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Transcript
UNIT I
DC MACHINES
Three phase circuits, a review. Construction of DC machines – Theory of operation of DC
generators – Characteristics of DC generators- Operating principle of DC motors – Types of DC
motors and their characteristics – Speed control of DC motors- Applications
DC GENERATOR
A dc generator is an electrical machine which converts mechanical energy into direct current
electricity. This energy conversion is based on the principle of production of dynamically
induced emf.
CONSTRUCTION:
Cross Sectional View of DC Generator
Above figure shows the constructional details of a simple 4-pole DC generator. A DC generator
consists of two basic parts, stator and rotor.
Basic constructional parts of a DC generator are described below:
Yoke: The outer frame of a generator or motor is called as yoke. Yoke is made up of cast iron or
steel. Yoke provides mechanical strength for whole assembly of the generator (or motor). It also
carries the magnetic flux produced by the poles.
Poles: Poles are joined to the yoke with the help of screws or welding. Poles are to support field
windings. Field winding is wound on poles and connected in series or parallel with armature
winding or sometimes separately.
Pole shoe: Pole shoe is an extended part of the pole which serves two purposes, (i) to prevent
field coils from slipping and (ii) to spread out the flux in air gap uniformly.
Armature core: Armature core is the rotor of a generator. Armature core is cylindrical in shape
on which slots are provided to carry armature windings.
Commutator and brushes: As emf is generated in the armature terminals, it must be taken out
to make use of generated emf. But if we can't directly solder wires to Commutator conductors as
they rotate. Thus commutator is connected to the armature conductors and mounted on the same
shaft as that of armature core. Conducting brushes rest on commutator and they slides over when
rotor (hence commutator) rotates. Thus brushes are physically in contact with armature
conductors hence wires can be connected to brushes.
WORKING PRINCIPLE OF A DC GENERATOR:
According to Faraday's law of electromagnetic induction, when a conductor moves in a
magnetic field (thereby cutting the magnetic flux lines), a dynamically induced emf is produced
in the conductor. The magnitude of generated emf can be given by emf equation of DC
generator. If a closed path is provided to the moving conductor then generated emf causes a
current to flow in the circuit.
Thus in DC generators, when armature is rotated with the help of a prime mover and field
windings are excited (there may be permanent field magnets also), emf is induced in armature
conductors. This induced emf is taken out via commutator-brush arrangement.
EMF EQUATION OF A DC GENERATOR:
Let Ø = flux/pole in Wb (weber)
Z = total no. of armature conductors
P = no. of generator poles
A = no. of parallel paths in armature
N = rotational speed of armature in revolutions per min. (rpm)
E = emf induced in any parallel path in armature
Now,
Generated e.m.f Eg = e.m.f generated in any one of the parallel paths i.e E.
Average e.m.f geneated /conductor = dΦ/dt volt (n=1)
Now, flux cut/conductor in one revolution dΦ
= ΦP Wb
No.of revolutions/second
= N/60
Time for one revolution,
dt
= 60/N second
Hence, according to Faraday's Laws of Electroagnetic Induction,
E.M.F generated/conductor is
For a simplex wave-wound generator, No.of parallel paths = 2
No.of conductors (in series) in one path
= Z/2
E.M.F. generated/path is
For a simplex lap-wound generator
No.of parallel paths
No.of conductors (in series) in one path
=P
= Z/P
E.M.F.generated/path
In general generated e.m.f
where
A = 2 - for simplex wave-winding
A= P - for simplex lap-winding
CHARACTERISTICS OF DC GENERATOR:
In a separately excited DC generator, the field winding is excited by an external
independent source. There are generally three most important characteristic of DC generator:
Magnetic or Open Circuit Characteristic of Separately Excited DC Generator
The curve which gives the relation between field current (If) and the generated voltage (E0) in
the armature on no load is called magnetic or open circuit characteristic of a DC generator.
The plot of this curve is practically same for all types of generators, whether they are separately
excited or self-excited. This curve is also known as no load saturation characteristic curve of
DC generator. Here in this figure below we can see the variation of generated emf on no load
with field current for different fixed speeds of the armature. For higher value of constant speed,
the steepness of the curve is more. When the field current is zero, for the effect residual
magnetism in the poles, there will be a small initial emf (OA) as show in figure.
Let us consider a separately excited DC generator giving its no load voltage E0 for a constant
field current. If there is no armature reaction and armature voltage drop in the machine then the
voltage will remain constant. Therefore, if we plot the rated voltage on the Y axis and load
current on the X axis then the curve will be a straight line and parallel to X-axis as shown in
figure below. Here, AB line indicating the no load voltage (E0).
When the generator is loaded then the voltage drops due to two main reasons1) Due to armature reaction,
2) Due to ohmic drop ( IaRa ).
Internal or Total Characteristic of Separately Excited DC Generator
The internal characteristic of the separately excited DC generator is obtained by
subtracting the drops due to armature reaction from no load voltage. This curve of actually
generated voltage ( Eg ) will be slightly dropping. Here, AC line in the diagram indicating
the actually generated voltage (E_g ) with respect to load current. This curve is also called
total characteristic of separately excited DC generator.
External Characteristic of Separately Excited DC Generator
The external characteristic of the separately excited DC generator is obtained by subtracting
the drops due to ohmic loss ( Ia Ra ) in the armature from generated voltage ( Eg ).
Terminal voltage(V) = Eg - Ia Ra.
This curve gives the relation between the terminal voltage (V) and load current. The external
characteristic curve lies below the internal characteristic curve. Here, AD line in the diagram
below is indicating the change in terminal voltage(V) with increasing load current. It can be
seen from figure that when load current increases then the terminal voltage decreases
slightly. This decrease in terminal voltage can be maintained easily by increasing the field
current and thus increasing the generated voltage. Therefore, we can get constant terminal
voltage.
Separately excited DC generators have many advantages over self-excited DC generators.
It can operate in stable condition with any field excitation and gives wide range of output
voltage.
The main disadvantage of these kinds of generators is that it is very expensive of
providing a separate excitation source.
CHARACTERISTICS OF SELF EXCITED DC GENERATOR:
In shunt wound DC generators the field windings are connected in parallel with armature
conductors as shown in figure below. In these types of generators the armature current Ia divides
in two parts. One part is the shunt field current Ish flows through shunt field winding and the
other part is the load current IL goes through the external load.
Three most important characteristic of shunt wound dc generators are discussed below:
Magnetic or Open Circuit Characteristic of Shunt Wound DC Generator
This curve is drawn between shunt field current(Ish) and the no load voltage (E0). For a given
excitation current or field current, the emf generated at no load E0 varies in proportionally with
the rotational speed of the armature. Here in the diagram the magnetic characteristic curve for
various speeds are drawn. Due to residual magnetism the curves start from a point A slightly up
from the origin O. The upper portions of the curves are bend due to saturation. The external load
resistance of the machine needs to be maintained greater than its critical value otherwise the
machine will not excite or will stop running if it is already in motion. AB, AC and AD are the
slops which give critical resistances at speeds N1, N2 and N3. Here, N1 > N2 > N3.
Critical Load Resistance of Shunt Wound DC Generator
This is the minimum external load resistance which is required to excite the shunt wound
generator.
Internal Characteristic of Shunt Wound DC Generator
The internal characteristic curve represents the relation between the generated voltage Eg
and the load current IL. When the generator is loaded then the generated voltage is decreased due
to armature reaction. So, generated voltage will be lower than the emf generated at no load. Here
in the figure below AD curve is showing the no load voltage curve and AB is the internal
characteristic curve.
External Characteristic of Shunt Wound DC Generator
AC curve is showing the external characteristic of the shunt wound DC generator. It is
showing the variation of terminal voltage with the load current. Ohmic drop due to armature
resistance gives lesser terminal voltage the generated voltage. That is why the curve lies below
the internal characteristic curve.
The terminal voltage can always be maintained constant by adjusting the of the load terminal.
When the load resistance of a shunt wound DC generator is decreased, then load current
of the generator increased as shown in above figure. But the load current can be increased to a
certain limit with (up to point C) the decrease of load resistance. Beyond this point, it shows a
reversal in the characteristic. Any decrease of load resistance, results in current reduction and
consequently, the external characteristic curve turns back as shown in the dotted line and
ultimately the terminal voltage becomes zero. Though there is some voltage due to residual
magnetism.
Now, when IL increased, then terminal voltage decreased. After a certain limit, due to heavy load
current and increased ohmic drop, the terminal voltage is reduced drastically. This drastic
reduction of terminal voltage across the load, results the drop in the load current although at that
time load is high or load resistance is low.
That is why the load resistance of the machine must be maintained properly. The point in which
the machine gives maximum current output is called breakdown point (point C in the picture).
CHARACTERISTICS OF DC SERIES GENERATOR:
In these types of generators the field windings, armature windings and external load circuit all
are connected in series as shown in figure below.
Therefore, the same current flows through armature winding, field winding and the load.
Let, I = Ia = Isc = IL
Here, Ia = armature current
Isc = series field current
IL = load current
There are generally three most important characteristics of series wound DC generator which
show the relation between various quantities such as series field current or excitation current,
generated voltage, terminal voltage and load current.
Magnetic or Open Circuit Characteristic of Series Wound DC Generator
The curve which shows the relation between no load voltage and the field excitation
current is called magnetic or open circuit characteristic curve. As during no load, the load
terminals are open circuited, there will be no field current in the field since, the armature, field
and load are series connected and these three make a closed loop of circuit. So, this curve can be
obtained practically be separating the field winding and exciting the DC generator by an external
source. Here in the diagram below AB curve is showing the magnetic characteristic of series
wound DC generator. The linearity of the curve will continue till the saturation of the poles.
After that there will be no further significant change of terminal voltage of DC generator for
increasing field current. Due to residual magnetism there will be a small initial voltage across the
armature that is why the curve started from a point A which is a little way up to the origin O.
Internal Characteristic of Series Wound DC Generator
The internal characteristic curve gives the relation between voltage generated in the
armature and the load current. This curve is obtained by subtracting the drop due to the
demagnetizing effect of armature reaction from the no load voltage. So, the actual generated
voltage ( Eg) will be less than the no load voltage (E0). That is why the curve is slightly
dropping from the open circuit characteristic curve. Here in the diagram below OC curve is
showing the internal characteristic or total characteristic of the series wound DC generator.
External Characteristic of Series Wound DC Generator
The external characteristic curve shows the variation of terminal voltage (V) with the
load current ( IL). Terminal voltage of this type of generator is obtained by subtracting the
ohomic drop due to armature resistance (Ra) and series field resistance ( Rsc) from the actually
generated voltage ( Eg).
Terminal voltage V = Eg - I(Ra + Rsc)
The external characteristic curve lies below the internal characteristic curve because the
value of terminal voltage is less than the generated voltage. Here in the figure OD curve is
showing the external characteristic of the series wound DC generator.
It can be observed from the characteristics of series wound DC generator, that with the
increase in load (load is increased when load current increases) the terminal voltage of the
machine increases. But after reaching its maximum value it starts to decrease due to excessive
demagnetizing effect of armature reaction. This phenomenon is shown in the figure by the dotted
line. Dotted portion of the characteristic gives approximately constant current irrespective of the
external load resistance. This is because if load is increased, the field current is increased as field
is series connected with load. Similarly if load is increased, armature current is increased as the
armature is also series connected with load. But due to saturation, there will be no further
significance raises of magnetic field strength hence any further increase in induced voltage. But
due to increased armature current, the effect of armature reaction increases significantly which
causes significant fall in load voltage. If load voltage falls, the load current is also decreased
proportionally since current is proportional to voltage as per Ohm's law. So, increasing load,
tends to increase the load current, but decreasing load voltage, tends to decrease load current.
Due these two simultaneous effects, there will be no significant change in load current in dotted
portion of external characteristics of series wound DC generator. That is why series DC
generator is called constant current DC generator.
CHARACTERISTICS OF DC COMPOUND GENERATOR:
In compound wound DC generators both the field windings are combined (series and
shunt). This type of generators can be used as either long shunt or short shunt compound wound
generators as shown in the diagram below. In both the cases the external characteristic of the
generator will be nearly same. The compound wound generators may be cumulatively
compounded or differentially compounded. Differentially compound wound generators are very
rarely used. So, here we mainly concentrate upon the characteristic of cumulatively compound
wound generators.
In series wound DC generators, the output voltage is directly proportional with load current and
in shunt wound DC generators, output voltage is inversely proportional with load current.
The electric current in the shunt field winding produces a flux which causes a fall in
terminal voltage due to armature reaction and ohmic drop in the circuit. But the current in the
series field also produces a flux which opposes the shunt field flux and compensates the drop in
the terminal voltage and try to operate the machine at constant voltage.
