Physics Investigatory Project on transformers. Class 12th

Physics investigatory project on transformers with required details for 12thes. with index, theory, types of transformers (with relevant images), procedure, sources of error, aim n apparatus along with bibliography🗃️📜. Please try to add your own imagination rather than just copy paste... Hope you all guys friends n juniors' like it. peace out✌🏻✌🏻

1. * PHYSICS INVETIGATORY *
PROJECT



TRANSFORMERS
Submitted by- Pratikshya Moharana
Class- XII ‘A’
Roll No.- 16

3. Acknowledgement
In the accomplishment of this project successfully, many
people have bestowed their blessings upon me and given
their precious time and support, I’m utilizing my thanks
to all the people concerned with this project
Primarily I would thank our physics teacher Mr.
Sanjaya Kumar Parida, whose valuable guidance has
been the one that has helped me patch this project and
make it a full proof success. His suggetions and
instructions have served as the major contributor
towards the completionof the project.
I would like to thank my parents and friends who have
helped me with suggetions and advisory comments to
enhance an rectify mistakes done in this project.
Last but not not he least, a heartful thanks goes to my
classmates whose assistance are of a huge success in the
completion of this project.

4. INDEX













6. Transformer is a device used for in the power
transmission of electric energy. The transmission current is
AC. It is commonly used to increase or decrease the supply
voltage without a change in the frequency of AC between
circuits. The transformer works on the basic principles
of electromagnetic induction and mutual induction.

 Types of Transformers- Transformers are used in various fields like power
generation grid, distribution sector, transmission and electric energy
consumption. There are various types of transformers which are classified
based on the following factors:

o Working voltage range
o The medium used in the core
o Winding arrangement
o Installation location


 Step-up Transformer: used between the
power generator and the power grid. The
secondary output voltage is higher than the
input voltage.
 Step-down Transformer: These transformers are
used to convert high-voltage primary supply to
low-voltage secondary output.

 Air Core Transformer: The flux linkage between
primary and secondary winding is through the
air. The coil or windings wound on the non-
magnetic strip.
 Iron Core Transformer: Windings are wound on
multiple iron plates stacked together, which
provides a perfect linkage path to generate
flux.
 Autotransformer: It will have only one winding
wound over a laminated core. The primary
and secondary share the same coil. Auto
means “self” in the Greek language.

 Power Transformer: It is used at power
generation stations, as they are suitable
for high voltage application
 Distribution Transformer: It is mostly used
at distribution lanes for domestic
purposes. They are designed for carrying
low voltages. It is very easy to install and
characterized by low magnetic losses.
 Measurement Transformers: They are
mainly used for measuring voltage,
current and power.
 Protection Transformers: They are used for
component protection purposes. In
circuits, some components must be
protected from voltage fluctuation, etc.
Protection transformers ensure
component protection.

1. In voltage regulators for T.V, refridgerators , computers, air
conditioners, etc.
2. A step- down transformer is used for welding purposes & obtaining
large current
3. A step- up transformer is used for production of X- rays and neon
advertisement.
4. Used in voltage regulators & stabilized power supplies.
5. Transmission of A.c currnet over long distances.

Uses of Transformers

8. Theory
Principle: a transformer works on the principle of mutual induction
Principles of mutual induction:
Faraday’s Law of Electromagnetic Induction:
Changing magnetic fields generate an EMF in a conductor.
Mutual inductance:
When two coils are close together, the magnetic field from one links with the other, inducing
a voltage in the second coil. Mutual inductance modifies current and voltage in a
secondary coil
Construction: A transformer essentially consists of 2 coils of insulated copper wire having
different no. of turns and wound on the same soft iron core. The coil P to which electric
energy is supplied is called the primary coil and the coil from which energy is drawn or
output is obtained is called the secondary coil.
When an alternating E.M.F is supplied to the primary coil P1 P2, an alternating current starts
falling in it. The alternating current in the primary produces a changing magnetic flux linked
with primary is also linked with the secondary is equal to that induced in each turn of the
primary. Thus if Ep & Es be the instantaneous values of the E.M.F’s induces in the primary and
secondary and Np & Ns are the no. of turns of the primary secondary coils of the transformer
and
ⅆϕ
ⅆt
=�??????�� �� ??????ℎ??????��� �� �??????�?????? ??????� �????????????ℎ ���� �� �ℎ� ??????�????????????
At this instant we have,
E
P=−N
p
ⅆϕ
ⅆt
…(1)
And
??????
�=−??????
�
ⅆ??????
ⅆ�
…(2)
Since the above relations are true at every instant, so by dividing (2) by (1), we get
Es
E
P
=−
Ns
Np
…(3)
As Ep is the instantaneous value of back E.M.F induced in the primary coil P1, so the
instantaneous current in primary coil is due to the difference (E – Ep) in the instantaneous
values of the applied & back E.M.F further if Rp is the resistance O, P1, P2 coil then the
Instantaneous current Ip in the primary coil is given by,
I
p=E−
E
P
Rp
 E−E
p=I
pR
p
when the resistance of the primary is small, Rp, Ip can be further neglected. Therefore,
E−E
p=0  E
??????=E (Thus back E.M.F = Input E.M.F) -
For ideal transformers
Es
E
P
=
Es
E
=
??????��??????�� �.??????.�
Input E.M.F
=
Ns
Np
=??????
Es
E
P
=
Is
Ip
=
Ns
Np
= Transformation Ratio

9. Efficiency
Efficiency of a transformer is defined as the ratio of output power to the input power.
η=
Output power
Input power
=
E
sI
s
E
pI
p

Thus in an ideal transformer where there’s no power loss, ??????� η=1, But in actual practice
there are many power losses
η=
Output power
Input power
x 100


Step- Up Transformer
The formula is used to design a step up
transformer is-
??????
??????
??????
??????
=
??????
??????
??????
??????

