ELECTRICAL TECHNOLOGY WORK BOOK print

ELECTRICAL TECHNOLOGY

WORK BOOK

FAIZAL MOHAMAD TWON TAWI
NURUL HIDAYAH AHMAD SHAIRAZI

MOHD NAJIB SHUIB
JUNAINAH ABD. KADIR
HIDAYAH JAMALUDDIN

02

ELECTRICAL TECHNOLOGY WORK BOOK
ISBN: 9789670783550
First Print: 2019
Copyright 2019 © Politeknik Seberang Perai
Hak cipta terpelihara. Tidak dibenarkan mengeluar ulang mana-mana bahagian artikel, ilustrasi
dan isi kandungan buku ini dalam apa jua bentuk dan dengan apa cara sekalipun, sama ada
secara elektronik, fotokopi, mekanik, rakaman atau cara lain sebelum mendapat izin bertulis
daripada Politeknik Seberang Perai.

Review and edited by:

Unit Penyelidikan & Penerbitan Perpustakaan
Politeknik Seberang Perai.
Publisher:
Perpustakaan,
Politeknik Seberang Perai,
Jalan Permatang Pauh, 13500 Permatang Pauh,
Pulau Pinang.
Tel: 04-5383322 Fax: 04-5389266
Email: [email protected].
Muka Taip Teks: Franklin Gothic
Saiz Taip Teks: 11
Editor:
Faizal Mohamad Twon Tawi
Penyemak:
Amir Abu Bakar
Pencetak:
AZIMAT ADVANCE VENTURES
H/P: 019-5702881/ 012-4652881

03

PREFACE

Electrical Technology is a technical subject for Electrical Engineering students during their first
semester in Polytechnic Malaysia. This subject is very important for students because the
knowledge of this subject will be used throughout the entire period of study in engineering diploma.
Besides, this subject is prerequisite to a subject Electrical Circuit in second semester in Polytechnic
Malaysia.

Even though today have many reference books on electrical circuits for the electrical engineering
students, there are still no coexisting of work books. Moreover, exercise book for electrical
engineering is very important in order to help students in improving their knowledge and
understanding of the subject. Therefore, we as lecturers try to produce a workbook that contains
a collection of selected questions. These questions would help students to practice doing the
exercises.

This book consists collection of questions related to the electrical circuit. It is sorted the sub-topics
based on the syllabus in Polytechnic Malaysia so that it would be easier for students to study.
Students are able to practice the knowledge that they have acquired in class through this book. It
is hoped that this book would contribute towards achieving the objectives of this subject in which
is to increase knowledge in the electrical circuit.

04

ACKNOWLEDGMENTS

Alhamdulillah. Thank God enables us to produce a book that aims to help students to improve their
knowledge of the subject Electrical Technology in Politeknik Malaysia.

We would like to thank many people for their help in completing this book, but my foremost
appreciation goes to the lecturers and students who have used “Work Book Electrical Technology”.

We also would like to thank our colleagues in Electrical Engineering Department from Politeknik
Seberang Perai that help us to complete this book. Finally, we would like to thank the UPPen staffs
for their helpful efforts during this project. The cheerful professionalism of the staffs were highly
appreciated.

CONTENTS

PREFACE .................................................................................................................................................3
ACKNOWLEDGMENTS............................................................................................................................4
CHAPTER ONE: INTRODUCTION TO ELECTRICAL CIRCUIT ...................................................................6

