Woldia University

Introduction To Electrical Machine

A Non ideal Transformer

A practical transformer differs from the ideal transformer in many respects. The
practical transformer has (i) iron losses (ii) winding resistances and (iii) magnetic
leakage, giving rise to leakage reactances.
1. Winding Resistances
However small it may be, each winding has some resistance. Nonetheless, we can
replace a non ideal transformer with an idealized transformer by including a
lumped resistance equal to the winding resistance of series with each winding. The
resistor R1 and R2 are the winding resistances of the primary and the secondary,
respectively. The inclusion of the winding resistances dictates that (a) the power
input must be greater than the power output, (b) the terminal voltage is not equal to
the induced emf, and (c) the efficiency (the ratio of power output to power input)
of a non ideal transformer is less than 100%.
2. Leakage Fluxes
Not all of the flux created by a winding confines itself to the magnetic core on
which the winding is wound. Part of the flux, known as the leakage flux, does
complete its path through air.
The primary leakage flux set up by the primary does not link the secondary.
Likewise, the secondary leakage flux restricts itself to the secondary and does not
link the primary. The common flux that circulates in the core and links both
windings is termed the mutual flux.

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3. Iron losses
Since the iron core is subjected to alternating flux, there occurs eddy current and
hysteresis loss in it. These two losses together are known as iron losses or core
losses. The iron losses depend upon the supply frequency, maximum flux density
in the core, volume of the core etc. It may be noted that magnitude of iron losses is
quite small in a practical transformer.

Practical Transformer losses

We will consider two cases (i) when such a transformer is on no load and (ii)
transformers on load
1. Transformer on No-load

Consider a practical transformer on no load i.e., secondary on open-circuit as

shown in the figure below. The primary will draw a small current I0 to supply (i)
the iron losses and (ii) a very small amount of copper loss in the primary. Hence
the primary no load current I0 is not 90° behind the applied voltage V1 but lags it by
an angle 0 < 90° as shown in the phasor diagram below.

No load input power, W0 = V1 I0 cos 0

From the above phasor diagram the no-load primary current I0 can be resolved into
two rectangular components:
1. The component IW in phase with the applied voltage V1. This is known as
active or working or iron loss component and supplies the iron loss and a
very small primary copper loss.
IW = I0 cos 0

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2. The component Im lagging behind V1 by 90° and is known as magnetizing

component. It is this component which produces the mutual flux in the
core.
Im = I0 sin 0
Clearly, I0 is phasor sum of Im and IW,
Io= 𝐼𝑚 2 + 𝐼2
𝑤
Iw
No load p.f. = cos 𝜑𝑂 =
Io
It is emphasized here that no load primary copper loss (i.e.I2 R) is very small and
may be neglected. Therefore, the no load primary input power is practically equal
to the iron loss in the transformer i.e., No load input power, W0 = Iron loss

Example-1: A 2,200/200-V transformer draws a no-load primary current of 0.6A

and it absorbs 400 watts. Find the magnetising and iron loss currents. (b) A
2,200/250-V transformer takes 0.5 A at a p.f. of 0.3 on open circuit. Find
magnetizing and working components of no-load primary current.
Solution:

Now: I02 =IW2+Im2 then Im= 𝐼𝑂2 − 𝐼𝑤2 = 0.62 − 0.1822 =0.572A

(b)
Im= 𝐼𝑂2 − 𝐼𝑤2 = 0.52 − 0.152 =0.476A
2. Practical Transformer on Load
We shall consider two cases (i) when such a transformer is assumed to have no
winding resistance and leakage flux (ii) when the transformer has winding
resistance and leakage flux.
1. No winding resistance and leakage flux
As shows a practical transformer with the assumption that resistances and leakage
reactances of the windings are negligible, with this assumption, V2 = E2 and V1 =
E1. Let us take the usual case of inductive load which causes the secondary current
I2 to lag the secondary voltage V2 by 2. The total primary current I1 must meet two
requirements viz.