The combination of a series generator and a shunt generator gives the characteristic of a
cumulative compound wound generator.
At no load condition there is no current in the series field because the load terminals are
open circuited. But the shunt field current helps to produce field flux and excite the machine.
When the dc generator supplies load, the load current increases and current flows through the
series field. Therefore, series field also provides some field flux and emf is also increased. The
voltage drop in the shunt machine is therefore compensated by the voltage rise in the series
machine.
Characteristics of DC Compound Generator:
For small distance operation the flat compounded generators are generally used because the
length of the feeder is negligible. But to maintain constant voltage over a long period, the over
compounded generators are used. It works as a generator and a booster (boost the terminal
voltage).
External characteristic of DC compound wound generator is drawn between the terminal
voltage and the load current. By adjusting the no. of amp-turns in the series field winding we can
get following external characteristics:



If the series turns are so adjusted that with the increase in load current the terminal
voltage also increases, then the generator is called over compounded. The curve AB in
the figure showing this characteristic. When the load current increases then the flux
provides by the series field also increases. It gives the additional generated voltage. If the
increase in generated voltage is greater than the voltage drops due to armature reaction
and ohmic drop then, terminal voltage of the generator is increased.
If the series turns are so adjusted that with the increase in load current the terminal
voltage remains constant, then the generator is called flat compounded. The curve AC in
the figure showing this characteristic. When the load current increases then the flux
provides by the series field also increases and gives the additional generated voltage. If
the increase in generated voltage is equal to the voltage drops due to armature reaction
and ohmic drop then, rated terminal voltage of the generator remains same as no load
voltage.
If the series field winding has lesser no. of turns then the rated terminal voltage becomes
less than the no load voltage, then the generator is called under compounded. Because,
the increase in generated voltage is lesser than the voltage drops due to armature reaction
and ohmic drop. Curve AD in the figure is showing this characteristic.
DC MOTOR:
A motor is a device which converts an electrical energy into the mechanical energy . The
energy conversion process is exactly opposite to that involved in a d.c. generator. In a generator
the input mechanical energy is supplied by a prime mover while in a d.c. motor, input electrical
energy is supplied by a d.c. supply. The construction of a d.c. machine is same whether it is a
motor or a generator.
Principle of Operation of a D.C. Motor
The principle of operation of a d.c. motor can be stated in a single statement as 'when a
current carrying conductor is placed in a magnetic field' it experiences a mechanical force'. In a
practical d.c. motor, field winding produces a required magnetic field while armature conductors
play a role of a current carrying conductors and hence armature conductors experience a force.
As a conductors are placed in the slots which are in the periphery, the individual force
experienced by the conductors acts as a twisting or turning force on the armature which is called
a torque. The torque is the product of force and the radius at which this force acts. So overall
armature experiences a torque and starts rotating. Let us study this motoring action in detail.
Consider a single conductor placed in a magnetic field as shown in the Fig .1(a). The
magnetic field is produced by a permanent magnet but in a practical d.c. motor it is produced by
the field winding when it carries a current.
Fig. 1
Now this conductor is excited by a separate supply so that it carries a current in a particular
direction. Consider that it carries a current away from an observe as shown in the Fig. 1(b). Any
current carrying conductor produces its own magnetic field around it. hence this conductor also
produces its own flux, around. The direction of this flux can be determined by right hand thumb
rule. For direction of current considered, the direction of flux around a conductor is clockwise.
For simplicity of understanding, the main flux produced by the permanent magnet is not shown
in the Fig. 1(b).
Now there are two fluxes present,
1. The flux produced by the permanent magnet called flux.
2. The flux produced by the current carrying conductor.
There are shown in the Fig.2(a). Form this, it is clear that on one side of the conductor, both
the fluxes are in same direction. In this case, on the left of the conductor there is gathering of the
flux lines as two fluxes help each other. As against this, on the right of the conductor, the two
fluxes are in opposite direction and hence try to cancel each other. Due to this, the density of the
flux lines in this area gets weakened. So on the left, there exists high flux density area while on
the right of the conductor there exists low flux density area as shown in the Fig. 2(b).
Fig. 2
This flux distribution around the conductors acts like a stretched rubber band under tension.
This exerts a mechanical force on the conductor which acts from high flux density area towards
low flux density area. i.e. from left to right for the case considered as shown in the Fig. 2(b).
Fig. 3
Key point : In the practical d.c. motor, the permanent magnet is replaced by a field winding
which produces the required flux called main flux and all the armature conductors, mounted on
the periphery of the armature drum, get subjected to the mechanical force. Due to this, overall
armature experiences a twisting force called torque and armature of the motor starts rotating.
1. Direction of Rotation of Motor
The magnitude of the force experienced by the conductor in a motor is given by,
F = B l I Newton (N)
B = Flux density due to the flux produced by the field winding.
l = Active length of the conductor.
I = Magnitude of the current passing through the conductor.
The direction of such force i.e. the direction of rotation of a motor can be determined by
Fleming's left hand. So Fleming's right hand rule is to determine direction of induced e.m.f. i.e.
for generating action while Fleming's left hand rule is to determine direction of force experienced
i.e. for motoring action.
1.1 Fleming's left hand rule
The rule states that, 'Outstretch the three fingers of the left hand namely the first finger,
middle finger and thumb such that they are mutually perpendicular to each other. Now point the
first finger in the direction of magnetic field and the middle finger in the direction of the current
then the thumb gives the direction of the force experienced by the conductor'.
The Fleming's left hand rule can be diagramatically shown as in the Fig. 1.
Fig. 1
Apply the rule to crosscheck the direction of force experienced by a single conductor, placed
in the magnetic field, shown in the Fig. 2(a), (b), (c) and (d).
Fig. 2
It can be seen from the Fig. 2 that if the direction of the main field in which current carrying
conductor is placed, is reversed, force experienced by the conductor reverses its direction.
Similarly keeping main flux direction unchanged, the direction of current passing through the
conductor is reversed. The force experienced by the conductor reverses its direction. However if
both the directions are reversed, the direction of the force experienced remains the same.
Key point : So in a practical motor, to reverse its direction of rotation, either direction of main
field produced by the field winding is reversed or direction of the current passing through the
armature is reversed.
The direction of the main field can be reversed by changing the direction of current passing
through the field winding, which is possible by interchanging the polarities of supply which is
given to the field winding . In short, to have a motoring action two fluxes must exist, the
interaction of which produces a torque.
Significance of Back E.M.F.
It is seen in the generation action, that when a conductor cuts the lines of flux, e.m.f. gets
induced in the conductor. The question is obvious that in a d.c. motor, after a motoring action,
armature starts rotating and armature conductors cut the main flux. So there is a generating
action existing in a motor.
After a motoring action, there exists a generating action. There is an induced e.m.f. in the
rotating armature conductors according to Faraday's law of electromagnetic induction. This
induced e.m.f. in the armature always acts in the opposite direction of the supply voltage. This is
according to the Lenz's law which states that the direction of the induced e.m.f. is always so as to
oppose the cause producing it. In a d.c. motor, electrical input i.e. the supply voltage is the cause
and hence this induced e.m.f. opposes the supply voltage. This e.m.f. tries to set up a current
through the armature which is in the opposite direction to that, which supply voltage is forcing
through the conductor.
So as this e.m.f. always opposes the supply voltage, it is called back e.m.f. and denoted as
Eb. Though it is obtained as Eb, basically it gets generated by the generation action which we
have seen earlier in case of generation. So its magnitude can be determined by the e.m.f.
equation which is derived earlier. So,
where all symbols carry the same meaning as seen earlier in case of generators.
Fig. 1
This e.m.f. is shown schematically in the Fig. 1(a). So if V is supply voltage in volts and R a
is the value of the armature resistance, the equivalent electric circuit can be shown as in the Fig.
1(b).
TYPES OF D.C. MOTORS
Similar to the d.c. generators, the d.c. motors are classified depending upon the way of
connecting the field winding with the armature winding. The difference types of d.c. motors are ;
1. Shunt motor
2. Series motors
3. Compound motors
The compound motors are further classified as ;
1. Short shunt compound
2. Long shunt compound
DC SHUNT MOTOR:
In this type, the field winding is connected across the armature winding and the combination is
connected across the supply, as shown in the Fig. 1.
Fig. 1 D.C. shunt motor
Let Rsh be the resistance of shunt field winding.
Ra be the resistance of armature winding.
The value of Ra is very small while Rsh is quite large. Hence shunt field winding has more
number of turns with less cross-sectional area.
1.1 Voltage and Current Relationship
The voltage across armature and field winding is same equal to the supply voltage V.
The total current drawn from the supply is denoted as line current IL.
IL = Ia + Ish
Ish = V/Rsh
and
V = Eb + Ia Ra + Vbrush
Vbrush is generally neglected.
Now flux produced by the field winding is proportional to the current passing through it i.e.
Ish.
Note : As long as supply voltage is constant, which is generally so in practice, the flux produced
is constant. Hence d.c. shunt motor is called constant flux motor.
DC SERIES MOTOR:
In this type of motor, the series field winding is connected in series with the armature and the
supply, as shown in the Fig. 1.
Fig. 1 D.C. series motor
Let Rse be the resistance of the series field winding. The value of R se is very small and it is
made of small number of turns having large cross-sectional area.
1.1 Voltage and Current Relationship
Let IL be the total current drawn from the supply.
So
IL = Ise = Ia
and
V = Eb + Ia Ra + Ise Rse + Vbrush
V = Eb + Ia (Ra + Rse) + Vbrush
Supply voltage has to overcome the drop across series field winding in addition to Eb and
drop across armature winding.
Note : In series motor, entire armature current is passing through the series field winding. So flux
produced is proportional to the armature current.
for series motor
DC COMPOUND MOTOR:
The compound motor consists of part of the field winding connected in series and part of
the field winding connected in parallel with armature. It is further classified as long shunt
compound and short shunt compound motor.
1.1 Long Shunt Compound Motor
In this type, the shunt field winding is connected across the combination of armature and the
series field winding as shown in the Fig. 1.
Fig. 1 Long shunt compound motor
Let Rse be the resistance of series field and Rsh be the resistance of shunt field winding. The
total current drawn from supply is IL.
So
IL = Ise + Ish
But Ise = Ia
...
IL = Ia+ Ish
And Ish = V/Rsh
And V = Eb + Ia Ra + Ise Rse + Vbrush
But as
Ise = Ia ,
.
..
V = Eb + Ia (Ra + Rse) + Vbrush
1.2 Short Shunt Compound Motor
In this type, the shunt field is connected purely in parallel with armature and the series field
is connected in series with this combination shown in the Fig. 2.
Fig. 2 Short shunt compound motor
...
...
IL = Ise
The entire line current is passing through the series field winding.
and
IL = Ia + Ish
Now the drop across the shunt field winding is to be calculated from the voltage equation.
So
V = Eb + Ise Rse + Ia Ra + Vbrush
but
Ise = IL
V = Eb + IL Rse + Ia Ra + Vbrush
Drop across shunt field winding is,
= V - IL Rse = Eb + Ia Ra + Vbrush
Apart from these two, compound motor can be classified into two more types,
i) Cumulatively compound motors and ii) Differential compound motors.
Note : If the two field windings are wound in such a manner that the fluxes produced by the two
always help each other, the motor is called cumulatively compound. If the fluxes produced by
the two field windings are trying to cancel each other i.e. they are in opposite direction, the
motor is called differential compound.
A long shunt compound motor can be of cumulative or differential type. Similarly short
shunt compound motor can be cumulative or differential type.
TORQUE EQUATION OF A D.C. MOTOR
It is seen that the turning or twisting force about an axis is called torque. Consider a wheel of
radius R meters acted upon by a circumferential force F newtons as shown in the Fig. 1.
Fig. 1
The wheel is rotating at a speed of N r.p.m. Then angular speed of the wheel is,
ω = (2πN)/60 rad/sec
So workdone in one revolution is,
W = F x distance travelled in one revolution
= F x 2 R joules
And P = Power developed = Workdone/Time
= (F x 2πR) / (Time for 1 rev) = (F x 2πR) / (60/N) = (F x R) x (2πN/60)
...
P = T x ω watts
Where T = Torque in N - m
ω = Angular speed in rad/sec.
Let Ta be the gross torque developed by the armature of the motor. It is also called armature
torque. The gross mechanical power developed in the armature is Eb Ia, as seen from the power
equation. So if speed of the motor is N r.p.m. then,
Power in armature = Armature torque x ω
...
...
Eb Ia = x (2N/60)
but Eb in a motor is given by,
Eb = (ΦPNZ) / (60A)
(ΦPNZ / 60A) x Ia = Ta x (2πN/60)
This is the torque equation of a d.c. motor.
