Summary Notes of Electrical Technology

Inductor and Inductance

When needed, it may release 01 Unit = Henry (H),
the stored energy in the form of Symbols =
a magnetic field.
06

Within a conductor, Factors that affect
magnetic flux lines are
formed every time the inductance of a coil.
conductor is put under • Number of turns in
any kind of load. The
induced e.m.f. causes 02 05 the coil
this magnetic flux to • Coil Area
resist any change in • Coil Length
current • Core Material

The core of an inductor is made

up of a ferromagnetic substance

We call this resistance to current 03 04 that is wrapped in copper wire.
flow inductance, and the device The capacity of a substance to
that generates it is known as an support the creation of a
inductor.
magnetic field inside itself is

measured by its permeability.

86

Electromotive Force

= ( − , ) Webers (Wb) are units for measuring
magnetic flux.

A coil's electromotive Symbols : phi @ ɸ
force is the sum of the
number of turns and the Flux density, B and
current flowing through measured in tesla (T).
it.

As the number of turns and current through 1 tesla = 1T = 1 Wb/ 2
the coil increase, the flux density of the coil
decreases.

87

Types of Inductors

Air core 01
- Air

Non – magnetic core 02 Air core

- Ceramics

Solid
core

Soft Ferrites

01 - Magganese - Zinc
02

Hard Ferrites
- Cobalt

88

There is no core in this Types of Inductors
arrangement. There is The wire is coiled around a
extremely low distortion due Air magnetic material in these
to the absence of a metal Core inductors. Due to their
core, but the coil must be circular design, inductors of
quite long in order to carry
high inductance, hence the Toroidal this kind have minimal
inductor is very huge.
losses in magnetic flux
Ferromagnetic materials, because of the symmetry
attained throughout the
such as ferrite or iron, are whole shape of the inductor.
These inductors are often
used in the construction of used in AC circuits.

these inductors. Due to their

high magnetic permeability,

core materials such as these

are useful for increasing When it comes to inductors,

inductance. Materials' ceramic cores are unique

permeability is a measure of because they are made from

how easily magnetic fields a dielectric ceramic core;
this means that they can't
may be generated inside the Iron Ceramic store much energy but have
Core
material. An Iron-core Core low distortion.

Inductor is seen in the

following illustration.

89

The core of these inductors Types of Inductors The core material of these

consists of magnetic materials inductors is laminated steel

with a little amount of air space sheets, such as stacks. The energy

between them. However, the core Powered losses in an inductor often grow
benefits from this sort of Iron
architecture by storing a higher Laminatedwhen the loop area is expanded
degree of energy than with other Core Core for the current to flow. In contrast,

the laminated core Inductors'

forms. A Powdered Iron core stacks of thin steel sheets reduce

Inductor is seen in the following loop action by preventing eddy

illustration. Low eddy current currents. The primary benefit of

losses and hysteresis losses are these inductors is their structure,

provided by these inductors. which minimises energy loss. This

These are the cheapest and most kind of laminated core inductor is

stable inductance products on most often employed in

the market. transformers and other electrical

devices.

Plastic or ceramic insulation is Multi-layered coils are twisted
used to mould these inductors.
Windings may be either around a core to create this
cylindrical or bar-shaped, having
terminations at either end, and Molded Multi-layer inductor type. Multi-layer
are often used in circuit boards. ceramic inductors have a large inductance
because of the several layers and

the insulation that separate them.

90

Construction of Inductors

ɸ
=

µ
=

µ = Permeability (Wb/A · m) = µ ∗ µ
N = number of turns (t)
A = , 2
l = length, m
L = henry (H)
µ = 4Π x 10−7

91

Inductors in Series and Parallel Circuit

Series Parallel

Inductance Total Inductance Total

= + +
= + +

92

Example 1 : Series and Parallel Circuit Parallel

Calculate the equivalent inductance of the following circuit :
Series

= + + + = = µ
= + + + + + = =
=

= + = + + = + +
= +
=
= .

= .

93

Example 2 : Series-Parallel Circuit

Calculate the equivalent inductance of the following circuit

= + = +
= + = +
= =


= + + = + +

=
= .