SHORT NOTE ......................................................................................................................................6
EXERCISE......................................................................................................................................... 10
CHAPTER TWO: DC EQUIVALENT CIRCUIT.......................................................................................... 27
SHORT NOTE ................................................................................................................................... 27
EXERCISE......................................................................................................................................... 30
CHAPTER THREE: CAPACITOR AND CAPACITANCE............................................................................ 48
SHORT NOTE ................................................................................................................................... 48
EXCERCISE ...................................................................................................................................... 51
CHAPTER FOUR: INDUCTOR AND INDUCTANCE ................................................................................ 70
SHORT NOTE ................................................................................................................................... 70
EXCERCISE ...................................................................................................................................... 72
CHAPTER FIVE: MAGNETIC CIRCUIT ELECTROMAGNETISM & ELECTROMAGNETIC INDUCTION ... 90
SHORT NOTE ................................................................................................................................... 90
EXERCISE......................................................................................................................................... 92
BIBLIOGRAPHY .................................................................................................................................. 102
ANSWERS .......................................................................................................................................... 103

06 Electrical Technology Work Book

CHAPTER ONE: INTRODUCTION TO ELECTRICAL CIRCUIT

SHORT NOTE

Six SI unit based on quantities and units

Table 1 SI unit

Quantity Symbol Unit Abbreviation
Length l meter m
Mass m kilogram kg
Time t seconds s
Electric current I ampere A
Thermodynamic temperature T kelvin K
Amount of substance n mole mol
Luminous instance Iv candela cd

Types of Cells and Batteries

1. Carbon-zinc cell
2. Alkaline cell
3. Nickel Cadmium cell
4. Edison cell
5. Mercury cell

Formula of Resistance and Resistivity

=

Where

= (Ω )

= ℎ ( )

= − ( 2)

Four factors that affect the electrical resistance

Any material with a uniform cross sectional area is determined by the following four factors:

1. ( ) − ℎ ℎ ℎ , ℎ
ʹǤ ℎ ( ) − ℎ ℎ, ℎ
3. − ( ) − ℎ , ℎ
ͶǤ − ℎ ℎ ℎ , ℎ

Electrical Technology Work Book 07

Ohm’s Law = ×
Where = ( )
Series Circuit = ℎ (Ω)

Parallel circuit = 1 + 2 + 3
= 1 + 2 + 3
= 1 = 2 = 3

= 1 + 1 + 1
1 2 3

= 1 = 2 = 3

Voltage Divider Rule = 1 + 2 + 3

1 = ( 1 1 2)
+
2
2 = ( 1 + 2)

08 Electrical Technology Work Book
Current Divider Rule

1 = ( 1 2 2)
+
1
2 = ( 1 + 2)

Delta-Star Transformation

1 =
+ +

2 = + +

3 =
+ +
Star-Delta Transformation

= 1 2 + 2 3 + 3 1
3
1 2 + 2 3 + 3 1
= 2

= 1 2 + 2 3 + 3 1
1

Electrical Technology Work Book 09

Formula of Electrical Power and Energy

Power =
Where

= ( )
= ( )

= ( )

Power Dissipated = 2
Where
= ( )
Alternatively = ( )
Where
= (Ω)
Alternatively
Where =

= ( )
= ( )

= ( )

= 2


= ( )
= ( )

= (Ω)

10 Electrical Technology Work Book

EXERCISE

Question 1

Gives Four (4) types of cells and batteries

Answer

1. Carbon-zinc cell
2. Alkaline cell
3. Nickel Cadmium cell
4. Edison cell
5. Mercury cell

Question 2

A direct current (d.c.) electric motor consumes 36 MJ when connected to a 250 V supply for 1 hour.
Find the power rating of the motor and the current taken from the supply.

Answer

Given

= 36 , = 1ℎ

Find the power rating of motor

( ) = = 36 × 106 = 10 × 103
60 × 60

Find the current taken from the supply using this formula

= ×

=

= 10 × 103 = 40
250

Electrical Technology Work Book 11

Question 3

TWELVE cells (12), each with an internal resistance of 0.24 Ω and an e.m.f. of 1.5 V are connected:
a) In series
b) In parallel
Find the total emf and total internal resistance for both connection.

