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Good morning. In the last lecture, we have
seen that a practical single phase transformer
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like this supplied from a single phase constant
frequency ac source, and connected to a load
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on the other side
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drawing current I 1 from the primary and supplying
a current I 2, on the secondary can be modeled
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using an ideal transformer along with other
passive components that represents the non-ideality
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of the transformer. This is the ideal transformer.
Non idealities to be included are
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a core loss resistance, a magnetizing resistance,
the leakage reactance in the primary side
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and the winding resistance of the primary
side; similarly, the leakage reactance on
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the secondary site and the winding resistance
of the secondary side.
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This is the load; this is the output voltage
V 2. This is secondary resistance r 2; this
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is secondary leakage reactance j X 2, the
primary leakage reactance j X 1, primary winding
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resistance r 1, core loss resistance R 0,
magnetizing reactance j X m. So, it draws
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it is connected to a constant voltage constant
frequency source V 1 draws a current I 1 from
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the primary side, and no load current is I
0. It sends out a current I 2 on the secondary
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side; the turns ratio being N 1 is to N 2.
This is the model of a practical single phase
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transformer incorporating an ideal transformer;
this is the ideal transformer.
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We have also mentioned that this model is
not very convenient for applying all the circuit
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analyses tools that we have learnt in our
electrical technology class. Since, there
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is a isolation barrier at the ideal transformer
which means we cannot for example, choose
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a loop current across the primary and secondary;
since, we do not know the voltage drop between
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the primary and the secondary winding of the
ideal transformer. Therefore, we said that
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this can be converted to a convenient equivalent
circuit by either referring whatever is there
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on the secondary side to the primary or referring
the primary to the secondary side which results
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in two equivalent circuit which are very closely
related as follows. The equivalent circuit
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referred to primary will look like this.
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The elements on the primary side do not change
so that they retain their value; however,
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the elements on the secondary side has to
be referred to primary with a given turns
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ratio. So, primary voltage will still be V
1, primary winding resistance will be r 1,
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leakage reactance will be j X 1, core loss
resistance and magnetizing reactance are also
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generally referred to the primary. So, this
will remain R 0, this will be j X m. The primary
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current will be I 1, no load current I 0.
The secondary current I 2 will now be replaced
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by I 2 dash given by N 2 by N 1 into I 2.
The secondary leakage reactance j X 2 dash
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is given by N 1 by N 2 square j X 2.
Similarly, the referred secondary winding
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resistance r 2 dash will be given by N 1 by
N 2 square r 2. The load impedance Z L dash
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the referred loading impedance Z L dash will
be given by N 1 by N 2 square Z L, and the
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output voltage referred output voltage V 2
will be V 2 dash will be N 1 by N 2 into V
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2 . So, this is the exact equivalent circuit,
why we are using the term exact we will come
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later on, of the single phase transformer
referred to the primary side. Of course, this
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can be this equivalent circuit can be drawn
referred to the secondary side as well, and
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that circuit also we have shown the other
day. It is in that case the secondary elements
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are not changed, but all the primary elements
are referred to the secondary by the turns
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ratio.
Now the secondary load impedance Z L remain
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as it is, winding resistance remain as it
is, leakage reactance remains as it is, the
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current I remains I 2, the load voltage remain
V 2; however, the primary side elements for
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example, the magnetizing reactance now j X
m dash will be equal to N 2 by N 1 whole square
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j X m. R 0 dash will be equal to N 2 by N
1 whole square R 0. I 0 dash will be equal
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to N 1 by N 2 into I 0. The primary winding
resistance r 1 dash will be equal to N 2 by
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N 1 whole square r 1. Primary leakage reactance
j X 1 dash will be given by N 2 by N 1 whole
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square X 1, and the source voltage V 1 dash
will be equal to N 2 by N 1 V 1. So, these
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are the two equivalent circuits.
The first one is referred to the primary with
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number of turns N 1; the second one is referred
to the secondary with number of turns N 2.
