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The
equivalent circuit of an electric transformer is its theoretical representation
by using standard linear active and passive electrical elements, while
retaining all the electrical properties of an actual transformer.Â

An
equivalent circuit of a transformer, or any other electrical machine, is the
simplest circuit of that machine, which is used to make various calculations
easier and to analyze the performance of machine without actually loading it

In
transformers, the equivalent circuit helps us to calculate their core losses,
copper losses, magnetizing current and other performance parameters in a
straightforward manner.

Since,
the equivalent circuit of any electrical machine retains all the electrical
properties of that machine, Therefore, it is assumed that one should know all
the basics of that particular machine.Â

So,
before moving on to the main concept of the equivalent circuit of an electrical
transformer, it is recommended that first go through the **basics of electric transformer**, such as: What is a transformer? Working principle of transformer,
types of transformers, and constructional features of transformer, etc.
Studying these topics will make it easier to grasp the equivalent circuit of an
electric transformer.

**Equivalent
Circuit of an Electrical Transformer**

In
the basics of transformers, we study that there are two sets of windings, the
primary and the secondary, which are mounted on the magnetic core, as shown in
the given figure below.Â

In
that article, we first explain the working of transformers for an ideal
transformer and then move on to the practical one.Â

There we
discussed that in an ideal transformer, there are no losses, meaning the
resistance of its windings is zero, and the reluctance of the magnetic core is
also zero. Therefore, all the magnetic flux passes through the core, resulting
in no leakage of flux. However, in practical transformers, these conditions do
not hold true. A practical transformer has losses, leakage flux and also its
magnetic core offers some reluctance to the magnetic flux.

In
practical transformers, both the primary and secondary winding have finite
resistance which is associated with the copper losses (I^{2}R) of the
transformer. Let us say for the given example, R_{1} and R_{2}
is the resistance for both the primary and secondary winding respectively,
which are distributed over the length.

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In
an ideal transformer, the magnetic core has zero reluctance, meaning all the
magnetic flux passes through the core. However, it is not possible in practical
transformers. In reality, while a major part of a total flux is confined to the
core as mutual flux É¸ linking both the primary and secondary, a small amount of
flux does leak through paths, which lie mostly in air and link separately to
the individual winding.Â

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Below
the given figure shows the schematic of a practical transformer on load. In the
given figure É¸_{L1} and É¸_{L2} is the leakage flux caused by
primary MMF and secondary MMF respectively and the path of the leakage is shown
by the green dotted line.

É¸_{L1}
Is the leakage flux caused by the primary MMF which links with the primary
winding itself, similarly É¸_{L2} is the leakage flux caused by
secondary MMF which links to secondary winding, thereby causing self-linkages
of two windings. The self-linkage caused by leakage flux and the winding MMF
are linearly related with each other in each winding. Therefore, contributing
constant leakage inductance or leakage reactance corresponding to the frequency
at which the transformer operates. Let us say the leakage reactance offered by
the primary winding is X_{L1} and the leakage reactance offered by the
secondary winding is X_{L2}.

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In practical transformers, both the resistance and the leakage reactance of the winding are considered series effects. Therefore, the practical transformer can now be represented with the help of their winding resistance and leakage reactance as shown in the given figure. Theoretically, this representation is referred to as a semi-ideal transformer because it accounts for the effect of winding resistance (i.e. copper losses) and leakage reactance, but does not include the effect of resistance and reluctance of the magnetic core.Â

Since
the magnetic core practically has some resistance and reluctance, so when the
transformer is on no load, connected to the AC supply then the primary winding
draws some no-load current (I_{o}). This no-load current is the
combination of the two components of current: the one component of current (I_{c})
is dissipated in the magnetic core resistance (core losses) and the other
component of current (I_{m}) is used to establish the magnetic field in
the magnetic core.Â

Â

And
when the transformer is loaded, the load draws the current I_{2} from
the transformer's secondary winding. This secondary current I_{2}
produces its own MMF that opposes the main flux. So, in order to maintain main
flux constant and independent of load, the primary winding draws more current
which counterbalances the secondary current requirement.Â

Â

So,
in a practical transformer under load condition, the current drawn by primary
winding is the sum of no load current and the secondary current referred to the
primary side.

