Theory and Working Completed

7/30/2019 Theory and Working Completed

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THEORY AND

WORKINGLet the alternating emf supplied by

the a.c. source connected to primary be

E= E0 Sin wt

As we have assumed the primary to be a pure

inductance , the sinusoidal primary currentIp lags the voltage Ep by 90 degree . The

primarys power factor,

Cos= cos 90 = 0.

Therefore , no power is dissipated in primary .The alternating primary current induced an

alternating magnetic flux B inthe iron core because the core extendsthrough the secondary winding, the induced

flux also extends through the turns ofsecondary .


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According to faradays law of induction,

the induced emf per turn( Eturn) is

same for the primary and secondary . Also,the voltageEP across the primaryis equal to the emf induced in the primary,

and the voltage Es across the

secondary is equal to the emf inducedin the secondary . Thus ,

Eturn=d B = Ep =Es

dt NP Ns

Here , Np;Ns represent total no of turns

in primary and secondary coils

respectively.

Or, Es =Ep * NsNp

** Ns = K ( TRANSFOMATION RATIO)Np


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Note that this relation is based on threeassumption

1) The primary resistance and current are small.2) There is no leakage of flux . The same flux

links both, the primary and secondary coils.3) The secondary current is small .

Now , the rateat which generator / source transfersenergy to the primary = IpEp.

The rate which the primary then transfers energy tothe secondary( via the alternating magnetic fieldlinking the two coils) is IsEs.

As we assume that no energy is lost along the

way , conservation of energy requires that

Ip*Ep = Is*Es

Or , Is=Ip*EpEs

Or, Is = Ip* Np =Ip.(1)Ns K


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Types of Transformer

** Set up transformer-A transformer which increases the a.c.voltages is called a step up transformer.Here,N s > Np ; Es > Ep

Hence, K > 1 & Is < Ip

** Set down transformer-A transformer which decreases the a.c.voltages is called step down transformer.Here ,

N s < Np ; Es < EpHence , K < 1 & Is > Ip


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From equation1

Ip= Is(Ns/Np)=(Es/R)(Ns/Np)

Ip= (1/R)* Ep(Ns/Np)(Ns/Np)

Ip= (1/R) * Ep{(Ns/Np)^2}

This equation has the form

Ip=Ep/Req

, where the equivalent resistance Req is

Req={(Np/Ns)^2}*R

thus Req is the value of loadresistance as seen by the source /generatori.e. the source / generator producescurrent Ip & voltage Ep as if it whereconnected to a resistance Req .


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EFFICIENCY OF A TRANSFORMER

= the ratio of output power to

the input power.

i.e.n = output power = EsIs

input power EpIp

For an ideal transformer,n=1( i.e. 100%).

However , practically there are many energy losses.Hence efficiency in practice is


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Energy Losses inTransformer

In practice, the output energy of atransformer is always less than the input energy,because energy losses occurdue to a number of reasons as explainedbelow.

1.Loss of Magnetic Flux:The coupling between the coils is seldomperfect. So, whole of the magnetic flux producedby the primary coil is not linked up with thesecondary coil.

2. Iron Loss:In actual iron cores inspite oflamination , eddy currents are produced. Themagnitude of eddy current may, however besmall. And a part of energy is lost as the heatproduced in the iron core.


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2.Copper Loss:In practice , the coils of the transformer possessresistance. So a part of the energy is lost dueto the heat produced in the resistance of thecoil.

4. Hysteresis Loss:The alternating current in the coil tapes the ironcore through complete cycle of magnetization.So Energy is lost due to hysteresis.

5. Magneto restriction:The alternating current in the Transformermay be set its parts in to vibrations and soundmay be produced. It is called humming. Thus, apart of energy may be lost due to humming.

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Theory and Working Completed