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Now the major application of this equivalent
circuits are in determining the different
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variables; for example, if V 1 is given, the
load parameter is given, and the circuit parameter
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is given, then you can find out all the currents,
the voltage, etcetera. Let us look at a concrete
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example. Let us say that this is a single
phase transformer. The ratings are 230 volt
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is to 115 volt, and the KVA rating is 3 KVA
which means N 1 is to N 2 for this case is
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2 is to 1. Also let us say the load impedance
is actually a load resistance of value 3.75
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ohm.
The other parameters r 1, say, it is given
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as 0.75 ohm. X 1 let us say it is j 0.8 ohm.
R0 given to be 600 ohm; j X m is given to
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be 300 ohm. Similarly X 2 dash, X 2 rather
is given to be 0.15 ohm and r 2 is given as
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0.125 ohm. The problem is to find out what
will be the load terminal voltage V 2, what
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will be the load current I 2, what will be
the current drawn from the source I 1 and
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what will be the no load current I 0? How
are you going to solve this problem? So, as
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a first step let us draw the equivalent circuit
referred to primary; the source voltage is
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given to be 230 volt 50 hertz single phase.
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So, let us draw the equivalent circuit of
the transformer referred to primary j 0.8
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ohm. Since we have referred it to the primary
it will be j X L 2 X 2 dash and its value
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will be 0.6. Since the actual value referred
to the secondary was 0.15 ohm, and N 1 by
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N 2 is 2; therefore, X 2 dash will be 0.15
into N 1 by N 2 square which is 4 and hence
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0.6 ohm. Similarly r 2 dashed will be 0.5
ohm, and R L dash will be equal to 15 ohm.
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The output voltage is V 2 dash; output current
is I 2 dash; input current is I 1; input voltage
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is 230 volt 50 hertz. So, let us take it to
be our reference phasor at 0 degree. The magnetization
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branch is referred to the primary. So, these
values do not change. This is 600 ohm; this
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is j 300 ohm. Now there are different ways
of solving this circuit.
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For example, the loop current method or the
node voltage method, however, for this particular
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problem we will apply the Thevenin’s theorem.
So, let us find out the Thevenin equivalent
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circuit at these two terminals looking back.
If V thevenin at these two terminal is given
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by
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impedance here divided by the total impedance
into the voltage; that is 600 parallel j 300
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divided by 600 parallel j 300 plus 0.75 plus
j0.8 into 230 angle 0 degree. Now 600 parallel
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j 300 is equal to 120 plus j 240; similarly,
the denominator becomes 600 parallel j 300
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plus 0.75 plus j 0.8 becomes 120.75 plus j
240.8 which is equal to 269.38 angle 63.37
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degrees. And this is 268.33 angle 63.43 degrees,
and V th is if we calculate comes to be 229.1
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angle 0.07 degrees. So, what is Z th? Z th
will be the equivalent impedance seen at this
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terminal when the input source is shorted.
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Therefore Z th is equal to 600 parallel j
300 parallel 0.75 plus j 0.8. In other words
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Z th equal to 120 plus j 240 parallel with
0.75 plus j 0.8, and if this is computed Z
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th comes to be 0.7461 plus j 0.7978. So, what
will be I 2 dash in this circuit? I 2 dash
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will be; so what is the Thevenin equivalent
circuit let us first see. The Thevenin equivalent
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circuit becomes V th in series with Z th in
series with rest of the circuit. Please note
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that if this voltage in the actual circuit
was E 1 the voltage at this terminal will
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also be E 1. Since E 1 equal to I 2 dash into
this complete impedance here also E 1 equal
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to I 2 dash into this impedance, and this
impedance do not change because only the input
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side has been changed.
