3 phase ac

Three-Phase Voltages
(b) Balanced three-phase voltages
(a) The three windings on a cylindrical drum
used to obtain three-phase voltages
2 cos
2 cos( 120 )
2 cos( 240 )
aa
bb
cc
v V t
v V t
v V t







  
  

Three-Phase Voltages
Generator with six terminals

Three-Phase Balanced Voltages
0
120
240 120
aa
bb
cc
V
V
V V



  
   
       
V
V
V
Phase sequence or phase rotation is abc
Positive Phase Sequence
0
120
240 120
a
c
b
V
V
V V
  
   
       
V
V
V
Phase sequence or phase rotation is acb
Negative Phase Sequence

(a) Y-connected sources (b) -connected sources
Two Common Methods of Connection
Phase Voltage

Phase and Line Voltages
The line-to-line voltage Vab of
the Y-connected source
0 120
( 0.5 0.866)
3 30
ab a b
p p
p p
p
V V
V V j
V
 
      
   
  
V V V
Similarly 3 90
3 210
bc p
ca p
V
V
   
   
V
V

The Y-to-Y Circuit
A four-wire Y-to-Y circuit

The Y-to-Y Circuit (cont.)
, , anda b c
aA bB cC
A B C
  
V V V
I I I
Z Z Z
nN aA bB cC  I I I I
The average power delivered by the three-phase source to the three-phase load
A B CP P P P  
When ZA = ZB = ZC the load is said to be balanced
0 120 120
, , and
, ( 120 ), and ( 120 )
a b c
aA bB cC
A B C
aA bB cC
V V V
Z Z Z
V V V
Z Z Z
  
  
      
     
  
             
V V V
I I I
Z Z Z
I I I
Four - wire

0nN aA bB cC   I I I I
The Y-to-Y Circuit(cont.)
2
cos( ) cos( ) cos( )
3 cos( )
A B CP P P P
V V V
V V V
Z Z Z
V
Z
  

  
     

The average power delivered to the load is
There is no current in the wire connecting the neutral node of the source to the
neutral node of the load.

A three-wire Y-to-Y circuit

Three - wire We need to solve for VNn
0
0 120 120
a Nn b Nn c Nn
A B C
Nn Nn Nn
A B C
V V V
  
  
         
  
V V V V V V
Z Z Z
V V V
Z Z Z
Solve for VNn
( 120 ) 120 0A C A B B C
Nn
A C A B B C
V V V        

 
Z Z Z Z Z Z
V
Z Z Z Z Z Z
, , anda Nn b Nn c Nn
aA bB cC
A B C
  
  
V V V V V V
I I I
Z Z Z

When the circuit is balanced i.e. ZA = ZB = ZC
( 120 ) 120 0
0
Nn
V V V        

 

ZZ ZZ ZZ
V
ZZ ZZ ZZ
The average power delivered to the load is
2
3 cos( )
A B CP P P P
V
Z

  


A three-wire Y-to-Y circuit with line impedances
Transmission lines

The analysis of balanced Y-Y circuits is simpler than the
analysis of unbalanced Y-Y circuits.
VNn = 0. It is not necessary to solve for VNn .
The line currents have equal magnitudes and
differ in phase by 120 degree.
Equal power is absorbed by each impedance.
Per-phase equivalent circuit

Example 1 S = ?
rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V
50 80
50
100 25
A
B
C
j
j
j
  
 
  
Z
Z
Z
Unbalanced 4-wire
rms
rms
rms
110 0
1.16 58 A
50 80
110 120
2.2 150 A
50
110 120
1.07 106 A
100 25
a
aA
A
b
bB
B
c
cC
C
j
j
j
 
     

  
    
 
     

V
I
Z
V
I
Z
V
I
Z

Example 2 (cont.)
*
*
*
68 109 VA
242 VA
114 28 VA
A aA a
B bB b
C cC c
j
j
j
  
 
  
S I V
S I V
S I V
The total complex power delivered to the three-phase load is
182 379 VAA B C j    S S S S

Example 3 S = ?
rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V
50 80
50 80
50 80
A
B
C
j
j
j
  
  
  
Z
Z
Z
Balanced 4-wire
rms
110 0
1.16 58 A
50 80
a
aA
A j
 
     

V
I
Z
*
68 109 VAA aA a j  S I V
3 204 326 VAA j  S S
Also rms rms1.16 177 A , 1.16 62 AbB cC      I I
68 109 VAB Cj  S S

Example 4 S = ?
rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V
50 80
50
100 25
A
B
C
j
j
j
  
 
  
Z
Z
Z
Unbalanced 3-wireDetermine VNn
rms
(110 120 ) 110 120 110 0
56 151 V
A C A B B C
Nn
A C A B B C
        

 
   
