Download DC Motors Introduction D.C. motors are seldom used in ordinary

DC Motors
Introduction
D.C. motors are seldom used in ordinary application because all electric supply caompaies
furnish altenating current. However, for special applications such as in steel mills mines and electric
trains, it is advantageous to convert alternation current into direct current i order to use d.c
motors. The reason is that speed /torque characteristic of dc motors are much more superior to
that of a.c motrors there for it is not suprising to note that for insustrial drives, d.c motors are also
of three types viz. Series -wound shunt -wound and compound -wound. The use of a particular
motor depends upon the mechanical load it ahas to drive.
D.C. MOTOR PRINCIPLE
A mechine that converts d.c. Power into mechanical powert is known as a d.c motor. Its operation is
based on the principle that when a current carrying conductor is placed in a magnetic field, the
conductor experiences a mechanical force. The direction of this force is given by Fleming's left had
rule and magnitude i given by:
F=BIl newtorns
Basically, there is no constructional difference between a d.c motor and a d.c generator. The same
d.c. Machine can be run as a generator or motor.
WORKING OF D.C MOTOR
Consider a part of a moltipolar d.c. Motor as shown in Fig. 5.1 . when the terminals of the motor
are connected to an external source of d.c. Supply
(i) the field magnets are excited developing alternate N and S poles.
(ii) The armature conductors carry * current . All conductors under N pole carry current in one
direction while all the conductors under S- pole carry currents in the opposite direction.
Suppose the conductorrs under N -pole carry currents into the plane of the paper and those under
S-pole carry currents out of the plane of the paper as shown in Fig. 5.1. Since each amatre
conductor is carrying current and is placed in the magnetic field, mechanical force acts on it.
Referring to Fig. 5.1 and applying Fleming's left hand rule, it is clear that force on each conductor
is tending to rotate the marmature in anticlockwise direction. All which sets the armature rotationg.
When the condutor moves from one side of a brush to the other the current in that conductor is
reversed and at the same time it comes under the influence of next pole which is of opposite
polarity. Consequently, the direction of force on the conductor remains the same.
BACK OR COUNTER E.M.F.
When the armature of a d.c. Motor rotates under the influence of the driving torque, the armature
conductors move through the magnetic field and hence e.m.f. Is induced in them as in a generator.
The induced e.m.f acts in opposite direction to the applied voltage V (*Lenz's law) and is known as
back or counter e.m.f. Eb. The back e.m.f. Eb (=p φ Z N / 60 A) is always less than the applied
voltage V. although this differnence is small when the motor is running under normal conditions.
Consider a shunt wound motor shown in Fig. 5.2. When d.c. Voltage V is appplied across the motor
terminals, the field magnets are excited and armature conductors are suppled with current,
Therefore, driving torque acts on the armature which begins to rotate. As the aarmature rotates ,
back e.m.f. Eb is induced which opposes the applied voltage V. The applied voltage V has to force
current through the armature against the back e.m.f. Eb. The electric work done in overcoming and
causing the current to flow against Eb is converted into mechanical energy developed in the
armature. It follows, therefore, that energy conversion in a d.c. Motor is only possible due to the
production of back e.m.f.. Eb.
Net voltage across armature circuit = V-Eb.
If Ra is the armature circuit resistance then,
Since V and Ra are usually fixed, the value of Eb will determine the current drawn by the motor. If
the speed of the motor is high, then back e.m.f. Eb (=p φ Z N / 60 A) is large and hence the motor
will draw less armature curren and vice-versa.
SIGNIFICANCE OF BACK E.M.F.
The presence of back e.m.f. Makes the d.c. Motor a self -regulationg mechine i.e., it makes the
motor to draw as much armature current as is just sufficient to develop the torque required by the
load.
Armature current ,
(i) When the motor is running on no load, small torque is required to overcome the friction and
windage losses. Therefore, the armature current Ia is small and the back e.m.f. Is nearly
equal to the applied voltage.
(ii) If the motor is suddenly loaded, the firest effect is to cause the armature to slow down.
Therefore, the speed at which the armature conductors move throug the field is reduced and
hence the back e.m.f. Eb. Falls. The decreased back e.m.f. Allows a larger current to flow
through the armature and larger current means increased driving torque. Thus the driving
torque increases as the motor slows down. The motor will stop slowing down when teh
armature current is just sufficient to produce the increased torque required by the load.
(iii) If the load on the motor is decreased, the driving torque is mimentarily in excess of the
requirement so that armature is accelerated. As the armature speed incteases, the back emf
Eb also increases and causes the armature current Ia to decrease. The motor will stop
accelerating when the armature current is just sufficient to produce the reduced torque
required by the load.
It follow, therefore , that back e.m.f in a d.c motor regulates the flow of armature current i.e, it
automatically changes the armature current to meet the load requirement.
VOLTAGE EQUATION OF D.C. MOTOR
Let in a d.c. Motor (See Fig. 5.3)
V = applied voltage
Eb= back e.m.f.
Ra = armatre resistance
Ia = armature current
Since back e.m.f. Eb acts in opposition to the applied voltage V, the net voltage across the
aarmature circuit is V-Eb. The armature current Ia is given by,
V=Eb+IaRa
or
This is known as voltage equation of the d.c. Motor.
POWER EQUUATION
If eq. (i) above is multiplied by Ia throughout, we get,
VIa = EbIa+I2aRa
This is know as power equation of the d.c. Motor.
VIa = electric power supplied to armature (armature input)
EbIa = * power developed by armature (armature output)
I2aRa = electric power wasted in armature (armature Cu loss)
Thus out of the armature input , a small portion (about 5%) is wasted as I2aRa and the
remaining portion EbIa is converted into mechanical power within the armature.
CONDITION FRO MAXIMUM POWER
The mechanical power developed by the motor is pm = EbIa
Pm = VIa - I2aRa
Now
Since , V and Ra are fixed, power developed by the motor depends upon armature current . Tor
maximum power, dPm/dIa should be zero.
Or IaRa = V/2
Now,
V= Eb + Ia Ra = Eb + V /2 [ IaRa = v/2]
Eb = V/2
Hence mechanical power devloped by the motor is maximum when back e.m.f. is equal to
hald the applied voltage.
Limitations : In practive, we never aim at achieving maximum power due to the following reasons:
(i) The armature current under this condition is ver large – much excess of rated current of the
machine.
(ii) Half of the input power is waseted in the armature circuit. In fact, if we take into account
other losses (iron and mechanical), the efficiency will be well below 50%.
TYPES OF D.C. MOTORS
Like generators, there are there types of d.c. Motors characterised by the connections of field
winding in relation to the armature viz.
(i) Shunt -wound motor in which the field winding is connected in parallel with the armature
[see Fig 5.4]. The current through the shunt field winding is not the same as thea armature
current Shunt Field windings are designed to produce the necessary m.m.f. By means of a
relatively large number of turn of wire having high resistance. Therefore, shunt field current
is relatively small compared with the armature current.
(ii) Series – wound motor in which the firld winding is connected in series with the armature
[See Fig. 5.5]. There fore , series field winding carries the armature current. Since the
current passing through a series field winding is the same as the armature current, series
field windings must be designed with much fewer turns than shunt field windings for the
same m.m.f. Therefore, a series field winding has a relatively small number of turns of
thick wire and , therefore, will possess a low resistance.
(iii) Compound – wound motor which has rwo field winding; one connected in parallel with
the armature and the other in series with it. There are two tyypes of compound motor
connections (like generators ). When the shunt field winding is directly connected acress
the armature terminals [See Fig. 5.6], it is called short – shunt connections. When the
shunt winding is so connected that it shunts the series combination of armature and
series field [see Fig. 5.7], it is called long shunt connection.
The compound machines (generators or motors ) are always designed so that the flux
produced by shunt field winding is considerably larger than the flux produced by the series field
winding. Therefore, shunt field in compound machines is the basic dominant factor in the production
of the magnetic field in the machine.
ARMATURE TORQUE OF D.C. MOTOR
Torque is the turning moment of a force about an axis and is measured by the product of force (F)
and radius(r) at right angle to which the force acts i.e.
