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Harmonics and Switching Transients in Single
Phase Transformers
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So far we have assumed that when you apply
a sine wave voltage to a single phase transformer;
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the flux in the core is sinusoidal, and the
current drawn by the transformer is also sinusoidal.
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However, in practice this is not the case;
it is mainly due to the nonlinearity of the
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magnetic core used in a transformer, which
is made up of ferromagnetic material.
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So, a practical transformer you will have
a core on which the windings will be placed
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and on the other winding the load may be connected.
The applied voltage we assumed to be sinusoidal,
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if we neglect the resistive drop in the winding,
then the counter induced emf must completely
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cancel the applied voltage; that is the counter
induced emf e should be equal to V i.
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Hence, the flux which is given by N 1 d phi
d t equal to e will be equal to this will
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be sine wave. So, the flux in the core will
be sine wave; and the flux will also lag the
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applied voltage by 90 degree. Now, had the
magnetic material be in linear; the current
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drawn the magnetizing current drawn by the
transformer; it also have been a sine wave.
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However, this is not the case in practical
transformer. In a practical transformer the
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no load current, the magnetizing current and
the flux are related by the B-H characteristic
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on the material; due to this non-linear B-H
characteristic we will soon see that the no
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load current drawn by the transformer will
not be a sine wave. But it will be distorted
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with different harmonic components. Let us
examine this a little more carefully.
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Let us say this is the applied voltage waveform.
The applied voltage waveform V i and the flux
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waveform which will lag the voltage by 90
degree; it will be somewhat like this the
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core flux. Now, let us try to find out what
will be the current due to in order to establish
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this flux in
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the core; for that we will have to consider
the B-H characteristic of the material. So,
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for the current let us see that at time t
equal to 0 when applied voltage was 0 and
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increasing; the flux voltage was at its negative
peak. This was the value of the flux and hence
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the corresponding current was this. So, current
was at negative maximum.
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Let us say this corresponds to the negative
maximum of the current. Now, as the voltage
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is increased now flux is increasing and while
increasing in a B-H curve the flux follows
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this path. So, we see the current when the
current become 0 the flux still has a negative
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component; that is the current will become
0 before the flux becomes 0. So, this is the
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point let us say where the flux will become
0; where some current still exist, some flux
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still exist. When the flux becomes 0 we find
that the current has become already small
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positive. So, this corresponds to the point;
again when the flux reaches its positive peak
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which coincides with the zero crossing of
the voltage the current also reaches its positive
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peak. So, this is the corresponding point
of the current. And since the B-H curve is
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symmetric around x axis; the positive peak
of the current wave form and the negative
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peak of the current waveform will be equal;
while returning though the flux takes a different
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path because of hysteresis.
So, we see that when the current becomes 0
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the flux still has some positive value. Therefore,
the corresponding point here will be let us
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say this. And when the flux becomes 0 the
current has already become negative and here
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again it is negative peak. Therefore, the
current waveform will look somewhat like this;
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which is not sinusoidal although it is symmetric
around y axis. But it is not sinusoidal unlike
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the applied voltage waveform V m; V i or the
flux waveform phi m. So, this current waveform
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is not sinusoidal; of course it can be divided
any into Fourier components it will have.
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So, the current drawn i m will have a fundamental
component i 1 m plus harmonic components.
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And because of its odd nature it
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will have only odd harmonics; that is where
i 1 m equal to I 1 m sine omega t cos omega
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t plus sine omega plus phi and i n m equal
to I n m sine n omega t plus n phi.
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So, this is the source of mostly the source
of harmonics in a single phase transformer.
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And mostly in the magnetization current that
this harmonic will be observable; there is
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one interesting thing to note here that the
zero crossing of the flux waveform and the
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current waveform do not coincide. In fact,
the zero crossing of the current waveform
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leads the zero crossing of the flux waveform
by some angle; although the peak of the current
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waveform coincides with the peak of the flux
waveform. If one decomposes the current waveform
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into its Fourier components at the fundamental
we will find that the current waveform has
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a component which is large component; which
is in phase with the flux waveform phi m.
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And a smaller component which is leads the
phi m and in fact it is in phase with the
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supply voltage V i.
