Main Textbook of Electrical Technology (Multicolour)

Textbook of Electrical Technology (Multicolour)

A Textbook Of Electrical Technology (Multi Colour) 25th Edition is a comprehensive book for electrical engineering students. The book comprises chapters, which cover the entire syllabus of electrical engineering in brief. In addition, the book has multicolor illustrations, which help the students in understanding the concepts better. This book is essential for electrical engineering students and educators alike.
Volume: Volume 2
Year: 2005
Publisher: S. Chand & Company Ltd
Language: english
Pages: 711
ISBN 10: 8121924413
File: PDF, 11.71 MB

You may be interested in


Most frequently terms

This site is like a god for studying
31 October 2019 (05:58) 


Introduction—Salient aspects of conversions—Energy-balance—Magnetic-fieldsystem:EnergyandCo-energy—Linear
System—A Simple Electromechanical System—In terms of
System—Energy stored in the coils—Different Categories—
One coil each on Stator and on Rotor—Vital Role of Air-gap—
Statically induced emf and Dynamically induced emf.



GeneratorPrincipal—SimpleLoopGenerator—PracticalGenerator—Yoke — Pole Cores and Pole Shoes—Pole Coils—
Two-layer Winding—Degree of Re-entrancy of an Armature
of Lap and Wave Windings—Types of Generators—Brush
Contact Drop—Generated E.M.F. or E.M.F. Equation of a
Generator—Iron Loss in Armature—Total loss in a D.C.
Generator—Stray Losses—Constant or Standing Losses—

AT per pole—Compensating Windings—No. of CompensatingWindings—Commutation—ValueofReactanceVoltage—
MethodsofImprovingCommutation—ResistanceCommutation—E.M.F. Commutation—Interpoles or Compoles—
for Paralleling D.C. Generators— Compound Generators in



... 890—915

CharacteristicsofD.C.Generators—Separately-excitedGenerator—No-load Curve for Self-excited Generator—How to
find Critical Resistance R? How to draw O.C.C. at Different
Speeds ?—Critical Speed—Voltage Build up of a Shunt Generator—Condition for Build-up of a Shunt Generator—Other
factors Affecting Voltage Building of a D.C. Generator—
External Characteristic—Voltage Regulation—Internal or TotalCharacteristic—SeriesGenerator—Compound-woundGenerator—How to calculate Required Series Turns?—Uses of
D.C. Generators Questions and Answers on D.C. Generators—Objective Tests.

29. D.C. Motor

... 916—949

Motor Principle—Comparison of Generator and Motor Action—Significance of the Back emf—Voltage Equation of a
Motor—Conditions for Maximum Power—Torque—Armature Torque of a Motor—Shaft Torque—Speed of a D.C.
Motor—Speed RegulationTorque and Speed of a D.C. Motor—Motor Characteristics—Characteristics of Series Motors—CharacteristicsofShuntMotors—CompoundMotors—
Performance Curves—Comparison of Shunt and Series Motors—LossesandEfficiency—PowerStages—ObjectiveTests.



Factors Controlling Motor Speed—Speed Control of Shunt
Motors—SpeedControlofSeriesMotors—MeritsandDemerits of Rheostatic Control Method—Series-Parallel Control—
Electric Braking—Electric Braking of Shunt Motor—Electric
Braking of Series Motors—Electronic Speed control Method
for D.C. Motors—Uncontrolled Rectifiers—Controlled Rectifiers—Thyristor Choppers—Thyristor Inverters—Thyristor
Speed Control of Separately-excited D.C. Motor—Thyristor
of a Shunt Motor—Thyristor Control of a Shunt Motor—
Thyristor Speed Control of a Series D.C. Motor—Necessity of
a Starter—Shunt Motor Starter—Three-point Starter—Fourpoint Starter—Starting and Speed Control of Series Motors—
Starters—Thyristor Controller Starters—Objective Tests.

31. Testing of D.C. Machines
Brake Test—Swinburnes Test—Advantages of Swinburnes
Test—Main Disadvantages—Regenerative or Hopkinson’s
Test—Alternative Connections for Hopkinson’s Test—Merits


of Hopkinson’s Test—Retardation or Running Down Test—
Field’s Test for Series Motors—Objective Tests—Questions
and Answers on D.C. Motors.



Working Principle of Transformer—Transformer Construction—Core-type Transform ers—Shell type Transformers—
with losses but no Magnetic Leakage—Transformer on Nolead—Transformer on Load—Transformer with Winding ResistancebutnoMagneticLeakage—EquivalentResistance—
Drop in Transformer—Exact Voltage Drop—Equivalent CircuitTransformerTests—Open-circuitorNo-loadTest—Separation of Core Losses—Short-Circuit or Impedence Test—
Why Transformer Rating in KVA?—Regulation of a Transformer—PercentageResistance,ReactanceandImpedance—
Efficiency of a Transformer—Condition for Maximum Efficiency—Variation of Efficiency with Power Factor—All-day
Efficiency—Auto-transformer—Conversion of 2-Winding
Transformer into Auto-transformer—Parallel Operation of


... 1123—1151

Three-phaseTransformers—Three-phaseTransformerConnections—Star/Star or Y/Y Connection—Delta-Delta or Connection—Wye/DeltaorY/Connection—Delta/Wyeor/YConnection—Open-DeltaorV-VConnection—PowerSuppliedby
V-V Bank—Scott Connection or T-T Connection—ThreephasetoTwo-PhaseConversionandvice-versa—ParallelOperationof3-phaseTransformers—InstrumentTransformers—
Current Transformers—Potential Transformers—Objective

Classification of AC Motors—Induction Motor: General Principal—Construction—Squirrel-cage Rotor—Phase-wound
Mathematical Proof—Why does the Rotor Rotate ?—Slip—
Frequency of Rotor Current—Relation between Torque and
Rotor Power Factor—Starting Torque—Starting Torque of a
Squirrel-cage Motor—Starting Torque of a Slip-ring Motor—


Condition for Maximum Starting Torque—Effect of Change in
Supply Voltage on Starting Torque—Rotor E.M.F and ReactanceunderRunningConditions—TorqueunderRunningCondition—Condition for Maximum Torque Under Running Conditions—RotorTorqueandBreakdownTorque—RelationbetweenTorqueandSlip—EffectofChangeinSupplyVoltageon
Torque and Speed—Effect of Change in Supply Frequency
Shape of Torque/Speed Curve—Current/Speed Curve of an
Plugging of an Induction Motor—Induction Motor Operating
as a Generator—Complete Torque/Speed Curve of a Threephase Machine—Measurement of Slip—Power Stages in an
Induction Motor—Torque Developed by an Induction Motor—Torque,MechanicalPowerandRotorOutput—Induction
Motor Torque Equation—Synchronous Watt—Variation in
Rotor Current—Analogy with a Mechnical Clutch—Analogy
with a D.C. Motor—Sector Induction Motor—Linear Induction
Motor—Properties of a Linear Induction Motor—Magnetic
Levitation—Induction Motor as a Generalized Transformer—
Rotor Output—Equivalent Circuit of the Rotor—Equivalent
Circuit of an Induction Motor—Power Balance Equation—
Maximum Power Output—Corresponding Slip—Objective



of the Approximate Equivalent Circle—Determination of G0
and B0— No-load Test—Blocked Rotor Test—Construction
of the Circle Diagram—Maximum Quantities—Starting of
Induction Motors—Direct-Switching or Line Starting of Induction Motors—Squirrel-cage Motors—Starting of Slip-ring
Control of Induction Motor—Three-Phase A.C. Commutator
Class C Motors—Class D Motors—Class E Motors—Class F
Motors—Questions and Answer on Induction Motors—ObjectiveTests.

Types of Single-phase Motors—Single-phase Induction Motor—Double-field Revoling Theory—Making Single-phase
Induction Motor Self-starting—Equivalent Circuit of Singlephase Induction Motor—without Core Loss—Equivalent


Cuicuit—With Core Loss—Types of Capacitors—Start Motors—Capacitor Start-and-Run Motor—Shaded-pole Singlephase Motor—Repulsion Type Motors—Repulsion Motor—
Repulsion Principle—Compensated Repulsion Motor—Repulsion-start Induction-run Motor— Repulsion Induction
Motor—A.C. Series Motors—Universal Motor—Speed Control of Universal Motors—Unexcited Single-phase SynchronousMotors—ReluctanceMotor—HysteresisMotor—QuestionsandAnswersonSingle-phaseMotors—ObjectiveTests.



Basic Principle—Stationary Armature—Details of Construction—Rotor—Damper Windings—Speed and Frequency—
Armature Windings—Concentric or Chain Windings—Twolayer Winding—Wye and Delta Connections—Short-pitch
Winding: pitch factor/chording factor—Distribution or Breadth
Factor or Winding Factor or Spread Factor—Equation of
Induced E.M.F.—Effect of Hormonics on Pitch and Distribution Factors—Factors Affecting Alternator Size—Alternator
Alternator—Voltage Regulation—Determination of Voltage
Regulation—Synchronous Impedence Method—Rothert's
Factor Method or Potier Method—Procedural Steps of Potier
Method—Operation of Salient Pole Synchronous Machine—
Phasor Diagram for a Salient Pole Synchronous Machine—
Calculations from Phasor Diagram—Power Developed by a
Synchonous Generator—parallel Operation of Alternators—
Synchronizing of Alternators—Synchronizing Current—Synchronizing Power—Alternators Connected to Infinite Busbars—SynchronizingTorqueTsy—EffectofLoadonSynchronizing Power—Alternative Expression for Synchronizing
Power—Parallel Operation of two Alternators—Effect of Unequal Voltages—Distribution of Load—Time-Period of Oscillation—MaximumPowerOutput—QuestionsandAnswerson

of Starting—Motor on Load with Constant Excitation—Power
Flow within a Synchronous Motor—Equivalent Circuit of a
Synchronous Motor—Power Developed by a Synchronous
Motor—Synchronous Motor with Different Excitations—Effect of increased Load with Constant Excitation—Effect of
Synchronous Motor—Power Developed by a Synchronous
Motor—Alternative Expression for Power Developed—Vari-


ous Conditions of Maxima—Salient Pole Synchronous Motor—Power Developed by a Salient Pole Synchronous Motor
— Effects of Excitation on Armature Current and Power
Procedure for Starting a Synchronous Motor—Comparison
between Synchronous and Induction Motors—Synchronous

Introduction—Stepper Motors—Types of Stepper Motors—
Motor—Low-inertia DC Motors—Shell-type Low-intertia DC
Motor—Printed-circuit (Disc) DC Motor—Permanent-Magnet Synchronous Motors—Synchros—Types of Synchros—
Applications of Synchros—Control Differential Transmitter—
Control Differential Receiver—Switched Reluctance Motor—
Comparison between VR Stepper Motor and SR Motor—The
Resolver—Servomotors—DCServomotors—ACServomotors—Objective Tests.