Answer
a. Total emf in series

= 12 × 1.5 = 18
Total internal resistance in series

( ) = × = 12 × 0.24 = 2.88Ω

b. Total emf in parallel

= 1.5

Total internal resistance in series

( ) = = 0.24 = 0.02 Ω
12

Question 4
Calculate total e.m.f. of the circuit in Figure 1

Figure 1

Answer
Total e.m.f.,

= 1 + 2 + 3 + 4 = 2 + 2 + 2 + 2 = 8

12 Electrical Technology Work Book
Question 5
Calculate total e.m.f. of the circuit in Figure 2

Answer Figure 2
Total e.m.f.
= 1 + 2 = 8 − 6 = 2

Question 6

Find the resistance of 800 m of copper cable of cross-sectional area 20 mm2. Take the resistivity
of copper as 0.02 μm.

Answer = 20 2, = 800 , = 0.02 Ω
Given
How to convert mm2 to m2 20 2 = _______ 2
Since
1 = 0.001
Therefore (1 )2 = (0.001 )2

1 = 0.0012 2

20 2 = 20 × 10−6 2

Find the resistance

= = 0.02 × 10−6Ω × 800 = 0.8Ω
20 × 10−6 2

Electrical Technology Work Book 13

Question 7

Calculate the cross-sectional area, in mm2, of a piece of aluminium wire 100 m long and having a
resistance of 2. Take the resistivity of aluminium as 0.03X10-6Ωm.

Answer = 100 , = 2Ω, = 0.03 × 10−6Ω
Given
Since, the resistance =

Then

= = 0.03 × 10−6Ωm × 100m = 1.5 2
2

Question 8

The resistance of 500 m of wire of cross-sectional area 2.6 mm2 is 5. Determine the resistivity of
the wire in μm

Answer

Given

= 2.6 2, = 500 , = 5Ω

Since, the resistance

=

Then, the resistivity,

= = 5Ω × 2.6 × 10−6 2 = 0.026 × 10−6Ωm = 0.026μΩm
500m

14 Electrical Technology Work Book

Question 9

When the switch in the circuit shown Figure 3 is closed the reading on voltmeter 1 is 30 V and that
on voltmeter 2 is 10 V. Determine the reading on the ammeter and the value of resistor RX

Figure 3

Answer

Voltage across 5  resistor

5Ω = 1 − 2 = 30 − 10 = 20

Hence, current in 5  resistor, i.e. reading on the ammeter

5Ω = 20 = 4
5 5

To find the total resistance,

Note that,

1 =
30
= = 4 = 7.5Ω

Hence

= 7.5 − 5 = 2.5Ω

Electrical Technology Work Book 15

Question 10

For the circuit in Figure 4, determine the value of V1. If the total circuit resistance is 36  determine
the supply current and the value of resistors R1, R2 and R3

Figure 4

Answer
Given Rtotal = 36 and the supply voltage, Vsupply = 18V

18 = 1 + 5 + 3
1 = 18 – 5 – 3 = 10

Apply Ohm’s Law to find total current

Note that,

=

= = 18 = 0.5
36

Apply Ohm’s Law to find R1, R2 and R3

1 = 1 = 10 = 20Ω
0.5

2 = 2 = 5 = 10Ω
0.5

3 = 3 = 3 = 6Ω
0.5

16 Electrical Technology Work Book
Question 11
Based on Figure 5, determine (a) the reading on the ammeter, and (b) the value of resistor R.

Figure 5

Answer

Since the resistors are connected in parallel, the voltage for every junction is same value.

6Ω = 5Ω = 3 × 5Ω = 3 × 5 = 15
a. Apply Ohm’s Law to find the reading on the ammeter

= 6Ω = 15 = 2.5
6Ω 6

b. Apply Kirchhoff’s Law to find IR

11.5 − − 3 − 2.5 = 0
= 11.5 − 3 − 2.5 = 6

Therefore = = 15 = 2.5Ω
Question 12 6

Find the total resistance between terminals A and B of the circuit in Figure 6 below.

