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CHAPTER # 3 THREE-PHASE CIRCUITS
1. Introduction
Circuits or systems in which the ac sources operate at the same frequency but different
phases are known as poly phase. Fig. 1 shows a three-phase four-wire system. As
distinct from a single-phase system, a three-phase system is produced by a generator
(alternator), whose cross-sectional view is shown in Fig. 2(a). The alternator basically
consists of a rotating magnet (called the rotor) surrounded by a stationary winding
(called the stator). Three separate windings or coils with terminals a-a', b-b', and c-c'
are electrically placed 120◦ apart around the stator. Because the coils are placed 120◦
apart, the induced voltages in the coils are equal in magnitude but out of phase by
120◦ as shown in Fig. 2 (b). Since each coil can be regarded as a single-phase
generator by itself, the three-phase generator can supply power to both single-phase
and three-phase loads.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Generator (Source) side
Load side
Fig. 1, Three-phase, 4-wire circuit
(a) 3-phase alternator
(b) 3-phase voltages
Fig. 2, Generating 3-phase voltages
A typical three-phase system consists of three voltage sources connected to loads by
three or four wires (or transmission lines) using both step-up and step-down
transformers is shown in Fig. 3. The connection to a particular home is shown in Fig.
4. Finally the connection inside rooms is shown in Fig. 5.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 3, Three-phase system
Fig. 4, Connection to a particular home
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 5, Connection to a particular room
2. Balanced three-phase circuits
The three-phase sources, explained before, can be connected either wye (Y) or delta
(∆) as shown in Fig. 6.
(a) Wye (Y) Connection
(b) Delta (∆) Connection
Fig. 6, Three-phase connection (Generator side)
Consider the wye-connected voltages in Fig. 6(a). The voltages Van, Vbn, and Vcn are
respectively between lines a, b, and c, and the neutral line n. These voltages are called
phase voltages. If the voltage sources have the same amplitude and frequency ω and
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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are out of phase with each other by 120◦, the voltages are said to be balanced. This
gives
There are two possible combinations. One possibility is called positive (abc) sequence
This sequence is produced when the phasor diagram rotates counterclockwise. The
phase sequence is determined by the order in which the phasors pass through a fixed
point in the phase diagram. This can be expressed mathematically as:
The other possibility is called negative (acb) sequence and expressed mathematically
as:
Like the generator connections, a three-phase load can be either wye-connected or
delta-connected, depending on the end application as shown in Fig. 7.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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(a) Wye (Y) Connection
(b) Delta (∆) Connection
Fig. 7, Three-phase connection (Load side)
For a balanced wye-connected load,
whereZY is the load impedance per phase. For a balanced delta-connected load,
where Z∆ is the load impedance per phase.
Y-connected load can be transformed into a ∆-connected load, or vice versa as:
Since both the three-phase source and the three-phase load can be either Y or ∆, we
have four possible connections:
• Y-Y connection (i.e., Y-connected source with a Y-connected load)
• Y-∆ connection (i.e., Y-connected source with a ∆-connected load)
• ∆-∆ connection (i.e., ∆-connected source with a ∆-connected load)
• ∆-Y connection (i.e., ∆-connected source with a Y-connected load)
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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It is appropriate to mention here that a balanced delta-connected load is more common
than a balanced Y-connected load. This is due to the ease with which loads may be
added or removed from each phase of a delta-connected load. This is very difficult
with a Y-connected load because the neutral may not be accessible. On the other hand,
delta connected sources are not common in practice because of the circulating current
that will result in the delta-mesh if the three-phase voltages are slightly unbalanced.
2.1 Balanced Y-Y three-phase circuits
Any balanced three-phase system can be reduced to an equivalent Y-Y system.
Therefore, analysis of this system should be regarded as the key to solving all
balanced three-phase systems.
Consider the 3-phase, Y-Y circuit shown in Fig. 8. Assuming the positive sequence,
the phase voltages (or line-to-neutral voltages) are:
∠0,
∠
120 ∠ 120
Fig. 8, Y-Y three-phase circuit
The line-to-line voltages or simply line voltages Vab, Vbc, and Vca are related to the
phase voltages. mathematically as:
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Similarly, we can obtain
The relation between phase and line voltages can be proved Vectorially as in Fig. 9.
Fig. 9, Relation between phase and line voltages in Y connected circuits
From phasor diagram shown above, it is clear that the magnitude of the line voltages
VL is √3 times the magnitude of the phase voltages Vp, or
√3 where
and
Also the line voltages lead their corresponding phase voltages by 30◦.
