METU
Three Phase Systems
by
Prof. Dr. Osman SEVAİOĞLU
Electrical and Electronics Engineering Department
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 1
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Three Phase Systems
Why Three Phase ?
The reasons for using
Three Phase
Single Phase
Three Phase
The main reasons for using three
phase systems are;
1.
The kVA rating of the three
phase equipment (i.e.
machinary or transformer) is
is 150 % greater than that of
a single phase equipment
with the same frame
structure (weight , size)
This part of the metal work
is not utilized (wasted)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 2
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Three Phase Systems
Why Three Phase ?
The Reasons for using Three Phase
Single Phase
Please note that most of the iron part in the stator and
rotor are not utilized (wasted)
This part of the metal work
is not utilized (wasted)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 3
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Three Phase Systems
Why Three Phase ?
The Reasons for using Three Phase
Three Phase
Please note that most of the iron part in the stator and
rotor are now fully utilized (not wasted)
This part is now fully
utilized (not wasted)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 4
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Three Phase Systems
Why Three Phase ?
The Reasons for using Three Phase
The main reasons for using three phase systems are;
2. Conductor volume in a three phase system is
about 25-40 % less than that of a single phase
two-wire system with the same kVA rating.
Current: I
= 1.000.000 VA / (34.500 V x 0.85)
= 34,10 Amp
no. of conductors
Cross section = 6 mm2
Cond. volume = 1000 m x 2 x 6 x 10-6 = 0,012 m3
= 1.000.000 VA / (  3 x 34.500 V x 0.85)
= 19,69 Amp
no. of conductors
Cross section = 2,5 mm2
Cond. volume = 1000 x 3 x 2.5 x 10-6 = 0.0075 m3
Current: I
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 5
Three Phase Systems
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Three Phase Voltages
Circuit - a
Va(t) = Vmax cos wt
Ia(t) = Imax cos( wt - )
Circuit - b
Vb(t) = Vmax cos( wt -120o )
Ib(t) = Imax cos( wt -120o- )
Ia
+
Za = 3 + j4 
Va
=  =Tan-1 ( 4 / 3 )
= 53.13 o
Vc(t) = Vmax cos( wt -240o )
Ic(t) = Imax cos( wt -240o- )
Ic
Ib
+
Vb
n
Za
Circuit - c
Zb = 3 + j4 
+
Vc
n
Zb
=  = Tan-1 ( 4 / 3 )
= 53.13 o
Zc = 3 + j4 
n
Zc
=  = Tan-1 ( 4 / 3 )
= 53.13 o
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 6
Three Phase Systems
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Three Phase Voltages
Circuit - a
Va(t) = Vmax cos wt
Ia(t) = Imax cos( wt -53.13o )
Circuit - b
Vb(t) = Vmax cos( wt -120o )
Ib(t) = Imaxcos(wt -120o-53.13o)
Circuit - c
Vc(t) = Vmax cos( wt -240o )
Ic(t) = Imax cos(wt -240o-53.13o)
Va(t) 25,0
+
Ia(t)
Va(t) = Vmax cos wt
20,0
15,0
10,0
5,0
0,0 0
-5,0
-10,0
-15,0
-20,0
2
4
6
8
10
12
20
14 16 18
Time (msec)
2
4
6
8
10
12
16 18
20
14 Time (msec)
2
4
6
8
10
12
18 20
14 16
Time (msec)
Vb(t) 25,0
+
Ib(t)
Vb(t) = Vmax cos(wt-120o)
20,0
15,0
10,0
5,0
0,0 0
-5,0
-10,0
-15,0
-20,0
Vc(t) 25,0
+
Ic(t)
Vc(t) = Vmax cos(wt-240o)
20,0
15,0
10,0
5,0
0,0 0
-5,0
-10,0
-15,0
-20,0
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 7
Three Phase Systems
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Three Phase Voltages
Circuit - a
Va(t) = Vmax cos wt
Ia(t) = Imax cos( wt -53.13o )
Circuit - b
Vb(t) = Vmax cos( wt -120o )
Ib(t) = Imaxcos(wt -120o-53.13o)
Circuit - c
Vc(t) = Vmax cos( wt -240o )
Ic(t) = Imax cos(wt -240o-53.13o)
Va(t) 25,0
20,0
15,0
10,0