Z Z Z Z Z Z
V
Z Z Z Z Z Z
, , anda Nn b Nn c Nn
aA bB cC
A B C
  
  
V V V V V V
I I I
Z Z Z

Example 3(cont.)
1.71 48 , 2.45 3 , and 1.19 79aA bB cC         I I I
* *
* *
* *
( ) 146 234 VA
( ) 94 VA
( ) 141 35 VA
A aA a aA aA A
B bB b bB bB B
C cC c cC cC C
j
j
j
   
  
   
S I V I I Z
S I V I I Z
S I V I I Z
287 364 VAA B C j    S S S S

Example 4
S = ? Balanced 3-wire
rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V
50 80
50 80
50 80
A
B
C
j
j
j
  
  
  
Z
Z
Z
rms
110 0
1.16 58 A
50 80
a
aA
A j
 
     

V
I
Z
*
68 109 VAA aA a j  S I V
3 204 326 VAA j  S S

The -Connected Source and Load
rms
rms
rms
120 0 V
120.1 121 V
120.2 121 V
ab
bc
ca
  
   
  
V
V
V
( )
3.75 A
1
ab bc ca 
  
V V V
I
Total resistance around the loop
Circulating
current
Unacceptable
Therefore the  sources connection is seldom used in practice.

The -Y and Y-  Transformation
AZ BZ
CZ
1Z
3Z
2Z
1 3
1 2 3
2 3
1 2 3
1 2
1 2 3
A
B
C


 
 

 
Z Z
Z Z Z
Z Z
Z Z Z
Z Z
Z Z
Z
Z
Z
Z
1
2
3
A B B C A C
B
A B B C A C
A
A B B C A C
C
 
 





Z Z Z Z Z Z
Z
Z Z Z Z Z Z
Z
Z
Z
Z
Z Z Z Z Z Z
Z

The Y-  Circuits
aA AB CA
bB BC AB
cC CA BC
 
 
 
I I I
I I I
I I I
where
3
1
2
AB
AB
BC
BC
CA
CA



V
I
Z
V
I
Z
V
I
Z

The Y-  Circuits (cont.)
cos sin cos( 120 ) sin( 120 )
3 ( 30 )
aA AB CA
I j I j
I
   

 
       
   
I I I
or 3 3aA L pI I I  I

Example 6 IP = ? IL = ?
rms
rms
rms
220
30 V
3
220
150 V
3
220
90 V
3
a
b
c
   
   
  
V
V
V
The -connected load is balanced with
10 50   Z
rms
rms
rms
220 0 V
220 120 V
220 240 V
AB a b
BC b c
CA c a
    
     
     
V V V
V V V
V V V
rms
rms
rms
22 50 A
22 70 A
22 190 A
AB
AB
BC
BC
CA
CA



   
    
    
V
I
Z
V
I
Z
V
I
Z

The line currents are
22 3 20 , 22 3 100 , 22 3 220aA AB CA bB cC          I I I I I

The Balanced Three-Phase Circuits
3
Y


Z
Z
Per-phase equivalent circuit
Y-to- circuit
Equivalent Y-to-Y circuit

Instantaneous and Average Power in BTP Circuits
One advantage of three-phase power is the smooth flow of
energy to the load.
 
 
2 2 2
2
2 2
2
( )
1 cos2 1 cos2( 120 ) 1 cos2( 240 )
2
3
cos2 cos(2 240 )cos(2 480 )
2 2
0
3
2
ab bc cav v v
p t
R R R
V
t t t
R
V V
t t t
R R
V
R
  
  
  
         
      

1 4 4 4 4 4 4 44 2 4 4 4 4 4 4 4 43
2
cos , cos( 120 ),
and cos( 240 )
(1 cos2 )
cos
2
ab bc
ca
v V t v V t
v V t
t
t
 



   
  


The instantaneous power

The total average power delivered to the balanced
Y-connected load is
,aA L AI AN P AVI V    I V
Phase A

The total average power delivered to the balanced
delta-connected load is

Two-Wattmeter Power Measurement
cc = current coil
vc = voltage coil
W1 read
1 1cosAB AP V I 
W2 read
2 2cosCB CP V I 
For balanced load with abc phase sequence
1 230 and 30a a        
is the angle between phase current and phase voltage of phasea a

1 2
2 cos cos30
3 cos
L L
L L
P P P
V I
V I


 
 

To determine the power factor angle
1 2 2cos cos30L LP P V I   
1 2 ( 2sin sin30 )L LP P V I    
1 2
1 2
2cos cos30 3
( 2sin sin30 ) tan
L L
L L
P P V I
P P V I

 
  
 
  
11 2 1 2
1 2 1 2
tan 3 or tan 3
P P P P
P P P P
     
    
  
Two-Wattmeter Power Measurement

Summary
 Three-Phase voltages
 The Y-to-Y Circuits
The -Connected Source and Load
 The Y-to-  Circuits
 Balanced Three-Phase Circuits
 Instantaneous and Average Power in Bal. 3 Load
 Two-Wattmeter Power Measurement

3 phase ac

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3 phase ac