Transcript R - Note Khata

Single Phase Transformer
The efficiency of electrical power transmission has been improved by the use of
higher voltages.
This is one of the main reasons that alternating current (AC) has nearly entirely
replaced direct current (DC) for power transmission and distribution.
While it is true that the AC generator is better than the DC generator for
producing higher voltages.
The transformer is the device or machine primarily responsible for the wide use
of AC today.
Basically, the transformer is a device for transferring electrical energy from one
circuit to another circuit without a change in frequency.
The transformer accomplishes the change in voltage without use of moving
parts, and therein lies its great advantage.
The cost per kilowatt is comparatively low, and the efficiency is high.
As a matter of fact, the transformer is the most efficient piece of electrical
machinery, and efficiencies of 98 and 99% are not at all uncommon.
Since there are no moving parts, maintenance is simpler and cheaper, and the
required insulation for the extremely high voltages obtained can more easily be
constructed.
Working Principle of a Transformer
A transformer is a static (or stationary) piece of apparatus by means of
which electrical power in one circuit is transformed into electric power
of the same frequency in another circuit.
It can raise (step up) or lower (step down) the voltage in a circuit but
with a corresponding decrease or increase in current.
Physically, a transformer is mutual induction between two circuits
linked by a common magnetic flux.
It consists of two inductive coils which are electrically separated but
magnetically linked through a path of low reluctance (Fig. 32.1).
The two coils possess high mutual inductance.
If one coil is connected to a source of AC
voltage, an AC flux is set up in the
laminated core, most of which is linked with
the other coil in which it produce mutually
induced emf (according to the faraday’s law
of electromagnetic induction e=Mdi/dt).
If the second coil circuit is closed, a current flows in it and so electric
energy is transferred (entirely magnetically) from the first coil to the
second coil.
The first coil, in which electrical energy is fed from the AC supply
main, is called primary winding and the other from which energy is
drawn out, is called secondary winding.
In brief, a transformer is a device that
1. transfers electrical power from one circuit to another
2. it does so without a change of frequency
3. it accomplishes this by electromagnetic induction and
4. where the two electric circuits are in mutual inductive
influence of each other.
Transformer Construction
Constructionally, the transformers are two general types.
The two types are known as: Core-type, and Shell-type.
In the so called core-type transformers, the windings surround a
considerable part of the core (as shown schematically in Fig. 30.3a)
whereas in shell type transformers, the core surrounds a considerable
portion of the windings (as shown schematically in Fig. 30.3b).
For the core type transformers, the
primary and secondary winding are
located on the opposite legs (or limbs) of
the core, but in actual construction, these
are always interleaved to reduce leakage
flux.
As shown in Fig. 32.4, half the primary and half the
secondary winding have been placed side by side or
concentrically on each limb, not primary on one limb (or
leg) and the secondary on the other.
In both core and shell-type transformers, the individual laminations
are cut in the form of long strips of L’s, E’s and I’s as shown in Fig.
32.5.
The assembly of the compete core for
the two types of transformer is shown
in Fig. 35.6 and Fig 32.7.
In order to avoid high
reluctance at the joints
where the laminations
are butted against each
other, the alternate layer
are stacked differently to
eliminate these joints.
Core-type Transformers
In small size core-type transformers, a simple rectangular core is used
with cylindrical coils which are either circular or rectangular in form.
But for large-size core-type transformers, round or circular cylindrical
coils are used.
The core is made up of silicon-steel laminations which are either
rectangular or L-shaped.
With the coils wound on the two legs, the appearance is that of Fig.
14.20.
In order to provide maximum linkage between the windings, the group
of each leg is made up of both high-tension and low-tension coils.
This may be seen in Fig. 14.21, where a cross-sectional cut is taken
across the legs of the core.
By placing the high-voltage winding around the low-voltage winding,
only one layer of high voltage insulation is required.
If the high-voltage coils were adjacent to the core, an additional high
voltage insulation layer would be necessary between the coils and the
iron core.
Thus the arrangement of
high-voltage
winding
around the low-voltage
winding reduces the cost
of insulation.
Shell-type Transformers
As can be seen in Fig. 14.22, the iron almost entirely surrounds the copper in the
shell-type construction.
The core is made up of E-shaped or F-shaped laminations which are stacked to give
a rectangular figure eight.