Â

I_{1}Â = Â I_{o}
Â + Â I_{2}â€™

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where,

Â

- I
_{1}Â = Â Primary Current - I
_{o}Â = Â No load current - I
_{2}â€™ Â = Â Secondary current referred to primary sideÂ

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The
concept of referring is discussed further in this article.

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So
based on the above discussion, the above equivalent circuit of transformer can
now be redrawn as,

Above the given figure shows the equivalent circuit of transformer, in which:

- R
_{1}and R_{2}represent the resistance of primary and secondary winding, respectively. - X
_{L1}and X_{L2}represent the leakage reactance of primary winding and secondary winding, respectively. - R
_{c}represents the core loss resistanceÂ - And,
X
_{m}represents the magnetizing reactance of the core.

In
the given figure above, we clearly see that the current drawn by the primary
winding I1 is divided into two components, i.e. the no load current Io and the
current I_{2}â€™ current which counterbalances the effect of secondary
current. The no load current Io is further divided into two components: one is
passed through the branch R_{C}, which represents the core loss branch
of the transformer and the other component of the current is passed through the
branch X_{m}, which represents magnetizing reactance of the
transformer.Â

However,
there is a discrepancy in the above equivalent circuit that is transformer
action, complicates the calculations. So, to remove this transformer action or
that ideal transformer in the above equivalent circuit there is a certain step
of referring the secondary side element to the primary side.Â Â

The
equations supporting the process of referring the secondary side elements to
the primary side is shown below.

Secondary
side quantities referred to primary

- X
_{L2}â€™Â =Â a^{2}* X_{L2} - R
_{2}â€™Â =Â a^{2}* R_{2} - V
_{2}â€™Â = Â a^{2}* V_{2} - I
_{2}â€™Â =Â (1/a)^{2}* I_{2}Â

where, a is the transformation ratio of the transformer

These
secondary side quantities referred to the primary side represent the image of
the secondary side quantities which can be used in the primary side to make
calculations easier without altering their characteristics. So, after referring
secondary side quantities to the primary side, the equivalent circuit now can
we redrawn as follows:

In
the equivalent circuit of the transformer shown above, there are no parameters
on the secondary side therefore, there is no need to show that ideal
transformer. After removing that ideal transformer, the equivalent circuit of
the transformer can now be redrawn as shown in the given figure.

Above
the given figure represents the final equivalent circuit of a transformer
referred to the primary side. This equivalent circuit can also be redrawn by
referring the primary elements to the secondary side, resulting in the
equivalent circuit of a transformer referred to the secondary side.

The
equations supporting the process of referring the primary element to the
secondary side are as follows:

- X
_{L1}â€™Â =Â (1/a)^{2}* X_{L2} - R
_{1}â€™Â =Â (1/a)^{2}* R_{2} - V
_{1}â€™Â = (1/a)^{2}* V2 - I
_{1}â€™Â =Â a^{2}* I_{2}Â - R
_{c}â€™Â =Â (1/a)^{2}* R_{c} - X
_{m}â€™Â =Â (1/a)^{2}* X_{m}

With
the understanding that all the quantities have been referred to a particular
side; either primary side or secondary side, Equivalent circuit of the
transformer can be drawn referred to either side, depending on the convenience.

The
circuit deduced in the above discussion is the final equivalent circuit of the
transformer and is adequate for the most power and radio frequency
transformers. In transformers operating at higher frequencies, such as pulse
transformers and other electronics transformers, the interwinding capacitances
are often significant and must be included in the equivalent circuit.

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