A three-phase system uses three sinusoidal voltages that are 120 degrees out of phase to provide constant power output. A balanced three-phase system can be analyzed using a single-phase equivalent circuit. Power in a three-phase system is measured using three wattmeters connected in a Y configuration for a Y-connected load or two wattmeters connected between line voltages for a three-wire system. The total power is the sum of the wattmeter readings.

1. 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
( )
( ) ,240tcosVv
120tcosVv
tcosVv
mc
mb
ma


−ω=
−ω=
ω=
(1)
whereas the corresponding phasors are
.eVV
eVV
VV
240j
mc
120j
mb
ma


−
−
=
=
=
(2)
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
I
V
Z
I
V
Z
I
V
Z
a
a
b
b
c
c
= = = .
1
In
Ia ZaVa
Ic ZcVc
Ib ZbVb

2. Using KCL, we have
( )I I I I
Z
V V Vn a b c a b c= + + = + +
1
,
(4)
where
( )
( ) .0
2
3
j
2
1
2
3
j
2
1
1V240sinj240cos120sinj120cos1V
ee1VVVV
mm
240j120j
mcba
=







+−−−=−+−+=
=++=++ −−


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)
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
n’n
Ia ZVa
Ic ZVc
Ib ZVb

3. Fig.3
The phasors of the phase voltages can be generally written as follows
V V V e
V Ve
V Ve
a m
j
b
j
c
j
= =
=
=
−
−
α
120
240


.
(6)
We determine the line voltages Vab, Vbc, Vca ( see Fig.3). Using KVL, we obtain
.e3Ve
2
3
2
3
V
2
3
j
2
3
V
2
3
j
2
1
1VVVV
30j
a
3
3
tanj
22
a
aabaab
1

=







+





=
=







+=







++=−=
−
Thus,
V V eab a
j
= 3 30
.
(7)
holds and similarly we obtain
V V ebc b
j
= 3 30
(8)
V V eca c
j
= 3 30
.
(9)
3
Vca
c
Vbc
Vab
n
a
Va
Vc
bVb

4. The phasor diagram showing the phase and line voltages is shown in Fig.4.
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.
V V V Vab bc ca a= = = 3 . (10)
The same conclusion is valid in the Y connected load ( see Fig.5).
Fig.5
4
Va
c
b
a
Vbc
Vb
Vc
Zc=Z
Za=Z
Zb=Z
Vab
Vca
Vc
Vbc
Vb
30° Va
Vab
30°
30°
Vca

5. 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).
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
I
V V
Z Z
a
a n
a p
=
−
+
(11)
I
V V
Z Z
b
b n
b p
=
−
+
(12)
I
V V
Z Z
c
c n
c p
=
−
+
(13)
I
V
Z
n
n
n
= . (14)
Hence, we obtain the node equation
V
Z
V V
Z Z
V V
Z Z
V V
Z Z
n
n
a n
a p
b n
b p
c n
c p
−
−
+
−
−
+
−
−
+
= 0
Solving this equation for Vn, we have
V
V
Z Z
V
Z Z
V
Z Z
Z Z Z Z Z Z Z
n
a
a p
b
b p
c
c p
n a p b p c p
=
+
+
+
+
+
+
+
+
+
+
+
1 1 1 1
. (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:
5
c
b
a
Vn
Zn
c’
a’
b’
Zp
Zp
Zp
In
n’
n
Ia Za
Va
Ic ZcVc
Ib
ZbVb

6. ( i ) Determine Vn using (15)
( ii ) Calculate the currents Ia, Ib, Ic and In applying (11) - (14).
( iii ) 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 Zn →∞ into (15)
V
V
Z Z
V
Z Z
V
Z Z
Z Z Z Z Z Z
n
a
a p
b
b p
c
c p
a p b p c p
=
+
+
+
+
+
+
+
+
+
+
1 1 1
. (16)
The balanced system can be considered as a special case of the unbalanced system, where Za
= Zb = Zc = Z. Using (16), we obtain
( )
V
Z Z
V V V
Z Z
n
p
a b c
p
=
+
+ +
+
=
1
3
0 . (17)
Consequently, the relationships (11)-(13) reduce to
I
V
Z Z
a
a
p
=
+
(18)
I
V
Z Z
b
b
p
=
+
(19)
I
V
Z Z
c
c
p
=
+
. (20)
Since

120j
ab eVV −
= and

240j
ac eVV −
= , we have

120j
ab eII −
= and

240j
ac eII −
= .
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.
Fig.7
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
6
n’
aV′
n
Z
Zp Ia
Va

7. 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 jp = +2 2 Ω and the phase impedances of the load are ( )Ω+= 4j2Za ,
( )Ω−= 2j4Zb , ( )Ω+= 4j2Zc . We wish to determine the line currents.
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
( )
( ) .V44.269j56.155
2
3
j
2
1
2220eVV
V44.269j56.155
2
3
j
2
1
2220eVV
V2220V
240j
ac
120j
ab
a
+−=