So, the out of the phase that is component
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of i m which is in phase with phi m. And in
quadrature with that is at 90 degree lagging
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V i is responsible for setting up the flux
in the core. While the in phase component
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of I m that is the component which is in phase
with the applied voltage V i supplies the
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hysteresis loss; of course this magnetization
current waveform will also be modified by
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the eddy current component. However, this
component is sinusoidal because the flux is
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sinusoidal which introduces a sinusoidal voltage
in the core. And sinusoidal current in the
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core material which will be compensated by
drawing sinusoidal current from the source
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which will be in again in phase with V i.
So, the in phase component sinusoidal component
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of V i will increase that will make the waveform
less distorted.
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Therefore, the no load current of a transformer
will have a in phase component; which is due
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to in phase component which is in phase with
the voltage which is due to the hysteresis
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and eddy current loss. And a component which
is perpendicular which is lagging V i by 90
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degree and in phase with the flux waveform
which establishes the flux. In addition to
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these two fundamental components there will
be harmonics; odd harmonics that is third
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harmonic, fifth harmonic, seventh harmonic
etcetera of the fundamental frequency; which
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are due to the saturation of the core material.
The deeper into saturation the transformer
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walks, the more will be harmonics and the
nature of the current waveform will become
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very peaky.
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However, when you since the magnetization
current is non sinusoidal one would expect;
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if one assumes the winding resistance that
the applied voltage is sine wave; the current
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drawn i 0 has harmonic. Then one should accept
the induced voltage e also to have harmonics.
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Hence, even the flux waveform will possibly
not be exactly a sine wave which may or may
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not increase the harmonics in i 0 depending
on the nature of the B-H characteristics.
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However, since i 0 magnitude is very small
and also the state resistance is small e and
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phi m can be for all practical purpose assume
to be sine wave. So, in a transformer if the
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applied voltage is sinusoidal then the core
flux is almost sinusoid. But the magnetization
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current drawn is distinctly non sinusoidal;
it has fundamental components and odd harmonics.
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However, when we put a load on the transformer;
I have already mentioned that there will be
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a reflection current i L dash flowing the
transformer. Since, the core flux is a sine
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wave the induced voltage on the secondary
side is also a sine wave. And if the load
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is a resistive or inductive a linear load;
then the current I L will be a sine wave.
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And hence the, its reflection i L dash will
also be a sine wave.
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Now, i 0 is much smaller in a normal transformer
i 0 is much smaller compare to i L dash. So,
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when transformer is loaded I L will dash will
dominate over i 0. Therefore, the waveform
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will look very much sinusoidal. And although
there will be harmonics of the magnetization
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current present. But it will not be apparent
since the magnetization current consistute
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only a very small fraction of the total output
current of the transformer.
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So, a transformer single phase transformer
at no load draws predominate draws a non sinusoidal
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current from the supply consisting of fundamental
and odd harmonics. However, on load, the fundamental
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component also include the reflected load
currents of the fundamental component increases
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but the harmonics do not increase. Therefore,
on load transformer current appears to be
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more or less sine wave. At this point one
should be careful about how one should connect
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a transformer for performing certain test.
In this circuit we have assumed r 1 to
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be small. However suppose we want to perform
a no load test on this transformer. And the
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available supply voltage is larger than the
transformer voltage rating. Then one possibility
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could have been to connect the transformer
like this. And then connecting a resistance
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of large value and connecting a transformer
across the supply.
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Since, we are doing a no load test then the
current drawn by the transformer will be non
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sinusoidal. And the voltage; actual voltage
across the transformer will also be non sinusoidal
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this will make the flux non sinusoidal with
predominantly third harmonic component.
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The flux waveform, instead of looking sinusoidal
may look something like this; this is the
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sine wave component of the flux, this is third
harmonic component of the flux. So, the resultant
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flux waveform near show predominantly peaky
nature; additional resistance is connected
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here; in which case because of saturation
this will be significantly higher than the
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fundamental magnitude of the flux.
So, the current drawn by the circuit may be
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significantly large. Therefore, it is not
desirable to make a circuit connection like
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this; hence for doing no load test under such
situation it will be preferable instead of
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using a resistance to drop the voltage. One
should use an auto transformer, variac, in
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order to reduce the voltage to the desired
level. Because the induced voltage in the
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auto transformer will be almost sinusoidal.
