Basics of electrical and electronics engineering

Transformer

1. TRANSFORMER
•Transformer works on the principle of
electromagnetic induction
•It has no moving parts
•It by mutual induction transfers electric
energy from one circuit to another at the
same frequency

2. Transformer
An A.C. device used to change high voltage low
current A.C. into low voltage high current A.C. and
vice-versa without changing the frequency
In brief,
1. Transfers electric power from one circuit to another
2. It does so without a change of frequency
3. It accomplishes this by electromagnetic induction
4. Where the two electric circuits are in mutual
inductive influence of each other.

3. Principle of operation
It is based on
principle of MUTUAL
INDUCTION.
According to which
an e.m.f. is induced
in a coil when
current in the
neighbouring coil
changes.

4. Working of a transformer
1. When current in the primary coil
changes being alternating in
nature, a changing magnetic field
is produced
2. This changing magnetic field gets
associated with the secondary
through the soft iron core
3. Hence magnetic flux linked with
the secondary coil changes.
4. Which induces e.m.f. in the
secondary.

5. CLASSIFICATION
DUTY THEY PERFORM
CONSTRUCTION
VOLTAGE OUTPUT
APPLICATION
COOLING
INPUT SUPPLY

6. Constructional detail : Shell type
• Windings are wrapped around the center leg of a
laminated core.

7. Core type
•Windings are wrapped around two sides of a laminated square
core.

8. Sectional view of transformers
Note:
High voltage conductors are smaller cross section conductors
than the low voltage coils

9. Construction of transformer from
stampings

10. Core type
Fig1: Coil and laminations of
core type transformer
Fig2: Various types of cores

11. Shell type
•The HV and LV
windings are split
into no. of sections
•Where HV winding
lies between two LV
windings
•In sandwich coils
leakage can be
controlled
Fig: Sandwich windings

12. Cut view of transformer

13. Transformer with conservator and
breather

15. Ideal Transformers
•Zero leakage flux:
-Fluxes produced by the primary and secondary
currents are confined within the core
•The windings have no resistance:
- Induced voltages equal applied voltages
•The core has infinite permeability
- Reluctance of the core is zero
- Negligible current is required to establish magnetic
flux
•Loss-less magnetic core
- No hysteresis or eddy currents

16. Ideal transformer
V
1
– supply voltage ; I
1
- noload input current ;
V
2-
output voltgae; I
2
- output current
I
m- magnetising current;
E
1-self induced emf ; E
2- mutually induced emf

17. EMF EQUATION OF A TRANSFORMER
Let,
N
1
=numbers of primary turns
N
2
=Number of secondary turns
Ø
m
=Maximum Value of flus in the core in wb
B
m
=Maximum value of the flux density in the core in wb/m
2

A=Area of the core in m
2
f
=Frequency of the AC supply in Hz
V
1
=Supply voltage across primary in volts
V
2
=Terminal voltage across secondary in volts
I
1
=Full load primary current in amperes
I
2
=Full load secondary current in amperes
E
1
=Emf induced in primary in volts
E
2
=Emf induced in secondary in volts
Fig (1)
AC
supply
V
1
N
1
E
1
N
2
E
2
Primary Secondary
V
2
L
O
A
D

18. Since applied voltage is alternating in nature the flux established is
also an alternating one
From it is clear that the flux is attaining its maximum value in one
quarter of the cycle i.e. T/4 seconds, where ‘T’ is time period in
seconds.
Fig (2)
T/4
T/2
T
0
π
2π Time
Flux

19. If we assume single turn coil, then according to Foraday’s laws of electromagnetic
induction, the average value of emf induced / turn
RMS value = Form factor x Average Value

20. Transformation Ratio (k)
For an ideal transformer
V
1
= E
1

V
2 v
= E
2
and
V
1
I
1
= V
2
I
2
Note:
If N
2
> N
1
i.e. K > I, then transformer is a step up transformer
If N
2 < N
1, i.e. K < I, then transformer is a step down transformer
From EMF equation, we know
Where ‘K’ is called transformation ratio

21. Transformer on load
Fig. a: Ideal transformer on load
Fig. b: Main flux and leakage
flux in a transformer

22. Transformer Voltage Regulation
and Efficiency
Electrical Machines
The output voltage of a transformer varies with the load even if the input
voltage remains constant. This is because a real transformer has series
impedance within it. Full load Voltage Regulation is a quantity that compares
the output voltage at no load with the output voltage at full load, defined by
this equation:
%100down Regulation
%100up Regulation
,
,,
,
,,






nlS
flSnlS
flS
flSnlS
V
VV
V
VV 

%100
/
down Regulation
%100
/
up Regulation
V
V
k noloadAt
,
,
,
,
p
s
x
V
VkV
x
V
VkV
nlS
flSP
flS
flSP





Ideal transformer, VR = 0%.

23. voltageload-no
voltageload-fullvoltageload-no
regulationVoltage











1
2
12
N
N
VV
p
s
p
s
N
N
V
V

recall
Secondary voltage on no-load
V
2 is a secondary terminal voltage on full load


















1
2
1
2
1
2
1
regulationVoltage
N
N
V
V
N
N
V
Substitute we have

24. Transformer Phasor Diagram
09/19/24 24Electrical Machines
 Aamir Hasan Khan
To determine the voltage regulation of a transformer, it is necessary
understand the voltage drops within it.

25. Transformer Phasor Diagram
09/19/24 25Electrical Machines
 Aamir Hasan Khan
Ignoring the excitation of the branch (since the current flow through the
branch is considered to be small), more consideration is given to the series
impedances (R
eq 
+jX
eq).
Voltage Regulation depends on magnitude of the series impedance and the
phase angle of the current flowing through the transformer.
Phasor diagrams will determine the effects of these factors on the voltage
regulation. A phasor diagram consist of current and voltage vectors.
Assume that the reference phasor is the secondary voltage, V
S. Therefore
the reference phasor will have 0 degrees in terms of angle.
Based upon the equivalent circuit, apply Kirchoff Voltage Law,
SeqSeqS
P
IjXIRV
k
V


26. Transformer Efficiency
Electrical Machines
Transformer efficiency is defined as (applies to motors, generators and
transformers):
%100
in
out
P
P

%100


lossout
out
PP
P

Types of losses incurred in a transformer:
Copper I
2
R losses
Hysteresis losses
Eddy current losses
Therefore, for a transformer, efficiency may be calculated using the following:
%100
cos
cos
x
IVPP
IV
SScoreCu
SS




