Transformer class 12 cbse project 2019.

1. PHYSICS PROJECT
INVESTIGATORY PROJECT
ON
“TRANSFORMER”
Guided by: - Mr.B.K. Rath (Physics department)
NAME: - ASHWINI KUMAR SAH
CLASS: - XII (A)
ROLL NO. :-13
SUBMITTED AS PER THE REQUIREMENT OF
AISSCE- 2019-20

2. Acknowledgement
I would like to extend my deepest gratitude to
Mr. B.K. Rath (Physics department) for his constant
support and encouragement during the making of this
project.
I should also not forget Mr.N.R. Pattanaik (LabAssistant)
for his morale boosting words as well as his help with the
laboratory apparatus during our practicals. I am also
greatly thankful to our principal Dr.B.K.Mishra for his
moral support and guidance.
ASHWINI KUMAR SAH
Class-XII(A)

3. CERTIFICATE-1
This is to certify that ASHWINI KUMAR SAH of class XII-‘A’
has successfully completed his project titled –
“TRANSFORMER” under the guidance and
supervision of Mr. B.K. Rath (Physics department) as a
part of fulfilment for the requirement of
AISSCE 2019-20.
Mr.N.R. Pattanaik
Mr. B.K. Rath
(Lab Assistant)
(Physics department)
Nalco Nagar,Angul Nalco
Nagar,Angul

4. EXTERNAL EXAMINER
CERTIFICATE-2
This is to certify that ASHWINI KUMAR SAH of class XII-A
has successfully completed his project titled –
“TRANSFORMER” under the guidance and supervision of
Mr. B.K. Rath of Physics department as a part of fulfilment
for the requirement of
AISSCE 2019-20.
Dr.B.K.Mishra
Principal
DELHI PUBLIC SCHOOL

5. Nalco Nagar, Angul
CONTENT
1. Aim of the Experiment
2. Introduction
3. Theory
4. Apparatus Required
5. Procedure
6. Efficiency
7. Energy Loses
8. Uses of the Transformer
9. Precaution
10. Source of Error
11.Conclusion
12.Bibliography

6. AIM OF THE PROJECT
To investigate the relation between the ratio of-
1. Input and output voltage
2. Number of turnings in
The secondary coil and the primary coil of self made transformer:

7. INTRODUCTION
The transformer is a device used for converting a low alternating
voltage to a high alternatingvoltage or vice-versa.
A transformer is an electrical device which is used for changing
the A.C. voltages. A transformer is most widely used device in
both low and high current circuit. As such transformers are built
in an amazing strength of sizes. In electronic, measurement and
control circuits, transformer size may be so small that it weights
only a few tens of grams whereas in high voltage power circuits,
it may weight hundredsof tons
A Transformer based on the principle of mutual induction
according to this principle, the amount of magnetic flux linked
with the coil changing, an emf is inducedin the neighboringcoil.
In a transformer, the electrical energy transfer from one circuit to
another circuit takes place without the use of moving parts
A transformer which increases the voltages is called a step-up
transformer. A transformer which decreases the A.C. voltages is
called a step-down transformer.
Transformer is, therefore, an essential piece of apparatus both
for high and low current circuits.

8. THEORY
When an altering emf is suppliedto the primary coil P1P2, an
alternatingcurrent 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 linkedwith primary
is also linkedwith the secondary, then the induced emf induced
in each turn of the secondary is equalto that inducedin each
turn of the primary. Thus if Ep and Es be the instantaneousvalues
of the emf induced in the primary and the secondary and Np and
Ns, are the no. of turns of the primary secondary coilsof the
transformer and
dɸb/dt =rate of change of flux in each turn off the coil at this
instant, we have
Ep =- Np dɸb/dt ----- (1)
And , Es= -Np d§b/dt ----- (2)
Since the above relationsare true at every instant, so by dividing
2 by 1, we get

9. Es/ Ep= - Ns/Np --------- (3)
As Ep is the instantaneousvalue of back emf inducedin the
primary coil P1 , so the instantaneouscurrent in primary coil is
due to the difference (E-Ep) in the instantaneousvalues of the
appliedand back emf further if Rp is the resistance O, P1P2 coil,
then the instantaneouscurrent lp in the primary coil is given by
Ip= E- Ep /Rp
E- Ep = IpRp
When the resistance of the primary is small, IpRp can be neglected
so therefore
E – Ep = 0 or Ep =E
Thus back emf = inputemf
Hence equation3 can be written as
Es / Ep = Es / E = output emf / input emf= Ns / Np = K
Where K is constant, called turn or transformation ratio.

