1) The document discusses balanced three-phase systems where the loads have the same impedance in each phase. 2) It provides examples of calculating currents in star-connected and delta-connected balanced three-phase load systems. The phase currents are equal but offset by 120 degrees, while the line currents are calculated using Kirchhoff's laws. 3) Power calculations for a star-connected balanced load show that total power is three times the power measured in one phase using a wattmeter.

1. Lecture 2
BALANCED THREE-PHASE
SYSTEMS

2. In the following discussion we are only going to consider three-phase
systems with balanced loads.
These are loads which have the same impedances in each phase.
They may be resistive or a combination of resistance and inductance
or capacitance. The only essential feature is that the same
impedance exists in each phase.
INTRODUCTION

3. Industrial loads are supplied with the three-phase system; the line
voltage is generally 400 V (so that the single-phase supply is
approximately 230 V). Several methods of connection are possible:
INDUSTRIAL LOADS
1. Three-Phase Four wire
2. Three phase Three wire
3. Balanced Delta

4. Three-phase four wire star-connected balanced loads
This is a star connected system in which the neutral makes the
fourth wire, as shown in Figure x.

5. Three-phase four wire star-connected balanced loads
Notes
1. We can make the assumption with a balanced load that what is
happening in each phase does not affect what is happening in
any other phase.
2. We can also say that whatever happens in phase A will occur in B
phase 120o later and again in C phase 240o later, assuming a
positive phase sequence.

6. Three-phase four wire star-connected balanced loads
Notes
3. The last condition that is taken is that there is no resistance in the
connecting conductors (including the neutral conductor). This
means that the two neutral points or star points are at the same
potential.

7. Calculations of currents in star-connected balanced
loads
We will now consider what the current values are in Figure 2.
If we consider each phase in isolation for the calculation of each
current, the result is as follows:
o
o
o
A
AN
A
Z
E
Z
E
Z
V
I









0

8. o
o
o
o
B
BN
B
Z
E
Z
E
Z
V
I











120
120
o
o
o
o
C
CN
C
Z
E
Z
E
Z
V
I











240
240
Calculations of currents in star-connected balanced
loads

9. Now, if we add, the resultant should flow in the neutral
conductor since all three phases are flowing to the neutral point of
the load.
C
B
A I
I
I 

0



 N
C
B
A I
I
I
I
N
C
B
A I
I
I
I 



0
240
120 












 o
o
o
o
o
C
B
A
Z
E
Z
E
Z
E
I
I
I 


This is indicated in the finicular diagram in Figure x.
Calculations of currents in star-connected balanced
loads

10. The voltage and current relationships for the balanced star-
connected load are
3
L
ph
V
V 
L
ph I
I 
Voltage- Current Relationships in star-connected
balanced loads

11. 1. IA, IB and IC form a balanced set of phasors of equal magnitude
and at 120o spacing since the angle of  is the same for each.
2. The resultant must be zero.
3. The condition of balanced currents can only exist for equal
impedances in the load.
4. The current in each phase is the same current in the line
supplying it so that the line currents are equal to the phase
currents.
Notes

12. A four-wire three-phase supply is connected to a balanced three-
phase load, the connections being three phases plus a neutral
conductor.
If VAN = 240 0o and ZA = 1030o and the phase sequence is ABC,
find IA, IB and IC and IN (the current in the neutral).
Example

13. Solution
o
o
o
o
A
o
AN
A
AN
A
Z
E
Z
V
I 30
24
30
10
0
240
0











o
o
o
o
o
o
B
o
BN
B
BN
B
Z
E
Z
V
I
150
24
30
120
24
30
10
120
240
120

















o
o
o
o
o
o
C
o
CN
C
CN
C
Z
E
Z
V
I
90
24
30
120
24
30
10
120
240
120
















14. N
C
B
A I
I
I
I 



N
o
o
o
I








 50
24
150
24
30
24
  N
I
j
j
j 






 0
24
12
78
.
20
12
78
.
20
Phasor diagram
Solution

15. • The power factor in a three-phase system must have the same
meaning as the power factor in a single-phase system.
• It must be the cosine of the angle which the current lags or leads
the voltage which causes that current.
• Because of the number of voltages both phase to neutral and
phase-to-phase in a three-phase circuit we must be careful that
we use the correct voltage with the current for which we are
trying to find the power factor. In all cases we use the phase
voltage and the respective phase to determine the angle.
POWER FACTOR

16. Notes
1. The power factor of any load can be determined by the
impedance angle of the load.
2. A positive impedance angle indicates an inductive load so the
power factor must be laggging.
POWER FACTOR

17. One-line equivalent circuits for star-connected balanced
loads
• For a star-connected balanced load all the currents can be
obtained by considering each phase separately.
• Since each current was equal in magnitude and only varied in
angles, if we are given one current and the phase sequence, we
can calculate the other two currents.

