1. Acknowledgements
I would like to express my sincere gratitude to my
physics mentor MR. V.K PATHAK, for his vital support,
guidance and encouragement, without which this
project would not have come forth. I would also like to
express my gratitude to the OTHER staff of the
Department of Physics for their support during the
making of this project.
I can’t forget to offer my sincere thanks to my class
mate who helped me to carry out this project work
successfully and for their valuable advice and support
which I received from them time to time.
SUSHIL NATH GUPTA

2. Certificate
THIS IS TO CERTIFY THAT SUSHIL NATH GUPTA OF CLASS 12 ‘C’ HAS
COMPLETED HIS PROJECT ON THE TOPIC TRANSFORMER.
HE HAS SUCCESSFULLY PREPARED THIS PROJECT REPORT IN THE
PHYSICS LABORATORY OF OUR SCHOOL.
THIS PROJECT MAY BE CONSIDERED AS PARTIAL FULLFILMENT OS
AISSCE 2014-2015 CONDUCTED BY THE CENTRAL BOARD OF
SECONDARY EDUCATION, DELHI.
……………………………………………….. …………………………………………………
INTERNAL EXAMINER EXTERNAL EXAMINER

3. INTRODUCTION
A transformer is an electrical device that transfers energy
between two or more circuits through electromagnetic induction.
A varying current in the transformer's primary winding creates a
varying magnetic flux in the core and a varying magnetic field
impinging on the secondary winding. This varying magnetic
field at the secondary induces a varying electromotive force (emf)
or voltage in the secondary winding. Making use of Faraday's
Law in conjunction with high magnetic permeability core
properties, transformers can thus be designed to efficiently
change AC voltages from one voltage level to another within
power networks.
Transformers range in size from RF transformers less than a
cubic centimetre in volume to units interconnecting the power
grid weighing hundreds of tons. A wide range of transformer
designs is encountered in electronic and electric power
applications. Since the invention in 1885 of the first constant
potential transformer, transformers have become essential for the
AC transmission, distribution, and utilization of electrical energy
PRINCIPLE

4. IT IS BASED ON THE PRINCIPLE OF MUTUAL INDUCTION THET IS IF A
VARYING CURRENT IS SET UP IN A CIRCUIT THEN INDUCED E.M.F. IS
PRODUCED IN THE NEIGHBOURING CIRCUIT. THE VARYING CURRENT IN
A CIRCUIT PRODUCE VARYING MAGNETIC FLUX WHICH INDUCES E.M.F.
IN THE NEIGHBOURING CIRCUIT.
CONSTRUCTION A
transformer consists of a rectangular shaft iron core made of laminated
sheets, well insulated from one another. Two coils p1 & p2 and s1 & s2
are wound on the same core, but are well insulated with each other.

5. Note that the both the coils are insulated from the core, the source of
alternating e.m.f is connected to p1p2, the primary coil and a load
resistance R is connected to s1 s2, the secondary coil through an open
switch S. thus there can be no current through the sec. coil so long as
the switch is open.
For an ideal transformer, we assume that the resistance of the primary
& secondary winding is negligible. Further, the energy loses due to
magnetic the iron core is also negligible.
 VP - is the Primary Voltage NP - is the Number of Primary Windings
 VS - is the Secondary Voltage NS - is the Number of Secondary Windings
 Φ (phi) - is the Flux Linkage
THEORY AND WORKING OF TRANSFORMER

6. When an altering e.m.f. is supplied to the primary coil p1p2, an
alternating current starts falling in it. The altering current in the primary
produces a changing magnetic flux, which induces altering voltage in
the primary as well as in the secondary. In a good-transformer, whole of
the magnetic flux linked with primary is also linked with the secondary,
then the induced e.m.f. induced in each turn of the secondary is equal to
that induced in each turn of the primary. Thus if Ep and Es be the
instantaneous values of the e.m.f.’s induced in the primary and the
secondary and Np and Ns are the no. of turns of the primary secondary
coils of the transformer and
Dфь / dt = rate of change of flux in each turnoff the coil at this instant,
we have
Ep = -Np dфь/dt -----------------(1) And
Es = -Ns dфь/dt ----------------- (2)
Since the above relations are true at every instant, so by dividing 2 by 1,
we get
Es / Ep = - 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 and back
e.m.f. further if Rp is the resistance o, p1p2 coil, then the instantaneous
current Ip in the primary coil is given by
Ip = E – Ep / Rp
E – Ep = Ip Rp