TORQUE SPEED EQUATIONS:
Before analysing the various characteristics of motors, let us revise the torque and speed
equations are applied to various types of motors.
...
T α Φ Ia from torque equation.
This is because, 0.159(PZ)/A is a constant for a given motor.
Now Φ is the flux produced by the field winding and is proportional to the current passing
through the field winding.
Φ α Ifield
But for various types of motors, current through the field winding is different. Accordingly
torque equation must be modified.
For a d.c. shunt motor, Ish is constant as long as supply voltage is constant. Hence Φ flux is
also constant.
...
T α Ia
for shunt motors
For a d.c. series motor, Ise is same as Ia. Hence flux Φ is proportional to the armature
current Ia.
...
T α Ia α Ia2
for series motors.
Similarly as Eb = (ΦPNZ)(60A), we can write the speed equation as,
Eb α Φ N
.
..
N α Eb/Φ
But V = Eb + Ia Ra
neglecting brush drop
.
..
Eb = V - Ia Ra
... Speed equation becomes,
N α (V-Ia Ra)/Φ
So for shunt motor as flux is constant,
...
N α V - Ia Ra
While for series motor, flux Φ is proportional to Ia.
FACTORS AFFECTING THE SPEED OF A D.C. MOTOR
According to the speed equation of a d.c. motor we can write,
The factors Z, P, A are constants for a d.c. motor.
But as the value of armature resistance Ra and series field resistance Rse is very small, the
drop Ia Ra and (Ra + Rse) is very small compared to applied voltage V. Hence neglecting these
voltage drops the speed equation can be modified as,
Thus the factors affecting the speed of a d.c. motor are,
1. The flux Φ
2. The voltage across the armature
3. The applied voltage V
depending upon these factors the various methods of speed control are,
1. Changing the flux Φ by controlling the current through the field winding called flux control
methods.
2. Changing the armature path resistance which in turn changes the voltage applied across the
armature called rheostatic control.
3. Changing the applied voltage called voltage control method.
D.C. Motor Characteristics
The performance of a d.c. motor under various conditions can be judged by the following
characteristics
i) Torque - Armature current characteristics (T Vs Ia ) :
The graph showing the relationship between the torque and the armature current is called a
torque-armature current characteristic. These are also called electrical characteristics.
ii) Speed - Armature current characteristics(N Vs Ia ) :
The graph showing the relationship between the speed and armature current characteristic.
iii) Speed - Torque characteristics(N Vs T) :
The graph showing the relationship between the speed and the torque of the motor is called
speed-torque characteristics of the motor. These are also called mechanical characteristic.
The nature of these characteristics can easily be obtained by using speed and torque
equations derived in previous post. These characteristics play a very important role in selecting a
type of motor for a particular application.
Characteristics of dc shunt motor:
i) Torque - Armature current characteristics
For a d.c. motor
T α Φ Ia
For a constant values of Rsh and supply voltage V, Ish is also constant and hence flux is also
constant.
...
Ta α Φ Ia
The equation represents a straight line, passing through the origin, as shown in the Fig. 1.
Torque increases linearly with armature current. It is seen earlier that armature current is decided
by the load. So as load increases, armature current increases, increasing the torque developed
linearly.
Fig. 1 T Vs Ia for shunt motor
Now if shaft torque is plotted against armature current, it is known that shaft torque is less
than the armature torque and the difference between the two is loss torque Tf as shown. On no
load Tsh = 0 but armature torque is present which is just enough to overcome stray losses shown
as Ta0. The current required is Ia0 on no load to produce Ta0 and hence Tsh graph has an intercept
of Ia0 on the current axis.
To generate high starting torque, this type of motor requires a large value of armature
current at start. This may damage the motor hence d.c. shunt motors can develop moderate
starting torque and hence suitable for such applications where starting torque requirement is
moderate.
ii) Speed - Armature current characteristics
From the speed equation we get,
N α (V - Ia Ra)/Φ
αV - Ia Ra as Φ is constant
So as load increases, the armature current increases and hence drop Ia Ra also increases.
Hence for constant supply voltage, V - Ia Ra decreases and hence speed reduces. But as Ra
is very small, for change in Ia from no load to full load, drop Ia Ra is very small and hence drop in
speed is also not significant from no load to full load.
Fig. 2 N Vs Ia for shunt motor
So the characteristics is slightly droping as shown in the Fig. 2.
But for all practical purposes these type of motors are considered to be a constant speed
motors.
iii) Speed - Torque characteristics
These characteristics can be derived from the above two characteristics. This graph is similar
to speed-armature current characteristics as torque is proportional to the armature current. This
curve shows that the speed almost remains constant through torque changes from no load to full
load conditions. This is shown in the Fig. 3.
Fig. 3 N Vs T for shunt motor
Characteristics of dc series motor:
i) Torque - Armature current characteristics
In case of series motor the series field winding is carrying the entire armature current. So
flux produced is proportional to the armature current.
...
Φ α Ia
Hence
Ta α Φ Ia α Ia2
Thus torque in case of series motor is proportional to the square of the armature current. This
relation is parabolic in nature as shown in the Fig. 1.
As load increases, armature current increases and torque produced increases proportional to
the square of the armature current upto a certain limit.
As the entire Ia passes through the series field, there is a property of an electromagnet called
saturation, may occur. Saturation means though the current through the winding increases, the
flux produced remains constant. Hence after saturation the characteristics take the place of
straight line as flux becomes constant, as shown. The difference between Ta and Tsh is loss torque
Tf which is also shown in the Fig. 2.
At start as Ta αIa2 , these types of motors can produce high torque for small amount of
armature current hence the series motors are suitable for the applications which demand high
starting torque.
ii) Speed - Armature current characteristics
From the speed equation we get,
N α (Eb/Φ) ) αV - Ia Ra - Ia Rse)/ Ia
as Φ α Ia in case of series motor
Now the values of Ra and Rse are so small that the effect of change in Ia on speed overrides
the effect of change in V - Ia Ra - Ia Rse on the speed.
Hence in the speed equation, Eb ≈ Vand can be assumed constant. So speed equation reduced
to,
N α 1/Ia
So speed-armature current characteristics is rectangular hyperbola type as shown in the Fig.
2.
iii) Speed - Torque characteristics
In case of series motors,
T α Ia2 and N α 1/Ia
Hence we can write,
N α 1/√T
Thus as torque increases when load increases, the speed decreases. On no load, torque is
very less and hence speed increases to dangerously high value. Thus the nature of the speedtorque characteristics is similar to the nature of the speed-armature current characteristics.
The speed-torque characteristics of a series motor is shown in the Fig. 3.
Fig. 3 N Vs T for series motor
DC Compound Motor Characteristics:
Compound motor characteristics basically depends on the fact whether the motor is
cumulatively compound or differential compound. All the characteristics of the compound motor
are the combination of the shunt and series characteristic.
Cumulative compound motor is capable of developing large amount of torque at low speeds
just like series motor. However it is not having a disadvantages of series motor even at light or
no load. The shunt field winding produces the definite flux and series flux helps the shunt field
flux to increase the total flux level.
So cumulative compound motor can run at reasonable speed and will not run with
dangerously high speed like series motor, on light or no load condition.
In differential compound motor, as two fluxes oppose each other, the resultant flux decreases
as load increases, thus the machine runs at a higher speed with increase in the load. This property
is dangerous as on full load, the motor may try to run with dangerously high speed. So
differential compound motor is generally not used in practice.
The various characteristics of both the types of compound motors cumulative and the
differential are shown in the Fig. 1(a), (b) and (c).
Fig. 1 characteristics of d.c. compound motor
The exact shape of these characteristics depends on the relative contribution of series and
shunt field windings. If the shunt field winding is more dominant then the characteristics take the
shape of the shunt motor characteristics. While if the series field winding is more dominant then
the characteristics take the shape of the series characteristics.
Why Series Motor is Never Started on No Load?
It is seen earlier that motor armature current is decided by the load. On light load or no load, the
armature current drawn by the motor is very small.
In case of a d.c. series motor, Φ α Ia and
on no load as Ia is small hence flux produced is also very small.
According to speed equation,
N α 1/Φ as Eb is almost constant.
So on very light load or no load as flux is very small, the motor tries to run at dangerously
high speed which may damage the motor mechanically. This can be seen from the speedarmature current and the speed-torque characteristics that on low armature current and low
torque condition motor shows a tendency to rotate with dangerously high speed.
This is the reason why series motor should never be started on light loads or no load
conditions. For this reason it is not selected for belt drives as breaking or slipping of belt causes
to throw the entire load off on the motor and made to run motor with no load which is dangerous.
STARTING METHODS OF A DC MOTOR
Basic operational voltage equation of a DC motor is given as
E = Eb + IaRa and hence Ia = (E - Eb) / Ra
Now, when the motor is at rest, obviously, there is no back emf Eb, hence armature current will
be high at starting.
This excessive current will1. Blow out the fuses and may damage the armature winding and/or commutator brush
arrangement.
2. Produce very high starting torque (as torque is directly proportional to armature current),
and this high starting toque will produce huge centrifugal force which may throw off the
armature windings.
Thus to avoid above two drawbacks, starters are used for starting of DC machine.
Starting Methods of a DC Motor
Thus, to avoid the above dangers while starting a DC motor, it is necessary to limit the starting
current. For that purpose, starters are used to start a DC motor. There are various starters like, 3
point starter, 4 point starter, No load release coil starter, thyristor starter etc.
The main concept behind every DC motor starter is, adding external resistance to the armature
winding at starting.
3 POINT STARTER:
The internal wiring of a 3 point starter is as shown in the figure.
When motor is to be started, the lever is turned gradually to the right. When lever touches point
1, the field winding gets directly connected across the supply, and the armature winding gets
connected with resistances R1 to R5 in series. Hence at starting full resistance is added in series
with armature. Then as the lever is moved further, the resistance is gradually is cut out from the
armature circuit. Now, as the lever reaches to position 6, all the resistance is cut out from the
armature circuit and armature gets directly connected across the supply. The electromagnet E (no
voltage coil) holds the lever at this position. This electromagnet releases the lever when there is
no (or low) supply voltage.
When the motor is overloaded beyond a predefined value, over current release electromagnet D
gets activated, which short circuits electromagnet E , and hence releases the lever and motor is
turned off.
4 POINT STARTER:
The main difference between a 3 point starter and a 4 point starter is that the no voltage coil
is not connected in series with field coil. The field gets directly connected to the supply, as the
lever moves touching the brass arc. The no voltage coil (or Hold on coil) is connected with a
current limiting resistance Rh. This arrangement ensures that any change of current in the shunt
field does not affect the current through hold on coil at all. This means that electromagnet pull of
the hold-on coil will always be sufficient so that the spring does not unnecessarily restore the
lever to the off position.
This starter is used where field current is to be adjusted by means of a field rheostat.
2 POINT STARTER: (DC Series motor starter)
Construction of DC series motor starters is very basic as shown in the figure. A start
arm is simply moved towards right to start the motor. Thus at first maximum resistance is
connected in series with the armature and then gradually decreased as the start arm moves
towards right. The no load release coil holds the start arm to the run position and leaves it at no
load.
SPEED CONTROL METHODS OF DC MOTOR:
We know, back emf of a DC motor Eb is the induced emf due to rotation of the armature
in magnetic field. Thus value of the Eb can be given by the EMF equation of a DC generator.
Eb = PØNZ/60A
(where, P= no. of poles, Ø=flux/pole, N=speed in rpm, Z=no. of armature conductors, A=parallel
paths)
Eb can also be given as,
Eb = V- IaRa
thus from above equations
N=
E 60A
/PØZ
b
but, for a DC motor A, P and Z are constant
Nα K
E
b /Ø
(where, K=constant)
thus, it shows speed is directly proportional to back emf and inversely proportional to the flux
per pole.
SPEED CONTROL METHODS OF DC MOTOR
Speed Control of Shunt Motor
1. Flux Control Method
It is seen that speed of the motor is inversely proportional to flux. Thus by decreasing flux
speed can be increased and vice versa.
To control the flux, a rheostat is added in series with the field winding, as shown in the
circuit diagram. Adding more resistance in series with field winding will increase the speed, as it
will decrease the flux. Field current is relatively small and hence I2R loss is small, hence this
method is quiet efficient. Though speed can be increased by reducing flux with this method, it
puts a limit to maximum speed as weakening of flux beyond the limit will adversely affect the
commutation.
2. Armature Control Method
Speed of the motor is directly proportional to the back emf Eb and Eb = V- IaRa.
That is when supply voltage V and armature resistance Ra are kept constant, speed is
directly proportional to armature current Ia. Thus if we add resistance in series with armature,
Ia decreases and hence speed decreases.Greater the resistance in series with armature, greater the
decrease in speed.
