Chapter 3 - 3Ø systems
A three phase load consists of 3 identical impedances, ZL are connected in Y configuration as
shown in Figure 1. This load is connected across a 50Hz three phase supply with line voltage
of 415V and consumes line current of 1A. The power factor measured is equal to one (unity
power factor). Justify the types of load in Figure 1. Calculate the total real power consumed
by the three phase load in Figure 1 and the value of ZL.
ZLoad
A
ZLoad
B
ZLoad
C
Figure 1. Three phase load in Y configuration.
Solution:
As the power factor is equal to 1, the voltage and current waveforms are in phase. The phase
angle of ZLoad (which is the phase shift between voltage and current waveforms) is 0o which
indicates that ZLoad consists of only real part. Therefore, Figure 1 shows a resistive load with
3 identical resistances connected in Y configuration.
ZLoad = R
VL  415V
I L  I p  1A
VL 415

 239.6V
3
3
S1  V p I p  (239.6)(1)  239.6VA
Vp 
P1  S1 ( PF )  239.6(1)  239.6W
P3  3P1  3(239.6)  718.8W
OR
S3  3VL I L  ( 3)(415)(1)  718.8VA
P3  S3 ( PF )  718.8(1)  718.8W
ZL  R 
Vp
Ip

239.6
 239.6
1
Chapter 5 - Transformer
Two tests were conducted on a transformer as shown in Figure 1. Referring to Figure 1,
answer the following questions:
Voltmeter
V = 230V
E
Secondary
Ammeter
I=?
Primary
Wattmeter
P = 80W
Transformer
a:1
(Open circuit)
Note: Current lags voltage by 70 degree
(a). Open circuit test.
Voltmeter
V = 10V
Secondary
E
Ammeter
I = 5A
Primary
Wattmeter
P = 20W
Transformer
a:1
(Short circuit)
(b). Short circuit test.
Figure 1. Measurement of transformer impedances.
(a). Calculate the open circuit current in Figure 1(a).
(b). Calculate the power factor during open circuit test and short circuit test.
(c). Sketch the approximate equivalent circuit of the transformer referred to primary side.
(d). Sketch the equivalent circuit for open circuit test and short circuit test.
(e). Calculate the following impedances: core resistance (Rc), magnetizing reactance (Xm),
equivalent winding resistance referred to primary side (ReqP), and equivalent leakage
reactance referred to primary side (XeqP).
(f). This transformer has total power loss of 100W at rated condition when supplying an
inductive load. The output current lags voltage by 50 degree. Given the apparent power rating
of this transformer is 1000VA, calculate the efficiency.
Solution:
(a).
Poc  Voc I oc cos oc
80  230( I oc ) cos 70
I oc  1.017 A
(b).
PFoc  cos oc  cos 70  0.342
Psc  Vsc I sc cos  sc  Vsc I sc ( PFsc )
20  (10)(5)( PFsc )
PFsc  0.4
(C).
ReqP
Ip
jXeqP
+
Vp
-
Is/a
+
jXm
Rc
aVs
Im
IRc
-
(d).
For open circuit test:
The equivalent circuit for open circuit test is given by
Ioc
+
Voc Rc
-
Zoc
For short circuit test:
The equivalent circuit for short circuit test is given by
jXm
Isc
ReqP
jXeqP
+
Vsc
-
Zsc
(e).
For open circuit test:
Poc  Voc I oc cos oc
I    oc
1
1
1


 oc
Z oc Rc jX m
Voc 0
1
j 1.017  70 1.017
cos(70 )  j sin(70 ) 



Rc X m
230
230 
By comparing real and imaginary parts of left and right hand sides,
1
Rc 
 661.23
1.5123 103
1
Xm 
 240.67
4.155 103
For short circuit test:
Psc  Vsc I sc cos  sc
 sc  cos 1
Psc
20
 cos 1
 66.42o
Vsc I sc
10  5
OR
PFsc  0.4  cos  sc
 sc  cos1 0.4  66.42o
Vsc 0
I sc    sc
By comparing real and imaginary parts of left and right hand sides,
10
10
ReqP  jX eqP 
 cos(66.42 )  j sin(66.42 ) 

