5. 1.1 CURRENT-CARRYING CONDUCTOR AND
( 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.
7. tarikan tolakan
( a) flow in the same direction ( b) opposite direction
8. 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
9. 1.2. MAGNETIC FIELD STRENGTH, H
• Magnetic field strength is defined as
magnetomotive force, Fm
H= = ampere turn / metre
N = bilangan lilitan pengalir
I = arus yang mengalir
l = panjang bahan magnet
10. 1.3. MAGNETIC QUANTITY AND THEIR
• 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
12. • 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
13. 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.
15. ● 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
= a constant
16. For air, free space and any other
non-magnetic medium, the ratio
= μ 0 = 4π x 10-7 H/m
For all media other than free pace,
17. 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
18. • 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
19. 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
20. • 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.
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
21. • 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.
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.
22. 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
23. Two laws of electromagnetic induction
i. Faraday’s law
Conductor cuts flux
27. 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
28. 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.
29. 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
40. • 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
41. • 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.
44. • 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
iv. the length of the conductor within the
61. 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
62. 1. Penjana Ujaan Berasingan
Penjana Ujaan Berasingan
66. 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,
(a) the terminal voltage
(b) the e.m.f. generated in the armature
67. 2.3. E.m.f generated ( Voltan janaan, dge )
2p Φ Zn
generated e.m.f, Eg =
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 )
69. 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
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 =
2(4)(20 × 10 )(
71. 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.
72. E = 240 V, Z = 50 x 16 = 800
c = 2p (for a lap winding), Φ = 30 × 10-3 Wb
Ans : ( 10 rev/s )
73. 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.
74. The principal losses of machine are:
• Copper loss ( I2R )
• Iron losses, due to hysteresis and eddy
• Friction and windage losses
• Brush and contact losses ( vB )
75. 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.
76. output power
efficiency, η = ( ) × 100 %
η =( ) × 100%
Vo + VD
VL I L
η= ( ) × 100 %
VL I L + I a R a + I f V f + C
77. 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.
(i) e.m.f generated.
(ii) copper losses
78. 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.
82. • Specific Objectives
• Explain the principle operation of DC
• 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
95. 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.
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.
107. 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
109. 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
110. 3.8 EFFICIENCY AND POWER LOSSES
efficiency, η = × 100%
VI − I R a − I f V − C
η =( ) × 100 %
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.
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.
114. 4.1 INTRODUCTION
An electric generator is a device used to
convert mechanical energy into electrical
The generator is based on the principle of
"electromagnetic induction" discovered by
Michael Faraday Law’s.
121. The amount of voltage generated
depends on the following:
• The strength of the magnetic field.
• The angle at which the conductor cuts the
• The speed at which the conductor is moved
• The length of the conductor within the
122. 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.
123. 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 ⎞
⎝ p ⎠
125. Example 4.1
A 3-phase, 50 Hz star-connected
alternator has 180 conductors per phase
and flux per pole is 0.0543 wb.
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.
126. 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.
130. 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 ?
To understand the basic principles of
a transformer, construction
principle, transformer ratio, current
and core, type of transformer and
144. At the end of the unit you will be able to :
• explain the operating principles of a
• describe transformer construction
• explain transformer ratio for voltage,
current and winding coil.
• calculating of the efficiency.
• describe auto transformer
147. 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.
159. • 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
160. 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 )
161. Example 5.1
A 2000/200V, 20kVA transformer has 66 turns in the
(i) primary turns
(ii) primary and secondary
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
b) the number of primary turns.
c) the maximum values of flux.
162. 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.
(i) primary and secondary currents
(ii) secondary e.m.f. and
(iii) the maximum core flux
166. Advantages and disadvantages of auto
- a saving in a cost
- less volume, hence less weight.
- higher efficiency
- a continuously variable output
- a smaller percentage voltage regulation.
167. 5.4 Efficiency and losses of a transformer
The losses which occur in the transformer :-
i. copper losses, I2R
ii. Core losses , Pc
- eddy current
168. output power
output power + losses
I sVs × p.f.
I sVs × p.f. + Pc + I p R p + I s Rs
169. 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