Applying KVL to each phase in Fig. 8, we obtain the line currents as:
We can readily infer that the line currents add up to zero,
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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so that
So that the voltage VnN is zero. Therefore, neutral line can thus be removed without
affecting the system. In fact, in long distance power transmission, conductors are used
with the earth itself acting as the neutral conductor. Power systems designed in this
way are well grounded at all critical points to ensure safety.
The line current is the current in each line, and the phase current is the current in each
phase of the source or load. In Y-Y, the line current is the same as the phase current.
An alternative way of analyzing a balanced Y-Y system is to consider the “per phase”
equivalent circuit given in Fig. 10. We look at one phase, say phase a, and analyze the
single-phase equivalent circuit. Then applying KVL for this circuit to get the same
values for the line currents.
Fig. 10, Equivalent single phase circuit
Once the current Ia is obtained, we use the phase sequence to obtain other line
currents.
EX. 1. Final Exam, First semester (1432/1433)
Three equal impedances, 60 + j30Ω each, are ∆-connected to a positive sequence 100V three-phase supply as shown in Fig. 11. Another three equal impedances, 40 + j10Ω
each, are Y-connected across the same circuit at the same points.
Determine:
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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a) The line currents
(b) The total complex power supplied to the two loads
(c) The power factor of the two loads combined
Fig. 11
Convert the delta connected load with impedance 60 + J 30 Ω to star connected on
for balanced load, ZY = Z∆ / 3 = 20 + J 10 Ω
This converted Y load becomes in parallel with the original Y load
ZY = (20+J10) // (40+J10) = 13.5+J5.5 Ω
for positive sequence supply
Va = 100∠0
Vb = 100∠− 120
Vc = 100∠120
The line current can be calculated as:
100∠0
6.86∠
13.5 5.5
100∠ 120
6.86∠
13.5 5.5
22.166 142.166 100∠120
6.86∠97.834 13.5 5.5
Total Complex power supplied to the two loads = 3×Va × Ia*= 3×100×6.86∠22.166
= 1905.9027 + J 776.4655 VA
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Power factor of the combined load is cosine the angle between the voltage and current
= cos(22.166) = 0.9261 Lagging
Report:
A Y-connected balanced three-phase generator with an impedance of 0.4+j0.3 Ω per
phase is connected to a Y-connected balanced load with an impedance of 24 + j19 Ω
per phase. The line joining the generator and the load has an impedance of 0.6 + j0.7
Ω per phase. Assuming a positive sequence for the source voltages and that Van =
120∠ 30◦ V, find: (a) the line voltages, (b) the line currents.
2.2 Balanced Y-∆
∆ three-phase circuits
The balanced Y-delta system is shown in Fig. 12, where the source is Y-connected and
the load is
∆-connected. There is no neutral connection from source to load for this
case. Assuming the positive sequence, the phase voltages are again:
Fig. 12. Y-∆ three-phase circuit
The line voltages are given as:
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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The line voltages are equal to the voltages across the load impedances for this system
configuration. From these voltages, we can obtain the phase currents as:
These currents have the same magnitude but are out of phase with each other by 120◦.
The line currents are obtained from the phase currents by applying KCL at nodes A, B,
and C. Thus,
This gives that
The relation between phase and line currents for ∆-connected circuits is shown in
Fig.13.
Fig. 13, Relation between phase and line currents in ∆-connected circuits
From phasor diagram above, the magnitude IL of the line current is √3 times the
magnitude Ip of the phase current, or
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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where
√3 and
Also, the line currents lag the corresponding phase currents by 30◦.
EX. 2. Final Exam, Second semester (1428/1429)
Determine the phase and line currents of the delta connected, +ve sequence load shown in
Fig. 14. Known that Z∆ = 60 ∠45˚ Ω
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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An alternative way of analyzing the Y-∆ circuit is to transform the ∆ -connected load
to an equivalent Y-connected load. Using the
∆-Y transformation formula
EX. 2. Final Exam, First semester (1430/1431)
For the three-phase Y-∆ circuit shown in Fig. 15, calculate the line and phase currents
of the load. Assume that ZL = 12 + j2 Ω
Fig. 15
Converting the 3-ph load from ∆ to Y then load impedance ZY =
"#$%#
&
4 '0.667 Ω
In that case the transmission line impedance (ZT) will be in series with ZY
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Overall load impedance = ZY + ZT = 4 + j0.667 + 1 + j2 = 5 + j2.667 Ω
IA (Line) =
()
*+ $ *,
"--∠-
.$%#.//0
17.647∠
28.075
IB (Line) =17.647∠
28.075 A
120 17.647∠
148.075 A
28.075 120 17.647∠ 91.925 A
IC (Line) =17.647∠
As we know, for ∆ connection, IPh is less than ILine by
IAB (Phase) =
"0./10
√&
∠
"
√&
and lead by 30◦
28.075 30 10.189∠1.925 A
IBC (Phase) = 10.189∠1.925
120 10.189∠
118.075 A
IBC (Phase) = 10.189∠1.925 120 10.189∠121.925 A
Report:
One line voltage of a balanced Y-connected source is VAB = 180∠ − 20◦ V. If the
source is connected to a ∆-connected load of 20∠ 40◦Ω, find the phase and line
currents. Assume the abc sequence.
2.3 Balanced ∆-∆
∆ three-phase circuits
The source as well as the load is delta-connected as shown in Fig. 16. Our goal is to
obtain the phase and line currents as usual. Assuming a positive sequence, the phase
voltages for a delta-connected source are:
Fig. 16, Delta-Delta 3-phase system
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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The line voltages are the same as the phase voltages. From Fig. 16, assuming there is
no line impedances, the phase voltages of the delta connected source are equal to the
voltages across the impedances; that is,
Hence, the phase currents are
The line currents Ia, Ib and Ic are greater than the phase current by √3 and lag the
corresponding phase current by 30°.
EX. 3. First Mid-term Exam, First semester (1431/1432)
A three-phase, positive sequence ∆- ∆ connected circuit with a line impedance of 1 +
j3 Ω. The line feeds a balanced load, which absorbs a total complex power of 12 + j5
kVA. If the line voltage VAB at the load side is 240∠0 V, calculate:
a) The line and phase currents at the load side,
b) The line voltage (Vab) at the source side
c) The source power factor.
Chapter Three: Three-Phase Circuits
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EE2020
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Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Report:
A positive-sequence, balanced ∆-connected source supplies a balanced ∆-connected
load. If the impedance per phase of the load is 18+j12 Ω and Ia = 22.5∠ 35◦ A, find
IAB and VAB.
2.4 Balanced ∆-Y three-phase circuits
Consider the ∆-Y circuit in Fig. 17. Again, assuming the abc sequence, the phase
voltages of a delta-connected source are:
Fig. 17, ∆-Y 3-phase system
At source side, the line voltages, given above, are also same as the phase voltages.
At load side the phase voltage VAN is less than the line voltage VAB by √3 and lags it
by 30°, therefore, the line current (which equals the phase current) Ia can be obtained
as:
we obtain the other line currents Ib and Ic using the positive phase sequence,
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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EX. 1. First Mid-Term Exam, First semester (1429/1430)
For the 3-phase, +ve sequence circuit shown in Fig. 18, if Vab = 440∠10˚. Calculate
the load currents, and then calculate the complex power at the source side.
Fig. 18, 3-phase ∆-Y circuit
Report:
In a balanced ∆-Y, negative sequence circuit, Vac = 240∠ 15◦ and ZY = (12 + j15) Ω.
Calculate the line currents.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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3. Power in Balanced 3-phase systems
For a Y-connected load, let the phase voltages are:
where Vp is the rms value of the phase voltage.
If the load impedance is ZY = Z ∠θ, then the phase currents lag behind their
corresponding phase voltages by θ. Thus,
where Ip is the rms value of the phase current.
The total instantaneous power in the load is the sum of the instantaneous powers in the
three phases; that is,
Based on the trigonometric relation:
Assuming that α = cos(2ωt - θ)
Chapter Three: Three-Phase Circuits
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Thus the total instantaneous power in a balanced three-phase system is constant. It
does not change with time as the instantaneous power of each phase does. This result
is true whether the load is Y- or ∆-connected.
-The active power (P) can be calculated as:
2 3 cos 6 √3 cos 6
-Also, The reactive power (Q) can be calculated as:
7 3 sin 6 √3 sin 6
- -The complex power (S) can be calculated as:
: 3 ∗ √3 ∗
First mid-term exam (Repeated), second semester 1433/1434
Four 3-phase, Y-connected, positive-sequence, balanced loads are connected to a 380V, 50-Hz line, as shown in Fig. 19-A. Rewrite Table 1 in your answer sheet and fill
the missing values.
Table 1
Load #1
Load#2
Load #3
Load #4
Apparent Power
Active Power
Reactive Power
Line Current
Power Factor
?
?
35 kVAR
?
0.6 lag
?
30 kW
?
?
0.8 lead
20 kVA
15 kW
?
?
? (lag)
35 kVA
?
20 kVAR
?
?
Combined
Load
?
?
?
?
?
