BOTSWANA INTERNATIONAL UNIVERSITY OF SCIENCE AND TECHNOLOGY
Electrical, Computer and Telecommunication Engineering Department
Test – 1
Electrical Machines I
Maximum Marks – 50 Marks
Time Duration: 1.30 Hrs
Answer All Questions
1. Explain the principle and working operation of transformer with its construction. Derive the EMF equation
expression for the transformer.
Define the term (i) Voltage Transformation Ratio. How it is applicable for Step up and Step down
Transformer.
•
[8M]
Static device. Electric power transformed from one circuit to another circuit without changing the
frequency. Can increase and reduce the voltage by changing the current value.
•
Physical basic of a transformer is Mutual induction between two circuits linked by common magnetic flux.
Two inductive circuits electrically separated but magnetically linked through a path of low. Two coils
possess high mutual inductance. One coil is connected to the alternating voltage.
•
Alternating flux is set up in the laminated core which is linked with the other coil which produces mutually
induced e.m.f. [Faradays law of mutual induction e=Mdi/dt]. If the secondary is closed then the current flows
so the electricity is transformed. Entirely magnetically from first coil to the second coil.
Briefly Transformer
1. Transfer electric power from one circuit to another.
2. It does so without a change of frequency.
3. Use electromagnetic induction.
4. Two electric circuits are in mutual inductive influence each other.
Parts
 Two winding coils having mutual inductance.
 Two coils insulated from each other and Laminated steel core.
 Suitable container
 Medium to insulate the core and winding from the container.
 Suitable bushing to bring out the terminal points from the tank.
EMF Equation
2. Derive the expression showing the saving of copper when a two winding transformer is converted into an
autotransformer. Prove that the latter can handle more power with the same amount of copper of the former?
[8M]
Auto transformer is kind of electrical transformer where primary and secondary shares same
common single winding. So basically it’s a one winding transformer.
Weight of copper of any winding depends upon its length and cross - sectional area. Again length of
conductor in winding is proportional to its number of turns and cross - sectional area varies with rated
current. So weight of copper in winding is directly proportional to product of number of turns and
rated current of the winding. Therefore, weight of copper in the section AC proportional to,
and similarly, weight of copper in the section BC proportional to,
Hence, total weight of copper in the winding of auto transformer proportional to,
In similar way it can be proved, the weight of copper in two winding transformer is proportional to,
N1I1 + N2I2
⇒ 2N1I1
(Since, in a transformer N1I1 = N2I2)
Let's assume, Wa and Wtw are weight of copper in auto transformer and two winding transformer
respectively,
∴ saving of copper in auto transformer compared to two winding transformer,
3. A Single phase, 50Hz transformer has 80turns on the primary winding and 400turns on the secondary
winding. The net cross sectional area of the core is 200cm2. If the primary winding is connected to a 240V,
50Hz supply
Determine. The EMF induced in the secondary winding and maximum value of the flux density in the core.
[5M]
E2 = 240 * 80/400 = 48 Volts
Bm = 240 / (4.44x50x400x200x10-4) = 0.1351wb/m2
4. A 25KVA Transformer has 500 turns on the primary winding and 50 turns on the secondary winding. The
primary is connected to 3000V, 50Hz supply. Find the full load primary and secondary currents in the
secondary EMF and maximum flux in the core.
[5M]
K = N2/N1 = 50/500 = 1/10
I1 = 25000/3000 = 8.33A, I2 = 10 x 8.33 = 83.3A
EMF per turn on primary side = 3000/500 = 6V
Secondary emf = 6x50 = 300V
E1 = 4.44x50x500xΦm ; Φm = 3000 / 4.44x500x50 = 27mwb
5. Explain in detail about the different types of losses in the transformer.
[5M]
In any electrical machine, 'loss' can be defined as the difference between input power and output power.
An electrical transformer is an static device , hence mechanical losses (like windage or friction losses) are absent
in it.
