1. A transformer transfers electrical energy between two stationary circuits through electromagnetic induction. It consists of two coils linked by a common magnetic core and operates without change in frequency. 2. An ideal transformer has negligible winding resistance and infinite core permeability, with no leakage flux or losses. A practical transformer model accounts for non-idealities like finite permeability and winding resistance. 3. Transformer tests determine losses and parameters for an equivalent circuit model. The open-circuit test measures core losses, while the short-circuit test measures copper losses at full load. Transformer regulation is the change in output voltage from no-load to full-load.

1. 1 –PHASE TRANSFORMER
SUBMITTED BY : AKSHAY KUMAR SUBMITTED TO : NITIN SUNDRIYAL

2. Content:-
• Introduction to transformer and its types
• Ideal transformer
• Practical transformer
• Short and open circuit test
• Voltage regulation
• Losses in transformer

3. INTRODUCTION
A transformer is a static device which
consist of two or more stationary
electrical circuit interlinked by a common
magnetic circuit for the purpose of
transferring electrical energy between
them. Without a change in frequency and
at a rated KVA .

4. Types of transformers :
Step up transformers – located at output of a
generator to step up the voltage level to
transmit the power. in which the primary
turns is less and secondary turns is more.
Step down transformers – Located at main
distribution or secondary level
transmission substations to lower the
voltage levels for distribution .in which the
primary turns is more and secondary turns
is less.

5. TWO WINDING TRANSFORMER
CONNECTION
•SINGLE PHASE

6. A) Core type B) Shell type
There are two types of transformer:-
i) Core form.
ii) Shell form

7. IDEAL TRANSFORMER
(SINGLE PHASE)
An ideal transformer is a imaginary transformer which
has following property:
1) The secondary and primary winding resistance are negligible.
2) The core has infinite permeablity.
3) The leakage flux and leakage inductance is zero.
4) There are no losses in ideal transformer like copper loss,
hystersis loss and eddy current loss.

8. Transformers at no load
The no load current If is needed to supply the no load
losses and to magnetize the transformer core.
fIu
Iw
v1
If
E1
IF
IF
IwIu
fo

9. for a no load transformer ,we have
f = fmsinwt
E1 = E1msinwt
E2 = E2msin(wt – 90)
Iw = Iocosfo
Iu = Iosinfo

10. Practical Transformer:-
An ideal transformer is useful in understanding the working
of a transformer. But it cannot be used for the computation
of the performance of a practical transformer due to the
non-ideal nature of the practical transformer. In a working
transformer the performance aspects like magnetizing
current, losses, voltage regulation, efficiency etc are
important. Hence the effects of the non-idealization like
finite permeability, saturation, hysteresis and winding
resistances have to be added to an ideal transformer to
make it a practical transformer.

11. REFFERAL VALUES
This is done here by replacing the secondary by a ‘hypothetical’
secondary having T1 turns which is ‘equivalent ’to the physical
secondary. The equivalence implies that the ampere turns, active and
reactive power associated with both the circuits must be the same.
Then there is no change as far as their effect on the primary is
considered.
Thus
V′2 = aV2, I′2 =I2a, r′2 = a2r2, x′l2 = a2xl2 , Z′L = a2ZL.
where a -turns ratio T1/T2

12. EQUIVALENT CIRCUIT OF A TRANSFORMER

13. EXACT EQUIVALENT CIRCUIT OF
TRANSFORMER REFFERED TO PRIMARY

14. APPROXIMATE EQUIVALENT CIRCUIT OF
THE TRANSFORMER REFFERED TO THE
PRIMARY

15. PHASOR OF TRANSFORMER AT FULL
LOAD

16. Open-circuit Test
• A voltmeter, wattmeter, and an ammeter are connected in
LV side of the transformer as shown in the figure below.

17. Equivalent Circuit

18. • The voltage at rated frequency is applied to that LV side
with the help of a variac of variable transformer.
• The HV side of the transformer is kept open. Now with help
of variac applied voltage is slowly increase until the
voltmeter gives reading equal to the rated voltage of the LV
side.
• After reaching at rated voltage, all three instruments reading
(Voltmeter, Ammeter and Wattmeter readings) are recorded.
• The ammeter reading gives the no load current I0 .
• As no load current I0 is quite small compared to rated
current of the transformer, the voltage drops due to this
electric current then can be taken as negligible.

19. • Since, voltmeter reading V can be considered equal to
secondary induced voltage of the transformer. The input
power during test is indicated by watt-meter reading.
• As the transformer is open circuited, there is no output
hence the input power here consists of core losses in
transformer and copper loss in transformer during no load
condition.
• The no load current in the transformer is quite small
compared to full load current so copper loss due to the
small no load current can be neglected.
• Hence the wattmeter reading can be taken as equal to core
losses in transformer.
• Therefore it is seen that the open circuit test on transformer
is used to determine core losses in transformer and
parameters of shunt branch of the equivalent circuit of
transformer.

20. Calculation

21. Short-circuit Test
• A voltmeter, wattmeter, and an ammeter are connected in
HV side of the transformer as shown in figure.

22. Equivalent Circuit

23. • The voltage at rated frequency is applied to that HV side
with the help of a variac of variable transformer
• The LV side of the transformer is short circuited . Now with
help of variac applied voltage is slowly increase until the
ammeter gives reading equal to the rated current of the HV
side
• After reaching at rated current of HV side, all three
instruments reading (Voltmeter, Ammeter and Watt-meter
readings) are recorded
• The ammeter reading gives the primary equivalent of full
load current IL .
• As the voltage, applied for full load current in short circuit
test on transformer, is quite small compared to rated primary
voltage of the transformer, the core losses in transformer
can be taken as negligible here.

24. • Let’s, voltmeter reading is VSC . The input power during test
is indicated by watt-meter reading.
• As the transformer is short circuited, there is no output
hence the input power here consists of copper losses in
transformer
• Since, the applied voltage VSC is short circuit voltage in the
transformer and hence it is quite small compared to rated
voltage so core loss due to the small applied voltage can be
neglected.
• Hence the wattmeter reading can be taken as equal to copper
losses in transformer.
• Therefore it is seen that the Short Circuit test on transformer
is used to determine copper loss in transformer at full load
and parameters of approximate equivalent circuit of
transformer.

25. Transformer Voltage Regulation
   
 
   
 
100
100







=
loadfullV
loadfullVloadnoV
loadfullV
loadfullVloadnoV
gulationReVoltage%
p
pp
s
ss
Because a real transformer has series impedance within
it, the output voltage of a transformer varies with the
load even if the input voltage remains constant. The
voltage regulation of a transformer is the change in the
magnitude of the secondary terminal voltage from no-
load to full-load.
Referred to the primary side

26. Transformer losses
• The transformer losses are divided into electrical losses (copper
losses) and Magnetic losses (Iron losses).
• Copper losses in both the primary and secondary windings.
• Magnetic losses, these losses are divided into eddy current losses
and hysteresis losses.
2
2
21
2
1 RIRI 
mhysteriseseddyi IVPPP 1=

27. Transformer Core losses
Eddy currents arise because of changing flux in core.
Eddy currents are reduced by laminating the core
Hysteresis losses are proportional to area of BH curve
and the frequency
These losses are reduced
by using material with a
“thin” BH curve
27

28. Copper loss
• Ideally copper loss is zero but practically
it is not possible however the winding is
made up of copper and it have some
resistance which is responsible for heat
loss or copper loss when the current is
flow through the winding