Figure 6

Answer

Total resistance between terminals A and B

= 1 + (6 ∕∕ 18) + 1.5
6 × 18
= 1 + 6 + 18 + 1.5 = 1 + 4.5 + 1.5 = 7Ω

Electrical Technology Work Book 17
Question 13
Determine the currents and voltages indicated in the circuit in Figure 7.

Figure 7

Answer

Firstly, find RTotal = 4 + +
The value of Ra
1 = 1 + 1 + 1 = 1
The value of Rb 2 3 6
1
Therefore = 1 = 1Ω

1 = 1 + 1 = 0.83
2 3
1
= 0.83 = 1.2Ω

= 4 + 1 + 1.2 = 6.2Ω

Then, find each current I1, I2, I3, I4, I5 and I6 by using Ohm’s Law

18 Electrical Technology Work Book

Find I1 and V1 =
Note that,
Find V2 and I2 I3 and I4 1 = = 30 = 4.84
6.2Ω
Find V3 and I5 and I6
1 = 1 × 4Ω = 4.84 × 4 = 19.36V

2 = 1 × = 4.84 × 1 = 4.84

2 = 2 = 4.84 = 2.42
2Ω 2Ω

3 = 2 = 4.84 = 1.61
3Ω 3Ω

4 = 2 = 4.84 = 0.81
6Ω 6Ω

3 = 1 × = 4.84 × 1.2 = 5.81
3 5.81
5 = 3Ω = 3Ω = 1.94

6 = 3 = 5.81 = 2.91
2Ω 2Ω

Question 14

Find the current I in the circuit in Figure 8.

Figure 8

Answer

The circuit is reduced step by step as shown in diagrams (a) to (d) below.

(a)
= 3 + 2 = 5Ω

Electrical Technology Work Book 19

(b)

= 5 ∕∕ 5 = 5 × 5 = 2.5Ω
5 + 5

(c)

= 6 ∕∕ 4 = 6 × 4 = 2.4Ω
6 + 4

(d) = 1.6 + 2.4 = 4Ω
From total current

= 24 = 24 = 6
4

20 Electrical Technology Work Book
Question 15

By referring to the circuit in Figure 9, E=20V, R1=4Ω, R2=6Ω, R3=8Ω calculate:
i) Total resistance of the circuit, Rtotal
ii) Current, I
iii) Voltage drop across resistor 8Ω, VR3
iv) Current through resistor 4Ω, IR1

Figure 9

Answer

Total resistance of the circuit, Rtotal

1 = 1 + 1 + 1 = 1 + 1 + 1 = 0.542
1 2 3 4 6 8

= 1 = 1.846Ω
0.542Ω
Current, I

= = 20 = 10.833
1.846Ω

Voltage drop across resistor 8Ω, VR3

Current through resistor 4Ω, IR1 3 = = 20

= = 20 = 5
1 4Ω

Electrical Technology Work Book 21

Question 16

By referring to the circuit in Figure 10, calculate:
i. Total resistance of the circuit, Rtotal
ii. Current, I
iii. Voltage drop across resistor 6Ω, VR2

Figure 10

Answer
Find the total resistance of the circuit, Rtotal

= 1 + 2 + 3 = 4 + 6 + 8 = 18Ω

Find the current, I using Ohm’s Law

= = 15 = 0.833
18Ω

Voltage drop across resistor 6Ω, VR2 using Ohm’s Law
2 = × 2 = 0.833 × 6 = 5

22 Electrical Technology Work Book
Question 17
By referring to the circuit in Figure 11, calculate:
i) Equivalent resistance of the circuit, Rtotal
ii) Current from supply, Is

Figure 11

Answer
Find the equivalent resistance of the circuit, Rtotal

= 4 + 8 + 6 = 18

Electrical Technology Work Book 23

1 = 1 + 1 = 1 + 1 = 1
18 18 18 9

Therefore, the total of resistance is = 9

= 2 + + 20 = 2 + 9 + 20 = 31
Find the current from supply, Is

= 20 = 645.16
31

Question 18

Transform the delta connected networks shown in Figure 12 to their equivalent star-connected
networks.

