1. LAB REPORT: 01 OPEN CIRCUIT AND SHORT
CIRCUIT TESTS
OF A SINGLE-PHASE TRANSFORMER
ECX 3232 ELECTRICAL POWER
Q. NO MARKS
NAME : M. S. D. PERERA.
REGNO : 311089590
CENTER : COLOMBO. TOTAL
DATE OF SUBMISSION: 07/11/2012 %

2. Experiment 1: Open circuit and short circuit tests of a single-phase
transformer.
Apparatus:
 500VA, 230V/230V 1:1 transformer.
 (0-250V,AC) voltmeter.
 (0-150V,AC) voltmeter.
 (0-1A,AC) Ammeter.
 Wattmeter.
 230/(0-250V) variac.
 Leads.
Theory
Working with a ideal transformer is easy. But life get complicated when it
comes to theoretical electrical engineering to real world electrical
engineering. Ideal model no longer useful in industrial electrical
engineering applications. So we have to come up with a model that represent
an normal non-ideal industry transformer.
The real transformer have following things to be included and modeled when in
when we drawing it’s equivalent circuit.
 Losses,
There are two main losses related to a power transformer. And they are,
* Core Losses
* Hysteresis loss.
* Eddy current loss.
* Copper loss.
 Leakage flux.
 No load flux(also known as magnetizing flux).
In an equivalent circuit we represent core losses as a parallel resistor
because it’s proportional to the number of turns in the winding. And
magnetizing flux could also represent as a parallel component as well as it’s
also proportional to the number of windings.
We represent copper loss as a serial resistive component , because it’s just
equal to a pure resistor passing current through it and disparaging energy.
And also we represent leakage flux as also a serial inductive component, we
could imagine it as a series inductor outside the transformer which is
blocking some potential difference across it so it will reduce the gross
potential difference among ideal transformer terminals.
Bellow figure depicts this model diagrammatically.
Since these two windings are magnetically coupled, we could get it’s thevean
equivalent circuit as we seen from the primary. (This could be done to the
secondary too).

3. Bellow figure depicts how we see it from primary side.
In the case of transforming secondary side to primary side we have to
multiply each inductive/resistive component by square of turns ration. Which
means,
When you transforming primary into the secondary side, you have to divide it
by square of turns ratio,
EXPREMENT:
PROCEDURE:
Part A: Open Circuit Test.
(a) The voltage ratings of the transformer is,
500VA, 230/230V 1:1 transformer.
So KVA rating is ½ KVA.
(b) Rated voltages,

4. (c) :
This is open circuit test. We log no load current and iron while changing the
voltage through variac device.
This is the data we have collected.
Impressed No Load Current Iron Loss(W)
Voltage(V) (I/A)
230 0.6115 15
180 0.224 12
160 0.195 10
140 0.116 8
120 0.142 6
100 0.121 5
80 0.101 4
60 0.081 3
40 0.060 2
Calculations:
Since there are no power desperation on the secondary side we have only
power desperations on the primary side. They are sum of copper loss+ core
loss. But in here, since we have very little current flowing through primary
winding, we could ignore copper loss and assuming that reading in the
wattmeter is equal to core losses. So through that we could find two
variables.

5. Graphs and Characteristics:
Part B: Short Circuit Test
Now we are going to short circuit the secondary side. We need to take caution
here, because there is a potential to burn the fuses in the learning panel if
we won’t be careful. So we keep the variac device at it’s lower position and
powering up the switches. Here we are getting a one reading only. It’s at
wattmeter and ammeter readings while variac kept at 9V.
We use such a very small (9V) potential thus because this is a short circuit
test and we are not supposing to burn that expensive learning panels.
Theory:

6. In here there are no power desperation on the secondary side. And just
because Rc and Xm are very large values, we could ignore them. So it’s safe
to assume that all the power desperation is now equal to the copper loss.
So,
Observations,
9Ammeter Voltmeter Wattmeter
2.17 9 32.5
So we get,
Discussion:
Why HV side is open circuited and LV side is short-circuited when performing
the practical?
Well as it term derives it’s meaning that HV side will generate high
voltages. So that will ramp up the short circuit current to a very large
value just because there are no any resistance to limit current flow on
secondary side. So that’s why we need to use LV side to be short circuited as
well as we should use very low voltage ( like 9V in our experiment ) to avoid
damaging or frying transformers, fuses or breakers.
Experiment 2 : Load Test Of Transformers
Apparatus
1. 500VA, 230V/230V single phase transformer
2. 0-250V,AC voltmeter
3. 0-5A,AC ammeter
4. Wattmeter
5. 230/(0-250V) variac
6. Resistor bank
7. Capacitor bank
8. Leads
Theory
Voltage regulation is a principle to keep voltage value independent of the
load. When it comes to voltage regulation we have to consider bellow facts
into consideration.
 Different loads will take different currents at same voltage.
 Different loads will have different leading/lagging reactive
components.

7.  A load may vary how much it will draw dynamically, take a washing
machine for a example, When it washing clothes it will have drive
motors and there will be a lagging current component and when it
switched to drying clothes it will turn it’s motors off and turn on
it’s heaters which will dynamically change gross load inductive load to
a resistive load.
Above facts are making voltage regulation a difficult subject. So it’s not
possible to get a ideal constant voltage, it will vary at least by a fraction
of a million when it’s load current changes.
By the way, we should have some standard index to measure how much bad or
good a particular device could regulate against varying load currents.
In transformers we use ,
And phasor diagram of a transformer when loaded with power factor load.
And the efficacy of the transformer is given by,
The first experiment is about resistive loads, so we could use as zero.
Observations And Calculations:

8. Graphs:
For a capacitive load
Here we can’t assume that power factor is 1, we have to calculate it.
Since ,

9. Graphs: