ELL 100 - Introduction to Electrical Engineering
LECTURE 17: THREE-PHASE AC CIRCUITS
Outline
 Disadvantages of the Single-Phase System
 Advantages of Three-Phase System
 Balanced Three-Phase System
 Y-Y connection (i.e., Y-connected source with a Y-connected load).
 Y-Δ connection.
 Δ-Δ connection.
 Δ-Y connection.
 Line and Phase Quantities in Three Phase Circuits
 Unbalanced Three-Phase System
2
INTRODUCTION
3-Phase Electric
Arc Furnace
Courtesy:http://metallurgymachine.com/big_img.html?etw_path=http://metallurgymachine.com/7-1-3-40ton-electric-arc furnace.html&big_etw_img
3
=product/7-1-3b.jpg/ (Available Online: 06 Dec. 2019)
INTRODUCTION
150 ton Industrial Water
Chiller Plant
4
INTRODUCTION
Industrial Fans (Vacuum
Pumps)
Courtesy: http://vactech.com.my/product/vacuum-pumps/industrial-fans-piller/ (Available Online: 06 Dec. 2019)
5
INTRODUCTION
Tesla Model S Rear Drive
Unit: The 3-phase 4-pole
induction motor
6
INTRODUCTION
Wind turbine 3-phase generator
7
INTRODUCTION
3-Phase CO2 Metal Laser
Cutting Equipment
Courtesy:
http://www.laserdiecuttingmachine.com/sale-6143645-1500w-3-phase-co2-metal-laser-cutting-equipment-for-die-cutting-factory.html
(Available Online: 06 Dec. 2019)
8
INTRODUCTION
IGBT based 3-Phase Locomotives
(by single-phase to 3-phase
conversion using power electronic
based technology)
Courtesy: https://indiarailinfo.com/faq/post/difference-between-igbt-and-gto-technology-of-3-phase-ac-electric-locomotives/1875 (Available
06 Dec. 2019)
Online:
9
INTRODUCTION
Single-Phase Systems:
• Single phase system consists of a generator connected through a pair of
wires (a transmission line) to a load.
• Vp is the rms magnitude of the source voltage.
• ϕ is the phase.
Single-phase two-wire system
10
INTRODUCTION
Single-Phase Systems:
• 3-wire system contains two identical sources (equal magnitude and
same phase) that are connected to two loads by two outer wires and
the neutral.
Single-phase three-wire system
11
INTRODUCTION
Polyphase Systems:
• Circuits or systems in which the AC sources operate at the same
frequency but different phases are known as polyphase.
• Examples:
• Two-phase systems.
• Three-phase systems, etc…
12
INTRODUCTION
Two-Phase Systems:
• It is produced by a generator consisting of two coils placed perpendicular
to each other so that the voltage generated by one lags the other by 90°.
Two-phase three-wire system
13
INTRODUCTION
Three-Phase Systems:
• It is produced by a generator consisting of three sources having the same
amplitude and frequency but out of phase with each other by 120°.
Three-phase four-wire system
14
DISADVANTAGES OF THE SINGLE-PHASE SYSTEM
• Initial application of AC supply was for heating the filaments of electric
lamps. For this, single-phase system was perfectly satisfactory.
• Few years later, AC motors were developed, and it was found that for this
application the single-phase system was not very satisfactory.
• For example, single-phase induction motor is not self-starting unless it is
fitted with an auxiliary winding.
• The single-phase induction motor is not self-starting and has poor
efficiency and power factor than the corresponding three-phase machine.
15
DISADVANTAGES OF THE SINGLE-PHASE SYSTEM
Single-phase induction motor
• Not self-starting
• Poor efficiency and power factor
Three-phase induction motor
• Self-starting
• Better efficiency and power factor
Courtesy: https://www.slideshare.net/mangatha1746/single-phase-induction-motor-50434369;
https://www.industrybuying.com/articles/how-does-a-3-phase-induction-motor-work/ (Available Online: 01 Dec. 2019)
16
ADVANTAGES OF THREE-PHASE SYSTEM
• Nearly, all electric power is generated and distributed in three-phase.
• When one-phase or two-phase inputs are required, they are taken from the
three-phase system rather than generated independently.
• Even when more than 3 phases are needed, they can be provided by
manipulating the available three phases.
Example: Aluminum industry, where 48 phases are required for melting
purposes.
17
ADVANTAGES OF THREE-PHASE SYSTEM
• The instantaneous power in a three-phase system can be constant (not
pulsating). This results in uniform power transmission and less vibration
of three-phase machines.
• For the same amount of power, the three-phase system is more
economical than the single-phase.
• The amount of wire required for a three-phase system is less than that
required for an equivalent single-phase system.
18
GENERATION OF BALANCED THREE-PHASE VOLTAGES
The generator basically consists of a
• Rotating magnet
(called the rotor).
• Stationary winding
(called the stator).
• Three separate windings or
coils with terminals a-a′, b-b′,
and c-c′ are physically placed
120° apart around the stator.
Three-phase generator
19
GENERATION OF BALANCED THREE-PHASE VOLTAGES
• As the rotor rotates, its magnetic field creates time-varying flux in the
three coils and induces voltages in the coils.
• Because the coils are placed
120°apart, the induced
voltages in the coils are
equal in magnitude but
out of phase by 120°.
The generated voltages are
120° apart from each other
20
BALANCED THREE-PHASE SYSTEM
• A typical three-phase system consists of three voltage sources connected
to loads by three or four wires (or transmission lines).
• The voltage sources can be either wye connected or delta-connected.
Y-connected source
Δ-connected source
21
BALANCED THREE-PHASE SYSTEM
• Phase voltages: voltages between lines a, b, and c and the neutral line n
(Van, Vbn, and Vcn).
• If the voltage sources have the same amplitude and frequency ω and are
out of phase with each other by 120°, the voltages are said to be balanced.
i.e.
Van  Vbn  Vcn  0
| Van || Vbn || Vcn |
Y-connected source
22
BALANCED THREE-PHASE SYSTEM
Phase Sequence:
• The phase sequence is the time order in which the voltages pass through
their respective maximum values.
1) abc sequence or positive sequence
2) acb sequence or negative sequence
Importance of Phase Sequence:
• It is important in three-phase power distribution because, it determines
the direction of the rotation of a motor connected to the power source.
23
BALANCED THREE-PHASE SYSTEM
abc Sequence or Positive Sequence:
• This sequence is produced when the rotor rotates
counterclockwise.
• Van leads Vbn, which in turn leads Vcn.
Van  V p 0
i.e.
Vbn  V p   120
Vcn  V p   240  V p   120
abc or positive sequence
where Vp is the effective or rms value of the phase voltages.
24
BALANCED THREE-PHASE SYSTEM
acb Sequence or Negative Sequence:
• It is produced when the rotor rotates in the
clockwise direction.
• Van leads Vcn, which in turn leads Vbn.
Van  V p 0
i.e. V  V   120
cn
p
Vbn  V p   240  V p   120
acb or negative sequence
25
NUMERICAL PROBLEMS
Q1. Determine the phase sequence of the set of voltages
Van  200 cos t  10
0