94

Example 3 : Series-Parallel Circuit

Calculate the equivalent inductance of the following circuit

= +
= +
=

= = +
+ = . +
= .


= +


= = .

95

Example 3 : (Continue) Series-Parallel Circuit

Calculate the equivalent inductance of the following circuit

=
+

.
= . +

. =
= . = . +

= + .
= . + = . +
= .
.
= . = .

96

Example 4 : Inductance

1. Calculate inductance if a metallic core with = 2000, cross-sectional area, A=25 2 and length, l = 50
mm is inserted inside a coil with N = 2000 turns

ANS :

L = 2 0A



L = (20002)(2000 0)(25 10−6) = 5.027 H
(50 10−3)

2. An e.m.f. of 3 kV is induced in a coil when a current of 8 A collapses uniformly to zero in 16 ms. Determine
the inductance of the coil

ANS :

e.m.f. = - L 3k = - L x 500
L =

3k = - L 8
16

97

Electromagnetic Induction

An electromagnetic induction occurs when a conductor is
subjected to a changing magnetic field and an
electromotive force is applied across it.

Faraday Law of
electromagnetic induction

Faraday’s First Law Faraday’s Second Law Lenz’s Law

When the magnetic field of a coil of Coils generate electromagnetic Henri Heinrich Lenz's theory claims
wire changes, an emf is generated fields that are proportional to the that, according to Faraday's Law,
in a coil. Induced electromagnetic rate at which the flux flowing an increase in magnetic flux
field There will be an induced through them changes. The coil's induces an electromotive force,
current flowing if the conductor flux linkage is calculated as the which, in turn, creates an opposing
circuit is sealed. product of the number of turns and magnetic field.
the flux associated with the coil.

98

Circuit with Resistive and Inductive Load

Rising Current : Rise of current in the inductor

Time constant, ɽ

01 = ( − − ɽ ) 03
ɽ = ( ∶ )

= ɽ
= ( ∶ )
02 ( ∶ ) 04

Maximum current of Energy stored in the Time for the current
inductance, inductor rising to the mazimum

= ( ∶ )


99

Circuit with Resistive and Inductive Load

Decaying Current : Rise of current in the inductor

Time constant, ɽ

01 = ( −ɽ ) 03
ɽ = ( ∶ )

S

= ɽ
= ( ∶ )
02 ( ∶ ) 04

Maximum current of Time for the current
inductance, rising to the mazimum

100

Exponential graph of inductor during
rising and decaying current

Rising Current Inductor Decaying Current

101

Resistive Load VS Inductive Load

Resistive Load vs Inductive Load

It transforms electrical energy into thermal energy. This device transforms electrical power into magnetic
power.
It does not have the ability to store electricity. A magnetic field may be used to store electrical
energy.
Changing the phase difference between voltage and
current is impossible with this device. It has the ability to alter the voltage and current
phase differences.
It generates a lot of heat, which is a waste of
electricity. Due to its low internal resistance, it consumes
extremely little electricity.
It is unable to provide the circuit with reverse power.
The circuit may be reversed using this device.
As a result, the power factor becomes one
Its same characteristics may be seen with an AC or DC It causes a drop in power factor.
It illustrates the differences between AC and DC
power supply. power supplies.

102

Example 5 : Rising Proccess b
a
Schematic diagram that consists of a resistive and inductive loads.
If the switch, S is switched to node A at t = 0 s, calculate the :
1. Time constant
2. Maximum current rises in the inductor
3. Instantaneous value of the current when t = 5 ms
4. Time taken to make the instantaneous current equal to 0.5 A
5. Maximum energy stored in the inductor

ANS : 3. Instantaneous value of the current when t = 5 ms

1. Time constant ( = ) = ( − −ɽ )

−
ɽ = = =
)
= ( − = − . −

2. Maximum current rises in the inductor = − −

= = = 1 A = .



= − −

103

Example 5 : (Continue) Rising Proccess

4. Time taken to make the instantaneous current equal to 0.5 A

= = − −ɽ − = − .



0.5 = − − - t = - 0.69 x 1 m

0.5 = − − t = 0.69 ms

− = − . 5. Maximum energy stored in the inductor
− = .
ln − = . =
− = − .

