DC MACHINES:

A DC machine is an electromechanical energy alteration device.

A DC Machine consists of a stator and a rotor with a uniform air gap in between them. The stator carries
the field windings and rotor carries the armature windings.
...
Here is a list of basic parts,
 Armature windings on rotor.
 Field windings(or magnets) on stator.
 Brushes.
 Slip rings.
 Rotor shaft.
 Stator core.
 Commutator.

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 STATOR

As the name signifies stator means the part which does not move or the stationary part of the DC
machine. Stator is mainly constituted of yoke, field poles, pole shoe, field windings etc.

 Yoke is the outer covering of the DC machine. Its main function is to provide the mechanical support and
the return path to flux. In large machines, yoke is composed in two halfs while for smaller machine in
single form.
 Field poles are usually laminated in order to reduce the eddy current loss. Its main function is setup the
necessary flux and to provide the support to the field windings.
 Poles shoes have surface area larger than the poles and are always laminated to reduce eddy current
losses. Pole shoes are jointed over the poles by of screws. It provide support to the field windings.
 Field windings are the wires or coils which remain wound over the poles in the DC machine. When
current flows through the windings it acts like an electromagnet. It remains connected with the
armature either in series or in parallel according to the type of DC machine.
 Brushes it connects the rotating armature and the external circuit. Brushes are employed to collect
current from the commutator and deliver it to the load. These are rectangular in shape and made of
carbon graphite.

ROTOR

The rotor is constituted of an armature core, armature windings and a commutator.

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 Armature core is the rotating part of the DC machine. It is made up of thin lamination having thickness
about 0.35–0.5 mm. In order to reduce eddy current losses the laminations are insulated from each
other by polish or varnish. Slots are cut on the upper periphery of the core. The slots are usually placed
parallel to the axis of armature but sometimes skewed at an angle to reduce the vibrations of the teeth.
 Armature windings are wires or coils which are wound on the armature core and placed inside the slots
of the core. The armature winding coil ends are connected to the commutator segments.

The place between two teeth is called armature slot, teeths are made by
armature stempings. The armature slots cover the whole magnetic frame.
These slots are done for wedges also. The main purpose of making these slot is
to place the winding properly over the ferro magnetic armature
 Commutator plays a very important role in the DC machine. Its main function is to convert the ac
voltage induced in the armature windings to the DC voltage in the external circuit, in case of generator
operation and to produce the unidirectional torque in case of motor operation. The ends of the armature
coils are joined to the commutator consisting of a small wedge-shaped segment. Commutator segments
are insulated from each other by the layers of mica strips. The segments are assembled side by side to
form a ring. The commutator is pressed onto the shaft.

The two types of winding mostly employed for the armatures of DC machines are known as Lap winding
and Wave winding.

The difference between the two is merely doe to different arrangement of the end connections at the
front or commutator end of armature. The coils of rotor part or armature are arranged in many options
which influence the performance of DC machines.

Lap winding:

Lap winding is the winding in which successive coils overlap each other. It is named "Lap" winding
because it doubles or laps back with its succeeding coils.

3
In this winding the finishing end of one coil is connected to one commutator segment and the
starting end of the next coil situated under the same pole and connected with same commutator
segment. Here we can see in picture, the finishing end of coil - 1 and starting end of coil - 2 are both
connected to the commutator segment - 2 and both coils are under the same magnetic pole that is
N pole here. Lap winding are of two type

(i) simplex Lap Winding

(ii) duplex Lap Winding
1. simplex Lap Winding
A winding in which the number of parallel path between the brushes is equal to the number of
poles is called simplex lap winding.

2. Duplex Lap Winding

A winding in which the number of parallel path between the brushes is twice the number of poles is
called duplex lap winding.

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Wave Winding:

Wave winding is another type of armature winding. In this winding the end of one coil is connected
to the starting of another coil of the same polarity as that of the first coil.

In this type of winding the coil side(A-B) progress forward around the armature to another coil side
and goes on successively passing through N and S pole till it returns to a conductor (A1-B1) lying
under the starting pole. This winding forms a wave with its coil, that’s why it is named as wave
winding. It is also called series winding because its coils are connected in series.

5
DEFINITIONS:

Coil span is defined as the peripheral distance between two sides of a coil, measured in term of
the number of armature slots between them. That means, after placing one side of the coil in a
particular slot, after how many conjugative slots, the other side of the same coil is placed on the
armature. This number is known as coil span.

Commutator pitch, Y, is defined as the distance between two commutator segments which two
ends of same armature coil are connected. We measure commutator pitch in term of
commutator bars or segment.

6
 c±1
For wave winding y= , C=¿ number of commutator segments, P=¿number of pairs of
P
poles.

Y must be an integer. For lap winding, commutator pitch, Y=1.

Example:

A six pole wave wound armature has 72 slots and 143 commutator segments. Find the suitable:

(i) Coil span

(ii) Commutator pitch.

SOLUTION:
72
(i) Number of slots per pole = =12
6
143 ± 1
(ii) y= =48.
3

TRIAL QUESTION:

A six pole wave wound armature has 78 slots and 155 commutator segments. Find the
suitable coil span and commutator pitch. [ Ans :13 , 52 ]

Back Pitch (Yb)

A coil advances on the back of the armature. This advancement is measured in terms of
armature conductors and is called back pitch. It is equal to the number difference of the
conductor connected to a given segment of the commutator.

Front Pitch (Yf)

The number of armature conductors or elements spanned by a coil on the front is called front
pitch.
Alternatively, we define the front-pitch as the distance between the second conductor of the
next coil which connects the front, i.e., commutator end of the armature. In other words, it is
the number difference of the conductors connected together at the back end of the armature.

