1.

Transformers

1. Transformers
1.1 Introduction
A transformer is a static electromagnetic device, which transfers electric power from one circuit to other
circuit without change in power and frequency. It can raise or lower the voltage in a circuit but with a
corresponding decrease or increase in current.
1.1.1 Faraday’s Laws of Electromagnetic Induction:
First Law: Whenever a conductor cuts magnetic flux, an EMF is induced in that conductor.
Second Law: The magnitude of induced EMF is equal to the rate of change of flux-linkages.
Suppose a coil has 𝑁 turns and the change of flux-linkages through it 𝑑𝜙 Wb in time interval 𝑑𝑡 seconds.
The total flux-linkages is the product of number of turns and the flux linked with the coil, i.e. total flux-
linkages = 𝑁𝑑𝜙.
Therefore, the induced EMF is given by,
𝑑 𝑑𝜙
𝑒= (𝑁𝜙) = 𝑁 Volt
𝑑𝑡 𝑑𝑡
Usually, a minus sign is given to the right-hand side expression to signify that the induced EMF sets up
current in such a direction that magnetic effect produced by it opposes the cause producing it. So,
𝑑𝜙
𝑒 = −𝑁 Volt
𝑑𝑡
1.1.2 Direction of Induced EMF and Current
The direction of induced current can be found by Lenz’s Law. Lenz’s law of electromagnetic induction
states that the direction of the current induced in a conductor by a changing magnetic field (as per
Faraday’s law of electromagnetic induction) is such that the magnetic field created by the induced
current opposes the initial changing magnetic field which produced it. The direction of this current flow
is given by Fleming’s Right hand rule.
1.1.3 Induced EMF
It can be either (i) Dynamically Induced EMF or (ii) Statically Induced EMF.
(i) Dynamically Induced EMF: In this case, usually the field system is stationary and conductors are
rotating (e.g. DC generators). The magnitude of induced EMF is given by 𝑒 = 𝐵𝑙𝑣 sin 𝜃 and its
direction is given by Fleming’s Right hand rule.
(ii) Statically Induced EMF: In this case, usually the conductors remain stationary and flux linked with
it (or field system) is changing by changing the current producing this flux (e.g. transformers).
Statically induced EMF can be further sub-divided into (a) Self Induced EMF and (b) Mutually Induced
EMF.
(a) Self Induced EMF: This is the induced
EMF in a coil due to the change of its own
flux linked with it. If current through the
coil is changed, then the flux linked with its
own turns will also change, which will
produce self-induced EMF.
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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
1.2 Necessity of Transformer
(b) Mutually Induced EMF:
Induced EMF in one coil by
the influence of the other
coil is called as mutually
induced EMF. Consider two
magnetically coupled coils
A and B having 𝑁1 and 𝑁2
turns respectively, lying
close to each other as shown in figure. Coil B is connected to a battery, a switch and a variable
resistance 𝑅 whereas coil A is connected to a sensitive voltmeter V. When current through coil B is
established by closing the switch, its magnetic field is set up which partly links with the coil A. As
current through V is changed, the flux linked with A is also changed. Hence, mutually induced EMF
is produced in A.
1.2 Necessity of Transformer
Transformers help to improve power systems safety and efficiency by increasing and lowering voltage
levels as and when needed. These are used in a wide range of residential and industrial applications,
mainly and perhaps most significantly in long distance power distribution and regulation.
The power that is generated at the power stations need to be transported to the electrical grids and from
there to consumers. This process needs to be done so that there is minimal loss of energy and also the
process is as cheap as possible. For that purpose, transformers are used.
Usually, electrical power is generated at 11 KV. For economic reasons, AC power is transmitted at very
high voltages say 220 kV or 440 kV over long distances. Therefore, a step-up transformer is applied at
the generating stations. Now for safety reasons the voltage is stepped down to different levels by step
down transformer at various substations to feed the power to different locations and thus the utilisation
of power is done at 400/230 V.
In industries, transformers are used to ensure appropriate, safe, and careful usage of power. It will help
to prevent the damage of industrial equipment and also to protect the people working around it.
Depending on their use, modern appliances have different requirements of voltage, which is supplied to
them by transformers. Additionally, transformers isolate the electrical device from the main power