Figure 12

Answer
The delta-connected network is shown labelled, together with the equivalent star-connected
network is shown below.

24 Electrical Technology Work Book

1 = = 4 4×1 5 = 0.4Ω
+ + +1+

2 = = 4 4×5 5 = 2Ω
+ + +1+

3 = = 4 1×5 5 = 0.5Ω
+ + +1+

Question 19

Transform the delta connected networks shown in Figure 13 to their equivalent star-connected
network.

Figure 13

Answer
The delta-connected network is shown labelled below, together with its equivalent Star-connected
network.

1 = = 100 × 100 = 33.33Ω 3 = = 100 × 100 = 33.33Ω
+ + 100 + 100 + 100 + + 100 + 100 + 100

2 = = 100 × 100 = 33.33Ω
+ + 100 + 100 + 100

Electrical Technology Work Book 25
Question 20
Calculate the total resistance, Rxy of the circuit in Figure 14.

Figure 14

Answer

Converting from Delta to Star connection

= 4 4 x 8 6 = 1.78Ω
+8+

= 4 8 x 6 6 = 2.67Ω
+8+

= 4 4 x 6 6 = 1.33Ω
+8+

= 1.33 + 12 = 13.33
= 2.67 + 10 = 12.67

26 Electrical Technology Work Book

= 13.33 ∕∕ 12.67 = 13.33 × 12.67 = 6.5Ω
13.33 + 12.67

Therefore

= 1.78 + 6.5 = 8.28

Electrical Technology Work Book 27

CHAPTER TWO: DC EQUIVALENT CIRCUIT

SHORT NOTE

Kirchhoff’s Voltage Law (KVL)

The sum of all voltages or potential differences in an electrical circuit loop is 0. (i.e. See Figure 15)

Figure 15

Σ = 0
E − 1 − 2 = 0

Or, in any closed loop, the algebraic sum of the e.m.f applied is equal to the algebraic sum of

the voltage drops in the elements (i.e. See Figure 15)

Kirchhoff’s Current Law (KCL) = ΣV
E = 1 + 2

In any electrical circuit, the algebraic sum of incoming currents to a point and outgoing currents
from that point is Zero. (See Figure 16)

Figure 16

Σ = 0
1 − 2 − 3 + 4 − 5 − 6 = 0

Or the entering currents to a point are equal to the leaving currents of that point.

Figure 17

28 Electrical Technology Work Book

Mesh Analysis =
1 + 4 = 2 + 3 + 5 + 6

It is a technique to solve problem for the current each closed loop of the circuit using Kirchhoff’s
Voltage Law (KVL) in electrical circuit.

Nodes Analysis

It is a technique to solve problem for the voltage each common node of the circuit using Kirchhoff’s
Voltage Law (KVL) in electrical circuit.

Thevenin’s Theorem

Any linear circuit consisting several energy source and resistances or complex circuit can be
replaced by just a single voltage (VTH) in series with a single resistor (RTH). (see Figure 18)

Figure 18

Step to apply the Thevenin’s Theorem
1. Terminate the load resistor, RL and find the Thevenin source, VTH
2. Find the Thevenin resistance, RTH by removing all power sources (voltage source is
shorted and current source is opened)
3. Draw the Thevenin circuit
4. Find the the ITH

Norton’s Theorem
Any linear circuit consisting several energy sources and resistance or complex circuit can be
replaced by a single constant current generator in parallel with a single resistor. (see Figure 5)

Electrical Technology Work Book 29

Figure 19

Step to apply the Norton’s Theorem

1. Terminate RL and find IN
2. Find RN by removing all power sources (shorted voltage and opened current)
3. Draw the Norton’s equivalent circuit
4. Find IL

Load Power

Where = 2

1. PL = Load Power
2. IL = Load Current
3. RL = Load Resistance

30 Electrical Technology Work Book

EXERCISE

Question 1

Based on Figure 20, determine the voltage drop, V3 at resistance, R3 by using Kirchhoff’s Voltage
Law (KVL)

