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Basic electrical engineeringUNIT-2

Q-1: what is a generator?

Ans: Generator is a machine which converts mechanical energy in to electrical energy.

Q-2: State the principle by which generators convert mechanical energy to

electrical energy.

Ans: It works on the principle of electromagnetic induction.

Whenever a conductor is moved within a magnetic field in such a way that the

conductor cuts across magnetic lines of flux, voltage is generated in theconductor.

The AMOUNT of voltage generated depends on

(1) the strength of the magnetic field,

(2) the angle at which the conductor cuts the magnetic field,(3) the speed at which the conductor is moved, and

(4) the length of the conductor within the magnetic field.

Q-3: What is the polarity of the generated voltage?

Ans: The POLARITY of the voltage depends on the direction of the magnetic

lines of flux and the direction of movement of the conductor. To determine the

direction of current in a given situation, the LEFT-HAND RULE FOR

GENERATORS is used. This rule is explained in the following manner.Extend the thumb, forefinger, and middle finger of your left hand at right

angles to one another, as shown in figure 1-1. Point your thumb in the directionthe conductor is being moved. Point your forefinger in the direction of magneticflux (from north to south). Your middle finger will then point in the direction of

current flow in an external circuit to which the voltage is applied.

Left-hand rule for generators.

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Q-4: Draw sketch of an Elementary Generator ?

Ans:

Q-5: Give the sectional view a D.C.Generator ?

Ans:

Construction of a dc generator (cutaway drawing).

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Q-6: What are the main parts of a d.c.generator?

Ans: Following are the principle parts of a d.c.generator 1. Magnetic frame or Yoke.

2. Pole cores & pole-shoe.

3. Pole coils or field coils.4. Armature .

5. Armature conductors or Windings.

6. Commutator.7. Brushes & Bearings.

8. Legs or support stand.

Q-7: What is a Yoke?

Ans: The outer most frame is called yoke.

It serves double purpose.1. it provides mechanical support for poles .

2. it provides less reluctance path for the flow of flux.

Q-8: What is stator & rotor?

Ans: Stator is the stationary part of the machine.Where as rotor is the rotating part of the machine.

Q-9: What is an Armature?

Ans: The rotating part of a d.c.generator which is a assembly of

conductor,commutator & the shaft is called as an armature.

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Q-10: State the purpose of slip rings.

Ans: The ends of the armature loop are connected to rings called SLIP RINGS.They rotate with the armature. The brushes, with wires attached to them, ride

against these rings. The generated voltage appears across these brushes which is

then given to the external circuit.

Q-11: What is a Commutator?

Ans:

A commutator is a physical rectifier which converts alternating voltage to

pulsating dc voltage .

A single-loop generator with each terminal connected to a segment of a

two-segment metal ring is shown in figure . The two segments of the split metal

ring are insulated from each other. This forms a simple COMMUTATOR. Thecommutator in a dc generator replaces the slip rings . This is the main difference

in their construction. The commutator mechanically reverses the armature loopconnections to the external circuit. This occurs at the same instant that the polarity

of the voltage in the armature loop reverses.

Q-12: What is commutation?

Ans: Commutation is the process by which a dc voltage output is taken from an

armature that has an ac voltage induced in it. .As commutator mechanicallyreverses the armature loop connections to the external circuit. This occurs at the

same instant that the voltage polarity in the armature loop reverses. A dc voltageis applied to the load because the output connections are reversed as eachcommutator segment passes under a brush. The segments are insulated from each

other.

Q-13: How field strength can be varied in a dc generator?

Ans: Nearly generators use electromagnetic poles instead of the permanent

magnets used in our elementary generator. The electromagnetic field poles consistof coils of insulated copper wire wound on soft iron cores, as shown in figure 1-6.

The main advantages of using electromagnetic poles are

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(1) Increased field strength and

(2) A means of controlling the strength of the fields. By varying the input voltage,

the field strength is varied. By varying the field strength, the output voltage of thegenerator can be controlled.

Four-pole generator (without armature).

Q-14: What is a Field Winding?

Ans: The windings which provided on the poles for excitation are called as field

windings.

Field Windings On A Pole Piece.

Q-15: What is armature winding?

Ans: The conductors are connected to commutator in a definite manner is

known as armature windings.

Q-16: What are the types of armature windings?

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Ans: There are two types of windings.

a) Lap winding

b) Wave winding.

The winding are lapped over one another & which are used for high current &

low voltage applications is known as lap winding.

The winding which are in the form of a wave & are used for low current & high

voltage applications is known as wave winding.

This is shown by the fig below.

Q-17: What is a parallel path?

Ans: The number of conductors per pole are connected in series . such type ofseries conductors are connected across two brushes which constitutes the parallel

path. It is represented by A.

For a lap winding number of parallel paths = number of poles- P

For a wave winding number of parallel paths = 2 always irrespective of number

of poles.

Q-18: What is field excitation?

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Ans: When a dc voltage is applied to the field windings of a dc generator,

current flows through the windings and sets up a steady magnetic field. This is

called FIELD EXCITATION.

Q-19: What is self excitated generator?

Ans: A generator that supplies its own field excitation is called a SELF-

EXCITED GENERATOR. Self-excitation is possible only if the field pole pieceshave retained a slight amount of permanent magnetism, called RESIDUAL

MAGNETISM. When the generator starts rotating, the weak residual magnetism

causes a small voltage to be generated in the armature. This small voltage applied

to the field coils causes a small field current. Although small, this field currentstrengthens the magnetic field and allows the armature to generate a higher

voltage. The higher voltage increases the field strength, and so on. This process

continues until the output voltage reaches the rated output of the generator.

