Electrical Circuit.pdf

ELL 100 - Introduction to Electrical Engineering
LECTURE 17: THREE-PHASE AC CIRCUITS

 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
Outline

INTRODUCTION
3
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
=product/7-1-3b.jpg/ (Available Online: 06 Dec. 2019)

INTRODUCTION
4
150 ton Industrial Water
Chiller Plant

INTRODUCTION
5
Industrial Fans (Vacuum
Pumps)
Courtesy: http://vactech.com.my/product/vacuum-pumps/industrial-fans-piller/ (Available Online: 06 Dec. 2019)

INTRODUCTION
6
Tesla Model S Rear Drive
Unit: The 3-phase 4-pole
induction motor

INTRODUCTION
7
Wind turbine 3-phase generator

INTRODUCTION
8
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)

INTRODUCTION
9
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 Online:
06 Dec. 2019)

INTRODUCTION
10
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

INTRODUCTION
11
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

INTRODUCTION
12
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…

INTRODUCTION
13
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

INTRODUCTION
14
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

DISADVANTAGES OF THE SINGLE-PHASE SYSTEM
15
• 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.

DISADVANTAGES OF THE SINGLE-PHASE SYSTEM
16
Single-phase induction motor Three-phase induction motor
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)
• Self-starting
• Better efficiency and power factor
• Not self-starting
• Poor efficiency and power factor

ADVANTAGES OF THREE-PHASE SYSTEM
17
• 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.

ADVANTAGES OF THREE-PHASE SYSTEM
18
• 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.

GENERATION OF BALANCED THREE-PHASE VOLTAGES
19
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

GENERATION OF BALANCED THREE-PHASE VOLTAGES
20
• 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

BALANCED THREE-PHASE SYSTEM
21
• 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

22
• 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.
Y-connected source
0
an bn cn
V V V
  
| | | | | |
an bn cn
V V V
 

23
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.

24
abc Sequence or Positive Sequence:
• This sequence is produced when the rotor rotates
counterclockwise.
• Van leads Vbn, which in turn leads Vcn.
i.e.
where Vp is the effective or rms value of the phase voltages.
abc or positive sequence
0
120
240 120
an p
bn p
cn p p
V V
V V
V V V
 
  
     

25
acb Sequence or Negative Sequence:
• It is produced when the rotor rotates in the
clockwise direction.
• Van leads Vcn, which in turn leads Vbn.
i.e.
acb or negative sequence
0
120
240 120
an p
cn p
bn p p
V V
V V
V V V
 
  
     

NUMERICAL PROBLEMS
26
Q1. Determine the phase sequence of the set of voltages
Ans: The voltages can be expressed in phasor form as
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 , find Van and Vcn, assuming a positive (abc)
sequence.
Answer:
 
0
200cos 230
bn
V t

 
 
0
200cos 10
an
V t

   
0
200cos 110
cn
V t

 
0 0 0
200 10 ; 200 230 ; 200 110
an bn cn
V V V
       
0
220 30 V
bn
V  
0 0
220 150 V, 220 90 V.
  

27
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.

28
For a balanced wye-connected load,
where ZY is the load impedance per
phase.
1 2 3 Y
Z Z Z Z
  
Y-connected load

29
For a balanced delta-connected load,
where ZΔ is the load impedance per
phase.
Impedance relation between Y and Δ
connected load:
a b c
Z Z Z Z
  
Δ-connected load
1
3 or
3
Y Y
Z Z Z Z
 
   

30
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.

31
Balanced Wye-Wye Connection:
A balanced Y-Y system is a three-
phase system with a balanced Y-
connected source and a balanced
Y-connected load.
A balanced Y-Y system, showing the
source, line, and load impedances

32
• By lumping the impedances
together,
• 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
Y S l L
Z Z Z Z
  

33
Assuming the positive sequence, the phase voltages (or line-to neutral
voltages) are,
The line-to-line voltages or simply line voltages
Vab, Vbc, and Vca are related to the phase voltages as,
0 , 120 , 120
an p bn p cn p
V V V V V V
       
0 120
1 3
1 3 30
2 2
ab an nb an bn p p
p p
V V V V V V V
V j V
        
 
    
 
 

34
Similarly, one can obtain:
Thus, the magnitude of the line voltages is,
where
3 90
bc bn cn p
V V V V
    
3 210
ca cn an p
V V V V
    
3
L p
V V

| | | | | | and
| | | | | |
p an bn cn
L ab bc ca
V V V V
V V V V
  
  