T= Fxr
In a d.c. Motor, each conductor is acted upon by acircumferential force F at a distance r, the
radius of the armature (Fig 5.11). Therefore each conductor exerts a torque, tending to rotate the
armature. The sum of the torques due to all armature conductors is known as gross or armature
torque (Ta).
Let in a d.c. Motor
r = average radius of armature in m
l = effective length of each conductor in m
Z = total number of armature conductors
A = number of parallel paths
i = current in each condutor = Ia/A
B = average flux desity in Wb/m2
Φ = flux per pole in Wb
P = number of poles
Force on each conductor, F = B i l newtons
Torque due to one conductor = F X r newtons – metre
Total armature torque, Ta = Zfr newtons – metre
= Zbilr
Now i = Ia/A, B = Φ/a where a is the x-sectional area of flux path per pole at radius r.
Clearly a= 2∏rl/P.
Ta = ZX( Φ/a)X(Ia/A)XlXr
or
Ta = 0.159 Z Φ Ia (P/A)N-m
Since Z, P and A are fixed for a given machine,
Ta α Φ I a
Hence torque in a d.c. Motor is directly proportional to flux per pole and armature current.
(i) For a shunt motor , flux Φ is practically constant.
Ta α I a
(ii) For a series motor, flux Φ is directly proportional to armature current Ia provided magnetic
saturation does not take place.
T a α I2 a
Alternative expression for Ta
Eb = p ΦZN/60A
p ΦZ/A = 60 x Eb/N
From equation (i) , we get the expression of Ta as :
Ta = 0.159x(60 X Eb/N)xIa
Ta = 9.55 x EbIa/N (N-m)
or Note that developed torque or gross torque means armature torque Ta.
SHIFT TORQUE (Tsh)
The torque which is vailable at the motor shift for doing useful work is known as shaft torque. It is
represented by Tsh. Fig. 5.12 illustrates the concept of shaft torque. The total or gross torque T a
developed in the armature of a motor is not available at the shaft because a part of it is lost in
overcoming the iron and frictional losses in the motor. Therefore, shaft torque Tsh is somw what less
than the armature torque Ta. The difference Ta-Tsh is called lost totque.
Clearly Ta – Tsh = 9.55 x Iron and frictional losses/N
For example, if the iron and frictional losses in a motor are 1600 W and the motor runs at 800
r.p.m, then,
Ta – Tsh = 9.55 x 1600/800 = 19.1(N-m)
As stated above, it is the shaft torque Tsh that produces the useful output. If the speed of the
motor is N r.p.m., then,
Outpurtin watts = 2∏NTsh/60
or Tsh = (output in watts /2∏N/60 )(N-m)
or Tsh = 9.55x Output in watts /N (N-m)
(60/2∏=9.55)
LOSSES IN A D.C. MOTOR
The losses occuring in a d.c. Motor are the same as ina d.c. Genetator [See Art 2.27]. These are
[See Fig. 5.21]:
(i) copper losses
(ii) iron losses or magnetic losses
(iii) mechnanical losses
As in a generator, these losses cause (a) an increase of machine temperature and (b) reduction in
the efficiency of d.c. Motor.
The following points may be noted:
(i) Apart form armature Cu loss, field Cu loss and brush contact loss, Cu losses also occur in
interpoles (commutationg poles ) and compensationg windings. Since these windings carry
armature current (Ia),
Loss in interpole winding = I2a x Resistance of interpole winding
Loss in compensation winding = I2a x Resistance oc compensating winding
(ii) Since d.c. Machines (generators or motors ) are generally operated at constant flux density
and constant speed, the iron losses are nearly constant.
(iii) The mechnanical losses (i.e. friction and windage) vary as the cube of the speed of rotation
of the d.c. Machine (generator or motor). Science d.c. Machines are generally operated at
constant speed, mechanical losses are considered to be constant.
EFFICIENCY OF A D.C. MOTOR
Like a d.c. Genetator , the efficiency of a d.c motor is the ratio of output power to the input power
i.e,
Efficiency, η = output/input x100
= (output/(output+losses ))x100
As for a generator (see Art 2.30), the efficiency of a d.c. Motor will be maximum when:
variable losses = Constant losses
Therefore , the efficiency curve of a d.c. Motor is similar in shape to that of a d.c. Generator (Refer
back to Fig. 5.57).
POWER STAGES
The power stages in a d.c. Motor are represented diagrammatically in Fig. 5.22.
A-B = copper losses
B- C – Iron and friction losses
Overall efficiency , ηc = C/A
Electrical efficiency , ηe = B/A
Mechanical efficiency , ηm = C/B
D.C. MOTOR CHARACTERISTICS
There are three principal types of d.c. Motors viz. Shunt motors, series motors and compound
motors. Both shunt and series types have only one field winding wound on the core of each pole of
the motor. Both shunt and series types have only one field winding wound on the core of each pole
of the motor. The compound type has two separate field windings wound on the core of each pole.
The perfoemance of a d.c. Motor can be judged from its characteristic cureves known as mitor
characteristics. Following are the three important characteristics of a d.c. Motor:
(i) Torque and Armature current characteristic (Ta/Ia)
It is the cureve between armature torque Ta and armature current Ia of a d.c motor. It is
also known as electical characteristics of the motor.
(ii) Speed and armature current characteristiv (N/Ia)
It is the cureve between speed N and aramture curret Ia of a d.c. Motor. It is very important
characteristic as it is often the deciding factor in the selectionof the motor for a particular
application.
(iii) Speed and torque characteristic (N/Ta)
It is the curve between speed N and armature torque Ta of ad.c. Motor. It is also known as
mechanical characteristic.
CHARACTERISTICS OF SHUNT MOTORS
Fig. 5.30 shows the connection of a d.c shunt motor . The field current I sh is constant since the
field winding is directly connected to the supply voltage V which is assumed to be constant. Hence,
the flux in a shunt motor is * approximately constant.
(i) Ta/Ia Characteristic. We know that in a d.c. Motor,
Ta α φ Ia
Since the motor is perating from a constant supply voltage, flux φ is constant (neglecting armature
reaction).
Ta α Ia
Hence Ta/Ia characteristic is a straight line passing through the origin as shown in Fig. 5.31. The
shaft torque (Tsh) is less than Ta and is shown by a dotted line. It is clear from the cureve that a
very large current is required to start a heavy load. Therefore , a shunt motor should not be started
on heavy load.
(ii) N/Ta Characteristics. The speed N of a d.c. Motor is given by:
N α Eb/φ
The flux φ and back e.m.f. Eb in ashunt motor are almost constant under normal conditions
. Therefore, speed of a shunt motor will remain constant as the armature current varies (dotted line
AB in Fig. 5.32 ). Strictly speaking, when load is increased , Eb (= V-IaRa) and φ decreases due
to the armatur resistance drop and armature reaction respectively. However, Eb decreases slightly
more than φ so that the * speed of the motr decreases slightly with load .(line AC in Fig. 5.32).
(iii) N/Ta Characteristic. The cureve is obtained by plotting the valued of N and Ta for various
armature current (See Fig. 5.33). It may be seen that speed falls somewhat as the load
torque increases.
Conclusion : Following two important conclusions are drawn from the slight above characteristics.
(i) There is a slight change in the speed of a shunt motor form no-load to full-load. Hence , it is
essentially a costant -speed motor.
(ii) The starting towque is not high because Ta α Ia
CHARACTERISTICS OF SERIES MOTORS
Fig. 5.34 shows the connectiond of a series motor. Note that current passing through the field
winding is the same as that in the armature. If the mechnanical load on the motor increases, the
armature current also increases. Hence , the flux in a series motor increases with the increase in
armaure current and vice -versa.
(i) Ta/Ia Characteristics. We know that
Ta α φ Ia
Upto magnetic saturation. φ α Ia so that Ta α I2a
After magnetiv saturation φ is constant so that Ta α Ia
Thus upto magnetiv saturation, the armature torque is directly proportional to the sequare of
armature current. If Ia is doubled, Ta is almost quadrupled. Theredor , Ta/Ia curve upto magnetiv
saruration is a parabola( portion OA of the curve in Fig. 5.35). However, after magnetic saturation,
torque is directly proportional to the armature current. Therefore, Ta/Ia curve after magnetic
saturation is straight line (portion AB of the curve in Fig. 5.35).
