Three-phase systems
1. Introduction
Three-phase systems are commonly used in generation, transmission and distribution of
electric power. Power in a three-phase system is constant rather than pulsating and three-phase
motors start and run much better than single-phase motors. A three-phase system is a
generator-load pair in which the generator produces three sinusoidal voltages of equal
amplitude and frequency but differing in phase by 120 from each other.
The phase voltages va(t), vb(t) and vc(t) are as follows
v a  Vm cos t


cost  240  ,
v b V m cos t  120 
v c  Vm
(1)

whereas the corresponding phasors are
Va  Vm

Vb  Vm e  j120
(2)

Vc  Vm e  j240 .
Va
Ia
Za
Vb
Ib
Zb
Vc
Ic
Zc
In
Fig.1
A three-phase system is shown in Fig 1. In a special case all impedances are identical
Za = Zb = Zc = Z .
(3)
Such a load is called a balanced load and is described by equations
V
Ia  a
Z
V
Ib  b
Z
Using KCL, we have
1
V
Ic  c .
Z
I n  Ia  I b  Ic 
1
Va  Vb  Vc  ,
Z
(4)
where




Va  Vb  Vc  Vm 1  e  j120  e  j240 
 1
3 1
3
  0.
 Vm 1  cos120   j sin 120   cos 240   j sin 240   Vm 1   j
 j

2
2
2
2




Setting the above result into (4), we obtain
In  0 .
(5)
Since the current flowing though the fourth wire is zero, the wire can be removed (see
Fig.2)
Va
Ia
Z
Vb
Ib
Z
Vc
Ic
Z
n’
n
Fig. 2
The system of connecting the voltage sources and the load branches, as depicted in Fig. 2, is
called the Y system or the star system. Point n is called the neutral point of the generator and
point n’ is called the neutral point of the load.
Each branch of the generator or load is called a phase. The wires connecting the supply to
the load are called the lines. In the Y-system shown in Fig. 2 each line current is equal to the
corresponding phase current, whereas the line-to-line voltages ( or simply line voltages ) are
not equal to the phase voltages.
2 Y-connected systems
Now we consider the Y-connected generator sources ( see Fig. 3).
2
Va
a
Vb
b
Vc
c
Vab
n
Vca
Vbc
Fig.3
The phasors of the phase voltages can be generally written as follows
Va  V  Vme j
Vb  Ve  j120
Vc  Ve  j240
o
.
(6)
o
We determine the line voltages Vab, Vbc, Vca ( see Fig.3). Using KVL, we obtain
 1
3
3
3
Vab  Va  Vb  Va 1   j   Va   j  
2 
2 
 2
2
2
2
 3   3  j tan 1
 Va    
e
 2   2 
3
3

 Va 3e j30 .
Thus,
o
Vab  Va 3 e j30 .
(7)
holds and similarly we obtain
o
Vbc  Vb 3 e j30
(8)
o
Vca  Vc 3 e j30 .
3
(9)
The phasor diagram showing the phase and line voltages is shown in Fig.4.
Vab
Vc
30
Va
30
Vca
30
Vb
Vbc
Fig.4
Thus, the line voltages Vab, Vbc, Vca form a symmetrical set of phasors leading by 30 the set
representing the phase voltages and they are 3 times greater.
Vab  Vbc  Vca  3 Va .
(10)
The same conclusion is valid in the Y connected load ( see Fig.5).
a
Za=Z
Va
Vab
Vca
Zc=Z
Zb=Z
Vc
Vb
b
Vbc
c
Fig.5
4
3. Three-phase systems calculations
When the three phases of the load are not identical, an unbalanced system is produced. An
unbalanced Y-connected system is shown in Fig.1. The system of Fig.1 contains perfectly
conducting wires connecting the source to the load. Now we consider a more realistic case
where the wires are represented by impedances Zp and the neutral wire connecting n and n’ is
represented by impedance Zn ( see Fig.6).
Va
n
Zp
a
Ia
Vb
b
Ib
Vc
c
Zp
Zp
Ic
Zn
a’
Za
b’
Zb
c’
Zc
n’
In
Vn
Fig.6
Using the node n as the datum, we express the currents Ia, Ib, Ic and In in terms of the node
voltage Vn
Ia 
Va  Vn
Za  Zp
(11)
Ib 
Vb  Vn
Zb  Zp
(12)
Ic 
Vc  Vn
Zc  Zp
(13)
In 
Vn
.
Zn
(14)
Hence, we obtain the node equation
Vn Va  Vn Vb  Vn Vc  Vn