Answer Figure 20
Apply KVL Theorem
Σ = 0
Question 2 20 + 10 + 3 − 45 = 0
3 = 45 − 30 = 15

Refer to Figure 21, determine the voltage drop, V2 at resistance, R2 by using Kirchhoff’s Voltage
Law (KVL)

Answer Figure 21
Apply KVL Theorem
Σ = 0
20 + 2 + 15 − 45 = 0
3 = 45 − 35 = 10

Electrical Technology Work Book 31
Question 3
Find the value of voltage drop Vx for a circuit in Figure 22 using KVL Theorem.

Answer Figure 22
Apply KVL Theorem
Σ = 0
Question 4 20 + + 15 − 45 = 0
= 45 − 35 = 10

Calculate the voltage drop, V3 at resistance R3 in Figure 23 by using KVL theorem.

Answer Figure 23
Apply KVL Theorem
Σ = 0
20 + 10 + 3 − 45 = 0
3 = 45 − 30 = 15

32 Electrical Technology Work Book
Question 5
Use Kirchhoff’s Current Law (KCL) to find Current Supply, Is in Figure 24.

Figure 24

Answer

Apply KVL Theorem Σ = 0
Question 6 − 1 − 2 = 0
− 8 − 4 = 0
= 8 + 4 = 12

By using Kirchhoff’s Current Law (KCL) to determine I3 in Figure 25.

Answer Figure 25
Apply KVL Theorem
Σ = 0
3 + 4 − 1.5 − 3 = 0
3 = 7 − 1.5 = 5.5

Electrical Technology Work Book 33

Question 7

Find the current in each junction of the circuit in Figure 26 by using Kirchhoff’s Current Law (KCL).

Figure 26

Answer

To calculate I1, we need to find Rtotal.

1 = 1 + ( 2 1 3)
1 +

1 = 1 + (1 1 2) = 1 + 1
2 + 2 3

Then, we use Ohm’s Law to find I1 = 1.2Ω

=

1 = = 8 = 6.67
1.2

To find, I2 we use Ohm’s Law

2 = = 8 = 4
1 2

Apply Kirchhoff’s Current Law (KCL) to find I3

Σ = 0
1 − 2 − 3 = 0
6.67 − 4 − 3 = 0

3 = 2.67

34 Electrical Technology Work Book

Question 8

Figure 27 shows an electric circuit. Calculate every current, I1, I2, I3, I4, I5 and I6 using Kirchhoff’s
Current Law (KCL).

Figure 27

Answer Σ = 0
To find I1, we can apply KCL at node a 6 + 2 − 1 = 0
To find I2, we can apply KCL at node b
1 = 8
To find I5, we can apply KCL at node c
To find I3, we can apply KCL at node f Σ = 0
To find I4, we can apply KCL at node e 1 − 2 − 1 = 0
8 − 2 − 1 = 0
To find I5, we can apply KCL at node d
2 = 7

Σ = 0
1 − 3 + 5 = 0

5 = 2

Σ = 0
3 + 3 − 12 = 0

3 = 9

Σ = 0
2 + 4 − 3 = 0
7 + 4 − 9 = 0

4 = 2

Σ = 0
6 − 4 − 2 = 0
6 − 2 − 2 = 0

6 = 4

Electrical Technology Work Book 35
Question 9
Find the current flow through each resistor using mesh analysis for the circuit in Figure 28

Figure 28

Answer
Step 1:
Assign a direction of current to each closed loop of the circuit

Step 2:

Apply KVL around each closed loop of the circuit

Loop 11

Loop 12 20 − 3 2 − 4( 2 + 1) = 0

20 − 3 2 − 4 2 − 4 1 = 0
20 − 7 2 − 4 1 = 0
7 2 + 4 1 = 20 … (1)