So, what will be I 2 dash? I 2 dash will be
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equal to V Thevenin divided by Z Thevenin
plus sum of these impedances which is 15.5
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plus j0.6. This will be equal to 229.1 angle
0.07 degrees; that’s what we have found
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for V th previously, and the total impedance
will be 16.2461 plus j1.3978, and this comes
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to be 14.05 amperes at an angle of minus 4.85
degree with respect to the primary applied
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voltage. Now I 2, how much is I 2? I 2 equal
to N 1 by N 2 into I 2 dash; hence, I 2 is
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28.1 ampere at an angle of minus 4.85 degree
with respect to the input applied voltage.
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Therefore, we have found answer to one of
the problems; that is we have found what is
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the output current? So, what is the output
voltage? For that we will calculate what is
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V 2 dash? V 2 dash is nothing but I 2 dash
into load resistance.
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And this comes to 210.75 volts at an angle
of minus 4.85 degrees with respect to the
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input applied voltage. What is V 2? The actual
load voltage V 2 equal to N 2 by N 1 into
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V 2 dash; hence it is, 105.375 volt angle
minus 4.85 degrees. So, this is the answer
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to the second part of our question what is
the load voltage. How do we find out the magnetizing
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current? Please note that we have already
argued that if the voltage at the air gap,
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this is for the air gap voltage is E 1, in
the actual equivalent circuit then in this
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Thevenin equivalent circuit also the voltage
will be E 1. So, let us try to find out what
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is E 1 first. Now E 1 from this Thevenin equivalent
circuit is I 2 dash multiplied by the total
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impedance of the secondary side. These are
all phasors.
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And if we calculate this comes to 217.94 volts
at an angle of minus 2.633 degrees with respect
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to the primary applied voltage V 1. So, what
is no load current I 0? I 0 is E 1 divided
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by 600 parallel j X m that is from this actual
equivalent circuit. Hence if we compute I
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0 will come to be 0.8122 ampere at an angle
of minus 66.07 degrees. This is phase angle
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with respect to the primary applied voltage
V 1; since that is what we have assumed to
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be our reference phasor with angle 0. What
is I 1 then? Obviously, I 1 is the phasor
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sum of I 2 dash and I 0. It should be noted
that the current supplied by the Thevenin
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source is not I 1; it is I 2 dash.
So, therefore, to find out the source current
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I 1 we should go back to the original equivalent
circuit where before we took the Thevenin
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equivalent circuit. Therefore I 1 is given
by I 2 dash plus I 0, phasor sum of I 2 dash
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and I 0; that is we have found out I 2 dashed
to be 14.05 ampere angle minus 4.85 degrees,
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and I0 to be 0.8122 ampere angle at minus
66.07 degrees. Therefore, if we do the computation
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I 1 will come as 14.458 amperes angle minus
7.67 degrees. So, that shows how to use this
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equivalent circuit to compute different variables,
voltages and currents at different parts of
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the transformer. And it will be interesting
to solve the same problem using this equivalent
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circuit referred to the secondary, but this
is left as an exercise.
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However, the purpose of solving this problem,
was not just to show how this problem can
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be solved. I would like to draw your attention
to a few salient features of this equivalent
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circuit. For example, let us see the applied
voltage; the applied voltage was 230 volt
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at an angle of 0 degree while when we computed
the Thevinin voltage it comes to
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something like 229.1 volt angle 0.07 degrees.
Now is this very different from the actual
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applied voltage? Similarly the Thevinin impedance
Z th comes to 0.7461 plus j 0.7978. If you
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look at the actual primary impedance it was
0.75 ohm and 0.8 ohm here; therefore, it possibly
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make sense in order to reduce the computational
complexity to obtain this thevinin equivalent
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circuit by simply omitting this part and connecting
this right across the supply voltage.
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In fact, that is what is done in most practical
cases; since, the results that are obtained
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by that approximation is not very different
from this exact equivalent circuit; we will
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shortly see that. The rational is this impedance
of the magnetizing branch being much higher
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than the impedance of the primary winding;
that is winding resistance and leakage reactance,
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the inter applied primary voltage appears
almost across the magnetization branch. The
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drop across the primary impedance is very
small even at a rated current; with that assumption
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the exact equivalent circuit referred to the
primary can be simplified by the following
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approximate equivalent circuit.