+ 0 ) 2 6 - 4

Learning Objectives
➣ Introduction
➣ Salient aspects of
➣ Energy-balance
➣ Magnetic-field system:
Energy and Co-energy
➣ Linear System
➣ A Simple Electromechanical System
➣ In terms of Field Energy
➣ In terms of field Coenergy
➣ Energy in terms of
Electrical parameters
➣ Rotary Motion
➣ Description of Simple
➣ Energy stored in the coils
➣ Different Categories
➣ One coil each on Stator
and on Rotor
➣ Vital Role of Air-gap
➣ Statically induced emf
and Dynamically induced





An electric motor is a machine which converts
electrical energy into mechanical (rotational
or kinetic) energy


Electrical Technology

25.1. Introduction
“Energy can neither be created nor be destroyed”. We can only change its forms, using
appropriate energy-conversion processes.
An interesting aspect about the energy in “Electrical form” is that neither it is so available
directly from nature nor it is
required to be finally consumed in
that form. Still, it is the most
popular form of Energy, since it
can be transported at remote
Load-locations, for optimum
utilization of resources. Further,
technological progress has now
made it possible to device
controllable that modern systems
tend to be energy-efficient, with
increase in life-span of main
equipment and the associated
switches, connecting cables,
contactors, etc.) since it is now
possible to avoid overstrain (= over-currents or over-voltage) on the system. This means a lot for
the total production process (for which electrical energy is being used) since the quality of the
production improves, and plant-maintenance is minimal. Energy-conversion systems then assume
still higher importance.
Energy conversion takes place between well known pairs of forms of Energy: Electrical ↔
Chemical, Electrical ↔ Thermal, Electrical ↔ Optical, Electrical ↔ Sound, and Electrical ↔
Mechanical are the common forms with numerous varieties of engineering - applications. Electrical
↔ mechanical conversion is the focus of discussion in this chapter.
The elements of electro-mechanical energy conversion shall deal with basic principles and
systems dealing with this aspect. Purpose of the study is to have a general approach to understand
to design, and later to modify the system with the help of modern technologies, for overall
It is necessary to be aware about:
(a) basic conditions to be fulfilled by the conversion system.
(b) methods for innovating the conversion systems.
Electromechanical energy-conversion finds applications in following categories of systems:
(a) transducers: Devices for obtaining signals for measurement / control,
(b) force-producing devices: Solenoid-actuators, relays, electromagnets,
(c) devices for continuous-energy-conversion: Motors / Generators
These systems have different configurations. But the principles of their working are
common. Understanding these principles enables us to analyze / design / improvise / innovate such
systems. As a result of such development, newer types of motors and the associated modern power
controllers have recently been manufactured and become popular. Controllers using power-

Elements of Electromechanical Energy Conversion


electronics switching devices offer energy-efficient, user-friendly, and high-performance drives.
Their initial investment may be larger but two important parameters justify their use: (i)
Considerable energy is saved, resulting into payback periods as short as 18-24 months, (ii) The
controllers ensure to limit the currents to pre-set values under conditions of starting /overload/
unbalanced supply. Hence, the entire system enjoys longer life. Both these effects lead to better
production-process and hence these are readily acceptable by industries.

25.2 Salient Aspects of Conversions
Purpose of electro-mechanical conversion device is to change the form of energy. Here, for
simpler discussion, only rotary systems will be dealt with. When it is converting mechanical input
to electrical output the device is “generating”. With electrical input, when mechanical output is
obtained, the device is motoring.
Some simple aspects of an electrical machine (motor / generator) have to be noted at this






Electrical machine has a Stator, a Rotor, and an air-gap in between the two. For a flux path,
the magnetic circuit has these three parts in series. In general, magnetic poles are established
in Stator and in Rotor.
Magnetic effects of following types can be
(a) Electromagnetic: Due to currents passed
through windings on Stator and / or Rotor,
producing certain number of poles on these
(b) Permanent Magnets: One side (Stator or
Rotor) can have permanent magnets.
(c) Reluctance variation: Surface of Rotor
near the air-gap can be suitably shaped to
Stator and Rotor
have a particular pattern of Reluctancevariation so as to control the machine behaviour as per requirements.
Basic conditions which must be satisfied by such devices are:
(a) Equal number of poles must be created on the two sides.
(b) In some cases, reluctance-variation is primarily used for machine-action. The Stator
side must accommodate a winding carrying current for the electromagnetic effect, when
rotor surface is shaped so as to have the desired pattern of reluctance variation. Or,
non-cylindrical rotor cannot have the current -carrying winding for machine action.
Out of stator, rotor and air-gap, maximum energy-storage at any angular position takes
place in the air-gap, since its reluctance is highest out of the three members.
Stored energy must depend on rotor-position and the device tends to occupy that angular
position which corresponds to maximum stored energy. If this position varies as a function
of time, the device produces continuous torque.
Ideal output of a motor is a constant unidirectional torque with given currents through its
windings. In some cases, the output torque (as a compromise) is an average value of a
cyclically varying torque.
Where current-switching is done for motor-control, as in modern controllers, instantaneous
effect has to be understood to conclude on any of the points mentioned above.
A device can work either as a generator or as a motor, provided pertinent conditions are
satisfied for the concerned mode of operation.


Electrical Technology

25.3. Energy – Balance
For an electro-mechanical system, following terms are important:
Electrical port (= armature terminals): receiving / delivering electrical energy.
Mechanical port (= shaft): delivering / receiving mechanical energy.
Coupling field: Magnetic field or Electric field.
Even though, theoretically, both the types
of fields mentioned above are able to convert the
energy, the magnetic medium is most popular since
the voltage levels required are not very high, and
the devices of given power rating are smaller in
size and are economical. Hence, only those will be
dealt with.
It is obvious that an electrical motor
receives energy at the electrical port and delivers it
at the mechanical port. While an electric generator Conversion of electrical energy into mechanical
receives the energy at the mechanical port and
delivers it at the electrical port. It is also known
that the following losses take place in such systems and are dissipated away as heat: (i) i2r losses
in the windings of the machines, (ii) friction and windage losses, (iii) core-losses.
These can be either neglected or attached to electrical port, mechanical port and coupling
magnetic field respectively, for simpler analysis. With this, the simple energy balance equation can
be written as:
Change in Electrical Energy = Change in Mechanical Energy + Change in Field-Energy

dWelec = dWmech + dWfld

. . .(25.1)

It is natural that this equation has +ve signs for electrical and mechanical-energy-terms
when the device is motoring. For generating mode, however, both the terms assume –ve signs.
In case no mechanical work is done, eqn. (25.1) reduces to eqn. (25.2) below indicating that
Electrical energy - input is stored in the magnetic field.
dWelec = dW fld

25.4. Magnetic-field System: Energy and Co-energy
25.4.1 Linear System

Fig. 25.1 (a) Magnetic circuit

. . . (25.2)

Elements of Electromechanical Energy Conversion


Fig. 25.1 (c) Energy and co-energy

Fig. 25.1 (b) Characteristic of a magnetic circuit

A simple magnetic current is shown in Fig 25.1 (a), with assumptions that air-gap length
at the joints is negligible, and the magnetic medium is not saturated. With A as the cross-sectional
area of the core and Lm as the mean length of the path, a coil with N turns carrying a current of i
amp has an mmf of F, establishing a flux of φ, related by
φ = F× m
where m = Permeance of the Magnetic circuit
= µoµr A/Lm
with µr = relative permeability of the magnetic medium,
This corresponds to the following relationships:
Coil Inductance, L = N2 m = Nφ / i = λ / i
where λ = flux-linkage of the coil, in weber-turns
Wfld = Energy stored in the coil = ½ Li2 = ½ N2
= ½ F(F

m i2


= ½ F2


= ½ Fφ

. . . (25.3)

In this eqn., m is the slope of the characteristic in Fig 25.1 (b). Hence, the inductance is
proportional to the slope of F-φ plot. In Fig., 25.1 (b), for the operating point A, the mmf is F1
and the flux φ1. At the point A, the energy stored in the field is given by eqn. below:
Wfld = ½ F1 φ1
F1 is due to the current i1. Wfld is given by area OATO in Fig 25.1(b).
In Fig 25.1(b), the origin refers to the system without magnetization.
The system can reach the point A, starting from O as the current in the coil is increased
from O to i1 .
Let us understand the intermediate events.
At point B, the flux is φ due to the mmf F.
An increment in coil current results into increase in mmf by dF. This increases the coreflux by dφ. New operating point is C.
Eqn. (25.3) is to be suitably re-written in terms of these incremental values.
Wfld =


= (1/

dWfld =






Fdφ =




φ . dφ = ½.φ1 . F1

(area of the strip NBB’Q) = area of triangle OAT
. . . (25.4)


Electrical Technology
Alternatively, we have the area of elemental strip kBB″M = φ . dF
Area of ∆ OAS =






(area of the strip KBB″M)


φ . dF =





F dF =


F12/2 = ½ F1 φ1

. . . (25.5)

In order to distinguish with respect to the terms in eqn. (25.4), this area is called as the
“Co-energy of the field” and is represented by W′fld. For a Linear system, however, for a given
operating point, say A, the two energy-terms are equal. Hence,
Wfld = W´fld = ½ F1φ1 = ½ Li12

. . . (25.6)

In order to have a simple and clear distinction between the two energy terms, it can be said
that the differential variable “Current” (or mmf ) is related to Co-energy and the differential
variable “flux - linkage” (or flux) is related to Energy.
This energy stored in the magnetic field comes from the electrical source connected to the
coil in Fig. 25.1(a).

25.4.2. A Simple Electromechanical System
A simple electro-mechanical system is shown in Fig. 25.2(a).

Fig. 25.2 (a)

Reference point O corresponds to the
unstretched spring. Energy stored in the spring is
then Zero. In position A of the movable member,
the spring is elongated by x, and the corresponding
energy stored in the spring is ½ Ks x2, where Ks is
“spring-constant” of the linear system in Nw / m. In
Fig 25.2 (a), the distance OA is x. The elemental
distance AB is dx, so that OB is x + dx. For simpler
analysis, it is assumed that magnetic material is
highly permeable and that the clearance at point M
(for movement of the member) is negligible. So
that the mmf of the coil is required to drive the flux
in the region OABC only. The flux-mmf

Fig. 25.2 (b)

A simple electromechanical system

Elements of Electromechanical Energy Conversion


relationships are plotted for these two positions in Fig 25.2 (b). In position A, the movable member
has moved a distance of x from its unstretched position or reference point. Let the operating point
be H, so that the coil-mmf OA (=F) establishes a flux OB (=φ). In this position, the movable
member experiences a force in such a direction that the energy stored in the field tends to increase.
It tends to reach B, so that an additional displacement of dx shifts the characteristic upwards and
final operating point in position B is C´. From H to C´, the operating point can move in any one
of the following ways:
(a) HC´ vertically, if the mechanical movement is too slow so that change of flux is slow
and induced emf in the coil is negligible. This corresponds to the coil-mmf remaining constant at
F during the transition. Constant mmf means vertical travel of the operating point from H to C´.
(b) H to K horizontally and then K to C´along the characteristic corresponding to (x + dx)
as the displacement of the movable part. This is possible when the motion is very fast, resulting
into flux remaining constant till the operating point traverses from H to K. Then, from K to C´,
the flux increases, an emf in induced the coil and the mmf finally reaches its value of F, at the
point C´.
(c) In reality, the transition from H to C´ will be somewhere in between these two
extremes mentioned above.
However, for simplicity, one of these extreme conditions has to be accepted. In (a) above,
the mmf remains constant. In (b) above, the flux (and hence the flux-linkage) remains constant. Let
us take the case of constant-mmf. If the process has taken a time of dt,
Electrical-energy input during the process = dWelec
= (voltage applied to the coil) × current × dt = eidt
= dλ /dt × i × dt = idλ = i N dφ = Fdφ = area of rectangle B′HC′D
In this case, coil-resistance has been neglected. In terms of Field Energy
At the previous operating point H, the energy stored in the magnetic field,
Wfld1 = area of ∆ OHB´
At the new position corresponding to the operating point C′, the field energy stored is given
by Wfld2 = area of the ∆ OC´D
The difference of these two is the change in the energy stored in the
magnetic field
= dWfld = ½ [OA´ × A´C´ – OA´ × A H]
= ½ OA [A C´ × A´ H] = ½ OA´ HC´
= ½ . F . dφ = ½ dWelec
Out of the energy delivered by the source, half is stored in the magnetic field. Where has
the remaining half been utilized? Obviously, this must have been transformed into the mechanical
work done. In this case, neglecting losses, it is finally stored in the stretched spring due to its
elongation by dx.
Comparing this with the equation (25.1),
dWelec = dWmech + dWfld
= dWmech + ½ dWelec

dWmech = dWfld = ½ dWelec

Consider that a force F is operative at the displacement of x. This force is in such a
direction that x increases or the movable member is attracted towards D. In the same direction, a