Step 3: 10 − 1 − 4( 2 + 1) = 0
10 − 1 − 4 2 − 4 1 = 0

10 − 4 2 − 5 1 = 0
4 2 + 5 1 = 10 … (2)

Solving the simultaneous linear equations (1) and (2) using matrix

7 2 + 4 1 = 20 … (1)
4 2 + 5 1 = 10 … (2)

36 Electrical Technology Work Book

Matrix form

[45 47] [ 12] = [1200]
∆= |45 47| = 16 − 35 = −19
∆ 1 = |2100 47| = 80 − 70 = 10
∆ 2 = |54 1200| = 40 − 100 = −60
∆ 2 −60
2 = ∆ = −19 = 3.16

1 = ∆ 1 = 10 = −0.53
∆ −19
Apply KCL to find I3

3 = 1 + 2 = −0.53 + 3.16 = 2.63
Alternative solution for simultaneous linear equations

Equations (1) X 5 (7 2 + 4 1 = 20) × 5
Equations (2) X 4 (4 2 + 5 1 = 10) × 4

Equations (3) - equations (4) 35 2 + 20 1 = 100 … (3)
16 2 + 20 1 = 40 … (4)

Replace I2 = 3.16A into equation (1) 35 2 + 20 1 = 100
16 2 + 20 1 = 40

19 2 = 60
2 = 3.16

7(3.16) + 4 1 = 20

4 1 = 20 − 22.12
−2.12
1 = 4 = −0.53

Apply KCL to find I3,

3 = 1 + 2 = 3.16 + (−0.53) = 2.63

Electrical Technology Work Book 37
Question 10
Find the current for every junction in Figure 29 using Nodal Analysis

Figure 29

Answer

Apply KCL at Va 2 − 3 + 1 = 0
Nodal at Va

20 − − + 10 − = 0
3 4 1
20 10
3 − 3 − 4 + 1 − 1 = 0

50 − (13 + 1 + 1) = 0
3 4
(1129) 50
= 3

= 50 × 12 = 600 = 10.53
3 × 19 57
10 − 10 − 10.53
1 = 1 = 1 = −0.53

2 = 20 − = 20 − 10.53 = 3.16
3 3
10.53
3 = 4 = 4 = 2.63

38 Electrical Technology Work Book
Question 11
Find the current in the circuit in Figure 30 using Mesh Analysis

Figure 30

Answer

Apply KVL around each closed loop of the circuit

Loop 11

Loop 12 10 − 10 1 − 40( 1 − 2) = 0

10 − 10 1 − 40 1 + 40 2 = 0
10 − 50 1 + 40 2 = 0
50 1 − 40 2 = 10 … (1)

−20 2 − 30 2 − 40( 2 − 1) = 0

−50 2 − 40 2 + 40 1 = 0
40 1 − 90 2 = 0 … (2)

Solving equation 1 and equation 2 using matrix

Matrix Form 50 1 − 40 2 = 10 … (1)
40 1 − 90 2 = 0 … (2)

Determinant [4500 −−9400] [ 12] = [100]

∆= |4500 −−4900| = −4500 − (−1600) = −2900

Electrical Technology Work Book 39

Determinant for I1 ∆ 1 = |100 −−9400| = −900 − (0) = −900
Determinant for I2
Therefore, ∆ 2 = |5400 100| = 0 − 400 = −400

Apply KCL to find I3 2 = ∆ 2 = −400 = 0.138
Question 12 ∆ −2900
∆ 1 −900
1 = ∆ = −2900 = 0.310

3 = 1 − 2 = 0.310 − 0.138 = 0.172

Calculate the current flow, I3 through R3 in Figure 31 using Thevenin's Theorem.