3. Voltage Control Method
a) Multiple voltage control: In this method the, shunt filed is connected to a fixed exciting
voltage, and armature is supplied with different voltages. Voltage across armature is changed
with the help of a suitable switchgear. The speed is approximately proportional to the voltage
across the armature.
b) WARD-LEONARD SYSTEM:
Ward Leonard control system is introduced by Henry Ward Leonard in 1891. Ward Leonard
method of speed control is used for controlling the speed of a DC motor. It is a basic armature
control method. This control system is consisting of a dc motor M_1 and powered by a DC
generator G. In this method the speed of the dc motor (M_1) is controlled by applying variable
voltage across its armature. This variable voltage is obtained using a motor-generator set which
consists of a motor M_2(either ac or dc motor) directly coupled with the generator G. It is a very
widely used method of speed control of DC motor.
Principle of Ward Leonard Method
Basic connection diagram of the Ward Leonard speed control system is shown in the figure
below.
The speed of motor M1 is to be controlled which is powered by the generator G. The
shunt field of the motor M1 is connected across the dc supply lines. Now, generator G is driven
by the motor M2. The speed of the motor M2 is constant. When the output voltage of the
generator is fed to the motor M1 then the motor starts to rotate. When the output voltage of the
generator varies then the speed of the motor also varies. Now controlling the output voltage of
the generator the speed of motor can also be controlled. For this purpose of controlling the output
voltage, a field regulator is connected across the generator with the dc supply lines to control the
field excitation. The direction of rotation of the motor M1 can be reversed by excitation current
of the generator and it can be done with the help of the reversing switch R.S. But the motorgenerator set must run in the same direction.
Advantages of Ward Leonard System
1. It is a very smooth speed control system over a very wide range (from zero to normal
speed of the motor).
2. The speed can be controlled in both the direction of rotation of the motor easily.
3. The motor can run with a uniform acceleration.
4. Speed regulation of DC motor in this ward Leonard system is very good.
Disadvantages of Ward Leonard System
1. The system is very costly because two extra machines (motor-generator set) are required.
2. Overall efficiency of the system is not sufficient especially it is lightly loaded.
Application of Ward Leonard System
This Ward Leonard method of speed control system is used where a very wide and very
sensitive speed control is of a DC motor in both the direction of rotation is required. This speed
control system is mainly used in colliery winders, cranes, electric excavators, mine hoists,
elevators, steel rolling mills and paper machines etc.
Speed Control Of Series Motor
1. Flux Control Method
a) Field diverter :
A variable resistance is connected parallel to the series field as shown in fig. This variable
resistor is called as diverter, as desired amount of current can be diverted through this resistor
and hence current through field coil can be decreased. Hence flux can be decreased to desired
amount and speed can be increased.
b) Armature divertor:
Diverter is connected across the armature as in fig .
For a given constant load torque, if armature current is reduced then flux must increase. As,
Ta α ØIa
This will result in increase in current taken from the supply and hence flux Ø will increase and
subsequently speed of the motor will decrease.
c) Tapped field control:
As shown in fig field coil is tapped dividing number of turns. Thus we can select different value
of Ø by selecting different number of turns.
d) Paralleling field coils:
In this method, several speeds can be obtained by regrouping coils as shown in fig
2. Variable Resistance In Series With Armature
By introducing resistance in series with armature, voltage across the armature can be
reduced. And hence, speed reduces in proportion with it.
3. Series-Parallel Control
This system is widely used in electric traction, where two or more mechanically coupled
series motors are employed. For low speeds, motors are joined in series, and for higher speeds
motors are joined in parallel.
When in series, the motors have the same current passing through them, although voltage
across each motor is divided. When in parallel, voltage across each motor is same although
current gets divided.
REVIEW QUESTIONS
Construction details
1. Explain the constructional details of a DC Generator. Describe its working principle and
derive its emf equation.
(16) (N/D-12)
2. Explain in detail, the functions of various parts of a DC motor.
(16)
Emf equation
1. A six pole lap connected DC generator has a useful flux per pole of 0.045 Wb. If the no-load
voltage at 400 rpm is 300 V, find the conductors on the armature periphery.
(8) (N/D-09)
2. Derive the emf equation from fundamentals and also explain why the lap winding is preferred
compared to wave winding for low voltage high current machines.
(8)
3. Derive an expression for emf generated in a DC machine.
(8)
Self and separately excited generators
1. A separately excited generator, when running at 1000 rpm supplies 200 A at 125 V. What
will be the load current when the speed drops to 800 rpm, if the field current is unchanged?
Assume the armature resistance as 0.04  and the brush drop is 2 V.
(16) (N/D-09)
Characteristics of series, shunt and compound generators
1. Explain in detail, the characteristics of a DC generator with neat diagrams.
(16) (N/D-09)
2. A DC shunt generator has a terminal voltage of 160 V and a no-load induced emf of 168 V.
The resistances of armature and field are 0.03 Ω and 20 Ω respectively. Find the armature
current, field current and load current. Neglect armature reaction.
(6)
Principle of operation of a DC motor
1. Explain the principle of operation of a DC motor with relevant diagrams.
(8) (N/D-11)
Back emf and torque equation
1. Explain the significance of back emf.
(6)
2. Prove that the torque developed by a DC motor is proportional to flux and armature current.
(8)
Characteristics of series, shunt and compound motors
1. A 25 kW, 250 V, DC shunt generator has an armature and a field resistance of 0.06  and
100  respectively. Determine the total armature power developed during operating
condition.
(6) (A/M-10)
2. A 4 pole, 220 V DC shunt motor has 540 lap-wound conductors. It takes 32 A from the
supply mains and develops output power of 5.595 kW. The field winding takes 1 A. The
armature resistance is 0.9  and the flux per pole is 30 mWb. Calculate the speed in rpm and
the torque developed in Nm.
(6) (A/M-10)
3. Explain in detail, the different classifications of DC motors with respect to their equivalent
circuits.
(8) (N/D-09)
4. Estimate the percentage reduction in speed of a separately excited DC generator working with
constant excitation of 400 V, bus bar voltage is decreasing as its load varies from 600 kW to
400 kW. The machine resistance is 0.02 . Neglect the armature reaction.
(8) (A/M11)
5. The armature winding of a 200 V, 4 pole series motor is lap connected. There are 280 slots
with 4 conductors per slot. The current is 45 A and the flux per pole is 18 mWb. The field
resistance is 0.3 Ω, armature resistance 0.5 Ω and the total iron and the friction losses are 800
W. The pulley diameter is 0.41 m. Find the pull in newton at the rim of the pulley.
(10)
6. Explain the speed-current, speed-torque and torque-current characteristics of a DC shunt and
series motor.
(8)
Types of starters
1. Draw the schematic diagram of a 3 point starter and explain its working principle. (8)
(M/J-12)
2. Explain the operation of a 4 point starter with a neat diagram and also write its advantages
over a three point starter.
(10) (A/M-10)
Speed control of DC shunt motors
1. (i) A 4 pole DC motor runs at 600 rpm on full load and takes 25 A , 450 V the armature is lap
wound with conductors and flux per pole is given by the equation  = (1.7 X 10-2) x I0.5 Wb
where I is the motor current. If the supply voltage and torque are halved, calculate the speed
at which the motor will run. Neglect stray losses.
(8)
(ii) Explain the various types of speed control techniques of DC Machines.
(8) (M/J-12)
2. Explain the procedure to predetermine the efficiencies of a DC Machine as a motor and as a
generator.
(8) (N/D-12)
3. Explain in detail, the Ward-Leonard system of speed control of a DC motor.
4. Explain in detail, the different methods of speed control of a DC motor.
(16) (N/D-12)
UNIT II
TRANSFORMER
Introduction – Single phase transformer construction and principle of operation – EMF equation
of transformer-Transformer no–load phasor diagram –– Transformer on–load phasor diagram ––
Equivalent circuit of transformer – Regulation of transformer –Transformer losses and
efficiency-All day efficiency –auto transformers.
SINGLE PHASE TRANSFORMER
Working Principle of 1-Phase Transformer
1. Introduction
The main advantage of alternating currents over direct current is that, the alternating currents
can be easily transferable from low voltage to high voltage or high voltage to low. Alternating
voltages can be raised or lowered as per requirements in the different stages of electrical network
as generation, transmission, distribution and utilization. This is possible with a static device
called transformer. The transformer works on the principle of mutual induction. It transfer an
electric energy from one circuit to other when there is no electrical connection between the tow
circuits. Thus we can define transformer as below :
Key point : The transformer is a static piece of apparatus by means of which an electrical power
is transformed from one alternating current circuit to another with the desired change in voltage
and current, without any change in the frequency.
The use of transformers in transmission system is shown in the Fig 1.1.
Fig. 1.1 Use of transformer in transmission system
2. Principle of working
The principle of mutual induction states that when tow coils are inductively coupled and if
current in one coil is changed uniformly then an e.m.f. gets induced in the other coil. This e.m.f
can drive a current, when a closed path is provided to it. The transformer works on the same
principle. In its elementary form, it consists of tow inductive coils which are electrically
separated but linked through a common magnetic circuit. The two coils have high mutual
inductance. The basic transformer is shown in the Fig 1.2.
One of the two coils is connected to source of alternating voltage. This coil in which
electrical energy is fed with the help of source called primary winding (P). The other winding is
connected to load. The electrical energy transformed to this winding is drawn out to the load.
Fig.1.2 Basic transformer
Fig 1.3 Symbolic representation
This winding is called secondary winding (S). The primary winding has N1number of turns
while the secondary winding has N2 number of turns. Symbolically the transformer is indicated
as shown in the Fig 1.3.
When primary winding is excited by an alternating voltage, it circulates an alternating
current. This current produces an alternating flux (Φ)which completes its path through common
magnetic core as shown dotted in the Fig 1.2. Thus an alternating, flux links with the secondary
winding. As the flux is alternating, according to Faraday's law of an electromagnetic induction,
mutually induced e.m.f. gets developed in the secondary winding. If now load is connected to the
secondary winding, this e.m.f. drives a current through it.
Thus through there is no electrical contact between the two windings, an electrical energy
gets transferred from primary to the secondary.
Key point : The frequency of the mutual induced e.m.f. is same as that of the alternating source
which is supplying energy to the primary winding.
3. Can D.C. Supply be used for Transformer?
The d.c. supply can not be used for the transformers.
The transformer works on the principle of mutual induction, for which current in one coil
must change uniformly. If d.c. supply is given, the current will not change due to constant supply
and transformer will not work.
Practically winding resistance is very small. For d.c., the inductive reactance XL is zero as
d.c. has no frequency. So total impedance of winding is very low for d.c. Thus winding will draw
very high current if d.c. supply is given to it. This may cause the burning of windings due to
extra heat generated and may cause permanent damage to the transformer.
There can be saturation of the core due to which transformer draws very large current from
the supply when connected to d.c.
Thus d.c. supply should not be connected to the transformers.
CONSTRUCTION OF TRANSFORMER
There are two basic parts of a transformer i) Magnetic Core ii) Winding or Coils.
The core of the transformer is either square or rectangular in size. It is further divided into tow
parts. The vertical position on which coils are wound is called limb while the top and bottom
horizontal portion is called yoke of the core. These parts are shown in the Fig.1(a).
Core is made up of lamination. Because of laminated type of construction, eddy current
losses get minimised. Generally high grade silicon steel laminations (0.3 to 0.5 mm thick) are
used. These laminations are insulated from each other by using insulation like varnish. All
laminations are varnished. Laminations are overlapped so that to avoid the air gap at joints. For
this generally 'L' shaped or 'I' shaped laminations are used which are shown in the Fig 1(b).
Fig. 1 Construction of transformer
The cross-section of the limb depends on the type of coil to be used either circular or
rectangular. The different cross-section of limbs, practically used are shown in the Fig. 2.
Fig. 2 Different cross-sections
Types of Windings
The coils used are wound on the limbs and are insulated from each other. In the basic
transformer shown in the Fig 1.2 (see post : Working Principle of 1-Phase Transformer ) the two
windings wound are shown on two different limbs i.e. primary on one limb while secondary on
other limb. But due to this leakage flux increases which effects the transformer performance
badly. Similarly it is necessary that the windings should be very closes to each other to have high
mutual inductance. To achieve this, the tow windings are split into number of coils and are
wound adjacent to each other on the same limb. A very common arrangement is cylindrical coils
as shown in the Fig. 3.
Fig. 3 Cylindrical concentric coils
Such cylindrical coils are used in the core type transformer. Theses coils are mechanically
strong. These are wound in the helical layers. The different layers are insulated from each other
by paper, cloth or mica. The low voltage winding is placed near the core from ease of insulating
it from the core. The high voltage is placed after it.