= ( ) Did you know,
Ln e = 1

=

104

Example 6 : Decaying Proccess

A schematic diagram that consists of resistive and inductive load. If 200V
the switch SW is switched to node b after connecting with node a
initially :
1. Time constant
2. Maximum current rises in the inductor
3. Instantaneous value of the current when t = 1 ms
4. Time taken
5. Maximum energy stored in the inductor

ANS : 3. Instantaneous value of t(h e−cɽ )urrent when t = 1 ms

1. Time constant ( = ) =

= ( − )
ɽ = = =

2. Maximum current rises in the inductor = − .
= .
= = = 2 A = .



105

Example 6 : (Continue) Decaying Proccess

4. Time taken − = − .

= . − −ɽ
1.21 = . − −
1.21 = 0.74 − − - t = - 0.45 x 1 m
. − = − − t = 450 µs
0.64 = −
ln 0.64 = − 5. Maximum energy stored in the inductor

=


= ( ) Did you know,
Ln e = 1

=

106

06

Magnetic circuit
electromagnetism

and
electromagnetic

Induction

107

Principles of Magnetism and Characteristic

Magnetism

The magnetic field's force is generated by charged particles (electrons).
An object that creates a magnetic field is called a magnet.

Permanent Substance that attracts On the exterior of the magnet, a Flux direction
magnet additional bits of flux line extends from the north
ferromagnetic material (such pole to the south pole, and it is
as iron, nickel, or cobalt) When expected to continue back to the
left to float, a permanent north pole in the same way. It is
magnet will find its own north therefore impossible to cross such
and south poles. Magnets lines of flux, since they always
have north and south poles, form full closed loops or routes
which are referred to as N and that never meet. Two bar magnets
S, respectively. may be used to show the
principles of magnetic attraction
and repulsion.

108

Magnetic Field

The magnetic field is the region surrounding a magnet where the effects of the magnet's magnetic force
may be measured. It is difficult to portray a magnetic field since it cannot be seen, touched, smelled, or
heard. By picturing the magnetic field to be made up of lines of magnetic flux, Michael Faraday proposed

that the field might be shown pictorially, allowSinoglaarnSyasntaelymsis of the field's distribution and density to be

carried out.

Plot : Characteristics :

● Compass a. Completing the circuit
● Iron dust method
b. Did not come into contact with each other

a b c. Thirdly, it has an established path
e d. The act of repelling one another

e. Has a tendency to make its distances as

short as possible throughout its length

dc
109

Magnetic Field

Magneto Reluctance, S Magnetic field Magnetic flux Permeability and
motive force, strength, H B-H curve for
and flux density
Fm Magnetizing different magnetic
force. A measure of the materials
magnetic flux and the
It is the density of the As the ability of a
magnetic flux. There
Drives a resistance of a are a finite number of magnetic circuit to
magnetic lines of force
magnetic circuit magnetic circuit, that may be generated create magnetic
by a magnetic source.
to generate measured in The magnetic flux flux lines in a
symbol is. The Weber,
magnetic flux, ohms, to Wb, is the standard substance
unit of magnetic flux.
which in turn magnetic flux, The perpendicular produced by a
quantity of flux flowing
generates that is known as across a given region magnetic force
is known as the
electricity. reluctance. magnetic flux density. known as

'permeability' is

defined.

110

Magnetic Field Unit

Formula At – Ampere turns

Magneto motive force, , = 1/H or A/Wb
Fm
= = = = = A/m - Ampere per
Reluctance, S ɸ metre
ɸ
T – Tesla
Magnetic field strength, H
H = µ µ Wb/AT -
Magnetic flux and flux Weber/ampere turns
density ɸ
=
Permeability and B-H
curve for different µ = @µ µ =
magnetic materials
111

Comparison between Electrical and Magnetic Quantities

Electrical Circuit Magnetic Circuit

01 e.m.f. E (V) Mmf, Fm (A) 01

vs

02 Current, I (A) Flux, ɸ (Wb) 02

03 Resistance, R (Ω) vs Reluctance, S ( −1) 03
vs
04 04
= vs ɸ = 05

05 ⍴ vs
= = µ µ
112

Electromagnetism

The electromagnetic force is the subject of
electromagnetism, a field of physics. Interaction
between electrically charged particles is what this
phenomenon is all about.

