EE 8 BASIC ELECTRICAL ENGINEERING 1

1 | P a g e URS-IM-AA-CI-0465 Rev 00 Effective Date: August 24, 2020

BASIC ELECTRICAL ENGINEERING EE 8 UNIT

1

BASIC

ELECTRICAL ENGINEERING

Author

Joshua P. Tejada, REE, RME Michael L. Pascua, REE, RME Marvin P. Amoin, ECE, Ph.D

Marlon A. Bautista, REE, RME, MAT John Neil B. Herrera, REE Rene A. Ariston, REE, RME Roy C. Andrada, REE, MIT



1

To the College of Engineering student: Welcome to EE 8 (Basic Electrical Engineering). This course guide will be the alternative assessment for the knowledge and skills you attained from the electrical engineering lessons and practical guides. This course guide covers the requirements, schedule of assignments, SAQs and Activities throughout the semester.

I am Engr. Joshua P. Tejada your course coordinator for EE 8. I am currently a Part Time Instructor in the College of Engineering of this university, a freelance Electrical System Designer and Consultant for different companies and was a solo book author of Electrical Shop Theory and Practice.

The other members of the course are: Engr. Michael L. Pascua is a professor in the college of engineering and an

engineer in University Physical Facilities Development Unit of this university. He was a Registered Electrical Engineer and a Registered Master Electrician at the same time and authored books Electrical Circuits I along with Engr. Bautista and Basic Electrical Engineering along with Engr. Tejada.

Dr. Marvin P. Amoin is the current campus director of the University of Rizal System Morong Campus. He was a graduate of BS in Electronics and Communications Engineering in Rizal Technological University and graduates his Masters and Doctoral degree in University of Rizal System. He authored the book entitled Advance Engineering Mathematics and Engineering Economy.

Engr. Marlon A. Bautista is the former dean of College of Engineering and a former Program Head for Electrical Engineering Department for 15 years. He authored many books such as Mathematics in the Modern World, Integral Calculus, Discrete Mathematics, and many more.

Engr. John Neil Herrera is the current president of the college of engineering faculty. At present, he takes his graduate studies in EARIST. He was also a former advisor for the student organization of electrical engineering,

Engr. Roy C. Andrada is a Master in Information Technology and a Registered Electrical Engineer. He is also the research coordinator for electrical engineering program. Engr. Andrada was a former Program head of the Electrical Engineering Department.

Engr. Rene A. Ariston is a Part Time Instructor which link the industry knowledge with academe. Currently, he was working in Solid Cement Corporation- Cemex Philippines which he works as Cement Mill Operations at CCR Shift Supervisor. He also was a Professional Electrical Engineer Aspirant 2020 of the IIEE Metro East Chapter.



1 Course Description

EE 8 (Basic Electrical Engineering) deals with the general principles and methods in electrical circuits, Ohm’s law, as well as the use of instruments for electrical applications such as multimeter/tester, and resistor laboratory activities.

Course Objectives General Objectives: Apply the principles and methods of Electrical Engineering

Circuits in different situations concerning electricity.

Specific Objectives:

At the end of the course, the student should be able to:

1. Analyze the characteristics, components, quantities regarding electrical circuits in direct current application.

2. To understand the flow of electricity in different methodology such as: 2.1 kirchhoff’s Law 2.2 Maxwell’s Mesh 2.3 Superposition Theorem 2.4 Millman’s Theorem 2.5 Nodal Node Voltage Analysis

3. To differentiate the procedures and considerations in Alternating Current to the Direct Current.

4. Be able to apply the basics of electrical engineering to the different fields of engineering.



1 Course Structure

The course EE8 consists of three (3) Instructural Units divided into fifteen (15) Modules namely:

Unit 1 – Basics of Electrical Engineering Circuits

Module Writers

1. Introduction - Michael L. Pascua, REE, RME 2. The Power, Energy and Ohm’s Law - Roy C. Andrada, REE, MIT 3. The Multimeter - Joshua P. Tejada, REE, RME 4. The Resistor - John Neil Herrera, REE 5. Batteries - Marlon A. Bautista, REE, RME, MAT

Unit 2 – Network Theorems Laws and Methods

Module Writers

1. Kirchhoff’s Law - John Neil Herrera, REE 2. Maxwell Mesh - Rene A. Ariston, REE, RME 3. Superposition Theorem - Michael L. Pascua, REE, RME 4. Millman’s Theorem - Joshua P. Tejada, REE, RME 5. Nodal Node Voltage Analysis - Marvin P. Amoin, ECE, Ph.D

Unit 3 – AC Circuits

Module Writers

1. AC Circuit Introduction - Marvin P. Amoin, ECE, Ph.D 2. Impedance and Admittance - Michael L. Pascua, REE, RME 3. Network Theorems in AC Circuits - Marlon A. Bautista, REE, RME, MAT 4. Resonance - Joshua P. Tejada, REE, RME 5. Power factor Correction - Roy C. Andrada, REE, MIT



1 Course Requirements

The following are the major course requirements:

1. The circuit familiarization activities

Each student is required to conduct activities regarding electrical circuits. This activity is equivalent to one laboratory activity. This activity may be one of the following.

1.1 Create your own electrical circuit design 1.2 Breadboard activities 1.3 Simulation of circuits using simulation applications

2. Examination, quizzes and self-assessment questions. 3. Other requirements enumerated by the instructor.

GRADING SYSTEM Basic Electrical Engineering is a three (3) unit subject. Final grade will be based

on your performance in the written assignments and reports for the lecture (40% or 1 units) and laboratory activities (60% or 2 units).

Lecture (40%)

• Examination 40% • Quizzes 30% • Activities 20% • Class Participation / Recitation 10%

Laboratory (60%)

• Examination 40% • Activities 40% • Class Participation / Recitation 20%



1 RUBRIC FOR ASSESSMENT

CRITERIA EXEMPLARY

4

SATISFACTORY

3

DEVELOPING

2

BEGINNING

1

RATING

Comprehensive Ability

The score of the SAQ and Activities is around 90% to 100% Correct.




