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MECHANICAL ENGINEERING DEPARTMENT DJJ 20053

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SITI HAJAR BINTI MOHD NOH K.M.WONG HELLY SUHAILLA BT SOLAIMAN

Published by:

Politeknik Port Dickson

KM14, Jalan Pantai,

71050 Si Rusa,

Negeri Sembilan

SYNOPSIS ELECTRICAL TECHNOLOGY exposes students to basic electrical

circuit concepts, the application of electromagnetism in electrical

machines and transformers. The course focuses on the different

types of electrical circuits, the relationship between current and

voltage including the resistance. It also provides the skills on the

methods of constructing basic circuits and operation of electrical

machines and transformers. This course also exposes the students

to the demonstration of experiments in Electrical Engineering.

eISBN NO:

AUGUST 2020

DJJ20053 – ELECTRICAL TECHNOLOGYCopyrighted 2020 *Port Dickson Polytechnic Mechanical Engineering Department * All Right Reserved

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LOGOTitle Page

1. INTRODUCTION TO ELECTRICAL

CIRCUITS

2. INDUCTOR, CAPACITOR, ALTERNATING

CURRENT CIRCUITS & 3 PHASE SYSTEM

5. AC ELECTRICAL MACHINES

3. BASIC PRINCIPLES OF

ELECTROMAGNETISM

4. TRANSFORMER

6. PRACTICAL TASK

1-33

34-82

83-98

99-121

122-139

140-168

CHAPTER 1INTRODUCTION TO ELECTRICAL CIRCUIT

(CLO 1)

Explain the principles and fundamental of electrical

circuits, electromagnetism, transformers and

electrical machines (C2,PLO1)

(CLO2)

Solve the problem related to electrical circuits,

electromagnetism, transformers and electrical

machine. (C3,PLO1)

(CLO3)

Organize appropriately experiments in groups

according to the Standard Operating Procedures.

(P4,PLO5)

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1

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LOGO1.0 WHAT IS ELECTRIC ?

Electric - an energy cannot be seen but can be felt

- can create impact from action as friction, heat

and electromagnetic field

Electric energy can change into :

a. Light energy – lamp

b. Heat energy – Iron

c. Sound energy – Radio

d. Kinetic energy – Motor

2 Types of Electric:

a. Static electricity

– no electron movement

b. Dynamic electricity

– with electron movement

2

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LOGO1.1 BASIC ELECTRICAL QUANTITIES

Electromotive

Force (e.m.f.)

Unit: Volt

(V)

Electrical Units

Definitions

Electric

Charge (Q)

Unit: Coulomb

(C)

Electric

Current (I)

Unit: Ampere

(A)

Voltage (V)

Unit: Volt

(V)

Resistance (R)

Unit: Ohm

(Ω)

Force or electric pressure that cause the flow of

electrons or the flow of current in given circuit.

The movement of Electric

Charge by free electrons

movement from (+) terminal

to (-) terminal.

2 type of Electric Charges :

Positive Charge & Negative Charge

The potential different

between two points in a

circuit.

Property of material by

which it oppose the flow

of current through it.

3

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LOGO1.2 FACTORS AFFECT RESISTANCE

It depends on the physical properties of the

material. Four factor that influence the value

of resistance:

Formula of resistance (R):

4

Symbol Unit

Area (A) Meter square (m2)

Resistivity (ρ) Ohm meter (Ωm)

Current (l) Meter (m)

Resistance (R) Ohm (Ω)

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LOGO1.2 PROBLEM SOLVING

Example 1 :

Calculate the resistance for the Aluminum wire that has

1.5km length, the diameter is 10mm and the resistivity is

0.025 .m.

Solution :

Given,

Formula,

Where,

477.06

1054.78

)3

105.1)(6

10025.0(

x

xxR

mxd 31010

mx 3105.1

mx 610025.0

AR

261054.78

2)

2

31010

(2

)2

( mxxd

A

5

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LOGO 1.3 TYPES OF ELECTRIC CIRCUITS

A combination of conductor / cable makes current flow

from voltage sources to electrical components / load.

2 types:

i. Complete electric circuit (close circuit)

- simple basic circuit must have voltage supply (V),

electric current (I) and resistance (R). A closed end

connection makes current go through completely

which the current flow from source and back flow to

sources again.

ii. Non Complete electric circuit

- Current flow will never happen with perfect

- open circuit / short circuit

6

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1.3.1 ELECTRICAL CIRCUITS

Types of Electrical Circuits

Definition Schematic diagram

Simple Circuit

A simple circuit may include two components: a battery

and a lamp. The circuit allows current to flow from the

battery to the lamp, through the lamp, then back to the

battery. Thus, the circuit forms a complete loop.

Complex Circuit

The load can be as simple as a single light bulb. In

complex circuits, the load is a combination of

components, such as resistors, capacitors, transistors,

and so on.

Open Circuit

Open circuit is most often used to describe a circuit that

has become broken, either on purpose (by the use of a

switch) or by some error, such as a loose connection or

a damaged component.

Short Circuit

Short circuit refers to a circuit that does not have a load.