The pole pitch is defined as peripheral distance between Centre of two adjacent poles in DC machine.
This distance is measured in term of armature slots or armature conductor come between two adjacent
pole centers.
Pole Pitch is naturally equal to the total number of armature slots divided by the number of poles in the
machine. If there are 96 slots on the armature periphery and 4 numbers of poles in the machine, the
numbers of armature slots come between two adjacent poles centres would be 96/4 = 24. Hence, the

7
pole pitch of that DC machine would be 24.
As we have seen that, pole pitch is equal to total numbers of armature slots divided by total numbers of
poles, we alternatively refer it as armature slots per pole.

METHODS OF EXCITATION •

Separately excited machine: field flux is produced by an external source.

Self-excited machine: The field flux is produced by connecting the field winding with the
armature. A self-excited machine requires residual magnetism for operation.

Self-excited machine

• Shunt machine • Series machine • Compound machine

1. Shunt machine • The field winding consisting of large number of turns of thin wire is
usually excited in parallel with armature circuit and hence the name shunt field winding.
• This winding will be having more resistance and hence carries less current.

8
2. Series machine • The field winding has a few turns of thick wire and is connected in
series with armature.
3. Compound machine • Compound wound machine comprises of both series and shunt
windings and can be either short shunt or long shunt.

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In DC motor or DC generator two windings are use:

 Field winding
 Armature winding.
Why we use Field winding and Armature winding
In a d.c. generator, the purpose of field winding is to produce magnetic field (called main flux) whereas the
purpose of armature winding is to carry armature current. Although the armature winding is not provided for
the purpose of producing a magnetic field, despite of that the current in the armature winding will also
produce magnetic flux (called armature flux). The armature flux distorts and weakens the main flux posing
problems for the proper operation of the d.c. generator. The action of armature flux on the main flux is called
armature reaction.

Armature reaction:

The distortion of the magnetic flux in a machine as the load is increased is called the armature reaction.

UNDESIRABLE EFFECTS OF ARMATURE REACTION:

1. Armature reaction causes a net reduction in the field flux per pole. Due to this net flux decrease,
induced armature e.m.f decreases and also the torque decreases.
2. Distortion of the main field flux along the air gap i.e. MNA axis shifted. Due to this there is a problem
of commutation which results in copper losses, iron losses, sparking etc.

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What is GNA and MNA?

The weakening of flux due to armature reaction in a dc generator also depends on the position of the
brushes.

 GNA (Geometrical Neutral Axis) may be defined as the axis which is perpendicular to the stator
field axis.
 EMF is induced in the armature conductors when they cut the magnetic field lines. But, there is
an axis along which armature conductors move parallel to the flux lines. MNA (Magnetic Neutral
Axis) may be defined as the axis along which no e.m.f is generated in the armature conductors as
they move parallel to the flux lines. Brushes are always placed in MNA because reversal of
current in the armature conductor takes place along this MNA axis.

This non uniform field distribution due to armature reaction in dc generator produces two effects:

1. Demagnetizing Effect (when main flux is weakened)

2. Cross-magnetizing Effect (when main Flux is Distorted)

Fig: (i) : No current is flowing in the armature conductors and only the field winding is energized. In this
case, magnetic flux lines due to the field poles are uniform and symmetrical to the polar axis. The MNA
(Magnetic Neutral Axis) coincides with the GNA (Geometric Neutral Axis).

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fig:(ii): armature flux lines due to the armature current. when flux is distort MNA takes a new position
and shift Θ degree to its rotating direction.

fig:(iii): in case the machine is running, both the fluxes (flux due to the armature conductors and flux due
to the field winding) will be present at a time. The armature flux superimposes with the main field flux
and, hence, disturbs the main field flux.

Due to Armature Reaction, the rotating direction of the magnetic field of the poles is distorted. Thus the
neutral plane is also altered. Therefore, the armature conductors are not at zero potential when they
come in contact with the brushes. This leads to sparking across the brushes and loss of power.

Three approaches have been developed to partially or completely correct the problems of armature reaction:

 Brush shifting.
 Compensating windings
 Commutating poles or inter-poles

1. Interpoles

 Interpoles are similar to the main field poles and located on the yoke between the main
field poles.
 They have winding in series with the armature winding.
 Interpoles have the function of reducing the armature reaction effect in the
commutating zone. They eliminate the need to shift the brush assembly.

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2. Compensating Winding

 The compensating winding are basically used to neutralize the armature flux in the pole
arc region which will otherwise cause severe distortion of main field flux.
 By using high reluctance pole tips, reduction in armature flux and using strong main field
flux poles we neutralize the armature reaction.
 These compensating winding are of concentric type and are placed in axial slots in the
pole faces.
 A compensating winding is an auxiliary winding embedded in slots in the pole faces .It is connected in
series with armature in a manner so that the direction of current through the compensating
conductors in any one pole face will be opposite to the direction of the current through the adjacent
armature conductor.

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In case of large machines having large fluctuations in load, the effect of armature reaction is
produce and sudden shift of flux oscillate depending on change in the load. This will cause due
to statically induced e.m.f. and dynamically induced e.m.f. in the armature coil.

In worst conditions these two e.m.f.s may become additive. This will occur when load is
increased on motor and decreased from generator.

 If this e.m.f. is more than the breakdown voltage across adjacent commutator segments,
a spark over may occur which can easily spread over as conditions near commutator are
favorable for flash over.
 The maximum allowable voltage between the segment is 30 to 40 V. Thus there is always
danger of short circuiting the whole armature if armature flux is not compensated.
 These winding are connected in series with the armature.
 The current in this winding flow in opposite direction to that in armature conductors
below the pole shoes.
 This will counterbalance the cross magnetizing effect of armature reaction which may
cause flash over between the segments.

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Commutation in DC Machine:
The voltage generated in the armature, placed in a rotating magnetic field, of a DC generator is
alternating in nature. The commutation in DC machine or more specifically commutation in DC
generator is the process in which generated alternating current in the armature winding of a dc
machine is converted into direct current after going through the commutator and the
stationary brushes.
Again in DC Motor, the input DC is to be converted in alternating form in armature and that is
also done through commutation.

