Here are the steps to solve Example 2.1:(a) Terminal voltage = Generated EMF - IaRa Given: Load = 20 kW, Vt = 200 VUsing P = VI, Ia = 20,000/200 = 100 AEg = Vt + IaRa = 200 + 100 x 0.04 = 204 V(b) Generated EMF = Terminal voltage + IaRf = 204 + 100 x 0.05 = 204.5 VSo the answers are: (a) Terminal voltage = 200 V(b) Generated EMF = 204.5 V

Unit 1
ELECTROMAGNETISM
( Keelektromagnetan )

General Objective

To understand the basic principles of
electromagnetism.

1.0 INTRODUCTION
Arah uratdaya magnet

The pattern of magnetic field of bar a magnet.

Tarikan

Tolakan

Attraction and Repulsion
( tarikan dan tolakan )

1.1 CURRENT-CARRYING CONDUCTOR AND
ELECTROMAGNETISM
( keelektromagnetan ke atas pengalir
yang membawa arus )

• A flow of current through a wire
produces a magnetic field in a circular
path around the wire.The direction of
magnetic line of flux around the wire
is best remembered by the screw
rule or the grip rule.

Arus masuk

Arus keluar

The field pattern of current flowing in the wire

tarikan tolakan

( a) flow in the same direction ( b) opposite direction

If two closed current-carrying conductors
flow in the same direction, magnetic flux
around that conductor will combine to
create attraction between them. If closed
current-carrying conductors flow in
opposite direction, these two conductors
will repulse each other

1.2. MAGNETIC FIELD STRENGTH, H
(MAGNETISING FORCE)

• Magnetic field strength is defined as
magnetomotive force, Fm

Fm NI
H= = ampere turn / metre
l l
N = bilangan lilitan pengalir
I = arus yang mengalir
l = panjang bahan magnet

1.3. MAGNETIC QUANTITY AND THEIR
RELEVANT FORMULAE

• 1.3.1 Magnetic Flux and Flux density
-Magnetic flux,Φ is the amount of magnetic
filed produced by a magnetic source.

- Flux density,B is the amount of flux
passing through a defined area

unit for flux is the weber, wb

Φ
- Flux density, B = Tesla
A

• Example 1.1
• A current of 500mA is passed through a
600 turn coil wound of a iron of mean
diameter 10cm. Calculate the magnetic
field strength.

Example 1.2
• An iron ring has a cross-sectional area of
400 mm2 and a mean diameter of 25 cm.
it is wound with 500 turns. If the value of
relative permeability is 250, find the total
flux set up in the ring. The coil resistance
is 474 Ω and the supply voltage is 20 V.

● 1.3.2 Permeability ( ketelapan )
( Kebolehan sesuatu bahan magnet untuk
menghasilkan uratdaya magnet )
the ratio of magnetic flux density to
magnetic field strength is constant

B
= a constant
H

For air, free space and any other
non-magnetic medium, the ratio
B
= μ 0 = 4π x 10-7 H/m
H

For all media other than free pace,
B
= μ0μr
H

Cast iron μr = 100 – 250
Mild steel μr = 200 – 800
Cast steel μr = 300 – 900
μr for a vacuum is 1

μ - absolute permeability
μr - relative permeability
μo - air permeability

where μ = μoμr

• 1.3.3 Reluctance ( Engganan )
Reluctance, S is the magnetic resistance
of a magnetic circuit
Fm Hl H l 1 l
S = = = =
Φ BA B A μ ομ r A
unit for reluctance is H-1

Perbadingan di antara Litar Elekrik
Dengan Litar magnet
Litar Elektrik Litar Magnet
1. Arus 1. Uratdaya ( Fluks )
2. Dge 2. Dgm
3. Rintangan 3. Engganan

• Example 1.3
A magnetic pole face has rectangular section
having dimensions 200mm by 100mm. If the
total flux emerging from the the pole is 150μWb,
calculate the flux density.
Example 1.4
A flux density of 1.2 T is produced in a piece of
cast steel by a magnetizing force of 1250 A/m.
Find the relative permeability of the steel under
these conditions.

