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In the last lecture, we have shown how to
relax two of the assumptions that we made
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regarding an ideal transformer that is of
lossless windings, and no leakage.
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And we found that, a practical transformer
model can be derived from an ideal transformer
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which is represented by a core and two windings,
by incorporating the resistance of the windings,
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and the effect of leakage flux can be incorporated
by assuming a leakage inductance in series
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with both the windings r 1, r 2, l 1, l 2,
v 1; the load you usually connect be connected
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here equal to this v 2. Current flowing in
coil one is i 1, flowing in coil two is i
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2; the induced voltage occurs coil one is
e1 while the induced voltage occurs coil two
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is e 2. Then we have seen v 1 equal to r 1
i1 plus l 1 d i 1 d t plus e 1, and e 2 equal
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to r 2 i2 plus l 2 d i2 d t plus v 2, and
as in the case of an ideal transformer e 1
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by e 2 equal to N 1 by N 2.
However this still does not take into account
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the magnetization current; that is the current
to be drawn by a practical transformer even
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when i2 is 0. This can be easily incorporated
by assuming a magnetization inductance across
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the coil l l. This is so because the induced
voltage e 1 leads the flux by 90 degree. So,
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an inductance connected across e 1 the current
flowing through the inductance connected across
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e 1 will lag e 1 by 90 degree and will be
in phase with the core flux phi m which is
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responsible for generating the core flux phi
m. So, the effect of finite permeability of
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a practical single phase transformer is taken
care of by considering a magnetizing inductance
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across the winding and leakage inductance
in series with the winding.
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The last detail that needs to be incorporated
in the model of a single phase transformer
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are the core losses. It is well known that
when a ferromagnetic material is subjected
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to an alternating flux then two type of losses
occur; one is the hysteresis loss, another
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is the eddy current loss. Both of them increase
with increase in the flux density and the
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frequency of the alternative current waveform;
although, these relations are not always linear;
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for the hysteresis loss the loss is proportional,
power loss is proportional to the frequency,
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however, not for the eddy current loss.
But in a normal application of a transformer
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normally the supply frequency and the voltage
will be constant; therefore, for a given application
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there will not be variation with frequency
since the frequency is constant. These losses
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can be approximated by losses occurring in
some resistances, fictitious resistances that
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are connected across the coil; therefore,
in order to represent the losses that occur
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in a practical transformer some core loss
resistance is connected across the coil that
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is parallel to the magnetization branch. Therefore
the complete model or a practical transformer
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incorporating an ideal transformer looks as
follows.
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First just put the winding resistance of coil
one, then the leakage inductance of coil one
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and then the coil one of the ideal transformer.
In order to take care of the finite magnetization
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current we connect a magnetization inductance
and to care of the core loss we connect a
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core loss resistance. This is the ideal transformer.
The secondary side similarly there will be
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a leakage inductance l 2 and the winding resistance
r 2. The applied voltage in the primary side
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is v 1, current through the primary is i1,
the magnetization current is im, the core
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loss component of the current i c together
they are called the no load current i 0.
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The current flowing through the ideal transformer
is the reflection of the load current i2 dash,
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and the current in the load circuit is i2,
load voltage be v 2. This is the complete
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model of a practical single phase transformer
taking into account the non idealities which
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also incorporates an ideal transformer. In
phasor form the equations can be written as
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V 1 equal to r 1 I 1 plus j X l 1 I 1 plus
E 1, and E 2 equal to r 2 I 2 plus j X l 2
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I 2 plus V 2. V 2 equal to Z L I 2; I 1 equal
to I2 dash plus I m plus I c equal to I 2
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dash plus I 0. I m equal to E 1 by j X m,
I c equal to E 1 by R 0; E 2 equal to N 2
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by N 1 E 1, and I 2 dash equal to N 2 by N
1 I 2. This set of equation defines the complete
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steady state model of a practical single phase
transformer. Normally the steady state phasor
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relationships are described by a phasor diagram.
Hence it will be interesting to draw the phasor
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diagram of a single phase transformer which
is shown next.
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Again we draw the phasor diagram taking the
core flux phi m to be the reference phasor.
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Both E 1 and E 2 will lead phi m by 90 degree.
Then the current I 2 is given by, which will
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lag assume to lag E 2 by an angle phi. The
phasor I 2 dash will be in phase with I 2
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and will be in fact a scaled version of I
2. The current I 1 can be obtained by phasor
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sum of I 2 dash with I m which is in phase
with phi m and I c which is in phase with
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E 1 which gives me the phasor I 1. The voltage
V 1 can be obtained on the first equation
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E 1 plus I 1 r 1 plus j X l 1 I 1, this is
I 1 r 1. So, this is the phasor V 1; similarly,
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phasor V 2 can be obtained by from the second
equation it is E 2 minus I 2 r 2 minus j X
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l 2 I 2 which can be used for solving any
problem related to single phase transformer.