This transformation of current from the rotating armature of a DC machine to the stationary
brushes needs to maintain continuously moving contact between the commutator segments
and the brushes. When the armature starts to rotate, then the coils situated under one pole (let
it be N pole) rotates between a positive brush and its consecutive negative brush and the
current flows through this coil is in a direction inward to the commutator segments.
Then the coil is short circuited with the help of a brush for a very short fraction of time (1⁄500
sec). It is called commutation period. After this short-circuit time the armature coils rotates
under S pole and rotates between a negative brush and its succeeding positive brush. Then the
direction is reversed which is in the away from the commutator segments. This phenomena of
the reversal of current is termed as commutation process. We get direct current from the brush
terminal.
The commutation is called ideal if the commutation process or the reversal of current is
completed by the end of the short circuit time or the commutation period.

If the reversal of current is completed during the short circuit time then there is sparking occurs
at the brush contacts and the commutator surface is damaged due to overheating and the
machine is called poorly commutated.

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 <

Physical Concept of Commutation in DC Machine

For the explanation of commutation process, let us consider a DC machine having an armature
wound with ring winding. Let us also consider that the width of the commutator bar is equal to
the width of the brush and current flowing through the conductor is I C.
Let the commutator is moving from left to right. Then the brush will move from right to left.
At the first position, the brush is connected the commutator bar b (as shown in fig 1). Then the
total current conducted by the commutator bar b into the brush is 2I C.

When the armature starts to move right, then the brush comes to contact of bar a. Then the
armature current flows through two paths and through the bars a and b (as shown in fig 2). The
total current (2IC) collected by the brush remain same.
As the contact area of the bar a with the brush increases and the contact area of the bar b
decreases, the current flow through the bars increases and decreases simultaneously. When
the contact area become same for both the commutator bar then same current flows through
both the bars (as shown in fig 3).

When the brush contact area with the bar b decreases further, then the current flowing
through the coil B changes its direction and starts to flow counter-clockwise (as shown in fig 4).

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When the brush totally comes under the bar a (as shown in fig 5) and disconnected with the bar
b then current IC flows through the coil B in the counter-clockwise direction and the short circuit
is removed.
In this process the reversal of current or the process of commutation is done.

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Methods of Improving Commutation
There are three methods of sparkles commutation:

1. Resistance Commutation
2. Voltage Commutation
3. Compensating Windings

E.M.F generated in an Armature Winding.

When an armature is rotated through one revolution, each conductor cuts the magnetic flux coming
from all the N poles and also that entering the S poles. As a result,
If ϕ=useful flux per pole∈webers , P=number of pole pairs∧¿
N=speed ∈revolutions /minute
60
Time for 1 revolution¿ seconds.
N
60 1
The time taken by a conductor to move one pole pitch is given by¿ × seconds.
N 2P

Therefore, average rate at which conductor cuts the flux ¿ ϕ ÷ ( 60N × 21P )
2 ϕNP
¿ webers / second .
60
2 ϕNP
Average e.m.f generated in each conductor ¿ volts .
60
If Z=total number of armature conductors
C=number of ∥paths through windingbetween positive∧negative brushes
Z
Then = number of conductors∈ series∈each path .
C
The number of conductors in series in each of the parallel paths between the brushes remains constant.
Hence the total e.m.f between the brushes is equal to the product of the average e.m.f per conductor
and the number of conductors in series per path.
2 ϕNP Z
E= ×
60 C
NZ
E ¿ 2 Pϕ volts .
60 C

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Example: An eight pole lap wound armature rotated at 350 rev /min is required to generate 260 V. The
useful flux per pole is about 0.05Wb. If the armature has 120 slots, calculate a suitable number of
conductors per slot.
SOLUTION.
C=8 ; E=260 V , P=4 , ϕ=0.05Wb , N=350 rev /min.
NZ
Substitute these in E ¿ 2 Pϕ to get Z=891.4
60 C
891.4
Number of conductors per slot ¿
120
¿ 7.43
This must be an even number, hence the suitable number of conductors per slot is 8.
Example: An eight pole lap wound armature has 96 slots with 6 conductors per slot and is driven at
500 rev /min . The useful flux per pole is about 0.09Wb. Calculate the generated e.m.f
SOLUTION.

C=8 , Z =( 96 ×6 )=576 ; N=500 rev /min ; ϕ=0.09Wb , P=4 , E=?

2 ϕNP Z
From E= ×
60 C
2 ×0.09 × 500× 4 × 576
E=
60 ×8
E=432 volts
Example: A four pole wave wound armature has 51 slots with 12 conductors per slot and is rotated at
900 rev /min . The useful flux per pole is about25 mWb . Calculate the value of generated e.mf.
SOLUTION.
C=2 , Z=( 51 ×12 ) =612; N =900 rev /min ; ϕ=0.025 Wb , P=2, E=?
2 ϕNP Z
From E= ×
60 C
2 ×0.025 × 900 ×2 ×612
E= ; E=459 volts
60× 2

Example: An eight pole wave connected armature has 600 conductors and is rotated at 625 rev /min .
The useful flux per pole is about20 mWb . Determine the value of generated e.m.f.
SOLUTION.

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 C=2 , Z=600 ; N =625 rev /min; ϕ=0.02 Wb , P=4 , E=?
2 ϕNP Z
From E= ×
60 C
2 ×0.02 × 625× 4 × 600
E=
60 ×2
E=500 volts
TRIAL QUESTION:
1. Calculate the flux in a four pole machine with 722 conductors generating 500V when running at
1000 rev /min , when the armature is :
(a) Lap connected.
(b) Wave connected. [ 41.55 mWb , 20.776 mWb ]
2. A four pole armature is lap-wound with 624 conductors and is driven at 1200rev/min. Calculate
the useful flux per pole required to generate an e.m.f of 250 V. [ 0.02 Wb ]

Example:

A four pole d.c machine has a lap wound armature with 50 slots with 15 conductors per slot. The useful
flux per pole is 30mWb. Determine the speed at which the machine must be driven to generate an e.m.f
of 240V.
SOLUTION:

: E=240 V , Z=( 50 ×16 )=800 , C=4 , ϕ=0.03 Wb , P=2 N=?