5,0
0,0 0
-5,0
-10,0
-15,0
-20,0
^
V
2
4
6
8
10
12
20
14 16 18
Time (msec)
-
Vb(t) 25,0
20,0
15,0
10,0
5,0
0,0 0
-5,0
-10,0
-15,0
-20,0
^
I
2
4
6
8
10
12
16 18
20
14 Time (msec)
Vc(t) 25,0
20,0
15,0
10,0
5,0
0,0 0
-5,0
-10,0
-15,0
-20,0
^
I
^
V
^I
2
4
6
8
10
12
18 20
14 16
Time (msec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 8
V^
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Three Phase Systems
Connection of Three Phase Voltages
Ia(t)
a
Now, re-draw the above three
phase circuits in an alternative
form as shown on the RHS
+
Va(t) = Vmax cos wt
Please note that the three
circuits are still electrically
unconnected
Vb(t) = Vmax cos(wt-120o)
+
Vc(t) = Vmax cos(wt-240o)
c
Ic(t)
+
b
Ib(t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 9
Three Phase Systems
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Connection of Three Phase Voltages
Phase Currents (Amp)
Ia(t)
a
Ia(t)
+
Va(t) = Vmax cos wt
0
2
4
6
8
10
12
16 18
Time (msec)
Ib(t)
Vb(t) = Vmax cos(wt-120o)
+
0
2
4
6
8
10
12
16 18
14 Time (msec)
Vc(t) = Vmax cos(wt-240o)
Ia + Ib + Ic
c
Ic(t)
+
b
0
2
4
6
8
10
12
14 16 18
Time (msec)
0
2
c
b
a
Ic(t)
Ib(t)
4
6
8
10
12
16
18
20
Time (msec)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 10
Three Phase Systems
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Three Phase Voltage Waveforms
Circuit - a
Circuit - b
Va(t) = Vmax cos wt
Circuit - c
Vc(t) = Vmax cos( wt -240o )
Vb(t) = Vmax cos( wt -120o )
Phase Voltages (Volts)
25
a
c
b
20
15
21.2 kV
10
5
0
-5
Time (msec)
0
2
4
6
8
10
12
14
14
16
18
-10
-15
-20
-25
20 / 3 = 6.67 msec
20 msec
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 11
20
Three Phase Systems
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Three Phase Voltage and Current Phasors
Three phase voltage and
current phasors may be drawn
as shown on the RHS
Tan-1 ( X/R ) = V – I = -53.13 o ↔ - 53.13 o /360 o x 20 sec.
Phase Currents (Amp)
30.0
^
Vc
^
Ic
b
c
20.0
10.0
120o
^
Va

^
Ib
a
Time (msec)
0.0
0,0
^
Ia
2,0
4,0
6,0
8,0
10,0
12,0
14,0
16,0
18,0 20,0
-10.0
^
Vb
-20.0
-30.0
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 12
Three Phase Systems
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Balanced Three Phase Circuits
^
Vc
^
Ic
Definition
120o
Please note that voltage and current phasors for
each phase are 120o displaced from each other

^
Ib
^
Ia
^
Vb
(a) │Va│ = │Vb│ = │Vc│
(b) Va - Vb = Vb – Vc = Vc - Va = 120o
Ia(t)
a
(c) │Ia│ = │Ib│ = │Ic│
(d) Ia - Ib = Ib – Ic = Ic -
^
Va
+
Ia =
Va(t) = Vmax cos wt
120o
or
(e) Va + Vb + Vc = 0
(f) Ia + Ib + Ic = 0
Vb(t) = Vmax cos(wt-120o)
+
Vc(t) = Vmax cos(wt-240o)
c
Ic(t)
+
b
Ib(t)
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 13
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Three Phase Systems
Balanced Three Phase Circuits
Definition
^
Vc
^
Ic
A three phase system satisfying the
above condition is said to be
“balanced”
120o
In a balanced three phase system,
• sum of phase currents is zero,
• sum of phase voltages is zero
Va + Vb + Vc = 0
Ia + Ib + Ic = 0
^
Va

^
Ib
^
Ia
^
Vb
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 14
Three Phase Systems
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Balanced Three Phase Circuits
Balance Condition
Please note that the central point does not move
Va + Vb + Vc = 0
Ia + Ib + Ic = 0
^
Vc
^
Ic
120o
120o
^
Va

^
Ib
^
Ia
^
Vb
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 15
Three Phase Systems
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Balanced Three Phase Circuits
In a balanced three phase system, sum of
currents at any instant is always zero, i.e.
