VTU EDUSAT PROGRAMME: 20
SUB: Basic Electrical Engineering
(15ELE15/25)
Notes on MODULE 2: DC Machines
Faculty:
Dr. Vidya H.A
Prof. & HoD, Dept. of E & EE,
Global Academy of Technology,
Ideal Homes Township, RR Nagar,
Bengaluru – 98.
vidyakrishna_ag@yahoo.co.in
hodeee@gat.ac.in
1
DC Machines
Direct Current machines deal with conversion of one form of energy to
another. A DC machine can be:
1. DC Generators – Converting Mechanical energy into electrical
energy.
2. DC Motors – Converting electrical energy into mechanical energy.
Working Principle of DC Machine as a Generator:
A generator works on the principles of Faraday’s law of electromagnetic
induction.
Whenever a conductor is moved in the magnetic field such that it cuts across
the lines of flux, dynamically inducede.m.f is produced according to
Faraday’s laws of electromagnetic induction. The magnitude of this induced
e.m.fin the conductor is given by the equation,
e = Blv sin θ
Where
l = length of the portion of the conductor within the magnetic field,
v = velocity of the conductor,
B = magnetic flux density and
θ = angle between direction of movement of the conductor and the
direction of magnetic flux.
The following diagram explains the principle operation of dc
machine as a generator.
The essential requirements are:
1. Conductor, 2. Magnetic field, and 3. Mechanical energy
2
This e.m.f causes a current to flow in the conductor if the conductor circuit
is closed. Thus, electrical power develops in the conductor. If the conductor
does not move or if it is moved parallel to the lines of flux, no e.m.f is
induced in it, and hence no power is generated. Hence it is clear that, for the
generation of e.m.f there should be relative motion between the conductor
and the magnetic field. This causes a change in the flux linking the
conductor which causes a voltage to be induced in the conductor.
The direction of the induced e.m.f (or the current) in the conductor in the
case of a dc generator is given by Fleming’s Right Hand Rule (or
Generator Rule)
According to Fleming, the right hand is held with the thumb, first
finger and second finger mutually perpendicular to each other (at right
angles), as shown in the diagram.



The thuMb is pointed in the direction of Motion of the conductor.
The First finger is pointed in the direction of the magnetic Field. (north
to south)
Then the seCond finger represents the direction of the induced e.m.f. or
Current (the direction of the induced current will be the direction of
conventional current; from positive to negative).
Principle of DC Generator
A DC generator produces direct power based on fundamental principle of
Faraday's laws of electromagnetic induction. According to these laws, when
a conductor moves in a magnetic field it cuts magnetic lines force, due to
3
which an e.m.f is induced in the conductor. The magnitude of this induced
e.m.f depends upon the rate of change of flux (magnetic line force) linkage
with the conductor. This e.m.f will cause current to flow if the conductor
circuit is closed.
Hence the two essential parts of a generator are
a) a magnetic field and
b) conductors which move inside that magnetic field.
The above figure shows, a single loop of conductor of rectangular shape is
placed between two opposite poles of magnet.
Consider, the rectangular loop of conductor is ABCD which rotates inside
the magnetic field about its own axis ab. When the loop rotates from its
vertical position to its horizontal position, it cuts the flux lines of the field.
During the movement two sides, i.e. AB and CD of the loop cut the flux
lines there will be an e.m.f induced in both sides (AB & DC) of the loop.
As the loop is closed there will be a current circulating through the loop. The
direction of the current can be determined by Fleming’s right hand Rule.
The waveform of the current through the load circuit is as shown in the
figure. This current is unidirectional.
4
Working Principle of DC Machine as a Motor:
A DC motor in simple words is a device that converts direct current
electrical energy into mechanical energy.
The very basic construction of a dc motor contains a current carrying
armature conductor, connected to the supply through commutator segments
and brushes and placed within the north south poles of a permanent or an
electro-magnet.
Fleming's Left Hand or Motor Rule determines the direction of rotation of
current carrying conductor in a fixed magnetic field.
Fleming’s left hand rule says that if we extend the index finger, middle
finger and thumb of our left hand in such a way that the current carrying
conductor is placed in a magnetic field (represented by the index Finger) is
perpendicular to the direction of current (represented by the SeCond or
middle finger), then the conductor experiences a force in the direction
5
(represented by the thuMb) mutually perpendicular to both the direction of
field and the current in the conductor.
Thumb = Direction of
Conductor Motion
Fore Finger = Direction of Fixed
Magnetic Field (N to S)
Middle Finger =
Conventional Current
Direction
For clear understanding the principle of DC motor we have to determine
the magnitude of the force, by considering the diagram below.
We know that when an infinitely
small charge dq is made to flow at
a velocity ‘v’ under the influence
of an electric field E, and a magnetic
field B, then the Lorentz Force dF
experienced by the charge is given
by:-
For the operation of dc motor, considering E = 0
i.e. it’s the cross product of dq, v and magnetic field B.
6
Where dL is the length of the conductor carrying charge dq
Constructional Features of DC Machine
Commutator
Sectional view of a DC machine: The sectional view of a D.C.
Machine is as shown below:
7
Construction of DC Generator: The main parts of DC machine are:
1. Frame or Yoke, 2. Field system,
3. Armature core and Armature winding
4. Commutator
5. Brushes
1. Frame or Yoke: The cross section of the yoke can be solid or it can be
fabricated. Small machines use solid yokes. For large machines it is
fabricated. Yoke protects the inner part of the machine from the atmosphere.
Yoke carry a unidirectional flux. The flux always moves from South Pole to
North Pole. Yoke flux is always half of the pole flux. Thus yoke provides a
return path for the pole flux. Therefore the cross section of the yoke can be
half of pole cross section. Since yoke flux is unidirectional no e.m.f is
induced in it.
2. Field system: Field system consists of the following parts: Pole core,
Pole shoes and Field coils. Pole core is the part of the field system where the
magnetic flux is set up. Pole core can be solid in cross section or it can be
laminated. Machines of smaller power rating uses pole core of solid cross
8
sections. Larger machines uses laminated pole core. In laminated pole,
laminations of steel sheets are stacked together to give the pole core.
Field windings are made of copper or aluminum conductor. They carry
current and produce magnetic field. All field windings on the poles are
connected in series and carry same current. Depending on the current
directions in the coil we get the polarity of the pole as north or south.
Field windings are of two types: Series field and shunt field windings.
Series field winding will have smaller number of turns of thick wire and
needs a large current to produce a required value of flux. The shunt field
winding will have large number of turns of fine wire and requires small
current to set up required flux. Series field winding will be connected in
series with the armature. And shunt field winding is connected in parallel
with the armature.
Armature core and Armature windings: The armature core is
cylindrical in shape. High permeability silicon steel stampings are used for
the armature core. The laminations are circular in shape with outer periphery
slotted to receive the armature conductors. The laminations of the core are
separated from each other by a layer of varnish coating. The varnish coating
acts as an insulator for the flux and limits the induced eddy currents to the
lamination. This reduces the eddy current loss.
The armature conductors are interconnected to form the coils. The coils are
interconnected to form the Armature winding. The armature windings will
have coils connected in parallel paths. All the coils in a path will be in series.
There are two types of winding: Lap and Wave winding.