To raise the combined load power factor to 0.98 lag, three capacitors ∆-connected in
parallel with the combined load as shown in Fig. 19-B. Calculate the kVAR rating of
each capacitor, then find the capacitance of each capacitor.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 19-A
Fig. 19-B
Load #1
Q1 = 35 kVAR
P.F.1 = 0.6 lag → φ1 = cos-1 0.6 = 53.13 → sin(φ1) = 0.8
S1 = Q1/ sin(φ1) = 35000/0.8 = 43.75 kVA
P1 = S1 cos (φ1) = 43750 × 0.6 = 26.25 kW
I1(Line) = S1/(√3 × VL) = 43750 / (√3 × 380) = 66.47 A
Load #2
P2 = 30 kW
P.F.2 = 0.8 lead → φ2 = cos-1 0.8 = 36.87 → sin(φ2) = 0.6
S2 = P2/ P.F.2 = 30000/0.8 = 37.5 kVA
Q2 = S1 sin (φ2) = 37500 × 0.6 = - 22.5 kVAR
I2(Line) = S2/(√3 × VL) = 37500 / (√3 × 380) = 56.975 A
Load #3
S3 = 20 kVA
P3 = 15 kW
p.F.3 = P3/S3 = 15/20 = 0.75 lag → φ3 = cos-1 0.75 = 41.41 → sin(φ3) = 0.6614
Q3 = S3 × sin(φ3) = 20000/0.6614 = 13.229 kVAR
I3(Line) = S3/(√3 × VL) = 20000 / (√3 × 380) = 30.3869 A
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Load #4
S4 = 35 kVA
Q1 = 20 kVAR
sin(φ4) =Q4/S4 = 20/35 =0.5714 → φ4 =sin-1 0.5714=34.85 → P.F.4=cos 0.5714 = 0.8207 lag
P4 = S4 cos (φ4) = 35000 × 0.8207 = 28.7245 kW
I4(Line) = S4/(√3 × VL) = 35000 / (√3 × 380) = 53.177 A
Combined Load
P (Total) = P1 + P2 + P3 + P4 = 99.9745 kW
Q (Total) = Q1 - Q2 + Q3 + Q4 = 45.729 kVAR
Complex power = 99.9745 + J 45.729 = 109.9365 ∟24.5797 kVA
S (Total) = 109.9365 kVA
P.F. (Total) = cos (24.5797) = 0.9094 Lag
I (Total) = S (Total) /(√3 × VL) = 109936.5 / (√3 × 380) = 167.0312 A
Table 1
Load #1
Load#2
Load #3
Load #4
Combined Load
Apparent Power
Active Power
Reactive Power
Line Current
Power Factor
43.75 kVA
26.25 kW
35 kVAR
66.47 A
0.6 lag
37.5 kVA
30 kW
–22.5 kVAR
56.975 A
0.8 lead
20 kVA
15 kW
13.229 kVAR
30.3869 A
0.75 (lag)
35 kVA
28.7245 kW
20 kVAR
53.177 A
0.82 Lag
109.936 kVA
99.9745 kW
45.729 kVAR
167.0312 A
0.9094 Lag
The new P.F = 0.98 → φ(new) =cos-1 0.98 = 11.4783
kVAR of each capacitor = P{tan(φold) – tan (φnew)}/3 = 8.4762 kVAR
<
7
8476.2
186.845 @A
2= > #
2= ? 50 ? 380#
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Final Exam, First semester 1429/1430
Load #1: 250 kVA, 0.8 P.F. lag
Load #2: 300 kVA, 0.95 P.F. lead
Load #3: 450 kVA, unity P.F.
The above three parallel-connected, three-phase loads are fed by a balanced threephase source. If the line voltage is 13.8 kV, calculate the line current and the power
factor of the source.
ST = 200 + j150 + 285 – j93.65 + 450
4. Unbalanced Three-phase systems
An unbalanced system is due to unbalanced voltage sources or an unbalanced load. To
simplify analysis, we will assume balanced source voltages, but an unbalanced load.
4.1 Y-Y four-wire and three-wire Unbalanced 3-phase systems
Figure 20 shows an example of an unbalanced 3-phase system.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 20, Unbalanced, 4-wire, Y-connected load
Since the load is unbalanced, ZA, ZB, and ZC are not equal. The line currents are
determined directly by Ohm’s law as
This set of unbalanced line currents produces current in the neutral line, which is not
zero as in a balanced system, and can be calculated as:
In a three-wire system where the neutral line is absent, we can still find the line
currents Ia, Ib, and Ic using mesh analysis.