Ia + Ib + Ic = 0
+
Va(t) = Vmax cos wt
Ia + Ib + Ic = 0
Tan-1 (X / R) = -53.13 o
Vb(t) = Vmax cos(wt-120o)
Phase Currents (Amp)
30.0
a
Ia(t)
a
+
b
Vc(t) = Vmax cos(wt-240o)
c
c
Ic(t)
+
b
Ib(t)
20.0
10.0
0.0
Time (msec)
0,0
2,0
4,0
6,0
8,0
10,0
12,0
14,0
16,0
18,0
20,0
-10.0
-20.0
-30.0
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 16
Three Phase Systems
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Balanced Three Phase Circuits
a
In a balanced three phase system, no current
will flow if we connect the ground wires of the
above three circuits
Ia + Ib + Ic = 0
Phase Currents (Amp)
30.0
a
Ia(t)
+
Va(t) = Vmax cos wt
Ia(t) + Ib(t) + Ic(t) = 0
Vb(t) = Vmax cos(wt-120o)
Ia + Ib + Ic = 0
b
c
20.0
Vb(t) = Vmax cos(wt-240o)
c
10.0
0.0
+
+
Ic(t)
b
Ib(t)
Time (msec)
0,0
2,0
4,0
6,0
8,0
10,0
12,0
14,0
16,0
18,0
20,0
-10.0
-20.0
-30.0
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 17
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Three Phase Systems
Three Phase Measurement-Energy Analzer
Three Phase Energy
Analyzer
Energy analyzer shown on the RHS is
capable of reading and recording
three phase voltages and currents in
rms, peak and time - waveform and
transmitting the resulting data to
computer
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 18
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Three Phase Systems
Three Phase Measurement-Energy Analzer
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 19
Three Phase Systems
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Three Phase Circuit Breaker
Three Phase Circuit Breaker
Three phase low voltage circuit breaker is a
device that breaks the three phases of power
service automatically or manually
This dashed line implies that poles
operate in “gang” manner
a
b
c
a
b
c
“Gang” Mechanism
Three Phase Load
a
b
c
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 20
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Three Phase Systems
Three Phase Measurement
Three Phase Power Analyzer
Device shown on the RHS is capable of
reading and recording three phase voltages
and currents in rms, peak and timewaveform and transmitting the resulting
data to computer
Clamp type Current Transformers
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 21
Three Phase Systems
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Three Phase Circuit
Phase - a
Va(t) = Vmax cos wt
Phase - b
Vb(t) = Vmax cos( wt -120o )
Ia
+
Va
Phase - c
Vc(t) = Vmax cos( wt -240o )
Ic
Ib
Load-a
+
Vb
Load-b
+
Vc
Load-c
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 22
Three Phase Systems
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Three Phase Circuit Connection
Phase - a
Phase - b
Va(t) = Vmax cos wt
Vb(t) = Vmax cos( wt -120o )
Phase - c
Vc(t) = Vmax cos( wt -240o )
Ia
Ib
Va
+
Vb
+
Vc
+
Ic
Loadb
Loadc
Loada
Ia(t) + Ib(t) + Ic(t) = 0
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 23
Three Phase Systems
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Three Phase Circuit Connection
n1
Basic Diagram
a
b
c
a
c
c
b
+
c
+
b
b
n2
a
a
+
Ia(t) + Ib(t) + Ic(t) = 0
n1, n2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 24
Three Phase Systems
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Three Phase Cable
a
Basic Diagram
b
c
n
a
b
c
a
+
c
c
b
+
b
a
+
Ia(t) + Ib(t) + Ic(t) = 0
n1, n2
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 25
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Three Phase Systems
Turkish 380 kV System
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 26
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Three Phase Systems
Part of Turkish 154 kV System
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 27
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Three Phase Systems
Ankara 154 kV Ring
Please note that the
lines form a closed ring
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 28
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Three Phase Systems
Istanbul Anatolian Side MV System
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 29
Three Phase Systems
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Three Phase Synchronous Generator
Phase Voltage (kolts)
25
20
15
10
5
0 0
2
4
6
8
10 12 14 16 18 20
-15
10
-20
5
-5
20
-10
15
0
25
-5
-25
0
2
4
6
8
10 12 14 16 18
Time ( msec )
20
-10
-15
-20
25
-25
Time ( msec )
b
a
Phase Voltage (kolts)
Phase Voltage (kolts)
c
20
15
10
5
0
-5
0
2
4
6
8
10 12 14 16 18 20
-10
-15
-20
-25
Time ( msec )
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 30
Three Phase Systems
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Three Phase Generation System
n
Three Phase Generator
a
c
c
b
a
a
b
c
b
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 31
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Three Phase Systems
Karakaya Hydroelectric Plant – 1800 MW
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 32
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Three Phase Systems
Hydroelectric Plant - Sectional View
Configuration
