Basic Electrical Engineering lab (BTEE-102-18)
CHANDIGARH ENGINEERING COLLEGE,
LANDRAN, MOHALI
DEPARTMENT OF APPLIED SCIENCES
EXPERIMENT NO. 10
OBJECTIVE:
To perform open and short circuit tests on a single phase transformer and calculate its
efficiency.
APPARATUS:
Open circuit/ short circuit test bench, Variac, Watt meter (qty 1), AC voltmeter (qty 2),
AC ammeter (qty 1), DPST Switch (qty 1) and connecting leads
DESCRIPTION:
A transformer is an AC machine that transfers electrical energy from one electrical
circuit to another. In the simplest version it consists of two windings wrapped around a
core: windings are coupled by the magnetic field and are note electrically connected.
There are mainly two kinds of losses in the transformer, the core and the ohmic losses.
Some of the power in the transformer that would ideally be transferred to the output is
lost in the core, resulting in heat and sometimes noise. These
lossesarecalledCoreLosses.Thecopperwindingsoftheprimaryandsecondaryofthetransfor
mer are (obviously) conductors, so some energy will be dissipated in them and these
losses are called OhmicLosses.
Open Circuit Test (no load test): The open circuit test is performed at the rated
voltage and the rated frequency. In this case one of the windings, usually secondary
winding, is kept open circuited and the rated voltage is applied to the primary winding
(figure 1). Under this condition, all the input current flows through the excitation
branch, and so, essentially, all the input voltage is dropped across the excitation
branch. With secondary open, the secondary current referred to the primary side is zero
as secondary current is zero. Full line voltage is applied to the primary of the
transformer and the input power to the transformer is measured. The primary loss
component is the core losses. The no-,in this test; core loss is found by subtracting the
ohmic loss in the primary, which is usually very small and may be neglected. Since the
no load current is less than 1% of the nominal current and hence the loss and drop that
takes place in the primary winding is negligible.
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Basic Electrical Engineering lab (BTEE-102-18)
CHANDIGARH ENGINEERING COLLEGE,
LANDRAN, MOHALI
DEPARTMENT OF APPLIED SCIENCES
Iron losses or no load losses = Input Power on no load = Wattmeter reading WI
Short Circuit Test (full load test): The short circuit test is performed at the rated
current. The purpose of this test is to determine full load copper losses. In this case the
secondary winding of the transformer is short circuited by thick wire or strip.
Refer figure 2. A variable low voltage is applied to the primary winding through an auto
transformer, the primary voltage is increased till the ammeter indicates the full load
rated current (IF). The supply
Voltage required to circulate the rated current is usually very small and is order of few
percent of the rated voltage (generally 5% to 10% of the rated value). Thus the induced
voltage on the secondary is small compared to the rated voltage. This leads to a low
core flux component and as such low core losses. The losses in this case are thus copper
losses. Copper losses or full load losses = Watt meter reading WSC. and IF = (KVA Rating) /
(Rated Voltage V)
Figure 2: Short Circuit Test
Transformer Efficiency: The efficiency of a transformer is typically in the range 95%
to 98%. This implies that the transformer losses are as low as 2% to 5% of the input
power. In calculating the efficiency, it is generally much better to determine the
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Basic Electrical Engineering lab (BTEE-102-18)
CHANDIGARH ENGINEERING COLLEGE,
LANDRAN, MOHALI
DEPARTMENT OF APPLIED SCIENCES
transformer losses rather than measuring the input and output powers directly. The
efficiency (h) of the transformer is defined as-
PROCEDURE:
1. Make the connection as per figure 1 to perform open circuit test(OCT).
2. Switch on the power supply. Apply the rated voltage (100%)to the transformer.
3. Note down wattmeter reading (WI), V1 and V2 for 40%, 60%, 80% and 120%
of the rated voltage value.
4. Switch off the power supply after recording the readings.
5. Make the connection as per figure 2 to perform short circuit test (SCT).
6. Adjust the autotransformer to minimum output voltage.
7. Increase the voltage fed to the primary of the transformer in steps so that
ammeter shows that rated current (full load rated current) is circulated in the
transformer. Allow a little for the system to stabilize and then record wattmeter
reading(WSC),short circuit current(ISC) and the short circuit voltage
8. Also take readings (WSC, ISC and VSC) for 40%, 60%, 80% and 120% of the
rated current value.
9. Switch off the supply after recording the readings.
10. Calculate the efficiency of transformer at full load.
OBSERVATIONS and CALCULATIONS:
Rated Voltage V = 220V and Rated Current I = 9.10A of the Transformer.