All the windings are placed on the center leg, and in order to reduce leakage, each
high-side coil is adjacent to a low-side coil.
The coils actually occupy the
entire space of both windows,
are flat in shape, and are usually
constructed of strip copper.
Again to reduce the amount of
high-voltage insulation required,
the low-voltage coils are placed
adjacent to the iron core.
Comparison Between Core-type and
Shell-type Transformers
In the core-type, the magnetic-circuit path is longer, than
in the shell-type transformer.
In the core-type, the average length per turns is shorter,
than in the shell-type transformer.
High voltage coil insulation is more easily and
economically in the core-type transformer, and therefore
its use is often favored for high-voltage low and
medium-capacity circuits.
For multi-winding design, shell-type construction is
preferred by many manufacturers rather than core-type.
In order to obtain circular coils that are used
are easier to wound and provide more
mechanical strength, especially when short
circuit occur, the core of transformer is made
by
different
sizes
of
laminations.
Laminations are used to minimized the eddy
current loss. This type of design is called
cruciform type as shown in Fig. 14.23.
In order to obtain a shorter magnetic circuit
and a shorter average length of turn per
coils a special type design of core for shell
type is shown in Fig. 14.24. This type of
design is called distributed shell-type
transformer. The distributed core has four
returns legs for the magnetic field i.e. there
are two more legs at right angles to the
original ones. The additional advantage of
this type of design is that it makes for a
more compact transformer per KVA of
capacity.
Transformers are generally housed in tightly-fitted sheet-metal;
tanks filled with special insulating oil.
This oil has been highly developed and its function is two fold.
By circulation, it not only keeps the coils reasonably cool, but
also provides the transformer with additional insulation not
obtainable when the transformer is left in the air.
Another mean of classifying the transformer is according to the
type of cooling employed.
The following types are in common use:
(a) oil-filled self-cooled,
(b) oil-filled water cooled, and
(c) air-blast type
Elementary Theory of an Ideal Transformer
An ideal transformer is one which has no losses i.e. its winding have
no ohmic resistance, there is no magnetic leakage and hence which has
no I2R and core losses.
Consider an ideal transformer as shown in Fig. 32.13(a) whose
secondary is open and whose primary is connected to sinusoidal
alternating voltage V1.
This potential difference causes an alternating current to flow in the
primary.
The function of this current is merely to magnetize the core; it is small
in magnitude and legs V1 by 90o.
This alternating current Im
produces an alternating flux
 which is, at all times,
proportional to the current
and hence, is in phase with it.
This changing flux is linked both with the primary and the secondary
windings.
Therefore, it produces self-induced emf in the primary.
This self-induced emf E1 is, at every instant, equal to and in opposition
to V1.
It is also known as counter emf or back emf of the primary.
Similarly, there is produced in the secondary an induced emf E2 which
is known as mutually induced emf.
The instantaneous values of applied voltage, induced emfs, flux, and
magnetizing current are shown by sinusoidal waves in Fig. 32.13 (b).
Fig.
32.13(c)
shows
the
vectorial
representation of
the
effective
values of the
above quantities.
EMF Equation of a Transformer
Let,
N1 = No. of turns in primary
N2 = No of turns in secondary
m = maximum flux in core in webers
=BmA
f = frequency of AC input in Hz
As shown in Fig. 32.14, flux increases from its zero value to maximum value m in
one quarter of the cycle i.e. in (1/4f) second .
m
 4 f m wb/s or volt
So average rate of change of flux=
1/ 4 f
Now rate of change of flux per turn means induced emf in volts.
So average emf turn =4fm volt
If flux  varies sinusoidally, then rms value of induced emf is obtained by
multiplying the average value with form factor.
rms value
Form factor 
 1.11
average value
So rms value of emf/turn =1.11 4fm = 4.44fm volt
Now rms value of the induced emf in the whole of primary winding
= (induced emf/turn)(No. of primary turns)
E1  4.44 fN1 m  4.44 fN1Bm A
(i)
Similarly, rms value of the induced emf in the secondary is,
E2  4.44 fN 2 m  4.44 fN 2 Bm A
(ii)
E1 E2

 4.44 f m
It is seen from (i) and (ii) that
N1 N 2
It means that emf turns is the same
in both the primary and secondary
windings.
In an ideal transformer on no load
V1=E1 and V2=E2, where V2 is the
secondary terminal voltage.
