DC_Motors

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

Now Pm = VIa - I2aRa

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 Ia

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 Ia

(ii) For a series motor, flux Φ is directly proportional to armature current Ia provided magnetic

saturation does not take place.

Ta α I2a

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 Ta

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 no-load.

(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 V1 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. E2 is included in the secondary. This inuced e.m.f. E 2 in the

secondary causes a secondary current I2. Consequently, terminal voltage V 2 will appear acrose the load.

If V 2 > V 1, it is called a step up-transformer. On the other hand, if V 2 < V 1, it is called a step-down

transformer.

Working. When an alternating voltage V1 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 E1 and E2 in them according to

Faraday’s laws of electromagnetic induction. The e.m.f. E1 is termed as primary e.m.f. and e.m.f. E2 is

termed as secondary e.m.f.

Clearly, E1 = - N1dф

Note that magnitudes of E1 and E2 *depend up on the number of turns on the secondary and primary

respectively. If N 2 > N 1, then E2 > E1 (or V 2 >

V 1) and we get a step-up transformer. On the other hand, if

N 2 > N 1, then E2 > E1 (or V 2 >

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 R0-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 no-load

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 + I2

1R'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 -R2

01

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.