7. When the resistance of the primary is small, Rp Ip can be neglected so
therefore
E – Ep = 0 or Ep = E
Thus back e.m.f = input e.m.f
Hence equation 3 can be written as
Es / Ep = Es / E = output e.m.f / input e.m.f = Ns / Np = K
Where K is constant, called turn or transformation ratio.
In a step up transformer
Es > E so K > 1, hence Ns > Np
In a step down transformer
Es < E so K < 1, hence Ns < Np
If Ip = value of primary current at the same instant t
And Is = value of sec. current at this instant, then
Input power at the instant t = Ep Ip and
Output power at the same instant = Es Is
If there are no losses of power in the transformer, then
Input power = output power Or
Ep Ip = Es Is Or
Es / Ep = Ip / Is = K
In a step up transformer

8. As k > 1, so Ip > Is or Is < Ip
i.e. current in sec. is weaker when secondary voltage is higher.
Hence, whatever we gain in voltage, we lose in current in the same
ratio.
Similarly it can be shown, that in a step down transformer, whatever we
lose in voltage, we gain in current in the same ratio.
Thus a step up transformer in reality steps down the current & a step
down transformer steps up the current.
ENERGY LOSSES:-
Following are the major sources of energy loss in a transformer:

9. 1. Copper loss is the energy loss in the form of heat in the copper coils
of a transformer. This is due to joule heating of conducting wires.
2. Iron loss is the 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.
3. Leakage of magnetic flux occurs inspite of best insulations.
Therefore, rate of change of magnetic flux linked with each turn of S1S2
is less than the rate of change of magnetic flux linked with each turn of
P1P2.
4. Hysteretic loss is the loss of energy due to repeated magnetization
and demagnetization of the iron core when A.C. is fed to it.
5. Magneto striation i.e. humming noise of a transformer.
EFFECIENCY OF TRANSFORMER
A transformer does not require any moving parts to transfer energy.
This means that there are no friction or windage losses associated

10. with other electrical machines. However, transformers do suffer
from other types of losses called “copper losses” and “iron losses”
but generally these are quite small.
Copper losses, also known as I2R loss is the electrical power which is
lost in heat as a result of circulating the currents around the
transformers copper windings, hence the name. Copper losses
represents the greatest loss in the operation of a transformer. The
actual watts of power lost can be determined (in each winding) by
squaring the amperes and multiplying by the resistance in ohms of
the winding (I2R).
Iron losses, also known as hysteresis is the lagging of the magnetic
molecules within the core, in response to the alternating magnetic
flux. This lagging (or out-of-phase) condition is due to the fact that it
requires power to reverse magnetic molecules; they do not reverse
until the flux has attained sufficient force to reverse them.
Their reversal results in friction, and friction produces heat in the
core which is a form of power loss. Hysteresis within the
transformer can be reduced by making the core from special steel
alloys.
The intensity of power loss in a transformer determines its
efficiency. The efficiency of a transformer is reflected in power
(wattage) loss between the primary (input) and secondary (output)
windings. Then the resulting efficiency of a transformer is equal to
the ratio of the power output of the secondary winding, PS to the
power input of the primary winding, PP and is therefore high.
An ideal transformer is 100% efficient because it delivers all the
energy it receives. Real transformers on the other hand are not
100% efficient and at full load, the efficiency of a transformer is
between 94% to 96% which is quiet good. For a transformer
operating with a constant voltage and frequency with a very high
capacity, the efficiency may be as high as 98%. The efficiency, η of a
transformer is given as:
Transformer Efficiency

11. where: Input, Output and Losses are all expressed in units of power.
Generally when dealing with transformers, the primary watts are
called “volt-amps”, VA to differentiate them from the secondary
watts. Then the efficiency equation above can be modified to:
It is sometimes easier to remember the relationship between the
transformers input, output and efficiency by using pictures. Here the
three quantities of VA, W and η have been superimposed into a
triangle giving power in watts at the top with volt-amps and
efficiency at the bottom. This arrangement represents the actual
position of each quantity in the efficiency formulas.
USES OF TRANSFORMER

12. A transformer is used in almost all a.c. operations
· In voltage regulator for T.V., refrigerator, computer, air
conditioner etc.
In the induction furnaces.
· A step down transformer is used for welding purposes.
· A step down transformer is used for obtaining large current.
· A step up transformer is used for the production of X-Rays and
NEON advertisement.
· Transformers are used in voltage regulators and stabilized
power supplies.
· Transformers are used in the transmissions of a.c. over long
distances.
· Small transformers are used in Radio sets, telephones, loud
speakers and electric bells etc.
BIBLIOGRAPHY

13. 1. Physics for class XII by pradeep
2. www.yahoo.com
3. www.google.com
4. www.live.com
5. www.rediffmail.com

14. SESSION: - 2014-2015
PHYSICS PROJECT
ON
TRANSFORMER
NAME – SUSHIL NATH GUPTA
CLASS – XII ‘C’
BOARD ROLL –
ROLL - 56