Figure 31

Answer

Terminate the load resistor, R3 and find the Thevenin’s source, VTH

= 1 (1 + 2) × 8
+ (1 + 2)

= 6

Find the Thevenin resistance, RTH by removing all power source (voltage source is shorted)

40 Electrical Technology Work Book

= 1 ∕∕ ( 2 + 4)

= 1 ∕∕ (1 + 2)
1 1 1
= 1 + 3

= 1 × 3 = 3 = 0.75Ω
1 + 3 4
Draw the Thevenin equivalent circuit

Find the the ITH (Use Ohm’s Law)

= 3 = 6 2 = 2.18
+ 0.75 +

Electrical Technology Work Book 41
Question 13
Find the current load, IL using the Thevenin’s Theorem for the circuit in Figure 32.

Figure 32

Answer

Terminate the load resistor, RL and find the Thevenin source, VTH

= 4 4 1 × 10
+
= 8

Find the Thevenin resistance, RTH by removing all power sources (voltage source is shorted)

= 2 + ( 3|| 1)

42 Electrical Technology Work Book

= 3 + (4||1)
(4 × 1)
= 3 + (4 + 1) = 3.8Ω

Draw the Thevenin circuit

Find the the ITH (Use Ohm’s Law)

= = 8 2 = 1.38
+ 3.8 +

Question 14

Find the current flow in R4 for the circuit in Figure 33 using Thevenin’s Theorem.

Figure 33

Answer
Terminate the load resistor, RL and find the Thevenin source, VTH

Electrical Technology Work Book 43

Apply Ohm’s Law to find I3 = 3

3 = 1 = 3
3 = 3 3 = 3 × 2 = 6

Find the Thevenin resistance, RTH by removing all power source (current source is opened)

Draw the Thevenin circuit = 3 + 2
= 2 + 3 = 5Ω

Find the Thevenin’s current ITH (Use Ohm’s Law)

= = 5 6 4 = 0.67
+ +

44 Electrical Technology Work Book
Question 15
Find the current, I2 in the load resistance, RL in Figure 34 using Norton’s Theorem

Figure 34

Answer
Terminate RL and find IN

Firstly, find the Rtotal

= 1 + ( 3 ∕∕ 2)

= 1 + (3 ∕∕ 2)
3 × 2
= 1 + 3 + 2 = 2.2Ω

Next, find I1 by using Ohm’s Law

1 = = 8 = 3.64
2.2

To find IN, apply Current Divider Rule (CDR)

= 2 = 3 3 2 × 3 = 3 3 2 × 3.64 = 2.184
+ +

Find RN by removing all power sources (shorted voltage and opened current)

Electrical Technology Work Book 45

= 2 + ( 3 ∕∕ 1)

= 2 + (3 ∕∕ 1)
3 × 1
= 2 + 3 + 1 = 2.75Ω

Draw the Norton’s equivalent circuit

Find IL
Apply CDR to find IL

= × = 2.75 4 × 2.184 = 0.89
+ 2.75 +

46 Electrical Technology Work Book
Question 16
Apply Norton’s Theorem to find the current flow, I2 through resistor 4Ω in Figure 35.

Figure 35

Answer
Terminate RL and find IN

Apply Current Divider Rule (CDR) to find IN

= 2 = 3 × 3 = 2 2 3 × 3 = 1.2
3 + 2 +

Find RN by removing all supply (shorted voltage and opened current)

= 2 + 3

Electrical Technology Work Book 47

Draw the Norton’s equivalent circuit = 2 + 3 = 5Ω

Find IL
Apply CDR to find IL

= × = 5 5 4 × 1.2 = 0.67
+ +

48 Electrical Technology Work Book

CHAPTER THREE: CAPACITOR AND CAPACITANCE

SHORT NOTE

Formula Capacitance

, =

= ( )

= ( )

= ℎ

Capacitor Connected In Series

1 = 1 + 1 + 1
1 2 3
Capacitor Connected In Parallel

= 1 + 2 + 3

Electrical Technology Work Book 49

Charging Process In Capacitor

Discharging Process In Capacitor

50 Electrical Technology Work Book

Fully Discharge Is

5 = ×
= ( )

= ℎ (Ω)

Energy Stored In A Capacitor

= 1 × 2( )
2
= ( )

= ( )

= ( )