18. Consider the previous example as a one-line equivalent circuit.
Example
Solution
o
o
o
o
A
o
AN
A
AN
A
Z
E
Z
V
I 30
24
30
10
0
240
0












19. Since the phase sequence is positive then
Solution
o
BN E
V 120



o
CN E
V 120



then the currents
o
o
o
B
I 150
24
30
120
24 






o
o
o
C
I 90
24
30
120
24 





20. If the load connected to the four-wire star connected system is
balanced and the supply voltages and currents are symmetrical, then
the current in the neutral wire is zero, and the neutral wire may be
omitted. The resulting connection is the three-wire system as shown
in the figure below.
Three-phase Three wire star-connected balanced loads

21. A 300 V three-phase three-wire star-connected load has an
impedance of 18-20  in each phase. Find the line currents and
draw the phasor diagram. The phase sequence is ABC.
Example

22. Solution
We must use phase-to-neutral voltages.
The 300 volts given is the phase-to-phase value.
V
o
o
AN
V 0
173
3
0
300




A
o
o
o
AN
A
Z
V
I 20
6
.
9
20
18
0
173









A
o
o
o
BN
B
Z
V
I 100
6
.
9
20
18
120
173











A
o
o
o
CN
C
Z
V
I 140
6
.
9
20
18
120
173









Phasor diagram

23. All the examples given earlier have been using the positive
sequence of ABC. We will now consider the effect on a three-phase
four wire star-connected load with a negative phase sequence of
CBA.
Effect of phase reversal on star load currents

24. Consider the three-phase system below with a negative phase
sequence. The supply voltage is VAN = 240 0 volts and ZA =
1030o .
Example

25. A
o
o
o
AN
A
Z
V
I 30
24
30
10
0
240









A
-
o
o
o
o
o
BN
B
Z
V
I
90
24
30
120
24
30
10
120
240










Solution
VAN = 2400o
VCN = 240-120o
VBN = 240120o
ACB phase sequence

26. A
-
o
o
o
o
o
CN
C
Z
V
I
150
24
30
120
24
30
10
120
240













Solution
IB and IC are in the reverse positions which is consistent with the voltage
VBN and VCN being in the reverse positions.
The conclusion is that reversing the phase sequence on a balanced
star-connected load does not alter the magnitudes of the currents, only
the currents appear in the same order as the voltage sequence (see
figure).

27. Calculation of power in star-connected balanced loads
The figure shows a wattmeter is connected in one leg of the star load
to measure the power in that leg.
Since we have assumed that the load is balanced,
then the power dissipated in ZA (called Pph) will be
3
total
ph
P
P 
Then the wattmeter uses IA the line current, which
is equal to the phase current and VAN the phase-to-
neutral voltage.

28. Then Z
line
ph
ph I
V
P 
cos

Since phase
to
phase
ph V
V 


3
1
3
line
ph
V
V 
or
Then
Z
line
line
ph I
V
P 
cos
3

Calculation of power in star-connected balanced loads

29. The total true or real power for the three-phase load
ph
total P
P 3

Then
z
line
line
total I
V
P 
cos
3
1
3 


z is the impedance angle of load in A phase. This is the angle
between the current in A phase, IA and the phase to neutral
voltage VAN.
where
z
line
line
total I
V
P 
cos
3

or
W
Calculation of power in star-connected balanced loads

30. The total apparent power is
line
line
total I
V
S 
 3
and the total reactive power is
z
line
line
total I
V
Q 
sin
3

VA
VAr
Calculation of power in star-connected balanced loads

31. THREE-PHASE DELTA-CONNECTED BALANCED LOADS
Consider the figure below.
The most significant difference with voltages for the balanced delta
load is that there no neutral point, so that there are only phase-to-
phase voltages. Each voltage is across one leg of the delta but it is
also across two other legs in parallel with the first leg. For example,
VAB is across ZAB which is in parallel with ZCA and ZBC.
Supply voltage phasor diagram

32. Calculation of currents in delta-connected balanced loads
When we consider the delta-connected load, there is a
significant difference to the star-connected load. In the star-
connected load the line currents and the phase currents were
equal to as they were the same currents.
We now compare the star case with the delta by looking at the
following figure.

33. Consider first the supply voltages which cause currents to flow
through the impedances.
Consider VAB which causes current IAB to flow where
Calculation of currents in delta-connected balanced loads
AB
AB
AB
Z
V
I  amperes
Now consider VCA which causes ICA to flow where
CA
CA
CA
Z
V
I  amperes
These phase current also form a balanced set of phasors as they did
in the star connection. They have equal magnitudes and are out of
phase with each other by 120o.

34. The line currents are obtained by using Kirchhoff’s current laws at
each node of the delta.
At Node A:
CA
AB
A I
I
I 

At Node B:
At Node C:
AB
BC
B I
I
I 

BC
CA
C I
I
I 

Calculation of currents in delta-connected balanced loads

35. Calculation of currents in delta-connected balanced loads
We have said that as this is also a balanced load as in the
connection considered, then
0


 C
B
A I
I
I
Then
0








 BC
CA
AB
BC
CA
AB
C
B
A I
I
I
I
I
I
I
I
I
To find the relationship between IA and IAB, that is the relationship
between the line current and the phase current in a delta load,
consider the following.
If we let


 I
IAB amperes
then in the ABC sequence
o
BC I
I 120


  amperes and
o
CA I
I 120


  amperes

39. Solution
Let the voltage VAN be the reference phasor.

40. Solution

41. Solution

42. Solution

43. Solution

44. Solution

46. Exercises