Q-20: What is armature reaction?

Ans: The current-carrying conductors produce magnetic fields. The

magnetic field produced by current in the armature of a dc generator affects the

flux pattern and distorts the main field. This distortion causes a shift in the neutral

plane, which affects commutation. This change in the neutral plane and thereaction of the magnetic field is called ARMATURE REACTION.

Armature reaction.

Q-21: What are Armature losses?

Ans: In dc generators, as in most electrical devices, certain forces act to decrease theefficiency. These forces, as they affect the armature, are considered as losses and

may be defined as follows:

1. 2I R or copper loss in the winding.

2. Eddy current loss in the core.

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3. Hysteresis loss (a sort of magnetic friction).

Q-22: What causes copper losses & how it can be minimized?

Ans: The power lost in the form of heat in the armature winding of a generator isknown as COPPER LOSS. Heat is generated any time current flows in a

conductor. Copper loss is an I2 R loss, which increases as current increases. The

amount of heat generated is also proportional to the resistance of the conductor.The resistance of the conductor varies directly with its length and inversely with

its cross sectional area.

Copper loss is minimized in armature windings by using large diameter wire.

Q-23: What are eddy currents?

Ans: The core of a generator armature is made from soft iron, which is a conductingmaterial with desirable magnetic characteristics. Any conductor will have currents

induced in it when it is rotated in a magnetic field. These currents that are inducedin the generator armature core are called EDDY CURRENTS.

Q-24: What are eddy current losses?

Ans: The power dissipated in the form of heat, as a result of the eddy currents, isconsidered a loss.

Q-25: How can eddy current be reduced?

Ans: Eddy currents, just like any other electrical currents, are affected by the

resistance of the material in which the currents flow. The resistance of anymaterial is inversely proportional to its cross-sectional area. Figure 1-11, view A,

shows the eddy currents induced in an armature core that is a solid piece of soft

iron.

Figure below, view B, shows a soft iron core of the same size, but madeup of several small pieces insulated from each other. This process is called

lamination. The currents in each piece of the laminated core are considerably less

than in the solid core because the resistance of the pieces is much higher.(Resistance is inversely proportional to cross-sectional area.) The currents in the

individual pieces of the laminated core are so small that the sum of the individual

currents is much less than the total of eddy currents in the solid iron core.

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Eddy currents in dc generator armature cores.

Q-26: What is a Hysteresis Loss?

Ans: Hysteresis loss is a heat loss caused by the magnetic properties of the

armature. When an armature core is in a magnetic field, the magnetic particles of

the core tend to line up with the magnetic field. When the armature core isrotating, its magnetic field keeps changing direction.

The continuous movement of the magnetic particles, as they try to alignthemselves with the magnetic field, produces molecular friction. This, in turn,

produces heat. This heat is transmitted to the armature windings. The heat causesarmature resistances to increase. This results in losses.

Q-27: How Hysteresis Loss can be minimized?

Ans: To compensate for hysteresis losses, heat-treated silicon steel laminations

are used in most dc generator armatures. After the steel has been formed to theproper shape, the laminations are heated and allowed to cool. This annealing

process reduces the hysteresis loss to a low value.

Q-28: What is a Field Excitation?

Ans: When a dc voltage is applied to the field windings of a dc generator,

current flows through the windings and sets up a steady magnetic field. This iscalled FIELD EXCITATION.

Q-29: What is a separately excited d.c.generator?

Ans: This excitation voltage can be supplied by an outside source, such as a

battery. Hence it is called as separately exited.

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Q-30: What are the three types of self excited dc generators?

Ans: Self-excited generators are classed according to the type of fieldconnection they use. There are three general types of field connections

a) SERIES-WOUND

b) SHUNT-WOUND (parallel), andc) COMPOUND-WOUND.

COMPOUND-WOUND Generators are further classified as cumulative-compound and differential-compound.

Q-31: What is a D.C.MOTOR?

Ans: A d.c.motor is a machine which converts electrical energy to mechanical energy.

Q-32: What is the Principle of Operation of d.c.motor?

Ans: The operation of a dc motor is based on the following principle:

A current-carrying conductor placed in a magnetic field, perpendicular to the linesof flux, tends to move in a direction perpendicular to the magnetic lines of flux.

Q-33: State the right-hand rule for motors.

Ans: There is a definite relationship between the direction of the magnetic field,the direction of current in the conductor, and the direction in which the conductor

tends to move. This relationship is best explained by using the RIGHT-HAND

RULE FOR MOTORS .

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Ans: While a dc motor is running, it acts somewhat like a dc generator. There is

a magnetic field from the field poles, and a loop of wire is turning and cutting this

magnetic field. As the loop sides cut the magnetic field, a voltage is induced inthem, the same as it was in the loop sides of the dc generator. This induced

voltage causes current to flow in the loop. This voltage developed in the opposite

direction as that of supply voltage is called as back emf.

Q-38: What is the load on a dc motor?

Ans: Motors are used to turn mechanical devices, such as water pumps,

grinding wheels, fan blades, and circular saws. For example, when a motor is

turning a water pump, the water pump is the load. The water pump is the

mechanical device that the motor must move. This is the definition of a motorload.

Q-39: What is the main advantage of a series motor?

Ans: In a series dc motor, the field is connected in series with the armature. The

field is wound with a few turns of large wire, because it must carry full armaturecurrent.