Electrical Technology

displacement by dx results into the increase in the energy stored by the spring. Relating the
concerned terms,
F = ks.x
dWmech = mechanical work done against the force of the stretched spring
= – F dx = dWfld
or F
= – dWfld / dx, in this case
= – δWfld / δx, in general
Alternatively, the difference in the energy stored in the spring also gives a very useful
In the position corresponding to x + dx, the energy stored in the spring
= ½ ks (x + dx)2. Similarly, at x, the energy = ½ ks x2
Difference = ½ ks [(x + dx)2 – (x)2]
= ½ ks [ x2 + 2.x.dx + (dx)2 – x2]
= ½ ks [2. x. dx], neglecting (dx)2
= ks. x. dx = F dx
This difference is nothing but dWmech, which is equal in magnitude to dWfld and confirms the
relationship obtained earlier. In terms of field Co-energy
Proceeding along lines similar to those while dealing with field-energy above, following
relationships exist. For simpler discussion, the transition is assumed to be along HKC´. Neglecting
area of the small triangle HKC´, we have
at x, W´fld1 = area of ∆ OA´H
at x + dx, W´fld2 = area of ∆ OA´C´ (neglecting ∆ HKC´)
dW´fld = W´fld2 – W´fld1 = φ. dF where dF = MA
= Co-energy in the field F dx
F = + dW´fld / dx, in this case
= + δW´fld / δx, in general

25.5. Energy in Terms of Electrical Parameters
In the preceding article, the energy and force were related in terms of magnetic-system
parameters, namely flux and mmf, through the third parameter, the permeance.
It is at times convenient to relate these things in terms of electrical-system-parameters,
namely, the inductances and currents. That is being dealt with here only for linear systems. Let
be the permeance of the magnetic circuit and L be the coil-inductance.
Wfld = Field-energy = ½ F φ = ½ Ni (Ni ) = ½ (N2 )i2
= ½ Li2
F = – dWfld /dx = – ½ . i2. dL/dx
Thus, a force exists if the coil-inductance is dependent on x. Such analysis is more suitable
when the system has more than one coils coupled through the magnetic circuit. If two such coils
are considered, following data should be known for evaluation of the force, in case of linear
L11 = self inductance of coil - 1
L22 = self inductance of coil - 2

Elements of Electromechanical Energy Conversion


L12 = Mutual inductance between two coils, 1 and 2
i1, i2 = currents through the two coils.
Wfld = Total energy stored in the field = ½ L11 i12 + ½ L22 i22 + L12 i1 i2
Magnitude of Force, F = dWfld /dx = ½ i21 dL11 / dx + ½ i22 dL22 /dx + i1i2 dL12 /dx
From the right-hand side of this equation, it is noted that the inductance-term which is
dependent on x contributes to the force.

25.6. Rotary Motion

Ring winding →
Ring rotor →
Ring gear →

Most popular systems for electromechanical energy are Generators and
→ Planet gears
Motors. The preceding discussion dealt with
Sun gear →
& cage
the Linear motions, wherein x represented the
displacement parameter, and force was being
→ Sun rotor magnets
→ Sun windings
Now we shall deal with the rotary
systems, wherein angular displacement
parameters (such as θ) and corresponding
torque developed by the system will be
correlated, through a systematic procedure for a typical rotary machine.
Rotary motion

25.6.1. Description of Simple System
A simple rotary system has a
‘stator’ and
a ‘rotor’. Air-gap
separates these two. Stator has two
similar coils ‘a’ and ‘b’ located at
90 0 electrical, with respect to each
other. Inner surface of the stator is
cylindrical. Outer surface of the rotor
is also cylindrical resulting into
uniform air-gap length for the
The diagram represents a twopole machine. Axis of coil ‘a’ may be
taken as reference, with respect to
which the rotor-coil axis makes an
angle of θ, at a particular instant of
time. For a continuous rotation of the
rotor at ω radians / sec, θ = ωt. Coil‘b’-axis is perpendicular to the
reference, as shown. Due to the
uniform air-gap length, and due to the
perpendi-cularity between coils ‘a’
Fig. 25.3. Simple rotary system
and ‘b’, inductance-parameters
exhibit the following patterns:
Let x represents the inductance parameter as a function of θ. The subscripts indicate the
particular parameter. xaa = self-inductance of coil ‘a’ , xab = mutual inductance between coils a and
b and so on. x represents value of the particular inductance parameter, which will help in knowing
the variation of inductance with θ.


Electrical Technology

(a) Self inductances of coils ‘a’ and ‘b’ are not dependent on rotor position.
xaa = Laa and xbb = Lbb, at all values of θ.
(b) Mutual inductance between stator coils ‘a’ and ‘b’ is zero, due to perpendicularity.
xab = zero, for all values of θ.
(c) Self-inductance of rotor-coil is constant and not dependent on θ.
xra= Lrr (= constant), at all values of θ.
(d) When ‘r’ and ‘a’ coils have their axes aligned, at θ = 0, the mutual inductance between
them is maximum, which is denoted by xar. At θ = 900, their axes are perpendicular resulting into
no coupling or zero mutual inductance. When θ = 1800, r and a coils are aligned in anti-parallel
way and hence maximum mutual inductance exists between them with negative sign.
Further, whatever happens to coupling between r and a at a value of θ happens to that
between r and b with a delay of 900. All these are mathematically represented as :
xra = Lra cos θ = Lra cos (ωrt)
xrb = Lra sin θ = Lra sin (ωrt) = Lra cos (ωrt – 900)
Since ‘a’ and ‘b’ coils are alike, the maximum mutual inductance is represented by the same
term Lra.

25.6.2. Energy stored in the coils
Energy stored in the magnetic field can either be expressed in terms of mmf and flux or be
expressed in terms of inductance-terms and coil currents. If ia, ib and ir are the coil-currents, storedenergy-terms are as given below:
(i) W1 = Energy in Self-ind. of coil ‘a’: ½ Laa i2a
(ii) W2 = Energy in Self-ind. of coil ‘b’: ½ Lbb i2b
(iii) W3 = Energy in Self-ind. of coil ‘r’: ½ Lrr i2r
(iv) W4 = Energy in mutual inductance between ‘a’ and ‘r’: xra ir ia = xra ir ia = xra .cos θ.ir. ia
(v) W5 = Energy in mutual inductance between ‘b’ and ‘r’ : xrb ir . ib = xra sin θ. ir . ib
W = Total energy stored in the system = Sum of all the energy-terms cited above
= W1 + W2 + W3 + W4 + W5
T = Torque produced = δW / δθ
If ia, ib, ir are assumed to be constant currents, for simplicity, so that their derivatives with
respect to θ (and hence with respect to time t) are zero, the energy-terms which include constant
inductances do not contribute to torque. W1, W2, and W3 thus cannot contribute to torque. W4 and
W5 contribute to torque related by:
= δW / δθ = δ / δθ [W4 + W5] = δ / δθ [Lra iria cos θ + Lra irib sin θ ]
= Lra ir [ – ia sin θ + ib cos θ ]
If ia = ib = is, T = Lra ir is [– sin θ + cos θ ]
For such a system, the torque is zero at θ = + 45o, and the torque is maximum at θ = – 45o.
If one of the stator currents is reversed, the result differs. For this, let ia = – is and ib = + is

T = Lra ir is [sin θ + cos θ ]
And the maximum torque occurs at θ = 450. This is a position for rotor, which is midway
between the two stator coils.

Elements of Electromechanical Energy Conversion

875 Different Categories
From the torque expressions above, it is clear that the torque exists only when stator and
rotor-coils carry currents. When only stator-coils (or only rotor coil) carry current, torque cannot
be produced.
(a) One coil each on Stator and on Rotor
In the above mentioned case, let us excite only one stator-coil. Let ia = is, ib = 01 and ir
maintained as before.
T = – Lra iris sin θ
Following observations are made for such a case:
(i) At θ = 0, T = 0, Mutual inductance xra is maximum (= Lra) and hence the stored energy
in mutual inductance is maximum, but torque is zero.
(ii) At θ = 900, T is maximum. xra is zero, hence the concerned stored energy is zero.
(iii) As seen earlier, W4 = Lra iria cos θ = Lra iris cos θ
Torque = dW4 / dθ = iris d (Lra cos θ) / dθ
Contributed by W4, the power is related as follows:
Power = rate of change of energy with time
= dW4 / dt = dW4 / dθ. dθ / dt = T. ωr = iris Lra (– sin θ ) ωr
Magnitudes of these terms are maximum for θ = 900. If θ can be set at 900, at all instants
of time, torque obtained is maximum. Such a situation
does exist in a d. c. machine in which rotor carries an
armature winding which is a lap- or wave-connected
commutator winding. The brushes are so placed on the
commutator that rotor-coil-axis satisfies the abovementioned condition of θ = 900, irrespective of the
rotor-position or rotor speed. Such an equivalence of a
rotating armature coil with such an effectively
stationary coil is referred to as a quasi-stationary
coil. It means that a rotating coil is being analyzed as
a stationary coil due to its typical behaviour for
electro-mechanical energy conversion purposes.
(b) Two stator coils carrying two-phase
currents and rotor-coil carrying d. c.:
two stator coils carry Permanent magnet synchronous motor for
two-phase alternating
washing machine
currents, a synchronously
rotating mmf is established. If the rotor-coil carries direct current, and
the rotor is run at same synchronous speed, a unidirectional constant
torque is developed. Mathematically, similar picture can be
visualized, with a difference that the total system is imagined to rotate
at synchronous speed. Such a machine is Synchronous machine, (to
be discussed in Later chapters). It can be understood through the
simple system described here.
(c) Machines with Permanent Magnets.
With suitable interpretation, the field side of the simple
system can be imagined to be with permanent magnets in place of
Permanent magnet
coil-excited electromagnets. All the interpretations made above are


Electrical Technology

valid, except for the difference that in this case there is no scope for controlling the rotor-coilcurrent-magnitude.
(d ) Machines with no rotor coil, but with premeance variation.
Smooth cylindrical rotor surfaces do not exist in such cases. There are no rotor-coils. Due
to geometry of the rotor surface, stator-coil- self inductances vary with rotor position. Thinking on
lines of relating energy terms and their derivatives for torque-calculations, the working principles
can be understood. With simple construction,
Reluctance motors belong to this category.
(e) Switched currents in Stator
In yet another type, stator coils are
distributed and properly grouped. One group
carries currents during certain time interval.
Then, this current is switched off. Another
group carries current in the next time interval
and so on. The rotor surface is so shaped that
it responds to this current switching and
torque is produced. Even though stator-coilinductances are complicated functions of rotor
position, the method of analysis for such
Current in stator coils
machines is same. Prominent types of
machines of this type are: switched reluctance motors, stepper motors, etc. Vital Role of Air-gap
Magnetic circuit of an electrical machine has a flux established due to coil-mmfs. This flux
is associated with stator core, rotor core and air-gap. An important point for understanding is to
know which out of these three stores major portion of the field energy. Through an illustrative case,
it will be clear below, in example 25.1.
Example 25.1. Let a machine with following
data be considered.
Calculate the energy stored in the air-gap and
compare the same with that stored in the cores.
Stator-core outer diameter = 15 cm
Stator-core inner diameter = 10.05 cm
Rotor-core outer diameter = 10.00 cm
Rotor-core inner diameter = 5 cm
Axial length of the machine = 8 cm
Effect of slotting is neglected. The core
volumes and air-gap volume for the machine shown
in Fig 25.4 have to be calculated.
Fig. 25.4
Volume of Stator-core = (π / 4) × (152 – 10.052) × 8 cm3 = 779 cm3
Volume of Rotor-core = (π / 4) × (102 – 52) × 8 cm3 = 471 cm3
Volume of air-gap in the machine = (π / 4) × (10.052 – 102) × 8 cm3
= 6.3 cm3

Elements of Electromechanical Energy Conversion


Let the relative permeability of the core material be 1000. If the flux density is B Wb / m2,
and µ is the permeability, the energy-density is ½ x B2 / µ Joules / m3. Let the flux density be
1.20 Wb / m2. Energy density in air-gap = ½ × 1.202 / (4 π × 10–7) = 572350 Joules / m3
= 0.573 Joules / cm3
Energy stored in air-gap = 0.573 × 6.3 = 3.6 Joules
Energy-density in Magnetic medium = ½ × 1.202 / (4π × 10–7 × 1000) = 573 J / m3
It is assumed only for simplicity that the flux density is same for the entire core of stator
and of rotor.
Energy stored in stator-core = 573 × 779 × 10–6 = 0.45 Joule
Energy stored in rotor-core = 573 × 471 × 10–6 = 0.27 Joule
It is worth noting that even though the ratio of volumes is 198, the ratio of energies is 0.2,
since, for the present case,

Kv =

volume of (Stator - core + Rotor - core) 779 + 471
Volume of air - gap

kE = Energy stored in cores / Energy stored in air-gap = (0.45 + 0.27) / 3.6 = 0.2
The ratio are like this due to µr being 1000, and kv / kE = 198 /0.2 = 1000
Alternatively, an air-gap of volume 6.3 cm3, [surrounded by the magnetic medium of
µr = 1000] is equivalent to the magnetic medium of volume 6.3 × 1000 cm3.

