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UNIT-3
MARKS
1. Explain the operating principle of a transformer?
Answer: Transformer works on the principle of Mutual Induction.
Mutual Induction: It is an electrical phenomenon where by an Electromotive Force (EMF) is
generated in a closed circuit by rate of change of flux. It is given by Faraday’s law of
electromagnetic induction as,
Emf = dϕ/dt
Where dϕ= change in flux linking the conductor
dt= change in time (seconds)
2. Explain the principle of indicating instruments?
Answer: The principle is
When a current carrying conductor is placed in a magnetic field, it experiences force by which
the conductor starts rotating in the magnetic field and the direction of force is found by using
Fleming’s left hand rule. This force is given by
F= BIL sinϴ
3. What is a transformer?
Answer: A transformer is a static electrical device which transfers electrical energy from one
circuit to another circuit without change in frequency. This energy transfer is done by mutual
induction.
4. Classify different electrical measuring instruments?
Answer: In general, many of the electrical instruments depend on magnetic affect to measure a
given electrical quantity. These electrical instruments are classified as:
Indicating Instruments: These are the instruments which indicate the instantaneous value of
electrical quantity, which is to be measured .Their indications are given by a pointer
moving over a calibrated scale.
Ex: Ammeter, Voltmeter.
(ii)
Recording Instruments: These instruments give a continuous record of variations in
the electrical quantity which is going to be measured.
Ex: Used in power stations &sub-stations.
(iii)
Integrating Instruments: These are the instruments which measure the electrical
quantity or total amount of electrical energy supplied to a circuit in a given time.
Ex: Energy meter (Kwhr).
5. Define transformation ratio of a transformer?
Answer: The transformation ratio of the transformer is defined as the ratio of number of turns of
on the secondary (output side) of the transformer to the number of turns of on the primary side
(input side) of the transformer. This ratio then directly corresponds to the ratio of secondary voltage
(output) to primary voltage (input).
Transformer ratio (K) =
3 Marks
What are the losses in a transformer?
Answer: The power losses in a transformer are of two types, they are:Core loss (or) Iron loss
Copper loss
These losses appear in the form of heat and produce an increase in temperature and drop in
efficiency.
Core or Iron losses (Pi):- These are the losses because of the material of the core in use and
reversal of magnetisation. These are of hysteresis and eddy current losses and occur in the
transformer core due to the alternating flux. These can be determined by open-circuiting test.
Hysteresis losses = kh f Bm(1.6)
Eddy current losses = ke f2Bm 2t2
b. Copper losses:These losses occur in both the primary and secondary windings due to their ohmic resistance.
These can be determined by short-circuit test.
Total Cu losses, Pc = I12R1 + I22R2
=I12R01 (OR) I22R02
Total losses in a transformer = Pi + Pc
= Constant losses + Variable losses
It may be noted that in a transformer, copper losses account for about 90% of the total losses.
2. What are the applications of moving coil instrument?
Answer: Moving coil instruments are acknowledged to be the best type for all dc measurements.
They are very sensitive and maintain a high degree of accuracy over long periods. The main
applications of moving coil instruments are:(¡) Used in the measurement of direct currents and voltages.
(¡¡) In dc galvanometers to detect small currents
(¡¡¡) In ballistic galvanometers used mainly for measuring changes of magnetic flux linkages.
3. What are the different electrical measuring instruments?
Answer: The instruments that are used to measure electrical quantities such as current, voltage,
power, energy etc are called electrical measuring instruments. According to their function these
instruments are classified as
(¡) Indicating instruments:- These are the instruments which indicate the instantaneous value
of electrical quantity which is to be measured . Their indications are given by a pointer moving
over a calibrated scale.
Eg. :- Ammeter, voltmeter.
(¡¡) Recording instruments:- These instruments give a continuous record of variations in the
electrical quantity which is going to be measured. These instruments are generally used in power
stations and sub-stations.
(¡¡¡) Integrating instruments:- These are the instruments which measure the electrical quantity
(or) the total amount of electrical energy supplied to a circuit in a given time.
Eg. :- Energy meters(Kwhr)
4. What is step-down and step-up transformer?
Answer: There are two types of transformers namely, they are as follows
(¡) Step-down transformer:- In a step down transformer the secondary windings are lesser than
the primary windings i.e., the secondary voltage is less than the primary voltage. So, step down
transformer is used to convert high voltage, low current power into a low voltage high current
power. It is mainly used in domestic consumption.
(¡¡)
Step-up transformer:- A step up transformer has more number of turns in the
secondary winding than in the primary winding. Voltage in the secondary winding is greater than
the voltage across the primary winding. So, a step up transformer is used to converts low voltage,
high current to high voltage low current. These are also used for local applications such as x-ray
machines which require about 50,000 volts. Even a micro-wave oven requires a small step up
transformer to operate.
10 Marks
Explain the construction and working principle of transformer?
Answer: Transformer – Working Principle
A transformer is a static device which helps in the transformation of electric power in one
circuit to electric power of the same frequency in another circuit. The voltage can be raised or
lowered in a circuit, but with a proportional increase or decrease in the current ratings.