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(a) It must supply the no-load current I0 to meet the iron losses in the transformer
and to provide flux in the core.
(b) It must supply a current I'0 to counteract the demagnetizing effect of secondary
currently I2. The magnitude of I'2 will be such that:
N1I'2 N2I2
N2I2
I’= =KI2
N1
The total primary current I1 is the phasor sum of I'2 and I0 i.e.,
I1 I'2 I0, where:

2. Transformer with resistance and leakage reactance

As shown in the figure below, a practical transformer having winding resistances
and leakage reactances. These are the actual conditions that exist in a transformer.
There is voltage drop in R1 and X1 so that primary e.m.f. E1 is less than the applied
voltage V1. Similarly, there is voltage drop in R2 and X2 so that secondary terminal
voltage V2 is less than the secondary e.m.f.

I1 = I'2 + I0
Where: I'2 = KI2

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V1=E1+I1 (R1+X1)
V2 = E2 I2 (R2 + jX2)
= E2 I2Z2
Po= Re (V2I2*) and pin =Re (V1I1*)
Example: A 23-kVA, 2300/230-V, 60 HZ step-down transformer has the following
resistance and leakage-reactance values: R1 = 4Ω, R2 = 0.04 Ω, X1 = 12 Ω, and
X2 = 0.12 Ω. The transformer is operating at 75% of its rated load. If the power
factor of the load is 0.866 leading, determine the efficiency of the transformer.

Solution: Since the transformer is operating at 75% of its rated load, the effective
value of the secondary winding current is:

Assuming the load voltage as a reference, the load current at a leading power factor
of 0.866, in phasor form, is:

The secondary winding impedance is:

𝑧2 = R1 + jX2 = 0.04 + j0.12 Ω
The induced emf in the secondary winding is:

2300
Since the transformation ratio is: K or a = = 10
230
We can determine the induced emf and the current on the primary side as:

The primary winding impedance is: 𝑍1 = R1 + jX1 = 4 + j12 Ω

Hence, the source voltage must be:

The power supplied to the load is:

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The power input is:

Impedance Transformation through a Transformer

Consider the following figures that represent the impendence ratio:

N1 V1 I2
Since from the voltage transformation ratio = = =K
N2 V2 I1

V2 𝑉1
Z2= 𝑎𝑛𝑑 𝑍1 = 𝑡ℎ𝑒𝑛
𝐼2 𝐼1

𝑍1 𝑉1 𝐼2 𝑍1
= 𝐼1
= 𝐾2 : = 𝐾2
𝑍2 𝑉2 𝑍2
𝑅1 𝑋1
Similarly: = K2 𝑎𝑛𝑑 =K2
𝑅2 𝑋2

 Note the importance of above relations. We can transfer the parameters from
one winding to the other. Thus:
 A resistance R1 in the primary becomes R1/ K2 when transferred to the
secondary.
 A resistance R2 in the secondary becomes K2 R2 when transferred to the
primary.
 A reactance X1 in the primary becomes X1/ K2 when transferred to the
secondary.
 A reactance X2 in the secondary becomes K2 X2 when transferred to the
primary.
It is important to remember that:

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 When transferring resistance or reactance from primary to secondary, divide

it by K2.
 When transferring resistance or reactance from secondary to primary,
multiply it by K2.
 When transferring voltage or current from one winding to the other, only K
is used
Note: .But the shunt value (RC and XM) are always divided by K2.

Shifting impedances in a transformer

The following figure shows a transformer where resistances and reactances are
shown external to the windings. The resistance and reactance of one winding can
be transferred to the other by appropriately using the factor K2. This makes the
analysis of the transformer a simple affair because then we have to work in one
winding only.
𝑁1
𝑁2

1. Referred to primary
When secondary resistance or reactance is transferred to the primary, it is
multiplied by K2. It is then called equivalent secondary resistance or reactance
referred to primary and is denoted by R'2 or X'2.
Equivalent resistance of transformer referred to primary
R01=R1+R’2=R1+K2R2
Equivalent reactance of transformer referred to primary
X02=X1+X’2=X1+K2X2
Equivalent impedance of transformer referred to primary
Z01= R201 + X012

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𝑁1
𝑁2
K 2 R2 𝐾 2 𝑋2

Referred to secondary
When primary resistance or reactance is transferred to the secondary, it is
multiplied by K2. It is then called equivalent primary resistance or reactance
referred to the secondary and is denoted by R'1 or X'1.