Converted equivalent volume of air-gap
Volume of (Stator + Rotor)
779 + 471
Energystored in air-gap
Energystored in (Stator + Rotor) (0.45 + 0.27)
This correlates the various parameters and confirms that the stored energy is maximum in the
Or, one can now say that in the process of electro-mechanical energy-conversion, the air-gap
plays a very vital role.
However, the stator-core and rotor-core help in
completing the flux-path in a well defined manner for
effective and efficient working of a rotary machine.
Example 25.2. An electromagnetic relay has
an exciting coil of 800 turns. The coil has a crosssectional area of 5 cm × 5 cm. Neglect reluctance of
the magnetic circuit and fringing.
(a) (i) Find the coil inductance if the air-gap
length is 0.5 cm.
(ii) Find the field energy stored for a coil
current of 1.25 amp.
(b) Coil-current remaining constant at 1.25 A,
find the mechanical energy output based on fieldenergy changes when the armature moves to a
position for which x = 0.25 cm. Assume slow
movement of armature.
Fig. 25.5 Electro-magnetic relay


Electrical Technology
(c) Repeat (b) above based on force-calculations and mechanical displacement.
(d) What will be change in above results of mechanical work done, if the mechanical
movement is fast, keeping the flux initially constant ?

µ 0 5 × 5 × 10 −4
(a) (i) Permeance at air-gap = 0.5 10−2 = 4π × 10 × 10 = 6.28 ×10
Coil Inductance = N2 m = 800 × 800 × 6.28 × 10–7 = 0.402 H
(ii) Energy stored in magnetic field = 1/2 Li2 = 1/2 × 0.402 × 1.252
= 0.314 joule

 H 2 µ0 A  1/ 2 × 800 × 800 × 4π × 10−7 × 5 × 5 × 10−4 2
(iii) W ′ fd = 1/ 2 L( x ) i 2 =1/ 2 
 lx 
1.005 × 10 −3 2


Ff =

δ [i 2 × (1.005 ×10−3 )]
= [1.005 ×10 −3 ] × i 2 × 2

x = 0.5 × 10–2

This is to be evaluated at


−1.005 × 10 −3 × 1.25 ×1.25
= − 62.8 NW
(0.5 ×10 −2 )2

This force has to be balanced by the spring-tension.
(b) Energy-computations : Inductance for x = 0.25 cm is first calculated. L(x2) = N2
= 800 × 800 × 2 × 6.28 ×



= 0.804 Henry

If the mechanical movement is slow, net mmf
remains unchanged and the operating point moves
along HC vertically upwards and settles at C. Added
Electrical Energy input during change-over of the
operating point from H to C,
= area of rectangle BDCH = (φ2 – φ1) F1

 L( x2 ). i − L( x1 )i 
 Ni
= i2 [L (x2) – L (x1) ] = 1.252 × [0.804 – 0.402]
= 0.628 joule
Out of this, the additional stored energy in field,

Fig. 25.6 Graphical correlation of
energy-terms for the relay

dWfld = ½ [L(x2) – L(x1)]
= 0.314 joule

The remaining 0.314 joule is transformed into mechanical form and is related to the workdone. This is obtained when the force on moving member is multiplied by the displacement.




dWmech = 0.314 joule

(c) As in a (iii) above,
F(x) = [1.005 × 10–3 × 1.252] [–1/x2]

Elements of Electromechanical Energy Conversion


dWmech = F(x).dx
Wmech =



F(x)dx = k. [ ∫x


–1/ x2dx] = – k.




1 
1.005 ×1.252
 1
=1.005 ×10−3 ×1.252 × 
×102 =
× 2 = 0.314 joule
 0.25 0.5 
This agrees with answer obtained in (b) above.
With fast movement of armature, the operating point will move from H to K first, then
follow the path KC.
This means that the energy represented by the area of the triangle KHC corresponds to the
reduced consumption of energy.
OC has a slope of m 2 = 12.56 × 10–7
OH has a slope of m 1 = 6.28 × 10–7
BK = mmf required for establishing a flux of φ1 with an air-gap of 0.25 cm
BK = ½ OA = ½ × (800 × 1.25) = 500 amp-turns = HK in the present case.
Area of the triangle KHC

= ½ × KH × HC
= ½ × 500 × [φ2 – φ1]
= ¼th of area of rectangle BDCH, in this case
= ¼ × 0.628 = 0.157 joule.

Hence, Electrical energy fed during this process = area BKCD
= area BDCH – area KHC = 0.628 – 0.157 = 0.471 joule
Increase in field energy stored ∆Wfld = area OKH
= area OHC – area KHC = 0.314 – 0.157 = 0.157
Mechanical Energy output = 0.471 – 0.157 = 0.314 Joule
It indicates that with fast movement, the electrical energy-input and the field-stored energy
have decreased by 0.157 J each but the mechanical-energy-term remains unaffected by fast or slow
movements of armature.

25.7. Dynamic Equations and System-model of a Simple System
It is quite necessary to analyze electro-mechanical conversion system for predicting the
performance and/or for monitoring the system. A simple system is being taken up here to deal with
dynamic equations and a simple model with its components is being related to the system. The
details will vary from system to system, and accordingly the equations will vary.
Fig. 25.7 shows different components of such a system meant for electrical to mechanical
conversion. On one side, an electrical source feeds the device at the ‘electrical port’. On the other
side, a force fe is developed at the ‘mechanical port’. Mechanical load is connected to this port.
(a) At Electrical Port : A voltage source is shown to feed the device. r is its effective
internal resistance. At the electrical port, the inputs are λ (= flux linkage with the coil) and i. From
λ , the voltage induced in the coil can always be evaluated.
(b) Role of the Conversion device: With these inputs, the device converts the energy into
mechanical form, and is available as a force fe (in case of linear motions), and, displacement x
measured from a suitable reference.


Electrical Technology

(c) At the Mechanical Port: The possible items are: spring, damper, mass and an applied
mechanical force. Their natural and simple dependence on displacement x and its derivatives are
indicated below:

Fig. 25.7 Linear motion: MODEL


Force required to overcome spring elongation is proportional to the
displacement x.
(ii) Damper: Force required to overcome damping action in the system is proportional to
derivative of x.
(iii) Mass: Force required to overcome acceleration of mass is proportional to second
derivative of x.
(iv) Applied force, fo: This has to be overcome by fe . In terms of an equation, these terms
are related as follows:
fe = ks (x – xo) + Bx + M x + fo
ks = spring constant
xo = value of x for unstretched spring
B = damping constant
M = Mass to be accelerated
fo = External mechanical force applied to the system.

25.8. Statically induced emf and Dynamically Induced emf :
In Fig. 25.7 source voltage is ν o. Let L (x) be the coil inductance as a function of
displacement x. In a very general case,
νo (t) = ri + dλ /dt
= ri + d /dt [L (x) × i]
= ri + L (x). di / dt + i. dL (x)/dx. dx/dt
The second term on the right-hand side is statically induced emf (or transformer-emf), since
change of current with time is responsible for it. This cannot produce any force (or torque) and
hence cannot convert energy from electrical to mechanical form (or vice-versa).
The third term on the right hand side includes the speed (= dx/dt) and dependence of L (x)
on x. Any of these, if non-existent, will mean that third term reduces to zero. This term relates
dynamically induced emf (= speed emf) and is the main indicator of the process of electro-

Elements of Electromechanical Energy Conversion


mechanical energy conversion. So, for conversion, there must be an inductance which varies with
the system position, and a motion must be there. In addition, coil must carry a current.
Having understood the linear-motion-system, it is easier to understand the system with
rotary motion, with due modifications.
Example 25.3. A doubly excited rotating machine has the following self and mutual
rs = 40 Ω , Ls = 0.16 H
rr = 2 Ω ,
Lr = 0.04 + 0.02 cos 2θ
Msr = 0.08 cos θ
where θ is the space-angle between axes of rotor-coil and of stator-coil. The rotor is revolving
at a speed of 100 radians/sec. For is = 10 Amp d. c., and ir = 2 Amp d. c., obtain an expression
for torque and corresponding electrical power.
[Rajiv Gandhi Technical University, Bhopal, Summer 2001]
Solution. Wfld = Total energy stored
= ½ Ls i2s + 1/2 Lr i2r + Msr isir
= ½ (0.16) i2s + ½ [0.04 + 0.02 cos 2θ]i2r
+ [0.08 cos θ] isir
since is and ir are direct currents of constant magnitudes, there is no variation with φ or with t.
Relating torque with Wfld and substituting current-magnitudes,
Torque, T =

−dW fld

= – [0 + ½ × 0.02 × 22 (– 2sin 2 θ) + 0.08 (–sin θ ) (10 × 2)]
= 0.08 sin 2 θ + 1.6 sin θ
= 1.6 sin θ + 0.08 sin 2 θ Nw-m

On the right hand side, the first term is
electromagnetic Torque which is dependent on
both the currents. Second term is dependent only
on one current, and is of the type categorized as
Reluctance-torque which depends on noncylindrical shape,in this case, on the stator side,
as shown in Fig. 25.8
Starting from Wfld, electrical power can
be expressed, since it is well -known that

Fig. 25.8

Power = time rate of change of energy
Electrical power,

p =

dW fld


d 1
 2 L i 2 s + 1 2 Lr i 2 r + M ST is ir 
dt  s




i 2S

( Ls ) + 1 2 i 2r ( Lr ) + is ir ( M sr )


Electrical Technology
 dθ 1 2  d
 dθ
 dθ
= 1 2 i 2 S  ( Ls ) 
+ 2 i r  + ( Lr ) 
+ is ir  ( M st ) 
 dθ
 dt
 dθ
 dt
 dθ
 dt
= 1 2 i 2 S [zero] ×100 + 1 2 ir [− 0.02 × 2 × sin 2θ]×100 + is ir [− 0.08sin θ]×100

Substituting numerical values of currents,the electrical power is expressed as a function of
θ, as below :
p = [zero + (– 8) sin 2θ + (–160) sin θ] watts
= [– 160 sin θ – 8 sin 2θ ] watts
Proper interpretation of sign of power (as dependent on θ) is important. Positive power is
received by the coils, while negative power is received by the source.
Example 25.4. An inductor has an inductance which varies with displacement x as
L = 2 Lo / [1 + (x / xo)]
Where Lo = 50 mh, xo = 0.05cm, x = displacement in cm,
The coil –resistance is 0.5 ohm.
(a) The displacement x is held constant at 0.075 cm, and the current is increased from 0 to 3
amp. Find the resultant magnetic stored energy in the inductor.
(b) The current is then held constant at 3 amp and the displacement is increased to 0.15 cm. Find
the corresponding change in the magnetic stored energy.
Note. Assume that all electrical transients are negligible.
Solution. (a) Inductance at x = 0.075 cm is calculated first.