supply, thus protecting them from any damage or risk.
Apart from the above, small size transformers are used in communication circuits, radio and TV circuits,
telephone circuits, instrumentation and control systems.
1.3 Types of Transformer
(a) According to magnetic circuit: (i) Core Type Transformer (ii) Shell Type Transformer
(b) According to voltage level: (i) Step-up Transformer (ii) Step-down Transformer
(c) According to number of phases: (i) Single Phase Transformer (ii) Three Phase Transformer
(d) According to cooling methods: (i) Oil Cooled Transformer (ii) Dry TypeTransformer
(e) According to windings: (i) Two Winding Transformer (ii) Autotransformer
(f) According to application: (i) Power Transformer (ii) Distribution Transformer (iii) Instrument
Transformer (iv) Isolation Transformer
Step up transformers converts the low voltage (LV) and high current from the primary side of
transformer to the high voltage (HV) and low current value on the secondary side of transformer.
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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
 1. Transformers
Step down transformers converts the high voltage (HV) and low current from the primary side of the
transformer to the low voltage (LV) and high current value on the secondary side of the transformer.
A three phase transformer is generally used in three phase system as it is more cost effective than single
phase transformers. However, when size matters, it is preferable to use a bank of three single phase
transformer vs a three-phase transformer, as it is easier to transport than one single three-phase
transformer unit.
In oil cooled transformers, the cooling medium is transformer oil. Whereas in the dry type transformer,
air cooling is used instead.
A two-winding transformer (conventional transformer) has a fixed number of turns. It has two separate
winding namely primary and secondary winding, which are electrically insulated from each other. An
autotransformer shares the same turns between input and output connections. The primary and
secondary winding are not electrically insulated. Two winding transformers are generally used where
ratio of high voltage and low voltage is greater than 2. It is cost effective to use auto transformer, where
the ratio between high voltage and low voltage is less than 2.
Power transformers are generally used in transmission network for stepping up or down the voltage
level. It operates mainly during high or peak loads and has maximum efficiency at or near full load.
Distribution transformer steps down the voltage for distribution purpose to domestic or commercial
users. It has good voltage regulation and operates 24 hrs a day with maximum efficiency at 50% of full
load.
Instrument transformers include current transformer (CT) and potential transformer (PT), which are
used to reduce high voltages and current to lesser values. These lesser values can be measured by
conventional instruments.
The isolation transformer is a type of transformer used to isolate a device or a circuit from the power
supply. It has a turns ratio of 1:1 meaning that the primary and secondary of an isolation transformer
contains an equal number of windings.
1.4 Construction of Transformer
A single-phase transformer basically consists of two parts (i) windings and (ii) core. There are two
windings wounded on the two limbs of core and are insulated from each other as well as from the core
as shown in figure. The windings are made of copper, so that they carry a very small resistance. The
windings connected to the supply is called primary winding and the winding connected to the load is
called secondary winding. The windings are electrically separated but magnetically linked to each other.
The core provides magnetic path for the flux, to get linked with the secondary winding. The core is made
of silicon steel sheets, which has a high relative permeability and low hysteresis losses. The vertical
portions of core are termed as limbs, and the top and bottom portions are called as yokes. The core is
laminated to reduce eddy current losses. For small transformers, each lamination is a single piece. For
large transformers, each lamination is made of two or more sections, which are joined together to form
a complete lamination. The joints of such laminations are staggered while forming the core, so that the
reluctance offered by these joints is minimum.
Other different parts and accessories are also fitted on the transformer for its efficient work as well as
for longer life and better services of the transformer. They are as follows: (i) Conservator (ii) Breather
(iii) Explosion Vent (iv) Radiator (v) Bushings