Equivalent resistance of transformer referred to secondary

R02 R2 R'1 R2  R1K2
Equivalent reactance of transformer referred to secondary
X02 X2 X'1 X2 X1K2
Equivalent impedance of transformer referred to secondary
Z02  𝑅022
+ 𝑅022

 𝑋1
𝑅1
𝐾2
𝐾2
𝑁1
𝑁2


Exact Equivalent Circuit of a Loaded Transformer
The following figure shows the exact equivalent circuit of a transformer on load.
Here R1 is the primary winding resistance and R2 is the secondary winding
resistance. Similarly, X1 is the leakage reactance of primary winding and X2 is the
leakage reactance of the secondary winding. The parallel circuit R0 X0 is the no-
load equivalent circuit of the transformer. The resistance R0 represents the core
losses (hysteresis and eddy current losses) so that current IW which supplies the
core losses is shown passing through R0. The inductive reactance X0 represents a
loss-free coil which passes the magnetizing current Im.

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1. Equivalent circuit referred to primary

If all the secondary quantities are referred to the primary, we get the equivalent
circuit of the transformer referred to the primary as in the figure below. Note that
when secondary quantities are referred to primary,
2 2
resistances/reactances/impedances are multiplied by K or a , voltages are
multiplied by K or a’ and currents are divided by K or a.
R2’=a2× R2; X2’=a2× X2; ZL’=a2× 𝑍L; V2’ = a×V2 and I2’=I2/a

Equivalent circuit referring to the primary side of the transformer


2. Equivalent circuit referred to secondary.
If all the primary quantities are referred to the secondary, we get the equivalent
circuit of the transformer referred to the secondary as in the below. Note that when
primary quantities are referred to secondary, resistances/reactances/impedances are
divided by K2 or a2, voltages are divided by K or a and currents are multiplied by
K.
R1’=R1/a2; X1’=X1/ a2; V1’ =V1/a and I1’=I1a

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Equivalent transformer referring to the secondary side of the transformer

Approximate Equivalent Circuits

The figure shown below is equivalent circuit of the transformer referring to the
primary side of transformer.
In which the equivalent value of impendence is given below:

Similarly the figure below shows the approximate equivalent circuit as referred to
the secondary side of the transformer.

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A 23-kVA, 2300/230-V, 60 HZ step-down transformer has the following

resistance and leakage-reactance values: R1=4Ω R2=0.04 Ω, X1=12 Ω, and X2=
0.12 Ω. The transformer is operating at 75% of its rated with the load resistor of
30Ω. If the power factor of the load is 0.866 leading and the equivalent core-loss
resistance and the magnetizing reactance on the primary side of the transformer are
20 kΩ and 15 kΩ respectively. Draw the equivalent circuit referred to (a) the high-
voltage side and (b) the low-voltage side, and label the impedances numerically.
SOLUTION:
I 2 23000
IP = × 75% = 75∠30°/10 = 7.5∠30°
a 230
ZL’=a2ZL= 10× 30=3000Ω

Example: A 50-kVA 2400:240-V 60-Hz distribution transformer has a leakage

impedance of 0.72 + j0.92 Ω in the high-voltage winding and 0.0070 + j0.0090 Ω
in the low-voltage winding. At rated voltage and frequency, the impedance Z~ of
the shunt branch (equal to the impedance of Re and j Xm in parallel) accounting
for the exciting current is 6.32 + j43.7 Ω when viewed from the low-voltage side.
Draw the equivalent circuit referred to (a) the high-voltage side and (b) the low-
voltage side, and label the impedances numerically.