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 33
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Three Phase Systems
Hydroelectric Plant - Sectional View
Typical Hydoelectric Plant
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 34
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Three Phase Systems
Atatürk Hydroelectric Plant; 8 x 300 = 2400 MW
Penstock
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 35
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Three Phase Systems
Hydroelectric Plant - Water Turbine
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 36
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Three Phase Systems
Generation of AC Voltage - Synchronous Generator
Bagnell Dam on Ozarks Lake
Turbine - Generator Set
Water Flow
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 37
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Three Phase Systems
Atatürk Dam Generator Room
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 38
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Three Phase Systems
Itaipu Power Plant - 12500 MW
Stator Mounting Ceremony
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 39
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Three Phase Systems
Combined Cycle Power Plant
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 40
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Three Phase Systems
Thermal (Coal) Power Plant
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 41
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Three Phase Systems
Parallel Operation of Plants (Double Bus Configuration)
Units
Open circuit breaker
Main Bus
Spare Bus
Load and Feeder Group
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 42
Three Phase Systems
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Phase and Line Voltages
n1
Definition
• Voltage between a phase conductor and
ground is called phase voltage,
• Voltage between two phase conductors
is called line voltage
a
b
c
a
+
+
Phase Voltage
c
b
a
Line Voltage
c
c
b
n2
b a
+
Ia(t) + Ib(t) + Ic(t) = 0
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 43
Three Phase Systems
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Phase and Line Voltages
Definition
Birds prefer neutral wire since the voltage is zero
• Voltage between a phase conductor and
ground is called phase voltage,
• Voltage between two phase conductors
is called line voltage
a
b
c
a
+
c
c
b
+
b a
+
Ia(t) + Ib(t) + Ic(t) = 0
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 44
Three Phase Systems
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Phase and Line Voltages
Definition
Netral wire acts as lightning arrestor
• Voltage between a phase conductor and
ground is called phase voltage,
• Voltage between two phase conductors
is called line voltage
a
b
c
a
+
c
c
b
+
b a
+
Ia(t) + Ib(t) + Ic(t) = 0
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 45
Three Phase Systems
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Phase and Line Voltages
Definition
Van = Phase Voltage
Vab = Line Voltage
a
• Voltage between a phase conductor and
ground is called phase voltage,
• Voltage between two phase conductors
is called line voltage
b
c
n2
n1
Van = Phase Voltage
c
b
a
a
+
c
c
b
+
b
a
+
Vab = Line Voltage
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 46
Three Phase Systems
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Relation between Phase and Line Voltages
Definition
^
Vc
Vab = Va – Vb = Va + ( -Vb )
Van = Phase Voltage
^
-Vb
Vab = Line Voltage
30o
a
^
Va
b
c
a
b
+
c
c
+
b
a
^
Vb
+
n
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 47
Three Phase Systems
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Relation between Phase and Line Voltages
Definition
Vab / 2 = Va + (-Vb)/ 2
Vab = Va – Vb = Va + ( -Vb )
^
Vab
-Vb
Vab / 2 =  Va  cos 30 =  Va   3 / 2
Vab =  Va   3
Line Voltage =  3
^
Vc
x Phase Voltage
^
-Vb
30o
30o
^
Va
^
Va
^
Vb
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 48
Three Phase Systems
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Example - Turkish HV System
n1
Example (Turkish HV System)
Line Voltage =  3
n2
x Phase Voltage
Phase voltage (rms) = 220 kV
Line voltage (rms) = 220 x  3 = 380 kV
c
b
a
Phase Voltage =220 kV
Vab / 2 = Va + (-Vb)/ 2
Vab
^
-Vb
30o
Line Voltage = 380 kV
^
Va
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 49
Three Phase Systems
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Star (Y) Connection
Definition
Alternative Representation
Star-connected configuration is the one
with all the neutral points are connected to
a common ground point
Line Voltage =  3 x Phase Voltage
c
+
a
b
c
+
c
c
b
a
n
+
+
+
b
Van = Phase Voltage
n
b
+
a
Vac = Line Voltage
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 50
Three Phase Systems
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Star (Y) Connected Generator
Definition
Alternative Representation
Star-connected generator is the one with
all the generator neutral points are