+−==
−−=







−−==
=
−
−


Using (16), we find
( ) ( )
( ) .V2.61j5.97
6j4
1
6
1
6j4
1
6j4
44.269j56.155
6
44.269j56.155
6j4
2220
Vn −=
+
++
+
+
+−
+
−−
+
+
=
Next, we compute the line currents using (11)-(13)
( )
( )
( ) .A63.54j68.18
2j24j2
2.61j5.9744.269j56.155
ZZ
VV
I
A70.34j18.42
2j22j4
2.61j5.9744.269j56.155
ZZ
VV
I
A94.19j49.23
2j24j2
2.61j5.972220
ZZ
VV
I
pc
nc
c
pb
nb
b
pa
na
a
+=
+++
+−+−
=
+
−
=
−−=
++−
+−−−
=
+
−
=
−=
+++
+−
=
+
−
=
7
c
a
b
Zp
Zp
Zp
Vn
caV′
bcV′
abV′
n’n
Ia ZaVa
Ic ZcVc
Ib
ZbVb

8. 4 Power in three-phase circuits
In the balanced systems, the average power consumed by each load branch is the same and
given by
φ= cosIVP
~
effeffav (21)
where Veff is the effective value of the phase voltage, Ieff is the effective value of the phase
current and φ 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
φ== cosIV3P
~
3P effeffavav (22)
In the balanced Y systems, the phase current has the same amplitude as the line current
( )Leffeff II = , whereas the line voltage has the effective value ( )LeffV which is 3 times
greater than the effective value of the phase voltage, ( ) effLeff V3V = . Hence, using (22), we
obtain
( ) ( ) ( ) ( ) φ=φ= cosIV3cosI
3
V
3P LeffLeffLeff
Leff
av (23)
Similarly, we derive
( ) ( )P V Ix eff L eff L
= 3 sinφ . (24)
In the unbalanced systems, we add the powers of each phase
( ) ( ) ( ) ( ) ( ) ( )P V I V I V Iav eff a eff a a eff b eff b b eff c eff c c= + +cos cos cosφ φ φ (25)
( ) ( ) ( ) ( ) ( ) ( )P V I V I V Ix eff a eff a a eff b eff b b eff c eff c c= + +sin sin sinφ φ φ . (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
( ) ( ) ( ) ( ) ( )P V I V I V I PW a a m a m a a eff a eff a a a1
1
2
1
2
= = = =∗
Re cos cosφ φ .
8

9. 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).
Fig.10
The indications of the wattmeters are
( )P V IW ac a1
1
2
= ∗
Re , (27)
( )P V IW bc b2
1
2
= ∗
Re . (28)
9
*
*
*
*
c
Vbc
Vac
b
a
Ia
Ic
Ib
LoadW2
W1
n’
*
*
*
*
*
*
Icc’
b’
a’
Vc
Vb
Zb
Ia Za
Va
Zc
Ib
W2
W1
W3

10. The load is shown in Fig.11.
Fig.11
Since Vac = Va - Vc and Vbc = Vb - Vc, we obtain
( )( ) ( )
( )( ) ( ).IVIV
2
1
IVVRe
2
1
P
,IVIV
2
1
IVVRe
2
1
P
bcbbbcbW
acaaacaW
2
1
∗∗∗
∗∗∗
−=−=
−=−=
The sum of PW1 and PW2 gives
( )[ ]∗∗∗∗
+−+=+ bacbbaaWW IIVIVIVRe
2
1
PP 21
. (29)
Currents Ia, Ib, Ic satisfy KCL
Ia + Ib + Ic = 0
Hence, it holds
I + I + I = 0a b c
∗ ∗ ∗
,
or
I + I = - Ia b c
∗ ∗ ∗
. (30)
Substituting (30) into (29) we have
[ ] avccbbaaWW PIVIVIVRe
2
1
PP 21
=++=+ ∗∗∗
. (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 v t i t v t i t v t i ta a b b c c= + + , (32)
where
10
c
Vbc
Vac
b
a
Vc
Vb
Zb
Ia
Za
Va
Ic
Zc
Ib

11. ( )
( ) ( )
( ) ( )
v t V t
v t V t
v t V t
a m
b m
c m
=
= −
= −
cos
cos
cos
ω
ω
ω
120
240
o
o
(33)
and
( ) ( )
( ) ( )
( ) ( ).240tcosVti
120tcosVti
tcosVti
mc
mb
ma
φ−−ω=
φ−−ω=
φ−ω=
o
o
(34)
where ( ) ( ) ( )v t v t v ta b c, , are the voltages of the load branches, ( ) ( ) ( )i t i t i ta b c, , are the
currents of the load branches and φ is the angle of the load impedance. We substitute (33)-(34)
in (32)
( ) ( ) ( ) ( )
( ) ( )
p t V I t t t t
t t
m m= − + − − − +
+ − − −
[cos cos cos cos
cos cos ]
ω ω φ ω ω φ
ω ω φ
120 120
240 240
o o
o o
and use the trigonometric identity
( ) ( )[ ]cos cos cos cosx y x y x y⋅ = − + +
1
2
,
finding
( ) ( ) ( ) ( )p t V I t t tm m= + − + − − + − −



1
2
3 2 2 240 2 480cos cos cos cosφ ω φ ω φ ω φo o
.
Since
( ) ( ) ( )cos cos cos2 2 240 2 480 0ω φ ω φ ω φt t t− + − − + − − =o o
we obtain
( ) aveffeffmm PcosIV3cosIV
2
3
tp =φ=φ= (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