Vbn  200 cos t  2300 
Vcn  200 cos t  1100 
Ans: The voltages can be expressed in phasor form as
Van  200100 ; Vbn  200  2300 ; Vcn  200  1100
We notice that Van leads Vcn by 120° and Vcn in turn leads Vbn by 120°.
Hence, we have an acb (negative) sequence.
Q2. Given that Vbn  22030 V , find Van and Vcn, assuming a positive (abc)
sequence.
0
Answer: 2201500 V, 220  900 V.
26
BALANCED THREE-PHASE SYSTEM
Possible Three-Phase Load Configurations:
• Depending on the end application, a three-phase load can be either
• Wye-connected (or)
• Delta-connected.
• A balanced load is one in which the phase impedances are equal in
magnitude and in phase.
• However, a wye- or delta-connected load is said to be unbalanced if the
phase impedances are not equal in magnitude or phase.
27
BALANCED THREE-PHASE SYSTEM
Possible Three-Phase Load Configurations:
For a balanced wye-connected load,
Z1  Z 2  Z 3  ZY
where ZY is the load impedance per
phase.
Y-connected load
28
BALANCED THREE-PHASE SYSTEM
Possible Three-Phase Load Configurations:
For a balanced delta-connected load,
Z a  Zb  Zc  Z 
where ZΔ is the load impedance per
phase.
Impedance relation between Y and Δ
connected load:
1
Z   3  ZY or ZY   Z 
3
Δ-connected load
29
BALANCED THREE-PHASE SYSTEM
Possible Connections:
There exists four possible connections because, the three-phase source and
the three-phase load can be either Y or Δ-connected.
• Y-Y connection (i.e., Y-connected source with a Y-connected load).
• Y-Δ connection.
• Δ-Δ connection.
• Δ-Y connection.
30
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
A balanced Y-Y system is a threephase system with a balanced Yconnected source and a balanced
Y-connected load.
A balanced Y-Y system, showing the
source, line, and load impedances
31
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
• By lumping the impedances
together,
ZY  Z S  Z l  Z L
• Zs and Zl are often very small
compared with ZL, so one can
assume that ZY = ZL if no
source or line impedance is
given.
Balanced Y-Y connection with ZY = Zs + Zl + ZL
32
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
Assuming the positive sequence, the phase voltages (or line-to neutral
voltages) are,
Van  V p 0 , Vbn  V p   120 , Vcn  V p   120
The line-to-line voltages or simply line voltages
Vab, Vbc, and Vca are related to the phase voltages as,
Vab  Van  Vnb  Van  Vbn  V p 0  V p   120
 1
3
 V p 1   j
  3V p 30
2 
 2
33
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
Similarly, one can obtain:
Vbc  Vbn  Vcn  3V p   90
Vca  Vcn  Van  3V p   210
Thus, the magnitude of the line voltages is,
VL  3V p
where
V p | Van || Vbn || Vcn | and
VL | Vab || Vbc || Vca |
34
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
Phasor diagrams illustrating the relationship
between line voltages and phase voltages
35
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
By applying KVL to each phase,
line currents are obtained:
Van
Ia 
ZY
Vbn Van   120
Ib 