It may be seen that in the initial portion of the curve (i.e. upto magnetiv saturation). Ta α
I2a. This means tht starting torque of a d.c. Series motor will be very high as compared to a shunt
motor (wher Ta α Ia).
(ii) N/Ia Characteristic. Thhe speed N of a series motor is given by ;
N α Eb/ φ where Eb = V- Ia(Ra+Rse)
When the armature current increases, the back e.m.f. Eb decreases dur to Ia(Ra+Rse) drop while
the flux φ increases. However, Ia(Ra+Rse) drop is quite small under normal conditions and may be
neglected.
N α I/ φ
α I / Ia upto magnetic saturation
Thus, upto magnetic saturation, the N/Ia curve follows the hyperbolic path as shown in Fig 5.36.
After the flux become constant and so does the speed.
(iii) N/Ta characteristic. The N/Ta characteristic of a series motor is shown in Fig. 5.37. It is
clear that series motor develops high torque at low speed and vice-versa. It is because and
increase in torque requires an increase in armature current, which is also the field current.
The result is that flux is strengthened and hence the speed drops ( N α I/ φ). Reverse
happens should the torque be low.
Conclusion : Follwing three imprtant conclusions are drawn from the above characteristics of series
motors:
(i) It has a high starting torque because initially Ta α I2a
(ii) It is a variable speed motor (see N/Ia curve in Fig. 5.36) i.e. it automatically adjusts
the speed as the load changes. Thus if the load decreases, its speed is automatically
raised and vice-versa.
(iii) At no-load, the armature current is very small ans so is the flux. Hence, the speed
rises to an excessive high value( N α I/ φ). This is dangerous for the machine
which may be destroyed dur to centrifugal forces set up in the rotationg parts.
Therefore, a series motor should never be started on no-load. However, to start a
series motor, mechanical load is first put and then the motor is started.
Note. The minimum load on a d.c. Motor should be grea enough to keep hte speed within
limits. If the speed becomes dangerously high, then motor must be disconnected from the supply.
COMPOUND MOTORS
a compound motor has both series field and shunt field. The shunt field is always stronger than the
series field. Compound motors are of two types:
(i) Cumulative – compound motors in which series field aids the shunt field.
(ii) Differential -compound motors in which series field opposes the shunt field.
Differential compound motors are rarely used due to their poor torque characteristics at
heavy loads.
CHARACTERISTICS OF CUMULATIVE COMPOUNDS MOTORS
Fig. 5.38 shows the connections of a cumulative compound motor. Each pole carries a series as well
as shunt field winding: the series field aiding the shunt field.
(i) Ta/Ia charecteristic. As the load increases, the series field increases bbut shunt field strangh
remains constant. Consequently, total flux is increased and hence the armature torque (Ta
α φ Ia). It may be noted thtat torque of a cumulative – compound motor is greate than that
of shunt motr for a given armature current due to seires field [See Fig. 5.39].
(ii) N/ Ia Characteristic. As explained above, as the load increases, the flux per pole also
increases. Consequently, the speed ( N α I/ φ) of the motor falls as the load increases
(see Fig. 5.40). It may be noted that as the load is added, the increased amount of flux
causes the speed to decrease more than does the speed of a shunt motor. Thus the speed
regulation of a cumulative compound motor is poorer than that of a shunt motor.
Note : Due to shunt field, the motor has a definite no load speed and can be operated safely
at
no – load.
(iii) N/Ta Characteristic. Fig 5.41 shows N/Ta charecteristic of a cumulative compound motor
for a given armature current, the torque of a cumulative compound motor is more than that
of a shunt motor but less than of a series motor.
Conclusions : a cumulative compound motor has characteristics intemediate between series
and shunt motors.
(i) Due to the presence of shunt field, the motor is pevented form running away at noload.
(ii) Due to the people of series field, the starting torque is increased.
COMPARISON OF THREE TYPES OF MOTORS
(i) The speed regulation of a shunt motor is better than that of a series motor.
However, speed regulation of a cumulative compound motor lies between shunt and series
motors (see Fig. 5.42)
(ii) For a given armature current, the starting torque of a series motor is more than that of a
shunt motor. However, the starting torque of a cumulative compound motor lies between
series and shunt motors (See Fig. 5.43).
(iii) Both shunt and cumulative compound motor have definite no-load speed. However, a series
motor has dangerously high speed at no-load.
APPLICATIONS OF D.C. MOTORS
1. Shunt motors. The characteristics of a shunt motor reveal that it is an approximately
constant speed motor. It os, therfore, used.
(i) where the speed is required to remain almost constant from no-load to full-load
(ii) where the load has to be driven at a number of speed and any one of which is required to remain
nearly constant
Insustrial use: Lathes, drills, boring mills, shapers, spinning and weaving machines etc.
2. Series motors. It is a variable speed motor i.e. speed is low at high torque and vice- vers.
However, at light or no-load, the motor tends to attain dengerusly high speed. The motor
has a high starting torque. It is, therfore, used
(i) wher large starting torque is required e.g. in elevatiors and electirc traction
(ii) where the load is subjected to heavy fluctuatuins and the speed is automatically
required to reduce at high torques and vice-versa.
Industrial use : Electric traction, cranes, elevators, air compressors, vacuumm cleaners hair
drier, sewing machines etc.
3. Compound motors. Diffenrential-compound motors are rarely used because of their poor
torque chareacteristics. However cumulative -compound motors are used where a fairly
constant speed is required with irregular loads or suddenly applied heavy loads.
Industrial use : Presses, shears, reciprocating machines etc.
TROUBLES IN D.C. MOTORS
Several toubles may arise in a d.c. Motor and afew of them are discussed below.:
1. Failure to start. Tis may be due to (i) ground fault (ii) open or short -circuit fault (iii) wrong
connections (iv) too low supply voltage (v) frozen bearing or (vi) ecessive load.
2. Sparking at brushes. This may be due to (i) troubles in brushes (i) troubles in commutator
(iii) troubles in armature or (iv) excessive load
(i) Brush troubles may arise due to insuffiecient contact surface, too short a brush , too
little spring tension or wrong brush setting.
(ii) Commutator troubles may be dur to dirt on the commutator, high mica, rough surface
or eccentricity.
(iii) Armature troubles may be due to an pen armature coil. An open armature coil will
cause sparking each tim the open coil passes the brush. The location of this open coil
is noticeable by a burnt line between segments connection the coil.
3. Vibrations and pounding noises. These may be due to (i) worn bearing (ii) loose parts
(iii) rotating parts hitting stationary parts (iv) armature unbalanced (v) misalignment
of machine (vi) loose coupling etc.
4. Overheating. The overheating of motor may be due to (i) overloads (i) sparking at
the brushes (iii) short- circuited armature or field coils (iv) too frequent starts or
reversals (v) poor ventilations (vi) incorrect voltage.
Speed control of D.C. Motors
Introduction
Although a fro greter percentage of electric motors in service are a.c. Motors, the d.c. Motor is of
considerable industrila importance. The principal advantage of a d.c. Motor is that its speed can be
changed ove a wide range by a varietyof simple methods. Such a fine speed control is generally not
possible with a.c. Motors. In fact fine speed control is one of the reason for the strong competitive
position of d.c. Motors in the modern industrial application. In this chapete , we shall discuss the
various methods of speed control of d.c. Motors.
SPEED CONTROL OF D.C. MOTORS
The speed of a d.c. Motor is given by :
N α I/ φ
N = K ((v-IaR)/Q) r.p.m
where R = Ra
= Ra + Rse
for shunt motor
From exp. (i) it is clear that there are three main methods of controlling the speed of a d.c. Motor,
namely:
(i) By varying the flux per pole (φ). This is known as flux control method.
(ii) By varying the resistance in the armature circuit. This is know as armature control method.
(iii) By varying the applied voltage V. This is known as voltage control method.