In Lap winding, the number of parallel paths (A) equal to number of pole (P)
and is used in high current, low voltage machines
In Wave winding, the number of parallel paths is always two. Thus each
path will have more coils compared to lap winding. Therefore wave winding
is used for high voltage, low current machines,
Commutator: As the coil rotates in the magnetic field, the e.m.f induced
in the conductors varies sinusoidally and results in sinusoidally alternating
current in the conductors. This sinusoidally alternating current must be made
unidirectional so that the output current is a direct current. Commutator
converts ac current into unidirectional current. Commutator is cylindrical in
shape, made of V - shaped copper or brass segments insulated from each
other by a mica layer.
9
Carbon brushes: Carbon brushes are used in DC machines to collect the
current from the rotating armature. These brushes, under spring force always
make contact with the commutator surface. These brushes are placed in
brush holders supported on brush arm.
EMF equation
Consider a dc generator. Let,






Ø = flux per pole in Weber.
Z = Total number of conductor.
P = Number of poles.
A = Number of parallel paths.
N = armature speed in rpm.
Eg = emf generated in any of the parallel path.
A conductor when rotated, in one revolution it moves under P poles.
Flux cut by a conductor in one revolution = Flux per pole x No. of poles
=φ*P
Time for one revolution = 1/ N minutes = 60 / N seconds.
In one revolution a conductor moves under all P poles.
As the conductor moves under one pole it cuts a flux of φ Webers.
Time taken by the conductor to move under one pole = Time for one
revolution / number of poles.
dt = (60/N)/ P = 60 / NP
Flux cut by the conductor when moved under one pole = φ
Average e.m.f induced in a conductor =(d φ/dt)
= φ / (60 /NP)
= P φ N /60
If Z is the number of conductors on the armature, they are connected A
parallel paths, with each parallel path having Z / A conductor in a path
all connected in series.
Emf generated between the terminals of a parallel path, Eg = e.m.f induced
in one conductor x No. of parallel paths.
= ( PφN/60 ) x (Z / A)
10
For Lap wound armature, A = P, no. of poles
Eg , e.m.f induced = ( PφN/60 ) x (Z / P)
= ( φN/60 ) x (Z )
=( φZN/60 )
For wave wound armature, A=2.
Eg , e.m.f induced = ( PφN/60 ) x (Z / 2)
= (P φN/60 ) x (Z/2 )
= ( φZN/60 ) x (P/2)
For a given number of poles, P we can say that e.m.f generated in the wave
winding = e.m.f generated in the lap winding x (P/2)
Problems on EMF equation
P1. In a DC machine, if P=8, Z=400, N=300 rpm, and φ=100 mWb,
Calculate Eg = with winding (i) Lap connected, (ii) Wave connected.
Soln: WKT, A=P for Lap wound connected dc machine & A=2 for Wave
connected dc machine.
Emf generated between the terminals
Eg = PφNZ/60A volts
(i) Lap-connected machine, Eg = 200 volts
(ii) Wave-connected machine, Eg = 800 volts
P2. An 8 pole lap-connected armature has 960 conductors, a flux of 40 mWb
per pole and a speed of 400 rpm. Calculate the emf generated. If the
armature were wave-connected, at what speed must it be driven to
generate 400 V.
Soln: Given P=8, Z=960, N=400 rpm, and φ=40 mWb.
11
WKT, A=P for Lap-connected armature & A=2 for Wave-connected
armature.
Emf generated between the terminals
Eg = PφNZ/60A volts
(i)
(ii)
Lap-connected armature, Eg = 250 volts
Given Eg = 400 volts for Wave-connected armature.
Therefore, N = 156 rpm.
P3. An 8 pole lap-connected armature driven at 400 rpm is required to
generate 250 V. The useful flux per pole is 0.05 Wb. If the armature has
150 slots, calculate a suitable number of conductors per slot.
Soln: Given A=P=8, N=400 rpm, and φ=0.05 Wb, Eg=250 V,
Eg = PφNZ/60A volts
Therefore, N=750
No. of conductors per slot = Z/S = 750/150 = 5
P4. A 4-pole generator with wave wound armature has 51 slots, each having
24 conductors. The flux per pole is 0.01 weber. At what speed must the
armature rotate to give an induced emf of 220 V? what will be the
voltage developed if the winding is lap and the armature rotates at the
same speed?
Soln: WKT, A=P for Lap wound connected dc machine & A=2 for
Wave connected dc machine.
Given P=4, S=51, Z=24x51, φ=0.01 Wb, and Eg=220 V, N=?
Eg = PφNZ/60A volts
(i) For Wave-wound machine, A=2
N = 540rpm
(ii) Lap-connected machine, A=P=4, N=540rpm
Eg = 110volts
12
Voltage equation of generator: The induced e.m.f Eg of the
generator while allowing a current of Ia through the armature
conductors causes an armature resistance drop IaRa and due to the
bush contact with commutator a small amount of voltage is also
dropped (usually around 2 volts). The remaining voltage is
available at the terminals of the generator as terminal voltage, V.
Thus terminal voltage is given by
Eg= V + IaRa+ brush drop
R
A
+
+
DC Source
+
A
-
-
V
E
g
A = Armature
-
Field Coil
Types of DC Generators
 Separately Excited DC Generators: In this generator, the field
winding is connected to a separate source. The current in the field
winding is independent of the generated voltage.
Ia
L+
If
f
f
+
Vf
-
Ia
R
a
E
g
L
O
A
D
V
-
Field
Coil
13
Armature current Ia = IL
Terminal Voltage V = Eg - IaRavolts
Power developed P = EgIaWatts
Power delivered to the load = EgIa -IaIaRa
= Ia (Eg – IaRa )
= V Ia
 Self Excited DC Generators: The field winding is connected to the
armature of the generator. The current required for producing the
magnetic flux is given by the induced voltage of the machine.
Therefore the generator is called self excited.
Following are different types of self excited generators
 Shunt wound DC generator – Afield winding with large number of
turns of fine wire is connected in parallel with the armature. It takes a
small current from the armature to produce required value of the
magnetic flux. The flux produced will almost remains constant.
 Series wound DC generator - Afield winding with small number of
turns of thick wire is connected in series with the armature. It always
carries the armature current. The flux produced will be proportional to
the armature current. The flux will vary with the armature current.
 Compound wound DC generator – This machine will have both series
field and shunt field windings. Based upon the connection of the field
windings, the compound generator can be short shunt or long shunt.
