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Transformer
DC Machines
AC Machines
DC and AC Meters
DC and AC Bridges
Sensor and Transducers
Lab 0 – Lab Introduction
Lab 1 – Single Phase Transformer; Voltage and Current Ratio
Lab 2 – DC Series Motor
Lab 3 – Three Phase AC Induction Motor
Lab 4 – d’Arsonval Galvanometer
Lab 5 – The Basic Voltmeter Design
Lab 6 – The Wheatstone Bridge
A transformer is a device that changes ac electric power at one voltage level
to another voltage level through the action of a magnetic field.
Power Distribution
System
DC System
Thomas A. Edison
For incandescent bulb
Low Voltage
(120-V)
High Current
Huge Power Loss in
Transmission Line
AC System
(Transformer)
High Voltage
(12 to 25-kV)
Step Up Voltage
(for Transmission
over Long Distance)
110 to 1000-kV
Decrease
Current
Step Down Voltage
(for Final Use)
12 to 34.5-kV
Very Low Power Loss in
Transmission Line
Basic design of a transformer consists of
one core and two coils.
The core can be air, soft iron or steel.
(a)
The core types of a transformer:
1. Core form
2. Shell form
 Core form is a construction of
two-legged laminated core with
the coils wound on two legs.
(b)
 Shell form is a construction of
three-legged laminated core with
the coils wound on center legs.
 Windings of shell form are arranged
concentrically to minimize the flux
leakages.
Fig.1-1 The transformer construction of
(a) Core-form and (b) shell-form
(a)
(b)
Figure 1-2 Core construction of (a) hollow type and (b) shell-type.
Air core
Iron core
Figure 1-3 Schematic symbols of transformer
~
Primary windings
RL
Secondary windings
Figure 1-4 Simple circuit of transformer
Table 1-1: Principle Parts of a Transformer
Part
Function
Core
Provides a path for the magnetic lines of flux
Primary winding
Receives energy from the ac source
Secondary winding
Receives the energy from primary winding and delivers
it to the load
Enclosure
Protects the above components from dirt, moisture, and
mechanical damage
►
Core
~
RL
AC
Source
►
Primary
Windings
Secondary
Windings
The flux produced by a transformer
is divided into two components:
►
►
N2
►
►
►
N1
►
~
►
RL
ΦLS
► ►
Leakage flux:
the portion of flux that goes
through one of the windings but
not the other one. It returns
through the air.
ΦM
►
ΦLP
►
Mutual flux:
the portion of flux that goes from
the primary winding to secondary
winding through the core.
It remains in the core and links
both windings.
►
Fig.1-5 Paths of mutual and leakage flux
The total average primary flux:
 P  M   LP
(1-1)
where ФM = flux component linking both primary and secondary coils
ФLP = primary leakage flux
The total average secondary flux:
 S   M   LS
where ФLP = secondary leakage flux
(1-2)
Total power losses in a transformer are a combination of four types of losses:
1. Leakage flux: fluxes which escape the core and pass through only one of
the transformer windings.
2. Copper losses: resistive heating losses in the primary and secondary
windings of the transformer. They are proportional to the square of the
current in the windings.
3. Eddy current losses: resistive heating loss in the core of transformer. They
are proportional the square of the voltage applied to the transformer.
4. Hysteresis losses: losses caused by the rearrangement of the magnetic
domains in the core.
By definition, an ideal transformer is a device that has no core loss and no leakage flux
due to its core is infinitely permeable. Any flux produced by the primary is completely
linked by the secondary, and vice versa.

No current is(t) flowing out of the
secondary side.
Relationship
between
voltage
+
applied to the primary side of the
transformer, VP(t) and the voltage
produced on the secondary side, v (t)
P
VS(t) is equal to the ratio of the
numbers of turns on the primary
and on the secondary.
iP(t) NP
►
~
vP (t ) N P

a
vS (t ) N S
where a is defined to be the turns
ratio of the transformer.
(1-3)
NS iS(t) = 0
►
+
VS vS(t)
VP
►
-
►
Fig.1-6 Schematic symbol of an ideal
transformer at no-load.
-
There is current iS(t) flowing out of the secondary side. The relationship between the current
iP(t) flowing into the primary side of the transformer and the current iS(t) flowing out of the
secondary side is
►
(1-4)
where, iP(t) = primary current (A)
iS(t) = secondary current (A)
~
In terms of phasor quantities, the equation 1-2
and 1-3 can be written as follow:
VP I S N P
 
a
VS I P N S
+
(1-5)
iP(t) NP
NS iS(t)
►
►
+
VP
VS
-
-
►
iP (t ) N S 1


iS (t ) N P a

RL
Fig.1-7 Schematic symbol of an ideal
transformer under load.
Phase angle of VP is the same as the angle of VS and the phase angle of IP is the same as
the phase angle of IS.