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However, if we take a closer look at the circuit
model of the single phase transformer that
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we have so derived, we will find that this
is not very suitable for applying the known
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circuit analysis techniques since there is
an ideal transformer between the two parts.
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Hence we cannot apply for example a KCL or
KVL where the loop current is expected to
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flow across the ideal transformer. This path
we cannot write a loop equation involving
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a path like this since we do not know the
voltage drop across the ideal transformer;
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therefore, it will be more convenient if we
can eliminate this ideal transformer.
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This is done by referring the secondary of
the ideal transformer to the primary or vice
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versa; the primary can also be referred to
the secondary; let us see the first option.
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By referring we mean we connect an equivalent
circuit across the terminals of the ideal
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transformers so that the current drawn by
the equivalent circuit is the same i2 dash
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as it is drawn by the ideal transformer. Let
us see what should be that circuit.
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The current I 2 is given by E 2 divided by
Z L plus r 2 plus j X l 2 and I 2 dash equal
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to N 2 I 2 by N 1; therefore, I 2 dash equal
to N 2 by N 1 E 2 divided by Z L plus r 2
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plus j X l 2 or I can write I 2 dash to be
N 2 by N 1 square into N 1 by N 2 E 2 divided
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by Z L plus r 2 plus j X l 2, but N 1 by N
2 E 2 is equal to E 1. Therefore I 2 dash
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equal to E 1 divided by N 1 by N 2 square
into Z L plus r 2 plus j X l 2. Now let us
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have a look at the model circuit model that
we have obtained. The voltage across these
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two terminals is E 1 and the current drawn
from these terminals is I 2 dash. Now the
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same current I 2 dash will flow provided an
impedance of this form is connected across
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these terminals; therefore, we can replace
the ideal transformer and the secondary circuit
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after that by an equivalent impedance given
by this formula.
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Hence the original circuit model I 0 I 1 I
0 I m I c I 2 dash I 2 and v 2. This was the
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original circuit model of the practical single
phase transformer which incorporated an ideal
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transformer with primary turns N 1 and secondary
turns N 2. This circuit model in argue can
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be replaced by an equivalent circuit model
where upto this point there is no change,
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but the rest of the circuit is replaced by
an equivalent circuit where the elements are
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given by, so that the current drawn from this
branch is I 2 dash as in the case of the original
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model. The required equivalent impedances
are related to the actual impedances as follows;
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r 2 dash equal to N 1 by N 2 whole square
r 2 X l 2 dash equal to N 1 by N 2 square
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X l 2 and Z l 2 dash equal to N 1 by N 2 square
Z l 2.
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That dashed quantities are called the referred
impedance of the secondary side referred to
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the primary side, and the equivalent circuit
thus obtained is called the exact equivalent
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circuit of a single phase transformer referred
to the primary side; that is the side with
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suffix 1 and number of turns N 1. Now let
us see the actual secondary current was I2.
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The power loss in the resistance was I 2 square
r 2. This was the actual power loss in the
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secondary circuit; in the equivalent circuit
the power loss is I 2 dash square r 2 dash,
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but I 2 dash itself is equal to N 2 by N 1
I 2, and r 2 dash equal to N 1 by N 2 square
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r 2. Therefore, I 2 dash square r 2 dash is
same as I 2 square r 2; therefore, we see
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that the power consumed by the secondary resistance
is not changed by referring this circuit.
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So, it preservers the power relations; what
will be the referred secondary voltage V 2
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dash? We call this V 2 dash.
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When V 2 dash equal to I 2 dash Z l 2 dash,
but I 2 dash equal to N 2 by N 1 I 2, and
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Z l 2 dash equal to N 1 by N 2 whole square
Z l 2. Therefore, V 2 dash equal to I 2 dash
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Z l 2 dash is equal to N 2 by N 1 I 2 into
N 1 by N 2 whole square Z l 2 equal to N 1
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by N 2 V 2. So, we have I 2 dash equal to
N 2 by N 1 into I 2 and V 2 dash equal to
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N 1 by N 2 into V 2; therefore, V 2 dash I
2 dash equal to V 2 I 2. So, the kVA consumed
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by the referred circuit referred load circuit
is same as the kVA consumed by the original
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load circuit. So, you have seen that the load
circuit can be referred to the source side
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by multiplying the load side parameter values
that is the impedances by the turns ratio
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square. Similar thing can be done for the
source side; that is instead of referring
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the rest of the circuit across this terminal
it is possible to refer the left of the circuit
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across this terminal; in which case the secondary
side impedances voltages and current will
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remain same while the impedances of the primary
side will change.