Workmanship The neatness of the solution for the SAQ and Activities is very good quality.

The neatness of the solution for the SAQ and Activities is good quality.

The neatness of the solution for the SAQ and Activities is standard quality

The neatness of the solution for the SAQ and Activities needed improvement

Accuracy The submitted work manifests qualities which go beyond the requirements

The submitted work manifest the required qualities

The submitted work partially manifest the required qualities. Certain aspects are either incomplete or incorrect.



1 ACADEMIC INTEGRITY

Due to COVID – 19 pandemic, alternative learning sessions must be produced. Even the course is delivered online, as a student you are expected to observe intellectual honesty at all times. This means doing your assignments and term papers yourself, acknowledging indebtedness to sources of original ideas and quoted materials which you used, and taking examinations yourself.

SCHEDULE

The schedule making depends on the schedule given by the memorandum from the university. The time allotment from the syllabus can be also a factor but the huge factor for the scheduling is on the instructor. The instructor can give his/her schedule of dates for submission of different requirements.



1

Unit 1 Basics of Electrical

Engineering Circuits



1

Module 1: Introduction Table of Contents Introduction

Lesson 1: Electron Theory

Lesson 2: Circuit Elements

Lesson 3: Circuit Parameters

Lesson 4: The Resistance



1 Lesson 1: Electron Theory

Matter is whatever has mass and volume that occupies space. The fundamental unit of matter is the atom. The atom is basically consisting of the Nucleus and orbits. The orbit contains electron, while the Nucleus contains proton and neutron. Proton and Neutron are often called nucleons because they are found inside the nucleus. The number of protons usually gives the atomic number of an element, while the sum of protons and neutrons gives the atomic mass of an element.

𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵𝑵 𝒐𝒐𝒐𝒐 𝑵𝑵𝒆𝒆𝑵𝑵𝒆𝒆𝒆𝒆𝑵𝑵𝒐𝒐𝒆𝒆𝒆𝒆 𝒊𝒊𝒆𝒆 𝒂𝒂𝒆𝒆𝒂𝒂 𝑶𝑶𝑵𝑵𝑵𝑵𝒊𝒊𝒆𝒆 = 𝟐𝟐𝒆𝒆𝟐𝟐

Where: n = the number of the desired orbit of the material

Example 1. Find the maximum number of electron in the 4th orbit

Solution. N = 2n2 = 2 x 42 = 32 electrons.

Example 2. If the maximum number of electron in the orbit is 8, what orbit contains those electron?

Solution. n = sqrt(N/2) = sqrt(8/2) = 2. It means it is in the 2nd Orbit



1 Self-Assessment Questions

Answer the following neatly

Find the maximum number of electron in the 3rd orbit

If the maximum number of electron in the orbit is 2, what orbit contains those electron?

Activity 1

On a clean sheet of paper, answer the following neatly. Send a photograph of your work to your designated GDrive

1. Find the maximum number of electron in the 5th orbit 2. Find the maximum number of electron in the 4th orbit 3. If the maximum number of electron in the orbit is 72, what orbit contains those

electron?



1 Answers to Self-Assessment Questions 1

Find the maximum number of electron in the 3rd orbit

18 electrons

If the maximum number of electron in the orbit is 2, what orbit contains those electron?

1st Orbit



1 Lesson 2: Circuit Elements

Figures are from

book “ Basic Electrical Engineering by Joshua Tejada and Michael Pascua ”

Source – Provides the necessary voltage to force the current to flow throughout the circuit.

Conductors – The path where the current flows though out the circuit. It may be copper, aluminium, silver or any low resistance materials.

Protection – It is used to protect the load and the circuit from the sudden change in current.

Load – The purpose of electrical energy is to be used in many ways, to convert into other kinds of energy.

Control Device – The control of the circuit when and when not to let current pass though the circuit.



1 Self-Assessment Questions 2

Identify the following symbols

Activity 2

On a clean sheet of paper, draw the symbols with their identification for familiarization of electrical symbols. Send your output to your designated GDrive



1

Answers to Self-Assessment Questions 2

Source: Basic Electrical Engineering by J. Tejada and M. Pascua



1 Lesson 3: Circuit Parameters Electric Charge

The electric charge represents the volume of electricity. The electric charge is denoted as “Q” or “q”. the lowercased letter denotes time varying quantities while uppercased letters are not. The unit for Electrical Charge is Coulomb (C).

𝟏𝟏 𝑪𝑪𝒐𝒐𝑵𝑵𝒆𝒆𝒐𝒐𝑵𝑵𝑵𝑵 = 𝟔𝟔.𝟐𝟐𝟐𝟐 𝒙𝒙 𝟏𝟏𝟏𝟏𝟏𝟏𝟐𝟐 Electrons

Example. If the total number of charge pass through a conductor is 13µ Coulomb. How many electrons it contains?

Solution. 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 = 𝑄𝑄 𝑥𝑥 6.28𝑥𝑥1018 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 𝑝𝑝𝑁𝑁𝑁𝑁 𝐶𝐶𝑜𝑜𝑁𝑁𝐸𝐸𝑜𝑜𝑁𝑁𝑁𝑁

𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 = 13𝑥𝑥10−6 𝐶𝐶𝑜𝑜𝑁𝑁𝐸𝐸𝑜𝑜𝑁𝑁𝑁𝑁 𝑥𝑥 6.28𝑥𝑥1018 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 𝑝𝑝𝑁𝑁𝑁𝑁 𝐶𝐶𝑜𝑜𝑁𝑁𝐸𝐸𝑜𝑜𝑁𝑁𝑁𝑁

𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 = 8.164𝑥𝑥1013 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸

Electric Current

The electric current is the rate of flow of current at a point per unit of time and is denoted as “I” or “I” because it is also called Current Intensity. The unit of electrical current is Ampere (A). To produce a one (1) ampere of current, it requires a one coulomb or 6.28 x 1018 Electrons to pass a point in a second. The steady state of current is given by the formula:

𝑰𝑰 = 𝑸𝑸/𝒆𝒆

Where: I – Electric Current

Q – Electric Charge

t - Time (seconds)

Example. When checking through a circuit, it observes that a current of 200 mA is passing through a circuit in 1.2 seconds. Find the total charge in the circuit.