First mid-term exam, second semester 1432/1433
A Y-Y, 4-wire, Three-Phase circuit with load shown in Fig. 21. The line voltages all
have the same magnitude and are in a positive phase sequence. If Vab =250 V;
a) Calculate the line currents and the neutral current
b) Calculate the complex power for the 3-phase load
c) Calculate the dissipated power at the load side
d) Find the power factor of each load
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 22
the Y-Y circuit can be as follows:
a) Since positive sequence then:
Vab = 250∠0
V
→ Van = 144.3376∠-30 V
Vbc = 250∠-120 V
→ Vbn = 144.3376∠-150 V
Vbc = 250∠+120 V
→ Vcn = 144.3376∠+90 V
Since 4-wire circuit, then
144.3376∠ 30
3.60844∠
40∠60
144.3376∠ 150
2.4056∠
60∠ 45
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
90 #
105 #
EE2020
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144.3376∠ 90
7.2169∠90 #
20∠0
the neutral current In is
In = - (Ia + Ib + Ic) = - 1.4277∠115.8544 A #
The complex power for load at phase #a =
(3.60844)2 ×40∠60 = 520.8336∠60=260.4168+j451.0551 VA
The complex power for load at phase #b =
(2.4056)2 ×60∠-45 = 347.2147∠-45 = 245.5179 - j245.5179 VA
The complex power for load at phase #c =
(7.2169)2 ×20∠0 = 1041.6729∠0 = 1041.6729VA
b) The total complex power = 1561.1966∠7.5652 = 1547.6076 + j 205.5373 VA ##
c) The dissipated power at load side is the real part of the complex load power =
1547.6076 W
###
d) the power factor of load at phase #a = cos(-90-(-30)) = cos (-60) = 0.5 lag
the power factor of load at phase #b = cos(-105-(-150)) = cos (45) = 0.707 lead
the power factor of load at phase #c = cos(90-90) = cos (0) = 1.0
First mid-term exam, first semester 1433/1434
For the three-phase circuit with negative sequence shown in Fig. 23, if Vb = 120∠30
determine: a) The line and phase currents,
b) The active power dissipated in phase A load,
c) The complex power delivered by the source.
Chapter Three: Three-Phase Circuits
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Fig. 23
the Y-Y circuit can be as follows:
a) Since negative sequence and Vb=120∠30 then:
Va = 120∠-90
V,
Vc = 120∠150 V
Taking KVL for both current I1 and I2 as follows:
for loop I1
120∠
90
207.864∠
120∠30 " C40∠60 60∠
120 " C62.91∠
7.11D
45D
# C60∠
# C60∠
45D
45D
C1D
for loop I2
120∠30
120∠150 Chapter Three: Three-Phase Circuits
" C60∠
45D # C60∠
By Dr. Ahmed Mustafa Hussein
45 20∠0D
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" C60∠
207.846∠0 45D # C75.48∠
34.2D C2D
Equations (1) and (2) can be arranged in matrix form as follows:
207.846∠ 120
62.91∠
E
FE
207.846∠0
60∠
62.91∠
G "H E
#
60∠
7.11
60∠
45
75.48∠
7.11
60∠
45
75.48∠
45
F G "H
34.2 #
45 I" 207.846∠ 120
F
F E
207.846∠0
34.2
1
207.846∠ 120
75.48∠ 34.2
60∠ 45
E
F E
F
G "H #
207.846∠0
60∠ 45
62.91∠ 7.11
3597∠7.434
4.594∠ 116.1
F
G "H E
#
1.3718∠ 86.49
Since the load is Y connected, so that the phase and line currents are equal and are
calculated as follows:
" 4.594∠
#
116.1 #
" 3.468∠52.63 #
# 1.3718∠93.51 #
b) The Active power for load at phase #a =
(4.5931)2 ×20 = 421.931 W
The complex power for load at phase #a =
(4.5931)2 ×40∠60 = 843.863∠60 VA
The complex power for load at phase #b =
(3.468)2 ×60∠-45 = 721.6214∠-45 VA
The complex power for load at phase #c =
Chapter Three: Three-Phase Circuits
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(1.3718)2 ×20∠0 = 37.637∠0 VA
c) The total complex power = 994.592∠12.811 VA ##
First mid-term exam, second semester 1433/1434
In the unbalanced Y-Y, three-phase circuit shown in Fig. 24, calculate the phase and
line currents if the neutral wire is a) connected b) not connected. Take Vp = 240 V
rms.
