6  (0.01  j0.01)S
RP
100
Line current
I L  I P  VPYP  240(0.01  j0.01)  2.4  j 2.4  3.3945
Volt-ampere per phase



240

3
.
39

45

814
.
4

45
PVA  VP I P
Active power per phase
PPA  814.4 cos 45  576
Total active power
PA  3  576  1.728 kW
(b)

P

j
814
.
4
sin
45
 j576
Reactive power per phase PR
Total reactive power
(c)
(d)
Total volt-ampere
PR  j3  576  j1.728 kW
 3  814 .4  2.44 kVA
Power Factor = cos = cos 45 = 0.707 (leading)
A three phase star-connected system having a phase voltage of
230V and loads consist of non reactive resistance of 4 , 5 
and 6 respectively.
Calculate:(a) the current in each phase conductor
(b) the current in neutral conductor
and
(c) total power absorbed.
230
I 4 
 57.5A
4
230
I 5 
 46A
5
230
I 6 
 38.3A
6
38.3 A
57.5 A
(b)
46 A
X-component = 46 cos 30 + 38.3 cos 30 - 57.5 = 15.5 A
Y-component = 46 sin 30 - 38.3 sin 30 = 3.9 A
Therefore
(c)
I N  15.52  3.9 2  16A
P  230 57.5  46  38.3  32.61kW