By supplying a current to conductors,
electromagnetism generates magnetic fields.
Conductor magnetic force lines are generated when it
is electrically charged.

Right-hand Rule may be used, for example, to
calculate the direction of magnetic lines and force
when current, i.e. positive charges, are travelling along
a wire.
Many household equipment are powered by
electromagnetism, which is a basic operating
principle. Communication systems employ
electromagnetic radiation to transport data from one
location to another.

113

Electromagnetic Induction

Induction occurs when a conductor is "Faraday's Law of
subjected to a strong magnetic field Electromagnetism" refers to this as
and an electromotive force is applied the most fundamental principle of

across it. electromagnetism.

1 3

2 4

Faraday's Law induction, developed - Faraday’s First Law
in 1831 by English physicist - Faraday’s Second Law

Michael Faraday, confirms this. - Lenz’s Law.

114

Direction of the Magnetic Field

COMPASS THE GRIP RULE

The direction of a magnetic field may The magnetic field within the solenoid
be determined with a compass. One of may be seen if the coil is held in the palm
the basic components of every of the right hand with the fingers facing
compass is a tiny, self-magnetizing in the direction of current.
metal needle. As a result, when a
magnetic field is present, the needle
may align itself with the field.

SCREW RULE

Right-hand thread screws may be used
in the direction of the magnetic field if
they are put into a conductor in a typical

right-hand thread.

115

Electromagnetic Induction

Faraday’s Any change in the coil's magnetic 1
First Law field will cause an e.m.f. to be 2
generated. It is known as induced 3
Faraday’s emf. A secondary current will
Second flow if the conductor circuit is
closed.
Law
The coil's induced e.m.f. is
Lenz’s proportional to the flux's rate of
Law change across its links. The flux
linkage of a coil is the product of
116 the number of turns in the coil
and the flux associated with the
coil.

To put it another way, when a
change in magnetic flux induces
an e.m.f., the polarity of the
resulting current generates an
opposing magnetic field to the
one caused by the flux change,
according to Lenz's law.

Faraday’s Law & Lenz’s Law
Rate of change of
flux linkage
Formula ɸ

Faraday’s =

Law
Induced e.m.f.

ɸ
= =

Lenz’s Lenz’s Law
Law
ɸ
Formula = − =

117 Where ;

ɸ = B.A ;

B = magnetic field
strength

A = area of the coil

Fleming’s Right-Hand Rule

Second finger will point to the direction of
the induced e.m.f. if first finger points in
direction of magnetic field, thumb points in
direction of motion of conductor relative to
magnetic field.

It is important that the right hand's Conductors in an electrical circuit pass
first and second fingers be positioned through a magnetic field in a generator.
at right angles to each other. EMF is generated by Faraday's law in the
conductors, hence there is a source of EMF.
Mechanical energy is transformed into
electrical energy via a generator.

118

Example 1

1. A 300 mm long conductor travels at a constant 4 m/s in a 1.25 T flux density uniform magnetic field.
Calculate the conductor's current when its ends are either open-circuited or connected to a 20
resistance load.

ANS :
It is possible for an e.m.f. to be generated when a conductor travels in the magnetic field but only if the
circuit is closed can this e.m.f. create current.
E = Blv =(1.25)(300/1000)(4) = 1.5 V
▪ Even if 1.5 V has been produced, no current will flow if the conductor's ends are open circuited.

▪ = = 1.5 =

20

119

Example 1

2. At what speed does the conductor need to cut through the magnetic flux density (0.6 T) to induct 9 V ?
Perpendicular to each other in all three dimensions: conductor, the field, and motion

ANS :

e.m.f. E = Blv, hence velocity, = = 0.6 9 = 9 10−3 = /
75 10−3
0.6 75

3. There are three possible directions in which the conductor might go through a magnetic field: straight
ahead, (a) 90 degrees, (b) 60 degrees and (c) 30 degrees away from the field generated by two 2 cm
square-faced poles. In each scenario, determine the magnitude of the induced e.m.f. if the flux exiting a
pole face is 10 Wb.