Shunt Excited DC Generator
I
a
a
If
fs
h
+
Ia
Ra
Shunt Field Coil
Rs
h
E
- g
a
a
L
+
L
O
A
D
V
-
14
Field current If = V / Rs
Armature current Ia = If + IL
Generated e.m.f Eg = V + IaRa + Brush drop = (ΦZN/60)(P/A)
Terminal Voltage V = Eg - IaRa – brush drop
Power developed P = EgIaWatts
Power delivered to the load = Power developed – power loss = EgIa –Ia2 Ra
= Ia (Eg – IaRa )
= V Ia
I
Series Excited DC Generator
L
+
+
Ia
Ra
-
Eg
L
O
A
D
Field Coil
V
Rse
-
Armature current Ia =Ise= IL
Volts
Terminal Voltage V = Eg - Ia(Ra+Rse)
Power developed P = EgIa
watts
Compound Excited DC Machine
Short shunt
15
Series Field Coil
If
s
h
Shunt Field CoilSeries
I
+
R
se
+
Rs
h
Ia
Ra
L
O
A
D
E
- g
VVf
-
Shunt Field current Ish = V / Rsh
Series field current Ise = IL
Armature current Ia = Ish+ IL
Generated e.m.f Eg = V + IaRa + IL Rse +Brush drop = (Φt ZN/60)(P/A)
Terminal Voltage V = Eg - IaRa – brush drop – IL Rse
Power developed P = EgIaWatts
Power delivered to the load = Power developed – power loss
= EgIa – Ia2Ra– IL2Rse
Long shunt
Is
h
I
L
Series Field Coil
Rs
e
Shunt Field Coil
Rs
h
+
Ia
Ra
E
- g
+
L
O
A
D
V
-
16
Shunt Field current Ish = (V – IaRse) / Rsh
Series field current Ise = Ia
Armature current Ia = Ish+ IL
Generated e.m.f Eg = V + Ia (Ra + Rse)+Brush drop = (ΦtZN/60)(P/A)
Terminal Voltage V = Eg - Ia (Ra + Rse) – brush drop – IL Rse
Power developed P = EgIaWatts
Power delivered to the load = Power developed – power loss
= EgIa – Ia2 (Ra + Rse)
Problems
P1. A 110 V DC Shunt generator delivers a load current of 50 amps. The
armature resistance is 0.2 ohm and the field circuit resistance is 55
ohms. The generator, rotating at a speed of 1800 rpm has 6 poles lap
wound and has a total a total of 360 conductors. Calculate
(i) The no-load voltage in the armature
(ii) The flux per pole.
Soln: Load Current I =50 A, I
L
I = I + I = 50 + 2 = 52 A
a
Sh
= 110 / 55 = 2A
I R = 52 x 0.2 = 10.4 V
a
a
E = V + I R = 110 + 10.4 = 120.4 V
g
a
I
L
I
s
h
sh
+
a
E = ZN φP / 60A
g
φ = 0.011 Wb
+
Ia
Ra
Shunt Field Coil
R
s
h
L
O
A
D
E
- g
V
-
P2. A 4 pole lap-wound 750 rpm shunt generator has an armature resistance
of 0.4 ohm and field resistance of 200 ohms respectively. The armature
17
has 720 conductors, and the flux per pole is 0.3 mwb. If the load
resistance is 10 ohms, determine the terminal voltage.
I
L
I
s
h
Soln:
+
+
Ia
Ra
Shunt Field Coil
R
s
h
E
- g
L
O
A
D
V
-
E = ZN φP / 60A = 270 V
g
In Fig. the shunt & load resistances are in parallel with each other
Therefore, their combined resistance = 10 x 200 / 210 = 9.52 Ω
The total resistance = R + 9.52 = 0.4 + 9.52 = 9.92 Ω
a
Armature Current I =270 / 9.92 = 27.2 A
a
Armature drop = 27.2 x 0.4 = 10.88 volts
Therefore, terminal voltage = 270 – 10.88 = 259.12 V
P3. A shunt generator supplies a load of 10 kW at 200V through a pair of
feeders of total resistance 0.05 ohm. Armature resistance is 0.1 ohm.
Shunt field resistance is 100 ohms. Find the terminal voltage and
generated emf of the generator.
Soln:
Load current I = 10000 / 200 = 50 A
Drop in the feeders = 0.05 x 50 = 2.5 V
Therefore, terminal voltage, V = 200 + 2.5 = 202.5 V
I = 202.5 / R = 202.5 / 100 = 2.025 A
sh
sh
I = I + I = 50 + 2.025 = 52.025 A
a
sh
Generated emf = E = V + I R = 202.5 + (52.025 x 0.1) = 207.7 V
g
a
a
P4. A 4-pole lap wound shunt generator delivers 200 amperes at terminal
voltage of 250 volts. It has a field and armature resistance of 50 ohms
and 0.05 ohm respectively. Neglecting brush drop determine
18
(i) Armature current, (ii) the current per armature parallel path, (iii) emf
generated, (iv) power developed.
Soln: The current through the shunt field winding is
Ish = 250 / 50= 5 A
(i)
Load current I = 200 A
Armature current, Ia = I + Ish= 200 + 5 = 205 A
(ii)
In lap – wound generator, the number of parallel paths = No.
of poles = 4
Current per armature parallel path is Ia / A = 205 / 4 = 51.25 A
Armature voltage drop = IaRa = 205 x 0.05 = 10.25 A
(iii)
Emf generated, Eg = terminal voltage + armature voltage drop
(iv)
= 250 + 10.25 = 260.25 V
Power developed, P = EgIa = 260.25 x 205 = 53351 W
P5. A separately excited generator when running at 1200 rpm supplies 200 A
at 125 V to a circuit of constant resistance. What will be the current
when the speed is dropped to 1000 rpm if the field current is unaltered:
Ra = 0.04 ohm? Total drop at brushes is 2 V, ignore changes in armature
reaction.
Soln: Given, N = 1220 rpm, I = 200 A, V = 125 V, N = 1000 rpm,
1
L1
t1
R = 0.04 Ω, Total brush drop = 2 V
2
a
Initial load: I = I = 200 A
a1
L1
E = V + I R + Brush drop = 125 + 200 x 0.04 + 2 = 135 V
g1
t1
a1
a
As I is constant, E α N, i.e. E / E = N / N
f
g
g1
g2
1
2
Therefore, E = 135 x 1000 / 1200 = 112.5 V
g2
But E = V + I R + Brush drop
g2
t2
a2
a
Now load resistance R is constant. According to ohm’s law for
L
constant resistance,
R = V / I = V / I = 125 / 200 = 0.625 Ω
L
t1
L1
t2
L2
V = I x R = I x 0.625
t2
L2
L
L2
E = 0.625 I + I x 0.04 + 2
g2
L2
L2
112.5 – 2 = 0.665 I
L2
I = 166.1654 A
L2
19
P6. A 4-pole, DC shunt generator has 386 wave connected conductors. The
armature and shunt field resistances are 1 ohm and 100 ohm
respectively. The flux per pole is 25 mWb and the speed is 1000 rpm. If
the load resistance is 40 ohm, calculate the armature current and the
power output.
Soln: Given, P = 4, Z = 386, A = 2 as wave, R = 1 Ω, R = 100 Ω,
a
sh
φ = 25 m Wb, N = 1000 rpm, R = 40Ω
L
E = ZNφP / 60A = 321.667 V
g
I =I +I
a
L
sh
I = V / R and I = V / R
L
t
L
sh
t
sh
And E = V + I R
g
t
a
a
Therefore, 321.667 = V + (I + I ) R
t
L
sh
a
321.667 = V + V R / R + V R / R
t
t
a
L
t
a
sh
321.667 = V + 0.025 V + 0.01 V
t
t
321.667 = 1.035 V
t
t
This gives V = 310.79 V
t
Calculating, I = 8.045 A and I = 3.1079 A
L
sh
I = 11.1529 A
a
P
out
= V x I = 310.79 x 8.045 = 2500.3 W
t
L
P7. A wave wound, 6 pole, long shunt compound dc generator has 600
armature conductors. The generator is driven at 300 rpm. Calculate the
emf generated if the flux / pole is 0.060 Wb. If now the generator is
required to produce emf of 550 V at a reduced value of flux / pole of
0.055 Wb, calculate the speed at which the armature of the generator
must be driven.