 I a   120
ZY
ZY
Vcn Van   240
Ic 

 I a   240
ZY
ZY
Balanced Y-Y connection
36
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
Now, one can readily infer that the line currents add up to zero,
I a  Ib  Ic  0
Therefore,
I n    I a  Ib  Ic   0
(or)
VnN  Z n I n  0
• The line current is the current in each line, the phase current is the current
in each phase of the source or load.
• In the Y-Y system, the line current is the same as the phase current.
37
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Wye Connection:
• An alternative way of analyzing a balanced Y-Y system is to do so on a
“per phase” basis.
• The single-phase analysis
yields the line current Ia as,
Van
Ia 
ZY
• The other line currents are
obtained from Ia, using the phase
sequence.

A single-phase equivalent
circuit of balanced Y-Y
connection
38
NUMERICAL PROBLEMS
Q1. Calculate the line currents in the three-wire Y-Y system of Fig. 1.
Fig. 1. For Q1.
39
NUMERICAL PROBLEMS
Ans. The three-phase circuit shown in the figure is balanced; we may replace
it with its single-phase equivalent circuit.
We obtain Ia from the single-phase
analysis as,

Van
Ia 
ZY
where
ZY   5  j 2   10  j8   15  j 6 
 16.15521.80
A single-phase equivalent
circuit of balanced Y-Y
connection
40
NUMERICAL PROBLEMS
Hence,
11000
0
Ia 

6.81


21.8
A
0
16.15521.8
In as much as the source voltages in Fig. 1 are in positive sequence, the line
currents are also in positive sequence:
I b  I a   1200  6.81  141.80 A
I c  I a   240  6.8198.2 A
0
0
41
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Delta Connection:
• A balanced Y-Δ system consists of a balanced Y- connected source
feeding a balanced Δ- connected load.
• This is perhaps the most practical three-phase system, as the three-phase
sources are usually Y-connected while the three-phase loads are usually
Δ-connected.
• It is to be noted that there is no neutral connection from source to load in
this system connection.
42
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Delta Connection:
Assuming the positive
sequence, the phase voltages
are again:
Van  V p 0 , Vbn  V p   120 ,
Vcn  V p   120
The line voltages are:
Vab  3V p 30  VAB
Vbc  3V p   90  VBC
Balanced Y-Δ connection
43
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Delta Connection:
Vca  3V p 150  VCA
The phase currents are:
I AB
VBC
VCA
VAB