connected to a common ground point
Line Voltage =  3 x Phase Voltage
c
+
+
c
Vac = Line Voltage
Van = Phase Voltage
c
n
+ a
a
b
+
+
+
b
b
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 51
n
Three Phase Systems
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Delta () Connected Generator
Definition
Configuration
Line Voltage = Phase Voltage
Delta-connected generator is the one
with the generator terminals are so
connected that they form a triangle
(delta)-configuration
c
c
Vac = Line Voltage
Phase Voltage
+
c
+
+
+
b
+
+
a
a
a
b
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 52
b
Three Phase Systems
METU
Currents in a Delta () Connected Generator
Definition
Configuration
Ic
Ia = Ica – Iab
Ib = Iab – Ibc
Ic = Ibc – Ica
c
c
Ibc
+
Ica = Phase Current
Ica
+
b
Ia = Line Current
Ia
+
a
Iab
a
Ib
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 53
Three Phase Systems
METU
Currents in a Delta () Connected Generator
Relation between Line and Phase Currents
Ia
Ia = Ica + ( – Iab )
Ia = Line Current
Iba = Phase Current
Ic
c
c
Ibc
^
Ica
^
- Iab
^
Vab
+
^
Ibc
+
Ica
+
b
^
Vca
Iab
a
Ia
Ib
a
^
Vbc
^
Iab
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 54
METU
Three Phase Systems
Currents in a Delta () Connected Generator
Definition
Ia = Ica + ( – Iab )
Relation between Line and
Phase Currents
^
Ia
^
^
^
Iline = Iba - Iac
^
Iline / 2
Iline  / 2 =  I phase  x cos 30o
=  / phase  x  3 / 2
Line Current =  3 x Phase Current
^
Vca
^
Ica
^
- Iab
^
Vab
^
Ibc
^
Vbc
^
Iab
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 55
Three Phase Systems
METU
Summary
Star (Y) Connected System
Line Voltage =  3 x Phase Voltage
Line Current = Phase Current
Delta () Connected System
Line Voltage = Phase Voltage
Line Current =  3 x Phase Current
Ia = Line Current
Iba = Phase Current
b
+
Ic
c
c
Vab = Line Voltage
Ibc
b
Van = Phase Voltage
+
+
+
n
+
Ica
+
a
b
Iab
a
Ia
Ib
a
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 56
Three Phase Systems
METU
Star-Star (Y-Y) Connected Systems
Definition
Configuration
A star-star connected system is the one with
both ends are star connected
Please note that neutral wire does not carry
any current for a balanced Y-Y Syatem
Ic
c
+
+
c
Please note that the neutral wire is not really essential,
since it carries no current. Hence it can be placed.
b
+
a
Ia
Ib
b
+
n
+
+
n
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 57
Three Phase Systems
METU
Star-Delta (Y- ) Connected Systems
Definition
A star-delta connected system is the one with
the source is star and load is delta connected
Configuration
Please note that neutral wire does not exist
in a system with one or both ends delta
c
c
+
Ibc
Ica
+
n
b
b
+
+
+
+ Iab
a
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 58
Three Phase Systems
METU
Delta-Star ( -Y) Connected Systems
Definition
A delta-star connected system is the one with
the source is delta and load is star connected
Configuration
Please note that neutral wire does not exist
in a system with one or both ends delta
Ic
c
+
c
+
+ Iab
b
n
Ica
a
Ib
b
+
+
+
Ibc
a
Ia
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 59
Three Phase Systems
METU
Three Phase Power in Star Connected Loads
Line Voltage =  3 x Phase Voltage
Line Current = Phase Current
Power Per Phase
c
+
Sph = Pph + j Qph
= Vph Iph*
Vac = Line
Voltage
Since we have three phases with
identical power consumption, total
power consumption becomes
+
S3-ph = 3 x Sph
= 3 Vph Iph*
= 3 (Vline /  3 ) Iline
Van = Phase
Voltage
n
b
+
Star (Y) Connected System
a
S3-ph =  3 Vline Iline
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 60
METU
Three Phase Systems
Three Phase Active Power in Star Connected Loads
Line Voltage =  3 x Phase Voltage
Line Current = Phase Current
Remember that;
c
+
Pph = Sph cos 
Since we have three phases with
identical power consumption, total
power consumption becomes
Vac = Line
Voltage
P3-ph = 3 x Sph cos 
= 3 Vph Iph cos 
= 3 (Vline /  3 ) Iline cos 
P3-ph =  3 Vline Iline cos 
Van = Phase
Voltage
+
n
b
+
Star (Y) Connected System
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 61
METU
Three Phase Systems
Three Phase Reactive Power in Star Connected Loads
Star (Y) Connected System
Line Voltage =  3 x Phase Voltage
Line Current = Phase Current
Remember that;
c
+
Qph = Sph sin 
Since we have three phases with
identical power consumption, total
power consumption becomes
Vac = Line
Voltage
Q3-ph = 3 x Sph sin 
= 3 Vph Iph sin 
= 3 (Vline /  3 ) Iline sin 
b
+
+
Q3-ph =  3 Vline Iline sin 
Van = Phase
Voltage
n
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 62
METU
Three Phase Systems
Three Phase Reactor
Three Phase Reactor
Line Voltage =  3 x Phase Voltage
Line Current = Phase Current
Remember that;
Q3-ph = 3 x Sph sin 
= 3 Vph Iph sin 
= 3 (Vline /  3 ) Iline sin 
Phase - a
Q3-ph =  3 Vline Iline sin 
Phase - b
Phase - c
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 63
METU
Three Phase Systems
Three Phase Reactor
Small-Size Three Phase
Reactor
Remember that;