• Example 1.5
Determine the reluctance of a piece of metal of
length 150mm, and cross sectional is 100mm2
when the relative permeability is 4000. Find also
the absolute permeability of the metal.
Exersice 1
The maximum working flux density of a lifting
electromagnet is 1.8 T and the effective area
of a pole face is circular in cross-section.
If the total magnetic flux produced is 353 mWb,
determine the radius of the pole.

1.4 ELECTROMAGNETIC INDUCTION

When a conductor is moved across a magnetic
field so as to cut through the flux,
an electromagnetic force (e.m.f.) is produced
in the conductor. This effect is known as
electromagnetic induction. The effect of
electromagnetic induction will cause
induced current.

Two laws of electromagnetic induction
i. Faraday’s law

Conductor cuts flux

This induced electromagnetic field is given by
θ°
Where ,

B = flux density, T
= length of the conductor in
the magnetic field, m
v = conductor velocity, m/s

ii. Lenz’z Law

Bar magnet move in and move out
from a solenoid

Example 1.6
A conductor 300mm long moves at a uniform
speed of 4m/s at right-angles to a uniform
magnetic field of flux density 1.25T. Determine
the current flowing in the conductor when
i. its ends are open-circuited
ii. its ends are connected to a load of 20 Ω
resistance.

Exersice 2

A conductor of length 0.5 m situated in and at
right angles to a uniform magnetic field of flux
density 1 wb/m2 moves with a velocity of 40
m/s. Calculate the e.m.f induced in the
conductor. What will be the e.m.f induced if the
conductor moves at an angle 60º to the field.

Solution to Example 1.3
Magnetic flux, Φ = 150 μWb = 150 x 10-6 Wb
Cross sectional area, A = 200mm x 100mm
= 20 000 x 10-6m2

Φ 150 × 10 − 6
Flux density, B = =
A 20000 × 10 − 6
= 7.5 mT


B = μ0μr H
B 1.2
μr = =
μ 0 H (4π × 10−2 )(1250)
= 764

l
Reluctance, S =
μ 0 μ r A

150 × 10 − 3
=
( 4π × 10 − 7 )( 4000 )(100 × 10 − 6 )

= H-1
Absolute permeability, μ = μ0μr
( 4π × 10 − 7 )( 4000 )
= 5.027 x 10-3 H/m

Solution To Example 1.6

i. If the ends of the conductor are open
circuited, no current will flow even though
1.5 V has been induced.

ii. From Ohm’s law
E 1 .5
I = = 75 mA
R 20

• OBJECTIVES

To apply the basic principle of DC
generator, construction
principle and types of DC generator.

2.0 Introduction

• A generator is a machine that converts
mechanical energy into electrical energy
by using the principle of magnetic
induction.

Penjana Arus Terus

Rajah Blok

Sumber Tenaga Tenaga
Tenaga Mekanikal Elektrikal

Rajah Blok Penjana

Pengalir Medan Magnet

The Princple of the DC Generator
( base on the Faraday’s Law )

The Princple of the DC Generator

The Princple of the AC Generator

• Whenever a conductor is moved within a
magnetic field in such a way that the
conductor cuts across magnetic lines of
flux, voltage is generated in the conductor.
- Faraday’s Law

• The POLARITY of the voltage depends on
the direction of the magnetic lines of flux
and the direction of movement of the
conductor. To determine the direction of
current in a given situation, the RIHGT-HAND
RULE FOR GENERATORS is used.

• The AMOUNT of voltage generated
depends on :
i. the strength of the magnetic field,
ii. the angle at which the conductor cuts
the magnetic field,
iii. the speed at which the conductor is
moved, and
iv. the length of the conductor within the
magnetic field.