10. In a step up transformer
Es > E, so K > 1, hence Ns > Np
As k> 1, so lp>Is
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.
Similarlyit can be shown, that in a step down transformer,
whatever we lose in voltage, we gain in current in the same ratio.

11. Thus a step up transformer in reality steps down the current & a
step down transformer step up the current.
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 = Eplp and
Output power at the same instant = Es Is

12. If there are no losses of power in the transformer, then
Input power = output power
Or Ep Ip = Es IsOr Es /Ep =Ip/Is = K
APPARATUS REQUIRED

13. 1. Iron rod
2. copper wire
3. voltmeter
4. ammeter
PROCEDURE

14. 1. Take thick iron rod and cover it with a thick paper and wind
large number of turns of thin Cu wire on thick paper (say 60). This
constitutes primary coil of the transformer.
2. Cover the primary coil with a sheet of paper and wound
relativelysmaller number of turns (say 20) of thick copper wire
on it. Thisconstitutes the secondary coil. It is a step down
transformer
3. Connect P1, P2 to A.C main and measure the input voltage and
current using A.C voltmeter and ammeter respectively.
4. Similarly,measure the output voltage and current through S1
and S2.
5. Now connect S1 andS2 to A.C main and again measure voltage
and current through primary and secondary coil of step up
transformer
6. Repeat all steps for other self made transformers by changing
number of turns in primary and secondary coil
EFFICIENCY

15. Efficiency of a transformer is defined as the ratio of output power
to the input power i.e.
ɳ = output power / input power = Es Is / Ep lp
Thus in an idealtransformer, where there is no power losses,
ɳ = 1. But in actual practice, there are many power losses;
therefore the efficiency of transformer is less than one.
ENERGY LOSSES
Followingare the major sources of energy loss in a transformer:
1. Copper loss is the energy loss in the form of heat in the copper
coils of a transformer, 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. Thisis due to formation of eddy currents in
iron core. It is minimized by taking laminatedcores.
3. Leakage of magnetic flux occurs, inspite of best insulations.
Therefore, rate of change of magnetic flux linked with each turn
of S1 S2 is less than the rate of change of magnetic flux linkedwith
each turn of P1P2
4. Hysteresis loss is the loss of energy due to repeated
magnetizationand demagnetizationof the iron core when A.C. is
fed to it.
5. Magneto striation i.e. humming noise of a transformer.
USES OF TRANSFORMER

16.  A transformer is used in almost all a.c. operations
 In voltage regulator for T.V., refrigerator, computer, air
conditioneretc., in the inductionfurnaces.
 A step down transformer is used for welding purposes.
 A step down transformer is used for obtaininglarge current.
 A step up transformer is used for the productionof 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 Radiosets, telephones,loud
speakers and electric bellsetc.
PRECAUTIONS
 Keep yourself safe from high voltage.
 While taking the readingsof current and voltage the A C
should remain constant.
SOURCES OF ERROR
 Values of current can be changed due to heating effect.
 Eddy current can change the readings.

17. CONCLUSION
i. The output voltage of the transformer across the secondary
coil dependsupon the ratio Ns/Np
with respect to the input voltage.
ii. There is loss of power between input and output
power of the transformer.

18. BIBLIOGRAPHY
i. www.google.com
ii. www.yahoo.com
iii. Physics Lab Manual book- Class XII(A)
iv. NCERT Physics text book- Class XII(A)
v. New simplified physics book- S.L. Arora