Unbalanced Three-phase Loads
A three-phase system
can be used to supply three-phase loads, or individual single-phase loads.
Loads having identical impedances per phase result in balanced loads. What are the chances of
such a condition occurring in actual practice? Three-phase equipment, such as motors , are
balanced loads, but when we add lights, appliances and other single phase loads on the system
the balance is lost. Homes are generally supplied with single-phase service. The power
company tries to obtain a balanced load by equalizing the number of homes on each phase of
the three-phase supply. Industrial consumers are generally supplied with all three lines and
neutral. The internal wiring is then planned to attempt to distribute the single-phase loads
equally between neutral and each of the three lines. However, any such measures, no matter
how carefully planned, can prevent only drastic unbalance. While dealing with unbalanced
circuits, careful attention must be paid to phase sequence, and realize that the solutions to the
problems are more involved and tedious.
We shall now work through a series of examples in increasing order of complexity:a) Unbalanced delta loads
b) Unbalanced 4-wire star-connected load (3lines and neutral)
c) Unbalanced 3-wire star-connected load
d) Unbalanced 4-wire star-connected load with impedance in the neutral
Unbalanced delta loads. Whenever dissimilar loads are connected across the lines of a
three-phase supply, an unbalanced delta load will result. The phase currents are the individual
load currents. Since the load currents in each load are not equal in magnitude and/or phase, the
three line currents will be unbalanced. Each load (phase) current can be found from the line
(phase) voltage and load impedance. The line currents can then be found from the phasor
addition of the appropriate load currents.
Example.
Three loads are connected across the lines of a 240 volt three-phase supply having a phase
sequence of L1 – L2 – L3.
Z1 = 31  resistance and 59  inductive reactance,
Z2 = 30  resistance and 40  capacitive reactance, and
Z3 = 80  resistance and 60  inductive reactance.
Find the line currents.
Solution:
1.
Write down the three impedances in both polar and rectangular format.
Rectangular
Polar
Z1 = 31 + j59 →
Z2 = 30 + j40 →
Z3 = 80 + j60 →
(Line 1 to Line 2)
(Line 2 to Line 3)
(Line 3 to Line 1)
→ 66.7  62.3⁰
→ 50 - 53.1⁰
→ 100 36.87⁰
Good idea!
At every stage in your calculations jot down answers in both rectangular and polar format so the
correct format can be grabbed straight away. This allows you to concentrate on the next stage of the
calculation without interruptions.
Underline these intermediate answers so that you can find them quickly.
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2. Draw a diagram of the circuit, and a phasor diagram of the line voltages. Note, the delta load
is labelled in a clockwise direction and the phasor diagram by convention rotates in an anticlockwise direction.
As the phase sequence is L1→ L2→ L3→ L1 the phase currents will flow
I1-2
I2-3
I3-1
from line terminal 1 to line terminal 2.
from line terminal 2 to line terminal 3.
from line terminal 3 to line terminal 1.
Circuit Diagram
Line Voltage Phasors
E3-1
240120
E1-2
2400
E2-3
240-
Calculate the phase currents.
3/ Calculate the delta phase currents. Use the phasor value of the voltages. See diagram
above









4/ Combine phase currents into line currents.
IL1
Positive flow for the line current is from supply into the load.
Therefore at terminal 1 of the load, phase current I1-2 will also be
positive as it flows in the same direction, but phase current I3-1 will
have a negative flow.
Therefore IL1 = I1-2 - I2-3
Continued over page.
Rectangular
1
I
→ 3-1
Polar
I1-2
2
1.67 - j 3.41
- 1.92 + j1.44
-0.25 - j1.97
Rectangular
(To subtract, change sign and add)
→ 1.99  -97.2
→
Polar
2.88 + j 3.84
-1.67 + j 3.41
1.21 + j 7.25
Rectangular
→ 7.35  80.5
→
Polar
1.92 – j 1.44
- 2.88 – j 3.84
- 0.96 – j 5.28
→ 5.37  100.3
Tutorial Exercise:
1. A delta-connected load is connected to a symmetrical 415-volt, 3-phase supply.
The impedances of the load are:Z1-2 = 100 +j0 
Z2-3 = 50 + j50 
Z3-1 = 100 –j 50 .
Determine the current in each line.
Answers: IL1 = 7.25 - j2.05 (7.53 A),
IL2 = -9.83 – j1.52 (9.96 A) IL3 = 2.58 + j3.57 (4.4A)
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