L1 =

2 L0
= 40 mH
1+ (0.075 / 0.05)

= L1 × current = 120 × 10–3, corresponding to point A in Fig. 25.9

Wfld = ½ L1 (3)2 = ½ (40 × 10–3 × 9) = 0.18 joule
(b) Inductance for x = 0.15 cm is to be calculated now.

L2 =

2 L0
= 25 mH
1 + (0.075 / 0.05)

With current held constant at 3amp, the flux-Linkage is now
λ2= (25 × 10-3) × 3 = 75 × 10-3
Since the current is constant at 3amp, magnetic stored
energy is reduced by the area of triangle OAB, in Fig. 25.9
Area of triangle OAB = ½ × 3 × (120 –75) 10-3 = 0.0675
Check : Stored-energy at B in terms of L2 and i, is given by
Wfld = ½ (25 x 10-3 ) × 32 =0.1125 joule
Wfld2 = Wfld1 – area of ∆ OAB
= 0.18 – 0.0675 = 0.1125 Joule

Fig. 25.9

Elements of Electromechanical Energy Conversion


Example 25.5. If the inductor in the previous case is
connected to a voltage source which increases from 0 to 3 V
[part (a)] and then is held constant at 3 V [part (b)],repeat
the problem, assuming that electrical transients are
Solution. Coil resistance is 0.5 ohm. When the
voltage reaches 3 V, the coil current is 6 amp. In part (a), L1
= 40 mH. Hence, Wfld3 = energy stored = ½ L1 i2l = 0.72
joule, at point C in Fig. 25.10. In part (b), L2 = 25 mH. The
current is held constant at 6 amp. Working on similar lines,
∆ W fld = change in the field energy stored = area of
triangle ODC
∆Wfld = Wfld3 – Wfld4

Fig. 25.10

Wfld = ½ × 25 × 10–3 × 36 = 0.45 Joule, at point D
Change in energy stored in the field = Wfld3 – Wfld4 = 0.72 – 0.45 = 0.27 joule
∆ Wfld = area of ∆ODC = ½ × 6 × (λ3 – λ4)


λ3 = 40 × 10–3 × 6, and λ4 = 25 × 10–3 × 6


∆ Wfld

= ½ × 6 × 6 × 10–3 (40 – 25) = 0.27 joule

Example 25.6. A coil of an electromagnetic relay is associated with a magnetic circuit
whose reluctance is given by
= a + bx
where a and b are positive constants decided by the details of the magnetic circuit, in which x is
the length of the air-gap between fixed and movable members. If the coil is connected to an A.C.
source where voltage is described by

ν = Vm sin ωt,
find the expression for the average force on armature, with air-gap held constant at x.
Solution. If φ = flux established, in Webers

= number of turns on the coil,

Wfld = ½


And force

δ W fld

λ = flux-linkage in Weber-turns


= 1 2 φ2

= − 1 2 b φ2

The current in the coil is given by
v=R+ L


for which, the steady-state solution for current with an a.c. voltage applied to the coil is
given by


R +ω L

2 2

∠ − θ where θ = tan −1

RMS voltage, V = (Vm) /


(ω L)


Electrical Technology
Instantaneous current i is expressed as


R + ω 2 L2

sin (ω t − θ)

Further, L = N2/

φ = Ni / =


Ff =

2 NV
sin (ωt − θ)
( R ) + ( N 2ω )2

−bN 2V 2
sin 2 (ω t − θ )
( R )2 + ( N 2ω )2

The last term sin2 (ωt – θ) has a time average (over a cycle) of ½.
Hence, average force, Ff (av) = − 1 2

bN 2V 2
( R ) 2 + ( N 2ω ) 2

Force is in such a direction that x will be reduced, or that the energy stored tends to
Example 25.7. Two coupled coils
have self- and mutual-inductances as
expressed below:
L11 = 1 + (1 / x),
L22 = 0.5 + (1 / x),
L12 = L21 = 1 / x
These expressions are valid over a
certain range of linear displacement x, in
cms. The first coil is excited by a constant
current of 20 A and the second one, by a
constant current of – 10 A. Find
(a) mechanical work done if x
changes from 0.5 to 1.0 cm
(b) energy supplied by the two
electrical sources in (a) above.

Two coupled coils

Solution. With data given, substituting the values of currents,

W fld = 1 2 L11i12 + L12i1i2 + 1 2 L22 i22
= 225 + (50 / x )
F f = − δ W fld / δ x = 50 / x 2
(a) ∆ Wmech = ∫ (50 / x )dx


= 50[ x −1 / − 1]1.0
0.5 = + 50 joules

Elements of Electromechanical Energy Conversion


At x = 0.5, Wfld = 325 joules, and
at x = 1.0, Wfld = 275 joules
Thus, increase in x from 0.50 to 1.0 cm decreases the stored energy in the field from 325
to 275 joules. The field-system, thus, releases an energy of 50 joules.
(b) Calculations of Energy input from electrical sources –
λ1 =

L11 i1 + L12 i2

= 20 [1 + (1/ x)] – 10 (1 / x) = 20 + (10 / x)
x = 0.5, λ1 = 20 + 20 = 40 Wb-turns


x = 1.0, λ1 = 20 + 10 = 30 Wb-turns
∆Welec1 = i1 [change in λ1 due to displacement]
= 20 × (–10) = – 200 Joules


= L12 i1 + L22 i2 = –5 + (10 / x)

At x = 0.5, λ2 = –5 + 20 = + 15 Wb-turns
x = 1.0, λ2

= –5 + 10 = + 5


∆Welec2 = (–10) (–10) = + 100 Joules.
Seeing the signs and numerical values, it can be seen that Source 1 receives an energy of
200 Joules, which comes from three constituents:
100 J from source 2,
50 J from field energy stored,
50 J from mechanical system.


Tutorial Problems 25.1

A magnetic circuit has a coil with 1000 turns. Its reluctance is expressed as
= [ 8.5 + 40 g] × 10+3 MKS units

where g = air- gap length in mm, between fixed and movable parts. For a coil current of 2.0
amp held constant and with slow movement, calculate the change in the field energy stored, if the length
of the air-gap changes from 0.20 to 0.15 cm. Calculate the mechanical force experienced by the system.
Hint: ∆ We = i ( λ2 – λ1), ∆ Wfld = ½ ∆ We
Force, F = – ∆ Wfld / ∆ x
[Ans. ∆Wfld = 6.60 J, Force = 13200 Nw]
(b) An electro-magnetic relay with an air-gap of x cm has the current and flux-linkage relationship as
i = λ2

+ λ (o.5 – x)2 amp, for x < 0.5 cm

Find the force on armature as a function of λ and x.

Wf (λ1 x) =




id λ

And Ff = – δWf / δx
[Ans. Wf ( λ , x) = ( λ3/3) + ( λ2/2) (0.5 – x)2
Ff = λ2 (0.5 – x)


Electrical Technology






Electromagnetic Relay

(c) For a rotary system, the stator-coil and the rotor-coil have self and mutual-inductances as
described below, with suffix 1 for stator and 2 for rotor:

= L22 = 4 – (6θ / π )


0< θ < π/2

= 1 + (6 / π) ( θ – 0.5 π) for π / 2 < θ < π
(Note: Self inductances cannot be negative.)
L12 = L21 = 6 (1 – 2θ/π) for 0 < θ < π
Evaluate the inductances and the torque for θ = π / 4 and the two coil currents of 5 amp constant
in magnitude.
Hint: δ L / δθ contributes to torque.
[Ans. L11 = L22 = 2.5 H
L12 = + 3 H

= 450 / π Nw]



+ 0 ) 2 6 - 4

ning Objectiv

Generator Principal
Simple Loop Generator
Practical Generator
Pole Cores and Pole Shoes
Pole Coils
Armature Core
Armature Windings
Bushes and Bearings
Conductor-Coil and
Winding Element
Coil-span or Coil-pitch
Pitch of a Winding
Back Pitch
Front Pitch
Resultant Pitch
Commutator Pitch
Single-layer Winding
Two-layer Winding
Degree of Re-entrancy of
an Armature Winding
Multiplex Winding
Lap and Wave Winding
Simplex-lap Winding
Numbering of Coils and
Commutator Segments
Simplex Wave Winding
Dummy or Idle Coils
Uses of Lap and Wave
Types of Generators
Brush Contact Drop
Generated E.M.F. or E.M.F.
Equation of a Generator
Iron Loss in Armature
Total loss in a D.C.
Stray Losses
Constant or Standing
Power Stages
Condition for Maximum





Generator converts mechanical energy into
electrical energy using electromagnetic


Electrical Technology

26.1. Generator Principle
An electrical generator is a machine
which converts mechanical energy (or
power) into electrical energy (or power).
The energy conversion is based on the
principle of the production of dynamically
(or motionally) induced e.m.f. As seen
from Fig. 26.1, whenever a conductor cuts
magnetic flux, dynamically induced e.m.f.
is produced in it according to Faraday’s
Laws of Electromagnetic Induction. This
Hence, two basic essential parts of an
electrical generator are (i) a magnetic field
and (ii) a conductor or conductors which
can so move as to cut the flux.

Cut-away view of dc generator

26.2. Simple Loop Generator
In Fig. 26.1 is shown a single-turn rectangular copper coil ABCD rotating about its own axis in a
magnetic field provided by either permanent magnet is or electromagnets. The two ends of the coil

Fig. 26.1

are joined to two slip-rings ‘a’ and ‘b’ which are insulated from each other and from the central shaft.
Two collecting brushes (of carbon or copper) press against the slip-rings. Their function is to collect
the current induced in the coil and to convey it to the external load resistance R.
The rotating coil may be called ‘armature’ and the magnets as ‘field magnets’.

Imagine the coil to be rotating in clock-wise direction (Fig. 26.2). As the coil assumes successive
positions in the field, the flux linked with it changes. Hence, an e.m.f. is induced in it which is

D.C. Generators


proportional to the rate of change of flux linkages (e = NdΦdt). When the plane of the coil is at right
angles to lines of flux i.e. when it is in position, 1, then flux linked with the coil is maximum but rate
of change of flux linkages is minimum.
It is so because in this position, the coil sides AB and CD do not cut or shear the flux, rather they
slide along them i.e. they move parallel to them. Hence, there is no induced e.m.f. in the coil. Let us
take this no-e.m.f. or vertical position of the coil as the starting position. The angle of rotation or
time will be measured from this position.

Fig. 26.2

Fig. 26.3

As the coil continues rotating further, the rate of change of flux linkages (and hence induced
e.m.f. in it) increases, till position 3 is reached where θ = 90º. Here, the coil plane is horizontal i.e.
parallel to the lines of flux. As seen, the flux linked with the coil is minimum but rate of change of
flux linkages is maximum. Hence, maximum e.m.f. is induced in the coil when in this position
(Fig. 26.3).
In the next quarter revolution i.e. from 90º to 180º, the flux linked with the coil gradually
increases but the rate of change of flux linkages decreases. Hence, the induced e.m.f. decreases
gradually till in position 5 of the coil, it is reduced to zero value.
So, we find that in the first half revolution of the coil, no (or minimum) e.m.f. is induced in it
when in position 1, maximum when in position 3 and no e.m.f. when in position 5. The direction of
this induced e.m.f. can be found by applying Fleming’s Right-hand rule which gives its direction
from A to B and C to D. Hence, the direction of current flow is ABMLCD (Fig. 26.1). The current
through the load resistance R flows from M to L during the first half revolution of the coil.
In the next half revolution i.e. from 180º to 360º, the variations in the magnitude of e.m.f. are
similar to those in the first half revolution. Its value is maximum when coil is in position 7 and
minimum when in position 1. But it will be found that the direction of the induced current is from D
to C and B to A as shown in Fig. 26.1 (b). Hence, the path of current flow is along DCLMBA which
is just the reverse of the previous direction of flow.
Therefore, we find that the current which we obtain from such a simple generator reverses its
direction after every half revolution. Such a current undergoing periodic reversals is known as alternating current. It is, obviously, different from a direct current which continuously flows in one and
the same direction. It should be noted that alternating current not only reverses its direction, it does
not even keep its magnitude constant while flowing in any one direction. The two half-cycles may be
called positive and negative half-cycles respectively (Fig. 26.3).
For making the flow of current unidirectional in the external circuit, the slip-rings are replaced
by split-rings (Fig. 26.4). The split-rings are made out of a conducting cylinder which is cut into two
halves or segments insulated from each other by a thin sheet of mica or some other insulating material (Fig. 26.5).