The Princple of the Generating AC Voltage

The Princple of the Generating DC Voltage

2.1 THE PATHS OF THE DC GENERATOR

1. Armature ( angker )

Cross sectional of the DC Generator

2.2. Types of DC Generator

Gelung Medan

Stator ( penetap )

(Gelung angker)

Armature ( Angker )

Bahagian Angker
( Gelung Angker )

Bahagian Stator
(Gelung Medan)

Types of DC Generator

• Separately-excited generators
• Self-excited generators
i. Shunt-wound generator
ii. Series-wound generator
iii. Compound-wound generator
a. Short compound generator
b. Long compound generator

1. Penjana Ujaan Berasingan
Angker
Medan

DC Power
Supply

Penjana Ujaan Berasingan

2. Penjana Ujaan Diri

1. Series-wound generator

Example 2.1

A shunt generator supplies a 20 kW
load at 200 V. If the field winding
resistance, Rf = 50Ω and the
armature resistance Ra = 40 mΩ,
determine
(a) the terminal voltage
(b) the e.m.f. generated in the armature

2.3. E.m.f generated ( Voltan janaan, dge )
2p Φ Zn
generated e.m.f, Eg =
c
Where ; Z = number of armature conductors,
Φ = useful flux per pole in Webers
Ρ = number of pairs of poles
n = armature speed in rev/s

( c=2 for a wave winding and
c= 2p for a lap winding )

Example 2.2.

An 8-pole generator, wave winding
connected armature has 600 conductor
and is driven 625 rev/min. If the flux per
pole is 20mWb, determine the generated
e.m.f.

Solution
Z = 600, c = 2 for a wave winding
P = 4 pairs, n = 625/60 rev/min, Φ = 30 × 10-3 Wb

2p Φ Zn
Dge, Eg =
c
625
2(4)(20 × 10 )(
-3
)
60
2

Example 2.3.
A 4-pole generator has a lap winding
armature, with 50 slots and 16 conductors
per slot. The useful flux per pole is
30mWb. Determine the speed at which the
machine must be driven to generate an
e.m.f. of 240 volts.

E = 240 V, Z = 50 x 16 = 800
c = 2p (for a lap winding), Φ = 30 × 10-3 Wb

Ans : ( 10 rev/s )

2.4 Power Losses and Efficiency

For any type of machine, output power is
different from input power. The difference is
caused by power losses that had happened
whenever one type of energy is converted
or delivered to the other type.

The principal losses of machine are:
• Copper loss ( I2R )
• Iron losses, due to hysteresis and eddy
current
• Friction and windage losses
• Brush and contact losses ( vB )

2.4.1. Efficiency of DC generator
The efficiency of an electrical
machine is the ratio of the output
power and input. The greek letter ‘η’
(eta) is used to signify efficiency, the
efficiency has no units.

output power
efficiency, η = ( ) × 100 %
input power

Vo
η =( ) × 100%
Vo + VD

VL I L
η= ( ) × 100 %
VL I L + I a R a + I f V f + C
2

Example 2.4

A shunt generator supplies 96 A at a terminal voltage
of 200 volts. The armature and shunt field resistances
are 0.1Ω and 50Ω respectively. The iron and frictional
losses are 2500 W.
Find :
(i) e.m.f generated.
(ii) copper losses
(iii) efficiency

Example 2.5
A 75 kW shunt generator is operated at 230
V. The stray losses are 1810 W and shunt
field circuit draws 5.35 A. The armature
circuit has a resistance of 0.035 Ω and
brush drop is 2.2 V. Calculate :
1. total losses
2. input of prime mover
3. efficiency at rated load.

DC motors are very
useful in many
applications of our
everyday life, for
example controlling, such
as crane, tape driver, lift
system and others.

OBJECTIVES

• General Objective
To apply the basic principles of DC motor
operation, types of DC motor and their
application

• Specific Objectives

• Explain the principle operation of DC
motor
• List the types of DC motor
• State the left-handed rule for motors
• List the advantages and disadvantages of
the different types of DC motors.
• State typical applications of DC motors

3.1 INTRODUCTION

DC Motor is a machine that converts
electrical energy into mechanical
energy.

Electrical Mechanical
Load
Energy Energy

Blok Diagram

3. 2 THE PARTS OF DC MOTOR

Stator

armature
( rotor )

Commutator Carbon Brush

shaft

Fan Stator Rotor

field coil

Cross sectional of the Stator

Armature coil

Commutator

shaft

Armature

TEST YOUR UNDERSTANDING

1 . What is a DC motor?