35
Phasor diagrams illustrating the relationship
between line voltages and phase voltages

36
By applying KVL to each phase,
line currents are obtained:
Balanced Y-Y connection
an
a
Y
V
I
Z

120
120
bn an
b a
Y Y
V V
I I
Z Z
 
    
240
240
cn an
c a
Y Y
V V
I I
Z Z
 
    

37
Now, one can readily infer that the line currents add up to zero,
Therefore,
(or)
• 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.
0
a b c
I I I
  
  0
n a b c
I I I I
    
0
nN n n
V Z I
 

38
• 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,
• The other line currents are
obtained from Ia, using the phase
sequence.
an
a
Y
V
I
Z

A single-phase equivalent
circuit of balanced Y-Y
connection


NUMERICAL PROBLEMS
39
Q1. Calculate the line currents in the three-wire Y-Y system of Fig. 1.
Fig. 1. For Q1.

NUMERICAL PROBLEMS
40
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,
where
A single-phase equivalent
circuit of balanced Y-Y
connection
an
a
Y
V
I
Z

     
0
5 2 10 8 15 6
16.155 21.8
Y
Z j j j
     
 


NUMERICAL PROBLEMS
41
Hence,
In as much as the source voltages in Fig. 1 are in positive sequence, the line
currents are also in positive sequence:
0
0
0
110 0
6.81 21.8 A
16.155 21.8
a
I

   

0 0
0 0
120 6.81 141.8 A
240 6.81 98.2 A
b a
c a
I I
I I
     
    

42
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.

43
Assuming the positive
sequence, the phase voltages
are again:
The line voltages are:
Balanced Y-Δ connection
0 , 120 ,
120
an p bn p
cn p
V V V V
V V
    
  
3 30
ab p AB
V V V
  
3 90
bc p BC
V V V
   

44
The phase currents are:
These currents have the same
magnitude but are out of phase
with each other by 120°.
3 150
ca p CA
V V V
  
, ,
BC CA
AB
AB BC CA
V V
V
I I I
Z Z Z
  
  

45
• Another way to get phase
currents is to apply KVL.
• Applying KVL around loop
aABbna gives:
(or)
0
an AB bn
V Z I V

   
an bn ab AB
AB
V V V V
I
Z Z Z
  

  

46
The line currents are obtained from the phase currents by applying KCL at
nodes A, B, and C. Thus,
Since
, ,
a AB CA b BC AB c CA BC
I I I I I I I I I
     
240 ,
CA AB
I I
  
 
 
1 1 240
= 1 0.5 0.866
3 30
a AB CA AB
AB
AB
I I I I
I j
I
     
 
  

47
The magnitude of line current (IL) relates the magnitude of phase current (Ip)
as:
where
3
L p
I I

| | | | | | and
| | | | | |
L a b c
p AB BC CA
I I I I
I I I I
  
  
Phasor diagram illustrating
the relationship between
phase and line currents

48
• An alternative way of
analyzing the Y-Δ circuit is
to transform the Δ-connected
load to an equivalent Y-
connected load.
• Using the Δ-Y
transformation formula:
Single-phase equivalent circuit
of a balanced Y-Δ circuit
3
Y
Z
Z 



NUMERICAL PROBLEMS
49
Q1. A balanced abc-sequence Y-connected source with is
connected to a Δ-connected balanced load (8 + j4) Ω per phase. Calculate the
phase and line currents.
Ans.
The load impedance is:
If the phase voltage , then the line voltage is
0
100 10 V
an
V  
0
8 4 8.944 26.57
Z j
     
0
100 10 V
an
V  
0 0
3 30 173.2 40 V=
ab an AB
V V V
   

NUMERICAL PROBLEMS
50
The phase currents are,
The line currents are,
0
0
0
0 0
0 0
173.2 40
19.36 13.43 A
8.944 26.57
120 19.36 106.57 A
120 19.36 133.43 A
AB
AB
BC AB
CA AB
V
I
Z
I I
I I


   

     
    
0 0
0 0
0 0
3 30 33.53 16.57 A
120 33.53 136.57 A
120 33.53 103.43 A
a AB
b a
c a
I I
I I
I I
     
     
    

PRACTICE NUMERICAL PROBLEM
51
Q2. One line voltage of a balanced Y-connected source is .
If the source is connected to a Δ-connected load of , find the phase
and line currents. Assume the abc sequence.
Answer:
Phase currents (IAB, IBC, ICA):
Line currents (IA, IB, IC):
0
120 20 V
AB
V   
0
20 40
 
0 0 0
0 0 0
6 60 A, 6 180 A, 6 60 A,
10.392 90 A, 10.392 150 A, 10.392 30 A.
    