Soln: Given, P = 6, Z = 600, N = 300 rpm, φ = 0.06 Wb, wave wound
1
1
N = 300 rpm, the emf generated is E = ZN φ P / 60A, A = 2 for wave
1
g1
1 1
= 540 V
Value of flux φ = 0.055 Wb, E = 550 V
2
g2
E = ZN φ P / 60A
g2
2 2
20
N2 = 333.33 rpm
It can be observed that, to increase the generated emf with reduced flux,
the speed must be increased.
Armature Reaction
Every current carrying conductor produces a magnetic field. Under
load conditions the armature conductors also carries a current. Therefore,
these conductors set up their own magnetic field. Definition of armature
reaction is: “The effect of the magnetic field produced by the current
carrying armature conductors on the distribution of main flux, i.e., the
flux produced by the poles”.
This armature reaction is proportional to the armature current and it
varies with the load on the generator. At light loads its effect is
negligible. At higher loads the effect is more. The armature reaction has
two effects:
1. Demagnetizing effect – Weakening of the main flux
2. Cross magnetizing effect – Shifts the direction of the magnetic axis.
It is not possible to eliminate the armature reaction completely.
21
P8. A DC series generator has armature resistance of 0.5 ohm and series
field resistance of 0.03 ohm. It drives a load of 50 A. If it has 6 turns /
coil and total 540 coils on the armature and is driven at 1500 rpm,
calculate the terminal voltage at the load. Assume 4 poles, lap type
winding, flux pole as 2 mWb and total brush drop as 2 V.
Soln:
= Ia
+
+
G
Ra
Series
Field Coil
L
O
A
D
Eg
Rse
-
Given, R = 0.5 Ω, R = 0.03 Ω, V
a
se
brush
= 2 V, N = 1500 rpm,
Total coils are 540 with 6 turns / coil,
Therefore, Total turns = 540 x 6 = 3240
Therefore, Total conductors Z = 2 x Turns = 2 x 3240 = 6480
E = ZNφP / 60A = 324 V
g
WKT, E = V + I (R + R ) + V
g
a
a
se
brush
Where I = I = 50 A, V = 295.5 V
a
L
P9. A short shunt compound DC generator supplies a current of 75 A at a
voltage of 225 V. Calculate the generated voltage if the resistance of
armature, shunt field and series field windings are 0.04 ohm, 90 ohm
and 0.02 ohm respectively.
Series Field Coil
Soln:
I
s
R
h
Shunt Field Coil
R
sh
-
+
L
se
+
Ia
Ra
I
E
g
L
O
A
D
V
22
Given, R = 0.04 Ω, R = 90 Ω, R = 0.02 Ω, V = 225 V
a
sh
se
I = 75 A
L
I =I +I
a
L
sh
Now E = V + I R + I R
g
a
a
L
se
And drop across armature terminal is,
E – I R = V + I R = 225 + 75 x 0.02 = 226.5 V
g
a
a
L
se
Therefore, I = (E – I R ) / R = 2.5167 A
sh
g
a
a
sh
Therefore, I = I + I = 75 + 2.5167 = 77.5167 A
a
L
sh
Therefore E = V + I R + I R = 229.6 V
g
a
a
L
se
P10. A long shunt dc compound generator drives 20 lamps, all are connected
in parallel. Terminal voltage is 550 V with each lamp resistance as 500
ohm. If shunt field resistance is 25 ohm, armature resistance is 0.06
ohm and series field resistance is 0.04 ohm, calculate the armature
current and the generated emf.
Soln:Given, Ra = 0.06 Ω, Rsh = 25 Ω, Rse = 0.04 Ω, V = 550 V
RLamp = 500 Ω, 20 lamps are connected in parallel
I
I
L Coil
Series Field
s
h
R
+
s
e
R
Shunt Field Coil
s
h
L
O
A
D
+
Ia
Ra
-
V
E
g
-
23
As all lamps are in parallel, the voltage across all of them is same which is
terminal voltage of generator V = 550 V. So current drawn by each lamp is
I=V/R
= 550 / 500 = 1.1 A
Lamp
Such 20 lamps are used as a load.
Therefore, I = 20 x I
= 20 x 1.1 = 22 A
L
Lamp
Now I = V / R = 550 / 25 = 22 A
sh
sh
Therefore, I = I + I = 44 A
a
L
sh
Therefore, E = V + I R + I R = 550 + 44 x 0.06 + 44 x 0.04 = 554.4 V
g
a
a
a
se
DC machine Operation as a DC Motor
DC Motors:
DC Motors convert electrical energy into mechanical energy. They work on
the principle that, “whenever the current carrying conductor is placed in the
magnetic field, a force is set up on the conductor”. Due to this force a torque
is produced, which is responsible for the rotation of armature of the
machine. This rotation produces the required mechanical power.
+
V
T
-
I
a
T
load
+
E

DC Motor
m
b
-
Mechanical
Load
(Pump,
Compressor)
T
dev
Eb is Back EMF
VT is Applied voltage
Tdev is the Torque developed by DC Motor
Tload is the opposing load torque
Input electrical power to dc motor =VIa
Electrical power to be converted to mechanical power = Eb Ia= Kam Ia
= Tdev m
Output mechanical power to load = Mechanical power developed - losses to
overcome the friction
24
Back e.m.f:
The current carrying armature conductors of a dc motor are placed in
the magnetic field produced by the main poles of the motor. The armature
conductors produce their own magnetic field.
Thus there are two
magnetic fields in the dc motor. The interaction between the two fields
causes the rotor to rotate. This rotation makes the armature conductors to
cut the magnetic field set up by the main poles. Therefore an e.m.f is
induced in the conductors. According to Lenz’s law this induced e.m.f
acts in a direction to oppose its cause of production, i.e., the rotation of
the armature. In turn it opposes the supply voltage given to the armature.
Therefore this induced e.m.f is called back e.m.f.
Back e.m.f is also an induced e.m.f. In case of dc motors it is also give by
the equation Eb = (ΦZN/60)(P/A)
Types of dc motors:
1. Separately excited dc motor – the armature and field windings are
connected to separate sources
2. Self excited dc motor – the armature and field winding are connected to
same supply.
Self excited motor are:
i) Series dc motor – Field winding will have small number of turns of thick
wire and is connected in series with the armature.
ii) Shunt dc motor - Field winding will have large number of turns of thin
wire and is connected in parallel with the armature
iii) Compound dc motor – This motor will have both series and shunt field
windings.
Compound motor can be short shunt or long shunt. They can be Differential
compound or Cumulative compounded.
Separately excited dc motor
E = ZNφP / 60A
b
Field equation: V =R I
f
se se
Voltage equation: E =V - I R ; E = K  
b
t
a
a
b
a
m
25
Ra
Rse
Ise
+
Vf
+
+
-
Armature
Eb
Field Coil
-
Vt
Ia
Self excited dc motors
1. Series motors
I
R
R
a
M
s
a
e
E
V
b
t (dc
supply)
Eb  K1 K 2 I a
Voltage equation V = E + I (R + R )
t
2. Shunt motors :
b
a
a
se
Voltage equation,
Vt = Eb + IaRa
Input current or Line current, IL = Ia+ Ish
Shunt field current, Ish = Vt / Rsh
I
R
a
a
I
M
E
R
b
sh
I
L
sh
V
t (dc
supply)
26
Multiplying voltage equation of a motor on both sides by I , we get
a
2
VI = E I + I R
t a
b a
a
a
V I = Electrical power input to armature
t a
E I = Electrical power converted to mechanical power developed by the
b a
armature (total armature output)
2
I R = Electrical power lost in the armature (armature copper loss)
a
a
2
The motor develops mechanical power given by, Pm = V I - I R
t a
a
a
For maximum mechanical power, dP / dI = V - 2 I R = 0
m
or I R = V / 2
a
a
a
t
a
a
t
Therefore, V = E + I R = E + V / 2 or E = V / 2
t
b
a
a
b
t
b
t
This says that when the motor works at maximum mechanical power
developed half of the input voltage is wasted as the drop across the armature
resistance.