SPEED CONTROL OF D.C. SHUNT MOTORS
The speed of shunt motor can be changed by (i) flux control method (ii) armature control method
(iii) voltage control method. The first method (i.e. flux control method) is frequently used because it
is simple and inexpensive.
1. Flux control method. It is based on the fact that by varying the flux φ, the motor speed (
N α I/ φ) can be changed and hence the name flux control method. In this method, a
variable resistance (known as shunt field rheostat) is placed in series with shunt field winding
as shown in Fig. 6.1.
The shunt field rheostat reduces the shunt field current Ish and hence the flux φ. Therefore , we
can only * raise the speed of the motor above the normal speed (ssee Fig. 6.2). Generally, the
mehtod permits to increase the speed in the ratio 3:1. Wider speed ranges tend to produce
instablity and poor commutation.
Advantages
(i) This is an easy and convenient method.
(ii) It is an inexpensive method since very little power is wasted in th shunt field rheostat due to
relatively small value of Ish.
(iii) The speed contrl expercised by this method is independent of load on the machine.
Disadvantages
(i) Only speeds higher than normal speed can be obtained since the total field circuit resistance
cannot be reduced below Rsh – the shunt field winding resistance.
(ii) There is a limit to the maximum speed obtained since the total field circuit much weakened,
commutation becomes poorer.
Note. The field of a shunt motor in operation shuld never be opened because its speed will
increase to an extremely highvalue.
2. Armature control method. This method is based on the fact that by varying the voltage
available across the armature , the back e.m.f and hence the speed of the motor can be
changed. This is done by inserting a variable resistance Rc (known as controller resistance)
in series with the armature as shown in Fig. 6.3.
where N α V-Ia(Ra+Rc)
Rc = Controller resistance
Due to voltage droop in the controller resistance, the back e.m.f (Eb) is decreased . Since N α Eb,
the speed of the motor is reduced. The highest speed obtainable is that corresponding to Rc =0 i.e.,
normal speed. Hence , this method can only provide speeds below the normal speed (see fig. 6.4).
Disadvantages
(i) A large amount of power is wasted in the controller resistance since it carreis full armature
curren Ia.
(ii) The speed varies widely with load since the speed depends upon the voltage drop in the
controller resistance and hence on the armature current demanded by the load.
(iii) The output and efficiency of the motor are reduced.
(iv) This method results in poor speed regulation.
Due to above disadvantages, this method is seldom use dto control the speed o shunt
motors.
Note. The armature control method is a very common method for the speed control of d.c. Series
motors. The disadvantages of poor speed regulation is not important in a series motor which is used
only where varying speed service is required.
3. Voltage control method. In this method, the voltage sourve supplying the field current is
different from that which supplies the armature. This method avoids the disadvantages of
poor speed regulation and low efficiency as in armature control method. However, it is quite
expensive. Therefore this method of speed control is employed for large size motors where
efficiency is of grat importance.
(i) Multiple voltage control, In this method, the shunt field of the motor is connected permanently
across a fixed voltage source. The armature can be connected acreoss several fiffernt voltage
through a suitable switchgear. In this way, voltage applied across the armature can be changed. The
speed will be approximately proportional to the voltage applied across the armature. Intermediate
speeds can be obtained by means of a shunt field regulator.
(ii) Ward -Leonard system. In this method, the adjustable voltage for the armature is obtained from
and adjustable voltage generator while the field circuit is supplied from a separate source. This is
illustrated in Fig. 6.5. The armature of the shunt motor M (whose speed is to be controlled ) is
connected direclty to a d.c. geneg=rator G driven by a constant – speed a.c. Motor A. The field of
the shunt motor is supplied from the exciter E. The voltage of the generator G can be varied by
means of its field regulator. By reversing the field current of generator G by controller FC, the
voltage applied to the motor may be reversed. Sometimes, a field regulator is included in the field
circuit of shunt motor M for additional speed adjustment. With this method , the motor may be
operated at any speed upto its maximum speed.
Advantages
(a) The speed of the motor can be adjusted through a wide range without resistance losses which
results in high efficiency.
(b) The motor can be brought to a standstill quikly, simply by rapidly reducing the voltage of
generator G. When the generator voltage is reduced below the back e.m.f. Of the motor, this back
e.m.f. Sends current through the generator armature, establishing dynamic banking. While this takes
place the generator G operated as a motor driving motor A which returns power to the line.
(c) This method is used for the speed control of large motors when a d.c. Supply is not available.
The disadvantage of the method is that a special motor generator set is required for each motor
and the losses in this set are high if the motor is operating under light loads for long periods.
Applications. The ward-Leonard system of speed control is expensive but is use where an unusually
wide and very sensitive speed control is desired. This arrangement offers an excellent stepless
speed control and is well suited for such applications as passenger elevators, electric excavators
etc.
SPEED CONTROL OF D.C. SERIES MOTORS
The speed control of d.c. Series motors can be obtained by (i) flux control method (ii) armature –
resistance control method. The latter method is mostly use
1. Flux control method. In this method, thhe fluux produced by the series motor is varied and
hence the speed. The variation of flux can be achieved in the following ways:
(i) Field diverters . In this method, a varoable resistance (called field diverter ) is connected in
parallel with series field winding as shown in Fig. 6.13. Its effect is to shunt some portion of the line
current from the series field winding , thus weakening the field and increasing the speed ( N α I/
φ). The lowest speed obtainable is that corresponding to zero curent in the diverter (i.e. diverte is
open ). Obviously, the lowest speed obtainable is the normal speed of the motor. Consequently, this
method can only provide speed above the normal speed. The series field diverter method is often
employed in traction work.
(ii) Armature diverter. In order to obtain speed below th normal speed, avariable resistance (called
armature diverter) is connected in parallel with the armature as shown in Fig. 6.14. The diverter
shunts some of the line current, thus reducing the armature current. Now for a giver load , if Ia is
decreased, the flux φ must increase ( T α φ Ia). Since N α I/ φ , the motor speed is
decreased. By adjusting the armature diverter, any speed lower than the normal speed can be
obtained.
(iii) Tapped field control . In this method, flux is reduced (a nd hence speed is increase)
by decreasing the number of turns of hte series field winding as shown in Fig. 6.15.
The switch S can short circuit any part of hte field winding , the motor runs at normal
speed and as the field turns are cut out, speed higher than normal speed are
achieved.
(iv) Paalleling field coils . This method is usually employed in the case of afan motors. By
regrouping the field coils as shown in Fig. 6.16. several ficed speed can be obtained.
2. Armature – resistanc control. In this method, a variable resistance is directly connected in
series with the supply to the complete motr as hsown in Fig. 6.17. This reduces the voltage
availabe acreoss the armature and hence the speed falls . By changing the value f variable
resistance, any speed below the normal speed can be obtained. This is the most common
mtthod employed to control the speed of d.c seried motors. Although this method has poor
speed regulation, this has no significance for seris motors because they are used in varying
speed applications. The loss of power in the series resistance for many applications of series
motors is not too serious since in these application, the control isutilised fro a large portion
of th etime for reducing the speed under light – load conditions and is only used
internuttently when the motor is carrying full load.
SERIES PARALLEL CONTROL
Another method used for the speed contol of d.c. Seris motors is teh series parallel method. In this
system which is widely used in traction system , two (or more ) similar d.c. Seris motors are
mehcanically coupled to the same load.
When the motor are connected in seris [See Fig. 6.23 (i)], each motor armature will recieve on half
the normal voltage. Therefor the speed will be low . When the notors are connected in parallel ,
each motor armature receives the normal voltage and the speed is high [See Fig. 6.23 (ii)]. Thus we
can obtain two speed. Note that forthe same load on the pair of motors, the system would run
appoximately four times the speed when the machines are in paralled as when they are in series
NECESSITY OF D.C. MOTOR STARTER
At starting when the kkotor is stationary, there is no back e.m.f int hearmatue. Conswquetly if the
motor is directly switched on to the mains, the armature will draw a hevy current (Ia=V/Ra)
beecause of small armature resistance. As an example, 5 h.p., 220 V shunt motor has a ful – load
current of 20 A and an armature resistance o fabout 0.5 Ω . If this motor is directly switched on to
supply, it would take an armarue current of 220/0.5=440 A which is 22 times the full – load curren .