Electrical Technology

As before, the coil ends are joined to these segments on which rest the carbon or copper brushes.
It is seen [Fig. 26.6 (a)] that in the first half revolution current flows along (ABMNLCD) i.e. the
brush No. 1 in contact with segment ‘a’ acts as the positive end of the supply and ‘b’ as the negative
end. In the next half revolution [Fig. 26.6 (b)], the direction of the induced current in the coil has
reversed. But at the same time, the positions of segments ‘a’ and ‘b’ have also reversed with the

Fig. 26.4

Fig. 26.5

result that brush No. 1 comes in touch with the segment which is positive i.e. segment ‘b’ in this
case. Hence, current in the load resistance again flows from M to L. The waveform of the current
through the external circuit is as shown in Fig. 26.7. This current is unidirectional but not continuous like pure direct current.

Fig. 26.6

Fig. 26.7

It should be noted that the position of brushes is so arranged that the change over of segments ‘a’
and ‘b’ from one brush to the other takes place when the plane of the rotating coil is at right angles to
the plane of the lines of flux. It is so because in that position, the induced e.m.f. in the coil is zero.
Another important point worth remembering is that even now the current induced in the coil
sides is alternating as before. It is only due to the rectifying action of the split-rings (also called
commutator) that it becomes unidirectional in the external circuit. Hence, it should be clearly understood that even in the armature of a d.c. generator, the induced voltage is alternating.

26.3. Practical Generator
The simple loop generator has been considered in detail merely to bring out the basic principle

D.C. Generators


underlying construction and working of an actual generator illustrated in Fig. 26.8 which consists of
the following essential parts :
1. Magnetic Frame or Yoke
2. Pole-Cores and Pole-Shoes
3. Pole Coils or Field Coils
4. Armature Core
5. Armature Windings or Conductors
6. Commutator
7. Brushes and Bearings
Of these, the yoke, the pole cores, the armature core and air gaps between the poles and the
armature core or the magnetic circuit whereas the rest form the electrical circuit.

Field Poles




Fig. 26.8

26.4. Yok
The outer frame or yoke serves double purpose :
(i) It provides mechanical support for the poles
and acts as a protecting cover for the whole machine
(ii) It carries the magnetic flux produced by the
In small generators where cheapness rather than
weight is the main consideration, yokes are made of
cast iron. But for large machines usually cast steel or
rolled steel is employed. The modern process of formYoke
ing the yoke consists of rolling a steel slab round a cylindrical mandrel and then welding it at the bottom. The feet and the terminal box etc. are welded to
the frame afterwards. Such yokes possess sufficient mechanical strength and have high permeability.

26.5. Pole Cor
es and P
ole Shoes
The field magnets consist of pole cores and pole shoes. The pole shoes serve two purposes


Electrical Technology

(i) they spread out the flux in the air gap and also, being of larger cross-section, reduce the reluctance
of the magnetic path (ii) they support the exciting coils (or field coils) as shown in Fig. 26.14.
There are two main types of pole construction.
(a) The pole core itself may be a solid piece made out of either cast iron or cast steel but the pole
shoe is laminated and is fastened to the pole face by means of counter sunk screws as shown in
Fig. 24.10.
(b) In modern design, the complete pole cores and pole shoes are built of thin laminations of
annealed steel which are rivetted together under hydraulic pressure (Fig. 26.11). The thickness of
laminations varies from 1 mm to 0.25 mm. The laminated poles may be secured to the yoke in any of
the following two ways :
(i) Either the pole is secured to the yoke by means of screws bolted through the yoke and
into the pole body or
(ii) The holding screws are bolted into a steel bar which passes through the pole across the
plane of laminations (Fig. 26.12).

Fig. 26.9

Fig. 26.10

Fig. 26.11

Fig. 26.12

D.C. Generators


26.6. Pole Coils
The field coils or pole coils, which consist of copper wire or strip, are former-wound for the
correct dimension (Fig. 26.13). Then, the former is removed and wound coil is put into place over the
core as shown in Fig. 26.14.
When current is passed through these coils, they electromagnetise the poles which produce the
necessary flux that is cut by revolving armature conductors.

26.7. Ar
e Cor
It houses the armature conductors or coils and causes them to rotate and hence cut the magnetic
flux of the field magnets. In addition to this, its most important function is to provide a path of very
low reluctance to the flux through the armature from a N-pole to a S-pole.
It is cylindrical or drum-shaped and is built up of usually circular sheet steel discs or laminations
approximately 0.5 mm thick (Fig. 26.15). It is keyed to the shaft.
The slots are either die-cut or punched on the outer periphery of the disc and the keyway is
located on the inner diameter as shown. In small machines, the armature stampings are keyed directly
to the shaft. Usually, these laminations are perforated for air ducts which permits axial flow of air
through the armature for cooling purposes. Such ventilating channels are clearly visible in the laminations shown in Fig. 26.16 and Fig. 26.17.
pole core


pole shoe

Fig. 26.13

Fig. 26.14

Up to armature diameters of about one metre, the circular stampings are cut out in one piece as
shown in Fig. 26.16. But above this size, these circles, especially of such thin sections, are difficult
to handle because they tend to distort and become wavy when assembled together. Hence, the circular laminations, instead of being cut out in one piece, are cut in a number of suitable sections or
segments which form part of a complete ring (Fig. 26.17).



Key way

Key way

Air holes
Fig. 26.15

Air holes
Fig. 26.16


Electrical Technology

A complete circular lamination is made up of four
or six or even eight segmental laminations. Usually, two
keyways are notched in each segment and are dove-tailed
or wedge-shaped to make the laminations self-locking in
The purpose of using laminations is to reduce the
loss due to eddy currents. Thinner the laminations, greater
is the resistance offered to the induced e.m.f., smaller the
current and hence lesser the I R loss in the core.

26.8. Ar
e Windings
Fig. 26.17
The armature windings are usually former-wound.
These are first wound in the form of flat rectangular coils and are then pulled into their proper shape
in a coil puller. Various conductors of the coils are insulated from each other. The conductors are
placed in the armature slots which are lined with tough insulating material. This slot insulation is
folded over above the armature conductors placed in the slot and is secured in place by special hard
wooden or fibre wedges.

26.9. Commutator
The function of the commutator is to facilitate collection of current from the armature conductors. As shown in Art. 26.2, it rectified i.e. converts the alternating current induced in the armature
conductors into unidirectional current in the external load circuit. It is of cylindrical structure and is
built up of wedge-shaped segments of high-conductivity hard-drawn or drop forged copper. These



Riser or



Fig. 26.18

Fig. 26.19

segments are insulated from each other by thin layers of mica. The number of segments is equal to the
number of armature coils. Each commutator segment is connected to the armature conductor by
means of a copper lug or strip (or riser). To prevent them from flying out under the action of centrifugal forces, the segments have V-grooves, these grooves being insulated by conical micanite rings. A
sectional view of commutator is shown in Fig. 26.18 whose general appearance when completed is
shown in Fig. 26.19.

26.10. Brushes and Bearings
The brushes whose function is to collect current from commutator, are usually made of carbon or


D.C. Generators

graphite and are in the shape of a rectangular block. These brushes are housed in brush-holders
usually of the box-type variety. As shown in Fig. 26.20, the brush-holder is mounted on a spindle and
the brushes can slide in the rectangular box open at both ends. The brushes are made to bear down on
the commutator by a spring whose tension can be adjusted by changing the position of lever in the
notches. A flexible copper pigtail mounted at the top of the brush conveys current from the brushes
to the holder. The number of brushes per spindle depends on the magnitude of the current to be
collected from the commutator.
Cu-pig tail




Fig. 26.20

Because of their reliability, ball-bearings are frequently employed, though for heavy duties, roller
bearings are preferable. The ball and rollers are generally packed in hard oil for quieter operation and
for reduced bearing wear, sleeve bearings are used which are lubricated by ring oilers fed from oil
reservoir in the bearing bracket.

26.11. Ar
e Windings
Now, we will discuss the winding of an actual armature. But before doing this, the meaning of the following terms used in connection with
armature winding should be clearly kept in mind.

26.12. Pole-pitch
It may be variously defined as :
(i) The periphery of the armature divided
by the number of poles of the generator
Armature winding
i.e. the distance between two adjacent
(ii) It is equal to the number of armature conductors (or armature slots) per pole. If there are 48
conductors and 4 poles, the pole pitch is 48/4 = 12.

26.13. Conductor
The length of a wire lying in the magnetic field and in which an e.m.f. is induced, is called a
conductor (or inductor) as, for example, length AB or CD in Fig. 26.21.


Electrical Technology

26.14. Coil and Winding Element
With reference to Fig. 26.21, the two conductors AB and CD along with their end connections
constitute one coil of the armature winding. The coil may be single-turn coil (Fig. 26.21) or multiturn coil (Fig. 26.22). A single-turn coil will have two conductors. But a multi-turn coil may have
many conductors per coil side. In Fig. 26.22, for example, each coil side has 3 conductors. The

Fig. 26.21

Fig. 26.22

Fig. 26.23

group of wires or conductors constituting a coil side of a multi-turn coil is wrapped with a tape as a
unit (Fig. 26.23) and is placed in the armature slot. It may be noted that since the beginning and the
end of each coil must be connected to a commutator bar, there are as many commutator bars as coils
for both the lap and wave windings (see Example 26.1).
The side of a coil (1-turn or multiturn) is called a winding element. Obviously, the number of
winding elements is twice the number of coils.

26.15. Coil-span or Coil-pitch (YS)
It is the distance, measured in terms of armature
slots (or armature conductors) between two sides of a
coil. It is, in fact, the periphery of the armature spanned
by the two sides of the coil.
If the pole span or coil pitch is equal to the pole
pitch (as in the case of coil A in Fig. 26.24 where polepitch of 4 has been assumed), then winding is called
full-pitched. It means that coil span is 180 electrical
degrees. In this case, the coil sides lie under opposite
poles, hence the induced e.m.fs. in them are additive.
Therefore, maximum e.m.f. is induced in the coil as a
whole, it being the sum of the e.m.f.s induced in the two
coil sides. For example, if there are 36 slots and 4 poles,
then coil span is 36/4 = 9 slots. If number of slots is 35,
then YS = 35/4 = 8 because it is customary to drop
If the coil span is less than the pole pitch (as in coil
B where coil pitch is 3/4th of the pole pitch), then the

Fig. 26.24

D.C. Generators


winding is fractional-pitched. In this case, there is a phase difference between the e.m.fs. in the two
sides of the coil. Hence, the total e.m.f. round the coil which is the vector sum of e.m.fs. in the two
coil sides, is less in this case as compared to that in the first case.

26.16. Pitch of a Winding (Y)
In general, it may be defined as the distance round the armature between two successive conductors which are directly connected together. Or, it is the distance between the beginnings of two
consecutive turns.
Y = YB − YF ........for lap winding
= YB + YF ........for wave winding
In practice, coil-pitches as low as eight-tenths of a pole pitch are employed without much serious
reduction in the e.m.f. Fractional-pitched windings are purposely used to effect substantial saving in
the copper of the end connections and for improving commutation.

26.17. Back Pitch (YB)
The distance, measured in terms of the armature conductors, which a coil advances on the back
of the armature is called back pitch and is denoted by YB.
As seen from Fig. 26.28, element 1 is connected on the back of the armature to element 8.
Hence, YB = (8 − 1) = 7.