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Here in the approximate equivalent circuit
the magnetization branch is connected right
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across the source, and the primary impedance
and secondary impedance are connected together
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V 2 dash load current I 2 dash. Please note
this was the exact equivalent circuit. In
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order to get the approximate equivalent circuit
what I have done? I have simply shifted this
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magnetization branch from this middle point
to the beginning that is right across the
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source to get the approximate equivalent circuit.
The underlying assumption being the drop across
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the across the primary winding impedance is
very small compared to the total applied voltage.
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Now this can be further simplified and secondary
the resistances can be clubbed together, and
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reactances can be clubbed together to give
a even simpler form of the same equivalent
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circuit.
It is equal to r 1 plus N 1 by N 2 square
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r 2, and j X eq equal to j X 1 plus X 2 dash;
that is equal to j X 1 plus N 1 by N 2 square
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X 2. This remains Z L dash, current is I 2
dash and voltage is V 2 dash. The primary
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current is I 1, magnetizing current is the
no load current is I 0; obviously, this circuit
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this approximate equivalent circuit is far
simpler compared to the exact equivalent circuit
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that has been shown earlier. Now how much
error do we incur if we use the approximate
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equivalent circuit instead of the actual equivalent
circuit? Let us try; let us try with the same
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problem.
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So, the approximate equivalent circuit for
the earlier problem will be this is 230 volt
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angle 0 degree. We have to find out this current
I 1, the no load current I 0. R eq from the
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given data is equal to 1.25 ohm; X eq from
the given data is j 1.4 ohm. The load resistance
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R L dash equal to 15 ohm. This is referred
load voltage V 2 dash; this is the referred
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load current I 2 dash. So, how do we compute
I 2 dash here? I 2 dash is simply equal to
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230 angle 0 degree divided by 16.25 sum of
these impedances plus j 1.4. This is given
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by 14.1 ampere angle minus 4.924 degrees.
So, I 2 equal to N 1 by N 2 I 2 dash is 28.2
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ampere angle minus 4.924 degrees. What is
V 2 dash? V 2 dash again is 15 I 2 dash, and
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the value is 211.5 volt angle minus 4.924
degrees, and V 2 equal to N 2 by N 1 V 2 dash
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equal to 105.75 volts angle minus 4.92 degrees.
The simplification in the calculation is very
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apparent; we do not have to find out the Thevenin
equivalent voltage or Thevenin impedance.
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We can simply calculate the variables.
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I 0 and I 1 can be computed; similarly, I
0 equal to 230 angle 0 degree divided by 600
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parallel j 300, and the value is 0.8571 amperes
at an angle of minus 63.435 degree with respect
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to the applied voltage phasor. The input current
I 1 is the phasor sum of I 0 plus I 2 dash
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and after computation if value comes to 14.566
ampere angle at minus 7.8 degrees. So, the
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amount of simplification in computation is
very apparent. So, let us compare the accuracy.
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So, this we can do by drawing this table.
So, let us see; this is the variable name.
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Next let us put the exact value from the exact
equivalent circuit. Then value from the approximate
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equivalent circuit
and percentage error let us say only in magnitude.
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First variable let us take the load voltage
V 2. The value you have got by exact equivalent
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circuit is 105.375 volt at an angle of minus
4.85 degrees, and from the approximate equivalent
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circuit we have got it to be 105.75 volts
at an angle of minus 4.924 degrees. The percentage
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00:44:50,380 --> 00:44:59,280
error in magnitude which is this value minus
this value divided by this comes to minus
187
00:44:59,280 --> 00:45:23,670
0.356 percent. For I 2 magnitude was 28.1
ampere, angle was minus 4.85 degrees. Here
188
00:45:23,670 --> 00:45:38,440
the magnitude was 28.2 amperes, angle was
minus 4.924 degrees. Again the percentage
189
00:45:38,440 --> 00:45:53,080
error in magnitude is just 0.356 percent.