The other type of coils which is very commonly used for the shell type of transformer is
sandwiching coils. Each high voltage portion lies between the two low voltage portion
sandwiching the high voltage portion. Such subdivision of windings into small portion reduces
the leakage flux. Higher the degree of subdivision, smaller is the reactance. The sandwich coil is
shown in the Fig. 4. The top and bottom coils are low voltage coils. All the portion are insulated
from each other by paper.
Fig. 4 Sandwich coils
CONSTRUCTION OF SINGLE PHASE TRANSFORMER
The various constructions used for the single phase transformers are,
1. Core type
2. shell type
and
3. Berry type
1. Core Type Transformer
It has a single magnetic circuit. The core rectangular having two limbs. The winding
encircles the core. The coils used are of cylindrical type. As mentioned earlier, the coils are
wound in helical layers with different layers insulated from each other by paper or mica. Both
the coils are placed on both the limbs. The low voltage coil is placed inside near the core while
high voltage coil surrounds the low voltage coil. Core is made up of large number of thin
laminations.
As The windings are uniformly distributed over the two limbs, the natural cooling is more
effective. The coils can be easily removed by removing the laminations of the top yoke, for
maintenance.
The Fig. 1(a) shows the schematic representation of the core type transformer while the Fig
1(b) shows the view of actual construction of the core type transformer.
Fig. 1 Core type transformer
2. Shell Type Transformer
It has a double magnetic circuit. The core has three limbs. Both the windings are placed on
the central limb. The core encircles most part of the windings. The coils used are generally
multilayer disc type or sandwich coils. As mentioned earlier, each high voltage coil is in between
tow low voltage coils and low voltage coils are nearest to top and bottom of the yokes.
The core is laminated. While arranging the laminations of the core, the care is taken that all
the joints at alternate layers are staggered. This is done to avoid narrow air gap at the joint, right
through the cross-section of the core. Such joints are called over lapped or imbricated joint.
Generally for very high voltage transformers, the shell type construction is preferred. As the
windings are surrounded by the core, the natural cooling does not exist. For removing any
winding for maintenance, large number of laimnations are required to be removed.
The Fig. 2(a) shows the schematic representation while the Fig. 2(b) shows the outaway
view of the construction of the shell type transformer.
Fig 2 Shell type transformer
3. Berry Type Transformer
This has distributed magnetic circuit. The number of independent magnetic circuits are more
than 2. Its core construction is like spokes of a wheel. Otherwise it is symmetrical to that of shell
type.
Diagramatically it can be shown as in the Fug. 3.
Fig. 3 Berry type transformer
The transformers are generally kept in tightly fitted sheet metal tanks. The tanks are
constructed of specified high quality steel plate cut, formed and welded into the rigid structures.
All the joints are painted with a solution of light blue chalk which turns dark in the presence of
oil, disclosing even the minutes leaks. The tanks are filled with the special insulating oil. The
entire transformer assembly is immersed in the oil. Oil serves two functions :i) Keeps the coil
cool by circulation and ii) Provides the transformers an additional insulation.
The oil should be absolutely free from alkalies, sulphur and specially from moisture.
Presence of very small moisture lowers the dielectric strength of oil, affecting its performance
badly. Hence the tanks are sealed air tight to avoid the contact of oil with atmospheric air and
moisture. In large transformers, the chambers called breather are provided. The breathers prevent
the atmospheric moisture to pass on to the oil. The breathers contain the silica gel crystal which
immediately absorb the atmospheric moisture. Due to long and continuous use, the sludge is
formed in the oil which can contaminate the oil. Hence to keep such sludge separate from the oil
in main tank, an air tight metal drum is provided, which is placed on the top of tank. This is
called conservator.
4. Comparison of Core and Shell Type Transformers
E.M.F EQUATION OF A TRANSFORMER
When the primary winding is excited by an alternating voltage V1, it circulates alternating
current, producing an alternating flux Φ. The primary winding has N1number of turns. The
alternating flux Φ linking with the primary winding itself induces an e.m.f in it denoted as E1.
The flux links with secondary winding through the common magnetic core. It produces induced
e.m.f. E2 in the secondary winding. This is mutually induced e.m.f. Let us derive the equations
for E1 and E2.
The primary winding is excited by purely sinusoidal alternating voltage. Hence the flux
produced is also sinusoidal in nature having maximum value of Φm as show in the Fig. 1.
Fig. 1 Sinusoidal flux
The various quantities which affect the magnitude of the induced e.m.f. are :
Φ = Flux
Φm = Maximum value of flux
N1 = Number of primary winding turns
N2 = Number of secondary winding turns
f = Frequency of the supply voltage
E1 = R.M.S. value of the primary induced e.m.f.
E2 = R.M.S. value of the secondary induced e.m.f.
From Faraday's law of electromagnetic induction the voltage e.m.f. induced in each turn is
proportional to the average rate of change of flux.
...
averagee.m.f. per turn = average rate of change of flux
...
averagee.m.f. per turn = dΦ/dt
Now
dΦ/dt = Change in flux/Time required for change in flux
Consider the 1/4 th cycle of the flux as shown in the Fig.1. Complete cycle gets completed in
1/f seconds. In 1/4 th time period, the change in flux is from 0 to Φm.
...
dΦ/dt = (Φm - 0)/(1/4f)
as dt for 1/4 th time period is 1/4f seconds
= 4 f Φm
Wb/sec
... Average e.m.f. per turn = 4 f Φm volts
As is sinusoidal, the induced e.m.f. in each turn of both the windings is also sinusoidal in
nature. For sinusoidal quantity,
From factor = R.M.S. value/Average value = 1.11
.
. . R.M.S. value of induced e.m.f. per turn
= 1.11 x 4 f Φm = 4.44 f Φm
There are number of primary turns hence the R.M.S value of induced e.m.f. of primary
denoted as is E1,
E1 = N1 x 4.44 f Φm volts
While as there are number of secondary turns the R.M.S values of induced e.m.f. of
secondary denoted is E2 is,
E2 = N2 x 4.44 f Φm
volts
The expression of E1 and E2 are called e.m.f. equation of a transformer.
Thus e.m.f. equations are,
E1 = 4.44 f Φm N1
volts
............(1)
E2 = 4.44 f Φm N2
volts
.............(2)
Ratios of a Transformer
Consider a transformer shown in Fig.1 indicating various voltages and currents.
Fig. 1 Ratios of transformer
1. Voltage Ratio
We known from the e.m.f. equations of a transformer that
E1 = 4.44 f Φm N1 and
E2 = 4.44 f Φm N2
Taking ratio of the two equations we
get,
This ratio of secondary induced e.m.f. to primary induced e.m.f. is known as voltage
transformation ratio denoted as K,
Thus,
1. If N2 > N1 i.e. K > 1, E2 > E1 we get then the transformer is called step-up transformer.
2. If N2 < N1 i.e. K < 1, we get E2 < E1 then the transformer is called step-down transformer.
3. If = i.e. K= 1, we get E2 = E1 then the transformer is called isolation transformer or 1:1
transformer.
2. Concept of Ideal Transformer
A transformer is said to be ideal if it satisfies following properties :
i) It has no losses.
ii) Its windings have zero resistance.
iii) Leakage flux is zero i.e. 100% flux produced by primary links with the secondary.
iv) Permeability of core is so high that negligible current is required to establish the flux in it.
Key point : For an ideal transformer, the primary applied voltage V1 is same as the primary
induced e.m.f. V2 as there are no voltage drops.
Similarly the secondary induced e.m.f. E2 is also same as the terminal voltage V2 across the
load. Hence for an ideal transformer we can write,
No transformer is ideal in practice but the value of E1 is almost equal to V1 for properly
designed transformer.
3. Current ratio
For an ideal transformer there are no losses. Hence the product of primary voltage V1 and
primary current I1, is same as the product of secondary voltage V2 and the secondary current I2.
So
V1 I1 = input VA
and
V2 I2 = output VA
For an ideal transformer,
V1 I1 = V2 I2
Key point : Hence the currents are in the inverse ratio of the voltage transformation ratio.
4. Voltage ampere rating
When electrical power is transferred from primary winding to secondary there are few power
losses in between. These power losses appear in the form of heat which increase the temperature
of the device.Now this temperature must be maintained below certain limiting values as it is
always harmful from insulation point of view. As current is the main cause in producing heat, the
output maximum rating is generally specified as the product of output voltage and output current
i.e.V2 I2. This always indicates that when transformer is operated under this specified rating, its
temperature rise will not be excessive. The copper loss (I2R) in the transformer depends on the
current 'I' through the winding while the iron or core loss depends on the voltage 'V' as frequency
of operation is constant. None of these losses depend on the power factor (cos Φ) of the load.
Hence losses decide the temperature and hence the rating of the transformer. As losses depend on
V and I only, the rating of the transformer is specified as a product of these two parameters VxI.
Key point : Thus the transformer rating is specified as the product of voltage and current called
VA rating.
On both sides, primary and secondary VA rating remains same. This rating is generally
expresses in KVA (kilo volt amperes rating).
Now
V1 /V2 = I2 /I1 = K
...
V1 I1 = V2 I2
If V1 and V2 are the terminal voltages of primary and secondary then from specified KVA
rating we can decide full load currents of primary and secondary, I1 and I2. This is the safe
maximum current limit which may carry, keeping temperature rise below its limiting value.
Key point : The full load primary and secondary currents indicate the safe maximum values of
currents which transformer windings can carry.
Ideal Transformer on No Load
Consider an ideal transformer on no load as shown in the Fig. 3. The supply voltage is and as it is
V1 an no load the secondary current I2 = 0.
The primary draws a current I1 which is just necessary to produce flux in the core. As it
magnetising the core, it is called magnetising current denoted as Im. As the transformer is ideal,
the winding resistance is zero and it is purely inductive in nature. The magnetising current is Im is
very small and lags V1 by 30o as the winding is purely inductive. This Im produces an alternating
flux Φ which is in phase with Im.
Fig. 1 Ideal transformer on no load
The flux links with both the winding producing the induced e.m.f.s E1 and E2 , in the primary
and secondary windings respectively. According to Lenz's law, the induced e.m.f. opposes the
cause producing it which is supply voltage V1. Hence E1 is in antiphase with V1 but equal in
magnitude. The induced E2 also opposes V1 hence in antiphase with V1 but its magnitude depends
on N2. Thus E1 and E2 are in phase.
The phasor diagram for the ideal transformer on no load is shown in the Fig. .2.
Fig. 2 Phasor diagram for ideal transformer on no load
It can be seen that flux Φ is reference. Im produces Φ hence in phase with Φ. V1leads Im by
90o as winding is purely inductive so current has to lag voltage by 90o.
E1 and E2 are in phase and both opposing supply voltage .
The power input to the transformer is V1 I1 cos (V1 ^ I1 ) i.e. V1 Im cos(90o) i.e. zero. This is
because on no load output power is zero and for ideal transformer there are no losses hence input
power is also zero. Ideal no load p.f. of transformer is zero lagging.
Practical Transformer on No Load
Actually in practical transformer iron core causes hysteresis and eddy current losses as it is
subjected to alternating flux. While designing the transformer the efforts are made to keep these
losses minimum by,
1. Using high grade material as silicon steel to reduce hysteresis loss.
2. Manufacturing core in the form of laminations or stacks of thin lamination to reduce eddy
current loss.
Apart from this there are iron losses in the practical transformer. Practically primary winding
has certain resistance hence there are small primary copper loss present.
Thus the primary current under no load condition has to supply the iron losses i.e. hysteresis
loss and eddy current loss and a small amount of primary copper loss. This current is denoted as
Io.
Now the no load input current Io has two components :
1. A purely reactive component Im called magnetising component of no load current required to
produce the flux. This is also called wattless component.
2. An active component Ic which supplies total losses under no load condition called power
component of no load current. This also called wattful component or core loss component of Io.
Thtotal no load current Io is the vector addition of Im and Ic.
In practical transformer, due to winding resistance, no load current Io is no longer at
90 with respect to V1. But it lags V1 by angle Φo which is less than 90o . Thus cos Φo is called no
load power factor of practical transformer.
o
Fig 1. Practical transformer on no load
The phasor diagram is shown in the Fig. 1. It can be seen that the two components Io are,
This is magnetising component lagging V1 exactly by 90o .
This is core loss component which is in phase withV1.
The magnitude of the no load current is given by,
While Φo = no load primary power factor angle
The total power input on no load is denoted as Wo and is given by,
It may be denoted that the current is very small, about 3 to 5% of the full load rated current.
Hence the primary copper loss is negligibly small hence Ic is called core loss or iron loss
component. Hence power input Wo on no load always represent the iron losses, as copper loss is
negligibly small. The iron losses are denoted as Pi and are constant for all load conditions.