26.18. Fr
ont Pitch (YF)
The number of armature conductors or elements spanned by a coil on the front (or commutator
end of an armature) is called the front pitch and is designated by YF. Again in Fig. 26.28, element 8
is connected to element 3 on the front of the armature, the connections being made at the commutator
segment. Hence, YF = 8 − 3 = 5.
Alternatively, the front pitch may be defined as the distance (in terms of armature conductors)
between the second conductor of one coil and the first conductor of the next coil which are connected
together at the front i.e. commutator end of the armature. Both front and back pitches for lap and
wave-winding are shown in Fig. 26.25 and 26.26.

Fig. 26.25

Fig. 26.26

Fig. 26.27

26.19. Resultant Pitch (YR)
It is the distance between the beginning of one coil and the beginning of the next coil to which it
is connected (Fig. 26.25 and 26.26).
As a matter of precaution, it should be kept in mind that all these pitches, though normally


Electrical Technology

stated in terms of armature conductors, are
also sometimes given in terms of armature
slots or commutator bars because commutator is, after all, an image of the winding.

YB = 7

26.20. Commutator Pitch (YG)
It is the distance (measured in commutator bars or segments) between the segments
to which the two ends of a coil are connected.
From Fig. 26.25 and 26.26 it is clear that for
lap winding, YC is the difference of YB and
YF whereas for wavewinding it is the sum of
YB and YF. Obviously, commutator pitch is
equal to the number of bars between coil
leads. In general, YC equals the ‘plex’ of the
lap-wound armature. Hence, it is equal to 1,
2, 3, 4 etc. for simplex-, duplex, triplex–and
quadruplex etc. lap-windings.

YF = 5


Fig. 26.28

26.21. Single-lay
er Winding
It is that winding in which one conductor or one coil side is placed in each armature slot as shown
in Fig. 26.27. Such a winding is not much used.

26.22. Two-lay
er Winding
In this type of winding, there are two conductors or coil sides per slot arranged in two layers.
Usually, one side of every coil lies in the upper half of one slot and other side lies in the lower half
of some other slot at a distance of approximately one pitch away (Fig. 26.28). The transfer of the
coil from one slot to another is usually made in a radial plane by means of a peculiar bend or twist at
the back end as shown in Fig. 26.29. Such windings in which two coil sides occupy each slot are
most commonly used for all medium-sized machines. Sometimes 4 or 6 or 8 coil sides are used in
each slot in several layers because it is not practicable to have too many slots (Fig. 26.30). The coil
sides lying at the upper half of the slots are numbered odd i.e. 1, 3, 5, 7 etc. while those at the lower
half are numbered even i.e. 2, 4, 6, 8 etc.

Fig. 26.29

Fig. 26.30

D.C. Generators


26.23. Degr
ee of R
e-entrant of an Ar
e Winding
A winding is said to be single re-entrant if on tracing
through it once, all armature conductors are included on
returning to the starting point. It is double re-entrant if
only half the conductors are included in tracing through
the winding once and so on.

26.24. Multiple
x Winding
In such windings, there are several sets of completely
closed and independent windings. If there is only one set
of closed winding, it is called simplex wave winding. If
there are two such windings on the same armature, it is
called duplex winding and so on. The multiplicity affects
a number of parallel paths in the armature. For a given
number of armature slots and coils, as the multiplicity
increases, the number of parallel paths in the armature
increases thereby increasing the current rating but
decreasing the voltage rating.


Multiplex Winding

26.25. La
p and Wav
e Windings
Two types of windings mostly employed for drum-type armatures are known as Lap Winding and
Wave Winding. The difference between the two is merely due to the different arrangement of the end
connections at the front or commutator end of armature. Each winding can be arranged progressively
or retrogressively and connected in simplex, duplex and triplex. The following rules, however, apply
to both types of the windings :

21 22 23 24 1
Lap winding






Wave winding

(i) The front pitch and back pitch are each approximately equal to the pole-pitch i.e. windings
should be full-pitched. This results in increased e.m.f. round the coils. For special purposes, fractional-pitched windings are deliberately used (Art. 26.15).
(ii) Both pitches should be odd, otherwise it would be difficult to place the coils (which are
former-wound) properly on the armature. For exmaple, if YB and YF were both even, the all
the coil sides and conductors would lie either in the upper half of the slots or in the lower


Electrical Technology

half. Hence, it would become impossible for one side of the coil to lie in the upper half.
Hence, it would become impossible for one side of the coil to lie in the upper half of one slot
and the other side of the same coil to lie in the lower half of some other slot.
(iii) The number of commutator segments is equal to the number of slots or coils (or half the
number of conductors) because the front ends of conductors are joined to the segments in
(iv) The winding must close upon itself i.e. if we start from a given point and move from one coil
to another, then all conductors should be traversed and we should reach the same point again
without a break or discontinuity in between.

26.26. Simplex Lap-winding*
It is shown in Fig. 26.25 which employs single-turn coils. In lap winding, the finishing end of
one coil is connected to a commutator segment and to the starting end of the adjacent coil situated
under the same pole and so on, till and the coils have been connected. This type of winding derives
its name from the fact it doubles or laps back with its succeeding coils.
Following points regarding simplex lap winding should be carefully noted :
1. The back and front pitches are odd and of opposite sign. But they cannot be equal. They
differ by 2 or some multiple thereof.
2. Both YB and YF should be nearly equal to a pole pitch.
Y + YF
3. The average pitch YA = B
. It equals pole pitch = .
4. Commutator pitch YC = ±1. (In general, YC = ± m)
5. Resultant pitch YR is even, being the arithmetical difference of two odd numbers, i.e., YR =
YB − YF.
6. The number of slots for a 2-layer winding is equal to the number of coils (i.e. half the
number of coil sides). The number of commutator segments is also the same.

Simplex lap winding


However, where heavy currents are necessary, duplex or triplex lap windings are used. The duplex lap
winding is obtained by placing two similar windings on the same armature and connecting the evennumbered commutator bars to one winding and the odd-numbered ones to the second winding. Similarly,
in triplex lap winding, there would be three windings, each connected to one third of the commutator bars.

D.C. Generators



The number of parallel paths in the armature = mP where m is the multiplicity of the winding
and P the number of poles.
Taking the first condition, we have YB = YF ± 2.**
(a) If YB > YF i.e. YB = YF + 2, then we get a progressive or right-handed winding i.e. a winding
which progresses in the clockwise direction as seen from the commutator end. In this case, obviously,
YC = + 1.
(b) If YB < YF i.e. YB = YF − 2, then we get a retrogressive or left-handed winding i.e. one which
advances in the anti-clockwise direction when seen from the commutator side. In this case, YC = −1.
(c) Hence, it is obvious that
Z −1 
YF =
 for progressive winding and YF = P + 1  for retrogressive winding
Z +1 
Z −1 
YB =
Obviously, Z/P must be even to make the winding possible.

26.27. Numbering of Coils and Commutator Segments
In the d.c. winding diagrams to follow, we will number the coils only (not individual turns). The
upper side of the coil will be shown by a firm continuous line whereas the lower side will be shown by
a broken line. The numbering of coil sides will be consecutive i.e. 1, 2, 3 ..... etc. and such that odd
numbers are assigned to the top conductors and even numbers to the lower sides for a two-layer
winding. The commutator segments will also be numbered consecutively, the number of the segments will be the same as that of the upper side connected to it.
Example 26.1. Draw a developed diagram of a simple 2-layer lap-winding for a 4-pole generator with 16 coils. Hence, point out the characteristics of a lap-winding.
(Elect. Engineering, Madras Univ. 1981)
Solution. The number of commutator segments = 16
Number of conductors or coil sides 16 × 2 = 32 ; pole pitch = 32/4 = 8
Now remembering that (i) YB and YF have to be odd and (ii) have to differ by 2, we get for a
progressive winding YB = 9 ; YF = −7 (retrogressive winding will result if YB = 7 and YF = −9).
Obviously, commutator pitch YC = −1.
[Otherwise, as shown in Art. 26.26, for progressive winding
Z − 1 = 32 − 1 = 7 and Y = Z − 1 = 32 + 1 = 9]
YF =
The simple winding table is given as under :
Back Connections
Front Connections
1 to (1 + 9) = 10
10 to (10 − 7) = 3
3 to (3 + 9) = 12
12 to (12 − 7) = 5
5 to (5 + 9) = 14
14 to (14 − 7) = 7
7 to (7 + 9) = 16
16 to (16 − 7) = 9
9 to (9 + 9) = 18
18 to (18 − 7) = 11
11 to (11 + 9) = 20
20 to (20 − 7) = 13
13 to (13 + 9) = 22
22 to (22 − 7) = 15
15 to (15 + 9) = 24
24 to (24 − 7) = 17
17 to (17 + 9) = 26
26 to (26 − 7) = 19
19 to (19 + 9) = 28
28 to (28 − 7) = 21
** In general, YB = YF ± 2m where m = 1 for simplex lap winding and m = 2 for duplex lap winding etc.


Electrical Technology

21 to (21 + 9) = 30
30 to (20 − 7) = 23
23 to (23 + 9) = 32
32 to (32 − 7) = 25
25 to (25 + 9) = 34 = (34 − 32) = 2 →
2 to (34 − 7) = 27
27 to (27 + 9) = 36 = (36 − 32) = 4 →
4 to (36 − 7) = 29
29 to (29 + 9) = 38 = (38 − 32) = 6 →
6 to (38 − 7) = 31
31 to (31 + 9) = 40 = (40 − 32) = 8 →
8 to (40 − 7) = 33 = (33 − 32) = 1
The winding ends here because we come back to the conductor from where we started.
We will now discuss the developed diagram which is one that is obtained by imagining the
armature surface to be removed and then laid out flat so that the slots and conductors can be viewed
without the necessity of turning round the armature in order to trace out the armature windings. Such
a developed diagram is shown in Fig. 26.31.

Fig. 26.31

Fig. 26.32

The procedure of developing the winding is this :
Front end of the upper side of coil No. 1 is connected to a commutator segment (whose number is also 1). The back end is joined
at the back to the 1 + 9 = 10th coil side in the lower half of 5th slot.
The front end of coil side 10 is joined to commutator segment 2 to
which is connected the front end of 10 − 7 = 3 i.e. 3rd coil side
lying in the upper half of second armature slot. In this way, by
travelling 9 coil sides to the right at the back and 7 to the left at the

D.C. Generators


front we complete the winding, thus including every coil side once till we reach the coil side 1 from
where we started. Incidentally, it should be noted that all upper coil sides have been given odd
numbers, whereas lower ones have been given even numbers as shown in the polar diagram (Fig.
26.32) of the winding of Fig. 26.31.
Brush positions can be located by finding the direction of currents flowing in the various conductors. If currents in the conductors under the influence of a N-pole are assumed to flow downwards (as
shown), then these will flow upwards in conductors under the influence of S-pole. By putting proper
arrows on the conductors (shown separately in the equivalent ring diagram), it is found that commutator bars No. 1 and 9 are the meeting points of e.m.fs. and hence currents are flowing out of these
conductors. The positive brushes should, therefore, be placed at these commutator bars. Similarly,
commutator bars No. 5 and 13 are the separating points of e.m.fs. hence negative brushes are placed
there. In all, there are four brushes, two positive and two negative. If brushes of the same polarity are
connected together, then all the armature conductors are divided into four parallel paths.










Fig. 26.33

Division of conductors into parallel paths is shown separately in the schematic diagram of Fig.
26.34. Obviously, if Ia is the total current supplied by the generator, then current carried by each
parallel path is Ia/4.
Summarizing these conclusions, we have
1. The total number of brushes is equal to the number of poles.
2. There are as many parallel paths in the armature as the number of poles. That is why such a
winding is sometimes known as ‘multiple circuit’ or ‘parallel’ winding. In general, number
of parallel paths in armature = mP where m is the multiplicity (plex) of the lap winding. For
example, a 6-pole duplex lap winding has (6 × 2) = 12 parallel paths in its armature.
3. The e.m.f. between the +ve and −ve brushes is equal to the e.m.f. generated in any one of the
parallel paths. If Z is the total number of armature conductors and P the number of poles,
then the number of armature conductors (connected in series) in any parallel path is Z/P.