ANS :

(a) E90 = Blv sin 90° = ɸ 900 = 10 10−6 0.02 15 1 = .
4 10−4

(b) E60 = Blv sin 60° = E90 sin 60° = 7.5 sin 60° = 3.25 mV

(c) E30 = Blv sin 30° = E90 sin 30° = 7.5 sin 30° = 1.875 mV

120

Example 2 ɸ
Given the value of the current = 0.3 A, coil = 500 turns, flux, ɸ = N

536 µWb, cross sectional area (A) = 1800 mm² and length, l = 150 A

mm. Calculate

a. Magnetomotive force (m.m.f)

b. Reluctance, S of the circuit

c. Magnetic flux density (B)

d. Permeability, µ a. Magnetomotive force (m.m.f) d. Permeability, µ
ANS :
m.m.f. = Fm = NI µ = ,

= 500 x 0.3

I = 0.3 A = 150 AT

N = 500 b. Reluctance, S of the circuit = =

ɸ = 536 µWb
A = 1800 mm² - m²
= = = 279,850 = 1000 AT/m
= 1800 x 10−⁶ m²
ɸ µ

S = 279.85k AT/Wb

l = 150 mm = = . = . −

c. Magnetic flux density (B) µ

= ɸ = µ = .
−

121

Example 3

The air gap in magnetic circuit is 3.5 mm long and 5000 mm² in cross sectional area. Calculate
a. The reluctance of the air gap
b. The m.m.f. required to set up a flux 800 µWb in the air gap

a. The reluctance of the air gap
ANS :
Reluctance, = µ
I = 3.5 mm
A = 5000 mm² - m² µ = µ µ = Π − = Π − µ =

= 5000 x 10−⁶ m² = Π . − − = . /
ɸ = 800 µWb −

b. The m.m.f. required to set up a flux 800 µWb in the air gap

= = ɸ = − − = .
µ µ − Π

. . . = = . . − = .

122

07

Questions

123

QUESTIONS

Q1 : The value of three (3) cells in parallel is 9 V and value of
three (3), calculate the total e.m.f. and in series and in parallel
of the circuit.

Q2 : Based on circuit, calculate the total resistance, RT on terminal A-B.
Use delta-star transformation to simplify the circuit in the correct order.

Q3 : Calculate the total equivalent capacitance, CT between terminal A-B based
on circuit below.

124

QUESTIONS

Q4 : Based on circuit below, a capacitor with capacitance of 4
µF is connected in series with a 0.2 MΩ resistor and 120 V DC
voltage supply. Calculate the initial current, the time constant
during charging, the time taken to be fully charged and
potential different across capacitor at t = 2ms.

Q5 : A steel magnetic circuit has a uniform cross-sectional area of 0.5 m2
and length of 0.4 m. A coil having 400 turns is wound uniformly over the
magnetic circuit. The current in the coil is 0.3 A and the total flux is 0.7
mWb. Based on that condition, calculate the magnetomotive force,
magnetic field strength and the relative permeability of steel.

Q6 : The simple magnetic shown has a cross-sectional area 100 cm² and mean
length of 4 m. The relative permeability of the core is 100. the coil has 250 turns
and the flux produced is 200 µWb. Find :
i. Reluctance of the magnetic circuit.
ii. Current flowing through the coil.

N = 250

A = 100 cm²

l=4m ɸ = 200 µWb

125

QUESTIONS

Q7 : Calculate the current flow through the 20Ω resistor, R5 = RL
for the circuit above by using Thevenin Theorem and Norton

Theorem.

Q8 : By using Kirchoff’s Law, calculate each current I1, I2, I3 and voltage drop
of R2 and power absorbed by the
resistor.

126

08

127

ANSWER

Q1 : The value of three (3) cells in parallel is 9 V, calculate the total e.m.f. in series and in parallel of the circuit.