Solution. 𝑄𝑄 = 𝐼𝐼𝐸𝐸 = 0.2𝐴𝐴 𝑥𝑥 1.2𝐸𝐸 = 𝟏𝟏.𝟐𝟐𝟐𝟐𝑪𝑪

The Voltage

are the force that pushes the electrons to move. It may be denoted as “V” or “E”. The unit for voltage is volts (V). To produce a voltage of 1 volt, you need an electrical energy of 1 Joule divided by 1 Coulomb of charge. Mathematically, it is expressed as:

𝑽𝑽 = 𝑾𝑾/𝑸𝑸



1 Where: V – Voltage

W = Work / Energy

Q = Electric Charge

Example. In a given circuit, the available energy is 2 joules where the system voltage is 1.2v. How many available electrons are in the circuit?

Solution. 𝑄𝑄 = 𝑊𝑊/𝑉𝑉 = 2𝐽𝐽 / 1.2𝑉𝑉 = 1.667𝐶𝐶

𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 = 𝑄𝑄 𝑥𝑥 6.28𝑥𝑥1018 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 𝑝𝑝𝑁𝑁𝑁𝑁 𝐶𝐶𝑜𝑜𝑁𝑁𝐸𝐸𝑜𝑜𝑁𝑁𝑁𝑁

𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 = 1.667 𝐶𝐶𝑜𝑜𝑁𝑁𝐸𝐸𝑜𝑜𝑁𝑁𝑁𝑁 𝑥𝑥 6.28𝑥𝑥1018 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 𝑝𝑝𝑁𝑁𝑁𝑁 𝐶𝐶𝑜𝑜𝑁𝑁𝐸𝐸𝑜𝑜𝑁𝑁𝑁𝑁

𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑜𝑜𝑜𝑜 𝐸𝐸𝐸𝐸𝑁𝑁𝐸𝐸𝐸𝐸𝑁𝑁𝑜𝑜𝐸𝐸𝐸𝐸 = 𝟏𝟏.𝟏𝟏𝟐𝟐𝟔𝟔𝟎𝟎𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑬𝑬𝒆𝒆𝑵𝑵𝒆𝒆𝒆𝒆𝑵𝑵𝒐𝒐𝒆𝒆𝒆𝒆





If the total number of charge pass through a conductor is 2.2 Coulomb. How many electrons it contains?

The total number of Electrons passes through is 2.164x1018. The total charge available is?

If the total number of charge pass through a conductor is 2.5 Coulomb in 3 seconds, find is the current passing in the conductor.

When checking through a circuit, it observes that a current of 200 mA is passing through a circuit in 1.2 seconds. Find the total charge in the circuit.

If the total energy available is 3 Joules, find the total number of charge if the potential difference available is 2 volts.

Activity 3

On a clean sheet of paper, answer the following neatly. Send a photograph of your work to your designated GDrive

1. If the total number of charge pass through a conductor is 2000 nC. How many electrons it contains?

2. If the total number of charge pass through a conductor is 12 mC. How many electrons it contains?

3. If the total number of charge pass through a conductor is 1.6 Coulomb in 10 seconds, find is the current passing in the conductor.

4. The total number of Electrons passes through is 9.62x1018 in 2 seconds. Find is the current passing in the conductor.

5. In a given circuit, the available energy is 3000 mJ where the system voltage is 3.6V. How many available electrons are in the circuit?

6. The total energy available is 300 Joules; find the total number of charge if the potential difference available is 230 volts.



1


If the total number of charge pass through a conductor is 2.2 Coulomb. How many electrons it contains?

𝟏𝟏.𝟑𝟑𝟐𝟐𝟏𝟏𝟔𝟔𝒙𝒙𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝑬𝑬𝒆𝒆𝑵𝑵𝒆𝒆𝒆𝒆𝑵𝑵𝒐𝒐𝒆𝒆𝒆𝒆

The total number of Electrons passes through is 2.164x1018. The total charge available is?

0.345C

If the total number of charge pass through a conductor is 2.5 Coulomb in 3 seconds, find is the current passing in the conductor.

𝟐𝟐𝟑𝟑𝟑𝟑𝑵𝑵𝟖𝟖

When checking through a circuit, it observes that a current of 200 mA is passing through a circuit in 1.2 seconds. Find the total charge in the circuit.

𝟏𝟏.𝟐𝟐𝟐𝟐𝑪𝑪

If the total energy available is 3 Joules, find the total number of charge if the potential difference available is 2 volts.

𝟏𝟏.𝟓𝟓𝑪𝑪

The total energy available is 3 Joules, find the total number of charge if the potential difference available is 2 volts.

𝟔𝟔.𝟏𝟏𝟏𝟏𝟗𝟗



1 Lesson 3: Circuit Parameters The Resistance

The resistance is the opposition given by a material to the flow of electricity. The resistance is denoted as “R” and has the unit of ohms (Ω).