The main principle of operation of a transformer is mutual inductance between two circuits which
is linked by a common magnetic flux. A basic transformer consists of two coils that are electrically
separate and inductive, but are magnetically linked through a path of reluctance. The working
principle of the transformer can be understood from the figure below.
Transformer Working:
As shown above the transformer has primary and secondary windings. The core
laminations are joined in the form of sheets in between the strips. These staggered joints are said
to be laminated. Both the coils have high mutual inductance. When an alternating voltage is given
to the primary winding, an alternating flux is produced in the core material of the transformer.
Most of the alternating flux developed by this coil is links the other secondary coil and thus
produces the mutual induced electro-motive force. The so produced electro-motive force can be
explained with the help of Faraday’s laws of Electromagnetic Induction as
Emf =M*dϕ/dt
Where M = mutual inductance
dϕ= change of flux
and
dt= change in time
If the second coil circuit is closed, a current flow in the secondary winding and thus electrical
energy is transferred magnetically from the first to the second coil.
The alternating current supply is given to the first coil and hence it can be called as the primary
winding. The energy is drawn out from the second coil and thus can be called as the secondary
winding.
Transformer Construction:
For the simple construction of a transformer, you must need two coils having mutual inductance
and a laminated steel core. The two coils are insulated from each other and from the steel core. In
order to insulate and to bring out the terminals of the winding from the tank, apt bushings that are
made from either porcelain or capacitor type must be used.
In all transformers that are used commercially, the core is made out of sheet steel laminations
assembled to provide a continuous magnetic path with minimum of air-gap included. The steel
should have high permeability and low hysteresis loss. For this to happen, the steel should be made
of high silicon content and must also be heat treated. By effectively laminating the core, the eddycurrent losses can be reduced. The lamination can be done with the help of a light coat of core
plate varnish or lay an oxide layer on the surface. For a frequency of 50 Hertz, the thickness of the
lamination varies from 0.35mm to 0.5mm for a frequency of 25 Hertz. Based on the construction
transformers are classified into two types. They are
Core type transformer: In this transformer the core is surrounded by the windings.
Shell type transformer: In this transformer the windings are surrounded by the core material.
2. Derive the Emf equation of a transformer?
Answer: When a sinusoidal voltage is applied to the primary winding of a transformer, alternating
flux ϕm sets up in the iron core of the transformer. This sinusoidal flux links with both primary and
secondary winding. The flux produced is an alternating flux represented by a sine function as ϕ =
ϕm sin(wt), where ϕm = maximum flux
The induced Emf in a transformer is given by Faradays law of EMI as E = dϕ/dt
Let the supply frequency be f(Hz)
N1 is the number of turns in the primary winding
N2 is the number of turns in the secondary winding.
The flux produced is represented as shown
Let us consider ¼ of the time interval where the flux is changing from zero to maximum value in
this interval.
When time t=o, the flux is zero and when time t= T/4 the flux is ϕm.
So we get dϕ = (ϕm-0)
which is dϕ = ϕm
dt = (T/4-0) which is dt = T/4
By Faradays law induced Emf = dϕ/dt
Therefore average induced Emf in a single turn is Eavg = 4ϕm/T
Eavg = 4ϕmf, where 1/T = frequency
The average induced Emf in the windings of ‘N’ turns is given by
Eavg = 4ϕmfN
We know that for a sinusoidal waveform the form factor value is 1.11, which is the ratio of RMS
value of a quantity to the Average value of a quantity.
Form factor = R.M.S value/Average value = 1.11
Erms/ Eavg = 1.11
Therefore RMS value of induced Emf is given by Erms = 1.11 * 4ϕmfN
Erms = 4.44 ϕm f N
The primary induced Emf is given by E1 = 4.44 ϕm f N1 and
Secondary induced Emf is given by E2 = 4.44 ϕm f N2, where N1 and N2 are primary and secondary
number of turns.
3. What are the losses in a transformer and explain them briefly?
Answer: The power losses in a transformer are classified into two types. They are
Core loss (or) Iron loss
Copper loss
These losses appear in the form of heat and produce an increase in temperature and drop in
efficiency.
Core or Iron losses (Pi):- These are the losses because of the material of the core in use
and reversal of magnetisation. These are of hysteresis and eddy current losses and occur in
the transformer core due to the alternating flux. These can be determined by open-circuiting
test.
When an alternating voltage is given to the primary winding of a transformer an
alternating field is produced in the core. Due to the reversal of magnetisation of the core
material, the loss occurred is termed as Hysteresis loss.
Hysteresis losses = kh f Bm(1.6)
During the operation of the transformer small currents flow in the core material
which are termed as eddy currents. The loss in the transformer because of these currents is
termed as Eddy current loss and it is given
Eddy current losses = ke f2Bm 2t2
ii) Copper losses:The losses in a transformer because of the ohmic resistance of both the primary and secondary
windings are termed as Copper losses. These can be determined by short-circuit test.
Total Cu losses, Pc = I12R1 + I22R2
=I12R01 (OR) I22R02
Total losses in a transformer = Pi + Pc
= Constant losses + Variable losses
It may be noted that in a transformer, copper losses account for about 90% of the total losses.