   

52
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

53
Assuming a positive sequence, the phase voltages for a delta-connected
source are:
Also,
Now, the phase currents are: Balanced Δ -Δ connection
0 , 120 , 120
ab p bc p ca p
V V V V V V
       
, ,
ab AB bc BC ca CA
V V V V V V
  
, ,
ab BC bc CA ca
AB
AB BC CA
V V V V V
V
I I I
Z Z Z Z Z Z
     
     

54
• The line currents are also obtained from the phase currents by applying
KCL at nodes A, B, and C, as:
• Also, each line current lags the corresponding
phase current by 30°. Moreover,
Balanced Δ -Δ connection
, ,
a AB CA b BC AB c CA BC
I I I I I I I I I
     
3
L p
I I


NUMERICAL PROBLEMS
55
Q1. A balanced Δ-connected load with impedance 20 − j15 Ω is connected to
a Δ-connected, positive-sequence generator having
Calculate the phase currents of the load and the line currents.
Ans: The load impedance per phase is
Since VAB = Vab, the phase currents are,
0
330 0 V.
ab
V  
0
20 15 25 36.57
Z j
      
0
0
0
330 0
13.2 36.87 A
25 36.87
AB
AB
V
I
Z

   
 
0 0
0 0
120 13.2 83.13 A
120 13.2 156.87 A
BC AB
CA AB
I I
I I
     
    

NUMERICAL PROBLEMS
56
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,
0 0
0 0
0 0
3 30 22.86 6.87 A
120 22.86 113.13 A
120 22.86 126.87 A
a AB
b a
c a
I I
I I
I I
    
     
    

57
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
I , find IAB and VAB.
Answer:
0
9.609 35 A
a
I  
0 0
5.548 65 A, 120 98.69 V.
 

58
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 delta-
connected source are:
Balanced Δ-Y connection
0 ,
120 ,
120
ab p
bc p
ca p
V V
V V
V V
 
  
  

59
• Line currents are obtained in many ways. One way is to apply KVL to
loop aANBba as:
(or)
Thus,
Here, Balanced Δ-Y connection
0
ab Y a Y b
V Z I Z I
   
  0
Y a b ab p
Z I I V V
   
 
0
p
a b
Y
V
I I
Z

 
120
b a
I I
  

60
• From this, the other line currents Ib and Ic are obtained using the positive
phase sequence, i.e.,
• Also note that phase currents are equal to the line currents.
  1 3
1 1 120 1 3 30
2 2
a b a a a
I I I I j I
 
         
 
 
 
30
0 0 3
3 30
p
p p
a b a a
Y Y Y
V
V V
I I I I
Z Z Z
 
 
      
120 , 120
b a c a
I I I I
     

NUMERICAL PROBLEMS
61
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
and the source voltage is
When the Δ-connected source is transformed to a Y-connected source,
0
40 25 47.17 32
Y
Z j
    
0
210 0 V
ab
V  
0 0
30 121.2 30 V
3
ab
an
V
V      

NUMERICAL PROBLEMS
62
which are the same as the phase currents.
0
0
0
0 0
0 0
121.2 30
2.57 62 A
47.12 32
120 2.57 178 A
120 2.57 58 A
an
a
Y
b a
c a
V
I
Z
I I
I I
 
    

     
    

UNBALANCED THREE-PHASE SYSTEM
63
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.

64
• 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,
Unbalanced three-phase Y-connected load
, ,
AN BN CN
a b c
A B C
V V V
I I I
Z Z Z
  

65
• 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,
• 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.
 
n a b c
I I I I
   

NUMERICAL PROBLEMS
66
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.
Fig. 1 for Q1.
0
0
100 0
6.67 0 A
15
a
I

  
0
0
100 0
8.94 93.44 A;
10 5
b
I
j

  

0
0
100 0
10 66.87 A
6 8
c
I
j

   

1200
-1200

NUMERICAL PROBLEMS
67
The current in the neutral line is:
   
0
6.67 0.54 8.92 3.93 9.2
10.06 0.28 10.06 178.4 A
n a b c
I I I I j j
j
         
    

NUMERICAL PROBLEMS
68
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.