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Tests of the transformer

The performance of a transformer can be calculated on the basis of its equivalent
circuit which contains four main parameters, the equivalent resistance R01 as
referred to primary (or secondary R02), the equivalent leakage reactance X01 as
referred to primary (or secondary X02), the core-loss conductance G0 (or resistance
R0) and the magnetising 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. These tests are very economical and convenient, because they
furnish the required information without actually loading the transformer. In fact,
the testing of very large A.C. machinery consists of running two tests similar to the
open and short-circuits tests of a transformer.

1. Open circuit test

The transformer’s secondary winding is open-circuited, and its primary winding is
connected to a full rated line voltage. In this test, the rated voltage is applied to the
primary (usually low-voltage winding) while the secondary is left open circuited.
The applied primary voltage V1 is measured by the voltmeter, the no-load current I0
by ammeter and no-load input power W0 by wattmeter. As the normal rated voltage
is applied to the primary, therefore, normal iron losses will occur in the transformer
core. Hence wattmeter will record the iron losses and small copper loss in the
primary. Since no-load current I0 is very small (usually 2-10 % of rated current).
Cu losses in the primary under no-load condition are negligible as compared with
iron losses. Hence, wattmeter reading practically gives the iron losses in the
transformer. It is reminded that iron losses are the same at all loads.
Iron losses, Pi = Wattmeter reading = W0
No load current = Ammeter reading = I0
Applied voltage = Voltmeter reading = V1
Input power, W0 = V1 I0 cos 0

Thus open-circuit test enables us to determine iron losses and parameters R0 and
X0 of the transformer.

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Short-Circuit or Impedance Test

This test is conducted to determine R01 (or R02), X01 (or X02) and full-load copper
losses of the transformer. In this test, the secondary (usually low-voltage winding)
is short-circuited by a thick conductor and variable low voltage is applied to the
primary. The low input voltage is gradually raised till at voltage V SC, full-load
current I1 flows in the primary. Then I2 in the secondary also has full-load value
since I1/I2 = N2/N1. Under such conditions, the copper loss in the windings is the
same as that on full load. Therefore, input power is all loss and this loss is almost
entirely copper loss. It is because iron loss in the core is negligibly small since the
voltage VSC is very small. Hence, the wattmeter will practically register the full-
load copper losses in the transformer windings.

Full load Cu loss, PC = Wattmeter reading = WS

Applied voltage = Voltmeter reading = VSC
F.L. primary current = Ammeter reading = I1

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Where:

Where R01 is the total resistance of transformer referred to primary.

V sc
Total impedance referred to primary, Z01 = I1
2 2
Total leakage reactance referred to primary, X01= 𝑍01 − 𝑅01
Pc
Short-circuit p.f, cos 
V sc I 1
Determining the Values of Components in the
Transformer Model
Since It is possible to experimentally determine the values of the inductances and
resistances in the transformer model An adequate approximation of these values
can be obtained with only two tests, the open-circuit test and the short-circuit
test. Open circuit is used to determine the shunt components of the transformer
such as Rc and Xm.
Therefore the conductance of the core-loss resistor is given by:
1
GC=
Rc
And the susceptance of the magnetizing inductor is given by:
1
Bm=
Xm
Since these two elements are in parallel, their admittances add, and the total
excitation admittance is:

And the angle

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The power factor is always lagging for a real transformer, so the angle of the
current always lags the angle of the voltage by θ degrees. Therefore, the
admittance YE is:

Similarly shot circuit is used to determine the series impendence such as resistance
and inductance of the transformers. Thus:

The power factor of the current is given by:

and is lagging. The current angle is thus negative, and the overall impedance angle
θ is positive:

Example: A l5-kVA, 2300/230-V transformer is to be tested to determine its

excitation branch components and its series impedances. The following test data
have been taken from the transformer:

(a) Find the equivalent circuit of the transformer referred to the high-voltage side.
(b) Find the equivalent circuit of this transformer referred to the low-voltage side.
Solution:
(a) The turns ratio of this transformer is a = 2300/230 = 10. The excitation branch
values of the transformer equivalent circuit referred to the secondary (low voltage)

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side can be calculated from the open-circuit test data, and the series elements
referred to the primary (high voltage) side can be calculated from the short-circuit
test data. From the open-circuit test data, the open-circuit impedance angle is:

The series elements referred to the primary side are:

The resulting simplified equivalent circuit referred to the primary side can be
found by convel1ing the excitation branch values to the primary side.