This type of motor develops a very large amount of turning force, called

torque, from a standstill. Because of this characteristic, the series dc motor can be

used to operate small electric appliances, portable electric tools, cranes, winches,hoists, and the like.

Q-40: What is the main advantage of a series motor?

Ans: The speed of d.c.series motor varies widely between no-load and full-load.

Series motors cannot be used where a relatively constant speed is required underconditions of varying load.

A major disadvantage of the series motor is related to the speed

characteristics. The speed of a series motor with no load connected to it increases

to the point where the motor may become damaged. Usually, either the bearingsare damaged or the windings fly out of the slots in the armature. There is a danger

to both equipment and personnel. Some load must ALWAYS be connected to a

series motor before you turn it on.

Q-41: What advantage does a shunt motor have over a series motor?

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Ans: A shunt motor is connected in the same way as a shunt generator. The

field windings are connected in parallel (shunt) with the armature windings.

Once you adjust the speed of a dc shunt motor, the speed remains

relatively constant even under changing load conditions. One reason for this isthat the field flux remains constant.

A constant voltage across the field makes the field independent of

variations in the armature circuit. If the load on the motor is increased, the motortends to slow down. When this happens, the counter emf generated in the

armature decreases. This causes a corresponding decrease in the opposition to

battery current flow through the armature. Armature current increases, causing the

motor to speed up.The conditions that established the original speed are reestablished, and

the original speed is maintained.Conversely, if the motor load is decreased, the motor tends to increase

speed; counter emf increases, armature current decreases, and the speed

decreases.

In each case, all of this happens so rapidly that any actual change in speed isslight. There is instantaneous tendency to change rather than a large fluctuation in

speed.

Q-42: What is a Starter & What is the necessity of a Starter?

Ans: A starter is a series resistance connected in series with the armatureconductors.

Because the dc resistance of most motor armatures is low (0.05 to 0.5

ohm), and because the counter emf does not exist until the armature begins toturn, it is necessary to use an external starting resistance in series with the

armature of a dc motor to keep the initial armature current to a safe value.

As the armature begins to turn, counter emf increases; and, since the

counter emf opposes the applied voltage, the armature current is reduced. Theexternal resistance in series with the armature is decreased or eliminated

as the motor comes up to normal speed and full voltage is applied across the

armature.

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Q-43: What is a Transformer & On what principle does it operates?

Ans: Transformer is a Static device which transfers electrical power from onecircuit to another circuit magnetically at a constant frequency.

In a two winding transformer it works on the principle of mutual induction but in

a single winding transformer it works on self induction.

Q-44: What are the primary parts of a transformer?

Ans: Transformer mainly consists ofa) Core

b) Windings

core

Windings

Windings are classified into,

a) Primary winding: The windings connected to the supply side of the transformer.

b) Secondary winding: The windings connected to the load side of the transformer

Q-45: How transformers are classified & What are they ?

Ans: Depending on the construction of the core transformer is classified into two types,

a) Core type

b) Shell type

Im

Vp

Ep

Es

= Vs

m

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A) Core type B) Shell type

Q-46: What is Step-up & Step-down transformer?

Ans: Let, N1 is the number of turns on the primary side.

And N2 is the number of turns on the secondary side,

Then ,if N1 > N2 It is called Step-Up transformer.

if N1 < N2 It is called Step-Down transformer.

if N1= N2 It is called one to one transformer.

Q-47: What is an Ideal transformer?

Ans: A transformer is said to be an ideal one in which

a) The resistance of windings are neglected.

b) There is no leakage flux &c) There are no iron losses in the core.

Q-48: What is a transformer on load?

Ans: Im

+ Ip

Vp

Ep

Es

Load

Vs

Is

sp

m

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Q-49:What is the efficiency of a transformer.

Ans: Efficiency of a transformer at a particular load and power factor is

defined as

Output

Input=

The output and input are measured in the same unit, either Watt or

Kilowatt.

As the transformer is a static device, it has no mechanical losses. That is

why its efficiency is high. A better way of expressing its efficiency is

Outputwhere

Output Losses

=

+

losses = Wi + Wc

Input Losses

Input

=

Q-50: What is Regulation of transformer?

Ans: When a transformer is loaded, its secondary terminal voltage

falls(for lagging power factor) provided the supplied voltage V1 is held constant.

The variation in secondary terminal voltage from no-load to full load is called the

voltage regulation of the transformer.

i.e., regulation = 0 2 2V - V

Where 0V2= No load secondary terminal voltage

V2 = Full load secondary terminal voltage.

therefore % Regulation =0 2 2

0 2

V - V100

V

Q-51: What is an Auto-Transformer?

Ans: This is also called as variac. It is the transformer with only one winding.

The secondary winding forms a part of the primary winding.

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EXAMINATION QUESTIONS AND ANSWERS

Q-1: Explain how a D.C.generators converts mechanical energy into electricalenergy. With suitable diagrams explain the principal parts of a

D.C.generator. -10 marks-

Ans: Principle of working:

The generator operates on the principle of the production of dynamically induced

EMF i.e. whenever flux is cut by the conductor, dynamically induced EMF is

produced in it according to the Faradays laws of electromagnetic induction,

which will cause a flow of current in the conductor if the circuit is closed.

For production of dynamically induced EMF, a magnetic field, a conductor and

motion of conductor with respect to field are necessary. In D.C. generators the

field is produced by the field magnets which are stationery. Permanent magnets

are used for large machines to create magnetic flux. The conductors are situated

on the periphery of the armature being rotated by the prime-mover.