2 ϕNP Z 60 × E × C
But from E= × , and making N the subject N=
60 C 2 ZP× ϕ
60 ×240 × 4
N= , and N=600 rev/min ,∨10 rev /s
2 ×2 ×800 × 0.03
TRIAL QUESTION:
A six pole armature is wave wound with 410 conductors. The flux per pole is 25 mWb . Find the speed at
which the armature must be driven to generate an e.m.f of 485V.[ 946.34 rev /min ].

D.C MOTORS

An electric motor is an electrical machine which converts electrical energy into mechanical
energy. In a motor the e.m.f generated is less than the terminal voltage. DC motors were the first
form of motor widely used, as they could be powered from existing direct-current lighting power
distribution systems.
They were often used in power stations to drive emergency stand by pump systems which come into
operation to protect essential equipment and plant should the normal a.c supplies or pumps fail.

Small DC motors are used in tools, toys, and appliances. The universal motor can operate on
direct current but is a lightweight brushed motor used for portable power tools and appliances.
Larger DC motors are currently used in propulsion of electric vehicles, elevator and hoists, and

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in drives for steel rolling mills. The advent of power electronics has made replacement of DC
motors with AC motors possible in many applications.

BACK E.M.F
When a d.c motor rotates, an e.m.f is induced in the armature conductors. By Lenz’s law, this
induced e.m.f E opposes the supply voltage V and is called back e.m.f.
The direction of this induced e.m.f is such that it opposes the armature current (I a). The circuit
diagram below illustrates the direction of the back e.m.f and armature current.

V =E + I a R a

where V=terminal voltage, E=e.m.f generated in the

armature , I a=armature current , Ra=armature resistance .

Example:

1. A d.c motor operates from a 350 V supply. If the armature resistance is 0.4 Ω , determine the back
e.m.f when the armature current is 60A.

SOLUTION:

V =350V , Ra =0.4 Ω , I a=60 A ,

From V =E + I a R a , E=350−( 60 × 0.4 ) =326 V

2. A d.c motor operates from a 240 V supply. If the armature resistance is 0.2 Ω, determine the back
e.m.f when the armature current is 50A.

SOLUTION:

V =240V , Ra =0.2Ω , I a=50 A ,

From V =E + I a R a , E=240−( 50 × 0.2 )=230 V

TRIAL QUESTIONS:

1.The armature of a d.c machine has a resistance of 0.1 Ω and is connected to a 230 V supply. Calculate
the generated e.m.f when it is running as a motor taking 60 A . [ Ans :224 V ] .

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2. A d.c motor operates from a 300 V supply. If the armature resistance is 0.3 Ω, determine the back
e.m.f when the armature current is 20A. [ Ans :294 V ]

For a generator, the back e.m.f or the generated e.m.f is given by the equation

V =E−I a R a.

Example:

1.The armature of a d.c machine has a resistance of 0.1 Ω and is connected to a 230 V supply. Calculate
the generated e.m.f when it is running as a generator taking 80 A .

V =230V , Ra =0.1Ω , I a=80 A

From V =E−I a R a , E=230+ ( 80 ×0.1 )=238 V .

TRIAL QUESTION:

1. The armature of a d.c machine has a resistance of 0.25 Ω and is connected to a 300 V supply.
Calculate the generated e.m.f when it is running as a generator taking 100 A . [ Ans :325 V ]
2. The armature of a d.c machine has a resistance of 0.4 Ω and is connected to a 415V supply.
Calculate the generated e.m.f when it is running as a generator taking 50 A . [ Ans : 435 V ]

Importance of back e.m.f:

Magnitude of back emf is directly proportional to speed of the motor. Consider the load on a dc
motor is suddenly reduced. In this case, required torque will be small as compared to the
current torque. Speed of the motor will start increasing due to the excess torque. Hence, being
proportional to the speed, magnitude of the back emf will also increase. With increasing back
emf armature current will start decreasing. Torque being proportional to the armature current,
it will also decrease until it becomes sufficient for the load. Thus, speed of the motor will
regulate.

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On the other hand, if a dc motor is suddenly loaded, the load will cause decrease in the speed.
Due to decrease in speed, back emf will also decrease allowing more armature current.
Increased armature current will increase the torque to satisfy the load requirement.

Hence, presence of the back emf makes a dc motor ‘self-regulating’.

Back EMF is a good and necessary phenomenon that makes running of motors possible, it
assumes menacing proportions in the operation of relays and solenoid. When the switch is
opened, the current from the battery stops flowing instantly. However, the energy in the relay
or solenoid “opposes the original change in magnetic flux”, which is now trying to collapse.

Speed of a motor.
The speed of a motor can be defined as the rate of rotation around an axis usually expressed
in radians or revolutions per second or per minute.
D.C. motors convert electrical energy into rotational energy. That rotational energy is then
used to lift things, propel things, turn things, etc... When we supply the specified voltage to a
motor, it rotates the output shaft at some speed.
Z NP
The relationship between the generated e.m.f E , speed N , flux ds is given by E=2 × ×Φ .
C 60
Z P
For a given machine, Z,C and P are constants. This means that E=k Nϕ where k =2 × .
C 60
From V =E + I a R a , we seethat V =kNϕ+ I a R a.
V −I a Ra
Therefore N= ,
kϕ
V
But the quantity I a R a is very small compared to V, and we can conclude that N ≈ .
kϕ
The speed of an electric motor ia approximately proportional to the voltage applied to the
armature and inversely proportional to the flux.
Torque of a d.c machine:

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The quantitative measure of the tendency of a force to cause or change rotational motion is
called torque.