0
Z n Za  Z p Z b  Z p Z c  Z p
Solving this equation for Vn, we have
Va
Vb
Vc


Za  Zp Zb  Zp Zc  Zp
Vn 
.
1
1
1
1



Zn Za  Zp Zb  Zp Zc  Zp
5
(15)
The above relationships enable us to formulate a method for the analysis of three-phase
systems. The method consists of three steps as follows:
(i)
( ii )
( iii )
Determine Vn using (15)
Calculate the currents Ia, Ib, Ic and In applying (11) - (14).
Find the phase and line voltages using Kirchhoff’s and Ohm’s laws.
When the neutral wire is removed, the system contains three connecting wires and is called
a three-wire system. In such a case we set Z n   into (15)
Va
Vb
Vc


Za  Zp Zb  Zp Zc  Zp
Vn 
.
(16)
1
1
1


Za  Zp Zb  Zp Zc  Zp
The balanced system can be considered as a special case of the unbalanced system, where
Za = Zb = Zc = Z. Using (16), we obtain
Vn 
1
 Va  Vb  Vc 
Z  Zp
3
Z  Zp
0 .
(17)
Consequently, the relationships (11)-(13) reduce to

Ia 
Va
Z  Zp
(18)
Ib 
Vb
Z  Zp
(19)
Ic 
Vc
.
Z  Zp
(20)



Since Vb  Va e j120 and Vc  Va e j240 , we have Ib  Ia e j120 and Ic  Ia e j240 .
Hence, we need to calculate Ia only using (18), which can be made applying the one-phase
circuit described by equation (18) shown in Fig.7.
Ia
Zp
Va
Va
Z
n’
n
Fig.7
6
This means that the analysis of a balanced three-phase system can be reduced to the analysis
of one-phase system depicted in Fig.7.
Example
Let us consider three-phase system shown in Fig.8. The system is supplied with a balanced
three-phase generator, whereas the load is unbalanced.
The effective value of the generator phase voltage is 220V, the impedance of any connecting
wire is Z p   2  j2 and the phase impedances of the load are Za  2  j4 ,
Zb  4  j2 , Zc  2  j4 . We wish to determine the line currents.
Va
Zp
Vb
Zp
Ia

Vab
Ib
b
n
Vca
Vc
Za
a
Zp
Ic
c
Zb
n’

Vbc
Zc
Vn
Fig.8
Since the circuit of Fig.8 is a three-wire system, we apply equation (16) to compute Vn. The
phase generator voltages are
Va  220 2 V
 1

3
   155.56  j269.44 V
Vb  Va e  j120  220 2    j

2
2


 1

3
   155.56  j269.44V .
Vc  Va e  j240  220 2    j

2
2


Using (16), we find
220 2  155.56  j269.44  155.56  j269.44


4  j6
6
4  j6
Vn 
 97.5  j61.2V .
1
1
1
 
4  j6 6 4  j6
Next, we compute the line currents using (11)-(13)
7
4
Ia 
Va  Vn 220 2  97.5  j61.2