Construction: Figure 6.7 shows various parts of D.C. machine.

It consists of mainly two parts:

(1) Field system and (2) Armature

1. Field system: It is a stationary part & its function is to produce the necessary

flux. It consists of the following parts: a) Yoke b) Pole core & pole shoe and c)

Field coils.

a) Yoke: It is a cast iron (for small machines) or cast steel (for large machines)

protective frame & houses all the parts of the generator. It is cylindrical in shape.

Its functions include (i) It provides mechanical support and (ii) It carries the

magnetic flux.

b) Pole core & pole Shoe: Pole core is made of cast steel 9wrought iron or steel

alloy) laminations. It is fixed to the yoke.

It is provided with the pole shoe for the following reasons:

i) To distribute the flux uniformly in the air gap.

ii) To reduce the reluctance path by a large cross sectional area.

iii) To support field coils.

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c) Field coils or Exciting coils: Filed coils are insulated copper wire, wound round

the poles and carry D.C.

Function of poles & field coils: As field coils carry D.C. the poles become

magnetized and produce the necessary flux. This process is called as Excitation or

Energization.

2. Armature: It is a rotating part and its shaft is mechanically coupled with a

prime mover such as turbine or diesel engine etc. It consists of the following

parts.

I. Armature Core,

II. Armature conductors or winding,

III. Commutator,

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IV. Brushes and

V. Bearings.

I. Armature Core: It is a cylindrical drum shaped body. It is made of high

permeability silicon steel laminations. The outer surface is provided with slots.

II. Armature Conductors or Windings: These are insulated copper conductors

placed in armature slots.

Function of armature core and armature conductors: As armature rotates in the

magnetic field, the armature conductors cut the magnetic flux and an EMF is

induced in it according to the Faradays Laws of EMI.

III. Commutator: It is of cylindrical structure and is made of wedge shaped

segments of hard drawn copper, insulated from each other by thin layers of mica.

Each wedge shaped segment is connected to the armature conductor by means of

a copper strip (riser).

Function of the Commutator: The nature of current is induced in the armature

conductor is alternating. The Commutator collects alternating current from the

armature conductors & converts into unidirectional current (i.e. direct current)

IV. Brushes: Brushes are made of carbon and are in the shape of rectangular

block. They are housed in brush holders. The function of the brushes is to collect

current from the commutators and supplies to external load circuit.

V. Bearings: For smooth rotation of armature, bearings are used. Usually ball

bearings are used. For heavy duty machines, roller bearings are used. Bearings

packed in lubricating oil for smoother operation and to reduce bearing wear.

Q-2: With usual notations obtain the expression for EMF generated in a

D.C.generator. -5 marks-

Ans: Let Z = Total number of armature conductors.

= Useful flux per pole in Weber.N = Speed of the armature in revolutions per minute (r.p.m)

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P = Number of poles

A = Number of parallel paths.

The flux cut by a conductor in one revolution = P=d.

The time taken by the conductor to make one revolution = 60/N Sec=dt.

Hence, the E.M.F induced in one conductor = d/dt =P/60/N=pn/60 volts.

The E.M.F induced per parallel path = E.M.F induced per conductor x number of

conductors per parallel path.= pn/60 x Z/A

E = znp / 60A volts.

The above equation represents E.M.F Equation of a D.C. generator.

Q-3: What is back EMF? Explain its significance. -5 marks -

Ans: When the armature current(Ia) it rotates in a magnetic field, cuts the flux

and hence an emf is induced in it according to Faradays laws of electromagnetic

induction, whose direction is given by Flemings right hand rule.

It is in opposition to the supply voltage and hence termed as back e.m.f(Eb)

Back em f is given by

Eb=V-Ia Ra

Ia=V-Eb/Ra

From the equation

if the mechanical load on the motor is reduced, then the armature speed

increases, Eb increases and hence Ia decreases.

if the mechanical load is applied on the motor, then the armature speed decreases,

Eb decreases and hence Ia increases.

Q-4: Derive an expression for the torque developed in a D.C motor.

-6 marks-Ans: Torque is the turning movement about an axis it is equal to the product of

the force and the radius at which it acts.

Consider the armature of the DC motor to have a radius r and let F be the force

acting tangential to its surface as shown in the fig(1).

The torque exerted by the force F on the armature is given by,

Ta = F X r Nm

The work done by this force F, in one revolution is given by,

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W = Force x distance covered in one revolution

=F x 2r Wsec

The power developed by he armature = Work done in one second

=F x 2r x number of revolutions per second

=F x 2rxN/60

=2N/60(Fxr)

=2NTa/60 watts

we know that, the electrical equivalent of the mechanical power developed by the

armature of the D.C motor is equal to EI.2NTa/60=EbIa=ZNP/60A x Ia

= 1/2 X ZIa (P/A)Nm

=0.159ZIa(P/A) Nm

Ta=0.0613ZIa (P/A)Kgm. This eqn. Represents torque eqn. of the D.C motor.

The above equation gives the gross torque developed by the armature,

which includes iron losses and mechanical losses of the motor . the actualtorque available at the shaft to do useful work , which is known as shaft

torque or useful torque TSH ,IS LESS THAN Ta by an amount of torque

which is equivalent to iron losses and mechanical losses in the dc motor .

Therefore Tsh = Ta TL

Where Tsh = Shaft torque or useful torque.