, I BC 
, I CA 
Z
Z
Z
These currents have the same
magnitude but are out of phase
with each other by 120°.
Balanced Y-Δ connection
44
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Delta Connection:
• Another way to get phase
currents is to apply KVL.
• Applying KVL around loop
aABbna gives:
Van  Z  I AB  Vbn  0
(or)
I AB
Van  Vbn Vab VAB



Z
Z Z
Balanced Y-Δ connection
45
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Delta Connection:
The line currents are obtained from the phase currents by applying KCL at
nodes A, B, and C. Thus,
I a  I AB  I CA , I b  I BC  I AB , I c  I CA  I BC
Since I CA  I AB   240 ,
I a  I AB  I CA  I AB 1  1  240

=I AB 1  0.5  j 0.866 
Balanced Y-Δ connection
 I AB 3  30
46
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Delta Connection:
The magnitude of line current (IL) relates the magnitude of phase current (Ip)
as:
I L  3I p
where
I L | I a || I b || I c | and
I p | I AB || I BC || I CA |
Phasor diagram illustrating
the relationship between
phase and line currents
47
BALANCED THREE-PHASE SYSTEM
Balanced Wye-Delta Connection:
• An alternative way of
analyzing the Y-Δ circuit is
to transform the Δ-connected
load to an equivalent Yconnected load.

• Using the Δ-Y
transformation formula:
Z
ZY 
3
Single-phase equivalent circuit
of a balanced Y-Δ circuit
48
NUMERICAL PROBLEMS
0
V

100

10
V is
Q1. A balanced abc-sequence Y-connected source with an
connected to a Δ-connected balanced load (8 + j4) Ω per phase. Calculate the
phase and line currents.
Ans.
The load impedance is:
Z   8  j 4  8.94426.57 0 
If the phase voltage Van  100100 V , then the line voltage is
Vab  Van 3300  173.2400 V=VAB
49
NUMERICAL PROBLEMS
The phase currents are,
I AB
VAB
173.2400
0



19.36

13.43
A
0
Z  8.94426.57
I BC  I AB   1200  19.36  106.57 0 A
I CA  I AB   1200  19.36133.430 A
The line currents are,
I a  I AB 3  300  33.53  16.57 0 A
I b  I a   1200  33.53  136.57 0 A
I c  I a   120  33.53103.43 A
0
0
50
PRACTICE NUMERICAL PROBLEM
0
V

120


20
V.
Q2. One line voltage of a balanced Y-connected source is AB
If the source is connected to a Δ-connected load of 20400  , find the phase
and line currents. Assume the abc sequence.
Answer:
Phase currents (IAB, IBC, ICA): 6  600 A, 6  1800 A, 6600 A,
Line currents (IA, IB, IC): 10.392  900 A, 10.3921500 A, 10.392300 A.
51
BALANCED THREE-PHASE SYSTEM
Balanced Delta-Delta Connection:
• A balanced Δ-Δ system is one in which both the balanced source and
balanced load are Δ-connected.
• Goal is to obtain
the phase and line
currents.
Balanced Δ -Δ connection
52
BALANCED THREE-PHASE SYSTEM
Balanced Delta-Delta Connection:
Assuming a positive sequence, the phase voltages for a delta-connected
source are:
Vab  V p 0 , Vbc  V p   120 , Vca  V p   120
Also,
Vab  VAB , Vbc  VBC , Vca  VCA
Now, the phase currents are:
I AB
Balanced Δ -Δ connection
VBC Vbc
VCA Vca
VAB Vab


, I BC 

, I CA 

Z Z
Z Z
Z Z
53
BALANCED THREE-PHASE SYSTEM
Balanced Delta-Delta Connection:
• The line currents are also obtained from the phase currents by applying
KCL at nodes A, B, and C, as:
I a  I AB  I CA , I b  I BC  I AB , I c  I CA  I BC
• Also, each line current lags the corresponding
phase current by 30°. Moreover,
I L  3I p
Balanced Δ -Δ connection
54
NUMERICAL PROBLEMS
Q1. A balanced Δ-connected load with impedance 20 − j15 Ω is connected to
0
a Δ-connected, positive-sequence generator having Vab  3300 V.
Calculate the phase currents of the load and the line currents.
Ans: The load impedance per phase is Z   20  j15  25  36.57 
0
Since VAB = Vab, the phase currents are,
I AB
VAB
33000
0