In a practical sense, the torque is the driving force that the output shaft will have, which is
completely independent of the time it takes to deliver this force.

It is measured in Newton metre (Nm).

From V =E + I a R a and multiplying each term by I ayields

2
V I a=E I a+ I a Ra

Where V I athe total electrical power is supplied to the motor, I a2 R a is the loss due to armature
resistance and E I a is the mechanical power developed by the armature.

The mechanical power developed is given by Tω watts.

Therefore Tω=2 πNT =E I a

E Ia
Torque T = Newton metre.
2π N

But the generated e.m.f E=

2 P Φ NZ
C
, It can be seen that2 πNT =E I a=(2 P Φ NZ
C )
I a.

Therefore2 πNT = ( 2 P Φ NZ
C )
I a and torqueT =
PϕZ I a
πC
.

For a given machine, Z, C and p are fixed values, hence torqueT ∝ ϕ I a.

Examples:
1. A four pole motor is fed at 440 V and takes an armature current of 50 A. the resistance
of the armature circuit is 0.28 Ω. the armature is wave –wound with 888 conductors and te
useful flux per pole is 0.023 Wb . Calculate its speed and the gross torque.

SOLUTION.
C=2 , Z=888; N =? , ϕ=0.023Wb , P=2 ,V =440 V , E=? Ra=0.28Ω . ; I a=50 A

24
 From V =E + I a R a , E=440−( 50 ×0.28 ) =426 V .

60 × E × C ( 60 ×426 × 2 )
But N= = =:625.7 rev /min ,.
2 ZP× ϕ ( 2× 888 ×2 ×0.023 )
PϕZ I a
Also Torque T = .
πC
( 2× 0.023 ×888 ×50 )
Then T = =325 Nm
( 3.142 ×2 )
2. A four pole motor has its armature lap-wound with 1040 conductors and runs at
1200rev/min when taking an armature current of 60 A from a 230 V supply. The
resistance of the armature circuit is 0.2 Ω. Calculate the useful flux per pole and the torque
developed.
SOLUTION.
C=4 , Z=1040 ; N =1200 rev/min , ϕ=? , P=2 , V =230 V , R a=0.2 Ω . ; I a =60 A

From V =E + I a R a , E=230−( 60 × 0.2 )=218 V .

60 × E × C ( 60 × 218× 4 )
But ϕ= = =10.48 mWb .
2 ZP× N ( 2× 1040 ×2 ×1200 )
PϕZ I a
Also Torque T = .
πC
( 2× 0.01048 ×1040 ×60 )
Then T = =104 Nm
( 3.142× 4 )

3. A six pole , 460 V shunt motor has its armature wave –wound with 936 conductors.The
useful flux per pole is 0.02Wb and the resistance of the armature circuit is 0.7 Ω.
Calculate the speed and the torque developed when the armature current is 30 A.

SOLUTION.
C=2 , Z=936 ; N =? , ϕ=0.02Wb , P=3 , V =460V , E=? Ra =0.7 Ω .; I a=30 A

From V =E + I a R a , E=460−( 30 ×0.7 )=439 V .

60 × E × C ( 60 × 439× 2 )
But N= = =469 rev /min ,.
2 ZP× ϕ ( 2× 936 ×3 × 0.02 )
PϕZ I a
Also Torque T = .
πC

25
 ( 3× 0.02 ×936 × 30 )
Then T = =268 Nm
( 3.142 ×2 )
4. An eight - pole d.c motor has a wave –wound with 900 conductors.The useful flux per
pole is25 mWb .Determine the torque exerted when a current of 30 A flows in each
armature conductor.
SOLUTION.
C=2 , Z=900; N =? , ϕ=0.025 Wb , P=4 , ; I a =30 A ,T =?
PϕZ I a ( 4 × 0.025 ×900 ×30 )
Torque T = . Therefore T = =429.66 Nm
πC ( 3.142× 2 )

5. A six pole lap –wound motor is connected to a 250V d.c supply.The armature has 500
conductors and a resistance of 1 Ω. The flux per pole is 20mWb. Calculate the speed
and the torque developed when the armature current is 40A.
SOLUTION.
C=6 , Z =500 ; N =? , ϕ=0.02Wb , P=3 , V =250 V , E=? R a=1 Ω .; I a=40 A

From V =E + I a R a , E=250−( 40 ×1 )=210 V .

60 × E × C ( 60 ×210 ×6 )
But N= = =1260 rev /min ,
2 ZP× ϕ ( 2× 500 ×3 ×0.02 )
PϕZ I a
Also Torque T = .
πC
( 3× 0.02 ×500 × 40 )
Then T = =63.654 Nm
( 3.142 ×6 )
TRIAL QUESTIONS:
1. A DC motor takes armature curent of 120A at 460V.The resistance of the armature
circuit is 0.2 Ω. The machine has 8 poles and the armature is lap connected with 960
conductors. The flux per pole is 0.08 Wb . calculate;
(i) Back e.m.f
(ii) The speed
2. An eight - pole wound d.c motor has a 200 V supply. The armature has 800 conductors
and a resistance of0.8 Ω .If the useful flux per pole is40 mWb .Calculate the speed and
the torque developed when the armature current is 30 A.

Types of DC Motors

A DC motor (direct current motor) has a lot of applications in today’s field of engineering and
technology. From electric shavers to automobiles – DC motors are everywhere. To cater to this

26
wide range of applications – different types of DC motors are used depending on the
application.

The types of DC motor include:

 Permanent Magnet DC Motor (PMDC Motor)

 Separately Excited DC Motor
 Self-Excited DC Motor
 Shunt Wound DC Motor
 Series Wound DC Motor
 Compound Wound DC Motor
 Short shunt DC Motor
 Long shunt DC Motor
 Differential Compound DC Motor

Separately Excited DC Motor

27
As the name suggests, in case of a separately excited DC motor the supply is given separately to
the field and armature windings. The main distinguishing fact in these types of DC motor is that,
the armature current does not flow through the field windings, as the field winding is energized
from a separate external source of DC current as shown in the figure beside.