Thsi hight starting current may result in :
(i) burning of armature due to excessive heating effect,
(ii) damaging the commutator and brushes dur to heavy aparking,
(iii) excessive voltage drop in the line to which the motor is connected. The result is that the
operation of other appliance connected to the line may be impaired and in particular cases,
they may refuse to work.
In order to order to avaoid excessive current at starting , variable resistance ( know n as startin
resistance) is inserted in series with the armature circuit. This resistance is gradually reduced as the
motor gains speed ( and hence Eb increases) and eventually it is cut out completely when the motor
has attained full speed. The value of starting resistanc eis generally such that starting current is
limited to 1.25 to 2 times the full -load current.
TYPES OF D.C. MOTOR STARTERS
The starting opertion of a d.c. Motor consists in the insertion of external resistance into the
armature circuit to limit the starting current taken by the motor and the removal of this resistance
is steps as the motor accelertes. When the motor attains the normal speed, this resistance is totally
cut out of hte armature circuit. It is very important and desirable to provide the starter with
protective devices to enable the starter srm to return to OFF position.
(i) When the supply fails thus preventing the rmature being directly across themains when this
voltage is restored. For this purpose , we use no-volt releae coil.
(ii) When the motor becomes overlaoded or develops a faul causing the motor to take an
excessive current. For this purpose , we use overload release coil.
There ar two principal types of d.c motor starter viz. Thrree point starter and four point starter. As
we shall see, the two types of strters differ only in the manner in which the no – volt release coi is
connected.
THREE POINT STARTER
This type of starter is widely used for starting shunt snd compound motors.
Schemativ diagram. Fig.6.28 shows the schematic diagram of a three point starter for a shunt motor
which protective device. It is so called because it has three terminals L,Z and A. The starter
consist of strting resistance dicided into several sections and connected in series with the armature.
The a=tapping points of he starting resistance are brought out to a number of studs. The three
terminals L.Z and A of the starter are conected respectively to the postitive line terminal, shunt
field terminal and armature teminal. The other terminals of the armature and shunt field windings are
connected to the negative terminal of the supply. The no voltrelease coil is connected in th shunt
field circuit. One end of hte hadle is connected toeh terminal L through the over load release coil.
The other end of the handle moves against a spiral spring and makes contact with each stud during
starting operation, cutting out more and more starting resistance as it passes over each stud in
clockwise direction.
Operation
(i) To start with , the d.c. Supply is switched on with handle in the OFFposition.
(ii) The handle is now moved clockwise to the first stud. As soon as it comes in contact with the
first stud, the dhunt field winding tis dierelctly connected across the supply while the whole
starting resistance is inserted in series with the armature circuit.
(iii) As the handle is gradually moved over to the final stud , the starting resistance is cut out of
the armature circuit in steps. The handle is now held magnetivally by the no volt release coil
whichis energised by shunt field current.
(iv) If the supply voltage is suddenly interrupted or if the field excitation is accidently cut,t he
no volt release coil is demagnetise d and the handle goes back to the OFF position under
the pull of the spring. If no volt relwase coil were not used , then in case of failure of supply,
the handle would remain on the final stud. If then supply is restored, the motor will be
directly connected across the supply , resulting in an excessive armature current.
(v) If the motor is over loade (or a fault occures , it wil draw excessive current formt he supply.
This current will increase the ampere turen of ht overload release coil and pull te rmaature
C, thus short circuiting the no- colt treleae coil. The no – volt coil is demagnetise d and the
handle is pulled to the OFFposition by the spring. Thus, the motor is automatiallly
disconnected rorm the supply.
Drawback. In a three point starter, the no volt release coil is connected in series tithe shunt field
circuit so that it carries the shunt field current. While excersing speed control through field
regulator , the field current may be weakened to such an extent that the no volt release coil may
not be able to keep the starter arm in the ON position. This may diconnected the motor from th
supply when it is not desired. This drewback is overcome in the four point starter.
TESTING OF D.C. MACHINES
INTROCCTION
There are several tests that are conducated upon a d.c. Machine ( generator or motor ) to judge its
performance. One important test is performed to measure the efficiency of a d.c. Machine . The
efficiency of a d.c. Machine depends upon its losses. The smaller the losses, the greater is the
efficiency of the machine and cice versa.The considerationof losses in a d.c. Machine is important
for two princeipal reasons. First, losses deeine the efficicncy of the mchine and appreciably influence
its operationg cost, Secondly, losses determine the heatingh of hte machine and hence the power
output that may be obtained withou undue deterioration of the indultation. In thi chapter, we shall
focus or attentionon the various methods for the determination of the efficiency of a d.c. Machine.
EFFICIENCY OF A D.C. MACHINE
The pwwer that a d.c. Machine receives is called the input ad the power it gives out is called the
outpuut. Therefore , the efficiency of a d.c. Machine like that of any energy tranferring device, is
given by:
Efficienvy = Output/Input
Output = Input – Loses and Input = Output Losses
Therefore, the efficiency of a d.c. Machine can also be expressed in the following forms:
Efficiency = Input – Losses /Input
Efficiency = Output / Output + Losses
The most obvious method of detemining the efficinecy of a d.c. Machine is to firectly load it and
measure the input power and putput powr . Then we can use eq (i) to detemine the ddefficiency of
hte machine . This method suffers form three main drawbacks. First , this method requires tha
application of load on the mahine. Secondlly for machine of large rating the load sof htrquired sizes
may not be availablle. Thirdly, even if it is possible to provide such loads, large power will be
dissipated, making it an expensive mathod.
The most common mehtod ofmeasuring the efficiency of a d.c. Machine is to deteminde its losses (
instead of measuringthe input and output on load) . We can then use eq. (ii) or eq (iii) to determine
the efficiency of themachine . This method has the obvious advantage of convenience and economy.
EFFICIENCY BY DIRECT LOADING
In this method , the d.c. Machine is loaded and output and input are measured to find the efficiency.
For this purpose two simple method can be used.
(i) Break test. In this mehtod a break is applied to a watr cooled pully mounted on the motor
shaft as hsown in Fig. 7.1. One end of hte rope is fixed to the floor via a spring balance S
and a known mass is sunspended at the other end. If the spring balance relading is S kg -wt
and th suspended mass has awheight to W kg-wt, then Net pull on the rope= (W-S)kg-wt=
(W-S)x9.81 newtons
If r is the radious of th pulley inmetrs , hten the shaft torque Tsh devleoped by the motor is
Tsh = (W-S)x9.81xr N-m
If the speed of th pully is N r.p.m. Then
Output power = 2лNtsh/60 =2лNx(W-S)xrx9.81/60 watts
Let V = supply voltage in volts
I current taken by the motr in amperes
Input to motor = VI watts
Efficiency = 2лNx(W-S)xrx9.81/60xVI
SWINBURNE'S METHOD FOR DETEMINING EFFICIENCY
In thsi method, the d.c. Machine (gentetor or motor ) is run as amotor at no load and losses of the
machine are detemined. Once the losses of ht emaahcine are known its efficiency at any desires load
machien are detemined. Once the losses of hte machineare known its efficieny at any desired load
can be determined in advance. It may noted taht this method is applicable t thsoe machine in which
fluex is practically constant at all loads eg. Shunt and computn machines. Let us see how the
efficiency of a d.c. Shunt machine( generator or motor) is determined by this method . The test
cosists of two steps:
(i) Detemination of hot resistance ofwindings. The aramture resistance and shunt field
resistance are measured using abattery vltmeter and ammetere. Since these resistance are
measured when te machine is cold, they must be converted to values corresponding tothe
temperture at which teh machine would work on full load. Generally thse value ate mesured
for a tempereture rise of 40 C above the room temperature .Ltt the hot resistance of
armaturse and shunt field be Ra and Rsh respectively.