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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
1.5 Working Principle of Transformer

1.4.1 Core Type and Shell Type Transformer

According to the magnetic
circuit, the construction of
transformers are of two types (i)
core type (ii) shell type. They
are mainly differentiated from
each other by the manner in
which primary and secondary
windings are wounded around
the laminated core. The
comparison between these two
types of transformers are as follows:
Basis for Comparison Core Type Transformer Shell Type Transformer
Definition The winding surrounds the core The core surrounds the winding
Lamination Shape The lamination is cut in the form Lamination are cut in the form of the
of the L strips long strips of E and L
Copper Required More Less
Number of Limbs Two Three
Insulation Required More Less
Flux Equally distributed on the side Central limb carries whole flux and
limbs of the core side limbs carries half of the flux
Winding Primary and secondary winding Primary and secondary windings are
are placed on the side limbs placed on the central limb
Magnetic Circuit Two One
Losses More Less
Maintenance Easy Difficult
Mechanical Strength Low High
Output Less More
Natural Cooling Does not Exist Exist
1.5 Working Principle of Transformer
A transformer works on the principle of Faraday’s law of electromagnetic induction between two
magnetically coupled coils. When the primary winding is connected to an alternating voltage of RMS
value 𝑉1 volts, an alternating current 𝐼1 flows through the primary winding and sets up an alternating

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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
 1. Transformers
flux 𝜙 in the core. This flux links with the
primary winding as well as secondary
winding. The flux links with the primary
winding produces self induced EMF 𝑒1 and the
flux links with the secondary winding
produces mutually induced EMF 𝑒2 , which are
given by,
𝑑𝜙 𝑑𝜙
𝑒1 = −𝑁1 and 𝑒2 = −𝑁2
𝑑𝑡 𝑑𝑡
𝑒2 𝑁2
So, =
𝑒1 𝑁1
Where, 𝑁1 and 𝑁2 are the number of turns of primary and secondary windings respectively.
When a load is connected to the secondary winding, an alternating current 𝐼2 flows through the load. 𝑉2
is the terminal voltage across the load. Thus, the power is transferred from primary to secondary side by
means of electromagnetic induction.
Since the power transferred from primary winding to secondary is same. Thus,
𝐸2 𝐼1
𝐸1 𝐼1 = 𝐸2 𝐼2 ⇒ =
𝐸1 𝐼2
The direction of both induced EMFs 𝐸1 and 𝐸2 in the primary and secondary windings are such that
they oppose the primary applied voltage 𝑉1.