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Let us see what will be the relations. The
actual circuit model which is given by this
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diagram, we now want to replace with by an
equivalent circuit with an equivalent network
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connected across this two terminal. This circuit
which is also an equivalent circuit referred
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to the primary side was obtained by replacing
the right hand side of the circuit after this
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point by an equivalent impedance, and this
circuit was called the exact equivalent circuit
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of the transformer referred to the primary
side since the impedance on the secondary
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side were referred by the turns ratio. Now
we want to derive an equivalent circuit of
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the transformer referred to the secondary
side where the secondary impedances will remain
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as it is; however, the primary source and
impedance will be referred to the secondary
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side. Although it is possible to connect different
types of impedances we would like to retain
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the structure or the topology of the circuit
on the primary side as far as possible; therefore,
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we assume the equivalent circuit to be as
follows.
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The secondary side does not change. We would
like to retain the structure of the primary
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circuit; therefore, you will have a equivalent
impedance in series equivalent inductance
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in series and call it j X l 1 dash. The shunt
branch consisting of a shunt reactor j X m
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dash and R 0 dash
and a series resistance r 1 dash connected
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to a source V 1 dash. Our objective is to
find out an expression of the dashed quantity
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so that
when the dashed quantities are applied to
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the circuit the secondary current remain I
2, and the secondary voltage becomes V 2 when
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the load connected is Z l 2. Let this current
be I 1 dash, the voltage E 2 equal to I 2
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into also E 2 equal to; therefore, multiplying
both sides by N 1 by N 2 we can write this.
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Or it can be written as but N 2 I 2 divided
by N 1 equal to I 2 dash
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and N 1 by N 2 square into this impedances
are. However, from our previous discussion
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we have seen that
this quantity I 2 dash into Z l 2 dash into
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r 2 dash plus j X l 2 dash this is equal to
E 1 and is given by V 1 minus I 1 r 1 plus
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j X l 1. Therefore, V 1 minus I 1 r 1 plus
j X l 1 equal to V 1 dash minus I 1 dash r
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1 plus j X l 1 into N 1 by N 2. Equating terms
we can say V 1 equal to N 1 by N 2 into V
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1 dash or V 1 dash equal to N 2 by N 1 into
V 1. Similarly I 1 r 1 plus j X l 1 equal
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to N 1 by N 2 I 1 dash into r 1 plus dash
j X l 1 dash.
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If we want the power relations power consumed
by the resistance r 1 should be same as the
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power consumed by the resistance r 1 dash
in the equivalent circuit, then we must have
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I 1 square r 1 equal to I 1 dash square r
1 dash. This will be ensured if we define
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I 1 dash equal to N 2 by N 1 into I 1 and
r 1 dash equal to N 2 by N 1 square into r
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1; I 1 dash equal to N 1 by N 2 square into
I 1 and r 1 dash square equal to N 2 by N
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1 square into r 1.
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Therefore we will say r 1 dash will be equal
to N 2 by N 1 square into r 1. Similarly X
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l 1 dash equal to N 2 by N 1 square into X
l 1 and we have already obtained V 1 dash
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equal to N 2 by N 1 into V 1. Regarding the
shunt in branches again if we want the impedances
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I mean the power balance to be maintained
then I should have E 2 square
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by R 0 dashed should be equal to E 1 square
by R 0, or R 0 dash should be equal to E 2
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by E 1 whole square into R 0, but E 2 by E
1 equal to N 2 by N 1; therefore, R 0 dash
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is equal to N 2 by N 1 square R 0. Similarly,
X m dashed equal to N 2 by N 1 whole square
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X m.
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Therefore, the equivalent circuit of the single
phase transformer refers to the secondary
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or the load side will be similar to the previous
circuit except the parameter values will now
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be different; it is given by a similar circuit.
Relationships are V 1 dash equal to N 2 by
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N 1 into V 1. I 1 dashed equal to N 1 by N
2 into I 1. I 0 dash equal to N 1 by N 2 into
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I 0. Ic dash equal to N 1 by N 2 into I c,
I m dash equal to N 1 by N 2 into I m. The
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parameters are r 1 dash equal to N 2 by N
1 square into r 1 X l 1 dash equal to N 2
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by N 1 square into X l 1.
X m dash equal to N 2 by N 1 square X m, and
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R 0 dashed equal to N 2 by N 1 square into
R 0. So, this is the equivalent circuit of
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a practical single phase transformer referred
to the secondary or the load side while this
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was the equivalent circuit of the same single
phase transformer referred to the primary
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or the source side. They are identical as
far as the circuit structure are concerned
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they are identical except that the parameter
values of the two circuits are different.