(iii)

+
V1
-
+
E1
-
+
E2
-
+
V2
-
Voltage Transformation Ratio (K)
From above equation, we get,
E2 N 2

 4.44 f m  K
E1 N1
This constant K is known as voltage transformation ratio.
if N2>N1 i.e. K>1, then trans former is called step-up transformer.
if N2<N1 i.e. K<1, then trans former is called step-down transformer.
Again for an ideal transformer, Input VA = output VA
I V N 1
2 1 1
V I V I
or
11 2 2
I V N K
1 2
2
Hence, currents are in
inverse ratio of the
voltage
transformation ratio.
Transformer with Losses but no Magnetic Leakage
We will consider two cases:
(i) when such a transformer is on no-load, and
(ii) When it is loaded.
Transformer on No-Load
The core losses and copper losses are neglected in the case of ideal
transformer.
But in practical conditions, when an actual transformer is put on load,
there is iron loss in the core and copper loss in the windings (both
primary and secondary) and these losses are not entirely negligible.
The primary input under no-load (secondary side open) condition has
to supply (i) iron losses in the core (hysteresis loss and eddy current
loss) and (ii) a very small amount of copper loss in primary.
No-load input power: W0=V1I0cos0
where, cos0 is primary power factor under no-load conditions.
No-load condition of an actual transformer is shown vectorially in Fig. 32.16.
As seen from Fig. 30.16, primary current I0 has two components:
1. one in phase with V1. This is known as active or working or iron loss component
Iw because it mainly supplies the iron loss plus small quantity of primary Cu loss.
2. The other component is in quadrature with V1 and is known as magnetizing
component Im because its function is to sustain the alternating flux in the core. It is
wattles.
2 I2
Obviously, I0 is the vector sum of Iw and Im, hence, I0  I m
w
The no-load primary current I0 is very small as compared
to the full-load primary current.
As I0 is very small, the no-load primary Cu loss is
negligibly small which means that no-load primary input
is practically equal to the iron loss in the transformer.
Since the core-loss is responsible for shift in the current
vector, angle 0 is known as hysteresis angle of advance.
The no-load current I0 is simulated by pure inductance X0 taking the
magnetizing component Im and non-inductive resistance R0 taking the
working component Iw connected in parallel across the primary circuit
as shown in the following Fig.1.
The value of X0=E1/Im and of R0=E1/Iw.
Transformer on Load
It has already seen that at no load condition only current I0 flows in the circuit as
shown in Fig. 32.17(a). This current sets up the flux .
When the secondary is loaded, the secondary current I2 is set up.
The magnitude and phase of I2 with respect to V2 determined by the characteristics
of the load.
The secondary current sets up its own mmf (=N2I2) and hence its on flux 2 which
is in opposition to the main primary flux  which is due to I0 as shown in Fig.
32.17(b).
The secondary ampere-turns N2I2 are known as demagnetizing amp-turns.
The opposing secondary flux 2 weakens the primary flux , hence primary
back emf E1 tends to be reduced.
Fig. 32. 17(a) Transformer without load
Fig. 32.17 (b) Transformer on load
For a moment V1 gains upper hand over E1 and hence causes more current to flow in
primary.
Let the additional primary current be I2’ as shown in Fig. 32.17(c).
It is known as load component of primary current. This current is anti-phase with I2 .
The additional primary mmf N1I2’ sets up its own flux 2’, which is in opposition to
2 (but is in the same direction as ) and is equal to it in magnitude.
Hence the two cancel each other out as shown in Fig. 32.17(d).
So, we find that the magnetic effects of secondary current I2 are immediately
neutralized by the additional primary current I2’, which is brought into existence
exactly at the same instant as I2.
Hence, whatever the load conditions, the net flux passing through the core is
approximately the same as at no-load.
Fig. 32.17(c)
Fig. 32.17(d)
An important deduction is that
due to the constancy of core flux
at all loads; the core loss is also
practically the same under all
load conditions.
When transformer is on load,
the primary winding has two
currents in it; one is I0 and the
other I2’ is which is anti-phase
with I2 and K times in
magnitude (i.e. I2’ = I2K).
The total primary current is the
vector sum of I0 and I2’.
Fig. 32.17
In Fig. 32.8 are shown the vector diagrams for a load transformer when load is noninductive and when it is inductive.
Voltage transformation ratio of unity is assumed so that primary vectors are equal to
the secondary vectors.