2. State the uses of DC motors.

3. What are the parts of DC motors?

3. 3 PRINCIPLE OF OPERATION

Fleming′s Left Hand Rule

When a wire carrying current sits into a magnetic field, a
force is created on the wire causing it to move
perpendicular ( tegak lurus ) to the magnetic field. The
greater the current in the wire, or the greater the magnetic
field, the faster the wire moves because of the greater
force created. If the wire sits parallel with the magnetic
field, there will be no force on the wire.

conductor

( field )

Left-hand rule for DC motors

field diretion

power supply

3.4 TYPES OF DC MOTOR

• The series DC motor
• The shunt DC motor
• The compound DC motor

Question

1. What are three major types of DC
motor?

2. Draw the schematic diagram of series,
shunt and compound of DC motors.

3.5 BACK ELECTROMOTIVE FORCE

2Φ N r p
Eb =
60

• Φ = useful flux per pole in webers
• Nr = the speed in revolution per minute
• P = the number of pairs of poles

• Torque ( Daya kilas )

60 E b I a
Ta =
2Π n

Example

A 350 V shunt motor runs at its normal speed of 12rev/s
when the armature current is 90 A. The resistance of
the armature is 0.3 Ω.
Find the speed when the armature current is 45 A and
a resistance of 0.4 Ω is connected in series with the
armature, the shunt field remaining constant.

3.6 FACTORS THAT INFLUENCE SPEED
CONTROL OF DC MOTOR

The speed of a dc motor is changed
by changing the current in the field or
by changing the current in the
armature.

3.7 REVERSE DIRECTION METHOD

• The direction of rotation of a dc motor
depends on the direction of the magnetic
field and the direction of current flow in the
armature.

3.8 EFFICIENCY AND POWER LOSSES

Kecekapan
output power
efficiency, η = × 100%
input power

VI − I R a − I f V − C
2

η =( ) × 100 %
a

VI

Example

A 320 V shunt motor takes a total current of 80 A
and runs at 1000 rev/min. If the iron, friction and
windage losses amount to 1.5 kW, the shunt
field resistance is 40 Ω and the armature
resistance is 0.2 Ω, determine the overall
efficiency of the motor.

OBJECTIVES

To analyze the basic principles of operation
of an AC generator and the differences
between DC generator and AC generator
by using commutator and slip ring.

4.1 INTRODUCTION

An electric generator is a device used to
convert mechanical energy into electrical
energy.
The generator is based on the principle of
"electromagnetic induction" discovered by
Michael Faraday Law’s.

The simple of Electric Generator

(AC waveform)

(DC waveform)

The amount of voltage generated
depends on the following:

• The strength of the magnetic field.
• The angle at which the conductor cuts the
magnetic field.
• The speed at which the conductor is moved
• The length of the conductor within the
magnetic field.

4.2 THE DIFFERENCES BETWEEN AC
GENERATOR AND DC GENERATOR

The difference between AC and DC generator is
that the DC generator results when you replace
the slip rings of an elementary generator with
commutator, changing the
output from AC to pulsating DC.
AC generator is also called Alternator.

4.3 E.M.F. Equation of an Alternator
E rms / phase = 2.22 KdKp f Φ Z volts
where, Z = No. of conductors or coil
Φ = Flux per pole in webers
P = Number of rotor poles
N = Rotor speed in r.p.m
Kd = Distribution factor
Kp = Pitch factor

⎛ 120 f ⎞
⎜N =
⎜ ⎟
⎟
⎝ p ⎠

Star-connected Delta-connected

Example 4.1
A 3-phase, 50 Hz star-connected
alternator has 180 conductors per phase
and flux per pole is 0.0543 wb.
Find:-
a) e.m.f. generated per phase
b) e.m.f. between line terminals.
Assume the winding to be full pitched
and distribution factor to be 0.96.

Exersice 1
Find the number of armature conductors
in series per phase required for the
armature of a 3-phase, 50Hz, 10-pole
alternator. The winding is star-connected
to give a line voltage of 11000. The flux
per pole is 0.16 wb. Assume Kp = 1 and
Kd = 0.96.

4.4 AC motor

Stator for an AC motor.

Differences between AC Motor and DC motor

In general, AC motors cost less than DC
motors. Some types of AC motors do not
use brush carbon and commutators.

What is the advantage of AC motor over DC motor ?