What about I 0? The magnitude by exact equivalent
190
00:45:53,080 --> 00:46:04,350
circuit came to be 0.8122 ampere at an angle
of minus 66.07 degrees.
191
00:46:04,350 --> 00:46:14,420
By approximate equivalent circuit it came
to 0.8571 amperes at an angle of minus 63.435
192
00:46:14,420 --> 00:46:35,980
degrees. The percentage error here comes to
be minus 5.53 percent. Lastly for I 1 the
193
00:46:35,980 --> 00:46:46,540
magnitude came to be exact equivalent circuit
it came to 14.458 ampere at an angle of minus
194
00:46:46,540 --> 00:46:57,510
7.67 degrees, and by approximate equivalent
circuit it came to 14.566 amperes and angle
195
00:46:57,510 --> 00:47:15,400
of minus 7.8 degrees. The percentage error
is just minus 0.747 percent. Now we have already
196
00:47:15,400 --> 00:47:24,600
seen how much simplification in computation
can be achieved if this approximate equivalent
197
00:47:24,600 --> 00:47:33,750
circuit is used, and this table shows the
kind of accuracy that we can expect.
198
00:47:33,750 --> 00:47:41,200
We see that in most cases the percentage error
is less than even 0.5 percent except in the
199
00:47:41,200 --> 00:47:45,410
value of the magnetization current. The magnetization
current itself is less than 5 percent of the
200
00:47:45,410 --> 00:47:53,240
rated current. That is why this value comes
slightly higher, but even then that is close
201
00:47:53,240 --> 00:48:08,470
to 5 percent. Hence for all practical purpose
in a transformer it is normally this approximate
202
00:48:08,470 --> 00:48:15,420
equivalent circuit which is used until and
unless it is absolutely necessary the exact
203
00:48:15,420 --> 00:48:20,150
equivalent circuit normally is not used. Now
the approximate equivalent circuit can also
204
00:48:20,150 --> 00:48:23,050
be referred to either primary or the secondary.
205
00:48:23,050 --> 00:48:30,400
For the primary we have already shown the
approximate equivalent circuit to look like
206
00:48:30,400 --> 00:50:20,110
this; however, this can also be referred to
the secondary side. They need to look like,
207
00:50:20,110 --> 00:51:07,820
where V 1 dash is equal to N 2 by N 1 into
V 1. V 2 dash equal to N 1 by N 2 V 2. I 2
208
00:51:07,820 --> 00:51:41,090
dash equal to N 2 by N 1 into I 2. I 1 dash
equal to N 1 by N 2 into I 1. For the variables
209
00:51:41,090 --> 00:52:04,570
R eq dash R eq itself is r 1 plus r 2 dash.
X eq equal to X 1 plus X 2 dash. R eq dash
210
00:52:04,570 --> 00:52:21,380
equal to N 2 by N 1 square R eq; hence, it
is equal to R 2 plus N 2 by N 1 square R 1
211
00:52:21,380 --> 00:52:33,420
equal to R 1 dash plus R 2.
Similarly, X e q dash equal to N 2 by N 1
212
00:52:33,420 --> 00:52:52,880
square X e q equal to X1 dash plus X 2. R
0 dash equal to N 2 by N 1 square R 0, and
213
00:52:52,880 --> 00:53:11,280
Xm dash equal to N 2 by N 1 square Xm. Therefore,
in order to solve a given transformer problem
214
00:53:11,280 --> 00:53:17,780
the approximate equivalent circuit will normally
be used. The approximate equivalent circuit
215
00:53:17,780 --> 00:53:24,060
can be either referred to the primary or referred
to the secondary; the relationship between
216
00:53:24,060 --> 00:53:31,260
the variables and the equivalent circuit parameters
are given by these equations.