With reference to Fig. 30.18(a), I2 is secondary current in phase with E2.
It causes primary current I2’ which is anti-phase with it and equal to it in magnitude.
Total primary current I1 is the vector sum of I0 and I2’ and lags behind V1 by an angle
1.
In Fig. 30.18(b) vectors are drawn for an
inductive load.
Here, I2 lags E2 (actually V2) by 2.
Current I2’ is again in anti-phase with I2
and equal to it in magnitude.
As before, I1 is the vector sum of I2’ and
I0 and lags behind V1 by 1.
In Fig. 32.8 are shown the vector diagrams for a load transformer when load is noninductive and when it is inductive.
Voltage transformation ratio of unity is assumed so that primary vectors are equal to
the secondary vectors.
With reference to Fig. 30.18(a), I2 is secondary current in phase with E2.
It causes primary current I2’ which is anti-phase with it and equal to it in magnitude.
Total primary current I1 is the vector sum of I0 and I2’ and lags behind V1 by an angle
1.
In Fig. 30.18(b) vectors are drawn for an
inductive load.
Here, I2 lags E2 (actually V2) by 2.
Current I2’ is again in anti-phase with I2
and equal to it in magnitude.
As before, I1 is the vector sum of I2’ and
I0 and lags behind V1 by 1.
Transformer with Winding Resistance but No Magnetic Leakage
An ideal transformer was supposed to possess no resistance, but in an actual
transformer, there is always present some resistance of the primary and secondary
windings as shown in Fig. 32.23.
Due to this resistance, there is some voltage drop in the two windings.
The result is that:
1. The secondary terminal voltage V2 is vectorially less than the secondary induced
emf E2 by an amount I2R2 where R2 is the resistance of the secondary winding.
Hence V2 is equal to the vector difference of E2 and resistive voltage drop I2R2.
V2=E2-I2R2
vector difference
2. Similarly, primary induced emf E1 is equal to the vector difference of V1 and I1R1
where R1 is the resistance of the primary winding.
E1=V1-I1R1
vector difference
Magnetic Leakage
In the case of ideal transformer it has been assumed that all the flux linked with
primary winding also links the secondary winding.
But in practice, all flux linked with primary does not link the secondary but part of
it i.e. completes its magnetic circuit by passing through air rather than around the
core, as shown in Fig. 32.26.
This flux is known as primary leakage flux and is proportional to the primary
ampere-turns alone because the secondary turns do not link the magnetic circuit of
L1.
This flux L1, which is in time phase with I1, induces an emf eL1 in primary but
none in secondary.
Similarly, secondary part sets up leakage flux L2 which linked with secondary
winding alone (and not with primary turns).
This flux L2, which is in time
phase with I2, produces a selfinduced emf eL2 in secondary (but
none in primary).
At no-load and light loads, the primary and secondary ampere-turns are small hence
leakage fluxes are negligible.
But when load is increased, both primary and secondary windings carry huge
currents.
The effect of induced emf due to the leakage flux is equivalent to a small inductive
coil in series with each winding such that voltage drop in each series coil is equal to
that produced by leakage flux as shown in Fig. 32.27 where X1=eL1/I1 and X2=eL2/I2.
The terms X1 and X2 are known as primary and secondary leakage reactances.
Following few points should be kept in mind:
1. The leakage flux links one or the other winding but not both, hence it in no
way contributes to the transfer of energy from the primary to secondary winding.
2. The primary voltage V1 will have to supply reactive drop I1X1 in addition to
I1R1. Similarly, E2 will have to supply reactive drop I2X2 in addition to I2R2.
Transformer with Resistance and Leakage Reactance
In Fig. 32.28 as shown the primary and secondary windings of a
transformer with reactances taken out of the windings.
The primary impedance and voltage equations are given by
X
Z  R  jX  R2  X 2 tan1 1 V1  E1 (R1 jX1)I1  E1 Z1I1
1 1
1
1
1
R
1
Similarly, secondary impedance and voltage equations are given by
X
2
2

1
2 E V  (R  jX )I V  Z I
Z  R  jX  R  X tan
2 2
2
2
2
2 2
2
2 2 2 2 2
R
2
The vector diagram for such a transformer for different kinds of loads
is shown in Fig. 32.29.
In these diagrams, vectors for resistive drops are drawn parallel to
current vectors whereas reactive drops are perpendicular to the
current vectors.