13.2

36.87
A
0
Z  25  36.87
I BC  I AB   1200  13.2  83.130 A
I CA  I AB   120  13.2156.87 A
0
0
55
NUMERICAL PROBLEMS
For a delta load, the line current always lags the corresponding phase current
by 30° and has a magnitude √3 times that of the phase current. Hence, the
line currents are,
I a  I AB 3  300  22.866.87 0 A
I b  I a   1200  22.86  113.130 A
I c  I a   1200  22.86126.87 0 A
56
PRACTICE NUMERICAL PROBLEM
Q2. A positive-sequence, balanced Δ-connected source supplies a balanced
Δ-connected load. If the impedance per phase of the load is 18 + j12 Ω and
II a  9.609350 A , find IAB and VAB.
Answer: 5.548650 A, 12098.690 V.
57
BALANCED THREE-PHASE SYSTEM
Balanced Delta-Wye Connection:
A balanced Δ-Y system consists of a balanced Δ-connected source feeding a
balanced Y- connected load.
• Again, assuming the abc
sequence, the phase
voltages of a deltaconnected source are:
Vab  V p 0 ,
Vbc  V p   120 ,
Vca  V p   120
Balanced Δ-Y connection
58
BALANCED THREE-PHASE SYSTEM
Balanced Delta-Wye Connection:
• Line currents are obtained in many ways. One way is to apply KVL to
loop aANBba as:
Vab  ZY I a  ZY I b  0
(or)
ZY  I a  I b   Vab  V p 0
Thus,  I a  I b  
V p 0
ZY
Here, I b  I a   120
Balanced Δ-Y connection
59
BALANCED THREE-PHASE SYSTEM
Balanced Delta-Wye Connection:
 1
3
I a  I b  I a 1  1  120   I a 1   j
  I a 330
2 
 2
 I a  Ib  
V p 0
ZY
 I a 330 
V p 0
ZY
Vp
 Ia 
3
  30
ZY
• From this, the other line currents Ib and Ic are obtained using the positive
phase sequence, i.e.,
I b  I a   120 , I c  I a   120
• Also note that phase currents are equal to the line currents.
60
NUMERICAL PROBLEMS
Q1. A balanced Y-connected load with a phase impedance of 40 + j25 Ω is
supplied by a balanced, positive sequence Δ-connected source with a line
voltage of 210 V. Calculate the phase currents. Use Vab as a reference.
Ans:
The load impedance is
ZY  40  j 25  47.17320 
and the source voltage is
Vab  21000 V
When the Δ-connected source is transformed to a Y-connected source,
Van 
Vab
3
  300  121.2  300 V
61
NUMERICAL PROBLEMS
The line currents are,
Van 121.2  300
0
Ia 