-A device in a circuit is said to be short circuit if it is by-

passed by a conductor of nearly zero resistance,

voltage drops to almost zero.

-When circuit is closed, the light bulb is short circuit

and it goes off. A relatively large current would flow

through the wire.

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LOGO 1.4 ELECTRICAL MEASURING

DEVICES

Voltmeter :

For measuring

electrical potential

difference

between two

points in an

electric circuit.

Ammeter :

Measure

the electric

current in a

circuit

Ohmmeter :

Measure the

electric

resistance in

a circuit

8

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MULTIMETER : Use to measure multiple characteristics

of electronic such as resistance, current and voltage

• Ohm meter

• Ammeter

• Voltmeter• Voltmeter

DCV ACV

ΩA

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LOGO1.5 OHM’S LAW

Ohm’s Law Formula Wheel

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LOGO1.5.1 DEFINITION OF OHM’S LAW

Ohm's Law states that, at constant temperature, the electric current(I) flowing in a conducting material is directly proportional to theapplied voltage (V), and inversely proportional to the resistance (R).

11

Symbol Unit

Votage (V) Volt (V)

Current (I) Ampere (A)

Resistance (R ) Ohm (Ω)

Ohm's Law Triangle :

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LOGO 1.5.2 DIFFERENCE OF LINEAR &

NON-LINEAR RESISTANCE

12

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Example 2:

Calculate the current value if the resistance is 10 and

the supply voltage is 15V. Then, calculate the new

current value if the resistance has changed to 10k.

Solution :

Given, R1 = 10 , V = 15V , R2 = 10k

Formula,

Where,

AR

VI 5.1

10

151

AxR

VI 0015.0

1010

15

32

22

1.5.3 PROBLEM SOLVING

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LOGO1.6 ELECTRICAL POWER

14

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The SI unit is the Watt (W).

15

1.6.1 DEFINITION OF ELECTRICAL

POWER

Electrical Power (P) is an indication of how much work (the

conversion of energy from one form to another) can be

done in a specific amount of time; that is, a rate of doing

work.

Symbol Unit

Power (P) Watt (W)

Current (I) Ampere (A)

Voltage (V) Volt (V)

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Electric power can be delivered or absorbed as defined by

the polarity of the voltage and the direction of the current.

Equation shows the relationship between the Power (P),

Current (I) and Voltage (V). It can also be connected to the

resistance through Ohm's Law equation.

Mathematical equation for Electric Power :

From the Ohm’s Law :

RIP

IRI

IVP

2

)(

R

VP

RR

V

IVP

2

)(

16

1.6.2 DERIVATION OF

ELECTRICAL POWER

a) Substitute V=IR

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Example 3:

A stereo receiver has a set of speakers, main & remote

for each channel. The speakers for each channel are

connected in parallel to the receiver. The AC voltage

across the speakers is 6.0 volts & the equivalent

resistance is 2.67Ω. Determine the total current supplied

by the receiver and the power dissipated in the set of

speakers.

Solution :

Given, V = 6V, R = 2.67Ω

Total current

Power dissipated

17

1.6.3 PROBLEM SOLVING

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LOGO1.7 ELECTRICAL ENERGY

18

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Electrical Energy (E) refers to energy which is converted

from electrical potential energy. This energy is supplied

by the combination of electric current and electrical

potential that is delivered by the circuit. Electrical Energy

lost or gained by any system is determined by:

19

1.7.1 DEFINITION OF ELECTRICAL

ENERGY

Symbol Unit

Energy (E) Joule (J)

Power (P) Watt (W)

Time (t) Second (s)

The SI unit is the Joule (J) or Watt Second (Ws).

1 Joule = 1Ws

4.18 Joules = 1 Calorie

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A Product of Power (P) and Time (t) :

Mathematically Electric Energy (E) is expressed as:

20

1.7.2 DERIVATION OF

ELECTRICAL ENERGY

Symbol Unit

Power (P) Watt (W)

Current (I) Ampere (A)

Voltage (V) Volt (V)

Resistance (R) Ohm (Ω)

Time (t) Seconds (s)

tR

VE

RtR

V

RtIE

tIRI

IVtE

tIV

PtE

2

2

2

)(

)(

)(

From the Ohm’s

Law:

From the Electrical

Power:

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21

1.7.4 USE OF WATTMETER , KILOWATT-

HOUR METER AND ITS SYMBOL

WATTMETER - measure value of Electric Power (or the

supply rate of electrical energy) in watts of any given

circuit.

Analog & Digital Wattmeter

WATTMETER SYMBOL

Symbol

KILOWATT-HOUR METER - Energy meter that measures

the amount of electric energy in kilowatt-hour (kWh)

consumed by a residence, business, or an electrically

powered device.

Analog & Digital Energy Meter

KILOWATT-HOUR METER

SYMBOL

Meter symbol

Meter symbol

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Law of Conservation of Energy :Energy can neither be created nor destroyed but it can be

converted from one form to another form.

22

1.7.4 RELATIONSHIP BETWEEN

ELECTRICAL & HEAT ENERGY

Electrical Energy to Heat

Energy :Ovens & heaters do the

opposite. They convert

electrical energy into thermal

energy. A thermo-electric

generator is a device based on

a thermocouple that converts

heat directly into electricity

without moving parts.