3. Compound motors:
a) Long shunt compound motor
Voltage equation Vt = Eb + Ia(Ra +Rse)
Line current
IL = Ia+ Ish
Shunt field current , Ish = V/ Rsh
Ia
Ra
Rse
IL
Ish
M
Eb
Rsh
Vt (dc
supply)
b) Long shunt compound motor
Voltage equation Vt = Eb + IaRa +ILRse
27
Line current
IL = Ia+ Ish
Shunt field current , Ish = (V - ILRse ) / Rsh
Compound motors are of two types
Differential compound: The flux produced by the series winding opposes the
flux produced by the Shunt field winding..
Cumulative compound: The flux produced by the series winding aid the flux
produced by the Shunt field winding.
If the series flux is proportional to armature current the motor is Long shunt
and if the series flux is proportional to line current the motor is short shunt
Production of Torque & Torque Equations
Rotation
The measure of causing the rotation of a wheel
or the turning or twisting moment of a force
about an axis is called the torque. It is measured
by the product of force and the radius at which
this force acts.
Consider a wheel of radius R meters acted
upon by a circumferential force F Newton, making it
rotate at n rps.
Torque = F x R Newton – metres
Work done by this force in one revolution = Force x Distance
= F x 2πR Joules
R
F
Work done per second, W = F x 2πR x n Joules / second
= (F x R) x (2πN/60) Joules / second
But 2πN/60 is angular velocity = ω radians / second
Also, torque, T = F x R
Therefore, Work done per second, W = T x ω Joules / second
Also, power developed, P = T x ω watts
Armature Torque
Let T = toque developed by the motor armature (N – m)
a
n = speed of rotation (r.p.s)
28
Angular velocity, ω = 2πN/60 radians / second
Therefore, power developed, P = T x ω
a
= T x ω = T x 2πN watts
a
a
The electrical power converted into mechanical power in the armature,
= E I watts
b a
Therefore, T x 2πN/60 = E I
a
b a
But, WKT, E = ZNφP / 60A
b
So, T x 2πN/60 = (ZNφP / 60A)I
a
a
or T = (φZI / 2π) x (P / A)
N-m
or T = 0.159φZI x (P / A)
N-m
a
a
a
a
or T = (0.159 / 9.81)φZI x (P / A)
a
kg-m
a
or T = 0.0162 φZI x (P / A)
a
a
T α φI
a
kg-m
a
For a series motor, T α φI
a
For a shunt motor, T α I
a
2
a
a
WKT, T x 2πN = E I
N-m
Ta = (1 / 2π) x E I / N
N-m
a
b a
b a
= 0.159x E I / N
N-m
b a
= (0.159 / 9.81) x E I / N
b a
= 0.0162x E I / N
b a
kg-m
kg-m
If N is in rpm, then armature torque equation is
29
Ta = E I / (2πN / 60) = 9.55 E I / N
b a
b a
N-m
Shaft Torque
The torque which is available at the motor shaft for doing useful work is
called shaft torque (T ). The total torque T developed in armature is not
sh
a
available at the shaft, as part of it is lost in overcoming the iron and
frictional losses. Therefore, shaft torque T is somewhat less than the total
sh
armature torque T
a.
Output = T x 2πN watts
sh
Where T is in N-m, and N is in rps.
sh
Therefore, T = Output in watts / 2πN
sh
N-m
If N is in rpm, then T = Output in watts / (2πN / 60)
sh
= 9.55 x Output in watts / N
N-m
N-m
Speed of a DC Motor
We know that, Eb = V - IaRa
or
ZNφP / 60A = V - IaRa
Therefore, N = ((V - IaRa) / φ) x (60A / ZP) rpm
Now, V - IaRa = Eb
Therefore, N = (Eb / φ) x (60A / ZP) rpm
As Z, A and P are constant for a particular machine,
N = K x (Eb / φ)
or
N α (Eb / φ)
Thus, speed is directly proportional to the back emf Eb and inversely
proportional to the flux φ.
For series motor,
Let N1, Ia1 and φ1 be the speed, armature current and flux per pole in the first
case.
30
Let N2, Ia2 and φ2 be the speed, armature current and flux per pole in the
second case.
Then using the relationship
N1 α (Eb1 / φ1) where Eb1 = V - Ia1Ra
And N2 α (Eb2 / φ2) where Eb2 = V - Ia2Ra
Therefore, N2 / N1 = (Eb2 / Eb1) x (φ1 / φ2)
Before saturation of the magnetic poles occurs, φ α Ia)
Therefore, N2 / N1 = (Eb2 / Eb1) x (Ia1 / Ia2)
For shunt motor,
In this case also same equation is used.
i.e., N / N = (E / E ) x (φ / φ )
2
1
b2
b1
1
2
But here, φ = φ
1
2
Therefore, N / N = E / E
2
1
b2
b1
Problems on back emf& torque equations
P1. A 4-pole, 250 volt, series motor has a wave – connected armature with
1254 conductors. The flux per pole is 22 mWb when the motor is taking 50
A. Armature resistance is 0.2 Ω and series field resistance is 0.2 Ω. Calculate
the speed.
Soln:
R
I
R
a
a
M
E
b
se
V
t (dc supply)
We know that,
E = V - I (R +R ) = 250 – 50(0.2 + 0.2) = 230 V
b
a
a
se
Also E = ZNφP / 60A
b
31
-3
Therefore, 230 = (1254 x N x 22 x 10 x 4) / (60 x 2)
Therefore, N = 250.1 r.p.m
P2. A 250 V shunt motor takes a total current of 20 A. Resistance of the
shunt field is 200 Ω and of the armature 0.3 Ω. Find the current in the
armature and the back e.m.f.
Soln:
I
I
R
a
a
I
E
M
L
sh
R
V
sh
b
t (dc
supply)
Line current, I = 20 A, V = 250 volts, R = 200 Ω, R = 0.3 Ω
sh
a
Therefore, shunt field current, I = V / R = 250 / 200 = 1.25 A
sh
sh
Armature current, I = I – I = 20 – 1.25 = 18.75 A
a
sh
Back e.m.f, E = V – I R = 250 – (18.75 x 0.3) = 244.375 volts
b
a
a
P3. A shunt generator delivers 100 kW at 250 V, when running at 400 r.p.m.
The armature resistance is 0.01 Ω and field resistance is 100 Ω. If the same
machine is run as a shunt motor with an input of 100 kW at 250 V, calculate
the speed of the machine as a motor. Contact drop per brush is 1 V.
Soln:
32
IL
Ish
+
+
Ia
Ra
-
L
O
A
D
Eg
Rsh
V
-
Ia
Generator
IL
Ra
Ish
M
Rsh
Eb
Vt
(dc
suppl
y)
Motor
As generator
3
Load current, I = 100 x 10 / 250 = 400 A
Shunt current, Ish = 250 / 100 = 2.5 A
Ia = I + Ish = 400 + 2.5 = 402.5 A
IaRa = 402.5 x 0.01 = 4 V
33
Brush drop = 2 x 1 = 2 V
Induced e.m.f in armature, Eg = V + IaRa + drop in brushes
= 250 + 4 + 2 = 256 V
It is apparent that, if this machine is to run as a motor at 400 r.p.m., it will
have a back e.m.f. of 256 V induced in its armature. Eb1 = 256 V, N1 = 400
r.p.m.
