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Equivalent circuit and Phasor diagram of transformers

Updated on August 28, 2014
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Electrical and Automation Engineer . Specialized in LV Switchgear Design and process automation

Ideal transformer

Ideal transformer

A transformer that possess the following properties are considered to be an ideal transformer.

  1. Primary and secondary resistances are assumed to be zero. Hence there is no power loss and voltage drop in an ideal transformer.
  2. Leakage flux is completely absent.
  3. The permeability of the core is infinite and so the magnetizing current is zero.
  4. The core losses are considered to be zero.

The figure shows the diagrammatic representation of an ideal transformer. It is represented in such a way that all the above conditions are satisfied.

Real transformer and Equivalent circuits

Transformer windings are made mainly of copper. Even though copper is a good conductor, it possess a finite resistance. Both the primary and the secondary have finite resistances R1 and R2.These resistances are uniformly spread throughout the windings. These resistances give rise to the copper losses (I2R). Consider that Φl1 be the leakage flux caused by the MMF I1N1 in the primary windings and Φl2 be the leakage flux caused by the MMF I2N2 in the secondary windings.

Both the resistance and the leakage reactance of the transformer windings are series effects and at operating frequencies, which is very low (50Hz / 60Hz) these can be regarded as the lumped parameters. Hence the transformer consists of lumped resistances R1 and R2 and reactance X l1 and X l2 in series with corresponding windings. Because of the presence of these lumped quantities the induced emf s E1 and E2 may vary from the secondary voltages V1 and V2 because of the small voltage drops in the winding resistances and leakage reactance.

The transformer ratio can be given by:

a= (N1/N2) = (E1/ E2) ≈( V1 / V2)

Real transformer

Equivalent circuit

The exciting current I0‾ can be resolved into two components, whose magnetizing component Im‾ creates mutual flux Φ‾ and whose core loss component Ii‾ provides the loss associated with alternation of flux.

I0‾ = Im‾ + Ii

Note: vector form is indicated by the symbol ‾.

Hence the equivalent circuit can be represented as shown in the figure.

Equivalent circuit of transformer


Gi = conductance

Bi = Susceptance

The impendence can be now referred to in the primary side resulting in the following circuit:

Equivalent circuit referred to primary side is as follows:(core is neglected)

Equivalent circuit referred to primary

X l2 = (N1/N2)2 X l2

R2 = (N1/N2)2 R2

The load voltage and currents referred to primary side are

V2’ = (N1/N2) V2

I2’ = (N1/N2) I2

Equivalent circuit referred to secondary side is as follows:

Equivalent circuit referred to secondary

The load voltage and currents referred to primary side are

V1’ = (N1/N2) V1

I1’ = (N1/N2) I1

Gi’= (N1/N2)2 Gi

Bi’= (N1/N2)2 Bi

X l1 = (N1/N2)2 X l1

R1 = (N1/N2)2 R1

Phasor diagram

Applying KVL on the primary and secondary side of the equivalent circuits



I1=I2’+I0’= I2’+ (Ii + Im)

Using these equations phasor diagram, can be drawn as follows:

Phasor diagram

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      Dharmendra Yadav 3 years ago (EN)


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      ashenafi 4 years ago

      in my opinion that is good for me.