2.57


62
A
0
ZY
47.1232
I b  I a   1200  2.57  1780 A
I c  I a   1200  2.57580 A
which are the same as the phase currents.
62
UNBALANCED THREE-PHASE SYSTEM
An unbalanced system is caused by two possible situations:
• The source voltages are not equal in magnitude and/or differ in phase by
angles that are unequal, or
• Load impedances are unequal.
Thus, an unbalanced system is due to unbalanced voltage sources or an
unbalanced load.
• Unbalanced three-phase systems are solved by direct application of mesh
and nodal analysis.
63
UNBALANCED THREE-PHASE SYSTEM
• To simplify analysis, balanced source
voltages, but an unbalanced load is
considered.
• Since the load is unbalanced, ZA, ZB,
and ZC are not equal.
• The line currents are determined by
Ohm’s law as,
VAN
VBN
VCN
Ia 
, Ib 
, Ic 
ZA
ZB
ZC
Unbalanced three-phase Y-connected load
64
UNBALANCED THREE-PHASE SYSTEM
• The set of unbalanced line currents produces current in
the neutral line, which is not zero as in a balanced
system.
• Applying KCL at node N gives the neutral line current
as,
I n    I a  Ib  Ic 
• In a three-wire system where the neutral line is absent, one can still find
the line currents Ia, Ib, and Ic using mesh analysis. At node N, KCL must
be satisfied so that Ia+Ib+Ic = 0.
• The same could be done for an unbalanced Δ-Y, Y-Δ, or Δ-Δ three-wire
system.
65
NUMERICAL PROBLEMS
Q1. The unbalanced Y-load of Fig. 1 has balanced voltages of 100 V and
the acb sequence. Calculate the line currents and the neutral current.
Take ZA = 15 Ω, ZB = 10 + j5 Ω, ZC = 6 − j8 Ω.
Ans.
The line currents are,
10000
Ia 
 6.6700 A
15
0 0
120
1000
Ib 
 8.9493.440 A;
10  j 5
100-120
00 0
Ic 
 10  66.87 0 A
6  j8
Fig. 1 for Q1.
66
NUMERICAL PROBLEMS
The current in the neutral line is:
I n    I a  I b  I c     6.67  0.54  j8.92  3.93  j 9.2 
 10.06  j 0.28  10.06178.40 A
67
NUMERICAL PROBLEMS
Q2. In a three-phase (RYB sequence) four-wire system the line voltage is
400 V and resistive loads of 10 kW, 8 kW and 5 kW are connected between
the three line conductors and the neutral as in Fig. 2. Calculate:
(a) the current in each line;
(b) the current in the neutral conductor.
Fig. 2 for Q2.
68
NUMERICAL PROBLEMS
(a) Voltage to neutral 
Vline
3
 230 V
If IR, IY and IB are the currents taken by the 10 kW, 8 kW and 5 kW loads
respectively,
1000
I R  10 
 43.5 A
230
1000
IY  8 
 34.8 A
230
1000
IB  5
 21.7 A
230
69
NUMERICAL PROBLEMS
(b) The current in the neutral is the phasor sum of the three line currents. In
general, the most convenient method of adding such quantities is to calculate
the resultant horizontal and vertical components thus: horizontal component
is
I H  IY cos 300  I B cos 300  0.866   34.8  21.7   11.3 A
and vertical component is
IV  I R  IY cos 600  I B cos 600  13 A
Current in neutral:
IN 
11.3
2
 13
2
  17.2 A
70
NUMERICAL PROBLEMS
Q3. A delta-connected load is arranged as in Fig. 3 with RYB sequence.
The supply voltage is 400 V at 50 Hz. Calculate:
(a) the phase currents;
(b) the line currents.
Fig. 3 for Q3.
71
NUMERICAL PROBLEMS
Ans:
(a) If I1, I2, I3 are the phase currents in
loads RY, YB and BR respectively:
I1 = 400/100 = 4.0 A, in phase with VRY
I2 
400
2
2
20

60


 6.32 A
 60 
I2 lags VYB by an angle 2  tan    71034'
 20 
1
I 3  2  3.14  50  30 106  400  3.77 A
leading VBR by 90°.
72
NUMERICAL PROBLEMS
(b) If the current IR in line conductor R is assumed to
be positive when flowing towards the load, the phasor
representing this current is obtained by subtracting I3 from I1.
I   4    3.77    2  4  3.77  cos 300   56.3
2
R
2
2
 I R  7.5 A
The current in line conductor Y is obtained by
subtracting I1 from I2, as shown separately in this Figure.
But angle between I2 and I1 reversed is
2  600  11034 '
73
NUMERICAL PROBLEMS
I   4    6.32    2  4  6.32  cos11034 '   105.5
2
Y
2
2
 IY  10.3 A
Similarly, the current in line conductor B is obtained
by subtracting I2 from I3, as shown in Figure.
Angle between I3 and I2 reversed is
180  30  11 34 '  138 26 '
0
0
0
0
I   6.32    3.77    2  3.77  6.32  cos138 26 '   18.5
2
B
2
 I B  4.3 A
2
0
74
PRACTICE NUMERICAL PROBLEM
Q4. The unbalanced Δ-load of Fig. 4 is supplied by balanced line-to-line
voltages of 440 V in the positive sequence. Find the line currents. Take Vab as
reference.
Answer.
39.71  41.060 A, 64.12  139.80 A,
70.1374.27 A.
0
Fig. 4. Unbalanced Δ-load, for Q4
75
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q1. 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.
Answer. (a) 231  300 , 231  1500 , 231900 V
(a ) 231300 , 2311500 , 231  900 V
76
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q2. What is the phase sequence of a balanced three-phase circuit for which
0
0
V

