project on transformers class xii

8/11/2019 Project on Transformers Class XII

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2014-2015


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Certificate

This is to certify that

student of Class XII, Kendriya Vidyalaya No.2 Uppal,

has completed the project titled Transformers

during the academic year 2014-2015 towards partial

fulfilment physics practical examination conducted by

AISSCE, New Delhi and submitted satisfactory report,

as compiled in the following pages, under mysupervision.

________________ _________________ _________________


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cknowledgement

I would like to express my special thanks

of gratitude to my teacher Mr N.V.N.G.K

Rao who gave me the golden opportunity

to do this wonderful project on the topic

Transformers, which also helped me in

doing a lot of Research and I came to

know about so many new things I am

really thankful to them.

Secondly I would also like to thank my

parents and friends who helped me a lot

in finalizing this project within the limited

time frame.


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P.DHARMA TEJA


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circuit. 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.

Close-up of single-phase pole mount transformer.


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It is based on the principle of mutual induction

that 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 whichinduces e.m.f. in the neighbouring circuit.


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A transformer consists of a rectangular shaft iron core

made of laminated sheets, well insulated from one another.

Two coils p1& p2and s1& s2are wound on the same core, but

are well insulated with each other. Note that the both thecoils 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


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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.

For operation at low frequency, we may have a soft iron.

The soft iron core is insulating by joining thin iron strips

coated with varnish to insulate them to reduce energy

losses by eddy currents. The input circuit is called primary.

And the output circuit is called secondary.

An ideal voltage step-down transformer. The secondary current arises from the action of the secondary EMF on the (not

shown) load impedance.


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The ideal transformer as a circuit element

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, and then the induced e.m.f. induced in each turn


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of the secondary is equal to that induced in each turn of

the primary.

Thus if Epand Esbe the instantaneous values of the e.m.f.s

induced in the primary and the secondary and Npand Nsare

the no. of turns of the primary secondary coils of the

transformer and, D / dt= rate of change of flux in each

turn of the coil at this instant, we have

Ep= -NpD/dt (1)

Es= -NsD/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 Epis the instantaneous value of back e.m.f induced in theprimary 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 Rpis the resistance

o, p1p2 coil, then the instantaneous current Ip in the

primary coil is given byI =E Ep/ Rp


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E Ep = IpRp

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 = EsIs


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If there are no losses of power in the transformer, then

Input power = output power or

EpIp = EsIs Or

Es/ Ep = Ip/ Is = K

In a step up transformer

As k > 1, so Ip> Isor 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.


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BASIC IDEA OF STEP DOWN TRANSFORMER

BASIC IDEA OF STEP UP TRANSFORMER


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Efficiency of a transformer is defined as the ratio of

output power to the input power i.e.

= output power / input power = EsIs/ EpIp

Thus in an ideal transformer, where there is no power

losses, = 1. But in actual practice, there are many power

losses; therefore the efficiency of transformer is less than

one.


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In practice, the output energy of a transformer is always

less than the input energy, because energy losses occur due

to a number of reasons as explained below.

1. Loss of Magnetic Flux: The coupling between the coils is

seldom perfect. So, whole of the magnetic flux produced by

the primary coil is not linked up with the secondary coil.

2. Iron Loss: In actual iron cores in spite of lamination,

Eddy currents are produced. The magnitude of eddy current

may, however be small. And a part of energy is lost as the

heat produced in the iron core.

3. Copper Loss: In practice, the coils of the transformer

possess resistance. So a part of the energy is lost due to

the heat produced in the resistance of the coil.

4. Hysteresis Loss: The alternating current in the coil

tapes the iron core through complete cycle of

magnetization. So Energy is lost due to hysteresis.5. Magneto restriction: The alternating current in the

Transformer may be set its parts in to vibrations and sound

may be produced. It is called humming. Thus, a part of

energy may be lost due to humming.


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1.Take thick iron rod and cover it with a thick paper and

wind a large number of turns of thin Cu wire on thickpaper (say 60). This constitutes primary coil of the

transformer.

2.Cover the primary coil with a sheet of paper and wound

relatively smaller number of turns (say 20) of thick

copper wire on it. This constitutes 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 s1and s2.5.Now connect s1and s2to A.C main and again measure


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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.

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.


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1. Values of current can be changed due to heating effect.

2.

Eddy current can change the readings.

1. The output voltage of the transformer across the secondary

coil depends upon the ratio (Ns/Np) with respect to the input

voltage2.The output voltage of the transformer across the secondary

coil depends upon the ratio (Ns/N p) with respect to the input

voltage

3.There is a loss of power between input and output coil

of a transformer.

1. Keep safe yourself from high voltage.

2. While taking the readings of current and voltage the A.C

should remain constant.


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A Big Transformer

NCERT Textbook Class 12

NCERT Physics Lab Manual Class 12

Google Website

The End


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project on transformers class xii

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