(ii) Determination of costant lossed . The machins is run as motor on noload with supply voltage
sdjustd to the reted voltage i.e. voltage stamped o the maeplate the speed ofthe motor is
adjusted to te rete dpeed with the help ofield regulator R as shown in Fig 7.2.
Let v = supply voltage
I0 = No – load currnetn read by mameter A1
Ish = Shunt field current read y ammetr A2
No load power input to armature = Via0 = V(I0-Ish)
Since the output of the motor is zero the no- load input power to the armature suppplies (a) iron
losses in the core (b) friction loss (c) windage loss (d) armature Culoss [I2a0 or (I0-Ish)2 Ra].
Constant losses Wc = Input to motor – Armature Cu loss
Wc = W I0-(I0-Ish)2Ra
Since constant lossssses are known the efficiency of th machine at any other load can be determined
suppose it is desired to determien the efficiency ofht machine at load current I then
Armature current Ia = I – Ish
= I + Ish
Efficinecy when running as motor
Input powr to motor = VI
Armature Cu loss = I2aRa= (I-Ish)2Ra
Constant losses = Wc
Total losses = (I-Ish)2Ra + Wc
Motor efficiency Mm = Input -Losses/Input = VI(I-Ish)2Ra-Wc/VI
Efficicncy when running as a generator
Out put of generator = VI
Armauture Cu loss = (I+Ish)2 Ra
Constant losses = wc
Total losses = (I+Ish)2 Ra+Wc
Generator efficiency Mg = Output/Output+Losses = VI/VI+(I+Ish)2Ra+Wc
Advantages of swinburne's test
(i) The power rewuire to carry out hte test is small because it is a no -load test. Therefor this
method is quite economical.
(ii) The efficiency can be detrmined at any load becaude constant losses are known
(iii) This test is ver vonvenient
Disadvantages of Swinburne's tst
(i) It does not take into account the stray load losses that occure when the machine is loaded.
(ii) This test does not aenable up the check the performane of the machine on full load. For
example it doed not indicte whether commutationon full load is satisfactory and whether the
temperature rise is within the specifice limits.
(iii) Thsi test does not give quite accurate efficency o fht machine. It is because iron losses
under actual load are greaer than those measured. This is manly due t armature reaction
distroting the field.
Transformer
INTRODUCTION
The transformer is probably one of the most useful electrical devices ever invented. It can change
the magnitude of alternating voltage or current from one valve to another. This useful property of
transformer is mainly responsible for the widespread use of alternating currents rather than direct
currents i.e., electric power is generated, transmitted and distributed in the form of alternating
current. Transformers have no moving parts, rugged and durable in construction, thus requiring
very little attention. Thy also have a very high efficiency-as high as 99%. In this chapter, we shall
study some of the basic properties of transformers.
8.1 TRANSFORMER
A transformer is a static piece of equipment used either for raising or lowering the voltage of an a.c.
supply with a corresponding decrease or increase in current. It essentially consists of two windings,
the primary and secondary, wound on a common laminated magnetic core as shown in Fig.8.1. The
winding connected to the a.c. sourse is called primary winding (or primary) and the one connected
to load is called secondary winding (or secondary). The alternating voltage
V
1
whose magnitude is to
be changed is applied to the primary. Depending up on the number of turns of the primary (N1) amd
secondary (N 2), an alternating e.m.f.
E
2
is included in the secondary. This inuced e.m.f.
secondary causes a secondary current I2. Consequently, terminal voltage
V
load. If V 2 >
V
V
, it is called a step up-transformer. On the other hand, if
1
E
2
in the
2
will appear acrose the
2
<
V
, it is called a
1
step-down transformer.
Working. When an alternating voltage
V
1
is applied to the primary, an alternating flux ф Is set up in
the core. This alternating flux links the windings and induces e.m.f.s
Faraday’s laws of electromagnetic induction. The e.m.f.
is termed as secondary e.m.f.
E
1
E
1
and
E
2
in them according to
is termed as primary e.m.f. and e.m.f.
E
2
E
Clearly,
1
Note that magnitudes of
E
1
= -
N
dф
and
E
*
>
E
respectively. If N 2 > N 1, then
E
hand, if N 2 > N 1, then E2 >
(or
E
1
2
1
2
V
depend up on the number of turns on the secondary and primary
1
2
(or
V
2
> V 1) and we get a step-up transformer. On the other
> V 1) and we get a step-down transformer. If load is connected
across the secondary winding, the secondary e.m.f. E2 will cause a current I2 to flow through the
load. Thus, a transformer enables us to transfer a.c. power from one circuit to another with a
change in voltage level.
The following points may be noted carefully:
(i)
The transformer action is based on the laws of electromagnetic induction.
(ii)
There is no electrical connection between the primary and secondary. The a.c. power is
transferred from primary to secondary through magnetic flux.
(iii)
There is no change in frequency i.e., output power has same frequency as the input
power.
(iv)
The losses that occur in a transformer are:
(a) core losses – eddy current and hysteresis losses
(b) Copper losses – in the resistance of the windings
In practice, these losses are very small so that output power is nearly equal to the input primary
power. In other words, a transformer has very high efficiency.
PRACTICAL TRANSFORMER
A practical transformer differs from the ideal transformer in many respects. The practical
transformer has (i) iron losses (ii) winding resistances and (iii) magnetic leakage, giving rise to
leakage reactances.
(i) Iron losses. Since the iron core is subjected to alternating flux, there occurs eddy current
and hysteresis loss in it. These two losses together are known as iron losses or core losses.
The iron losses depend upon the supply frequency, maximum flux density in the core, volume
of the core etc . It may be noted that magnitude of iron losses in quite small in a practical
transformer.
(ii) Winding resistance . Since the windigs consists of copper conductors, it immediately follows
that both primary and secondary will have winding resistance. The primary resistance R1 and
secondary resistance R2 act in series with the respective windings as shown in Fig. 8.6.
When current flows through the windings, there will be power loss as well as a loss in voltage
due to IR drop . This will affect the power factor and E1 will be less than V1 while V2 will be
less than E2.
(iii) Leackage reactances. Both primary and secondary current produce flux. The flux φ which
links both the windings is the useful flux and is called mutal flux. However, primary current
would produce some flux φ1 which would not link the secondary winding (See Fig. 8.7).
Similarly, secondary current would produce some flux φ2 that would not link the primary
winding. The flux such as φ1 or φ2 which links only one winding is called leakage flux.
The leakage flux paths are mainly throughthe air. The effect of these leakage fluxes would be
the same as though inductive reactances were connected in series with each winding of
transformer that had no loakage flux as ahown in Fig. 8.6. In other words , the effect of
primar leakage flux φ1 is to introduce an inductive reactance X1 in series with the primary
winding as shown in Fig. 8.6. Similarly, the secondary leakage flux φ2 introduces an
dinductive reactance X2 in series with the secondary winding. There will be no power loss
due to leakage reactane. However, the presence of lakage ractance in hte windings changes
the power factor as well as there is voltage loss due to IX drop.
Note : Although leakage flux in a transformer is quite small (about 5% of φ) compared to the
mutual flux φ yet it cannot be ignored. It is because leakage flux paths are through air of high
reluctance and hence require considerable e.m.f. It may be noted that energy is conveyed from the
primary winding to the secondary winding by mutual flux φ which links both the windings.
PRACTICAL TRANSFORMER ON LOAD
We shall consider two cases (i) when such a transformer is assumed to hav eno einding resisntance
and leakage flux (ii) when the transformer has winding resistance and leakage flux.
(i) No winding resisntace and leakage flux
Fig. 8.12 shows a practical transformer with the assumption that resistance and leakage reactances
of the windings are negligible. With this assumption, v2 = E2 and V1=E1. Let us take the usual case
of inductive load which causes the secondary current I2 to lag the secondary voltage V2 by ф2.
The total primary current I1 must meet two requirements viz.
(a) It must supply the no-load current l0 to meet the iron losses in the transformer and to provide
flux in the core.
(b) It must supply a current I2 to conteract the demagnetising effect to secondary current I2. The
magnitude of I2 will be such that N1I'2 = N2I2
or I'2 = (N2/N1) I2= KI2
The total primary current I1 is the phasor sumof I'2 and L0 ie,
I1 = I'2 + I0
where I'2 = -KI2
Note that I'2 is 180 0 out of phase with I2.