1.6 EMF Equation of a Transformer

Let 𝑁1 = Number of turns in primary
𝑁2 = No. of turns in secondary
𝜙𝑚 = Maximum core flux in webers
𝐵𝑚 = Maximum core flux density in webers/m2 or Tesla
𝐴 = Area of the core in m2
𝑓 = Frequency of AC input in Hz
When the primary winding is connected to an alternating sinusoidal voltage, the flux increases from its
zero value to maximum value 𝜙𝑚 in one quarter of the cycle i.e. in 𝑇⁄4 second as shown in figure.
Thus,
Average rate of change of flux
𝜙𝑚 − 0 4𝜙𝑚
= = = 4𝑓𝜙𝑚 Wb⁄sec or volt
𝑇 𝑇
− 0
4
Now, rate of change of flux per turn is the induced EMF in volts. So,
Average EMF per turn = 4𝑓𝜙𝑚 × 1 = 4𝑓𝜙𝑚 volt
If flux 𝜙 varies sinusoidally, then RMS value of induced EMF is obtained by multiplying the average
value with form factor.
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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
1.7 Losses in a Transformer
RMS value 0.707𝐼𝑚
Form Factor (𝐾𝑓 ) = = = 1.11
Average Value 0.637𝐼𝑚
Therefore, RMS value of EMF per turn = 1.11 × 4𝑓𝜙𝑚 = 4.44𝑓𝜙𝑚 volt
RMS value of the induced EMF in the primary winding = (induced EMF/turn) × No. of primary turns.
That is given by,
𝐸1 = 4.44𝑓𝜙𝑚 𝑁1 = 4.44𝑓𝐵𝑚 𝐴𝑁1 volt
Where, 𝜙𝑚 = 𝐵𝑚 𝐴.
Similarly, RMS value of the induced EMF in the secondary winding = (induced EMF/turn) × No. of
secondary turns. That is given by,
𝐸2 = 4.44𝑓𝜙𝑚 𝑁2 = 4.44𝑓𝐵𝑚 𝐴𝑁2 volt
Transformation Ratio (𝑲):
𝑁2 𝐸2 𝐼1
𝐾= = =
𝑁1 𝐸1 𝐼2
Turns Ratio (𝒏):
𝑁1 𝐸1 𝐼2
𝑛= = =
𝑁2 𝐸2 𝐼1
When the primary winding is connected to an alternating sinusoidal voltage (𝑒1 = 𝐸𝑚1 sin 𝜔𝑡), an
alternating current 𝐼1 flows through the primary winding and sets up an alternating flux 𝜙 in the core.
Thus, the equation for flux is given by,
𝜙 = 𝜙𝑚 sin 𝜔𝑡
The flux links with the primary winding produces self induced EMF 𝑒1 , which is given by,
𝑑𝜙 𝑑
𝑒1 = −𝑁1 = −𝑁1 (𝜙𝑚 sin 𝜔𝑡) = −𝜔𝑁1 𝜙𝑚 cos 𝜔𝑡
𝑑𝑡 𝑑𝑡
= −2𝜋𝑓𝑁1 𝜙𝑚 sin(𝜔𝑡 − 90°)
From above equation, we find that the induced EMF lags the flux by 90°.
The magnitude of maximum value of induced EMF in the primary winding is given by,
𝐸𝑚1 = 2𝜋𝑓𝑁1 𝜙𝑚
The RMS value of induced EMF in the primary winding is given by,
𝐸𝑚1 2𝜋𝑓𝑁1 𝜙𝑚
𝐸1 = = = 4.44𝑓𝜙𝑚 𝑁1 = 4.44𝑓𝐵𝑚 𝐴𝑁1 volt
√2 √2
Similarly, the RMS value of induced EMF in the secondary winding is given by,
𝐸𝑚2 2𝜋𝑓𝑁2 𝜙𝑚
𝐸2 = = = 4.44𝑓𝜙𝑚 𝑁2 = 4.44𝑓𝐵𝑚 𝐴𝑁2 volt
√2 √2
1.7 Losses in a Transformer