4.5 Types of AC Motor
1. Series AC Motor

Types of starting induction motor

1. Capasitor-Start

• Slip
The actual mechanical speed (nr) of the
rotor is often expressed as a fraction of
the synchronous speed (ns) as related by
slip (s), defined as

n s − n r
S=
n s

120 f
where ns =
P
n s − nr
Percent slip, %s = × 100%
ns
and fr = sf

fr = frequency rotor

Example 4.2

Determine the synchronous speed of the
six pole motor operating from a 220V,
50Hz source.

Example 4.3
The stator of a 3-phase, 4 pole induction motor is
connected to a 50 Hz supply. The rotor runs at 1455
rev/min at full load. Determine:
a) the synchronous speed
b) the slip

Example 4.4

The frequency of the supply to the stator of an 8-pole
induction motor is 50 Hz and the rotor frequency is 3 Hz.
Determine
i. the slip
ii. the rotor speed

Example 4.5

A 4-pole, 3 phase, 50 Hz induction motor runs at 1440
rev/min at full load. Calculate
a) the synchronous speed
b) the percent of slip
c) the frequency of the rotor.

OBJECTIVES

To understand the basic principles of
a transformer, construction
principle, transformer ratio, current
and core, type of transformer and
uses.

At the end of the unit you will be able to :
• explain the operating principles of a
transformer
• describe transformer construction
• explain transformer ratio for voltage,
current and winding coil.
• calculating of the efficiency.
• describe auto transformer

Primary Secondary
winding winding

Core
Coil /
Winding

transformer construction

5.1 Introduction

The basic transformer is an electrical
device that transfers alternating-current
energy from one circuit to another circuit
by magnetic flux of the primary and
secondary windings of the transformer.

A transformer circuit

Primary Secondary
winding winding

Primary Secondary

Simbol of the transformer

The uses of the transformer
A transformer is a device which used to
change the values of alternating voltages
or currents to step-up or step-down.

• Ep = 4.44 Np f Φm volts
• Es = 4.44 Ns f Φm volts

• If K < 1 i.e. Ns < Np : step-down
• If K > 1 i.e. Ns > Np : step-up
• If K = 1 i.e. Ns = Np : coupling

equations of ideal transformer
V p Np Is
= =
Vs Νs I p

Transformer rating :
The rating of the input power of the transformer.
example : 25 kVA ( kV x Arus )

Example 5.1
A 2000/200V, 20kVA transformer has 66 turns in the
secondary. Calculate
(i) primary turns
(ii) primary and secondary

Example 5.2
A 250 kVA, 1100 V / 400 V, 50 Hz single-phase
transformer has 80 turns on a secondary. Calculate :
a) the values of the primary and secondary
currents.
b) the number of primary turns.
c) the maximum values of flux.

Example 5.3

An ideal 25 kVA transformer has 500 turns on the
primary winding and 40 turns on the secondary winding.
The primary is connected to 3000 V, 50 Hz supply.
Calculate
(i) primary and secondary currents
(ii) secondary e.m.f. and
(iii) the maximum core flux

5.3 Auto-transformer

An auto-transformer is a transformer
having a part of its winding common to the
primary and secondary circuits

Advantages and disadvantages of auto
transformers

- a saving in a cost
- less volume, hence less weight.
- higher efficiency
- a continuously variable output
- a smaller percentage voltage regulation.

5.4 Efficiency and losses of a transformer

The losses which occur in the transformer :-
i. copper losses, I2R
ii. Core losses , Pc
- hysterises
- eddy current

output power
Efficiency =
input power

output power
=
output power + losses
I sVs × p.f.
=
I sVs × p.f. + Pc + I p R p + I s Rs
2 2

Example 5.4

The primary and secondary windings of a 500
kVA transformer have resistances of 0.42 Ω and
0.0019 Ω respectively. The primary and
secondary voltages are 11 000 V and 400 V
respectively and the core loss is 2.9 kW,
assuming the power factor of the load to be 0.8.
Calculate the efficiency on :
i. full load
ii. half load

‘Manusia tidak pernah merancang
untuk gagal, tetapi gagal untuk
merancang’

“Selamat maju jaya”

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