As motor
3
Input line current, I = 100 x 10 / 250 = 400 A
Shunt current, Ish = 250 / 100 = 2.5 A
Ia = I - Ish = 400 - 2.5 = 397.5 A
IaRa = 397.5 x 0.01 = 3.97 V , Brush drop = 2 x 1 = 2 V
Induced e.m.f in armature, Eb2 = V - IaRa - drop in brushes
= 250 – (3.97 + 2) = 244 V
N2= ?
Now,
N2 / N1 = (Eb2 / Eb1) x (φ1 / φ2)
However, as Ish is constant in both cases, φ1 = φ2
Therefore, N2 / N1 = (Eb2 / Eb1) or N2 / 400= 244 / 256
Therefore, N2= 381.25 r.p.m.
P4. A 120 V, DC shunt motor has an armature resistance of 0.2 Ω and a field
resistance of 60 Ω. It runs at 1800 r.p.m., when it takes a full load current of
40 A. Find the speed of the motor when it is operating with half full – load.
Soln:
Ia
IL
Ra
Ish
Eb
M
Rs
Vt
h
(dc
supp
Given, V = 120 V, R = 0.2 Ω, R = 60 Ω
a
ly)
sh
Shunt field current, I = V / R = 120 / 60 = 2 A
sh
sh
Full load,
34
Full load current, I = 40 A
Armature current, I = I – I = 40 -2 = 38 A
a
sh
Back e.m.f., E = V – I R =120 – (38 x 0.2) = 112.4 V
b
a
a
Half load,
Current on half load, I = 40 / 2 = 20 A
h
Armature current, I = I – I = 20 -2 = 18 A
ah
h
sh
Back e.m.f., E = V – I R = 120 – (18 x 0.2) = 116.4 V
bh
ah
a
Now, N / N = E / E
h
b
bh
1800 / N = 112.4 / 116.4
h
Therefore, speed on half load, N = 1864 r.p.m.
h
P5. A 4 – pole, 500 V, shunt motor has 720 wave connected conductors on its
armature. The full load armature current is 60 A, and the flux per pole 0.03
Wb. The armature resistance is 0.2 Ω and the contact drop is 1 volt per
brush. Calculate the full – load speed of the motor.
Soln:
Given P = 4, Z = 720, φ = 0.03 Wb / pole, A = 2, R = 0.2 Ω, I = 60 A, V
a
a
= 500 V
Therefore, EMF generated, E = ZNφP / 60A = 0.72N
b
WKT, E = V – (I R + Brush drop) = 500 - (60 x 0.2 + 2x1) = 486 V
b
a
a
Therefore, 486 = 0.72N
Therefore, N = 675 r.p.m.
35
P6. Find the useful flux per pole of a 250 V, 6 pole shunt motor having a two
circuit connected armature winding with 220 conductors. At normal working
temperature, the overall armature resistance including brushes is 0.2 Ω. The
armature current is 13.3 A at the no – load speed of 908 r.p.m.
Soln:
Given P = 6, Z = 220, A = 2, R = 0.2 Ω, I = 13.3 A, V = 250 V,
a
a
N = 908 r.p.m.
WKT, E = V – I R = 250 - (13.3 x 0.2) = 247.34 V
b
a
We have,
a
E = ZNφP / 60A
b
247.34 = φ x 220 x 908 x 6 / 60 x 2
Therefore, useful flux per pole, φ = 0.0247 Wb
P7. A DC shunt motor runs at 750 r.p.m. from 250 V supply and takes a full
load line current of 60 A. Its armature resistance is 0.4 Ω and the field
resistance is 125 Ω. Assuming 2 V brush drop and negligible armature
reaction effect, find the no – load speed for a no – load current of 6 amperes.
Soln:
Given, N = 750 r.p.m., R = 0.4 Ω, R = 125 Ω, I = 6 A, I = 60 A, N = ?
a
sh
o
The effect of armature reaction is negligible, so φ = φ
o
o
I = V / R = 250 / 125 = 2 A
sh
sh
Armature current under full load conditions,
I = I – I = 60 – 2 = 58 A
a
sh
I R = 58 x 0.4 = 23.2 V
a
a
Therefore, E = V – (I R + Brush drop) = 250 – (23.2 + 2) = 224.8 V
b
a
a
36
Armature current under no – load conditions,
I –I –I =6–2=4A
ao
o
sh
I R = 4 x 0.4 = 1.6 V
ao
a
Therefore, E = V – (I R + Brush drop) = 250 – (1.6 + 2) = 246.4 V
bo
ao
a
We have, N / N = E / E = 224.8 / 246.4
o
b
bo
Therefore, N = 822 r.p.m.
o
P8. A 230 V DC Shunt motor takes a no-load current of 2 A, and runs at
1100 r.p.m. If the full-load current is 40 A, find the speed at full load.
Assume the flux remains constant and armature resistance is 0.25 Ω.
Soln:
Given, V = 230 V, N = 1100 r.p.m., R = 0.25 Ω, I = 2 A, I = 40 A, φ = φ
o
WKT,
a
ao
a
o
1
E = V – I R = 230 – (40 x 0.25) = 220 V
b
a
a
E = V – I R = 230 – (2 x 0.25) = 229.5 V
bo
ao
a
We have, N / N = E / E
o
b
bo
Therefore, N / 1100 = 220 / 229.5
Therefore, speed at full load, N = 1054.5 r.p.m.
37
P9. Determine the total torque developed in a 250 V, 4-pole, DC shunt motor
with lap winding, accommodated in 60 slots, each containing 20 conductors.
The armature current is 50 A and the flux per pole is 23 mWb.
Soln:
Given, V = 250 V, P = 4, A = P = 4,
Z = No. of conductors in each slot x No. of slots = 20 x 60 = 1200
I = 50 A, φ = 23 mWb
a
WKT,
T = 0.159φZI x (P / A)
a
a
N-m
= 219.5 N-m
P10. A 250 V, DC shunt motor takes 6 amperes line current on no-load and
runs at 1000 r.p.m. The field resistance is 250 Ω and armature resistance 0.2
Ω. If the full-load line current is 26 amperes, calculate the full load speed,
assuming constant air gap flux.
Soln:
38
Given, V = 250 V, No = 1000 r.p.m., Rsh = 250 Ω, Ra = 0.2 Ω, I1 = 26 A, Io
= 6 A, φo = φ1
Ish = V / Rsh = 250 / 250 = 1 A
The armature current at no-load, Iao = Io – Ish = 6 – 1 = 5 A
The armature current at full – load, Ia1 = I – Ish = 26 – 1 = 25 A
Ebo = V – IaoRa = 250 – 5 x 0.2 = 249 V
Eb1 = V – Ia1Ra = 250 – 25 x 0.2 = 245 V
We have, N1 / No = Eb1 / Ebo
Therefore, N1 / 1000 = 245 / 249
Therefore, speed at full load, N = 984 r.p.m.