Line Voltage =  3 x Phase Voltage
Line Current = Phase Current
Phase - a
Phase - b
Phase - c
Q3-ph = 3 x Sph sin 
= 3 Vph Iph sin 
= 3 ( Vline /  3 ) Iline sin 
Q3-ph =  3 Vline Iline sin 
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 64
METU
Three Phase Systems
Three Phase Transformer
Primary Side (Delta)
Secondary Side (Star)
Primary Side Phase - a
Secondary Side Phase - a
Primary Side Phase - b
Secondary Side Phase - b
Primary Side Phase - c
Secondary Side Phase - c
Voltages
Primary Side
Vline = 34500 Volts
Vphase = Vline = 34500 Volt
Secondary Side
Vline = 380 Volts
Vphase = 380 / √ 3 = 220 Volts
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 65
Three Phase Systems
METU
Three Phase Transformer
Secondary Side (Star)
Primary Side (Delta)
Primary Side Phase - a
Secondary Side Phase - a
Primary Side Phase - b
Secondary Side Phase - b
Primary Side Phase - c
Secondary Side Phase - c
Please note that
almost all distribution
transformers are deltastar connected
c
Isec.-a
Ic
Vsec.-a
c
Ica
Ibc
Vsec.-a
+
n
b
Ib
a
Ia
Iab
Vsec.-b
a
b
Vsec.-c
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 66
METU
Three Phase Systems
Three Phase Transformer
Three-Phase Power
(Overview)
Pprim. - a
= Va Ia cos
Pprim. - b
= Vb Ib cos
Pprim. - c
= Vc Ic cos
+
Pprim. - Total = Va Ia cos + Vb Ib cos + Vc Ic cos
= 3 Vphase Iphase cos
= 3 V line Iline / √3 cos
= √3 V line Iline cos
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 67
METU
Three Phase Systems
Three Phase Transformer
Three-Phase Power
(Overview)
Power on the Primary Side
PPrim. - Total = √ 3 V Prim.- line IPrim.-line cos
QPrim. - Total = √ 3 V Prim.- line IPrim.-line sin
SPrim. - Total = √ 3 V Prim.- line IPrim.-line
Power on the Secondary Side
PSec. - Total
= √ 3 V Sec.- line ISec.-line cos
QSec. - Total
= √ 3 V Sec.- line ISec.-line sin
SSec. - Total
= √ 3 V Sec.- line Isec.-line
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 68
Three Phase Systems
METU
Example: Star (Y) Connected Load
Line Voltage =  3 x Phase Voltage
Line Current = Phase Current
135 MVA shunt reactor
delivered to Nevada Power
Company by VA TECH ELIN
Line Voltage
+
c
Vac = Line Voltage
Van = Phase Voltage
b
+
+
n
a
Phase Voltage
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 69
METU
Three Phase Systems
You do not seem to understand the line
voltage in a Y- Connected Load ...
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 70
Three Phase Systems
METU
Three Phase Power in Delta Connected Loads
Delta () Connected System
Power Per Phase
Sph = Pph + j Qph
= Vph Iph*
Line Voltage = Phase Voltage
Line Current =  3 x Phase Current
Ia = Line Current
c
Since we have three phases with
identical power consumptions, three
phase (total) power consumption
becomes
S3-ph = 3 x Sph
= 3 Vph Iph*
= 3 Vline (Iline /  3 )
b
S3-ph =  3 Vline Iline
a
Iba = Phase Current
c
Ic
Ibc
Ica
Ib
Ia
b
Iab
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 71
a
Three Phase Systems
METU
Three Phase Power in Delta Connected Loads
Delta () Connected System
Power Per Phase
Pph = Vph Iph cos 
Line Voltage = Phase Voltage
Line Current =  3 x Phase Current
Ia = Line Current
c
Since we have three phases with
identical power consumptions, three
phase (total) power consumption
becomes
P3-ph = 3 x Pph
= 3 Vph Iph cos 
= 3 Vline ( Iline /  3 ) cos 
P3-ph =  3 Vline Iline cos 
Iba = Phase Current
c
Ic
Ibc
Icb
Ib
b
Ia
b
Iab
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 72
a
Three Phase Systems
METU
Three Phase Power in Delta Connected Loads
Delta () Connected System
Power Per Phase
Qph = Vph Iph sin 
Line Voltage = Phase Voltage
Line Current =  3 x Phase Current
Ia = Line Current
c
Since we have three phases with
identical power consumptions, three
phase (total) power consumption
becomes
Q3-ph = 3 x Qph
= 3 Vph Iph sin 
= 3 Vline ( Iline /  3 ) sin 
Q3-ph =  3 Vline Iline sin 
Iba = Phase Current
c
Ic
Ibc
Ica
Ib
b
Ia
b
Iab
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 73
a
Three Phase Systems
METU
Three Phase Power - Summary
Star (Y) Connected System
Delta () Connected System
Three phase power expressions are identical for star and delta connections
S3-ph =  3 Vline Iline
P3-ph =  3 Vline Iline cos 
c
+
c
n
b
+
+
b
a
a
Q3-ph =  3 Vline Iline sin 
c
Ic
Ibc
Ica
Ib
Ia
b
Iab
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 74
Three Phase Systems
METU
Star - Delta Conversion
Formulation
zab = ( za zb + zb zc + zc za ) / zc
A star connected load can be converted
to a delta connected load as follows
c
zca = ( za zb + zb zc + zc za ) / zb
zbc = ( za zb + zb zc + zc za ) / za
c
c
Zc
c
Please note that the neutral
node is now eliminated
Zbc
Zca
n
Za
Zb
b
a
b
b
b
a
Zab
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 75
Three Phase Systems
METU
Star - Delta Conversion
If all impedances of star are identical,
then the formula reduces to the
following simple form
Simplification
z = ( zY2 + zY2 + zY2 ) / zY = 3 zY
Please note that the neutral
node is eliminated by transformation,
i.e. no. of nodes is reduced by one.