In the case of resistive load, the secondary voltage V2 and I2 are in
phase.
In the case of inductive
load, the secondary voltage
I2 legs V2 by an angle 2.
In the case of inductive
load, the secondary voltage
I2 leads V2 by an angle 2.
The angle 1 between V1
and I1 gives the power
factor
angle
of
the
transformer.
Equivalent Resistance, Reactance and Impedance
The resistances and reactance of the two windings of a transformer can be
transferred to any one of the tow windings.
The advantage of concentrating both resistances and reactances in one winding is
that it makes calculations very simple and easy because one has then to work in one
winding only.
It will be provided that the resistance R2, X2, in secondary is equivalent to R2/K2,
X2/K2 in primary.
The value R2/K2, X2/K2 will be denoted by R2’, X2’ the equivalent secondary
resistance as referred to primary.
The power loss of resistance R2 in secondary is= I22R2.
The power loss of resistance R2’ when R2 is referred to in secondary is= I12R2’.
2
I 
R
Equating the above two power, we obtain
I 2R'  I 2R or R'   2  R  2
1 2 2 2
2  I  2 K 2
 1
Similarly, equivalent primary resistance as referred to secondary is
R'  R K 2
1 1
Similarly, the leakage reactances can also be transferred from one
winding to the other in the same way as resistance. Thus
X '  X / K 2 and X '  X K 2
2
2
1 1
Total or effective resistance and reactance of the transformer as
referred to primary is
R  primary resistance equivalent secondary resistance as referred to primary
01
R  R  R'  R  R / K 2
01 1 2 1 2
Similarly, X  X  X '  X  X / K 2
01 1 2 1 2
Similarly total or effective resistance of the transformer as referred to
secondary is
R  secondary resistance equivalent primary resistance as referred to primary
02
R  R  R'  R  R K 2
02 2 1 2 1
Similarly, X  X  X '  X  X K 2
02
2 1
2 1
Total or effective impedance of the transformer as referred to
primary is
X
2
2

1
01
Z  R  jX  R  X tan
01 01
01
01 01
R
01
As shown in Fig. 32.30(a)
Similarly total or effective impedance of the transformer as referred to
secondary is
X
2
2

1
02
Z  R  jX  R  X tan
02 02
02
02 02
R
02
As shown in Fig. 32.30(b)
Equivalent Circuit
The transformer shown in Fig. 30.37(a) can be resolved into an
equivalent circuit in as shown in Fig. 30.37 (b).
To make transformer calculation simpler, it is preferable to transfer
voltage, current and impedance either to the primary or to the
secondary.
The primary equivalent of the secondary
induced voltage is E2’=E2/K=E1.
Similarly, primary equivalent of the secondary
terminal or output voltage is V2’=V2/K.
Primary equivalent of
secondary current is I2’=I2K.
the
For
transferring
secondary
impedance to primary, K2 is used.
R2’= R2/K2; X2’= X2/K2 ; ZL’= ZL/K2 ; E2’=E2/K=E1.
The secondary circuit is shown in Fig. 30.38(a)
and its equivalent primary values are shown in
Fig. 30.38(b).
The total equivalent circuit of the transformer
is obtained by adding in the primary impedance
as shown in Fig. 32.39.
This is known as the exact equivalent circuit but
it presents a somewhat harder circuit problem to
solve.
A simplification can be made by transferring the exciting circuit across
the terminal as in Fig. 32.40 or in Fig. 32.41 (a).
Further simplification may be achieved
by omitting I0 altogether as shown in Fig.
32.41(b).
From Fig. 32.39, it is found that total impedance between the input
terminal is
Zm(Z '  Z ' )
Zm is called
2 L
Z  Z  Zm (Z '  Z ' )  Z 
1
2 L
1 Z Z' Z'
m 2 L impedance of
the exciting
'
'
'
where, Zm  R jX ; Z  R  jX ; Z  R  jX . circuit.
0 0 1 1
1 2 2
2
The input voltage can be given by








Zm(Z '  Z ' ) 
2 L 
V I Z 
1 1 1 Z  Z '  Z ' 
m 2 L 
Example 32.1 The maximum flux density in the core of a 250/3000
volts, 50 Hz single phase transformer is 1.2 Wb/m2. If the emf per
turn is 8 volt, determine (i) primary and secondary turns, and (ii) area
of the core.