Ta = Armature torque

TL = Torque lost due to iron losses and mechanical losses.

The useful torque or shaft is given by

2 NTSHOutput of motor in watts = _____________

60

Output of motor in watts

Therefore Tsh = ________________________ Nm2 N / 60

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The output of the motor is usually expressed in H.P.

Tsh = output in H.P. 735.5 Nm

Q-5: What are the different types D.C. motor? -8 marks-

Ans: With respect to winding connection to the armature, d.c.motor is been classified

into 3 types. They are ,

1. Series motor

2. Shunt motor

3. compound motor

Compound motor can be further classified in to two

a) Cumulative compound

b) Differential compound

1. Series motor: fig(1) shows the series Field winding, which is connected in series with

the armature. it has many no turns of fine wires with high resistance.

2. Shunt motor: fig(2) shows the shunt Field winding, which is connected in parallel with

the armature. It has less no turns of fine wires with low resistance.

3.compound motor:FIG(3) shows, It is the combination of shunt and series windings.

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Fig (1)

Fig(2)

Fig (3)

Q-6: Explain the speed / load characteristics of DC shunt and series motor.

-6 marks-

Ans: N/Iload Characteristics of Shunt motor :

The equation for the back E.M.F. Eb is given by,

A60

ZNPE b

=

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bE

N

as the other quantities are constant.

For D.C. shunt motor, is constant

N Eb V- Ia Ra

When the load on the machine increases the speed will decreases. But the

drop of IaRa is very small compare to voltage and hence the decrease of speed as

the armature current increases is also small. The variation of speed with respect toload as shown in fig .

N/Iload Characteristics of Series motor :

We know that( )

b a a seE V I R R

N

+

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From the equation, we find that as the load on the motor increases, there are two

factors which influence the speed of the motor.

(i) Ia (Ra + Rse) increases and hence the speed decreased.

(ii) The flux also increases due to which the speed decreases.But it has been observed that the decrease of speed due to the first factor is

negligibly small as compared to the decrease in speed due to the second factor.Hence for all practical purpose, we can say,

1a

N but I

=

aI

1N

From the above characteristics we observe that, as the load increases, the

speed decreases over a wide range. Hence a D.C., series motor is considered as a

variable speed motor. At no load, Ia is very small and hence the speed will be

dangerously very high as per equation given by N 1 / Ia. Hence D.C. motor is

started without any load on it, the speed is very high and it may run out of thefoundation due to the centrifugal forces setup. Hence DC series motor never be

started without load.

Q-7: Write the characteristics of d.c series motor & its application. -6

marks-

Ans: 1. Armature Torque v/s Armature Current (Ta

/ Ia) or Electrical

Characteristics

For series motor, Ta Ia2 (because flux aI ) Therefore as Ia increases, Ta

increases as square of Ia

Hence the curve is aparabola, represented by od

After the situation* of poles, flux remains constant

a aTI

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Hence the curve is astraight line, represented by db,

(As Tsh < Ta; Tsh is represented by a dotted line)

Observation: Hence a d.c series motor exerts huge starting torque.

As T a Ia2;series motor can be started with heavy mechanical load.

2. Speed v/s Armature current ( N/Ia)

In general back e.m.f is given by

b

b

b

60

E 60 AN =

Zp

KE 60

EN

b

NZ P E

A

AwhereK

Zp

=

= =

If change is Eb is small and can be ignored

N 1

and for series motor, aI

Therefore as Ia increases increases and hence speed decreases or vice versa.

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Therefore as Ia increase, armature resistance drop (Ia Ra) increases and hence Eb

decreases. As Eb decreases the speed decreases and hence the curve slightly

droops.

Observation:

Shunt motor is considered as constant speed motor.

2. Speed v/s Armature Torque (n / Ta) or Mechanical characteristics

It is noticed from the above characteristics, as Ta increases, speed slightly decreases.

In short

1. Medium starting torque

2. Approximately constant speed or slightly drooping speed and

3. Adjustable speed

Applications

1. Driving constant speed line shafting

2. Centrifugal and Reciprocating pumps

3. Lathes and

4. Fans and blowers.

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A starter for a D.C., motor is nothing but a resistance connected in series

with the armature during starting and which is gradually cut out, so that, the motor

picks up its rated speed. When the motor is running, back E.M.F is developed andthe armature current will be within the limits and the starting resistance is not

required.

A three point starter with its internal wiring is shown . A,B and C are thethree terminals of he starter, to which external connections are made. Rs is the

starting resistance, which is divided into various sections of equal resistances and

these sections are connected to brass studs 1,2,3n. The +ve line of the supply is

connected to the point A. the field winding is connected to point B, along with avariable resistance R, ;which is used to control the speed of the motor. The +ve

terminals A of the armature is connected to point C.

To start the motor, the supply switch is closed and the starting brass arm L

is moved the right. As soon as the brass arm touches the stud number 1 of the

starting resistance Rs, it also touches the brass arc, thus giving supply to the shunt

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field winding as well as to the armature. The motor starts rotating. The starting

arm is moved gradually on the remaining studs till the starting resistance is

completely cut out. The speed of the motor gradually increases and almostreaches the rated speed. If the rated speed is not reached, then the resistance R

which is in series with the shunt field winding is slowly varied to bring the motor

to the rated speed. As long as the motor is running and the supply is on, the brassarm L is held in the ON position by the electromagnet E, which keeps the soft

iron piece which is attached to the arm, attracted to it. When he motor is running

at its speed, the starting resistance is completely cut out of the armature circuit.