P11. A four pole DC shunt motor takes 22.5 A from a 250 V supply. R =
sh
125 Ω, and R = 0.5Ω. The armature is wave – wound with 300 conductors.
a
If the flux per pole is 0.02 Wb, calculate (i) the speed, (ii) torque developed,
(iii) power developed.
Soln:
Given, V = 250 V, R = 125 Ω, R = 0.5 Ω, I = 22.5 A, φ = 0.02 Wb,
sh
a
A=2, Z = 300
I = V / R = 250 / 125 = 2 A
sh
sh
The armature current, I = I – I = 22.5 – 2 = 20.5 A
a
We have,
(i)
Speed,
sh
E = V – I R = 250 – 20.5 x 0.5 = 239.75 V
b
a
a
N = E x 60A/ ZφP = 1199 r.p.m
b
39
(ii) Torque, T = 0.159φZI x (P / A)
a
N-m
a
= 39.1 N-m
(iii) Power developed, E I = 239.75 x 20.5 = 4.915 kW
b a
P12. A 230 V shunt motor has an armature resistance of 0.6 Ω. If the full
load armature current is 30 A and no load armature current is 4 A, find the
change in back emf from no load to full load.
Soln:
Given, V = 230 V, R = 0.6 Ω, I = 30 A, I = 4 A
a
(i)
o
Back emf at no – load,
E = V – I R = 230 – 4 x 0.6 = 227.6 V
bo
(ii)
1
ao
a
Back emf on full load,
E = V – I R = 230 – 30 x 0.6 = 212 V
b1
a1
a
Therefore, change in back emf from no – load to full – load,
= 227.6 – 212 = 15.6 V
DC Motor Characteristics
DC Motor characteristics depict the relationships between following
quantities:
i) Torque & armature current or T / I characteristic (also called electrical
a
a
characteristic).
ii) Speed & armature current i.e., N / I characteristic.
a
iii) Speed & Torque or N / T characteristic (also called mechanical
a
characteristic). This can also be ascertained from (i) and (ii) above.
40
Important Equations
E = ZNφP / 60A volts
b
E =V - I R
b
t
a
a
T = 0.159 φ Z I x (P / A)
a
N-m
a
T = E I / (2πN / 60) = 9.55 E I / N
a
b a
b a
N-m
If N is in rpm, then T = Output in watts / (2πN / 60)
sh
= 9.55 x Output in watts / N
N-m
N-m
Characteristics of Series Motors
Ra
Ia
M
Rse
Vt (dc
Eb
supply)
We know that, E = ZNφP / 60A
b
For a series motor, T α I
a
2
a
N α (E / φ) The characteristics for the motor are as shown:
b
41
Applications of Series Motors: These motors are useful in applications
where starting torque required is high and quick acceleration. Like:
1.
2.
3.
4.
5.
Traction
Hoists and Lifts
Crane
Rolling Mills
Conveyors
Ia Ra
IL
Is
Characteristics of Shunt Motors
M
Eb
h
R
Vt
sh
(dc
sup
ply)
42
Applications of Shunt Motors
These motors are constant speed motors, hence used in applications
requiring constant speed. Like:
1.
2.
3.
4.
5.
Lathe machine
Drilling machine
Grinders
Blowers
Compressors
Characteristics of Compound Motors
43
Applications of Cumulatively Compound Motors
These motors have high starting torque. They can be operated even at no
loads as they run at a moderately high speed at no load. Hence these motors
are used for the following applications:
1. Elevators
2. Rolling mills
3. Punches
4. Shears
5. Planers
Applications of Differentially Compound Motors
 The speed of these motors increases as load increases which leads to
an unstable operation.
 Therefore, we cannot use this motor for any practical applications.
44
Need of a starter for DC Motors
At the time of starting a DC Motor, the Back EMF developed is zero as it's
speed is zero.
The starting current will be 5 – 6 times its full load current. Thus it is
required to reduce the voltage when the motor starts. This job is done with a
starter. In its simplest form, the starter of a dc motor works like a variable
resistance in series with the armature circuit. Its work is to reduce
the starting voltage up to such a value so that the current drawn does not
burn the armature windings. As the rotating armature of dc motor picks up
speed, the starter resistance is gradually reduced to almost zero. At full speed
the motor starts running normally, i.e. the job of starter finishes here.
In other words, the starter offers resistance during starting of dc motor only,
to provide large current.
At full speed, the starter is electrically out of armature circuit of the motor.
45
Problems
1. A 4-pole, 250 volt, series motor has a wave – connected armature with
200 conductors. The flux per pole is 25 mWb when the motor is
taking 60 A from the supply. Armature resistance is 0.15 Ω while
series field winding resistance is 0.2 Ω. Calculate the speed under this
condition.
Soln:
R
Ra
Ia
se
M E
b
Vt (dc
supply)
Given V = 250 V, P = 4, Z = 200, A = 2, φ = 25 mWb,
I = I = 60 A, R = 0.15 Ω, R = 0.2 Ω
a
L
a
se
WKT,
E = V - I (R +R ) = 250 – 60(0.15 + 0.2) = 229 V
b
a
a
se
Also E = ZNφP / 60A
b
-3
229 = (200 x N x 25 x 10 x 4) / (60 x 2)
Therefore, N = 1374 r.p.m
2. A DC series motor is running with a speed of 800 rpm while taking a
current of 20 A from the supply. If the load is changed such that the current
drawn by the motor is increased to 50 A, calculate the speed of the motor on
new load. The armature and series field winding resistances are 0.2 Ω and
0.3 Ω respectively. Assume the flux produced is proportional to the current.
Assume supply voltage as 250 V.
Soln:
For load 1, N = 800 rpm, I = I = I =20 A,
1
1
For load 2, I = I = I
2
a2
se2
a1
se1
= 50 A,
V = 250 volts, R = 0.2 Ω, R = 0.3 Ω
a
From voltage equation
se
V=E +I R +I
b1
a1
a
se1
R
se
46
Substituting the required values in the above equation, we get,
E = 240 V
b1
And
V=E +I R +I
b2
a2
a
se2
R
se
Substituting the required values in the above equation, we get,
E = 225 V
b2
From the speed equation,
NαE /φ
b
Now,
φαI αI
se
a
Therefore,
N / N = (E / E ) x (φ / φ )
Therefore,
N / N = (E / E ) x (I / I )
1
2
1
Therefore,
b1
2
b2
b1
2
b2
a2
1
a1
N = N x (E / E ) x (I / I )
2
1
b2
b1
a1
a2
Substituting the required values in the above equation, we get,
N = 300 rpm
2
3. A 220 V, DC motor has an armature resistance of 0.75 Ω. It is drawing an
armature current of 30 A, driving a certain load. Calculate the induced emf
in the motor under this condition.
Soln:
V = 200 V, I = 30 A, R = 0.75 Ω
a
For a motor,
a
V=E +IR
b
a
a
Therefore,
220 = E + 30 x 0.75
Therefore,
E = 197.5 volts
b
b
4. A 4 pole, DC motor has lap connected armature winding. The flux per
pole is 30 mWb. The number of armature conductor is 250. When connected
to 230 V DC supply it draws an armature current of 40 A. Calculate the back
emf and the speed with which motor is running. Assume armature resistance
is 0.6 Ω.