Fig. 24
If the neutral wire is connected,
given that Vph = 240 v (rms), then phase and line currents are:
240∠0
2.4∠
'100
240∠ 120
4.0∠
60
90 120 240∠120
3.0∠120 80
If the neutral wire is NOT connected,
given that Vph = 240 v (rms), then phase and line currents are:
loop I1
240∠0
240∠
Chapter Three: Three-Phase Circuits
120 " J60 '100K
By Dr. Ahmed Mustafa Hussein
# J60K
EE2020
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loop I2
415.6922∠30 " J60 '100K
240∠
120
# J60K
240∠120 # J140K
90 # J140K
415.6922∠
" J60K
" J60K
in matrix form
E
60 '100
415.6922∠30
FE
415.6922∠ 90
60
60 '100
G "H E
#
60
∆ C60 '100D ? 140
60 "
FG H
140 #
60 I" 415.6922∠30
F E
F
415.6922∠ 90
140
60 ? 60 14800∠71.0754
1 140
60
415.6922∠30
F
G "H G
HE
#
∆ 60 60 '100 415.6922∠ 90
3.147 ∠ 66.36
F
G "H E
#
4.351∠ 82.243
Therefore, the phase and line currents are:
" 3.147 ∠
#
" 4.351∠
Chapter Three: Three-Phase Circuits
66.36 # 4.351∠97.757
82.243
3.147 ∠
66.36 1.58∠
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EE2020
115.3 Page 31
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Report:
First Mid-term Exam, First semester (1432/1433)
For the three-phase, 3-wire, Y-Y
circuit shown, if the source has
positive sequence with phase voltage
va = 110∠0. Calculate:
a) The line and phase currents,
b) The source complex power,
c) The total dissipated power
through the transmission line,
d) The total complex power
supplied to the load.
4.2 ∆-∆
∆ Unbalanced 3-phase systems
Final exam, second semester 1433/1434
For the three-phase, positive-sequence circuit shown in Fig. 25, if the supply voltage
Vab is 100V, find the real power absorbed by the load.
Fig. 25
Since +ve sequence, Vab = 100 v, Vbc = 100∠–120 v and Vca = 100 ∠120 v.
For loop aABba:
100 = I1 (18 – J6) – I2 (5) – I3 (8 – J6) ………………………………………….(1)
Chapter Three: Three-Phase Circuits
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For loop cbBCc
100∠–120 = – I1 (5) + I2 (20) – I3 (10) …………………………………………(2)
For loop ABCA
0 = – I1 (8 – J6) – I2 (10) + I3 (22 – J3) …………………………………………(3)
Arrange the above 3 equations in matrix form as:
18 '6
100
5
M100∠ 120N M
C8 j6D
0
5
20
10
C8
j6D I"
10 N MI# N
22 j3 I&
The solution is:
18 '6
I"
5
M I# N M
I&
C8 j6D
5
20
10
C8
j6D I"
100
10 N M100∠ 120N
22 j3
0
Then the three currents can be obtained as:
I1 = 6.682∠-38.326 A
I2 = 8.725∠-92.935 A
I3 = 6.857∠-77.486 A
Another solution:
Multiplying Eqn. (3) by 2.0
0 = – I1 (16 – J12) – I2 (20) + I3 (44 – J6) ………………………………………(4)
Then add Eqn. (4) to Eqn. (2)
100∠–120 = I1 (–21+J12) + I3 (34 – J6) …………………………………………(5)
Multiplying Eqn. (1) by 4.0
400 = I1 (72 – J24) – I2 (20) – I3 (32 – J24) ..…………………………………….(6)
Then add Eqn. (6) to Eqn. (2)
360.56 ∠–13.9 = I1 (67 – J24) + I3 (–42 + J24) ………………………………… (7)
Chapter Three: Three-Phase Circuits
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From Eqn. (5), we can get an expression for I3
& 100∠
120 C 21 12D"
2.9∠
34 6
110
C0.7∠160.263D"
Substituting by the value of I3 in eqn. (7)
360.56 ∠–13.9 = I1 (67 – J24) + {2.9∠–110 – 0.7 ∠160.263 I1 }(–42 + J24)
360.56 ∠–13.9 = I1 (67 – J24) +140.283∠40.255 +33.86∠130.52 I1
360.56 ∠–13.9 – 140.283∠40.255 = 45.034 ∠ 2.22 I1
So that I1 = 6.678 ∠ –38.337 A
###
Substituting by the value of I1 to get I3
I3 = 6.855 ∠ –77.55 A
###
From Eqn. (1) we can get the value of I2 as
100=6.678 ∠ –38.337 (18–J6) – 5I2 – 6.855 ∠ –77.55(8–J6)
100= 126.706∠ –56.772 – 5I2 –68.55∠ –114.42
I2 = 8.726∠–92.93 A
###
The current through the 8Ω resistance of the branch AB = I1 – I3
= 6.678 ∠ –38.337 – 6.855 ∠ –77.55 = 4.544∠34.159 A
Then the power dissipated in 8Ω resistance (P1) = (4.544)2×8 = 165.196 W
The current through the 4Ω resistance of the branch CA = I3 =6.855 ∠ –77.55 A
Then the power dissipated in 4Ω resistance (P2) = (6.855)2×4 = 187.964 W
The current through the 10Ω resistance of the branch BC = I2 – I3
= 8.726∠–92.93 – 6.855 ∠ –77.55 = 2.79∠–133.593 A
Then the power dissipated in 10Ω resistance (P3) = (2.79)2×10 = 77.841 W
The real power absorbed by the load = P1 + P2 + P3 = 165.196 + 187.964 + 77.841
= 431 W
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Final Exam, Second semester (1432/1433)
Consider the three-phase, ∆-∆ system shown in Fig. 26, where Z1 = 8 + j6 Ω ,
Z2 = 4.2 – j2.2 Ω, and Z3 = 10 + j0Ω.
