This will help for transformer working principle

1. Chapter two
Transformer

2. Introduction
Transformer is a device that:
1. transfers electric power from one circuit to an other circuit.
2. It does so with out a change of frequency.
3. It accomplishes this by electromagnetic induction.

3. How The Transformer Works
A clear understanding of how transformers work is necessary in order
to wire them properly in an electrical system.
The electrical current flowing in a transformer is alternating current.
The current flows first in one direction, stops, then reverses and flows
in the other direction. The magnetic field around the winding is
constantly in motion.
Another property of electricity is important to the operation of a
transformer.
When a magnetic field moves across a wire, a voltage is induced into
the wire, If the wire forms a complete circuit, current will flow in the
wire.

4. Con’d
If a second coil of wire is placed in a moving magnetic field, then a
voltage will be induced in this second coil. This phenomenon is called
mutual induction.
Alternating current in one winding produces a moving magnetic field
that induces a voltage in a second winding.

5. How transformer is working

6. Turn ratio
It is important to know the ratio of the number of turns of wire on the
primary winding as compared to the secondary winding.
This is called the turns ratio of the transformer. The actual number of
turns is not important, just the turns ratio.

7. Schematic diagrams of step-down and step-up transformers

8. Ideal transformer

9. Con’d

10. Con’d

11. Exercise 1
A 25KVA, 7200 -240/120 center-tap single phase transformer operates
at rated voltage. It supplies a single phase load that has an equivalent
impedance of 7.2<+36.9˚ ohms. Assume ideal operation.
Find
a. turn ratio
b. secondary current
c. primary current
d. load Z as seen from primary side

12. Exercise 2
A 300KVA, 2400 - 120 60 Hz single phase transformer operates at
2300 volt on the primary side. It supplies 115KVA to a load that has a
power factor of 0.723 lagging. Assume ideal operation.
Find:
a. Secondary voltage at operating voltage
b. Secondary current
c. Impedance of the load as seen on the secondary side.
d. Impedance of the load as seen from the primary side.

13. Transformer Losses
Losses of transformer are divided mainly into two types:
1. Iron Loss
2. Copper Losses

14. Iron Loss
This is the power loss that occurs in the iron part.
This loss is due to the alternating frequency of the emf.
Iron loss in further classified into two other losses.
a) Eddy current loss
b) Hysteresis loss

15. Eddy current Loss
This power loss is due to the alternating flux linking the core, which
will induced an emf in the core called the eddy emf, due to which a
current called the eddy current is being circulated in the core.
 As there is some resistance in the core with this eddy current
circulation converts into heat called the eddy current power loss.
Eddy current loss is proportional to the square of the supply
frequency.

16. Hysteresis Loss
This is the loss in the iron core, due to the magnetic reversal of the
flux in the core, which results in the form of heat in the core.
This loss is directly proportional to the supply frequency.
Eddy current loss can be minimized by using the core made of thin
sheets of silicon steel material, and each lamination is coated with
varnish insulation to suppress the path of the eddy currents.
Hysteresis loss can be minimized by using the core material having
high permeability.

17. Copper Loss
This is the power loss that occurs in the primary and secondary coils
when the transformer is on load.
This power is wasted in the form of heat due to the resistance of the
coils.
This loss is proportional to the sequence of the load hence it is called
the Variable loss where as the Iron loss is called as the Constant loss as
the supply voltage and frequency are constants.

18. Transformer Efficiency

19. Con’d

20. Condition For Maximum Efficiency
In general for the efficiency to be maximum for any device the losses
must be minimum.
Between the iron and copper losses the iron loss is the fixed loss and the
copper loss is the variable loss.
When these two losses are equal and also minimum the efficiency will
be maximum.
Therefore the condition for maximum efficiency in a transformer is:
Copper loss = Iron loss

21. Exercise 3
 2 KVA transformer has an iron loss of 100 watt and a fill load copper
loss of 200 watt. What will be the maximum efficiency at unity power
factor?

22. End!
Any Question ?