-3
Soln: Given P = 4, Z = 250, φ = 30 x 10 Wb , A = 4, R = 0.6Ω, I = 40 A,
a
V = 230 V
WKT, For a motor,
V=E +IR
Therefore,
230 = E + 40 x 0.6
Therefore,
E = 206 volts
b
a
a
a
b
b
47
And, EMF generated, E = ZNφP / 60A
b
Substituting the required values in the above equation, we get,
N = 1648 rpm
5. A 4 pole DC motor takes a 50 A armature current. The armature has lap
connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross
torque developed by the armature of the motor.
-3
Soln: Given P = 4, Z = 480, φ = 20 x 10 Wb , A = 4, I = 50 A
a
WKT,
T = 0.159φZI x (P / A)
a
N-m
a
= 76.394 N-m
6. A 4 pole, lap wound DC motor has 540 conductors. Its speed is found to
be 1000 rpm when it is made to run light. The flux per pole is 25 mWb. It is
connected to 230 V DC supply. The armature resistance is 0.8 Ω. Calculate
induced emf and armature current.
-3
Soln: Given, P = 4, A = P = 4, Z = 540, φ = 25 x 10 Wb,
V = 230 V, R = 0.8 Ω
a
Running light load means it is on no load. N = 1000 rpm,
0
EMF generated, E = ZN φP / 60A
b0
0
Substituting the required values in the above equation, we get,
E = 225 V
b0
For a motor,
V=E +I R
b0
a0
a
Substituting the required values in the above equation, we get,
I = 6.25 A
a0
7. A DC series motor runs at 500 rpm on 220 V supply drawing a current of
50 A. The total resistance of the machine is 0.15 Ω, calculate the value of the
extra resistance to be connected in series with the motor circuit that will
reduce the speed to 300 rpm. The load torque being then half of the previous
value. Assume flux proportional to the current.
Soln: V = 220 volts, the total resistance R + R = 0.15 Ω
a
In first case, N = 500rpm, I = I = I
1
1
a1
se1
se
=50 A, and T = 0.5 T
2
1
In this case, the series field current is same as armature current,
φαI αI
se
a
48
2
According to torques equation, T α φI α I
a
2
a
Therefore, T / T = (I / I )
1
2
a1
a2
2
Therefore,
T / 0.5T = (50 / I )
Therefore,
I = 35.355 A
1
1
a2
a2
From voltage equation
E = V - I (R + R )
b1
a1
a
se
= 220 – 50 x 0.15 = 212.5 V
And
E = V - I (R + R + R )
b2
a2
a
se
x
= 220 – 35.355 (0.15 + R )
x
From the speed equation,
Therefore,
N α (E / φ) α (E / I )
b
b
a
N / N = (E / E ) x (I / I )
1
2
b1
b2
a2
a1
Substituting the required values in the above equation, we get,
R = 3.5225 Ω
x
8. A series motor having resistance of 1 Ω between its terminals drives a fan,
the torque of which is proportional to the square of the speed. At 230 V, its
speed is 300 rpm and takes 15 A. The speed of the fan is to be raised to 375
rpm by supply voltage control. Estimate the supply voltage required.
Soln: V = 230 volts, R + R = 1 Ω, N = 300 rpm, I = I = I
1
a
se
1
1
a1
se1
=15 A, and
N = 375 rpm
2
In this case, the series field current is same as armature current,
φαI αI
According to torques equation, T α φI α I
Therefore, T / T = (I / I )
1
Also
2
2
a1
2
a
se
2
a
a
a2
TαN
49
Therefore,
T / T = (N / N )
1
2
2
1
2
2
2
Therefore, (N /N ) = (I / I )
1
2
a1
a2
Substituting the required values in the above equation, we get,
I = 18.75 A
a2
From the speed equation,
Therefore,
N α (E / φ) α (E / I )
b
b
a
N / N = (E / E ) x (I / I )
1
2
b1
From voltage equation
b2
a2
a1
E = V - I (R + R )
b1
a1
a
se
= 230 – 15 x 1 = 215 V
In the second case, voltage is to be changed from V to V
1
Therefore,
2
E = V - I (R + R ) = V – 18.75
b2
2
a2
a
se
2
Therefore, substituting the required value in the equation,
N / N = (E / E ) x (I / I )
1
2
b1
b2
a2
a1
We get, V = 354.6875 V
2
This is the new supply voltage required to raise the speed of fan from 300
rpm to 375 rpm.
9. A 220 V shunt motor has an armature resistance of 0.25 Ω and a field
resistance of 170 Ω. Determine the back emf if output is 4.5 kW at 87 %
efficiency.
Soln: Given, V = 220 V, R = 0.25 Ω, R = 170 Ω, Output power = 4.5 kW,
a
sh
η = 87 %
We have, Efficiency, η = Output / Input
Therefore,
Input = (Output / η) = 4.5 / 0.87 = 5.172 kW
Input current, I = Input power / Voltage
L
3
= 5.172 x 10 / 220 = 23.51 A
Field current, I = V / R = 220 / 170 = 1.294 A
sh
sh
Armature current, I = I – I = 23.51 – 1.294 = 22.216 A
a
WKT,
L
sh
E = V – I R = 220 – 22.216 x 0.25 = 214.446 V
b
a
a
10. A DC shunt motor develops a back emf of 220 V at 1500 rpm. If the
armature current is 50 A, find its mechanical power developed and the
torque.
50
Soln: Given, E = 220 V, N = 1500 rpm, I = 50 A
b
a
Mechanical power = E I = 220 x 50 = 11 kW
b a
Torque,
Ta = 9.55 E I / N = 70.03
N-m
b a
11. A 220 V DC shunt motor runs at 1500 rpm, when the armature current is
50 A. If R = 0.12 Ω, calculate the speed if torque is doubled.
a
Soln: Given, V = 220 V, N = 1500 rpm, I = 50 A, R = 0.12 Ω
a
T α φI
a
a
But φ is constant for DC shunt motor, Therefore,
Therefore,
T / T = (I / I )
Given,
T / 2T
Therefore,
T / 2T = (50 / I )
Therefore,
I = 100 A
a1
a2
a2
a1
a1
a1
TαI
a
a2
(Torque is doubled)
a1
a2
a2
From voltage equation
And
E = V - I R = 220 – 50 x 0.12 = 214 V
b1
a1
a
E = V - I R = 220 – 100 x 0.12 = 208 V
b2
a2
a
From the speed equation,
Therefore,
N α (E / φ) α (E )
b
b
N / N = (E / E )
1
2
b1
b2
Substituting the required values in the above equation, we get,
N = 1457.9 rpm
2
12. A 4 – pole, 220 V shunt motor has 480 conductors, which are lap wound.
It takes 50 A from the mains supply and develops an output power of 6.2
kW. The field resistance is 220Ω. The flux per pole is 30 mWb and R = 0.15
a
Ω. Calculate (i) back emf, (ii) speed and (iii) shaft torque.
Soln: Given, V = 220 V, P = 4, Z = 480, A = 4, I = 50 A, R = 220 Ω,
L
sh
φ = 30 mWb, R = 0.15 Ω. Output power = 6.2 kW
a
Field current, I = V / R = 220 / 220 = 1 A
sh
sh
Armature current, I = I – I = 50 – 1 = 49 A
a
(i)
WKT,
(ii)
We have,
L
sh
E = V – I R = 220 – 49 x 0.15 = 212. 65 V
b
a
a
EMF generated, E = ZNφP / 60A
b
51
Substituting the required values in the above equation, we get,
N = 886.04 rpm
(i) WKT, Shaft torque, T = 9.55 x (Output / N)
sh
= 955 x (6200 / 886.04) = 66.82
N-m
52