The Cross Sectional Area of Wires

Wire also offers electrical resistance and the cross sectional area of wire gives a major factor to this. The size of wire is denoted by its cross sectional area such as mm2 for metric, Circular Mil for English and American Wire Gauge (AWG) for US. Mils and Circular mils are often used and the conversion for it is enumerated as follows

1000 𝑀𝑀𝑀𝑀𝐸𝐸 = 1 𝑀𝑀𝐸𝐸𝐸𝐸ℎ

(𝑑𝑑𝑀𝑀𝑀𝑀𝑀𝑀)2 = 1 𝐶𝐶𝑀𝑀𝑁𝑁𝐸𝐸𝑁𝑁𝐸𝐸𝐶𝐶𝑁𝑁 𝑀𝑀𝑀𝑀𝐸𝐸

1𝑀𝑀𝐶𝐶𝑀𝑀 = 1000 𝐶𝐶𝑀𝑀𝑁𝑁𝐸𝐸𝑁𝑁𝐸𝐸𝐶𝐶𝑁𝑁 𝑀𝑀𝑀𝑀𝐸𝐸 𝜋𝜋4

(𝐶𝐶𝑀𝑀𝑁𝑁𝐸𝐸𝑁𝑁𝐸𝐸𝐶𝐶𝑁𝑁 𝑀𝑀𝑀𝑀𝐸𝐸) = 1 𝑆𝑆𝑆𝑆𝑁𝑁𝐶𝐶𝑁𝑁𝑁𝑁 𝑀𝑀𝑀𝑀𝐸𝐸

Example 1. Convert 20 Mil wire to inches.

Solution. 20 𝑀𝑀𝑀𝑀𝐸𝐸 𝑥𝑥 1 𝑀𝑀𝐸𝐸1000 𝑀𝑀𝑀𝑀𝐸𝐸 = 𝟏𝟏.𝟏𝟏𝟐𝟐 𝒊𝒊𝒆𝒆

Example 2. Convert 3 inches diameter wire to mils.

Solution. 3 in 𝑥𝑥 1000 𝑀𝑀𝑀𝑀𝐸𝐸1 𝑀𝑀𝐸𝐸 = 𝟑𝟑𝟏𝟏𝟏𝟏𝟏𝟏 𝑴𝑴𝒊𝒊𝒆𝒆

The Resistance offered by Wires

It would be expected that a wire's electrical resistance would be greater for a longer wire, less for a wire of a larger cross sectional area, and would be expected to depend on the material from which the wire is produced. Mathematically, it is expressed as:

𝑅𝑅 = 𝜌𝜌𝐸𝐸𝐴𝐴 =

𝜌𝜌𝑉𝑉𝐴𝐴2 =

𝜌𝜌𝐸𝐸2

𝑉𝑉

Where: R – Resistance offered

𝜌𝜌 – Resistivity ( Ω-m, 𝛺𝛺−𝐶𝐶𝑀𝑀𝑓𝑓𝑓𝑓

)

𝐸𝐸 - Length (m, ft)

𝐴𝐴 - Area (𝑁𝑁2, ft2, CM)



1 𝑉𝑉 - Volume (𝑁𝑁3, ft3)

MATERIAL RESISTIVITY (at 20°C, in unit of 10-8 Ω-m)

Aluminum 2.83 Brass 6.16

Carbon (graphite) 800 Copper (annealed) 1.72

Copper (hard-drawn) 1.77 Iron (electrolytic) 9.95

Lead 22 Mercury 96 Silver 1.64 Gold 2.44

Tungsten 4.37 Table 1.1; Resistivity of Metals

Source: basic Electrical Engineering by J. Tejada and M. Pascua

Example 3. Find the resistance of an aluminium wire of length 1.5 km long and has a diameter of 0.1 m.

Solution. 𝐴𝐴 = 𝜋𝜋𝑑𝑑2

4= 𝜋𝜋(0.1𝑚𝑚)2

4= 𝜋𝜋

400𝑁𝑁2

𝑅𝑅 = 𝜌𝜌𝑀𝑀𝐴𝐴

= 2.83𝑥𝑥10−8Ω−m(1500𝑚𝑚)𝜋𝜋400𝑚𝑚

2 = 𝟓𝟓.𝟐𝟐𝒙𝒙𝟏𝟏𝟏𝟏−𝟑𝟑𝜴𝜴

The effect of Temperature in Resistance

Experiments have shown that resistance of all wires commonly used in electrical systems in operation is increasing as the temperature rises. The formulas for the effect of Temperature are enumerated as follows.

𝑅𝑅1/𝑅𝑅2 = (|𝑇𝑇| + 𝐸𝐸1) / (|𝑇𝑇| + 𝐸𝐸2) 𝑅𝑅2/𝑅𝑅1 = 1 + 𝛼𝛼𝑓𝑓1∆𝐸𝐸 𝛼𝛼𝑓𝑓1 = 1 / (|𝑇𝑇| + 𝐸𝐸1)

∆𝐸𝐸 = 𝐸𝐸2 − 𝐸𝐸1

Where: 𝑅𝑅1 – Initial Resistance

𝑅𝑅2 – Final Resistance

𝑇𝑇 - Inferred Zero, temperature when resistance of a given material is zero.

𝐸𝐸1 – Initial temperature

𝐸𝐸2 – Final temperature

∆𝐸𝐸– Change in temperature



1 𝛼𝛼 - temperature coefficient of resistance

Material

Inferred Absolute Zero (T), °C

Aluminum -236

Copper, annealed -234.5

Copper, hard-drawn -242

Iron -180

Nickel -147

Silver -243

Steel, soft -218

Tin -218

Tungsten -202 Zinc -250

Table 1.2; Inferred Absolute Temperatures for several Metals Source: basic Electrical Engineering by J. Tejada and M. Pascua

Example 1. A coil of copper wire has a resistance of 54Ω at a room temperature of 20ᵒC. What will be its resistance at 30ᵒC?

