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 THREE-PHASE POWER CIRCUITS
Prepared by:
•Meet Patel 140170119056
•Sindhi Parth 140170119057
•Soni Kunjan 140170119058
•Thakor Chetan 140170119059
•Thakor Milan 140170119060
 INTRODUCTION
Almost all electric power generation and most of the power
transmission in the world is in the form of three-phase AC
circuits. A three-phase AC system consists of three-phase
generators, transmission lines, and loads.
There are two major advantages of three-phase systems over
a single-phase system:
1) More power per kilogram of metal form a three-phase
machine;
2) Power delivered to a three-phase load is constant at all
time, instead of pulsing as it does in a single-phase system.
The first three-phase electrical system was patented in 1882
by John Hopkinson - British physicist, electrical engineer,
Fellow of the Royal Society.
1. GENERATION OF THREE-PHASE VOLTAGES AND
CURRENTS
A three-phase
generator consists of
three single-phase
generators with
voltages of equal
amplitudes and
phase differences of
1200.
Each of three-phase
generators can be
connected to one of three
identical loads.
This way the system would
consist of three single-phase
circuits differing in phase
angle by 1200.
The current flowing to each
load can be found as
V
I
Z

1. Generation of three-phase voltages and
currents
Therefore, the currents flowing in each phase are
0
0
0
0
120
120
240
240
A
B
A
V
I I
Z
V
I I
Z
V
I I
Z







  


   


   

(3.5.1)
(3.5.2)
(3.5.3)
We can connect the negative (ground) ends of the three single-
phase generators and loads together, so they share the common
return line (neutral).
The current flowing through a neutral can be found as
0 0
0 0 0 0
0 0 0 0
120 240
cos( ) sin( ) cos( 120 ) sin( 120 ) cos( 240 ) sin( 240 )
cos( ) cos( 120 ) cos( 240 ) sin( ) sin( 120 ) sin( 240 )
cos(
N A B CI I I I I I I
I jI I jI I jI
I jI
I
  
     
     

          
               
                     
  0 0 0 0
0 0 0 0
) cos( )cos(120 ) sin( )sin(120 ) cos( )cos(240 ) sin( )sin(240 )
sin( ) sin( )cos(120 ) cos( )sin(120 ) sin( )cos(240 ) cos( )sin(240 )jI
   
    
         
           
1 3 1 3
cos( ) cos( ) sin( ) cos( ) sin( )
2 2 2 2
1 3 1 3
sin( ) sin( ) cos( ) sin( ) cos( )
2 2 2 2
0
NI I
jI
    
    
 
          
 
 
          
 

Which is:
As long as the three loads are equal, the return current in
the neutral is zero!
Such three-phase power systems (equal magnitude, phase differences
of 1200, identical loads) are called balanced.
In a balanced system, the neutral is unnecessary!
Phase Sequence is the order in which the voltages in the individual
phases peak.
abc acb
VOLTAGES AND CURRENTS
There are two types of connections in three-phase circuits: Y and .
Each generator and each load can be either Y- or -connected. Any
number of Y- and -connected elements may be mixed in a power system.
Phase quantities - voltages and currents in a given phase.
Line quantities – voltages between the lines and currents in the lines
connected to the generators.
VOLTAGES AND CURRENTS
1. Y-connection
Assuming
a resistive
load…
VOLTAGES AND CURRENTS
1. Y-connection (cont)
0
0
0
0
120
240
an
bn
cn
V V
V V
V V



 
 
 
0
0
0
0
120
240
a
b
c
I I
I I
I I



 
  
  
Since we assume a resistive load:
VOLTAGES AND CURRENTS
1. Y-connection (cont 2)
The current in any line is the same as the current in the corresponding phase.
LI I
Voltages are:
0
0
0 1 3 3 3
0 120
2 2 2 2
3 1
3
2 2
303
aab bV V V V V V jV V j V
j V
V
V
     



 
              
 
 
  



Voltages and currents
1. Y-connection (cont 3)
3LLV V
Magnitudes of the line-to-line voltages and the line-to-neutral voltages are related as:
In addition, the line voltages are
shifted by 300 with respect to the
phase voltages.
In a connection with abc
sequence, the voltage of a
line leads the phase voltage.
Voltages and currents
1. -connection
0
0
0
0
120
240
ab
bc
ca
V V
V V
V V



 
 
 
0
0
0
0
120
240
ab
bc
ca
I I
I I
I I



 
 
 
assuming a
resistive load:
VOLTAGES AND CURRENTS
1. -connection (cont)
LLV V
The currents are:
0 0
0
1 3
0 240
2 2
3 3 3 1
3
2 2 2 2
3 30
ab caa I I I I I II
I
j I
I j I I j
    