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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
 1. Transformers
As the transformer is a static device and does not contain any rotating parts, there are no mechanical
losses (like friction and windage losses). The losses that occur in a transformer are (i) Core Loss and (ii)
Copper Loss.
(i) Core Loss or Iron Loss (𝑷𝒄 ): Iron losses are caused by the alternating flux in the core of transformer.
As this loss occurs in the core, it is also known as Core Loss. Core loss is further divided into two types
(a) hysteresis loss and (b) eddy current loss.
(a) Hysteresis Loss (𝑷𝒉 ): Hysteresis loss is due to the reversal of magnetization in the transformer core.
Power is dissipated in the form of heat known as Hysteresis Loss and can be given by, Steinmetz
empirical formula as follows:
1.6
𝑃ℎ = 𝐾ℎ 𝐵𝑚 𝑓𝑉 watts
Where, 𝐾ℎ = a constant that depends on the volume and quality of the core material
𝐵𝑚 = maximum or peak value of the flux density within the core in Wb/m2
𝑓 = frequency of supply in Hz
𝑉 = volume of core in m3
To reduce the hysteresis loss, the high-grade core material made of CRGO (Cold rolled grain oriented)
silicon steel can be used.
(b) Eddy Current Loss (𝑷𝒆 ): In a transformer, AC current is supplied to the primary winding which
sets up alternating magnetizing flux. When this flux links with secondary winding, it produces induced
EMF in it. But some part of this flux also gets linked with other conducting parts like steel core or iron
body of the transformer, which will result in induced EMF in those parts, causing small circulating
current in them. This current is called as eddy current, which is not responsible for doing any useful
work. It produces a power loss in the form of heat known as Eddy Current Loss and can be given by,
Steinmetz empirical formula as follows:
2 2 2
𝑃𝑒 = 𝐾𝑒 𝐵𝑚 𝑓 𝑡 𝑉 watts
Where, 𝐾𝑒 = a constant that depends on the volume and quality of the core material
𝐵𝑚 = maximum or peak value of the flux density within the core in Wb/m2
𝑓 = frequency of supply in Hz
𝑡 = thickness of lamination in meter
𝑉 = volume of core in m3
The eddy current loss is minimized by making the core with thin laminations of high permeability
magnetic material such as silicon steel and they are insulated from each other by coating them with
varnish or oxide layer.
Hence, total Core loss = Eddy current loss + Hysteresis loss
1.6 2 2 2
𝑃𝑐 = 𝑃ℎ + 𝑃𝑒 = 𝐾ℎ 𝐵𝑚 𝑓𝑉 + 𝐾𝑒 𝐵𝑚 𝑓 𝑡 𝑉 watts
From the above equation, we find that the core loss in a transformer depends upon the maximum value
of flux density (𝐵𝑚 ) and frequency of supply (𝑓); as the other quantities like volume and quality of the
core are constant. If the applied voltage remains constant, 𝐵𝑚 and 𝑓 remains constant. Therefore, core
loss in the transformer is considered to be constant at all loads including no load.
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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
1.8 Efficiency of a Transformer
(ii) Copper Loss or Ohmic Loss (𝑷𝒄𝒖 ): Copper losses occur because of the Ohmic resistance in the
windings of the transformer. If the primary and secondary winding resistances are 𝑅1 and 𝑅2
respectively; and the current flowing in these windings are 𝐼1 and 𝐼2 respectively. Then, the copper losses
occurred in these windings are 𝐼12 𝑅1 and 𝐼22 𝑅2 respectively. So, the total copper loss will be given by,
𝑃𝑐𝑢 = 𝐼12 𝑅1 + 𝐼22 𝑅2 watts
When the load in a transformer changes, the currents 𝐼1 and 𝐼2 will change. Hence, the copper loss will
also vary as the square of currents 𝐼1 and 𝐼2 . So, copper losses in the transformer are variable losses,
which depends on the load.
Copper loss can be minimized by minimizing the resistance of windings. To minimize the resistance of
windings, thick wires with considerably low resistance are used.
1.8 Efficiency of a Transformer
As is the case with other types of electrical machines, the efficiency of a transformer at a particular load
and power factor is defined as the ratio of output power at secondary winding to the input power in
primary winding.
Output Power
Efficiency, 𝜂 =
Input Power
Input Power, 𝑃𝑖𝑛 = 𝑉1 𝐼1 cos Φ1 , Output Power, 𝑃𝑜𝑢𝑡 = 𝑉2 𝐼2 cos Φ2 at full load
Where, 𝑉1 and 𝑉2 are primary applied voltage and secondary terminal voltage respectively
𝐼1 and 𝐼2 are primary and secondary current respectively
cos Φ1 and cos Φ2 are power factor of primary and load respectively
𝑉2 𝐼2 = Volt-ampere capacity of the transformer
Input Power = Output Power + Losses
Losses = Core Loss + Copper Loss = 𝑃𝑐 + 𝑃𝑐𝑢
Where, 𝑃𝑐 = Iron loss or Core loss and 𝑃𝑐𝑢 = Copper loss at Full Load
Output Power 𝑃𝑜𝑢𝑡
𝜂= =
Output Power + Losses 𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑃𝑐𝑢
Let 𝑥 is the percentage of full load then
Output Power = 𝑥𝑃𝑜𝑢𝑡 at 𝑥% of full load
Copper loss = 𝑥 2 𝑃𝑐𝑢 at 𝑥% of full load
𝑥𝑃𝑜𝑢𝑡 𝑥𝑉2 𝐼2 cos Φ2
𝜂𝑥 = =
𝑥𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑥 𝑃𝑐𝑢 𝑥𝑉2 𝐼2 cos Φ2 + 𝑃𝑐 + 𝑥 2 𝑃𝑐𝑢
2

Condition for Maximum Efficiency: The efficiency of transformer is maximum when

𝑑𝜂𝑥 𝑑 𝑥𝑃𝑜𝑢𝑡
=0⇒ { }=0
𝑑𝑥 𝑑𝑥 𝑥𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑥 2 𝑃𝑐𝑢
(𝑥𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑥 2 𝑃𝑐𝑢 )𝑃𝑜𝑢𝑡 − 𝑥𝑃𝑜𝑢𝑡 (𝑃𝑜𝑢𝑡 + 2𝑥𝑃𝑐𝑢 )
⇒ =0
(𝑥𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑥 2 𝑃𝑐𝑢 )2