c
c
c
c
ZY
3ZY
3ZY
ZY
a
b
ZY
a
b
a
b
3ZY
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 76
Three Phase Systems
METU
Delta - Star Conversion
Formulation
za = zba zac / ( zba + zac + zcb )
A delta connected load can be converted
to a star connected load as follows
zb = zcb zba / ( zba + zac + zcb )
zc = zac zcb / ( zba + zac + zcb )
c
c
c
c
Zc
Zac
Zcb
n
a
Za
a
b
Zba
b
a
Zb
a
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 77
b
Three Phase Systems
METU
Delta - Star Conversion
If all impedances of delta are identical,
then the formula reduces to the
following special simple form
c
c
Z/3
Z
a
Z/3
b
Z
b
zY = z  2 / ( z + z + z ) = z  / 3
c
c
Z
a
Simplification
a
a
Z/3
b
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 78
METU
Three Phase Systems
Solution Procedure for Three Phase Problems
Procedure
1. First convert all the -connected
loads, if any, to Y- connected loads by
employing the Delta - Star Conversion
Technique given in the previous
section,
2. Find the source voltages / phase by
dividing all the line voltages of the
sources by 3 for the Y-connected
sources
3. Decompose the given three phase
system into three independent
(electrically unconnected) single
phase systems
Phase Voltage =220 kV
Line Voltage = 380 kV
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 79
METU
Three Phase Systems
Solution Procedure for Three Phase Problems
Procedure (Continued)
4. Then solve one of these single phase
systems, i.e. in particular, the one
which corresponds to phase-a,
5. Calculate active and reactive powers
and power losses per phase,
Phase Voltage =220 kV
Line Voltage = 380 kV
6. Finally, multiply;
a) all these active and reactive
powers by three in order to find
the three phase powers,
b) all voltages by  3 in order to
find the resulting line voltages
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 80
Three Phase Systems
METU
Example
Problem
Solve the following three phase system for
line currents and line losses
c
r + j x = 1+j 2 
IL = Line current
Phase Voltage =220 kV
c
Line Voltage = 380 kV
n
Vs(line) = 6.3 kV (rms)
b
b
r + j x = 9 +j 12 
a
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 81
Three Phase Systems
METU
Example
1. First convert all the -connected loads to
Y- connected loads, if any, by employing
the Delta - Star Conversion Technique
given in the previous section
Z
a
Z/3
Z/3
b
a
c
b
c
Z
Z
b
c
c
c
n
n
Solution
a
Z/3
a
b
b
Z / 3 = (9 + j 12) / 3 = 3 + j 4 
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 82
Three Phase Systems
METU
Example
Problem
Solve the following three phase system for
line currents and line losses
c
IL = Line current
r +jx=1+j2
Phase Voltage =220 kV
Z/3 = 3 + j 4
Line Voltage = 380 kV
c
Neutral Wire
Vs(line) = 6.3 kV (rms)
n
a
+
b
+
n
a
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 83
Three Phase Systems
METU
Example
Solution
2. find the source voltages / phase
3. Divide all the line voltages by 3 for Yconnected sources in order to
c
Vs phase = Vs line /  3
= 6300 / 1.732
= 3637.41 Volts (rms)
+
c
b
c
b
a
+
Vs(line) = 6.3 kV
+
n
n
n
a
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 84
Three Phase Systems
METU
Example
n
n
Solution
3. Decompose the given three phase system
into three independent (electrically
unconnected) single phase systems
c
a
b
a
Ia(t)
+
Va(t) = Vmax cos wt
Vb(t) = Vmax cos(wt-120o)
+
Vc(t) = Vmax cos(wt-240o)
c
Ic(t)
+
Ib(t)
b
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 85
Three Phase Systems
METU
Example
4.