  
 
          
 
 
        
 

3LI IThe magnitudes:
VOLTAGES AND CURRENTS
For the connections with the abc phase sequences, the current of a
line lags the corresponding phase current by 300 (see Figure below).
For the connections
with the acb phase
sequences, the line
current leads the
corresponding phase
current by 300.
POWER RELATIONSHIPS
The instantaneous power is:
( ) ( ) ( )p t v t i t
Therefore, the instantaneous power supplied to each phase is:
0 0
0 0
( ) ( ) ( ) 2 sin( )sin( )
( ) ( ) ( ) 2 sin( 120 )sin( 120 )
( ) ( ) ( ) 2 sin( 240 )sin( 240 )
a an a
b bn b
c cn c
p t v t i t VI t t
p t v t i t VI t t
p t v t i t VI t t
  
  
  
  
    
    
Since
 
1
sin sin cos( ) cos( )
2
        
POWER RELATIONSHIPS
Therefore
 
0
0
( ) cos cos(2 )
( ) cos cos(2 240 )
( ) cos cos(2 480 )
a
b
c
p t VI t
p t VI t
p t VI t
  
  
  
  
     
     
The total power on the load
( ) ( )( ) 3 cos( )tot a b cp t p tp t Vp t I   
The pulsing components cancel each other because of
1200 phase shifts.
Power relationships
The instantaneous
power in phases.
The total power
supplied to the load is
constant.
Power relationships
Phase quantities in each phase of a Y- or -connection.
2
3 cos 3 cosP V I I Z    
2
3 sin 3 sinQ V I I Z    
2
3 3S V I I Z   
Real
Reactive
Apparent
Note: these equations are valid for balanced loads only.
Power relationships
Line quantities: Y-connection.
Power consumed by a load: 3 cosP V I  
Since for this load 3L LLI I and V V    
Therefore: 3 cos
3
LL
L
V
P I 
Finally: 3 cosLL LP V I 
Note: these equations are valid for balanced loads only.
Power relationships
Line quantities: -connection.
Power consumed by a load: 3 cosP V I  
Since for this load 3L LLI I and V V    
Note: these equations were derived for a balanced load.
Therefore: 3 cos
3
L
LL
I
P V 
Finally: 3 cosLL LP V I 
Same as for a Y-connected load!
Power relationships
Line quantities: Y- and -connection.
3 sinLL LQ V I 
3 LL LS V I
Reactive power
Apparent power
Note:  is the angle between the phase voltage and the phase
current – the impedance angle.
sky

