A
transformer is a static device which is used to transfer electrical energy from
one circuit to another by using the principle of mutual induction. It is
generally used to
transform voltage from one level to another. It means
transformer can be used to raise or lower down the voltage level by
corresponding decrease or increase in the current.
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| CIRCUIT SYMBOL OF TRANSFORMER |
Transformers
use the principle of mutual induction. The two circuits are electrically
isolated but are magnetically linked to each other by a mutual flux. The transfer
of energy takes place without any change in the frequency of the system.
CONSTRUCTION –
Consider the
circuit given below:
The
rectangular part represents the core, which is a material of high magnetic
permeability. The core is laminated in order to reduce eddy current losses. Two
coils are wound on the either side of core having turns N1 and N2.
The vertical portions of the core are called limbs and bottom portions are
called yokes.
The two
coils are insulated from each other and the steel core. The whole setup of core
and windings is kept in a container which is further insulated from the setup
by means of a suitable medium (generally called transformer oil). Transformer
oil also serves the purpose of cooling.
Electrical
transformer is one of the best examples of those devices which are simplest in
their construction, yet serve the most important function in our electrical
generation and transmission system.
OPERATION –
When an
alternating current is passed through coil 1(primary coil), it produces a
magnetic flux. Since, the current is alternating in nature a changing magnetic
flux is produced. This changing magnetic flux induces a voltage across the ends
of the coil 2 (secondary coil), which is governed by the relation, $e=-\frac{d\phi
}{dt}$ .
Simply
stating, changing current leads to changing magnetic field which in turn leads
to the voltage induced in the secondary circuit. Thus, energy is transferred
from one circuit to the other.
IDEAL TRANSFORMER –
To study
about transformers, we generally consider the ideal case. The ideal case of a
transformer should have the following properties:
- The magnetic permeability of the core should be infinite. I.e. no magneto motive force (mmf) is needed to setup the flux.
- Leakage flux should be zero. i.e. the flux is confined within the core and entirely links both the windings.
- Resistances of windings should be zero.
- There are no losses due to resistance, eddy currents and hysteresis.
In short, an
ideal transformer has no losses and stores no energy. The phasor diagram for an
ideal transformer is as follows:
However,
there is no such transformer that exists and is ideal in nature. The ideal case
is just to be used for study purpose. It is that no transformer has 100%
efficiency, but there are transformers which have efficiency of 99.75% (very
close to ideal one).
EMF EQUATION OF TRANSFORMER –
When an
alternating voltage is applied to the primary of a transformer, a varying
magnetic flux is setup. Let this varying flux be represented by
\[\phi
={{\phi }_{m}}\sin \omega t={{\phi }_{m}}\sin 2\pi ft\]
Where ${{\phi
}_{m}}$ = maximum flux
f= frequency of variation of
flux
By faraday’s
law of electromagnetic induction, the induced emf in winding of N turns is
given by,
\[E=-N\frac{d\phi
}{dt}=-N\frac{d}{dt}({{\phi }_{m}}sin\omega t)\]
\[=-N\omega {{\phi }_{m}}\cos
\omega t=N\omega {{\phi }_{m}}\sin (\omega t-\frac{\pi }{2})\]
Thus,
maximum value of induced emf, ${{E}_{m}}=N\omega {{\phi }_{m}}$
Rms value of
induced emf, ${{E}_{R}}=\frac{{{E}_{m}}}{\sqrt{2}}=\frac{N\omega {{\phi
}_{m}}}{\sqrt{2}}=\frac{2\pi fN{{\phi }_{m}}}{\sqrt{2}}$
\[{{E}_{R}}=4.44fN{{\phi
}_{m}}\]
TRANSFORMATION RATIO (K) –
Transformation
ratio is given by ratio of secondary voltage to the primary voltage. To obtain
this ratio:
The induced
emf’s in primary and secondary windings (rms values) is given by:
\[{{E}_{1}}=4.44f{{N}_{1}}{{\phi
}_{m}}\] and
\[{{E}_{2}}=4.44f{{N}_{2}}{{\phi
}_{m}}\]
Thus
transformation ratio is, $\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{4.44f{{N}_{2}}{{\phi
}_{m}}}{4.44f{{N}_{1}}{{\phi }_{m}}}$
\[\Rightarrow
\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{{{N}_{2}}}{{{N}_{1}}}=\frac{{{V}_{2}}}{{{V}_{1}}}=K\]
This is for
a two winding transformer. If the transformer has three windings (primary,
secondary and tertiary), then the ratio of emf’s is given by E1:E2:E3::
N1:N2:N3.