So, far in this figure we have assumed that
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the transformer is operating in steady state;
that is the given a sine wave applied voltage
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the flux waveform is also a sine wave. However,
when the voltage is first switched on in the
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transformer it under goes a transient before
the flux becomes settles down to its final
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sine wave value. During this period the transformer
can experience very larger current; let us
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see how?
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Again, let this is the transformer on no load.
If we assume neglect the transformer winding
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resistance; then we can write e equal to N
d phi m by d t equal to v i equal to V sine
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omega t or phi m equal to V by N minus V by
omega N cosine of omega t plus and constant
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of integration. At time t equal to 0 one would
expect that core the de energized. So, phi
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m at time t equal to 0 may be assumed
to be 0. So, at time equal to 0 or C equal
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to V divided by N omega or therefore, phi
m assuming at t equal to 0; the voltage was
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0 phi m will be V by N omega 1 minus cosine
of omega t. Let us plot this.
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So, this is the applied voltage sine omega
t applied voltage v i; we have switched on
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at t equal to 0 when the voltage was 0. Then
the flux waveform phi m is given by of the
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form phi m equal to phi max into 1 minus cos
omega t. At t equal to 0 it was 0; at omega
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t equal to pi by 2 it will become phi max.
And omega t equal to pi it will become 2 phi
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max. So, the flux waveform will cosine wave
which is shifted by phi max.
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Now, let us see what will be
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the consequences in term of currents? As we
have already mentioned a practical transformer
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core will exhibit a predominate saturation
characteristics along with hysteresis. Let
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us say transformer is usually designed to
operate at a flux density of phi max; maximum
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flux in flux phi max. However, we see that
because of switching initially the flux raises
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to almost double the value; due to non-linearity
of saturation the current will not be double
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it will be very large; incases it can be 50
to 100 times the normal no load peak current.
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The current waveform as we see at we neglect
the hysteresis at t equal to 0 it is 0; phi
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m reaches it at omega equal to pi by 2 reaches
rated value. And then at omega equal to pi
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it reaches a very high value and this waveform
repeats.
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So, if we look at a large scale; time scale
we will find right after switching the current
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raises to very high value. And is almost entirely
positive it does not become negative initially.
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Then slowly because of different losses in
the core; the flux slowly gradually reduces
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and settles down to its final value; accordingly
current also gradually reduces. So, this first
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peak can be as I said 50 to 100 times the
normal steady state peak magnetization current;
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which means it can be 5 to 10 times the full
load current
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of the transformer for first 1 or 2 cycles.
This is called the in-rush phenomenon in a
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single phase transformer; this occurs due
to the depending on the wrong switching instant
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of the voltage. It
is interesting to note the instead of switching
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the waveform; the voltage at zero crossing
had we switched it at the peak value; that
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is if the at t equal to 0 if the switched
it at the peak value.
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Then, V i would have been given by instead
of V sine omega it would have been equal to
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V cosine omega t that is
equal to V sine omega t plus pi by 2. In which
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case that it is if we would have switched
on at the positive peak, then the flux waveform
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would have been phi m would have been and
if at t equal to 0 phi m was 0. Then c also
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would have been then phi m waveform would
have been equal to V by N omega sine omega
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t equal to phi max sine omega t. And there
would not have been any in-rush phenomenon.
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So, depending on when the
transformer is switched on it is possible
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that there will be an magnetic in-rush in
a single phase transformer; due to which the
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no load current of the transformer can raise
to dangerously high level for initial few
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cycles. A protection system should be designed
to handle this. Because it is difficult to
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precisely determine exactly at which instant
a transformer will switched on. The situation
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will become worse since, a transformer
at t equal to 0 the flux in the core may not
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be 0. Because of previous excitation of
the transformer the
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there may be some flux remaining in the core
this is
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called retentivity.
So, if
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you have some remnant flux
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in the transformer, then the peak
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flux density may increase even more than twice
the steady state peak flux; in
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which case the peak of the in-rush current
will also increase accordingly; on load though
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this effect becomes somewhat less. Because
the
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flux decays fast and the peak value
of the
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current reduces faster. Therefore, in a single
phase transformer, a switching transient during
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turn on may cause magnetic in-rush; due to
which the no load current can increase to
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very high level this current slowly decays
because
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the losses in the transformer. And finally
settles downs to
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the normal magnetization current.