 23.49  j19.94A
Za  Z p
2  j4  2  j2
Ib 
Vb  Vn  155.56  j269.44  97.5  j61.2

  42.18  j34.70A
Zb  Zp
4  j2  2  j2
Ic 
Vc  Vn  155.56  j269.44  97.5  j61.2

 18.68  j54.63A .
Zc  Z p
2  j4  2  j2
Power in three-phase circuits
In the balanced systems, the average power consumed by each load branch is the same and
given by
~
Pav  Veff Ieff cos 
(21)
where Veff is the effective value of the phase voltage, Ieff is the effective value of the phase
current and  is the angle of the impedance. The total average power consumed by the load is
the sum of those consumed by each branch, hence, we have
~
Pav  3Pav  3Veff Ieff cos 
(22)
In the balanced Y systems, the phase current has the same amplitude as the line current
Ieff  Ieff L , whereas the line voltage has the effective value Veff L which is 3 times
greater than the effective value of the phase voltage, Veff L  3Veff . Hence, using (22), we
obtain
Pav  3
Veff L  
Ieff L cos  
3
3 Veff L Ieff L cos 
(23)
Similarly, we derive
Px  3 Veff  L  I eff  L sin  .
(24)
In the unbalanced systems, we add the powers of each phase
Pav   Veff  a I eff  a cos  a   Veff  b I eff  b cos  b   Veff  c I eff  c cos  c
(25)
Px   Veff  a  I eff  a sin  a   Veff  b  I eff  b sin  b   Veff  c  I eff  c sin  c .
(26)
In order to measure the average power in a three-phase Y-connected load, we use three
wattmeters connected as shown in Fig.9.
The reading of the wattmeter W1 is
PW1 


1
1
Re Va Ia   Vm  a  I m  a cos  a   Veff  a  I eff  a cos  a  Pa .
2
2
8
*
a’
Ia
Za
W1
*
Va
Ib
*
b’
Zb
W2
n’
*
Vb
*
*
Zc
Ic
c’
W3
Vc
Fig. 9
Similarly, W2 and W3 measure the average power of the load branch b and c, respectively.
Thus, the sum of the three readings will give the total average power. This method of the
average power measurement is valid for both balanced and unbalanced Y-connected loads.
Note that in the case of a balanced Y-connected load all three readings are identical and
therefore we use only one wattmeter.
For measuring average power in a three-phase three-wire system, we can use a method
exploiting two wattmeters. In this method two wattmeters are connected by choosing any one
line as the common reference for the voltage coils of the wattmeters. The current coils are
connected in series with the other two lines ( see Fig.10) and the asterisk terminals of each
wattmeter are short-circuited ( see Fig.10).
*
Ia
a
*
Vac
W1
*
b
Ib
W2
Load
*
Vbc
c
Ic
Fig.10
The indications of the wattmeters are


(27)


(28)
PW1 
1
Re Vac Ia ,
2
PW2 
1
Re Vbc Ib .
2
9
The load is shown in Fig.11.
Va
Ia
a
Vac b
Ib
Vb
Vbc
c
Ic
Vc
Za
Zb
Zc
Fig.11
Since Vac = Va - Vc and Vbc = Vb - Vc, we obtain
PW1 




PW2




1
1
Re Va  Vc Ia  Va Ia  Vc Ia ,
2
2
1
1
 Re Vb  Vc Ib  Vb Ib  Vc Ib .
2
2
The sum of PW1 and PW2 gives
PW1  PW2 



1
Re Va I a  Vb I b  Vc I a  I b .
2
(29)
Currents Ia, Ib, Ic satisfy KCL
Ia + Ib + Ic = 0
Hence, it holds
Ia + Ib + Ic = 0 ,
or
Ia + Ib = - Ic .
(30)
Substituting (30) into (29) we have
PW1  PW2 


1
Re Va I a  Vb I b  Vc I c  Pav .
2
(31)
Equation (31) says that the sum of the two wattmeters readings in a Y-connected system
equals the total average power consumed by the load.
Let us consider a balanced Y-connected load and calculate the instantaneous power
delivered by the generator to the load
p t   va  t  ia  t   vb t  i b t   vc t  ic  t  ,
where
10
(32)
va  t   Vm cos t


cost  240 
v b  t   Vm cos t  120o
vc  t   Vm
and
(33)
o
i a t   Vm cost  

cost  240

  .
i b t   Vm cos t  120  
i c t   Vm

(34)
where v a  t , v b  t , v c  t  are the voltages of the load branches, i a  t , i b  t , i c  t  are the
currents of the load branches and  is the angle of the load impedance. We substitute (33)-(34)
in (32)

 

p t   VmI m[cos t cost    cos t  120o cos t  120o   

 

 cos t  240o cos t  240o   ]
and use the trigonometric identity
cos x  cos y 


1
cos x  y  cos x  y ,
2
finding
p t  




1
VmI m 3cos   cos2t    cos 2t  240o    cos 2t  480o    .


2
Since




cos 2t    cos 2t  240o    cos 2t  480o    0
we obtain
pt  
3
Vm I m cos  3Veff I eff cos  Pav
2
(35)
Thus, the total instantaneous power p(t) delivered by a three-phase generator to the balanced
load is constant and equals the average power consumed by the load.
11