Solution. 𝑅𝑅1𝑅𝑅2

= (𝑇𝑇 + 𝑓𝑓1) (𝑇𝑇 + 𝑓𝑓2)

; 𝑅𝑅2 = 𝑅𝑅1(𝑇𝑇 + 𝑓𝑓2) (𝑇𝑇 + 𝑓𝑓1)

= 54 (234.5+ 30) (234.5+ 20)

= 𝟓𝟓𝟔𝟔.𝟏𝟏𝟐𝟐Ω

Example 2. A wire in a substation is made up of 2 inches round, 20 ft. long. If the resistance of the wire is 5.185x10-5 Ω, What is the material used in the wire?

Solution. 𝑅𝑅1𝑅𝑅2

= (𝑇𝑇 + 𝑓𝑓1) (𝑇𝑇 + 𝑓𝑓2)

; 𝐸𝐸2 = 𝑅𝑅2(𝑇𝑇 + 𝑓𝑓1)𝑅𝑅1

− T = (5+15)(236+ 20)5

− 236 = 𝟎𝟎𝟐𝟐𝟐𝟐ᵒ𝐂𝐂

Example 3. What is the temperature rise in Example 2?

Solution. ∆𝐸𝐸 = 𝐸𝐸2 − 𝐸𝐸1 = 788ᵒC − 20ᵒC = 𝟎𝟎𝟔𝟔𝟐𝟐ᵒ𝐂𝐂





1. Convert 160 Mil wire to millimeter. 2. Convert 52 mm diameter wire to mils. 3. Convert 54 Mil wire to MCM. 4. Convert 0.06 MCM wire to Mil. 5. Convert 12π wire to Circular Mil. 6. Convert 48π wire to MCM. 7. Convert 300 mm2 wire to CM. 8. Convert 500 mm2 wire to MCM.

Self-Assessment Questions 5

1. A wire in a substation is made up of 2 inches round, 20 ft. long. If the resistance of the wire is 5.185x10-5 Ω, What is the material used in the wire?

2. Find the length of a copper wire that has a cross sectional area of 0.00025m2. The resistance of the wire is 3.06 ohms.

3. What is the temperature coefficient of resistance of silver at 20ᵒC? 4. At the room temperature of 290.15 kelvin, find the temperature coefficient of

resistance of Zinc.

Activity 4

On a clean sheet of paper, answer the following neatly. Send a photograph of your work to [email protected]

1. A brass wire, that has a cross sectional area of 0. 12m2 and its resistance is 80µ ohms. Find the volume of the brass wire

2. If a 1 ohm copper wire of unknown length passes through drawing dies, the length of wire increased by 2 times of its original length. Determine its new resistance.

3. An iron wire has a resistance of 2.78Ω at room temperature of 20ᵒC. when placed near the engine of a vehicle, the resistance raised to 3.614Ω at its new temperature. What is the change in temperature given that the coefficient of temperature for iron at 20 degree Celsius is 0.005?




Convert 160 Mil wire to millimeter. 𝟐𝟐.𝟏𝟏𝟔𝟔𝟐𝟐𝑵𝑵𝑵𝑵

Convert 52 mm diameter wire to mils. 𝟐𝟐𝟏𝟏𝟐𝟐𝟎𝟎.𝟐𝟐𝟐𝟐 𝑴𝑴𝒊𝒊𝒆𝒆

Convert 54 Mil wire to MCM. 𝟐𝟐.𝟏𝟏𝟏𝟏𝟔𝟔 𝑴𝑴𝑪𝑪𝑴𝑴

Convert 0.06 MCM wire to Mil. 𝟎𝟎.𝟎𝟎𝟓𝟓 𝑴𝑴𝒊𝒊𝒆𝒆

Convert 12π wire to Circular Mil. 𝟐𝟐𝟐𝟐 𝑪𝑪𝒊𝒊𝑵𝑵𝒆𝒆𝑵𝑵𝒆𝒆𝒂𝒂𝑵𝑵 𝑴𝑴𝒊𝒊𝒆𝒆𝒆𝒆

Convert 48π wire to MCM. 𝟏𝟏.𝟏𝟏𝟏𝟏𝟐𝟐 𝑴𝑴𝑪𝑪𝑴𝑴

Convert 300 mm2 wire to CM. 𝟓𝟓𝟏𝟏𝟐𝟐𝟏𝟏𝟓𝟓𝟎𝟎.𝟓𝟓𝟎𝟎 𝑪𝑪𝒊𝒊𝑵𝑵𝒆𝒆𝑵𝑵𝒆𝒆𝒂𝒂𝑵𝑵 𝑴𝑴𝒊𝒊𝒆𝒆

Convert 500 mm2 wire to MCM. 𝟏𝟏𝟐𝟐𝟔𝟔.𝟎𝟎𝟔𝟔𝟐𝟐 𝑴𝑴𝑪𝑪𝑴𝑴


1. A wire in a substation is made up of 2 inches round, 20 ft. long. If the resistance of the wire is 5.185x10-5 Ω, What is the material used in the wire? Copper (annealed)

2. Find the length of a copper wire that has a cross sectional area of 0.00025m2. The resistance of the wire is 3.06 ohms. 44.47kΩ

3. What is the temperature coefficient of resistance of silver at 20ᵒC? 0.0038 4. At the room temperature of 290.15 kelvin, find the temperature coefficient of

resistance of Zinc. 𝟏𝟏.𝟏𝟏𝟏𝟏𝟑𝟑𝟎𝟎𝟐𝟐𝟓𝟓



1

End of Module 1



1

Module 2: The Power, Energy and Ohm’s Law Table of Contents Lesson 1: The Ohm’s Law

Lesson 2: Electrical Power

Lesson 3: Electrical Energy



1

Source: Basic Electrical Engineering By J. Tejada and M. Pascua

Lesson 1: The Ohm’s Law The Ohm’s Law

Ohm’s law is the relationship between Voltage, Current and Resistance. The ohm’s law can be applied in DC circuits as well as AC circuits. It states that “the current flowing in the circuit is proportional to the applied Voltage and inversely proportional to the circuit resistance”. Figure 2.1 shows the different formulas derived from the ohm’s law.