There are two protective devices in the starter. One is the electromagnet E,

which is also known as the HOLD ON COIL. When the power is failed and the

motor is running condition, the electromagnet E is de-energized and the spring Sattached to the other end of the brass arm, pulls it back to OFF position. When the

power supply is resumed again. The electromagnet E also prevents the motor

from reaching dangerously high-speed, when the field circuit is opened by

chance, when the motor is running.

The other protective device is the electromagnet M which is known as theOVER LOAD RELEASE. As long as the motor is drawing rated current and less,

the arm D is not attracted by M. But when the current increases beyond the rated

value, M attracts D, thereby short circuiting the electromagnet E. The

electromagnet E gets de-energized and hence the arm L is pulled back to OFFposition.

Q-11: Write the construction and working principal of a

transformer.

10marks-

Ans: It consists of two separate coils or windings called primary and

secondary placed on the core. The core is made up of silicon steel laminations,

which are assembled to provide a continuous magnetic path with minimum air

gap. core is laminated to reduce the eddy current loss, each lamination is insulated

from each other large rating transformers are kept in a suitable container, which

serves two purposes-provides insulation between transformer and tank and

cooling purpose. bushing are used to bring out the terminals of windings from the

tank.

A single phase transformer work on the principal of mutual induction b/w two

magnetically coupled coils when the primary winding is connected to an

alternating voltage of rms value V1 volts ,an alternating current flows through the

primary winding and sets of an alternating flux links not only primary winding

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but also the secondary windings. Therefore an emf E1is induced in the primary

winding & an emf E2 is induced in the secondary winding.

e1 & e2 are given by ,

1 1

de N

dt

=

2 2

de N

dt

=

and hence transformation ratio is given by,

1

2

E

E=

1 2

2 1

N I

N I= =K

where k is the transformation ratio.

Q-12: Derive the EMF equation of a transformer.

-6 marks-

Ans: When an alternating voltage v1= Vm sint of rms value

1 / 2mV V= is applied to the primary winding of the transformer the alternating

current flowing through the primary winding produces an alternating flux which

links both primary & secondary winding ,hence an emf e1 is induced in the

primary winding e2 is induced in the secondary winding.

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1 1

de N

dt

= as the primary induced emf & 2 2

de N

dt

= as the secondary

induced emf.

Let sinm t = be the flux which links both primary & secondary windings.

d(msint)

e1 = -N1dt

N1mcost=

N1msin(t -90)

= 2fN1msin(t - 90)

e1= Em1sin(t -90)

=

Where Em1 is the maximum value of emf is induced

1 2 1 Em fN m= The rms value emf induced is

( )

( )

2 1112 2

1 4.44 1 1

, 2 4.44 2 2

fN mEmEm

E f mN

lllly E f mN

= =

=

=

Divide eq2 by eq1 gives,

2 2

1 1

2 4.44

1 4.44

E f mN N K

E f mN N

= = =

As the power transferred from primary to secondary is same

1 1 2 2E I E I = & hence,

2

1

2 1

1 2

E I N K

E I N = = =

where K is the transformation ratio.

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Q-13: Give the constructional details of the core & shell type

transformer.-6 marks-

Ans: Depending upon the arrangement of the coil or winding on the limbs of

the transformer it is classified into

1.core type

2.shell type

1.core type : Coreis rectangular and is stacked by L section lamination it has two

limbs and one window .core is surrounded by the winding ,primary and secondary

windings wound over two limbs.

It has one magnetic path and is used for high voltage ratings.because of its simple

construction maintaince is very easy since most of winding is visible so easy to

insulate and repair .

2.shell type : core is made of E and I form of stamping .core has three limbs and

two windows and hence winding are surrounded by the core. Both primary and

secondary winding wound over the same central limbs .

It has two magnetic paths and is used for low voltage ratings.as most of the

winding are enclosed by the core it is difficult to maintained.

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Q-14: What are the different losses in a transformer? On what factor do

they depend? How are they minimized? -10

marks-

Ans: Transformer being a static machine does not contain any rotating part so there are

no mechanical losses ,there are only electrical losses associated with the device

namely

1. Iron loses or core losses.

2. Copper loses or 2I R loss.

` 1. Iron loss: it occurs in iron portion i.e., core of the transformer, this comprises

1. Eddy current loss

2. Hysterisis loss

Eddy current losses (We): this occurs due to the flow of eddy currents .the

empirical formula for this loss is

2 2 2eW Bm f t vwatts=

Where Bm is the maximum value of flux density in core in Weber /m2.

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f=frequency in Hz

v=volume of the core in m3

t= thickness of the lamination

=constant, depends on the materials used for making the core .

to keep the eddy current loss as small as possible ,the core has to be made of thin

laminations and they have to be insulated from one another by coating them with

varnish.

Hysteresis loss (Wh): this loss occurs because the core of the transformer is

subjected to cycles of magnetization. The expression for hysteresis loss is

Wh = Bm1.6 f v watts.

Where, =constant.

Iron loss= eddy current loss+ hysterisis loss

= We + Wh

The iron loss depends on BM, frequency of the supply, thickness of laminations, &

volume of the core. As long as supply voltage is constant flux density B M &

frequency are constant.

Therefore the iron loss is considered to be constant loss at all

loads.

2. Copper Loss: This loss is due to the resistances R1 & R2 of the primary &

secondary windings respectively.