Transformer on Load (M.M.F Balancing on Load)
When the transformer is loaded, the current I2 flows through the secondary winding. The
magnetic and phase of I2 is determined by the load. If load is inductive, I2 lags V2. If load is
capacitive, I2 leads V2 while for resistive load, I2 is in phase withV2.
There exists a secondary m.m.f. N2 I2 due to which secondary current sets up its own flux
Φ2. This flux opposes the main flux Φ which is produced in the core due to magnetising
component of no load current. Hence the m.m.f. is N2 I2 called demagnetising ampere-turns. This
is shown in the Fig.1(a).
The flux Φ2 momentarily reduces the main flux Φ, due to which the primary induced e.m.f.
also E1 reduces.
This additional current drawn by primary is due to the load hence called load component of
primary current denoted as I2' as shown in the Fig.1(b).
Fig. 1 Transformer on load
This current I2' is in antiphase with I2. The current sets up its own flux Φ2' which opposes
the flux Φ2 and helps the main fluxΦ. This flux Φ2' neutralises the flux Φ2produced by I2. The
m.m.f. i.e. ampere turns N2 I2' balances the ampere turns N2 I2. Hence the net flux in the core is
again maintained at constant level.
Key point : Thus for any load condition, no load to full load the flux in the core is practically
constant.
The load component current I2' always neutralises the changes in the loads. Hence the
transformer is called constant flux machine.
As the ampere turns are balanced we can write,
N2 I2=N2 I2'
...
I2' =(N2/N1) = K I2
..................(1)
Thus when transformer is loaded, the primary current I1 has two components :
1. The no load current Io which lags V1 by angle Φo. It has two components Im and Ic.
2. The load component I2' which in antiphase with I2. And phase of I2 is decided by the load.
Hence primary current I1 is vector sum of Io and I2'.
...
Ī1 = Īo + Ī2
...............(2)
Assume inductive load, I2 lags E2 by Φ2, the phasor diagram is shown in the Fig. 2(a).
Assume purely resistive load, I2 in phase with E2, the phasor diagram is shown in the
Fig.2(b).
Assume capacitive load, I2 leads E2 by Φ2, the phasor diagram is shown in the Fig. 2(c).
Note that I2' is always in antiphase with I2.
Fig. 2
Actually the phase of I2 is with respect to V2 i.e. angle Φ2 is angle between I2 and V2. For the
ideal case, E2 is assumed equal to V2 neglecting various drops.
The current ratio can be verified from this discussion. As the no load current Io is very small,
neglecting Io we can write,
I1 ~ I2'
Balancing the ampere turns,
N1 I1 = N1 I1 = N2 I2
.
..
N2 /N1 = I1 /I2 = K
Under full load conditions when Io is very small compared to full load currents, the ratio of
primary and secondary current is constant.
.
Effect of Winding Resistances
1. Effect OF Winding Resistances
A practical transformer windings process some resistances which not only cause the power
losses but also the voltage drops. Let us see what is the effect of winding resistance on the
performance of the transformer.
Let
R1 = primary winding resistance in ohms
R2 = secondary winding resistance in ohms
Now when current I1 flows through primary, there is voltage drop I1 R1 across the winding.
The supply voltage V1 has to supply this drop. Hence primary induced e.m.f. E1 is the vector
difference between V1 and I1 R1.
Similarly the induced e.m.f. in secondary is E2. When load is connected, current I2 flows and
there is voltage drop I2 R2. The e.m.f. E2 has to supply this drop. The vector difference between
E2 and I2 R2 is available to the load as a terminal voltage.
The drops I1 R1 and I2 R2 are purely resistive drops hence are always in phase with the respective
currents I1 and I2.
1.1 Equivalent Resistance
The resistance of the two windings can be transferred to any one side either primary or
secondary without affecting the performance of the transformer. The transfer of the resistances
on any one side is advantageous as it makes the calculations very easy. Let us see how to transfer
the resistances on any one side.
The total copper loss due to both the resistances can be obtained as,
total copper loss = I12 R1 + I22 R2
= I12{ R1 +(I22/I12) R2}
= I12{R1 + (1/K2) R2}
.......(3)
Where
I2/I1 = 1/K
neglecting no load current.
Now the expression (3) indicates that the total copper loss can be expressed as I12 R1 +
I12 .R2/K2. This means R2/K2 is the resistance value of R2 shifted to primary side which causes
same copper loss with I1 as R2 causes with. This value of resistance which R2 /K2 is the value of
R2 referred to primary is called equivalent resistance of secondary referred to primary. It is
denoted as R2'.
R2' = R2 /K2
........(4)
Hence the total resistance referred to primary is the addition of R1 and R2' called equivalent
resistance of transformer referred to primary and denoted as R1e.
= R1 + R2'= R1 + R2 /K2
.........(5)
This resistance R1e causes same copper loss with I1 as the total copper loss due to the
individual windings.
total copper loss = I12 R1e = I12 R1 + I22 R2
......(6)
So equivalent resistance simplifies the calculations as we have to calculate parameters on
one side only.
Similarly it is possible to refer the equivalent resistance to secondary winding.
total copper loss = I12 R1 + I22 R2
= I22 {(I12/I22) R1 + R2}
= I22 ( K2 R1 + R2)
........(7)
Thus the resistance K2 R1 is primary resistance referred to secondary denoted as R1'.
R1' = K2 R1
.......(8)
Hence the total resistance referred to secondary is the addition of R2 and R1'called equivalent
resistance of transformer referred to secondary and denoted as R2e.
R2e = R2 + R1' = R2 + K2 R1
.........(9)
2
total copper loss = I2 R2e
........(10)
The concept of equivalent resistance is shown in the Fig. 1(a), (b) and (c).
Fig. 1 Equivalent resistance
Key Point : When resistance are transferred to primary, the secondary winding becomes zero
resistance winding for calculation purpose. The entire copper loss occurs due to R1e. Similarly
when resistances are referred to secondary, the primary becomes resistanceless for calculation
purpose. The entire copper loss occurs due to R2e.
Important Note : When a resistance is to be transferred from the primary to secondary, it must
be multiplied by K2. When a resistance is to be transferred from the secondary to primary, it
must be divided by K2. Remember that K is N1 /N2.
The result can be cross-checked by another approach. The high voltage winding is always
low current winding and hence the resistance of high voltage side is high. The low voltage side is
high current side and hence resistance of low voltage side is low. So while transferring resistance
from low voltage side to high voltage side, its value must increase while transferring resistance
from high voltage side to low voltage side, its value must decrease.
Key point :
High voltage side → Low current side → High resistance side
Low voltage side → High current side → Low resistance side
Effect of Leakage Reactance
1. Effect of Leakage Reactance
Uptill now it is assumed that the entire flux produced by the primary links with the
secondary winding. But in practice it is not possible. Part of the primary flux as well as the
secondary flux completes the path through air and links with the respecting winding only. Such a
flux is called leakage flux. Thus there are two leakage fluxes present as shown in the Fig. 1.
Fig .1 Individual impedance
The flux ΦL1 is the primary leakage flux which is produced due to primary current I1. It is in
phase with I1 and links with primary only.
The flux ΦL2 is the secondary leakage flux which is produced due to current I2. It is in phase
with I2 and links with the secondary winding only.
Due to the leakage flux ΦL1 there is self inducede.m.f. eL1 in primary. While due to leakage
flux ΦL2 there is self inducede.m.f. eL2 in secondary. The primary voltage V1 has to overcome
this voltage eL1 to produce E1 while induced e.m.f. E2 has to overcome eL2 to produce terminal
voltage V2. Thus the self inducede.m.f.sare treated as the voltage drops across the fictitious
reactance placed in series with the windings. These reactances are called leakage reactance of the
winding.
So
X1 = Leakage reacatnce of primary winding.
and
X2 = Leakage reactance of secondary winding.
The value of X1 is such that the drop I1 X1 is nothing but the self inducede.m.f. eL1 due to
fluxΦL1. The value of X2 is such that the drop I2 X2 is equal to the self inducede.m.f. eL2 due to
flux ΦL1.
Leakage fluxes with the respective windings only and not to both the windings. To reduce
the leakage, as mentioned, int eh construction both the winding's are placed on same limb rather
than on separate limbs.
1.1 Equivalent Leakage Reactance
Similar to the resistances, the leakage reactances also can be transferred from primary to
secondary or viceversa. The relation through K2 remains same for the transfer of recatnaces as it
is studied earlier for the resistances.
Let X1 is leakage reactance of primary and X2 is leakage reactance of secondary.
Then the total leakage reacatance referred to primary is X1e given by,
X1e = X1 + X2' where X2' = X2/K2
While the total leakage reacatnce referred to secondary is given by ,
X2e = X2 + X1' where X1' = K2 X1
And
K = N2/N1 =transformation ratio
Equivalent Impedance
The transformer primary has resistance R1 and reactance X1. While the transformer
secondary has resistance R2 and reacatnce X2. Thus we can say that the total impedance of
primary winding is Z1 which is,
Z1 = R1 + j X1 Ω
..........(1)
And the total impedance of the secondary winding is which is ,
Z2 = R2 + j X2 Ω
...........(2)
This is shown in the Fig. 1.
Fig. 1 Individual impedance
The individual magnitudes of and are,
Z1 = √(R12 + X12)
...........(3)
2
2
and
Z2 = √(R2 + X2 )
.........(4)
Similar to resistance and reactance, the impedance also can be referred to any one side.
Let
Z1e = total equivalent impedance referred to primary
then
Z1e = R1e + j X1e
Z1e = Z1 + Z2' = Z1 + Z2/K2
............(5)
Similarly
Z2e = total equivalent impedance referred to secondary
then
Z2e = R2e + j X2e
Z2e = Z2 + Z1' = Z2 + K2 Z1
............(6)
The magnitude of Z1e and Z2e are,
Z1e = √(R1e2 + X1e2)
........(7)
and
Z2e = √(R2e2 + X2e2)
...........(8)
It can be denoted that,
Z2e = K2 Z1e
and Z1e = Z2e /K2
The concept of equivalent impedance is shown in the Fig. 2.
Fig 2 Equivalent impedance
Phasor Diagrams for Transformer on Load
Phasor Diagrams for Transformer on Load
Consider a transformer supplying the load as shown in the Fig. 1.
Fig. 1
The various transformer parameters are,
R1 = Primary winding resistance
X1 = Primary leakage reactance
R2 = Secondary winding resistance
X2 = Secondary leakage reactance
ZL = Load impedance
I1= Primary current
I2 = Secondary current = IL = Load current
now
Ī1 = Īo + Ī2'
where
Io = No load current
I2'= Load component of current decided by the load
= K I2 where K is transformer component
The primary voltage V1 has now three components,
1. -E1, the induced e.m.f. which opposes V1
2. I1 R1, the drop across the resistance, in phase with I1
3. I1 X1, the drop across the reactance, leading I1 by 90o
The secondary induced e.m.f. has also three components,
1. V2, the terminal voltage across the load
2.I2 R2, the drop across the resistance, in phase with I2
3. I2 X2, the drop across the reactance, leading I2 by 90o
The phasor diagram for the transformer on load depends on the nature of the load power
factor. Let us consider the various cases of the load power factor.
1.1 Unity power factor load, cosΦ2 = 1
As load power factor is unity, the voltage V2 and I2 are in phase. Steps to draw the phasor
diagram are,
1. Consider flux Φ as reference
2. E1 lags Φ by 90o. Reverse E1 to get -E1.
3. E1 and E2 are inphase
4. Assume V2 in a particular direction
5. I2 is in phase with V2.
6. Add I2 R2 and I2 X2 to to get E2.
7. Reverse I2 to get I2'.
8. Add Io and I2' to get I1.
9. Add I1 R1 and to -E1 to get V1.
Angle between V1 and I1 is Φ1 and cosΦ1 is primary power factor. Remember that I1X1 leads
I1 direction by 90o and I2 X2 leads I2 by 90o as current through inductance lags voltage across
inductance by 90o. The phasor diagram is shown in the Fig.2
Fig. 2 Phasor diagram for unity power factor load
Lagging Power Factor Load, cos Φ2
As load power factor is lagging cosΦ2, the current I2 lags V2 by angle Φ2. So only changes in
drawing the phasor diagram is to draw I2 lagging V2 by Φ2 in step 5 discussed earlier.
Accordingly direction of I2 R2, I2 X2, I2', I1, I1 R1 and I1X1 will change. Remember that whatever
may be the power factor of load, I2X2 leads I2 by 90o and I1X1 leads I1 by 90o.
The complete phasor diagram is shown in the Fig. 3.