Solution: (i) E1=N1induced emf/turn;
So N1= E1/(induced emf/turn)= 250/8=32
and N2= E2/(induced emf/turn)= 3000/8=375
(ii) We may use
E  4.44 fN Bm A
2
2
E
2
A
4.44 fN Bm
2
3000
A
 0.03m2
4.44503751.2
Example 32.3 A single-phase transformer has 400 primary and 1000
secondary turns. The net cross sectional area of the core is 60 cm2. If
the primary winding be connected to a 50 Hz supply at 520 V,
calculate (i) the peak value of the flux density in the core, and (ii) the
voltage induced in the secondary winding.
Solution:
N 1000
K 2
 2.5
N 400
1
E
520
1
(i) Bm 

 0.976 Wb/m2
4.44 fN A 4.445040060104
1
E
(ii) 2  K ; E  KE  2.5250 1300V
2
1
E
1
Example 32.4 A 25 KVA transformer has 500 turns on the primary
and 50 turns on the secondary winding. The primary is connected to
3000V, 50 Hz supply. Find the full-load primary and secondary
currents, the secondary emf and the maximum flux in the core.
Neglect leakage drops and no-load primary current.
N
Solution: K  2  50  1  0.1
N 500 10
1
Now, full load I1=2500/3000=8.33 A.
Full load I2=I1/K=108.33=83.3 A
EMF per turn on primary side= 3000/500= 6 V
So, secondary emf= 650= 300 V
E
3000
1 
Also, m 
 27 mWb
4.44 fN 4.4450500
1
Example 32.11 A single phase transformer has 1000 turns on the
primary and 200 turns on the secondary. The no load current is 3 A at
a pf 0.2 lagging. Calculate the primary current and pf when the
secondary current is 280 A at a pf of 0.8 lagging.
Solution: V2 is taken as reference cos-1(0.8)=36.87o sin(36.87o)=0.6;
K=200/1000=1/5.
I  280 36.87 A; I '  (280 / 5)36.87  5636.87 A
2
2
  cos1(0.2)  78.5; sin   0.98
1
1
I  I  I '  378.55636.87 A
1 0 2
I  I  I '  3(0.2  j0.98) 56(0.8 j0.6) A
1 0 2
I  45.4  j36.54  58.3 38.86 A
1
Thus I1 legs behind the supply by an angle 38.86o.
Transformer Tests
The performance of a transformer can be calculated on the basis of equivalent circuit
which contains four parameters, the equivalent resistance R01 as referred to primary
(or secondary R02), the equivalent leakage reactance X01 as referred to primary (or
reactance in secondary X02), the core loss conductance G0 (or resistance R0) and the
magnetizing susceptance B0 (or reactance X0).
These constants or parameters can be easily determined by two tests
(i) open-circuit test and
(ii) short-circuit test
Open-Circuit or No-Load Test
The purpose of this test is to determine no-load loss or core loss and no-load current
I0 which is helpful in finding X0 and R0.
Low voltage side connected with normal voltage and frequency and high voltage
side is left open.
A wattmeter W, voltmeter V and an ammeter A are connected in the low-voltage
winding i.e. primary winding in the present case as shown in Fig. 32.43.
The voltage V1 is measured using the voltmeter (V). With normal voltage applied to
the primary, normal flux will be set up in the core, hence normal iron losses will
occur which are recorded by the wattmeter (W). As the primary no-load current I0 (as
measured by ammeter, A) is small, Cu loss is negligibly small in primary and null in
secondary. Hence, the wattmeter reading represents practically the core loss under noload condition (and which is the same for all loads).
The no-load vector diagram is shown in Fig. 32.16.
If W0 is the wattmeter reading as shown in Fig. 32.43, then
W V I cos ;
Since the current is practically all0 10
0
exciting current when a transformer is
 cos W /V I
0
0 10
on no-load (i.e. I0=Im) and as the voltage
I m  I sin  ;
drop in primary leakage impedance is
0
0
small, hence the exciting admittance
I w  I cos
0
0
Y0(=1/Z0) of the transformer is given by
X V / I m ;
I0=V1Y0 or Y0=I0/V1.
0 1
R V / I w
0 1
The exciting conductance G0 is given
by W0=V12G0 or G0(=1/R0)= W0 /V12.
