PRACTICAL WORK ON THREE PHASE
VOLTAGE TRANSFORMER (PW1)
is based on the principles of a 3 phase grid power supply. In this experiment
we implement the method to nonlinear loads. Here we study about the relation
RMS Value of Voltage
RMS Value of Current
Relation between Phase Voltage (V) and Current (I)
Of The Experiment:
To calculate the current fluctuation in the case of Non-Linear loads.
of the Transformer:
Phase Current, to calculate the Power
P = Ö3 U I cos q
Where q = 0
constant voltage 250V the primary current In1 = 5.33A.
U = ?3 *2 * 63
= 218.23 V, where U = 63 is the voltage in secondary windings because it is a
step down transformer.
= ?3 *2 * 63
* In2 and In2 = 10.584 A, the current in the secondary
electrical instrument which increases and decreases the voltage of an
alternating current without making any changes in its frequency by means of
Types of transformer:
· Step-Up Transformer:
which converts low voltage into high voltage.
· Step-Down Transformer
which converts high voltage into low voltage.
Principle of Transformer:
consists of two isolated coils which works on the Faradays principle of mutual
induction, nothing but the electromagnetic induction.
Working of transformer:
A current varies in one coil of the transformer which
produces a varying magnetic field. This in turn induces a varying electromotive force (emf) in the secondary coil. Power is
transferred between the two coils through the magnetic field.
change of voltage between the primary and secondary coils depends on the ratio
of turns of the two coils. To increase the flux linkage, an
iron core provides a low reluctance path for the magnetic flux.
Parts of Transformers:
Ø Laminated core
Ø Insulating materials
Ø Tap changer
Ø Oil Conservator
Ø Cooling tubes
Ø Buchholz Relay
Ø Explosion vent
Types of losses in Transformers:
Core losses :
Eddy current and Hysteresis losses
Copper losses :
In the resistance of windings
Eddy Current Losses:
Eddy currents generate
resistive losses that
transform some forms of energy such as kinetic energy into heat.
The energy lost as heat is known as the hysteresis
loss. In reversing the magnetization of the material is proportional to the
area of the hysteresis loop.
the various parameters of an electrical power system.
Make connections according to the
circuit diagram given and checked it by the professor before switching it ON.
Make sure all the equipment’s are earth properly
and wear shoes throughout the laboratory time.
Liquid items should not be kept
near the equipment’s.
The control knob in the
electrodynamometer should be used gently and increased gradually to avoid
sudden rise in the current.
Do not cross the prescribed
maximum value, which cause damage to the transformer.
Once the data acquisition is
completed, switch off the main supply to avoid over heating of the instruments.
of a Transformer:
The combination of batteries and resistances with this
two terminals can be replaced using a single voltage and a single Resistor
= Rs + jLS
The value of the single Voltage source “Eth”
is the open circuit voltage at the terminals and the value of “Zth”
is “Eth” divided by the current with the terminals short circuited.
of resistance- Resistor
of reactance- Inductor
Zth= Rs + jLs?
3 x 400 V.
diagram of the circuit:
Star connection is processed with primary set up
and other connections are connected to the primary to the power analyser
through the clamps with different colours. 6 rounds are made and clamped with
the respective arrow mark in the clamp signifies the outlet of the wire to be
drawn from the clamps. Now we can measure a significant value of current.
Now in the secondary circuit we connect the two
loops from one end to another end to form the secondary coil and at the bottom
it will replicate the star connection which is similar to the connection made
in the primary coil. Finally, the voltmeter is connected to the secondary
circuit as well.
Phase Transformer in Star Connection:
11. Power Analyser Connection:
Experimental Circuit Connection:
i) Current Calculations
From the power analyser we can see that
= 2.2 A,
= 1.7 A and
= 2.3 A
To find the RMS current through each wire the
current has to be divided by 6 because we have used 6 turns of wire inside the
So using the RMS values from the above graph, we get
= 2.2/6 = 0.36666 A
= 1.7/6 = 0.28333 A (central column current)
= 2.3 /6 = 0.38333 A
All these currents are not equal because Iv2 magnetic field lines
acts as an intermediate and has more losses than Iv1 and Iv3.
Comparison between the primary no-load current with rated value of the nominal
current we can see that the primary no-load current is 0.36666 A which is less
than 10% of the nominal current which is 5.33 A.
We can use the primary coil of a transformer to get
an inductance, since the currents are not equal in their phases with each
. ii) Voltage Calculations
a) Compound Voltage
The compound voltage U1=
402.7 V, U2=403.0 V and U3
= 403.3 V are seems to be balanced and equal.
compound voltage are all 120•
out of phase to each other.
The Normal Voltage
V1 = 232.4 V,
= 231.9 V,
V3 = 232.9 V
are also balanced and equal.
The Normal voltage are all 120• out of phase to each
iii) Relation Between the voltage and current for
The current and voltage are +059°,+073° and +080° out of
phase with each other for Line 1, line 2 and Line 3.
From this experiment we observed the working of a
3-Phase transformer and the nature of the non-linear load shows the non-linear
characteristics of the current.