The dots appearing at one end of each winding in
Fig. 1-3 tell/”state” the polarity of the voltage and
current on the secondary side of the transformer.
This utilize the dot convention.
1. If the primary voltage is positive at the dotted
end of the winding with respect to the undotted
end, then the secondary voltage will be positive
at the dotted end also. Voltage polarities are
same with respect to the dots on each side of
the core.
2. If the primary current of the transformer flows
into the dotted end of the primary winding, the
secondary current will flow out of the dotted
end of the secondary winding.
iP(t) NP
NS iS(t)
+ ►
~
►
+
VP
VS
-
-
►
Dot convention:
►
RRL
Fig.1-7 Schematic symbol of an ideal
transformer under load.

►
Pin  VP I P cos  P
+
(1-6)
where θp is the angle between the primary voltage and
the primary current.
The power supplied by the secondary transformer to
its load is given by the equation:
Pout  VS I S cos  S
~
iP(t) NP
►
NS iS(t)
►
+
VP
VS
-
-
►
The power supplied by the ac generator to the
transformer is given by the equation:
RRL
Fig.1-7 Schematic symbol of an ideal
transformer under load.
(1-7)
where θs is the angle between the secondary voltage and the secondary current.
Cos θp = power factor of the primary windings and cos θs = power factor of the primary
windings.