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To find the equivalent circuit referred to the low-voltage side, it is simply

necessary to divide the impedance by a2, since a = Np/Ns = 10, the resulting values
are:

Advantages of Transformer Tests

The above two simple transformer tests offer the following advantages:
i. The power required to carry out these tests is very small as compared to the
full-load output of the transformer. In case of open-circuit lest, power
required is equal to the iron loss whereas for a short-circuit test, power
required is equal to full-load copper loss.
ii. These tests enable us to determine the efficiency of the transformer
accurately at any load and p.f. without actually loading the transformer.

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iii. The short-circuit test enables us to determine R01 and X01 (or R02 and X02).
We can thus find the total voltage drop in the transformer as referred to
primary or secondary. This permits us to calculate voltage regulation of the
transformer.
Performance of the Transformer
The performances of the transformer are determined by voltage regulation and
efficiency of the transformer.
Voltage regulation of the transformer
Voltage regulation is a quantity that compares the output voltage of the transformer
at no load with the output voltage at full load.

Where: V2NL and V2FL, are the effective values of no-load and full-load voltages at
the secondary terminal. The smaller the voltage regulation, the better the operation
of the transformer. The expressions for the percent voltage regulation for the
approximate equivalent circuits as viewed from the primary and the secondary
sides are:
Since at no load, Vs = VP/a, the voltage regulation can also be expressed as:

To determine the voltage regulation of a transformer, it is necessary to understand

he voltage drops within it. The effects of the excitation branch (Rc and Xm) on
transformer voltage regulation can be ignored, so only the series impedances need
be considered. The voltage regulation of a transformer depends both on the
magnitude of these series impedances and on the phase angle of the current
flowing through the transformer and apply Kirchhoff’s voltage law to the remain
equivalent circuit. Thus the voltage at no load on the secondary will be:

Depending on this formula the voltage regulation can be determine by the phasor
diagram. So, the phasor diagram for lagging, unity and leading power factor
respectively is:

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Lagging power factor

Unity power factor

Leading power factor

Transformer Efficiency
Transformers are also compared and judged on their efficiencies and it is the ratio
of output power to input power. It is expressed by percentage and denoted by
Greek letter "𝛈". The efficiency of a device is defined by the equation:
Input power
Efficiency =
Output power

Because the input power is the sum of output power and power losses.
The output power is given as:

The losses of the transformer are also core and copper losses:
power core loss: and power copper losses:

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Lastly the efficiency of the transformer is given as:

Example: A 15-kVA, 2300/230-V transformer is to be tested to determine its

excitation branch components, its series impedances, and its voltage regulation.
The following test data have been taken from the transformer:

a) Find the equivalent circuit of the transformer referred to the high-voltage side.
(b) Find the equivalent circuit of this transformer referred to the low-voltage side.
(c) Calculate the full-load voltage regulation at 0.8 lagging power factor, 1.0
power factor, and at 0.8 leading power factor using the exact equation for V p.
(d) Plot the voltage regulation as load is increased from no load to full load at
power factors of 0.8 lagging, 1.0, and 0.8 leading.
(e) What is the efficiency of the transformer at full load with a power factor of 0.8
lagging?
Solution:
(a) The turns ratio of this transformer is a = 2300/230 = 10

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To calculate Vp/a,:

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The phasor diagram in the current and voltage with respect to lagging,
unity and leading power factor respectively is:

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e) To find the efficiency of the transformer, first calculate its losses. The copper
(

and core losses are:

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