Conditions for Ohm’s Law

Ohm’s Law may be extended either to the entire

circuit or to a circuit section. Ohm’s Law can be applied to DC as well as AC

Circuits. However, in AC circuits, Impedance is used in place of Resistance.

When Ohm’s Law is applied to a branch of a circuit, the branch resistance, voltage or current should be used.

In Ohm’s Law, the temperature must be constant.

Example 1. Find the current I through a resistor of resistance R = 3 Ω if the voltage across the resistor is 6 V.

Solution. Current, 𝐼𝐼 = 𝑉𝑉/𝑅𝑅 = 6𝑉𝑉 / 3Ω = 𝟐𝟐 𝟖𝟖

Example 2. A resistor with a 100 kilo ohms resistance is connected to a 230 volts alternating current. Find the value of the current flowing in the resistor?

Solution. Current, 𝐼𝐼 = 𝑉𝑉/𝑅𝑅 = 230𝑉𝑉/100 Ω = 𝟐𝟐.𝟑𝟑 𝑵𝑵𝟖𝟖




1. Find the voltage V through a resistor of resistance R = 2.1 Ω if the current across the resistor is 6 A.

2. A resistor with a 100 ohms resistance is connected to a 5 amps alternating current. Find the value of the voltage across the resistor?

3. Find the resistance R through a circuit of voltage 12V if the current across the resistor is 6 A.

4. A resistor across a 230V AC is known to produce a 5 amps alternating current. Find the value of the resistor?

Activity 5

On a clean sheet of paper, create ten electrical energy questions and answer it. Send a photograph of your work to your designated GDrive




1. Find the voltage V through a resistor of resistance R = 2.1 Ω if the current across the resistor is 6 A.

2. A resistor with a 100 ohms resistance is connected to a 5 amps alternating current. Find the value of the voltage across the resistor?

3. Find the resistance R through a circuit of voltage 12V if the current across the resistor is 6 A.

4. A resistor across a 230V AC is known to produce a 5 amps alternating current. Find the value of the resistor?



1 Lesson 2: Electrical Power Electrical Power


The Electrical Power is the rate at which electrical energy is being transferred in a given time. It is denoted as “P” or “p” and has the unit of Watts (W). The Electrical Power is usually used as the rating of devices and machines such as motors, pumps and other devices.

Example 1. Find the Power dissipated through a resistor of resistance R = 3 Ω if the voltage across the resistor is 6 V.

Solution. Power, P = 𝑉𝑉2/𝑅𝑅 = (6𝑉𝑉) 2/ 3Ω = 𝟏𝟏𝟐𝟐 𝑾𝑾




1. A resistor with a 100 kilo ohms resistance is connected to a 230 volts alternating current. Find the value of the power dissipated in the resistor?

2. Find the Power dissipated through a resistor of resistance R = 2.1 Ω if the current across the resistor is 6 A.

3. A resistor with a 100 ohms resistance is connected to a 5 amps alternating current. Find the value of the power dissipated in the resistor?

4. Find the Power dissipated through a circuit of voltage 12V if the current across the resistor is 6 A.

Activity 6

On a clean sheet of paper, create ten electrical energy questions and answer it. Send a photograph of your work to your designated GDrive




1. A resistor with a 100 kilo ohms resistance is connected to a 230 volts alternating current. Find the value of the power dissipated in the resistor? 𝟏𝟏.𝟓𝟓𝟑𝟑𝟓𝟓𝑾𝑾

2. Find the Power dissipated through a resistor of resistance R = 2.1 Ω if the current across the resistor is 6 A. 𝟎𝟎𝟓𝟓.𝟔𝟔𝑾𝑾

3. A resistor with a 100 ohms resistance is connected to a 5 amps alternating current. Find the value of the power dissipated in the resistor? 𝟐𝟐.𝟓𝟓𝟓𝟓𝑾𝑾

4. Find the Power dissipated through a circuit of voltage 12V if the current across the resistor is 6 A. 𝟎𝟎𝟐𝟐𝑾𝑾



1

End of Module 2



1

Module 3: The Multimeter Table of Contents Introduction

Lesson 1: The Ohmmeter

Lesson 2: The Voltmeter

Lesson 3: The Ammeter



1 Introduction The Multimeter


A Multimeter or a tester is a necessary tool for an electrical practitioner and technicians. It is used to measure values and is used for trouble shooting. A common Multimeter has 5 parts known as ohmmeter, ammeter, DC voltmeter, AC voltmeter, and continuity/semiconductor tester.

Parts of a Multimeter

1. Pointer – It is the needle indicating the measured value that can be seen on the scale.

2. Scale – It is the list of values to be indicated by the pointer. It is usually consist of three parts, the ohmmeter, ammeter and the voltmeter.

3. Adjustment Screw – It is used to adjust the pointer to zero position.

4. Zero-Ohm adjustment knob – It is used for adjusting the pointer to zero for measuring resistance.

5. Ranges Selector Knob – It is used to choose from different function of the tester, and for selecting the proper multiplier to be used.

6. Test Probe – Are commonly in two colors. Red colored test probe is used for positive terminal while black colored test probe is used for common terminal.



1 Multimeter usage and Maintenance

First and Foremost, Always read the manual oh how to operate the multimeter.

Check if the battery connected is in good condition. Low battery level may affect

the precision of measured values, and also it can affect the internal setting of the multimeter.

Check if the multimeter is set to correct setting and multiplier. Ohmmeter measuring resistance, voltmeter measuring voltage, and ammeter measuring current.

When not in use, Range selector knob must be set on “OFF” or the highest voltage range available. It is to ensure that the battery will not drain.

Do not drop or break the multimeter. The multimeter is not designed as shockproof type of device.