5  66.42
5
ReqP  0.8
Z sc  ReqP  jX eqP 
X eqP  1.83
(f).
The efficiency is given by
P
  out  100%
Pin
where, Pin  Pout  Ploss
Pout

100%
Pout  Ploss
where, Pout  S ( PF )
Pout
S ( PF )
S (cos  )
100% 
100% 
100%
Therefore,  
Pout  Ploss
S ( PF )  Ploss
S (cos  )  Ploss
1000(cos 50)

100%  86.54%
1000(cos 50)  100
Chapter 6 - Electrical Machines
[Assignment and tutorial question]: Based on equivalent circuit of a shunt DC machine,
compare the operation in motor and generator mode.
Solution:
Shunt DC Motor
IA
Shunt DC Generator
IL
RA
RF
EA
VT
IF
LF
+
VT
IF
LF
-
Converts electrical to mechanical energy
VT  Ea  I a Ra
IF 
IL
RA
+
RF
EA
IA
-
Converts mechanical to electrical energy
VT  Ea  I a Ra
VT
RF
IF 
VT
RF
VT  Ea
E V
Ia  IL  IF  a T
Ra
Ra
Both shunt DC motor and shunt DC generator have same circuit diagram except that the
polarity of armature and line current is opposite. For operation in motor mode, the electrical
power is supplied across VT terminal and hence current flows from VT towards armature. In
contrast, operating the machine in generator mode produces electrical energy from rotating
armature and hence the electrical current generated flows from armature towards output
terminal at VT.
Ia  IL  IF 
[Similar to tutorial Q2]- A 4-pole DC shunt motor has armature and field resistances equal of
0.5Ω and 500Ω respectively. The input terminal is connected to a 200V DC supply. The
motor takes a line current of 3A at no load while running at 1500 rpm. Determine the line
current when the motor runs at 1300 rpm.
IA
IL
RA
RF
EA
VT
IF
LF
case 1: no load
EA1=?
IL1=3A
n1=1500rpm
+
-
E A1  VT  I a1 Ra  VT  ( I L1  I F ) Ra  VT  ( I L1 
E A1  VT  ( I L1 
VT
) Ra
RF
VT
200
) Ra  200  (3 
)0.5  198.7V
RF
500
case 2
EA2=?
IL2=?
n2=1300rpm
Important equation:
E A1 n1

E A 2 n2
198.7 1500

E A2
1300
EA2  172.2V
I L2  Ia2  I F 
VT  E A2 VT 200  172.2 200



 56 A
Ra
RF
0.5
500
Chapter 4 - Electricity & Magnetism
[Tutorial Q2] - Explain how voltage can be induced in a conductor surrounded by magnetic
field.
If a conductor is moved through a magnetic field so that it cuts magnetic lines of flux and a
voltage will be induced across the conductor. The magnitude of induced voltage is
proportional to the rate of changes of magnetic field (speed of movement) determined by
Faraday's Law.
Induced emf, e 
d
dt
[Tutorial Q3] - Define Right Hand Rule.
Case 1
Case 2
Case 1: - thumb: Direction of magnetic flux - other fingers: Direction of current flow
Case 2: - thumb: Direction of current flow - other fingers: Direction of magnetic flux
[Similar to Tutorial Q7] - Find the current I, in the Figure 1. Given Φ = 0.2x10-3Wb. Find the
relative permeability of each material.
Figure 1. Magnetic core energized by a current carrying conductor coil.
B-H curve
B-H curve -"Zoomed in"
Solution:
 0.2 103
B 
 0.625T
A 3.2 104
From B-H curve, the magnetizing force of cast iron is:
Hiron = 3180 At/m
From B-H curve -"zoomed in", the magnetizing force of cast steel is:
Hsteel = 360 At/m
  NI  H ironliron  H steel lsteel
(300) I  (3180)(0.06)  (360)(0.14)
I  0.804 A
B  r _ iron o H iron
0.625  r _ iron (4 107 )(3180)
r _ iron  156.4
B  r _ steel o H steel
0.625  r _ steel (4 107 )(360)
r _ steel  1381.6