a) Find the phase currents IAB, IBC, ICA.
b) Calculate line currents IaA, IbB, and IcC.
Fig. 26
5. Matlab Applications
In this section we will build a simulink model for balanced ∆-∆ system and a balanced
Y-Y system. We will present the phase and line quantities. Take R=5Ω, L=1 mH
Chapter Three: Three-Phase Circuits
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Three-phase, ∆-∆
∆, Balanced Circuit (Positive Sequence) (Va=50∠
∠0)
Oscilloscope 3
Oscilloscope 1
Oscilloscope 2
Oscilloscope 1
Chapter Three: Three-Phase Circuits
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EE2020
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Oscilloscope 2
Oscilloscope 3
Chapter Three: Three-Phase Circuits
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EE2020
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Three-phase, Y-Y, Balanced Circuit (Positive Sequence) (Va=50∠
∠0)
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Chapter Three: Three-Phase Circuits
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EE2020
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Configuring the simulation parameters
Chapter Three: Three-Phase Circuits
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EE2020
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Electric Circuit Analysis (EE 2020)
Sheet (2) Three-Phase Circuits
< <DNLNL<†ãÒE<íée†ãÓÖ]<†ñ]æ‚Ö]<Øé×
äqæù]<íémøm<†ñ]æ‚Ö]<DNE<ké
Problem #1
A) If Vab = 400 V in a balanced Y-connected three-phase generator, find the phase
voltages, assuming the phase sequence is:
(a) abc
(b) acb
B) What is the phase sequence of a balanced three-phase circuit for which Van =
160∠ 30◦ V and Vcn = 160 ∠ –90◦ V? Find Vbn.
C) Determine the phase sequence of a balanced three-phase circuit in which Vbn =
208∠ 130◦ V and Vcn = 208∠ 10◦ V. Obtain Van.
D) Assuming abc sequence, if Vca = 208∠ 20◦ V in a balanced three-phase circuit,
find Vab, Vbc, Van, and Vbn.
E) Given that the line voltages of a three-phase Y-connected circuit are:
Vab = 420∠ 0◦, Vbc = 420 ∠ –120◦ Vac = 420 ∠120◦ V
Find the phase voltages Van, Vbn, and Vcn.
Problem #2
A) For the Y-Y circuit shown in Fig. 1, find the line currents, the line voltages, and
the load voltages.
Fig. 1 Problem 2-A
B) Obtain the line currents in the three-phase circuit shown in Fig. 2.
Chapter Three: Three-Phase Circuits
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Fig. 2 Y-Y Problem 2-B
C) A balanced Y-Y, four-wire system has phase voltages Van = 120∠ 0◦, Vbn = 120
∠ –120◦Vcn = 120 ∠120◦ V. The load impedance per phase is 19 + j13 Ω, and
the line impedance per phase is 1 + j2 Ω. Solve for the line currents and neutral
current.
D) For the circuit shown in Fig. 3, determine the current in the neutral line.
Fig. 3 Y-Y Problem 2-D
Problem #3
A) For the positive-sequence, three-phase circuit shown in Fig. 4, IbB = 30 ∠60◦ A
and VBC = 220 ∠10◦ V. Find Van, VAB, IAC, and Z.
Chapter Three: Three-Phase Circuits
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EE2020
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Fig. 4. Problem 3-A
B) Find the line and phase currents at the load side in the Y-∆ circuit shown in Fig.
5. Take Z∆ = 60 ∠45◦.
Fig. 5. Problem 3-B
C) In a Y-∆, 3-ph circuit, the source is a balanced, positive sequence with Van =
120 ∠0◦ V. It feeds a balanced load with Z∆ = 9 + j 12 Ω per phase through a
balanced line with Zl = 1 + j 0.5 Ω per phase. Calculate the phase voltages and
currents at the load side.