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Basic Electrical Engineering lab (BTEE-102-18)
CHANDIGARH ENGINEERING COLLEGE,
LANDRAN, MOHALI
DEPARTMENT OF APPLIED SCIENCES
Open Circuit Test Short Circuit Test
Wattmete Voltmete Voltmete Wattmete Voltmeter Ammete Efficiency
Rated r Reading r r r Reading Reading r ()
value WI Reading Reading WSC VSC Reading
V1 V2 ISC
40% 15 88 88 20 5 3.58 90.6%
60% 20 132 132 43 8 5.44 91.6%
80% 32 176 176 78 12 7.25 91.2%
100% 45 220 220 120 15 9.18 90.7%
120% 50 227 227 170 18 10.84 89.9%
RESULT: Open Circuit and short Circuit test has been studied.
PRECAUTIONS:
1. All connection should be neat and tight.
2. While connecting the wattmeter make sure that the voltage coil must be
connected across
thesupplyandthecurrentcoilshouldbeconnectedinserieswiththetransformerwindi
ng.
3. Checktheconnectionbeforeswitchingonthecircuitandtakeallthereadingsproperly
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Basic Electrical Engineering lab (BTEE-102-18)
CHANDIGARH ENGINEERING COLLEGE,
LANDRAN, MOHALI
DEPARTMENT OF APPLIED SCIENCES
Q.1 What is the operating principle of a Transformer?
Ans A transformer works based on the principle of mutual induction. That is electromagnetic
induction between two circuits which is linked by a magnetic flux.
Q.2 What are the two types of transformer based on the manner in which the primary
and secondary coils are placed on the core?
Ans Core type transformer.
Shell type transformer.
2. What are the few main differences between the shell type and core type transformer?
In core-type transformer, the core is surrounded by the winding. Cores used are L-L
lamination. It has only one magnetic flux path. Flux leakage is high.
In shell-type transformer, the winding is surrounded by the core. E-I lamination cores are
used. It has two magnetic flux path. Low flux leakage.
Q.3What is the transformation ratio of a transformer?
Transformation Ratio K is defined as the ratio of the EMF in the secondary winding to that in
the primary winding.
Ep or VP – Primary voltage
Es or VS – Secondary voltage
IP – Primary current
IS – Secondary current
Transformation ratio, K = VS/VP = IP/IS
Q.4What are the two types of power losses in a transformer?
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Basic Electrical Engineering lab (BTEE-102-18)
CHANDIGARH ENGINEERING COLLEGE,
LANDRAN, MOHALI
DEPARTMENT OF APPLIED SCIENCES
Copper loss and Iron loss or core loss.
Q.5 What is a copper loss?
Copper loss is due to the power dissipation as heat in the primary and secondary windings
due to the winding resistance.
Q.6What is Iron loss?
Iron loss or core loss occurs in the core of the transformer due to alternating flux in the core.
It consists of eddy current loss and hysteresis loss which depends upon the core material,
supply frequency, and flux density.
Q.7What is hysteresis loss?
Hysteresis loss is caused due to the magnetization and demagnetization in the transformer
core. During the reversal of the magnetic field, some amount of energy is lost in the form of
heat in the core, called hysteresis loss.
Q.8What is eddy current loss?
The magnetic flux produces an induced EMF in the transformer core which forms small
circulating currents. Some amount of energy will be dissipated in the form of heat due to
these current loops and it is called an eddy current loss.
Q.9What is the method used to reduce the eddy current loss?
To reduce eddy current loss the core resistance needs to be increased to reduce the amount of
circulating current. So in order to increase core resistance, the transformer core will be made
of laminated thin sheets of steel.
Q.10What are the different methods used for transformer cooling?
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Basic Electrical Engineering lab (BTEE-102-18)
CHANDIGARH ENGINEERING COLLEGE,
LANDRAN, MOHALI
DEPARTMENT OF APPLIED SCIENCES
Air Cooling – Air cooling methods are used for dry-type transformers, it can be either air
natural or forced. Natural Air cooling where the air carries the heat produced by the
transformer. A forced-air also is used for better heat transfer.
Oil cooling – In an oil-cooled transformer, the primary and secondary windings of the
transformer are immersed in oil. Oil transfers the heat from the winding to surface or fins and
it also provides electrical insulation between windings. Both air and oil could be a natural or
forced type.
Oil-immersed Water Cooling – In this method water is used to dissipate heat from the oil
through heat exchangers.
Q.11Why power transformers windings are immersed in oil?
The main function of transformer oil is to provide insulation, suppress arcing, serve as a
coolant.
Q.12What is an autotransformer?
It is a type of transformer with a single winding on a core. That is the part of a winding is
shared common to both primary and secondary winding. The primary voltage will be
connected across the two end terminals of the total winding. And the secondary winding is
actually connected across a point tapped between the single winding and one of the end
terminal which is connected to the neutral. This is the case of a step-down autotransformer,
for a step up just the opposite the primary would be across the tapped side.