Total copper loss = copper loss in the primary + copper loss in the secondary.

2

2 21 2

1 2

2 2 11

1 1 2

2

2 21

1 2

2 21

01 2 02

I R I R

R II R I [ K]

K I

R= I [R ]

K

I R ORI R

= +

= + =

+

=

Q

Where,

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R01 = equivalent resistance referred to primary =2

1 2

RR

K+

R02 = equivalent resistance referred to secondary =2

1 2R K R+

Since copper loss vary as the square of the currents I 1 & I2 which vary with load

hence it is called variable loss .Total loss in a transformer is sum of constant &

variable loss.

Q-15: Explain why the core of a transformer is laminated? State why silicon

steel is selected for the core of the transformer. -4

marks-

Ans: The core of the transformer is made of silicon steel which has a

high relative permeability & low hysteresis coefficient which reduces the

hysteresis losses occurring in the transformer.

The core is laminated to reduce the eddy current losses. The core of the

transformer is either square or rectangular in shape . it has 2 parts , the vertical

portion on which coils are wound is called as the limb of the transformer. The top& bottom horizontal portion is called as the yoke of the core.

Q-16: Define the efficiency of a transformer. - 4

marks-

Ans: Efficiency of a transformer at a particular load and power factor is

defined as

Output

Input=

The output and input are measured in the same unit, either Watt or Kilowatt.

As the transformer is a static device, it has no mechanical losses. That is why its

efficiency is high. A better way of expressing its efficiency is

Outputwhere

Output Losses=

+losses = Wi + Wc

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Input Losses

Input

= Input Losses

Output

1

Losses

Input= Input = Output + Loses or Output = Input

Losses

Q-17: What is the condition for maximum efficiency?

-10 marks-

Ans: Let, the efficiency be given by

1 (1)Losses

Input=

where Losses = Iron loss + Copper Loss

Considering the primary side,

Let, V1= Primary voltage (or Supply voltage) Volt

I1= Input current, Ampere

Cos 1 = Power factor and

R01 = Equivalent resistance of transformer as referred to primary,

Iron loss = Wi Watt

Copper loss = Wc = Ii2 R01 Watt

Primary Input = V1 I1 cos 1 Watt

Substituting for losses and input in equation (1);

2

1 01

1 1 1

2

1 01

1 1 1 1 1 1

1 01

1 1 1 1 1

(1

cos

1cos cos

1cos cos

i

i

i

I R W

V I

I R W

V I V I

I R W

V V I

+=

=

=

Differentiate equation (2) both sides with respect to I1

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1 01

1 1 1 1 1 1

01

2

1 1 1 1 1

1cos cos

0cos cos

i

i

I R W d d

d dI V V I

R Wdor

d V V I

=

= +

For max efficiency, 0d

d=

I.e.,

01

2

1 1 1 1 1

01

2

1 1 1 1 1

2

1 01

2

2 01

0

, .,

i

i

i

i

R W

V COS V I COS

W R

V I COS V COS

W I R

i e W I R

+ =

=

=

=

Or Iron loss = Copper loss.

Hence this is the condition for maximum efficiency.

Output current corresponding to maximum efficiency is

2

02

iW

IR

=

Output KVA for maximum efficiency is,

=Full load KVAironloss

fullloadcopperloss

Q-18: What is Voltage Regulation of a transformer?

-3 marks-

Ans: When a transformer is loaded, its secondary terminal voltage falls

(for lagging power factor) provided the supplied voltage V1 is held constant.

The variation in secondary terminal voltage from no-load to full load is called the

voltage regulation of the transformer.

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i.e., regulation = 0 2 2V - V

Where 0V2= No load secondary terminal voltage

V2 = Full load secondary terminal voltage.

therefore % Regulation =0 2 2

0 2

V - V100

V

Q-29: Find the number of turns required on the HT side of 415 / 240 Volt,

50 Hz 1 phase transformer. If the cross- sectional area of the core is 25 cm 2

and maximum flux density is 1.3 Weber / m2. 6

marks-

Ans: Voltage:415 /230 V f = 50 Hz

Primary/Secondary A = 25 cm2 = 25 x 10-4 m2

High Tension (HT)/Low Tension (LT)

B = 1.3 Weber / m2

E.M.F. equation for primary or HT side is given by

E1 = 4.44 fBm A N1

N1

1

m

4

1

E

4.44fB A

415575.19

4.44 50 1.3 (25 10 )

N = 575

= =

Note: In this problem, V1 = E1 = 415 V, because the transformer is on no load

Q-30: A 125 kVA transformer has primary voltage of 2000 V at 60 Hz.

Primary turns are 182 and secondary turns are 40. Calculate neglecting

losses; (1) No load secondary e.m.f; (2) Full load primary and secondary

current; (3) Flux in core.

- 6 marks-

Ans: Rating = 125 kVA

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V1 = 2000 Volt

f = 60 Hz

N1 = 182N2 = 40

(1) No load secondary e.m.f

Turns ratio K =

2

1

2

1

2 1

N 40=

N 182

EAlsoK =

E

402000182

= 439.56volt

E KE = =

(2) To find I1 or I2

I1 =

1

kVA1000

V

1251000=

2000

= 62.5A

Given that losses can be neglected; i.e. the given transformer is an idealtransformer.