A transformer only consists of electrical losses (iron losses and copper losses).
Transformer losses are similar to losses in a DC machine, except that transformers do not have mechanical
losses. Core Losses Or Iron Losses : Eddy current loss and hysteresis loss depend upon the magnetic properties
of the material used for the construction of core.
Hence these losses are also known as core losses or iron losses.
Hysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in the transformer core. This
loss depends upon the volume and grade of the iron, frequency of magnetic reversals and value of flux density. It
can be given by, Steinmetz formula:
Wh=η Bmax1.6 f V(watts)
where, η=Steinmetz hysteresis constant
V = volume of the core in m3
Eddy current loss in transformer: In transformer, AC current is supplied to the primary winding which sets up
alternating magnetizing flux.
When this flux links with secondary winding, it produces induced emf in it.
But some part of this flux also gets linked with other conducting parts like steel core or iron body or the
transformer, which will result in induced emf in those parts, causing small circulating current in them.
This current is called as eddy current. Due to these eddy currents, some energy will be dissipated in the form of
heat.
Copper loss is due to ohmic resistance of the transformer windings.
Copper loss for the primary winding is I12R1and for secondary winding is I22R2.
Where, I1 and I2 are current in primary and secondary winding respectively, R1 and R2 are the resistances of
primary and secondary winding respectively.
It is clear that Cu loss is proportional to square of the current, and current depends on the load.
Hence copper loss in transformer varies with the load.
6. Briefly explain the parts of the transformer with neat sketch.
[5M]
Primary and secondary winding
Core medium
Conservator: This is a cylindrical tank mounted on supporting structure on the roof of the transformer's main
tank. When transformer is loaded, the temperature of oil increases and consequently the volume of oil in the
transformer get increased. The conservator tank of a transformer provides adequate space for expansion of
oil. Conservator tank of transformer also acts as a reservoir of oil.
Silica gel breather is the most commonly used way of filtering air from moisture.
Radiator it just increases the surface area for dissipating heat of the oil.
In case of electrical power transformer, due to the transport limitation, this unit are sent separately and
assembled at site with transformer main body.
7. Explain in detail about the Polarity test.
[5M]
A knowledge of polarity is also required to connect potential and current transformers to power metering
devices and protective relays.
Subtractive
polarity transformer
designs
are
shown
in
the
first
circuit
diagrams. Additive
polarity transformer designs are shown in the second circuit diagrams.
If the transformer is actually rated 480 - 120 volts, the transformer ratio is 4:1 (480 / 120 = 4).
Applying a test voltage of 120 volts to the primary will result in a secondary voltage of 30 volts (120 / 4 = 30).
If transformer is subtractive polarity, the voltmeter will read 90 volts (120 - 30 = 90).
If the voltmeter reads 150 volts, the transformer is additive polarity (120 + 30 = 150).
The red arrows indicate the relative magnitude and direction of the primary and secondary voltages
8. A 40KVA, 3300/240 V, 50Hz, single phase transformer has 660turns on the primary. Determine a) The
number of turns on the secondary b) The maximum value of flux in the core c) The approximate value of
primary and secondary full load currents.
9. What is an Ideal Transformer?
No losses [No Ohmic resistance], No magnetic leakage, No I2R and core losses.
[5M]
[2M]
Pure Inductive coils.
10. What are the different types of cooling methods used in the transformer?
[2M]
The transformer mounted and then purified high insulating oil is filled. The oil convey the heat from the core
and the windings where it is radiated out to the surroundings.
For small size tanks are usually smooth surfaced.
Larger size provided with radiators or pipes. [Cooling to be need more]
Construction of very large self cooled transformers are expensive, a more economical form of construction
for such large transformers is provided in the oil immersed, water cooled type.
ONAN Cooling of Transformer
ONAF Cooling of Transformer
OFAF Cooling of Transformer
OFWF Cooling of Transformer
ODAF Cooling of Transformer
ODWF Cooling of Transformer