120

30
V
Vcn =120
  −90°
90 VV? Find Vbn.
Van
120⧸
an = 120⧸ 30°and Vcn
Answer.
acb sequence, 1201500 V
77
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q3. Given a balanced Y-connected three-phase generator with a line-to-line
0
Vab =100

100

450 45
165
V ,determine the phase
V and V
voltage of Vab
100⧸
Vbc
=
100⧸
bc
sequence and the value of Vca.
Answer.
acb sequence, 100  750 V
78
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q4. For a Y-connected load, the time-domain expressions for three line-toneutral voltages at the terminals are:
vAN = 120 cos(ωt + 32°) V
vBN = 120 cos(ωt – 88°) V
vCN = 120 cos(ωt + 152°) V
Write the time-domain expressions for the line-to-line voltages vAB, vBC, and
vCA.
Answer. 207.8 cos(ωt + 62°) V, 207.8 cos(ωt − 58°) V,
207.8 cos(ωt −178°) V.
79
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q5. Obtain the line currents in the three-phase circuit of Figure shown.
Answer.
4453.130 A, 44  66.87 0 A,
44173.130 A.
Fig. For Q5
80
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q6. A balanced Y-Y four-wire system has phase voltages
Van  12000 V, Vbn  120  1200 V, Vcn  1201200 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.
Answer. 4.8  36.87 0 A, 4.8  156.87 0 A, 4.883.130 A.
81
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q7. In the Y-Δ system shown in Figure, the source is a positive sequence
0
V

440

0
with Vanan = 440⧸ 0°VV and phase impedance Zp = 2 – j3 Ω. Calculate the line
voltage VL and the line current IL.
Answer.
762.1 V, 366.1 A
Fig. For Q7
82
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q8. For the Δ-Δ circuit of shown Figure, calculate the phase and line
currents.
Answer.
13.915  18.430 A, 13.915  138.430 A,
13.915101.57 A;
0
24.1  48.430 A, 24.1  168.430 A,
24.171.57 0 A.
Fig. For Q8
83
MISCELLANEOUS PROBLEMS FOR PRACTICE
Q9. In the circuit of shown Figure, if
Vab  44010 V, Vbc  440  110 V, Vca  440130 V
0
0
0
Find the line currents.
Answer.
17.7424.780 A, 17.742  115.220 A,
17.742124.780 A.
Fig. For Q9
84
PRACTICE NUMERICAL PROBLEM
Q2. 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
0
sequence for the source voltages and that Van  12030 V , find:
(a) the line voltages, (b) the line currents.
Answer.
(a ) 207.860 V, 207.8  60 V, 207.8  180 V
0
0
0
(b) 3.75  8.660 A, 3.75  128.660 A, 3.75111.340 A
85
PRACTICE NUMERICAL PROBLEM
Q2. In a balanced Δ-Y circuit, Vab  440150 V
Calculate the line currents.
and ZY = (12 + j15) Ω.
Answer. 13.224  66.340 A, 13.224  173.660 A, 13.22453.660 A.
86
REFERENCES
[1]
Edward Hughes (revised by John Hiley, Keith Brown and Ian McKenzie
Smith),
Electrical
And
Electronic Technology. Pearson Education Limited, Edinburgh Gate, Harlow, Essex CM20 2JE,
England:10th Edition, 2008.
[2]
Charles K. Alexander, Matthew N. O. Sadiku, Fundamentals of Electric Circuits. 2 Penn Plaza,
York, NY, USA: McGraw-Hill Education, 6th Edition, 2017.
[3]
John Bird, Electrical Circuit Theory and Technology. Elsevier Science, Linacre House, Jordan Hill,
Oxford OX2 8DP, UK: Third Edition 2007.
[4]
Adrian Waygood, An Introduction to Electrical Science. Routledge (Taylor & Francis Group), 711 Third
Avenue, New York, NY: Second Edition 2019.
[5]
Allan R. Hambley, Electrical Engineering Principles and Applications. Pearson Higher Education, 1 Lake
Street, Upper Saddle River, NJ 07458: Sixth Edition 2014.
[6]
William H. Hayt, Jr., Jack E. Kemmerly, and Steven M. Durbin , Engineering Circuit Analysis. The
McGraw-Hill Companies, New York, NY: Eighth Edition 2012.
New
87