Where, Ns = no. of secondary turn
Np = no. of primary turn
Vs = voltage in secondary turn
Vp = voltage in primary turn
the number of turns in the secondary coils
should be always greater than the number of
turns in the primary coil. Np < Ns to work
transformer as step- up transformers
As the number of turns will be greater in
secondary coil, so total E.M.F is more &
hence the output voltage in secondary coil
will be more than primary input voltage.


Step- Down Transformer
The formula is used to design a step up
transformer is-
??????
??????
??????
??????
=
??????
??????
??????
??????

Where, Ns = no. of secondary turn
Np = no. of primary turn
Vs = voltage in secondary turn
Vp = voltage in primary turn
the number of turns in the secondary coils
should be always less than the number of
turns in the primary coil. Np > Ns to work
transformer as step- down transformers
As the number of turns will be less in
secondary coil, so total E.M.F is less &
hence the output voltage in secondary coil
will be less than primary input voltage.


(Ns > Np) (Ns < Np)

10.  Flux leakage – due to air gap in the core the
amount of magnetic flux linked with the secondary
is less than that of primary as the energy is last. It can
be reduced by winding the primary and secondary
coil over one another. Inspite of best insulaters,
therefore, rate of cahnge of magnetic flux linked
with each turn of S1 S2 is less than the rate of change
in magnetic flux linked with each turn of P1 P2
 Copper loss - energy is loss in the form of heat in the
copper coils of the transformer. This is due to joule
heating of conducting wires.
 Iron loss - energy loss in the form of heat in the iron
core of the transformer. This is due to formation of
eddy currents in iron core. It is minimized by taking
laminated cores.
 Hysteresis loss – loss of energydur to repeated
magnetization and de- magnetization of the iron
core when A.C. is fed to it.
 Magneto striation – humming noise of a transformer.
ENERGY LOSSES

11. Soft iron rod
(length- 15cm
diameter- 1.3 cm)
Enamelled copper
wire (20 or 22 SWG)
Enamelled copper
wire (16 or 18 SWG)
Step- down
transformer having
tappings 3, 6, 9
and 12 ac volts
2 ac voltmeters
(0 - 10 volts)
Thick sheet of
paper

12. Procedure
1. Take a soft iron rod of length 15 cm & diameter 1.3
cm wrap a thick sheet of paper on it.
2. Wind 200 turns of a coil P of enamellled copper wire
of 20 or 22 SWG.
3. Wind 50 turns of another coil of S enamelled copper
wire of 16 or 18 SWG.
4. Both coils are wound over same length of rod, so that
almost the entire flux produceed by the current in coil
P is linked to the coil S.
5. Connect a step- down transformer (having tappings of
3,6,9 & 12 ac volts) & an ac voltmeter (0 – 10 V)
across the ends of the coil P, as shown in the Fig.
6. Connect an ac voltmeter (0 – 10 V) across the ends of
the coil S.
7. Connect the step- down transformer to the ac mains
of 220 V. From it, supply 3 V to the coil P. Note the
readings in both the ac voltmeters.
8. Using output tappingof 6 V & 9 V of the step- down
transformer, supply 6V & 9 V to the coil P, in turn ,
& note down the readings of the both the AC
voltmeters in each case.

13. Observations
Sl.
No.
Input AC voltage
Vp (V)
Output AC
voltage Vs (V)
Ratio
V
&#3627408480;
V
p

1. 3 V 9 V
1
3

2. 6 V 6 V 1
3. 9 V 3 V 3


Conclusions
1. Within the limit of experimental errors, the ratio Vs / Vp is
constant & is equal to the ratio Ns /Np i.e., the ratio of output &
input voltages of a transformer is equal to the turns ratio Ns /
Np of the transformer.
2. By appropriate choice of turns ratio Ns / Np we can get a step-
down transformer.
3. The coils P & S are placed very close to each other but they
are insulated. The power in the primary is transferred to the
secondary through mutual induction.
4. A steady DC voltage can’t be stepped up or stepped down
by a transformer, because a steady DC current doesn’t
produce changing magnetic flux & therefore can’t produce
induced voltage.

14. Sources of Error
Values of current can be changed due to
heating effect or due to inaccurate counting of
turns of coil.
Eddy current can change the reading.

15. Bibliography
 Book References
SL Arora Physics Lab
Manual
 Internet References
- https://www.microsoft.com/
- https://www.google.com/
- https://www.slideshare.net/
- https://byjus.com/
- https://youtube.com/shorts/
- https://www.bing.com/
- https://in.pinterest.com/