NUMERICAL PROBLEMS
69
(a) Voltage to neutral
If IR, IY and IB are the currents taken by the 10 kW, 8 kW and 5 kW loads
respectively,
230 V
3
line
V
 
1000
10 43.5 A
230
1000
8 34.8 A
230
1000
5 21.7 A
230
R
Y
B
I
I
I
  
  
  

NUMERICAL PROBLEMS
70
(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
and vertical component is
Current in neutral:
 
0 0
cos30 cos30 0.866 34.8 21.7 11.3 A
H Y B
I I I
     
0 0
cos60 cos60 13 A
V R Y B
I I I I
   
   
 
2 2
11.3 13 17.2 A
N
I   

NUMERICAL PROBLEMS
71
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.

NUMERICAL PROBLEMS
72
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 lags VYB by an angle
leading VBR by 90°.
 
2
2 2
400
6.32 A
20 60
I  

1 0 '
2
60
tan 71 34
20
   
 
 
 
6
3 2 3.14 50 30 10 400 3.77 A
I 
      

NUMERICAL PROBLEMS
73
(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.
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
2 2 0
4 3.77 2 4 3.77 cos30 56.3
7.5 A
R
R
I
I
      
 
0 0
2 60 11 34'
  

NUMERICAL PROBLEMS
74
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
     
2
2 2 0
4 6.32 2 4 6.32 cos11 34' 105.5
10.3 A
Y
Y
I
I
      
 
     
2 2
2 0
6.32 3.77 2 3.77 6.32 cos138 26' 18.5
4.3 A
B
B
I
I
      
 
0 0 0 0
180 30 11 34' 138 26'
  

75
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.
Fig. 4. Unbalanced Δ-load, for Q4
0 0
0
39.71 41.06 A, 64.12 139.8 A,
70.13 74.27 A.
   


MISCELLANEOUS PROBLEMS FOR PRACTICE
76
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. 0 0 0
0 0 0
( ) 231 30 , 231 150 , 231 90 V
( ) 231 30 , 231 150 , 231 90 V
a
a
    
   

77
Q2. What is the phase sequence of a balanced three-phase circuit for which
Van = 120⧸ 30°and Vcn = 120⧸ −90° V? Find Vbn.
Answer.
0
120 30 V
an
V   0
120 90 V
cn
V   
0
sequence, 120 150 V
acb 

78
Q3. Given a balanced Y-connected three-phase generator with a line-to-line
voltage of Vab = 100⧸ 45 and Vbc = 100⧸ ,determine the phase
sequence and the value of Vca.
Answer.
0
100 45 V
ab
V   0
100 165 V
bc
V  
0
sequence, 100 75 V
acb  

79
Q4. For a Y-connected load, the time-domain expressions for three line-to-
neutral 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.

80
Q5. Obtain the line currents in the three-phase circuit of Figure shown.
Answer.
Fig. For Q5
0 0
0
44 53.13 A, 44 66.87 A,
44 173.13 A.
  


81
Q6. A balanced Y-Y four-wire system has phase voltages
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.
0
120 0 V,
an
V   0
120 120 V,
bn
V    0
120 120 V
cn
V  
0 0 0
4.8 36.87 A, 4.8 156.87 A, 4.8 83.13 A.
    

82
Q7. In the Y-Δ system shown in Figure, the source is a positive sequence
with Van = 440⧸ 0° V 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
0
440 0 V
an
V  

83
Q8. For the Δ-Δ circuit of shown Figure, calculate the phase and line
currents.
Answer.
Fig. For Q8
0 0
0
0 0
0
13.915 18.43 A, 13.915 138.43 A,
13.915 101.57 A;
24.1 48.43 A, 24.1 168.43 A,
24.1 71.57 A.
   

   


84
Q9. In the circuit of shown Figure, if
Find the line currents.
Answer.
Fig. For Q9
0
440 10 V,
ab
V   0
440 110 V,
bc
V    0
440 130 V
ca
V  
0 0
0
17.742 4.78 A, 17.742 115.22 A,
17.742 124.78 A.
  


85
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
sequence for the source voltages and that , find:
(a) the line voltages, (b) the line currents.
Answer.
0
120 30 V
an
V  
0 0 0
0 0 0
( ) 207.8 60 V, 207.8 60 V, 207.8 180 V
( ) 3.75 8.66 A, 3.75 128.66 A, 3.75 111.34 A
a
b
    
    

86
Q2. In a balanced Δ-Y circuit, and ZY = (12 + j15) Ω.
Calculate the line currents.
Answer.
0
440 15 V
ab
V  
0 0 0
13.224 66.34 A, 13.224 173.66 A, 13.224 53.66 A.
    

REFERENCES
87
[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, New
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.

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