Fig. 3 Phasor diagram for lagging power factor
Loading Power Factor Load, cos Φ2
As load power factor is leading, the current I2 leads V2 by angle Φ2. So change is to draw
I2 leading I2 by angle Φ2. All other steps remain same as before. The complete phasor diagram is
shown in the Fig. 4
Fig. 4 Phasor diagram for leading power factor
Equivalent circuit of Transformer
1. Equivalent circuit of Transformer
The term equivalent circuit of a machine means the combination of fixed and variable
resistances and reactances, which exactly simulates performance and working of the machine.
For a transformer, no load primary current has two components,
Im = Io sinΦo = Magnetizing component
Ic = Io cosΦo = Active component
Im produces the flux and is assumed to flow through reactance Xo called no load reractance
while Ic is active component representing core losses hence is assumed to flow through the
reactance Ro. Hence equivalent circuit on no load can be shown as in the Fig. 1. This circuit
consisting of Ro and Xo in parallel is called exciting circuit. From the equivalent circuit we can
write,
Ro = V1/Ic
and Xo= V1/Im
Fig. 1 No load equivalent circuit
When the is connected to the transformer then secondary current I2 flows. This causes
voltage drop across R2 and R2. Due to I2, primary draws an additional current
I2' = I2/ K. Now I1 is the phasor addition of Io and I2'. This I1 causes the voltage drop across
primary resistance R1 and reactance X1.
Hence the equivalent circuit can be shown as in the Fig. 2.
Fig. 2
But in the equivalent circuit, windings are not shown and it is further simplified by
transferring all the values to the primary or secondary. This makes the transformer calculation
much easy.
So transferring secondary parameters to primary we get,
R2'= R2/K2 ,
X2' = X2/K2' ,
Z2' = Z2/K2
While
E2' = E2/K'
I2' = K I2
Where
K = N2 /N1
While transferring the values remember the rule that
Low voltage winding High current Low impedance
High voltage winding Low current High impedance
Thus the exact equivalent circuit referred to primary can be shown as in the Fig. 3.
Fig. 3 Exact equivalent circuit referred to primary
Similarly all the primary value can be referred to secondary and we can obtain the equivalent
circuit referred to secondary.
R1' = K2 R1 ,
X1' = K2 X1,
Z1' = K2 Z1
E1'= K E1,
Io' = I1 /K' Io' = Io /K
Similarly the exciting circuit parameters also gets transferred to secondary as R o'and Xo '.
The circuit is shown in the Fig.4.
Fig. 4 Exact equivalent circuit referred to secondary
Now as long as no load branch i.e. exciting branch is in between Z1 and Z2', the impedances
can not be combined. So further simplification of the circuit can be done. Such circuit is called
approximate equivalent circuit.
1.1 Approximate Equivalent Circuit
To get approximate equivalent circuit, shift the no load branch containing Ro and Xo to the
left of R1 and X1. By doing this we are creating an error that the drop across R 1 and X1due to Io is
neglected. Hence such an equivalent circuit is called approximate equivalent circuit.
So approximate equivalent circuit referred to primary can be as shown in the Fig. 5.
Fig. 5 Approximate equivalent circuit referred to primary
In this circuit now R1 and R2' can be combined to get equivalent resistance referred to
primary R1e as discussed earlier. Similarly X1and X1' can be combined to get X1e. And equivalent
circuit can be simplified as shown in the Fig. 6.
Fig. 6
We know that, R1e = R1 + R2'= R1 + R2/K2
X1e = X1 + X2' = X1 + X2/K2
Z1e = R1e + j X1e
Ro = V1 /Ic and Xo = V1 /Im
Ic = Io cosΦo and Im = Io sinΦo
In the similar fashion, the approximate equivalent circuit referred to secondary also can be
obtained.
Voltage Regulation of Transformer
1. Voltage Regulation of Transformer
Because of the voltage drop across the primary and secondary impedances it is observed that
the secondary terminal voltage drops from its no load value (E2) to load value (V2) as load and
load current increases.
This decrease in the secondary terminal voltage expressed as a fraction of the no load
secondary terminal voltage is called regulation of a transformer.
The regulation is defined as change in the magnitude of the secondary terminal voltage,
when full load i.e. rated load of specified power factor supplied at rated voltage is reduced to no
load, with primary voltage maintained constant expressed as the percentage of the rated terminal
voltage.
Let
E2 = Secondary terminal voltage on no load
V2 = Secondary terminal voltage on given load
then mathematically voltage regulation at given load can be expressed as,
The ratio (E2 - V2 / V2 ) is called per unit regulation.
The secondary terminal voltage does not depend only on the magnitude of the load current
but also on the nature of the power factor of the load. If V2 is determined for full load and
specified power factor condition the regulation is called full load regulation.
As load current increases, the voltage drops tend to increase V2 and drops more and more. In
case of lagging power factor V2 < E2 and we get positive voltage regulation, while for leading
power factor E2 < V2 and we get negative voltage regulation.
The voltage drop should be as small as possible hence less the regulation better is the
performance of a transformer.
1.1 Expression for Voltage Regulation
The voltage regulation is defined as,
%R = (E2 - V2 /V2 ) x 100 = (Total voltage drop/V2) x 100
The expression for the total approximate voltage drop is already derived.
Total voltage drop = I2 R2e cos Φ ± I2 X2e sin Φ
Hence the regulation can be expressed as,
'+' sing for lagging power factor while '-' sing for leading power factor loads.
The regulation van be further expressed interms of I1 , V1, R1e and X1e.
V2 /V1 =I1 /I2 = K
...
V2 = KV1 , I2 = I1/K
while
R1e =R2e/K2 , X1e = X2e /K2
Substituting in the regulation expression we get,
1.2 Zero Voltage Regulation
We have seen that for lagging power factor and unity power factor condition V2 < E2 and we
get positive regulation. But as load becomes capacitive, V2 starts increasing as load increase. At a
certain leading power factor we get E2 = V2 and the regulation becomes zero. If the load is
increased further, E2 becomes less than V2and we get negative regulation.
...
for zero voltage regulation,
E2 = V2
...
E2 - V2 = 0
or
VR cos Φ - Vx sin Φ = 0 .......... -ve sing as leading power factor
where VR = I2 R2e /V2 = I1 R1e /V1
and Vx = I2 X2e /V2 = I1 X1e /V1
...
VR cos Φ = Vx sin Φ
...
tan Φ = VR /Vx
...
cos Φ = cos {tan-1(VR /Vx)}
This is the leading p.f. at which voltage regulation becomes zero while supplying the load.
1.3 Constants of a Transformer
From the regulation expression we can define constants of a transformer.
%R= (( I2 R2e cos Φ ± I2 X2e sin Φ )/ E2) x 100
= {(I2 R2e /E2) cos Φ ± (I2 X2e/E2 ) sin Φ} x 100
The ratio (I2 R2e /E2) or (I1 R1e /E1) is called per unit resistive drop and denoted as VR.
The ratio (I2 X2e/E2 ) or (I1 X1e/E1) is called per unit reactive drop and is denoted as Vx.
The terms VR and Vx are called constants of a transformer because for the rated output I2,
E2, R1e, X1e, R2e , X2e are constants. The regulation can be expressed interms of VR and Vx as,
%R = (VR cos Φ
iuu ±Vx sin Φ ) x 100
On load condition, E2 = V2 and E1= V1
where V1 and V2 are the given voltage ratings of a transformer. Hence VR and Vx can be
expressed as,
VR = I2 R2e/ V2 = I1 R1e/ V1
and
Vx =I2 R2e/ V2 = I1 X1e/ V1
where V1and V2 are no load primary and secondary voltages,
VR and Vx can be expressed on percentage basis as,
Percentage resistive drop = VR x 100
Percentage reactive drop = Vx x 100
Key Point : Note that and are also called per unit resistance and reactance respectively.
LOSSES IN A TRANSFORMER
1. Losses in a Transformer
In a transformer, there exists two types of losses.
i) The core gets subjected to an alternating flux, causing core losses.
ii) The windings carry currents when transformer is loaded, causing copper losses.
1.1 Core or Iron Losses
Due to alternating flux set up in the magnetic core of the transformer, it undergoes a cycle of
magnetisation and demagnetisation. Due to hysteresis effect there is loss of energy in this
process which is called hysteresis loss.
It is given by,
hysteresis loss = Kh Bm1.67 f v watts
where
Kh = Hysteresis constant depends on material.
Bm = Maximum flux density.
f = Frequency.
v = Volume of the core.
The induced e.m.f. in the core tries to set up eddy currents in the core and hence responsible
for the eddy current losses. The eddy current loss is given by,
Eddy current loss = Ke Bm2 f2 t2 watts/ unit volume
where
Ke = Eddy current constant
t = Thickness of the core
As seen earlier, the flux in the core is almost constant as supply voltage V1 at rated
frequency f is always constant. Hence the flux density Bm in the core and hence both hysteresis
and eddy current losses are constants at all the loads. Hence the core or iron losses are also called
constant losses. The iron losses are denoted as Pi.
The iron losses are minimized by using high grade core material like silicon steel having very
low hysteresis loop by manufacturing the core in the form of laminations.
1.2 Copper Losses
The copper losses are due to the power wasted in the form of I2 R loss due to the resistances
of the primary and secondary windings. The copper loss depends on the magnitude of the
currents flowing through the windings.
Total Cu loss = I12 R1 + I22 R2 = I12 ( R1 + R2' )= I22 ( R2 +R1' )
= I12 R1e = I22 R2e
The copper looses are denoted as. If the current through the windings is full load current, we
get copper losses at full load. If the load on transformer is half then we get copper losses at half
load which are less than full load copper losses. Thus copper losses are called variable losses.
For transformer VA rating is or. As is constant, we can say that copper losses are proportional to
the square of the KVA rating.
So, Pcu α I2α (KVA)2
Thus for a transformer,
Total losses = Iron losses + Copper losses
= Pi +Pcu
Key point : It is seen that the iron losses depend on the supply voltage while the copper losses
depend on the current. The losses are not dependent on the phase angle between voltage and
current. Hence the rating of the transformer is expressed as a product of voltage and current and
called VA rating of transformer. It is not expressed in watts or kilo watts. Most of the times,
rating is expressed in KVA.
EFFICIENCY OF A TRANSFORMER
Efficiency of a Transformer
Due to the losses in a transformer, the output power of a transformer is less than the input
power supplied.
...
Power output = Power input - Total losses
.
..
Power input = Power output + Total losses
= Power output + Pi +Pcu
The efficiency of any device is defined as the ratio of the power output to power input. So
for a transformer the efficiency can be expresses as,
η = Power output/power input
...
η = Power output/(power output + Pi + Pcu )
Now power output = V2 I2 cos Φ
where
cos Φ = Load power factor
The transformer supplies full load of current I2 and with terminal voltage V2.
Pcu = Copper losses on full load = I22 R2e
.
..
η = (V2 I2 cos Φ2 )/(V2 I2 cos Φ2 + Pi + I22 R2e)
But
V2 I2 = VA rating of a transformer
...
η = (VA rating x cos Φ) / (VA rating x cos Φ + Pi + I22 R2e)
This is full load percentage efficiency with,
I2 = Full load secondary current
But if the transformer is subjected to fractional load then using the appropriate values of
various quantities, the efficiency can be obtained.
Let n =Fraction by which load is less than full load = Actual load/Full load
For example, if transformer is subjected to half load then,
n = Half load/Full load = (1/2)/2 = 0.5
when load changes, the load current changes by same proportion.
.
. . new I2 = n (I2) F.L.
Similarly the output V2 I2 cosΦ2 also reduces by the same fraction. Thus fraction of VA
rating is available at the output.
Similarly as copper losses are proportional to square of current then,
new Pcu = n2 (Pcu ) F.L.
Key Point : So copper losses get reduced by n2.
In general for fractional load the efficiency is given by,
where n = Fraction by which load power factor lagging, leading and unity the efficiency
expression does not change, and remains same.
1. Condition for Maximum Efficiency
When a transformer works on a constant input voltage and frequency then efficiency varies
with the load. As load increases, the efficiency increases. At a certain load current, it achieves a
maximum value. If the transformer is loaded further the efficiency starts decreasing. The graph
of efficiency against load current I2 is shown in the Fig.1
Fig. 1
The load current at which the efficiency attains maximum value is denoted as I2mand
maximum efficiency is denoted as ηmax.
Let us determine,
1. Condition for maximum efficiency.
2. Load current at which ηmax occurs.
3. KVA supplied at maximum efficiency.
The efficiency is a function of load i.e. load current I2 assuming cos Φ constant. The
secondary terminal voltage V2 is also assumed constant. So for maximum efficiency,
Now
dη /d I2 = 0
η = (V2 I2 cos Φ2 )/(V2 I2 cos Φ2 + Pi + I22 R2e)
(V2 I2 cos Φ2 + Pi + I22 R2e)(V2 cos Φ2) - (V2 I2 cos Φ2)(V2 cos Φ2 + 2I2 R2e) = 0
Cancelling (V2 cos Φ2) from both the terms we get,
V2 I2 cos Φ2 + Pi +I22 R2e - V2 I2 Φ2 - 2I22 R2e = 0
...