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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE
 1. Transformers
⇒ (𝑥𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑥 2 𝑃𝑐𝑢 )𝑃𝑜𝑢𝑡 = 𝑥𝑃𝑜𝑢𝑡 (𝑃𝑜𝑢𝑡 + 2𝑥𝑃𝑐𝑢 )
⇒ 𝑥 2 𝑃𝑐𝑢 = 𝑃𝑐
That is, Copper loss = Core loss or Variable loss = Constant loss
Therefore, the maximum efficiency is given by,
𝑃𝑐
√
𝑃𝑐𝑢 𝑉2 𝐼2 cos Φ2
𝜂𝑚𝑎𝑥 =
𝑃𝑐
√
𝑃𝑐𝑢 𝑉2 𝐼2 cos Φ2 + 2𝑃𝑐
Example: A 250 KVA single phase 11000/415 V, 50 Hz transformer has 80 turns on the secondary.
Determine the (a) number of primary turns (b) maximum value of core flux (c) rated primary and
secondary currents (d) EMF induced per turn. [Ans: 𝑁1 = 2120.48, 𝜙𝑚 = 0.02336 Wb, 𝐼1 = 22.727 A,
𝐼2 = 602.41 A, 𝐸 ⁄𝑡𝑢𝑟𝑛 = 5.1875 volt]
Example: In a 25 KVA, 2000/200 V single phase transformer, the core and copper losses at full load
are 350 W and 400 W respectively. Determine its efficiency (a) at full load with UPF (b) at half load
with UPF (c) maximum efficiency at UPF (d) copper loss at maximum efficiency (e) load at which
efficiency is maximum. [Ans: 𝜂𝑓𝑙 = 97.08 %, 𝜂ℎ𝑙 = 96.52 %, 𝜂𝑚𝑎𝑥 = 97.09 %, 𝑃𝑐𝑢 (at 𝜂𝑚𝑎𝑥 ) = 350
W, load (at 𝜂𝑚𝑎𝑥 ) = 23.385 KVA]
Example: A single phase transformer has 400 primary and 1000 secondary turns. The net cross-
sectional area of the core is 60 cm2. The primary winding is connected to 500 V, 50 Hz supply. Find (a)
peak value of core flux density (b) EMF induced in the secondary winding. [Ans: 𝐵𝑚 = 0.938 Wb/m2,
𝐸2 = 1250 V]
Example: A 10 KVA single phase transformer has a maximum efficiency of 98 % at full load with 0.8
power factor. Determine (i) core loss (ii) copper loss at full load and (iii) copper loss at half load. [Ans:
𝑃𝑐 = 81.63 W, 𝑃𝑐𝑢 (at full load) = 81.63 W, 𝑃𝑐𝑢 (at half load) = 20.4075 W]
Example: A 600 KVA transformer has an efficiency of 92 % (a) at full load with UPF and (b) at half
load with 0.90 power factor. Determine its efficiency at 75 % of the full load with 0.90 power factor.
Also determine iron loss and copper loss at full load. [Ans: 𝜂𝑓𝑙 = 91.95 %, 𝑃𝑐 = 13913.04 W, 𝑃𝑐𝑢 =
38260.84 W]
Example: In a 100 KVA single phase transformer, the core loss is 1.2 kW and half load copper loss is
500 W. If the load power factor is 0.8 lagging, find the efficiency at (a) half load (b) full load. [Ans:
𝜂ℎ𝑙 = 95.92 %, 𝜂𝑓𝑙 = 96.15 %]
Example: A 11000/230 V, 150 kVA, single-phase, 50-Hz transformer has core loss of 1.4 kW and
copper loss of 0.9 kW at 3/4th load. Determine the (i) kVA load for maximum efficiency (ii) value of
maximum efficiency at unity power factor (iii) efficiency at half load with 0.8 power factor. [Ans: (i)
140.3 kVA (ii) 98.04% (iii) 97%]

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Dr. Dipesh Kumar, Asst. Prof., EEE, BMSCE