Phasor diagram Fig. 8.13 shows the phasor diagram for the usual caseof inductive load. Both E1 and
E2 lag behind the mutual flux ф by 90 0. The current I'2 represents the primary current to
nutralise the demagnetising effect of secondary current I2. Now I'2 = KI2 and is antiphase with I2. I0
is the noload current of the transformer. The phasor sum of I'2 and I0 gives the total primary
current I1. Note that in drawing the phasor diagram, the value of K is assumed to be unity so that
primary phasors are equal to secondary phasors.
Primar p.f = cos ф1
Secondary p.f. = cos ф2
Primary input power = V1I1 cos ф1
secondary output power = V2I2 cosф2
(ii) Transformer with resistance and leakage reactance
Fig. 8.17 shows a practical transformer having winding resistance and leakage reactances. These are
the actual conditions that exist in a transformer. There is voltage drop in R1 and X1 so that primary
e.m.f. E1 is less that the applied voltage V1. Similarly, there is voltage drop in R2 and X2 so that
secondary terminal voltage V2 is less thatn the secondary e.m.f. E2. Let us take the usual case of
inductive load which causes the secondary current I2 to lag behind the secondary voltage V2 by
ф2. The total primary current I1 must meet two requirements viz.
(a) It must supply the no-load current I0 to meet the iron losses in the transformer and to provide
flux in the core.
(b) It must supply a current I'2 to counteract the demagnetising effect of secondary current I2.
The magnitude of I'2 will be such that :
N1I'2 = N2I2
or I'2 = (N2/N1)I2 = KI2
The total primary current I1 will be the phasor sum of I'2 an I0 id,
I1 = I'2 + I0 where I'2 = -KI2
V1 = =E1+I1(R1+jX1) where I1 = I0 + (-KI2)
= -E1 +I1+Z1
V2 = E2 – I2 (R2 + jX2)
= E2 – I2 Z2
Phasor diagram . Fig 8.18 shows the phasor diagram of the practical trasformer fot he usual case of
inductive load. Both E1 and E2 lag the mutual flux ф by 900. The current I'2 represents the
primary current to neutralise the demagnetising effect of secondary current I2. Now I'2 = KI2 and is
opposite to I2. Also I0 is the no-load current of the transformer. The phasor sum of I2 and I0 gives
the total primary current I1.
Note that counter e.m.f that opposes the applied voltage V1 I1-E1. Therefore if we add I1R1 ( in
phase with I1) and I1X1 (900 of I1) to -E1, we get the applied primary voltage V1. The phasor E2
represents the induced e.m.f in the secondary by mutal flux ф. The secondary terminal voltage V2
will be what is left over after subtracting I2R2 and I2X2 from E2.
Load power factor = cos ф2
primary power factor = cos ф1
Input power to transformer, P1 = V1I1cos ф1
Output power of transformer, P2 = v2 I2 cos ф2
Note : The reader may draw the phasor diagram of a loaded trnsformer for (i) unit p.f and (ii) leading
p.f. As an exercise.
IMPEDANCE RATIO
Consider a transformer having impedance Z2 in the secodary as shown in Fig. 8.19
Z2 = v2/I2
Z1= V1/I1
N2/Z1 = (V2/V1) x(I1/I2)
Z2/Z1 = K2
ie, impedance ration (Z2/Z1) is equal to the square of voltage transformation ration. In other words,
an impedance Z2 in secondry becomes Z2/K2 when transferred to primary. Likewise, an impedance Z1
in the primary becomes K2Z1 when trensferred to the secondary.
Similarly, R2/R1 = K2 and X2/X1 = K2 s
Note the importance of above relations. We can transfer the parameters form one winding to the
other thus:
(i) A resistance R1 in the primary becomes K2R1 when transferred to the secondary.
(ii) A resistance R2 in the secondary becomes R2/ K2 when tranferred to the primary
(iii) A reactance X1 in the primary becomes K2 X1 when transferred to the secondary
(iv) A reactance X2 in the secondary becomes X2/ K2 when transferred to the primary.
Note : It is important to remember that :
(i)
When transferring resistance or reactance from primary to secondary, multiply it by K2
(ii) when transferring resistance or reactance from sedondary to primary , divide it by K2.
(iii) When transferring voltage or current from one winding the other only K is used.
EXACT EQUIVALENT CIRCUIT OF A LOADED TRANSFORMER
Fig. 8.28 shows the exact equivalent circuit of a transformer on load. Here R1 is the primary winding
resistance and R2 is the secondary winding resistance. Similarly, X1 is the leakage reactance of
primary winding and X2 is the leakage rectance of the secondary winding. The parallel circuit R 0-X0
is the no-load equivalent circuit of the transformer. The resistance R0 represents the core lossess
(hysteresis and eddy current losses) so thant current Iw which supplies the core losses is shown
passing through R0. The inductive reactance X0 represents a loss-free coil which passes the
magnetising current Im. The phasor sum of Iw and Im is the no-load current I0 of the transformer.
Note that in the equivaleeeent circuit shown in Fig. 8.28, the imperfections of the transformer have
been taken into account by various circuit elements. Therefore, the transformer is now the ideal
one. Note that equuivalent circuit has created two normal electrical circuits sseparated only by an
ideal transformer whose function is to change values according to the equation:
E2/E1 = N2/N1 = I'2/I1
The following points may be noted from the equivalent circuit:
(i) When the transformer is on no-load (ie, secondary terminals are open -circuited), there is
no current in the secondary winding. However, the primary draws a small no-load I0. The noload primary current I0 is composed of (a) magnetising current(Im) to create magnetic flux in
the core and (b) the current Iw required to supply the core losses.
(ii) When the secondary circuit of a transformer is closed through some external load ZL, the
voltage E2 induced in the secondary by mutual flux will produce a secondary current I2.
There will be I2R2 and I2 X2 drops in the secondary winding so that load voltage V2 will be less
than E2.
V2 = E2-I2 (R2+jX2) = E2 – I2Z
(iii) When the transformer is loaded to carry the secondary current I2, the primary current
consists of two components:
(a) The no-load current I0 to provide magnetising current and the current required to supply
the core losses .
(b) The primary current I'2 (-KI2) required to supply the load connected to the secondary.
Total primary current I1 = I0 + ( - KI2)
(iv) Since the transformer in Fig. 8.28 is now ideal, the primary induced voltage E1 can be
calculated from the relation :
E1/E2 = N1/N2
If we add I1R1 and I1X1 drops to E1 , We get the primary input voltage V1.
V1 = -E1 + I1 (R1 + j X1) = -E1 + I1Z1
V1 = -E1+I1Z1
SIMPLIFIED EQUIVALENT CIRCUIT OF A LOADED TRANSFORMER
The no-load current I0 of a transformer is small compared to the rated primary current. Therefore,
viltage drops in R1 and X1 due to I0 are negligible. The equivalent circuit shown in Fig. 8.28 above
can therefore , be simplified by transferring the shunt circuit R0 – X0 to the input terminals as shown
in Fig. 8.29. This modificarion leads to only slight loss of accuracy.
(i) Equivalent circuit referred to primary. If all the secondary quantities are referred to the
primary, we get the equivalent circuit of te transformer referred to the primary as shown in
Fig. 8.30. (i) This further reduces to Fig. 8.30 (ii). Note that when secondar quantities are
referred to primary, resistances/reactances/impedances are divided by K2, voltages are
divided by K and currents are multiplied by K.
R'2 = R2/K2 ; X'2 = X2/K2 ; Z'L = ZL/ K2 ; V'2 = V2/K ; I'2 = KI2
where R01 = R1 + R'2 ; X01 = X1 +X'2
(ii) Equivalent circuit referred to secondary. If all theprimary quantities are referred to
secondary, we get the equivalent circuit of the transformer referred to secondary as shown in
Fig. 8.32 (i) This further reduces to Fig. 8.32 (ii) Note that when primary quantities are
referred to secondary resistances/reactances/impedances are multiplied by K2, voltages are
divided by K and currents are divided by K.