Then, solve one of the resulting three single phase
systems, i.e. in particular, the one which corresponds
to phase-a, with zero phase angle, due to its
simplicity
Ia
+
n
n
Solution
c
b
a
r +jx=1+j2
Va = 3637.41 0o Volts
Z / 3 = 3 + j 4 
Iline = Va - phase / Z tot
= 3637.41 0o /  ( 1 + j2 ) + ( 3 + j4 )
= 3637.41 0o / (4 + j6) = 3637.41 0o / 8.9442 56.31o
= 406.67 -56.31o Amp
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 86
Three Phase Systems
METU
Example
n
n
Solution
5. Now, calculate;
a) Active and reactive power losses
r + jx = 1 + j2 
c
Ia = 406.67 -56.31o Amp
b
a
+
Va = 3637.41 0o Volts
Z / 3 = 3 + j 4 
Line Losses;
rline I2 = 1 x 406.67 2 = 165380 Watts / phase
= 165.38 kWs / phase
xline I2 = 2 x 406.67 2 = 330760 Vars / phase
= 330.76 kVARs / phase
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 87
Three Phase Systems
METU
Example
Solution
n
6. Finally, multiply;
a) Active and reactive power losses by
three in order to find the three phase
power losses
Ia = 406.67 -56.31o Amp
n
r + jx = 1 + j2 
b
+
Va = 3637.41 0o Volts
Z / 3 = 3 + j 4 
a
c
Three phase power losses;
Active power loss = 3 x active power loss /phase
= 3 x 165.38 = 496.14 kWs
Reactive Power loss = 3 x 330.76 = 992.28 kVARs
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 88
Three Phase Systems
METU
Example
n
n
Solution
7. Now, calculate the active and reactive
powers consumed by the load;
Ia = 406.67 -56.31o Amp
c
r + jx = 1 + j2 
+
Va = 3637.41 0o Volts
b
a
Z / 3 = 3 + j 4 
Active and reactive power consumptions;
rload I2 = 3 x 406.67 2 = 496140 Watts / phase
= 496.14 kWs / phase
xload I2 = 4 x 406.67 2 = 661520 Vars / phase
= 661.52 kVARs / phase
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 89
METU
Three Phase Systems
Example
Solution
8. Now, calculate the three phase active and
reactive powers consumed by the load;
a
b
Ia = 406.67 -56.31o Amp
r + jx = 1 + j2 
+
Va = 3637.41 0o Volts
Z / 3 = 3 + j 4 
c
Three phase active and reactive power
consumptions;
3 x rload I2 = 3 x 496.14 = 1488.42 kWs
3 x xload I2 = 3 x 661.52 = 1984.56 kVARs
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 90
METU
Three Phase Systems
Example
Solution
9. Now, calculate the load voltage / phase
Ia = 406.67 -56.31o Amp
r + jx = 1 + j2 
Ia = 406.67 -56.31o Amp
a
b
+
Va = 3637.41 0o Volts
Vload / phase
Z / 3 = 3 + j 4 
c
Load Voltage/phase (Y-connected load)
V Load/phase
= Z / 3 x Ia
= (3 + j4) x 406.67 -56.31o
= 5 53.13o x 406.67 -56.31o
= 2033.35 -3.18o Volts/phase
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 91
Three Phase Systems
METU
Example
Solution
Vab =
r + jx = 1 + j2 
Ia = 406.67 -56.31o Amp
+ 30o
^
-Vb
10. Now, calculate the load voltage / line
Ia = 406.67 -56.31o Amp
Va
30o
Vab
^
Va
+
Va = 3637.41
0o Volts
Vload / phase
Z / 3 = 3 + j 4 
Load Voltage / line (Y-connected load)
V Load/line
=  3 x 2033.35 -3.18o + 30o
= 3521.76 26.82o Volts / line
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 92
Three Phase Systems
METU
Example
Solution
Please note that phase voltage across the
delta is the same as line voltage, i.e.
c
Vload / line = 3521.76 26.82o Volts
b
a
c
Z
Z
b
a
Z
Load Voltage / line (Y-connected load)
V Load/line
= V Load / phase
= 3521.76 26.82o Volts/line
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 93
METU
Three Phase Systems
Did eveybody understand three
phase systems ?
EE 209 Fundamentals of Electrical and Electronics Engineering, Prof. Dr. O. SEVAİOĞLU, Page 94