Problem #4
A) Find the currents Ia , Ib, and Ic in the 3-ph network shown in Fig. 6. Take Z∆ =
12 − j15 Ω, ZY = 4 + j6 Ω, and Zl = 2 Ω.
Chapter Three: Three-Phase Circuits
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Fig. 6. Problem 4-A
B) For the ∆ - ∆ circuit shown in Fig. 7. Find the line and phase currents. Assume
that the load impedance Zl = 12 + j9 Ω per phase.
Fig. 7. Problem 4-B
C) A balanced delta-connected source has phase voltage Vab = 416 ∠30◦ V and a
positive phase sequence. If this is connected to a balanced delta-connected
load, find the line and phase currents. Take the load impedance per phase as
60 ∠30◦ Ω and line impedance per phase as 1 + j1 Ω.
Problem #5
A) In the circuit of Fig. 8, if Vab = 440 ∠10◦, Vbc = 440 ∠250◦, Vca = 440 ∠130◦
V, find the line currents and the load voltages.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 8. Problem 5-A
B) A delta-connected generator supplies a balanced Y-connected load with an
impedance of 30 ∠− 60◦ Ω. If the line voltages of the generator have a
magnitude of 400 V and are in the positive phase sequence, find the line currents
and the phase voltages at the load.
Problem #6
A) A balanced Y-connected load absorbs a 3-ph power of 5 kW at a leading power
factor of 0.6 when connected to a line voltage of 240 V. Find the impedance of
each phase and the total complex power of the load.
B) A balanced Y-load is connected to a 60-Hz three-phase source with Vab = 240
∠0°V. The load has pf = 0.5 lagging and each phase draws 5 kW. (a)
Determine the load impedance ZY . (b) Find Ia, Ib, and Ic.
C) A three-phase source delivers 4800 VA to a Y-connected load with a phase
voltage of 208 V and a power factor of 0.9 lagging. Calculate the source line
current and the source line voltage.
D) Given the circuit in Fig. 9 below, find the total complex power absorbed by the
load.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 9. Problem 6-D
E) The following three parallel-connected three-phase loads are fed by a balanced
three-phase source.
Load #1: 250 kVA, 0.8 pf lagging
Load #2: 300 kVA, 0.95 pf leading
Load #3: 450 kVA, unity pf
If the line voltage is 13.8 kV, calculate the line current and the power factor of the
source. Assume that the line impedance is zero.
F) Assume that the two balanced loads shown in Fig. 10 are supplied by an 840-V
rms 60-Hz line.
Load #1: Y-connected with 30+j40 Ω per phase,
Load #2: balanced three-phase motor drawing 48 kW at a power factor of 0.8
lagging.
Assuming abc sequence, calculate:
a) The complex power absorbed by the combined load,
b) The kVAR rating of each of the three capacitors ∆-connected in parallel with
the load to raise the power factor to unity,
c) The current drawn from the supply at unity power factor condition.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
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Fig. 10 Problem 6-F
Problem #7
A) Find the real power absorbed by unbalanced 3-phase load given in Fig. 11.
Fig. 11 Problem 7-A
B) A balanced, positive-sequence Y-connected source has Van = 240 ∠0°V rms
and supplies an unbalanced delta-connected load via a transmission line with
impedance 2 + j3 Ω per phase. (a) Calculate the line currents if ZAB = 40 + j15
Ω , ZBC = 60 Ω , ZCA = 18 – j12Ω. (b) Find the complex power supplied by
the source.
C) Consider the Δ - Δ system shown in Fig. 12. Take Z1 = 8 + j6 Ω , Z2 = 4.2 –
j2.2 Ω, Z3 = 10 + j0Ω.
(a) Find the phase current IAB, IBC, ICA.
(b) Calculate line currents IaA, IbB, and IcC.
Fig. 12, Problem 7-C
Chapter Three: Three-Phase Circuits
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D) In the Y-Y system shown in Fig. 13, loads connected to the source are
unbalanced.
(a) Calculate Ia, Ib, and Ic.
(b) Find the total power delivered to the load. Take Vp = 240 V rms.
Fig. 13, Problem 7-D
E) A balanced three-phase Y-source with VP = 210 V rms drives a Y-connected
three-phase load with phase impedance ZA = 80Ω, ZB = 60 + j90Ω, and ZC =
j80Ω. Calculate the line currents and total complex power delivered to the load.
Assume that the neutrals are connected.
Chapter Three: Three-Phase Circuits
By Dr. Ahmed Mustafa Hussein
EE2020
Page 48