1 Activity 8

Photograph an analog multimeter. Print it or edit it as a photo and identify the different parts of the multimeter. Send your work to your designated GDrive



1 Lesson 1: The Ohmmeter The Ohmmeter


Things to remember

For Analog ohmmeter, four things must be checked before taking any resistance test and enumerated as follows:

1. Select first the proper multiplier in the function selector. 2. Adjust the pointer to zero reading by shorting the test prods together as shown in

figure. It is necessary because without the zero adjustment, it may affect the accuracy of the measurement.

3. The component under test must be DE ENERGIZED during the test. This is to protect both yourself and the device from the electric current. Must be vigilant to ensure the safety of the device and the one using it.

4. You must not touch the metallic part of the test prods to prevent measuring your own body resistance. Although it is not harmful, the ohmmeter reading will be incorrect.

Reading exercise of Ohmmeter

There are many ways you can measure a specific resistance as long you have proper multiplier for the multimeter read resistance value. Suppose we have a 100 ohm resistor and we have to check the actual value of it, we can usex1, x10, and x100 multiplier as shown in the following figures.

Multimeter Display when measuring 100 ohm resistor in x1 multiplier



1 Activity 9

Answer the following values using analog multimeter. Send your work to your designated GDrive



1 Lesson 2: The Voltmeter The Voltmeter


A Voltmeter is used to measure electromotive force. The scale under the ohmmeter readings in the analog Multimeter is used for voltage readings. Common Multimeter has two voltmeter functions: one for AC and one is for DC. The figure shows the set of scales used for measuring electromotive force

Things to remember

For Analog voltmeter, four things must be checked before taking any resistance test and enumerated as follows:

1. Check first the polarity of the source being tested. An analog multimeter is incapable of testing a negative voltage, and if so, it may dislocate the needle of the tester.

2. It must be observed that the test probe will never touch each other when testing. It may cause short circuit.

3. When measuring an unknown voltage using a Voltmeter, it must be set in the highest range and then setting down one by one. This is used to protect the device from measuring values that exceeded the maximum value of the said range.

4. When measuring the voltage drop across a component, it must be tapped in parallel across the component.

Reading exercise of Voltmeter

When measuring voltage, the multiplier must be set in the proper range. Also, the proper range in the multiplier has its own scale in the multimeter screen. In the Philippines, we have a standard voltage of 230v in our outlet. We will try to test it using a multimeter.



1

Multimeter Display when measuring 230v AC in 250 Multiplier

The needle stops between 200 and 250 in the highest voltmeter scale. It means, the voltage is around 200 to 250. For more precise look, it stops one line after the middle that has a value of 225v. So it means it is measuring 230 volts potential.



1 Activity 10

Answer the following values using analog multimeter. Send your work to your designated GDrive



1 Lesson 3: The Ammeter The Ammeter


An Ammeter is used to measure Electrical Intensity or current. The scale under the ohmmeter readings in the analog Multimeter is used for voltage readings. The following figure shows the set of scales used for measuring electrical current.

Things to remember

For Analog ammeter, four things must be checked before taking any resistance test and enumerated as follows:

1. Check first the polarity of the source being tested. An analog multimeter is incapable of testing a negative current, and if so, it may dislocate the needle of the tester.

2. It must be observed that the test probe will never touch each other when testing. It may cause an open circuit.

3. When measuring an unknown current using an ammeter, it must be set in the highest range and then setting down one by one. This is used to protect the device from measuring values that exceeded the maximum value of the said range.

4. When measuring the current across a component, it must be tapped in series across the component.

Reading exercise of Ammeter

When measuring current, the multiplier must be set in the proper range. Also, the proper range in the multiplier has its own scale in the multimeter screen. A normal ammeter in the multimeter has only 0.25A capacity. If your goal is to measure a value above that, it is required to use a current transformer. We will try to test it using a multimeter.



1

Multimeter Display when measuring 10mA in 25DCmA Multiplier

The needle stops exactly at 1.0 in the scale. It terms of 25mA scale, stopping at 2.5 means it has exactly 25mA. Using ratio and proportion, 1.0 at 25mA scale means it has 10DCmA.



1 Activity 11

Read the following values using analog multimeter. Send your work to your designated GDrive



1

End of Module 3



1

Module 4: The Resistor Table of Contents Introduction

Lesson 1: Resistor Color Coding

Lesson 2: Series – Parallel Resistors

Lesson 3: Network Reduction (WYE-DELTA)



1 Introduction The Resistor

Source: Basic

Electrical Engineering By J. Tejada and M. Pascua

Resistors are electronic components made up of high resistive materials and is used to control and limit the current flow to a circuit

Different Types of Resistors

• Wire wound resistors • Metal film resistors • Thick film and Thin film resistors • Network and Surface Mount Resistors • Variable Resistors • Special resistors

Symbol Actual Component



1 Activity 12

In a document file (maybe hardcopy or softcopy), attach the different types of resistors and their identification. Send your work to your designated GDrive



1 Lesson 1: Resistor Color Coding Resistor Color Coding

Carbon Resistors are usually small in size depends on their wattage capacity. Color code is used to represent their resistance value in ohms. Note that BLACK color can never be the first band.


Color Digit Multiplier X10^(-)

Tolerance (%)

Mnemonics

Black 0 0 Bad Brown 1 1 ±1 ( F ) Boys Red 2 2 ±2 ( G ) Race Orange 3 3 Our Yellow 4 4 Young Green 5 5 ±0.5 ( D ) Girls Blue 6 6 ±0.25 ( C ) But Violet 7 7 ±0.1 ( B ) Violeta Gray 8 8 Gave White 9 9 Willingly with None ±20 No Silver -1 ±10 ( K ) Silver and Gold -2 ±5 ( J ) Gold



1 Find the resistance of the resistors

1.

2.