Pi - I22 R2e= 0
.
..
Pi = I22 R2e = Pcu
So condition to achieve maximum efficiency is that,
Copper losses = Iron losses
1.1 Load Current I2m at Maximum Efficiency
For ηmax,
I22 R2e = Pi but I2 = I2m
I2m2 R2e = Pi
I2m = √(Pi / R2e)
This is the load current at ηmax,
Let
(I2)F.L. = Full load current
...
I2m /(I2) F.L.= (1/(I2) F.L.)√(Pi / R2e)
...
I2m /(I2) F.L.= √(Pi )/({(I2) F.L.}2 R2e)
= √(Pi )/((Pcu) F.L.)
.
..
I2m = (I2 )F.L.√(Pi )/((Pcu) F.L.)
This is the load current at ηmax interms of full load current.
1.2 KVA supplied at maximum Efficiency
For constant V2 the KVA supplied is the function of.
KVA at ηmax = I2m V2= V2 (I2) F.L. x √(Pi) /((Pcu)F.L.)
KVA at ηmax = (KVA rating) x √(Pi) /((Pcu)F.L.)
Substituting condition for in the expression of efficiency, we can write expression for ηmax as,
...
ii)
KVA at = KVA rating x √(Pi) /((Pcu)F.L.)
Effect of Power factor on Efficiency
The efficiency of a transformer is given by,
Now,
input = output + losses = V2 I2 cos Φ + losses
Using in (1), η = 1 - (losses / (V2 I2 cos Φ + losses))
Let
losses/V2 I2 = x and using in (2),
Thus as the power factor of the load is more i.e. cos is higher, x/cosΦ is lesser. Hence the
second term in the equation (3) becomes lesser and efficiency will be more.
Key Point : As power factor increases, the efficiency increases.
Thus the family of efficiency curves are obtained as power factor increases, as shown in the
Fig.1.
Fig. 1 Effect of p.f. on efficiency
Effect of Frequency and Supply Voltage on Iron Losses
Effect of Frequency and Supply Voltage on Iron Losses
The iron losses of a transformer includes two types of losses,
1. Hysteresis loss
and
2. Eddy current loss
For a given volume and thickness of laminations, these losses depend on the operating
frequency, maximum flux density and the voltage.
It is known that for a transformer,
V = 4.44 f Φm N = 4.44 f Bm A N
Where
A = area
.
..
Bm α (V/f)
.......... For constant A and N
Thus as voltage changes, the maximum flux density changes and both eddy current and
hysteresis losses also changes. As voltage increases, the maximum flux density in the core
increases and total iron loss increases.
As frequency increases, the flux density in the core decreases but as the iron loss is directly
proportional to the frequency hence effect of increased frequency is to increase the iron losses.
Key Point : Thus iron loss increases as the voltage and frequency increases for the transformer.
ALL DAY EFFICIENCY OF A TRANSFORMER
For a transformer, the efficiency is defined as the ratio of output power to input power. This
is its efficiency. But power efficiency is not the true measure of the performance of some special
types of transformers such as distribution transformers.
Distribution transformer serve residential and commercial loads. The load on such
transformers vary considerably during the period of the day. For most period of the day these
transformers are working at 30 to 40 % of full load only or even less than that. But the primary
of such transformers is energised at its rated voltage for 24 hours, to provide continuous supply
to the consumer. The core loss which depends on voltage, takes place continuously for all the
loads. But copper loss depends on the load condition. For no load, copper loss is negligibly small
while on full load it is at its rated value. Hence power efficiency can not give the measure of true
efficiency of such transformers. in such transformers, the energy output is calculated in kilo watt
hour (kWh). Then ratio of total energy output to total energy input (output + losses) is calculated.
Such ratio is called energy efficiency or All Day Efficiency of a transformer. Based on this
efficiency, the performance of various distribution transformers is compared. All day efficiency
is defined as,
While calculating energies, all energies can be expressed in watt hour (Wh) instead of kilo
watt hour (kWh).
Such distribution transformers are designed to have very low core losses. This is achieved
by limiting the core flux density to lower value by using a relative higher core cross-section i.e.
larger iron to copper weight ratio. The maximum efficiency in such transformers occurs at about
60-70 % of the full load. So by proper designing, high energy efficiencies can be achieved for
distribution transformers.
The calculation of all day efficiency for a transformer are illustrated in the Ex. 1.
AUTOTRANSFORMER:
An Autotransformer has only one single voltage winding which is common to both
sides. This single winding is “tapped” at various points along its length to provide a percentage
of the primary voltage supply across its secondary load. Then the autotransformer has the usual
magnetic core but only has one winding, which is common to both the primary and secondary
circuits.
Therefore in an autotransformer the primary and secondary windings are linked together
both electrically and magnetically. The main advantage of this type of transformer design is that
it can be made a lot cheaper for the same VA rating, but the biggest disadvantage of an
autotransformer is that it does not have the primary/secondary winding isolation of a
conventional double wound transformer.
The section of winding designated as the primary part of the winding is connected to the
AC power source with the secondary being part of this primary winding. An autotransformer can
also be used to step the supply voltage up or down by reversing the connections. If the primary is
the total winding and is connected to a supply, and the secondary circuit is connected across only
a portion of the winding, then the secondary voltage is “stepped-down” as shown.
Autotransformer Design
When the primary current IP is flowing through the single winding in the direction of the
arrow as shown, the secondary current, IS, flows in the opposite direction. Therefore, in the
portion of the winding that generates the secondary voltage, VS the current flowing out of the
winding is the difference of IP and IS.
The Autotransformer can also be constructed with more than one single tapping point.
Auto-transformers can be used to provide different voltage points along its winding or increase
its supply voltage with respect to its supply voltage VP as shown.
Autotransformer with Multiple Tapping Points
The standard method for marking an auto-transformer windings is to label it with capital (upper
case) letters. So for example, A, B, Z etc to identify the supply end. Generally the common
neutral connection is marked as N or n. For the secondary tapping’s, suffix numbers are used for
all tapping points along the auto-transformers primary winding. These numbers generally start at
number 1 and continue in ascending order for all tapping points as shown.
Autotransformer Terminal Markings
An autotransformer is used mainly for the adjustments of line voltages to either change
its value or to keep it constant. If the voltage adjustment is by a small amount, either up or down,
then the transformer ratio is small as VP and VS are nearly equal. Currents IP and IS are also
nearly equal.
Therefore, the portion of the winding which carries the difference between the two
currents can be made from a much smaller conductor size, since the currents are much smaller
saving on the cost of an equivalent double wound transformer.
However, the regulation, leakage inductance and physical size (since there is no second
winding) of an autotransformer for a given VA or KVA rating are less than for a double wound
transformer.
Autotransformer’s are clearly much cheaper than conventional double wound
transformers of the same VA rating. When deciding upon using an autotransformer it is usual to
compare its cost with that of an equivalent double wound type.
This is done by comparing the amount of copper saved in the winding. If the ratio “n” is
defined as the ratio of the lower voltage to the higher voltage, then it can be shown that the
saving in copper is: n.100%. For example, the saving in copper for the two autotransformer’s
would be:
The Autotransformer have many advantages over conventional double wound
transformers. They are generally more efficient for the same VA rating, are smaller in size, and
as they require less copper in their construction, their cost is less compared to double wound
transformers of the same VA rating. Also, their core and copper losses, I2R are lower due to less
resistance and leakage reactance giving a superior voltage regulation than the equivalent two
winding transformer.
Disadvantages of an Autotransformer


The main disadvantage of an autotransformer is that it does not have the primary to
secondary winding isolation of a conventional double wound transformer. Then
autotransformer’s can not safely be used for stepping down higher voltages to much
lower voltages suitable for smaller loads.
If the secondary side winding becomes open-circuited, current stops flowing through the
primary winding stopping the transformer action resulting in the full primary voltage
being applied to the secondary circuit.


If the secondary circuit suffers a short-circuit condition, the resulting primary current
would be much larger than an equivalent double wound transformer due to the increased
flux linkage damaging the autotransformer.
Since the neutral connection is common to both the primary and secondary windings,
earthing of the secondary winding automatically earths the primary as there is no
isolation between the two windings. Double wound transformers are sometimes used to
isolate equipment from earth.
The autotransformer has many uses and applications including the starting of induction
motors, used to regulate the voltage of transmission lines, and can be used to transform voltages
when the primary to secondary ratio is close to unity.An autotransformer can also be made from
conventional two-winding transformers by connecting the primary and secondary windings
together in series and depending upon how the connection is made, the secondary voltage may
add to, or subtract from, the primary voltage.
The Variable Autotransformer
As well as having a fixed or tapped secondary that produces a voltage output at a specific
level, there is another useful application of the auto transformer type of arrangement which can
be used to produce a variable AC voltage from a fixed voltage AC supply. This type of Variable
Autotransformer is generally used in laboratories and science labs in schools and colleges and
is known more commonly as the Variac.
The construction of a variable autotransformer, or variac, is the same as for the fixed
type. A single primary winding wrapped around a laminated magnetic core is used as in the auto
transformer but instead of being fixed at some predetermined tapping point, the secondary
voltage is tapped through a carbon brush.This carbon brush is rotated or allowed to slide along
an exposed section of the primary winding, making contact with it as it moves supplying the
required voltage level.
Then a variable autotransformer contains a variable tap in the form of a carbon brush that
slides up and down the primary winding which controls the secondary winding length and hence
the secondary output voltage is fully variable from the primary supply voltage value to zero
volts.
The variable autotransformer is usually designed with a significant number of primary
windings to produce a secondary voltage which can be adjusted from a few volts to fractions of a
volt per turn. This is achieved because the carbon brush or slider is always in contact with one or
more turns of the primary winding. As the primary coil turns are evenly spaced along its length.
Then the output voltage becomes proportional to the angular rotation.
Variable Autotransformer
The variac can adjust the voltage to the load smoothly from zero to the rated supply
voltage. If the supply voltage was tapped at some point along the primary winding, then
potentially the output secondary voltage could be higher than the actual supply voltage. Variable
autotransformer’s can also be used for the dimming of lights and when used in this type of
application, they are sometimes called “dimmerstats”.
Variac’s are also very useful in Electrical and Electronics workshops and labs as they can
be used to provide a variable AC supply. But caution needs to be taken with suitable fuse
protection to ensure that the higher supply voltage is not present at the secondary terminals under
fault conditions.
PART B
Construction details
1. Explain in detail, the constructional features of a transformer with neat diagrams.
(16)
Principle of operation
1. Explain the working principle of a transformer.
(6)
Emf equation
1. Derive the emf equation of a transformer.
(6) (N/D-12)
2. Derive the emf equation of a transformer and prove that the number of turns on HV winding
and LV winding is in the ratio of their voltages.
(8) (A/M-10)
Transformer on no-load
1. Explain in detail, the operation of a transformer on no-load and load.
(10) (N/D-12)
2. The required no-load voltage ratio in a 1, 50 Hz, core type transformer is 6000/250 V. Find
the number of turns in each winding, if the flux is 0.06 Wb.
(10)
3. Draw and explain the no-load vector diagram of an ideal transformer and a practical
transformer.
(8)
Equivalent circuit
1. Draw and explain the equivalent circuit of a transformer on no-load and load.
2. Develop
(16)
the
equivalent
circuit
of
a
transformer
on
no-load
(8) (N/D-12)
and
load.
3. Calculate in terms of the primary, the equivalent resistance and the leakage reactance of a
transformer which gives the following data on test with the secondary terminals short
circuited:
Applied voltage = 60 V; Current = 100 A; Power input = 1.2 kW.
(10)
Transformer on-load
1. Draw and explain the phasor diagram of a single phase transformer supplying the following:
a) a UPF load
b) a lagging power factor load
(8) (A/M-10)
2. Explain the various losses in a transformer.
(8)
3. A single phase transformer is rated at 240/120 V, 50 Hz, find voltage and frequency of
secondary at no-load for the following conditions:
a) if primary voltage is 120 V at 25 Hz
b) if primary voltage is 240 V DC
(6)
4. Explain the principle of operation of a single phase transformer and its behaviour on load
with phasor diagrams.
(16)
Regulation
1. The maximum efficiency of a single phase 11000/400 V, 550 kVA transformer is 97.5% and
the efficiency occurs at 80% of full load at unity power factor. The percentage impedance is
3.5% and the load power factor is varied while the load current and the supply voltage are
held constant at their rated values. Determine the load power factor, at which the secondary
terminal voltage is minimum. Also find the value of the secondary voltage.
(16) (M/J-12)