Then,

2

1

1

2

II=K I

II =

K

62.5=

40182

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= 284.37 A

(3) To find flux in the core ( m )

e.m.f equation for primary is given by

1 m 1

1

m

1

m

E = 4.44f N Volt

E =

4.44fN

2000=

4.4460182

= 0.0412Wb = 41.2mWb

Q-31: In a 25 kVA, 2000 / 200 V transformer, the iron and full load copper

losses are 350 W and 400 W respectively. Calculate the efficiency at UPF at

full load & half full load. (Find also copper loss for maximum efficiency).

10marks-

Ans: Rating = 25 kVAVoltage: 2000 / 200Iron loss = Wi = 350 Watt

Copper loss = Wc = 400 Watt

Efficiency of a transformer is given by

Output =100

Input

To calculate efficiency at UPF and full load

Output at UPF and full load = (kVA x 1000)x Cos =(25 x 1000) x 1 =

=25000W

Input = Output + Losses

Where Losses = Iron loss + Copper loss = 350 + 400 = 750W Input = 25000 + 750 = 26750 W

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25000 =100

25750

= 97%

To Calculate efficiency at UPF and half full load

Output at UPF and half full load =

kVA1000 251000cos =1 =12500W

2 2

and Input = Output + Losses

where Losses = Iron loss + Copper loss

Iron loss = 350 W (This remains constant)

Copper loss at half full load =

2 2

1 1F.Lloss = 400 =100W

2 2

Losses = 350+100 = 450W

Input =12500+ 450 = 12950W

12500 =100

12950

= 96.52%

Condition for maximum efficiency is Copper loss = Iron loss

i.e. Copper loss = 350 W

Q-32: A 600 kVA single phase transformer has an efficiency of 92% both

at full load and half load at UPF. Determine its efficiency at 75% of full

load at 0.9 power factor lag. -10

marks-

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Ans: Rating: 600 kVA

Efficiency at full load = FL UPF

Efficiency at half full load = h = 92%

At first, Iron loss and Copper loss are to be determined:

Considering full load condition:

FL

Output =

Input

Output at full load & UPF = kVA x Cos = 600x 1 = 600 kW

Input =FL

Output 600

= = 652.173kW 0.92

Losses = Input Output = 652.173 600 = 52.173 kW

Let, X = Iron loss and Y = FL Copper loss

X + Y = Losses

i.e X + Y = 52.173 kW

Considering half full load condition

Output =

Input

Output at half full load & UPF =

h

600 = 300kW2

Output 300Input = = = 326.086kW

0.92

Losses = Input - Output = 326.086-300 = 26.086kW

Let X = Iron loss and

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Copper loss at half full load =2

2

1F.LCoopperloss

2

1= Y = 0.25Y

2

X + 0.25Y = Losses

i.e. X + 0.25 Y = 28.086 kW .(2)

Solving equations (1) & (2)

X + Y = 52.173

X = 0.25Y = 26.086(Substracting

0.75Y = 26.087eqn (2) from (1))

Or Y = 34.78

i.e. Copper loss = 34.78 kW

Substituting for Y in equation (1)

X + 34.8 = 52.173

Or X = 17.39

i.e Iron loss = 17.39 kW

To find efficiency at 75% full load and p.f = 0.9

Output = (0.75) kVA x p.f = (0.75) 600 x 0.9 = 405 kW

Copper loss = (0.75)2 x 34.78 = 19.56 kW

Iron loss = 17.39 kW

Input = Output + Losses = 405 + 19.56 + 17.39 = 441.95 kW

Input 405 = = = 91.63%

Output 441.95

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Q-33: A 50 kVA transformer has an efficiency of 98% at full load 0.8 p.f.

and 97% at full load 0.8 p.f. Determine the full load Copper loss and

Iron loss. Find the load at which maximum efficiency occurs and also

maximum efficiency.

10marks-

Ans: Range: 50 kVA

FL

h

FL efficiency = 98%

Efficiencyat1/2full - load = = 97%

=

p.f = 0.8

At first, Iron loss and Copper loss are to be determined:

Considering full load condition

FL

Output =

Input

Output at full load & 0.8 = kVA x Cos = 50 x 0.8 = 40 kW

FL

Output 40Input = = = 40.816kW 0.98


Let, X = Iron loss and Y = FL Copper loss

X + Y = Losses

i.e. X + Y = 0.816 kW

Considering half full load condition

Output =

Input

Output at half full load & 0.8 =40

= 20kW2

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Input =k

Output 20= = 20.61kW

0.97


Let, X = Iron loss

and Copper loss at half full load =

2 2

1 1 F.LCopperLoss = Y = 0.25Y

2 2

X + 0.25 Y = Losses

i.e X + 0.25 Y = 0.61 kW .(2)

Solving equation (1) & (2)

X + Y = 0.8.816

X + 0.25Y = 0.61(Substractingeqn(2)from(1)

0.75Y = 0.206

i.e Copper loss = 0.2746 kW = 274.66 W

Substituting for Y in equation (1),

X + 0.27466 = 0.816

Or X = 0.54133

i.e Iron loss = 0.5413 kW = 541.33 W

To find load at which maximum efficiency occurs

Load (i.e. output kVA) for maximum efficiency

Ironloss

= FullloadkVA FullloadCopperloss

0.54133= 50 = 70.19kVA

0.27466

To find efficiency

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Output = 70.19 X p.f. = 70.19 X 0.8 = 56.15 kW

Losses = Iron loss + Copper loss = 0.2746 + 0.5413 = 0.8159 kW

Input = Output + Losses = 56.15 + 0.8159 = 56.965 kW

Output 56.15 =100 = 100

Input 56.965

= 98.56%

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