Heat Energy to Electrical

Energy :The burning of fossil fuels (oil,

coal, or natural gas) generates

electrical energy.

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LOGO1.7.5 UNITS CONVERSION FOR ENERGY

Prefix PrefixSymbol

Prefixfactor

Example

pico p 10-12 1pF = 10-12F

nano n 10-9 1nF = 10-9F

micro μ 10-6 1μF = 10-6F

milli m 10-3 1mA = 10-3A

kilo k 10 3 1kΩ = 1000Ω

mega M 10 6 1MHz = 106Hz

giga G 10 9 1GHz = 109Hz

tera T 1012 1THz = 1012Hz

Units Prefix Table

kJ kcal kWh

1 kJ 1 0.2388 0.000278

1 kcal 4.1868 1 0.001163

1 kWh 3600 860 1

Kilowatt-hour

DefinitionKilowatt-hour is an energy unit

(symbol kWh or kW·h)

1 kWh = 1kW · 1h

1 kWh = 3.6106 J

1 J = 1 Ws

E(kWh) = P(kW) · t(h)

What is the energy consumed

when consuming 2kW for 3 hours?

Solution:

E(kWh) = 2kW · 3h = 6kWh

Unit Conversion Tables :

23

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Example 4:

Find the cost of running a 2000-watt heater for 12 hours if

electric energy costs RM0.25 per kilowatt-hour:

Solution :

Given, P = 2000W = 2kW, t = 12 hours,

cost = RM 0.25

Cost of Electrical energy = 2kW x 12 hours x RM 0.25

= RM 6.00

24

1.7.6 PROBLEM SOLVING

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Example 5:

A toaster take 5A from 240V supply within 15 minutes.

Calculate :

a. Power consumed. b. Electrical energy in kJ

Solution :

Given, I = 5A, V = 240V, t = 15x60 sec = 900 sec

a. Power consumed. P = IV = (5)(240) = 1200 W

b. Electrical energy in kJ. T = Pt = (1200)(900) = 1080000W

= 1080 kWj

= 1080 kJ

25

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Resistors can be connected in three different ways

which are series,parallel and combination circuits.

Series Circuit Parallel Circuit

Combination

Circuits

26

1.8 CHARACTERISTICS OF SERIES

AND PARALLEL CIRCUITS

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LOGO 1.8.1 FORMULAS USED IN

SERIES & PARALLEL CIRCUITS

Provides only ONE path for current Provides MORE THAN ONE path

for current

27

SERIES CIRCUIT PARALLEL CIRCUIT

Formulas : Formulas :

Voltage Drop: Current Divider Law :

i. Ohm’s Law

ii. Voltage Divider Law

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Example 6:

Based on figure, determine :

i. Total Resistance

ii. Total Current

iii. Voltage Drop at resistor R2 and R3

Solution:

28

1.8.2 PROBLEM SOLVING

(i)

A

R

VI

T

TT

4

3

105

1018

9

k

kkk

RRRRT

18

5103

321

(ii)

(iii)

V

k

RIV

5

10105 4

222

V

VR

RV T

T

5.2

91018

1053

3

33

Series : IT = I1 = I2 = I3

By using Voltage Divider Law

By using Ohm’s Law

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Example 6:

Based on figure, determine :

i. Total Resistance

ii. Total Current

iii. Voltage Drop at resistor R2 and R3

Solution:

29

(iii)

V

k

RIV

5

10105 4

222

V

VR

RV T

T

5.2

91018

1053

3

33

Series : IT = I1 = I2 = I3

By using Voltage Divider LawBy using Ohm’s Law

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Example 7:

Based on figure, determine :

i. Total Resistance

ii. Total Current

iii. Voltage drop at V3

iv. Current (I2) & (I3)

Solution:

30

(i)

A

R

VI

T

TT

021.0

1057.0

123

k

R

R

RRRR

T

T

T

57.0

7

4

4

71

4

1

2

1

1

1

1111

321

(ii)

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Example 7:

Voltage drop at V3

Current (I2) & (I3)

Solution:

31

(iv)

(iii)

VVVT 123

mA

A

IR

RI T

T

12

012.0

021.0101

1057.03

3

1

1

Parallel: VT =V1 =V2 =V3

mA

A

R

VI T

6

106

102

12

3

3

2

2

By using Current Divider LawBy using Ohm’s Law

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Example 8:

A 120V source is connected across resistors.

If R1 = 10Ω, R2 = 20Ω , R3 = 15Ω. Calculate,

i. Total Resistance (RT) ii. Total Current (IT)

iii. Current (I2) & (I3)

Solution :

i. ii. iii.

RT = R1 + R23

= 8.57Ω + 10 Ω

= 18.57Ω

I3 = IT - I2 = 6.46 – 2.79 = 3.67 A

32

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Example 9 :

Determine the

(a) total resistance of the circuit

(b) total current flow

(c) current of I1 and I2(d) voltage drop across R3 and R4

Solution :

33

(a) (b)

(c) (d)