From the torque equation of DC motor we know Tg = Ka φ Ia So the torque in this case can be
varied by varying field flux φ, independent of the armature current I a.

Permanent Magnet DC Motor

28
The permanent magnet DC motor (also known as a PMDC motor) consists of an armature
winding as in case of a usual motor, but does not necessarily contain the field windings. The
construction of these types of DC motor are such that, radially magnetized permanent magnets
are mounted on the inner periphery of the stator core to produce the field flux.

The rotor on the other hand has a conventional DC armature with commutator segments and
brushes. The diagrammatic representation of a permanent magnet DC motor is given below.

The torque equation of DC motor suggests

Here φ is always constant, as permanent magnets of required flux density are chosen at the
time of construction and can’t be changed thereafter.

For a permanent magnet DC motor

Where, Ka1 = Ka.φ which is another constant. In this case, the torque of DC Motor can only be
changed by controlling the armature supply.

Self Excited DC Motor

In case of self-excited DC motor, the field winding is connected either in series or in parallel or
partly in series, partly in parallel to the armature winding. Based on this, self-excited DC Motors
can be classified as:

1. Shunt wound DC motor

2. Series wound DC motor
3. Compound wound DC motor

Shunt Wound DC Motor

29
In case of a shunt wound DC motor or more specifically shunt wound self
excited DC motor, the field windings are exposed to the entire terminal
voltage as they are connected in parallel to the armature winding.

Series Wound DC Motor

In case of a series wound self excited DC motor or simply series wound DC motor, the entire
armature current flows through the field winding as it is connected in series to the armature
winding. The series wound self excited DC motor is diagrammatically represented below for
clear understanding.

30
Compound Wound DC Motor

The compound excitation characteristic in a DC motor can be obtained by combining the

operational characteristic of both the shunt and series excited DC motor. The compound wound
self excited DC motor or simply compound wound DC motor essentially contains the field
winding connected both in series and in parallel to the armature winding as shown in the figure
below:

The excitation of compound wound DC motor can be of two types depending on the nature of
compounding.

Cumulative Compound DC Motor

When the shunt field flux assists the main field flux, produced by the main field connected in
series to the armature winding then it is called cumulative compound DC motor.

Differential Compound DC Motor

In case of a differentially compounded self excited DC motor i.e. differential compound DC

motor, the arrangement of shunt and series winding is such that the field flux produced by the
shunt field winding diminishes the effect of flux by the main series field winding.

The net flux produced in this case is lesser than the original flux and hence does not find much
of a practical application.

31
Both the cumulative compound and differential compound DC motor can either be of short
shunt or long shunt type depending on the nature of arrangement.

Short Shunt DC Motor

If the shunt field winding is only parallel to the armature winding and not the series field
winding then it is known as short shunt DC motor or more specifically short shunt type
compound wound DC motor. The circuit diagram of a short shunt DC motor is shown in the
diagram below.

Long Shunt DC Motor

If the shunt field winding is parallel to both the armature winding and the series field winding
then it’s known as long shunt type compounded wound DC motor or simply long shunt DC
motor. The circuit diagram of a long shunt DC motor is shown in the diagram below.

32
SPEED AND TORQUE CHARACTERISTICS OF DC MOTORS:
(a) Shunt –Wound motor
In the shunt-wound motor, the field winding is in parallel with the armatire across the supply.

33
The Input Voltage or supply voltage , V =E + I a R a or the generated e.m.f E=V −I a R a.
From kirchoff’s law, the supply current I =I a + I f

where I f is the field current , I a is the armature current.

Example:

A 240 V shunt motor takesa total current of 80 A. If the field winding resistance is 120Ω and the
armature resistance is 0.4Ω, determine the current in the armature and the back e.m.f.

SOLUTION:

R f =120 Ω, Ra =0.4 Ω , I =80 A , V =240 V

V 240
Field current , I f = = =2 A but I a=I −I f =80−2=78 A
R f 120

Hencethe armature current I a=78 A .

Back e.m.f , E=V −I a R a=240−( 78 ×0.4 )=208.8 V .

TRIAL QUESTION:

1.A 240 V shunt motor takesa total current of 30 A. If the field winding resistance is 150Ω and the
armature resistance is 0.4Ω, determine the current in the armature and the back e.m.f.
[ Ans :28.4 A ,228.64 V ]
2.A 200 V shunt motor takesa total current of 40 A. If the field winding resistance is 80Ω and the
armature resistance is 0.6Ω, determine the current in the armature and the backe.m.f.
[ Ans :37.5 A , 177.5 V ]

CHARACTERISTICS:

34
(i) The theoretical torque Vs Armature current characteristic is derived from the expression
T =Φ I a. The field winding is connected in parallel with the armature circuit, and thus the
applied voltage gives a constant field current, The shunt wound motor isa constant flux
machine. Sice flus is a constant, it follows that T∝ I a and the characteristic is shown below.

(ii)The armature circuit of a d.c motor has resistance due the armature windings and brushes Ra
and when armature current I a is flowing through it, there is avoltage drop I a R a volts. Even
though the machine is a motor, because conductors are rotating in a magnetic field, a voltage
E
E ∝Φ ω. From the equation E ∝Φ N , it follows that N ∝ .
Φ
E V −I a Ra
The speed of rotaton N ∝ ∝ , For a shunt wound motor V,Φ∧Ra are constants.
Φ Φ
Hence as armature current I a increases, I a R a increases and V −I a Ra decreases and the speed is
proportional to the quantity which is decreasing.