APPROXIMATE EQUIVALENT CIRCUIT OF LOADED TRANSFORMER
The no-load current I0 in a transformer is only 1 – 3% of the rated primary current and may be
neglected without any serious error. The transformer can than be shown as in Fig. 8.36. This is an
approximate representation because no-load current has been neglected. Note that all the circuit
elements have been shown external so that the transformer is an ideal one.
As shown in Art. 8.11, if we refer all the quantities to one side (primary or secondary), the ideal
transformer stands removed and we ger the equivalent circuit.
(i) Equivalent circuit of transformer referred to primary. If all the secondary quantities are
referred to the primary, we get the secondary quatities are referred to primary, resistance
/resistance are divided by K2, voltages are divided by K and currents are multiplied by K.
The equivalent circuit shown in Fig. 8.37 is an electrical circuit and can be solved for various
currents and voltages. Thus if we find V'2 and I'2 , then actual secondary values can be
detrmined as under:
Actual secondary voltage V2 = K V'2
Actual secondary current I2 = I'2/K
(ii) Equivalent circuit of transformer referred to secondary. If all the primary quantities are
referred to secondary , we get the equivalent circuit of the transformer referred to secondary
as shown in Fig. 8.38. Note that when primary quantities are referred to secondary,
resistances/reactances are multiplied by K2, voltages are multiplied by K and currents are
divided by K.
The equivalent circuit shown in Fig. 8.38 is an electical circuit and can be solved for various
voltages and currents. Thus if we find V'1 and I'1, then actual primary values can be determined as
under:
Actual primary voltage , V1 = V'1/K
Actual primary current , I1 = KI'1
Note : The same final answers will be obtained where we use the equivalent circuit referred to
primary or secondary. The use of a particular equivalent circuit would depend upon the conditions
of the problem .
APPROXIMATE VOLTAGE DROP IN A TRANSFORMER
The approximate equivalent circuit of transformer referred to secondary is shown in Fig. 8.39. At
no-load , the secondary voltage is KV1. When a load having a lagging p.f. cos φ2 is applied, the
secondary carries a current I2 and voltage drops occur in (R2 + K2R1) and (X2 + K2X1). Consequently,
the secondary voltage falls from KV1 to V2. Referring to Fig. 8.39, we have,
V2 = K V1 -I2 [(R2 + K2 R1) + j(X2 +K2X1)]
= KV1 – I2 (R02 + j X02)
= KV1 – I Z02
Drop in secondary voltage = KV1 – V2 = I2X02
The phasor diagram is shown in Fig. 8.40. It is clear from the phasor diagram that drop in secondary
voltage is AC = I2Z02. It can be found as follows. With O as centre and OC as radius , draw an arc
cutting OA produced at M. then AC = AM = AN. From B, draw BD perpendicular to OA produced.
Draw CN perpendicular to OM and draw BL || OM.
Approcimately drop in secondary voltage = AN = AD+DN
= AD +BL
= I2 R02 cosφ2 + I2 X02 sin φ2
For a load having aleading p.f. cos φ2, we have,
Approximate voltage drop = I2R02cosφ2 – I2X02sinφ2
Note : If the circuit is referred to primary, then it can be easily established that:
Approximate voltage drop = I1R01cosφ2 ± I1X01 sinφ2
VOLTAGE REGULATION
The voltage regulation of a transformer is the arithmetic differnce (not phasor difference) between
the no-load secondary voltage (0V2) and the secondary voltage V2 on load expressed as percentage
of no-load voltage i.e,
% age volatage regulation = 0V2 -V2 /0V2 x 100
where 0V2 = No-load secondary voltage = KV1
V2 = Secondary voltage no load
As shown in Art 8.15, 0V2-V2 = I2 R02 cos φ2 ± I2 X02 sin φ2
The +ve sign is for lagging p.f. and -ve sign for lading p.f.
It may be noted that %age voltage regulation of athe transformer will be the same whether primary
or secondary side is considered.
TRANSFORMER TEST
The circuit constants , efficiency and voltge regultion ofa transformer can be determined by two
simple tests (i) open-circuit test and (ii) shor -circuit test. These tests are very convenient as tehy
provide the required information without actually loading the transformer . Further , the power
required to carry out these tests is very small as compared with full-load output of the transformer.
These tests consist of measuring the input voltage, current and power to the primary first with
secondary open -circuited (open -circuit test ) and then with the secondary short – circuited (short
circuit test).
OPEN-CIRCUIT OR NO-LOAD TEST
This test is conducted to determine the iron losses (or core losses) and parameters R0 and X0 of the
transformer. In this test, the rated voltage is applied to the primary (usually low-voltage winding)
while the secondary is left open -circuited. The applied primary voltage V1 is measured by the
voltmeter, the no-load current I0 by ameter and no -load input power W0 by wattmeter as iron
losses will occur in the transformer core. Hence wattmeter will record the iron losses and small
copper loss in the primary. Since no-load current I0 is very small (usually 2-10% of rated current),
Cu losses in the primary under no-load condition are negligible as compared with iron losses. Hence
wattmeter reading practically gives the iron lossees in the transforme. It is reminded that iron losses
are the same at all loads. Fig. 8.41 (ii) shows the equivalent circuit of transformer on no-load.
Iron losses Pi = Wattmeter reading = W0
No load current = Ammeter reading = I0
Applied voltage = Voltmeter reading = V1
Input power , W0 = V1 I0 cos φ0
N0-load p.f cos φ0 = W0/V1I0
Iw = I0cosφ0 ; Im = I0sin φ0
R0 = V1/Iw and X0 = V1/Lm
Thus open – circuit test enables us to detemine iron losses and parameters R0 and X0 of the
transformer.
SHORT CIRCUIT OR IMPEDANCE TEST
This test is conducted to determine R01 (orR02). X01 (or X02) and full-load copper loasses of the
transformer. In this test, the secondary (usually low-voltage winding) is short – circuited by a thick
conductor and variable low voltage is applied to the primary as shown in Fig.8.42 (i). The low input
voltage is gradually raised till at voltage Vsc, full-load current I1 flows in the primary. Then I2 in the
secondary also has full-load valllue since I1/I2 = N2/N1. Under such conditions, the copper loss in
the windings is the same as that on full load.
There is no ouput from the transforemer under short -circuit conditions. Therefore, input power is
all loss and this loss is almost entirely copper loss. It is because iron loss in the core is negligibly
small since the voltage Vsc is ver small. Hence the wattmeter will practically register the full load
copper losses in the transformer windings. Fig. 8.42 (ii) shows the equivalent corcuit of a
transformer on short circuit as referred to primary ; the no-load current being neglected due to its
smallness.
Full load Cu loss , Pc = Wttmeter winding = Ws
Applied voltage = voltmeter reading = Vsc
f.L primary current = Ammeter reading = I1
Pc = I21R1 + I21R'2 = I21R01
Where R01 is the total resistance of transformer referred to primary .
Total impedance referre to primary , Z01 = Vsc/I1
Total leakage rectance refrred to primary X01 = √Z201 -R201
short circuit p.f cost φs = Pc/VscI1
Thus short circuit test gives full load Cu loss, R01 and X01.
Note: The short circuit test will give full load Cu loss only if the applied voltage Vsc is such so as to
circulate full-load currents in the windings. If in a short-circuit test, current value is other than
full-load value, the Cu loss will be corresponding to that current value.
ADVANTAGES OF TRNSFORMER TESTS
The above two simple transformer test offer the following advantages:
(i) The power required to carry out these tests is very small as cmplared to the full load output
of the transformer . In case of open circuit test , power required is equal to the iron loss
whereas for a short circuit test , power required is equal to full-load copper loss.
(ii) These tests enable us to determine the efficiency of the transformer accurately at any load
and p.f. without actually loading the transformer.
(iii) The short- circuit enables us to determine R01 and X01 (or R02 and X02). We can thus find the
total voltage drop in the transformer as referred to primary or secondary. This permits us to
calculate voltage regulation of the transformer.