For an ideal transformer, θP = θS. Therefore,
applying the turns-ratio equations gives VS = VP/a
and IS = aIP, so the output power of an ideal
transformer is equal to its input power:
(1-8)
+
~
iP(t) NP
►
NS iS(t)
►
+
VP
VS
-
-
►
Pout = Pin
►
RRL
Fig.1-7 Schematic symbol of an ideal
transformer under load.
IP
The impedance of a device or an element is
defined as the ratio of the voltage across it to
the current flowing through it.
Thus, the impedance of the load connected to
the secondary winding of a transformer can be
expressed as:
Vs
ZL 
Is
VP
Z 
 a2ZL
IP
••
+
–
VP
VS
ZL
VAC
IP
(1-9)
The impedance of the primary circuit of the
transformer is given by:
'
L
IS
+
–
VP
Z L'  a 2 Z L
VAC
(1-10)
Fig.1-8 (a) Impedance scaling through a
transformer. (b) Definition of impedance.
V3
Z1
Objectives:
Z3
••
I1
+
–
•To solve for all the voltages
across each element.
I3
V2 Z2 I2
V4 Z4 I4
VAC
(a)
Fig.1-9
V3
+
–
VAC
a. The actual circuit showing the
actual voltages and currents on
the secondary side.
Z3
Z1
••
I1
a2Z2
aV2
I2/a
(b)
•To calculate the current flowing
each element.
I3
V4
Z4
b. Impedance Z2 is shifted to the
I4
primary side. Note the
corresponding change in V2
and I2.
aV3
I1
+
–
a2Z2
VAC
Fig.1-9
a2Z3
Z1
••
I3/a
aV2
V4
Z4
I4
I2/a
(c)
aV3
I=0
a2Z3
Z1
I1
+
–
a2Z2
VAC
••
I3/a
aV2
I2/a
a2Z4 aV4
I4/
a
(d)
I=0
c. Impedance Z3 is shifted to the
primary side. Note the
corresponding change in V3
and I3.
d. Impedance Z4 is shifted to the
primary side. Note the
corresponding change in V4
and I4. The currents in T are
now zero.
aV3
Z1
a2Z3
I1
I3/a
+
–
a2Z2
VAC
I2/a
Fi.1-9
a2Z4
aV2
aV4
• All the impedances are now
transferred to the primary side
and the transformer is no
longer needed.
I4/a
With shifting impedances from secondary to primary, we will find that:
The impedance values is multiplied by a2.
The real voltage across the transferred impedance increases by a factor a.
The real current flowing the transferred impedance decreases by a factor a.
V1
V3
Z1
Z3
Fig.1-10
I3
a. The actual circuit showing the
actual voltages and currents on
the primary side.
••
I1
+
–
V2 Z2 I2
V4 Z4 I4
VAC
(a)
••
V1/a
V3
Z1/a2
Z3
aI1
I3
+
–
V2 Z2 I2
VAC
(b)
V4 Z4 I4
b. Impedance Z1 is transferred to
the secondary side. Note the
corresponding change in V1
and I1.
••
V1/a
V3
Z1/a2
Z3
aI1
I3
+
–
V2 Z2 I2
Fig.1-10
V4 Z4 I4
VAC
(c)
+
–
V1/a
V3
Z1/a2
Z3
aI1
I3
V2 Z2 I2
VAC/a
(d)
V4 Z4 I4
c. The source is transferred to the
secondary side. Note the
corresponding change in VAC.
Note also that the currents in T
are zero.
d. All the impedances and even
the source are now on the
secondary side. The
transformer is no longer
needed because the currents
are zero.
With shifting impedances from primary to secondary, we will find that:
The impedance values is divided by a2.
The voltage across the transferred impedance is lower than the real voltage.
The real current flowing the transferred impedance is higher than the real
current.
T1 Iline
T2 Iload
••
Zline
••
IG
jX
R
+
–
VAC
3
Vload
1
2
+
Zload
–
Fig.1-11 AC power system (a) without and (b) with transformer
Based on the use in AC power system, a transformer is differed in a various names:
1. Unit transformer:
a transformer connected to the output of a generator and used to step up the voltage to
transmission line (110+ kV).
2. Substation transformer:
a transformer at the end of the transmission line, which steps down the voltage from
transmission levels to distribution levels (from 2.3 to 34.5 kV).
3. Distribution transformer:
a transformer that takes the distribution voltage and steps down it to the final voltage at
which the power is actually used (110, 208, 220 V, etc.).
All these transformers are essentially same—the only difference among them is their intended
use.
For your first exercise, open the text book of “Electrical Machinery
Fundamentals-Fourth Edition by Stephen J. Chapman, Pg. 73-76 (Example 21). You try to understand the example first and after that you just change the
values of V, Zload and Zline and then you answer all questions like in the
example. OK…
Remember….I’ll test you 
Iline
jX
R
Zline
IG
+
–
(a)
(b)
Zload
–
••
+
–
Vload
VAC
T1
IG
+
Iload
Iline
jX
R
Zline
T2
• •
Iload
+
Zload
VAC
Vload–
Fig.1-12 The power system (a) without and (b) with transformer
T1
Iline
jX
R
••
IG
+
–
•
Zline
'
Z load
VAC
•
•
IG
+
–
jX
R
'
Z line
VAC
Equivalent circuit
''
Z load
•
Equivalent circuit
Fig.1-13 (a) System referred to the transmission system voltage level. (b) System
and transmission line referred to the generator’s voltage level
I0
RP
IP
jXP
RS
jXS
(a)
IS = 0
ο
•►•
+
–
NP
VP
m
VS
NS
►
VAC
ο
I0
(b)
RP
IP
jXP
RS
jXS
IS ≠ 0
•►•
VAC
VP
NP
m
NS
VS
►
+
–
Fig. 1-14 Complete circuits of a real transformer for (a) open and
(b) short circuits
To construct a complete circuit of a real transformer:
each imperfection (core losses) and permeability are taken in to
account;
their effects are included in the transformer model.
Imperfection and permeability can be represented by the values of
resistance and reactance, respectively, by doing two tests:
open-circuit test
short-circuit test
I0
RP
IP = 0
jXP
RS
jXS
IS
ο
I0
+
–
RC
VAC
If
Im
•►•
jXM VP
m
VS
►
Excitation branch
ο
Fig.1-15 An imperfect core represented by a reactance Xm and a resistance Rm
The resistance, RC shows the core losses.
Magnetizing reactance, XM represents a measure of the permeability of
the transformer core.
The current, Im represents the magnetizing current needed to create the
magnetic flux Φm.
IOC
RP
IP = 0
jXP
RS
jXS
IS
ο
IOC
VAC
RC
If
Im
jXM VP
m
VS
►
+
–
•►•
Excitation branch
ο
Fig.1-15 An imperfect core represented by a resistance RC and a reactance XM
Because of RP and XP are too small in comparison to RC and XM to cause
a significant voltage drop, so all the input voltage is dropped across the
excitation branch.
Therefore, RC and XM can be obtained with measuring the magnitude of
input voltage (VOC), input current (IOC) and input power (POC).
ip (t)
A
(a)
v (t)
+
~-
+
Wattmeter
V
is (t)
vp (t)
ip (t)
A
(b)
v (t)
+
~-
V
is (t)
+
Wattmeter
vp (t)
Fig. 1-16 (a) open-circuit test and (b) short-circuit test
Using the input voltage, input current and input power of the open-circuit test ,
power factor (PF) can be determined as follows:
POC
PF  cos  
VOC I OC
(1-11)
The power factor angle (θ) is given by
POC
  cos
VOC I OC
1
(1-12)
The power factor is always lagging for a real transformer, so the angle of the
current always lags the angle of the voltage by θ degrees. Therefore, the
excitation admittance YE is
I OC
I OC
I OC
YE 
  