As the load on the shaft of the motor increases, I a increases and the speed drops slightly. In
practice , the sopeed falls by about 10 % between no-load and full load on many d.c shunt
wound motors.Because of this fact, d. c shunt wound motors is basically aconstant speed
machine and is used for driving lathes,lines of shafts, fans, conveyor belts , pumps, compressors
and driling machines.

35
(iii) Since torque is proportional to armature current, the speed/torque charactristic is shown
below.

Example 1:
A 200 V d.c shunt wound motor has an armature resistance of 0.4Ω and a certain load has an
armature current of 30 A and runs at 1350 rev/min. If the load on the shaft is increased so that the
armature current increases to 45 A , determine the speed of the motor assuming the flux remains
constant.

SOLUTION:

V =300V , I a 1=30 A , R a=0.4 Ω , N =1350 rev /min , I a 2=45 A .

From E=V −I a R a, E1=200−( 30 × 0.4 )=188 V ∧E2=200−( 45 × 0.4 ) =182V .

E1 Φ 1 N 1
Also from E ∝Φ N , it follows that = .
E2 Φ 2 N 2

Since flux is constant, Φ 1=Φ2.

188 Φ 1 × 1350
= . Therefore N 2=1306.9 rev /min ¿ 21.78 rev /s
182 Φ 2× N 2

Example 2:
A 350 V shunt motor runs at its normal speed of 12 rev/s when the armature current is 90 A.
The resistance of the armature is 0.3Ω. Determine the
(i) Speed when the current is 45A and aresistance of 0.4Ω is connected in series with the armature, the
shuntfield remaing constant.

(ii) Speed when the current is 45 Aand the shunt field is reduced to 75% of its normal value by increasing
resistance in the field circuit.

36
SOLUTION:

(i) V =350V , I a 1=90 A , Ra=0.3Ω , N =12rev / s, I a 2=45 A .

From E=V −I a R a, E1=350−( 90 × 0.3 )=323 V ∧E2=350−( 45 × 0.7 )=318.5 V .

E1 Φ 1 N 1
Also from E ∝Φ N , it follows that = .
E2 Φ 2 N 2

Since flux is constant, Φ 1=Φ2.

323 Φ 1 × 12
= . Therefore N 2=709.97 rev /min ¿ 11.832 rev /s
318.5 Φ2 × N 2

(ii) V =350V , I a 1=90 A , Ra=0.3Ω , N =12rev / s, I a 2=45 A .

From E=V −I a R a, E1=350−( 90 × 0.3 )=323 V ∧E2=350−( 45 × 0.3 )=336.5 V .

E1 Φ 1 N 1
Also from E ∝Φ N , it follows that = .
E2 Φ 2 N 2

S Φ 1=Φ , Φ2=0.75Φ .
323 Φ ×12
=
336.5 0.75Φ × N 2
. Therefore N 2=1000.1 rev /min ¿ 16.669rev / s

Example 3:
A 220 V d.c shunt –wound motor runs at 800 rev/min and the armature current is 30 A. The
armature circuit resistance is 0.4Ω. Determine
(i)The maximum value of the armature current if the flux is suddenly reduced by 10 %

(ii)The steady value of the armature current at the new value of flux , assuming that the shaft torque of
the motor remains constant.

SOLUTION:

(i) V =220V , I =30 A , R a=0.4 Ω , N =800 rev /s , Φ 1=Φ , Φ2=0.9Φ

From E=V −I a R a, E1=220−( 30 × 0.4 )=208V .

The generated e.m.f is such that E ∝Φ N , so at the instant the flux is reduced; the speed has not
changed.
0.9Φ
Hence E2=208× =187.2V and the voltage drop due to the armature resistance is given
Φ
by 220−187.2=32.8 V

37
 32.8
The instantaneous value of current is =82 A . This increase in current is about three times
0.4
the initial value and causes an increase in torque ( T ∝ Φ I a ). The motor accelerates because of
the large torque value until steady state conditions are reached.

(ii)From T ∝ Φ I a and since the torque is constant.Φ 1=Φ , Φ2=0.9Φ , I 1=30 A , I 2=?
30 1
It follows that Φ 1 ×30=0.9 Φ1 × I a 2 ; Therefore I a 2= =33 A .
0.9 3
(b)Series -wound motor

In the series –wound motor the field, the field winding is in series with the armature across the supply.

For the seires motor shown above; the armature current I a flows in the field winding and is
equal to the supply current I.

The supply voltage , V =E + I ( R a+ R f ) or generated e.m.f E=V −I ( R a + Rf )

Where Ra =armature resistance , R a=field resistance∨series resistance .

CHARACTERISTICS:

(i)Since the armature current I a that flows in the field winding and is equal to the supply
current, then from .T ∝ Φ I a thenT ∝Φ I over a limited range before magnetic saturatiuon of the
magnetic ctrcuit of the motor is reached.
Thus Φ ∝ I ∧T ∝ I 2. After magnetic saturation , Φ almost becomes a constant and T ∝ I .

The theoretical torque/current characteristic is shown below.

38
 ( V −IR )
(ii)In a series motor, I a=I ∧below the magnetic saturation level, Φ ∝ I . Therefore N∝ where
I
R is the combined resistance of the series field and armature circuit. Since IR is small compared to V ,
V 1
then an approximate relationship for the speed N ∝ ∝ since V is constant.
I I
Hence the theoretical speed/current characteristic is shown below.

The high speed at small values of current indicate that this type of motor must not be run on light loads.

(iii) The speed/torque characteristic is obtained by plotting values of torque and speed for various values
of current. This is shown below.

The d.c. series motor takes a large current on starting, it requires a large torque when current is large.

Such motors are used for traction (such as trains, milk delivery vehicles); driving fans and for cranes and
hoists, when a large initial torque is required.

Example 1:

A series motor has an armature resistance of 0.2 Ω and a series field resistance of0.3 Ω. It is connected
to a 240 v supply and at a particular load runs at 24 rev/s when drawing 15A from the supply.

39
(a)Determine the generated e.m.f at this load.