Electrical Technology

= eav ×
The total or equivalent armature resistance can be found as follows :
l = length of each armature conductor; S = its cross-section
A = No. of parallel paths in armature = P − for simplex lap winding
R = resistance of the whole winding then R =


Generated e.m.f. Eg = (Average e.m.f./conductor) ×

Fig. 26.34


Resistance of each path =
S× A
There are P (or A) such paths in parallel, hence equivalent resistance
= 1 ×
If Ia is the total armature current, then current per parallel path (or carried by each conductor) is Ia/P.

26.28. Simple
x Wav
e Winding*
From Fig. 26.31, it is clear that in lap winding, a conductor (or coil side) under one pole is
connected at the back to a conductor which occupies an almost corresponding position under the next
pole of opposite polarity (as conductors 3 and 12). Conductor No. 12 is then connected to conductor
No. 5 under the original pole but which is a little removed from the initial conductor No. 3. If,
instead of returning to the same N-pole, the conductor No. 12 were taken forward to the next N-pole,
it would make no difference so far as the direction and magnitude of the e.m.f. induced in the circuit
are concerned.

Like lap winding, a wave winding may be duplex, triplex or may have any degree of multiplicity. A
simplex wave winding has two paths, a duplex wave winding four paths and a triplex one six paths etc.

D.C. Generators


As shown in Fig. 26.35, conductor AB
is connected to CD lying under S-pole
and then to EF under the next N-pole. In
this way, the winding progresses, passing successively under every N-pole and
S-pole till it returns to a conductor A′B′
lying under the original pole. Because
the winding progresses in one direction
round the armature in a series of ‘waves’,
it is known as wave winding.
If, after passing once round the arFig. 26.35
mature, the winding falls in a slot to the
left of its starting point (as A′B′ in Fig. 26.35) then the winding is said to be retrogressive. If, however, it falls one slot to the right, then it is progressive.
Assuming a 2-layer winding and supposing that conductor AB lies in the upper half of the slot,
then going once round the armature, the winding ends at A′B′ which must be at the upper half of the
slot at the left or right. Counting in terms of conductors, it means that AB and A′B′ differ by two
conductors (although they differ by one slot).
From the above, we can deduce the following relations. If P = No. of poles, then
YB = back pitch
nearly equal to pole pitch
YF = front pitch
= average pitch ; Z = total No. of conductors or coil sides
YA =
Z ±2
YA × P = Z ± 2
YA =
Since P is always even and Z = PYA ± 2, hence Z must always be even. Put in another way, it


Z ±2
must be an even integer.
The plus sign will give a progressive winding and the negative sign a retrogressive winding.

means that

Points to Note :


Both pitches YB and YF are odd and of the same sign.
Back and front pitches are nearly equal to the pole pitch and may be equal or differ by 2, in
which case, they are respectively one more or one less than the average pitch.
Resultant pitch YR = YF + YB.
Commutator pitch, YC = YA (in lap winding YC = ±1).
No. of Commutator bars ± 1
YC =
No. of pair of poles
The average pitch which must be an integer is given by
Z +1
No. of Commutator bars ± 1
Z ±2
YA =
= 2
No. of pair of poles

It is clear that for YA to be an integer, there is a restriction on the value of Z. With Z = 32, this
winding is impossible for a 4-pole machine (though lap winding is possible). Values of Z = 30 or 34
would be perfectly alright.




Electrical Technology
The number of coils i.e. NC can be found from the relation.
PYA ± 2
NC =
This relation has been found by rearranging the relation given in (5) above.
It is obvious from (5) that for a wave winding, the number of armature conductors with 2
either added or subtracted must be a multiple of the number of poles of the generator. This
restriction eliminates many even numbers which are unsuitable for this winding.
The number of armature parallel paths = 2m where m is the multiplicity of the winding.

Example 26.2. Draw a developed diagram of a simplex 2-layer wave-winding for a 4-pole d.c.
generator with 30 armature conductors. Hence, point out the characteristics of a simple wave
(Elect. Engg-I, Nagpur Univ. 1991)
30 ± 2
Solution. Here, YA =
= 8* or 7. Taking YA = 7, we have YB = YF = 7

Fig. 26.36

As shown in Fig. 26.36 and 26.37, conductor No. 5 is taken to conductor No. 5 + 7 = 12 at the
back and is joined to commutator segment 5 at the front. Next, the conductor No. 12 is joined to
commutator segment 5 + 7 = 12 ( ∵ YC = 7) to which is joined conductor No. 12 + 7 = 19. Continuing
this way, we come back to conductor No. 5 from where we started. Hence, the winding closes upon

If we take 8, then the pitches would be : YB = 9 and YF = 7 or YB = 7 and YF = 9. Incidentally, if YA = YC is
taken as 7, armature will rotate in one direction and if YC = 8, it will rotate in the opposite direction.

D.C. Generators


The simple winding table is as under :
Back Connections
Front Connections
1 to (1 + 7) = 8
8 to (8 + 7) = 15
15 to (15 + 7) = 22
22 to (22 + 7) = 29
29 to (29 + 7) = 36 = (36 − 30) = 6 →
6 to (6 + 7) = 13
13 to (13 + 7) = 20
20 to (20 + 7) = 27
27 to (27 + 7) = 34 = (34 − 30) = 4 →
4 to (4 + 7) = 11
11 to (11 + 7) = 18
18 to (18 + 7) = 25
25 to (25 + 7) = 32 = (32 − 30) = 2 →
2 to (2 + 7) = 9
9 to (9 + 7) = 16
16 to (16 + 7) = 23
23 to (23 + 7) = 30
30 to (30 + 7) = 37 = (37 − 30) = 7
7 to (7 + 7) = 14
14 to (14 + 7) = 21
21 to (21 + 7) = 28
28 to (28 + 7) = 35 = (35 − 30) = 5
5 to (5 + 7) = 12
12 to (12 + 7) = 19
19 to (19 + 7) = 26
26 to (26 + 7) = 33 = (33 − 30) = 3
3 to (3 + 7) = 10
10 to (10 + 7) = 17
17 to (17 + 7) = 24
24 to (24 + 7) = 31 = (31 − 30) = 1
Since we come back to the conductor No. 1 from where we started, the winding gets closed at this

Brush Position
Location of brush position in wave-winding is slightly
difficult. In Fig. 26.36 conductors are supposed to be
moving from left to right over the poles. By applying
Fleming’s Right-hand rule, the directions of the induced
e.m.fs in various armature conductors can be found. The
directions shown in the figure have been found in this
manner. In the lower part of Fig. 26.36 is shown the
equivalent ring or spiral diagram which is very helpful in
understanding the formation of various parallel paths in the
armature. It is seen that the winding is electrically divided
into two portions. One portion consists of conductors lying
between points N and L and the other of conductors lying
between N and M. In the first portion, the general trend of
Fig. 26.37
the induced e.m.fs. is from left to right whereas in the second
portion it is from right to left. Hence, in general, there are only two parallel paths through the winding,
so that two brushes are required, one positive and one negative.
From the equivalent ring diagram, it is seen that point N is the separating point of the e.m.fs.
induced in the two portions of the winding. Hence, this fixes the position of the negative brush. But
as it is at the back and not at the commutator end of the armature, the negative brush has two
alternative positions i.e. either at point P or Q. These points on the equivalent diagram correspond
to commutator segments No. 3 and 11.
Now, we will find the position of the positive brush. It is found that there are two meeting
points of the induced e.m.fs. i.e. points L and M but both these points are at the back or non-commutator end of the armature. These two points are separated by one loop only, namely, the loop composed of conductors 2 and 9, hence the middle point R of this loop fixes the position of the positive
brush, which should be placed in touch with commutator segment No. 7. We find that for one
position of the +ve brush, there are two alternative positions for the −ve brush.
Taking the +ve brush at point R and negative brush at point P, the winding is seen to be divided
into the following two paths.


Electrical Technology











Fig. 26.38

In path 1 (Fig. 26.36) it is found that e.m.f. in conductor 9 is in opposition to the general trend
of e.m.fs. in the other conductors comprising this path. Similarly, in path 2, the e.m.f. in conductor
2 is in position to the direction of e.m.fs. in the path as a whole. However, this will make no
difference because these conductors lie almost in the interpolar gap and, therefore e.m.fs. in these
conductors are negligible.

Fig. 26.39

Again, take the case of conductors 2 and 9 situated between points L and M. Since the armature
conductors are in continuous motion over the pole faces, their positions as shown in the figure are
only instantaneous. Keeping in this mind, it is obvious that conductor 2 is about to move from the
influence of S-pole to that of the next N-pole. Hence, the e.m.f. in it is at the point of reversing.
However, conductor 9 has already passed the position of reversal, hence its e.m.f. will not reverse,

D.C. Generators


rather it will increase in magnitude gradually. It means that in a very short interval, point M will

Fig. 26.40

become the meeting point of the e.m.fs. But as it lies at the back of the armature, there are two
alternative positions for the +ve brush i.e. either point R which has already been considered or point
S which corresponds to commutator segment 14. This is the second alternative position of the positive brush. Arguing in the same way, it can be shown that after another short interval of time, the
alternative position of the positive brush will shift from segment 14 to segment 15. Therefore, if one
positive brush is in the contact with segment 7, then the second positive brush if used, should be in
touch with both segments 14 and 15.
It may be noted that if brushes are placed in both alternative positions for both positive and
negative (i.e. if in all, 4 brushes are used, two +ve and two −ve), then the effect is merely to shortcircuit the loop lying between brushes of the same polarity. This is shown in Fig. 26.40 where it will
also be noted that irrespective of whether only two or four brushes are used, the number of parallel
paths through the armature winding is still two.
Summarizing the above facts, we get
1. Only two brushes are necessary, though their number may be equal to the number of poles.
2. The number of parallel paths through the armature winding is two irrespective of the number
of generator poles. That is why this winding is sometimes called ‘two-circuit’ or ‘series’
3. The generator e.m.f. is equal to the e.m.f. induced in any one of the two parallel paths. If eav
is the e.m.f. induced/conductor, then generator e.m.f. is Eg = eav × Z/2.
4. The equivalent armature resistance is nearly one-fourth of the total resistance of the armature winding.
5. If Ia is the total armature current, then current carried by each path or conductor is obviously
Ia/2 whatever the number of poles.

26.29. Dummy or Idle Coils
These are used with wave-winding and are resorted to when the
requirements of the winding are not met by the standard armature
punchings available in armature-winding shops. These dummy coils
do not influence the electrical characteristics of the winding because
they are not connected to the commutator. They are exactly similar
to the other coils except that their ends are cut short and taped. They
are there simply to provide mechanical balance for the armature because an armature having some slots without windings would be out
of balance mechanically. For example, suppose number of armature
slots is 15, each containing 4 sides and the number of poles is 4. For
a simplex wave-windings,

Dummy coils


Electrical Technology

Z ± 2 60 ± 2
which does not come out to be an integer (Art. 26.28) as required by this winding. However, if we
make one coil dummy so that we have 58 active conductors, then
58 ± 2
YA =
= 14 or 15
This makes the winding possible.

YA =

26.30. Uses of La
p and Wav
e Windings
The advantage of the wave winding is that, for a given number of poles and armature conductors,
it gives more e.m.f. than the lap winding. Conversely, for the same e.m.f., lap winding would require
large number of conductors which will result in higher winding cost and less efficient utilization of
space in the armature slots. Hence, wave winding is suitable for small generators especially those
meant for 500-600 V circuits.
Another advantage is that in wave winding, equalizing connections are not necessary whereas in
a lap winding they definitely are. It is so because each of the two paths contains conductors lying
under all the poles whereas in lap-wound a