The exciting susceptance
B (1/ X )  Y 2  G2
0
0
0
0
Separation of Core Losses
The core loss of transformer depends upon the frequency and the maximum flux
density when the volume and the thickness of the core lamination are given.
The core loss is made up of two parts:
(i) Hysteresis loss: Wh=PBmax2f and (ii) Eddy current loss: We=QBmax2f 2
Where, P and Q are constant.
The total core loss is given by: Wi=Wh+We= PBmax2f + QBmax2f 2.
If we carry out two experiments using two different frequencies but the same
maximum flux density, we should be able to find the constants P and Q and hence
calculate hysteresis and eddy current losses separately.
If the maximum flux can be kept same value then the iron or core losses can be
written as follows:
Wi=Wh+We= Af +Bf 2;
where, A= PBmax2;
B= QBmax2.
From the measured the core loss in two different frequencies, the constant A and B
can be calculated.
Short-Circuit or Impedance Test
This is an economical method for determining the following:
(i) Equivalent impedance (Z01 or Z02), leakage reactance (X01 or X02) and total
resistance (R01 or R02) of the transformer as referred to the winding in which the
measuring instruments are placed.
(ii) Cu loss at full load. This loss is used in calculating the efficiency of the
transformer.
(iii) Knowing Z01 or Z02, the total voltage drop in the transformer as referred to
primary or secondary can be calculated and hence regulation of the transformer
determined.
In this test, one
winding, usually
the low-voltage
winding is solidly
short-circuited by
a thick conductor
as shown in Fig.
32.45.
A low-voltage (usually 5 to 10% of normal primary voltage) at correct
frequency is applied to the primary and is cautiously increased till fullload currents are flowing both in primary and secondary (as indicated
by the respective ammeters).
Since the applied voltage is a small percentage of the normal voltage,
the mutual flux  produced is also a small percentage of its normal
value.
Hence, core losses are very small with the result that the wattmeter
reading represents the full-load Cu loss or I2R loss for the whole
transformer i.e. both primary Cu loss and secondary Cu loss.
The equivalent circuit of the transformer under short-circuit condition
is shown in Fig. 32.46.
If VSC is the voltage required to circulate rated load currents.
Then Z01=VSC/I1. Also W=I12R01.  R01=W/I12 and
X
 Z 2  R2
01
01 01
If R1 and X1 can be measured, then knowing R01 and X01, we can find
R2’=R01-R1 and X2’=X01-X1.
Hence, the secondary resistance and reactance can be calculated by
using the following equation: R2= R2’K2; X2= X2’K2 .
Example 32.36 Obtain the equivalent circuit of a 200/400 V, 50 Hz, 1-phase
transformer from the following data:
O.C (Open Circuit) test: 200V, 0.7A, 70W on l.v (low voltage) side
S.C (Short Circuit) test: 15V, 10A, 85W on h.v (high voltage) side
Calculate the secondary voltage when delivering 5kW at 0.8 pf (power factor)
lagging, the primary voltage being 200V.
Solution: From O.C Test: V1I0cos0=W0. 200×0.7×cos0=70.
cos0=70/(200×0.7)=0.5 and sin0=0.866.
Iw=I0cos0= 0.7×0.5=0.35 A.
Im=I0sin0= 0.7×0.866=0.606 A.
R0=V1/Iw=200/0.35=571.4 .
X0=V1/Im=200/0.606=330 .
From S.C Test: It may be noted that in this test, instruments have been placed in the
secondary i.e high-voltage winding whereas the low voltage side i.e primary has
been short circuited. Where K=400/200=2;
Z02=VSC/I2=15/10=1.5 .
Z01=Z02/K2=1.5/4=0.375 .
Also, I22R02=W; R02=85/100=0.85 .
X
 Z 2  R2  (0.375)2  (0.21)2  0.31 
01
01 01
R01=R02/K2=0.85/4=0.21 .
X
 Z 2  R2  (1.5)2  (0.85)2 1.24 
02
02 02
The equivalent circuit is shown in Fig. 32.49.
The values of parameters are referred to primary i.e. low voltage side.
Output kVA=5/0.8=6.25; Output current, I2=6.25×1000/400=15.6 A
Total transformer drop as referred to secondary
=I2(R02cos2+X02sin2)
=15.6× (0.85×0.8+1.24×0.6)=22.2 V
V2= 400-22.2=377.8 V