(b)Calculate the speed of the motor when the load is changed such that the current is increased to 30 a.
assume that this causes a doubling in flux.

SOLUTION:

(a)Since E=V −I a ( Ra + R f ) and V =240V , I a=15 A , R a=0.2 Ω , R f =0.3 Ω then by substitution

E1=240−( 15 ) ( 0.2+0.3 )=232.5 V

(b) When the current is increased to 30 A, then I a 2=30 A and from E=V −I a ( Ra + R f ); it follows that
E2=V −I a 2 ( R a+ R f ) ;

This implies E2=240−( 30 )( 0.2+0.3 )=225 V .

E1 Φ 1 N 1
Also E ∝Φ N thus = , but Φ 1=Φ and Φ 2=2Φ , N 1=24 rev /s
E2 Φ 2 N 2
232.5 Φ ×24
Therefore =
232 2 Φ × N 2
, giving N 2=11.61rev / s=696.77 rev /min

Example 2:

A series motor has an armature resistance of 0.15 Ω and a series field resistance of0.25 Ω. It is
connected to a 220 V supply and at a particular load runs at 20 rev/s when drawing 20 A from the
supply. Determine the e.m.f generated at this load. Determine also the speed of the motor when the
load is changed such that the current increases to 25 A. Assume the flux increases by 25%.

SOLUTION: Since E=V −I a ( Ra + R f ) and V =220V , I a=20 A , Ra=0.15Ω , R f =0.25 Ω then by

substitution E1=220−( 20 )( 0.15+ 0.25 )=212 V

(b) When the current is increased to 25 A, then I a 2=25 A and from E=V −I a ( Ra + R f ); it follows that
E2=V −I a 2 ( R a+ R f ) ;

This implies E2=220−( 25 )( 0.15+ 0.25 )=210 V .

E1 Φ 1 N 1
Also E ∝Φ N thus = , but Φ 1=Φ and Φ 2=1.25 Φ , N 1=20 rev / s
E2 Φ 2 N 2
212 Φ × 20
Therefore, =
210 1.25 Φ × N 2
, giving N 2=15.85 rev /s=950.94 rev /min

Example 3:
A series wound motor is connected to a d.c supply and develops full load torque when the
current is 30 A and speed is 1000rev/min. If the flux per pole is proportional to the current
flowing, find the current and speed at half full load torque, when connected to same supply.
SOLUTION:
40
 1
Since Φ ∝ I , thenT ∝ I 2 and also N ∝ for a series wound motor .
I
1
But T 1=T , T 2= T , I 1=30 A , I 2=? , N 1=1000 rev /min , N 2=?
2
2
T1 I 1
2 T 30
= 2 2 900
Therefore, = 2 . This Implies that 1 I , And I 2 = , I 2=21. 2 A .
T2 I 2 T 2 2
2
1 1 N I
Also N 1 ∝ ∧N 2 ∝ . This means that 1 = 2 .
I1 I2 N 2 I1
N 1 × I 1 ( 1000 ) × ( 30 )
Therefore N 2= = =1415 rev /min
I2 ( 21.21 )

Example 4:
A series wound motor runs at 600 rev/min when taking 110 A from a 230 V supply. The
resistance of the armature circuit is 0.12 Ω and that of the series winding is 0.03 Ω. Calculate the
speed when the current has fallen to 50 A, assuming the useful flux per pole for 110 A to be 0.024 Wb
and that for 50 A to be 0.0155Wb.

SOLUTION:

Since E=V −I a ( Ra + R f ) and V =23 0V , I a 1=110 A , R a=0.12 Ω , R f =0.03 Ω then by substitution

E1=23 0−( 110 )( 0.12+0.03 )=213.5 V

(b) When the current has decreased to 50 A, then I a 2=5 0 A and from E=V −I a ( Ra + R f ); it follows
that E2=V −I a 2 ( R a+ R f ) ;

This implies E2=23 0−( 5 0 ) ( 0.12+0.03 )=222.5 V .

E1 Φ 1 N 1
Also E ∝Φ N thus = , but Φ 1=0.024 Wb and Φ 2=0.0155 Wb , N 1=600 rev /min
E2 Φ 2 N 2

213.5 0.024 × 600

Therefore, =
222.5 0.0155 × N 2
, giving N 2=16.14 rev /s=968.2 rev /min

Example 5:
A d.c motor has a speed of 900 rev/min when connected to a 460 V supply. Find the
approximate value of speed of the motor when connected to a 200V supply , assuming the flux
decreased by 30 % and neglecting armature voltage drop.
SOLUTION:
Since armature voltage drop can be neglected, then E=V .

41
 E1 Φ 1 N 1
From E ∝Φ N , it follows that = .
E2 Φ 2 N 2

But V 1=460 V ,V 2=200V , N 1=900 rev /min , N 2=? , Φ 1=Φ , Φ2=0.7 Φ

460 Φ ×900
Therefore =
200 0.7 Φ × N 2
, giving N 2=559 rev /min

(c) There are two types of compound- wound motor

(i) Cumulative compound wound motor: Is the one in which the series winding is o connected
that the field due to it assists that due to the shunt winding.
(ii) Differential compound wound motor: Is the one in which the series winding is so connected
that the field due to it opposes that due to the shunt winding.

The diagram below show a compound d.c motor, a long –shunt compound motor and a short –
shunt compound motor.

42
CHARACTERISTICS:
A compound-wound motor has both a series and a shunt field winding and is usually wound to
have characteristics similar in shape to a series wound motor. A limited amount of shunt
winding is present to restrict the no-load speed to a safe value. By varying the number of turns
on the series and shunt winding and the direction of the magnetic fields produced by these
windings, the characteristic graphs below are obtained.

43
Compound wound motors are used for heavy duties where sudden heavy loads may occur such
as for driving plunger pumps, presses, gear lifts, hoists, conveyors and etc.

44