ANS : - Parallel

- Series

Total e.m.f., EAB = 9 + 9 + 9 = 27 V

Total e.m.f., EAB = Eindcell = 9 V

128

ANSWER

Q2 : Based on circuit, calculate the total resistance, RT on terminal A-B. Use delta-star transformation to simplify the
circuit in the correct order.

ANS :

Rb Calculate the value of Ra, Rb and Rc
Ra
= = ( )( )
Rc + + + + = = . Ω

= = ( )( ) = = . Ω
+ + + +

= ( )( ) = . Ω
+ + = + + =

129

ANSWER (Continue)

Q2 : Based on circuit, calculate the total resistance, RT on terminal A-B. Use delta-star transformation to simplify the
circuit in the correct order.
ANS :

Calculate the value of Rb, R3 and Rc, R4
= + 3 = 3.03 + 15 = . Ω
= + 4 = 1.21 + 10 = . Ω

= 1 2 = 18.03 + 11.21
1 + 2 18.03 + 11.21 = . Ω

= + 3 = 4.85 + 6.91 = . Ω
130

ANSWER

Q3 : Calculate the total equivalent capacitance, CT between terminal A-B based on circuit below.

ANS : 1 1 11 1 1 1
1 = 4 + 5 + 6 = 4.7 µ + 1 µ + 47 µ

1
1 = 1.23 µ

= . µ

2 = 1 + 3 + 2 = 10 µ + 4.7 µ + 0.81 µ
= . µ

11 1 1 1
= 1 + 2 = 10 µ + 15.51 µ = 164. 47

= . µ

131

ANSWER

Q4 : Based on circuit below, a capacitor with capacitance of 4 µF is connected in series with a 0.2 MΩ resistor and 120
V DC voltage supply. Calculate the initial current, the time constant during charging, the time taken to be fully charged
and potential different across capacitor at t = 2 ms.

ANS :

Initial current,I0 = Imax = 1
1

120 Time taken to fully charge = 5τ
= 0.2 = 5 × 0.8
=
= .
Potential different across capacitor at t = 2ms,
Time constant during charging, τ = RC
−
τ = 0.2 M × 4 μ
= (1 − τ )
= .
−2

= 120 (1 − 0.8 )
= 120 (1 − 0.999 )

= .

132

ANSWER

Q5 : A steel magnetic circuit has a uniform cross-sectional area of 0.5 m2 and length of 0.4 m. A coil having 400 turns is
wound uniformly over the magnetic circuit. The current in the coil is 0.3 A and the total flux is 0.7 mWb. Based on that
condition, calculate the magnetomotive force, magnetic field strength and the relative permeability of steel.

ANS :

ℎ ,
ɸ
=
= 400 0.3 = =
0.7
ℎ
= 0.5
= = .

120
= 0.4 = µ µ =


µ = µ
1.4
µ = 300 4 10−7 = .

133

ANSWER

Q6 : The simple magnetic shown has a cross-sectional area 100 cm² and mean length of 4 m. The relative permeability

of the core is 100. the coil has 250 turns and the flux produced is 200 µWb. Find :

i. Reluctance of the magnetic circuit.

ii. Current flowing through the coil.

ANS : N = 250
. , = A = 100 cm²

= ° = 4 10−7 100 = 4 10−5 l = 4 m ɸ = 200 µWb


=

4 = . /
= 4 10−5 100 10−4

. . .
. ℎ ℎ ℎ , = =

=
∅ 200 10−6
= = = 100 10−4 4 10−5 = 159.15 /

159.15 4 134
= = 250 = .

ANSWER

Q7 : Calculate the current flow through the 20 Ω resistor, R5 = RL for the circuit above by using Thevenin Theorem and
Norton Theorem.

ANS : Thevenin Theorem

Step 1 :
Remove the Load, RL. Calculate Thevenin’s

Voltage is the open circuit voltage at terminal A-B.

VTH Step 2 : Calculate Thevenin’s Resistance which is the resistance seen
from terminal A-B. The power source is replaced by short circuit.

= 1 + 2 3 + 4
1 + 2 + 3
3
= 1 + 2 + 3 × 1 RTH 50 + 10 60
= 50 + 10 + 60 + 30
60
= 50 + 60 + 10 × 15 = 30 + 30 = Ω

= . 135