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AC CIRCUIT

  • 2.  THREE-PHASE POWER CIRCUITS Prepared by: •Meet Patel 140170119056 •Sindhi Parth 140170119057 •Soni Kunjan 140170119058 •Thakor Chetan 140170119059 •Thakor Milan 140170119060
  • 3.  INTRODUCTION Almost all electric power generation and most of the power transmission in the world is in the form of three-phase AC circuits. A three-phase AC system consists of three-phase generators, transmission lines, and loads. There are two major advantages of three-phase systems over a single-phase system: 1) More power per kilogram of metal form a three-phase machine; 2) Power delivered to a three-phase load is constant at all time, instead of pulsing as it does in a single-phase system. The first three-phase electrical system was patented in 1882 by John Hopkinson - British physicist, electrical engineer, Fellow of the Royal Society.
  • 4. 1. GENERATION OF THREE-PHASE VOLTAGES AND CURRENTS A three-phase generator consists of three single-phase generators with voltages of equal amplitudes and phase differences of 1200.
  • 5. Each of three-phase generators can be connected to one of three identical loads. This way the system would consist of three single-phase circuits differing in phase angle by 1200. The current flowing to each load can be found as V I Z 
  • 6. 1. Generation of three-phase voltages and currents Therefore, the currents flowing in each phase are 0 0 0 0 120 120 240 240 A B A V I I Z V I I Z V I I Z                        (3.5.1) (3.5.2) (3.5.3)
  • 7. We can connect the negative (ground) ends of the three single- phase generators and loads together, so they share the common return line (neutral).
  • 8. The current flowing through a neutral can be found as 0 0 0 0 0 0 0 0 0 0 120 240 cos( ) sin( ) cos( 120 ) sin( 120 ) cos( 240 ) sin( 240 ) cos( ) cos( 120 ) cos( 240 ) sin( ) sin( 120 ) sin( 240 ) cos( N A B CI I I I I I I I jI I jI I jI I jI I                                                                    0 0 0 0 0 0 0 0 ) cos( )cos(120 ) sin( )sin(120 ) cos( )cos(240 ) sin( )sin(240 ) sin( ) sin( )cos(120 ) cos( )sin(120 ) sin( )cos(240 ) cos( )sin(240 )jI                                1 3 1 3 cos( ) cos( ) sin( ) cos( ) sin( ) 2 2 2 2 1 3 1 3 sin( ) sin( ) cos( ) sin( ) cos( ) 2 2 2 2 0 NI I jI                                          Which is:
  • 9. As long as the three loads are equal, the return current in the neutral is zero! Such three-phase power systems (equal magnitude, phase differences of 1200, identical loads) are called balanced. In a balanced system, the neutral is unnecessary! Phase Sequence is the order in which the voltages in the individual phases peak. abc acb
  • 10. VOLTAGES AND CURRENTS There are two types of connections in three-phase circuits: Y and . Each generator and each load can be either Y- or -connected. Any number of Y- and -connected elements may be mixed in a power system. Phase quantities - voltages and currents in a given phase. Line quantities – voltages between the lines and currents in the lines connected to the generators.
  • 11. VOLTAGES AND CURRENTS 1. Y-connection Assuming a resistive load…
  • 12. VOLTAGES AND CURRENTS 1. Y-connection (cont) 0 0 0 0 120 240 an bn cn V V V V V V          0 0 0 0 120 240 a b c I I I I I I            Since we assume a resistive load:
  • 13. VOLTAGES AND CURRENTS 1. Y-connection (cont 2) The current in any line is the same as the current in the corresponding phase. LI I Voltages are: 0 0 0 1 3 3 3 0 120 2 2 2 2 3 1 3 2 2 303 aab bV V V V V V jV V j V j V V V                                    
  • 14. Voltages and currents 1. Y-connection (cont 3) 3LLV V Magnitudes of the line-to-line voltages and the line-to-neutral voltages are related as: In addition, the line voltages are shifted by 300 with respect to the phase voltages. In a connection with abc sequence, the voltage of a line leads the phase voltage.
  • 15. Voltages and currents 1. -connection 0 0 0 0 120 240 ab bc ca V V V V V V          0 0 0 0 120 240 ab bc ca I I I I I I          assuming a resistive load:
  • 16. VOLTAGES AND CURRENTS 1. -connection (cont) LLV V The currents are: 0 0 0 1 3 0 240 2 2 3 3 3 1 3 2 2 2 2 3 30 ab caa I I I I I II I j I I j I I j                                      3LI IThe magnitudes:
  • 17. VOLTAGES AND CURRENTS For the connections with the abc phase sequences, the current of a line lags the corresponding phase current by 300 (see Figure below). For the connections with the acb phase sequences, the line current leads the corresponding phase current by 300.
  • 18. POWER RELATIONSHIPS The instantaneous power is: ( ) ( ) ( )p t v t i t Therefore, the instantaneous power supplied to each phase is: 0 0 0 0 ( ) ( ) ( ) 2 sin( )sin( ) ( ) ( ) ( ) 2 sin( 120 )sin( 120 ) ( ) ( ) ( ) 2 sin( 240 )sin( 240 ) a an a b bn b c cn c p t v t i t VI t t p t v t i t VI t t p t v t i t VI t t                       Since   1 sin sin cos( ) cos( ) 2         
  • 19. POWER RELATIONSHIPS Therefore   0 0 ( ) cos cos(2 ) ( ) cos cos(2 240 ) ( ) cos cos(2 480 ) a b c p t VI t p t VI t p t VI t                         The total power on the load ( ) ( )( ) 3 cos( )tot a b cp t p tp t Vp t I    The pulsing components cancel each other because of 1200 phase shifts.
  • 20. Power relationships The instantaneous power in phases. The total power supplied to the load is constant.
  • 21. Power relationships Phase quantities in each phase of a Y- or -connection. 2 3 cos 3 cosP V I I Z     2 3 sin 3 sinQ V I I Z     2 3 3S V I I Z    Real Reactive Apparent Note: these equations are valid for balanced loads only.
  • 22. Power relationships Line quantities: Y-connection. Power consumed by a load: 3 cosP V I   Since for this load 3L LLI I and V V     Therefore: 3 cos 3 LL L V P I  Finally: 3 cosLL LP V I  Note: these equations are valid for balanced loads only.
  • 23. Power relationships Line quantities: -connection. Power consumed by a load: 3 cosP V I   Since for this load 3L LLI I and V V     Note: these equations were derived for a balanced load. Therefore: 3 cos 3 L LL I P V  Finally: 3 cosLL LP V I  Same as for a Y-connected load!
  • 24. Power relationships Line quantities: Y- and -connection. 3 sinLL LQ V I  3 LL LS V I Reactive power Apparent power Note:  is the angle between the phase voltage and the phase current – the impedance angle.
  • 25. sky