cos(  )  j
sin(  )
VOC
VOC
VOC
(1-13)
Beside that, the excitation admittance can be also obtained using the formula:
1
1
YE  GC  jBM 
j
RC
XM
(1-14)
where, GC = conductance of the core-loss resistor
BM = susceptance of the magnetizing inductor
RC = resistance in the transformer
XM = reactance in the transformer
Thus, we find
I OC
1

cos(  )
RC VOC
(1-15)
I OC
1

sin(  )
X M VOC
(1-16)
From the values of resistance (RC) and reactance (XM), we can determine the
core losses and reactive power following equations:
PM 
QM 
V p2
RC
V
2
p
XM
where, RM = resistance representing the core losses (Ω)
Xm = magnetizing reactance of the primary winding (Ω)
Vp = primary voltage (V)
PM = core losses (W)
QM = reactive power needed to set up the mutual flux Φm (W)
(1-17)
(1-18)
I0
RP
IP
jXP
RS
jXS
IS ≠ 0
•►•
VAC
VP
NP
m
NS
VS
►
+
–
Fig. 1-17 Complete circuit of a real transformer for short-circuit
Apart from the magnitude of input voltage (VSC), input current (ISC) and input
power (PSC), resistance and inductance values can be obtained with
constructing the equivalent circuit.
I0
RP
a2RS
jXP
ja2XS
IS/a
+
–
VAC
Fig. 1-18 Equivalent circuit of a real transformer for short-circuit
Equivalent resistance and reactance are
Reqp  Rp  a Rs
(1-19)
X eqp  X p  a 2 X s
(1-20)
2
The series impedance ZSE is equal to
Z SE  Req  jX eq
Z SE  ( Rp  a RS )  j ( X P  a X S )
2
2
(1-21)
The power factor of the current is given by:
PSC
PF  cos  
VSC I SC
(1-22)
and is lagging. The current angle is thus negative, and the overall impedance
angle θ is positive:
PSC
  cos
VSC I SC
1
(1-23)
Series impedance can be calculated as follows:
Z SE
VSC 0o
VSC
VSC
VSC
o


 
cos   j
sin 
o
I SC   
I SC
I SC
I SC
(1-24)
Therefore, we get
VSC
Req 
cos 
I SC
X eq
VSC

sin 
I SC
(1-25)
(1-26)
Per-unit (pu) system of measurements is an another method to solve
the circuits containing transformers. In this method, the impedance
transformations can be avoid. Thus, circuits containing many
transformers can be solved easily with less chance of error.
In pu system, the voltages, currents, powers, impedances, and other
electrical quantities are not measured in SI units (volts, amperes, watts,
ohms, etc.) but measured in decimal fraction.
Any quantity can be expressed on a per-unit basis by the equation:
Actual value
Quantity per unit 
base value of quantity
where actual value is a value in volts, amperes, ohms, etc.
(1-27)
To define a pu system, two quantities need to select. The ones usually
selected are voltage and real or apparent power.
In a single-phase system, these relationships are
Pbase, Qbase, or Sbase  VbaseI base
(1-28)
Vbase Vbase 
Z base 

I base
Sbase
(1-29)
I base
Ybase 
Vbase
(1-30)
2
Voltage regulation is a quantity that compares the secondary voltage of
a transformer at no load with the secondary voltage at full load. It is
defined by the equation:
VR 
VS ,nl  VS , fl
VS , fl
x100%
(1-31)
where, VS,nl = secondary voltage at no-load (V)
VS,fl = secondary voltage at full-load (V)
Since at no load, VS = VP/a, the voltage regulation can also be
expressed as
VR 
V p / a  VS , fl
VS , fl
x100%
(1-32)
If the transformer equivalent circuit is in the per-unit system, then voltage
regulation can be expressed as
VR 
VP , pu  VS , fl , pu
VS , fl , pu
For an ideal transformer, VR = 0.
x100%
(1-33)
The efficiency of a devices (motors, generators as well as transformers)
is defined by the equation:
Pout

x 100%
Pin
(1-34)
Pout

x 100%
Pout  Ploss
(1-35)
At full-load, a transformer has the total loss (Ploss) in Watts:
Ploss  Pcore  PCu
(1-36)
where,
Pcore = input power in Watts on the open-circuit test = core loss
PCu = input power in Watts on the short-circuit test with full-load currents
= I2R loss on full load
and the output power (Pout) in Watts is given by:
Pout  Vs I s cos  s
(1-37)
So, the efficiency of the transformer can be written as
Vs I s cos 

x 100%
Vs I s cos   Pcore  PCu
(1-38)