Identify the color codes of the resistor

Activity 13

On a clean sheet of paper, draw and color resistors with values as given. Send your work to your designated GDrive

VALUE 1ST 2ND 3RD 4TH 12 Ω ±10% 3.6 Ω ±1% 740 Ω ±5% 2600 Ω ±10% 100 Ω ±10%

160 kΩ ±10% 130 kΩ ±5% 2.65 Ω ±5%




1.

2.


VALUE 1ST 2ND 3RD 4TH 12 Ω ±10% brown red black silver

3.6 Ω ±1% orange blue gold brown 740 Ω ±5% violet yellow brown gold 2600 Ω ±10% red blue red silver 100 Ω ±10% brown black brown silver

12x10^(5) = 1.2MΩ

100 ohms



1 Lesson 2: Series – Parallel Resistors Series connection of resistors

Series connection of resistors is used to create a chain like circuit that is used to raise the circuit resistance while lowering the circuit current.

Example 1. Find the total Voltage in the Circuit shown

Solution. 𝐼𝐼𝑓𝑓 = 𝐼𝐼1 = 𝐼𝐼2 = 𝐼𝐼3 = 3𝐴𝐴

Voltage at 5Ω Resistor,

𝑉𝑉1 = 𝐼𝐼1𝑅𝑅1 = 3𝐴𝐴 𝑥𝑥 5𝛺𝛺 = 15𝑉𝑉

Voltage at 6Ω Resistor,

𝑉𝑉2 = 𝐼𝐼2𝑅𝑅2 = 3𝐴𝐴 𝑥𝑥 6𝛺𝛺 = 18𝑉𝑉 Voltage at 7Ω Resistor, 𝑉𝑉3 = 𝐼𝐼3𝑅𝑅3 = 3𝐴𝐴 𝑥𝑥 7𝛺𝛺 = 21𝑉𝑉

Getting the total voltage, 𝑉𝑉𝑓𝑓 = 𝑉𝑉1 + 𝑉𝑉2 + 𝑉𝑉3 = 15𝑉𝑉 + 18𝑉𝑉 + 21𝑉𝑉 = 𝟓𝟓𝟐𝟐𝑽𝑽



1 Parallel connection of resistors

is used to reduce the circuit resistance and to get the desired constant voltage per branch. It is commonly used in designing for all the ranch circuit have the same voltages especially in the Lighting outlet and convenience outlet designing

Example 1. Find the total Current in the Circuit shown

Solution. 𝑉𝑉𝑓𝑓 = 𝑉𝑉1 = 𝑉𝑉2 = 𝑉𝑉3

Current at 5Ω Resistor,

𝐼𝐼1 = 𝑉𝑉1/𝑅𝑅1 = 24𝑉𝑉/5𝛺𝛺 = 4.8𝐴𝐴

Current at 6Ω Resistor, 𝐼𝐼2 = 𝑉𝑉2/𝑅𝑅2 =24𝑉𝑉/6𝛺𝛺 = 4𝐴𝐴

Current at 7Ω Resistor, 𝐼𝐼3 = 𝑉𝑉3/𝑅𝑅3 = 24𝑉𝑉/7𝛺𝛺 = 3.43𝐴𝐴

Getting the total current, 𝐼𝐼𝑓𝑓 = 𝐼𝐼1 + 𝐼𝐼2 + 𝐼𝐼3 = 4.8𝐴𝐴 + 4𝐴𝐴 + 3.43𝐴𝐴 = 𝟏𝟏𝟐𝟐.𝟐𝟐𝟑𝟑𝟖𝟖



1 Series – Parallel connection of resistors

are the combination of series circuitry and parallel circuits that is commonly used by technicians to create the best possible circuit design.

Example 1. Find the resistance reading in the ohmmeter given in Figure 4.5A.

Solution. By analysis, The 6 ohms and the 10 ohms are in series with each other. That is also true with 5 ohm and 12 ohm resistors.

Using the equation of resistances in series circuit, the total resistance for the upper branch is

𝑅𝑅𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈 = 6𝛺𝛺 + 10𝛺𝛺 = 16 𝛺𝛺, And for lower Branch,

𝑅𝑅𝐿𝐿𝐿𝐿𝐿𝐿𝑈𝑈𝑈𝑈 = 5𝛺𝛺 + 12𝛺𝛺 = 17 𝛺𝛺.

The circuit may be written as a two parallel resistor circuit as shown in Figure 4.5B.

Solving for Total Resistance,

𝑅𝑅𝑓𝑓 = 1

1𝑅𝑅𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈

+ 1𝑅𝑅𝐿𝐿𝐿𝐿𝐿𝐿𝑈𝑈𝑈𝑈

= 11

16𝛺𝛺+1

17𝛺𝛺 = 8.2424Ω

Figure 4.5A




Answer the following

1. A two identical 15 kΩ resistors are connected in series with a fixed 34 volt power supply. Find the current being supplied in the circuit. Find also the voltage on each resistor.

2. Three 13 kΩ resistors and a 4 kΩ resistor are connected in parallel with a fixed 34 volt power supply. Find the current being supplied in the circuit. Find also the current on each resistor.

Activity 14

On a clean sheet of paper, create your own circuit and answer it. Send your work to your designated GDrive





1. A two identical 15 kΩ resistors are connected in series with a fixed 34 volt power supply. Find the current being supplied in the circuit. Find also the voltage on each resistor. I = 1A, V1 = 15 V V2 = 15 V

2. Three 13 kΩ resistors and a 4 kΩ resistor are connected in parallel with a fixed 34 volt power supply. Find the current being supplied in the circuit. Find also the current on each resistor. IT = 8.508A, IR1 = 2.615mA, IR2 = 2.615mA,

IR3 = 2.615mA, IR4 = 8.5A,



1 Lesson 3: Network Reduction (WYE-DELTA)


In certain circuits resistors are neither in series nor in parallel, so it is difficult to apply the rules for series or parallel circuits. Converting from one circuit type to another may be important for such circuits to simplify the solution.

Example. A circuit consist of three resistors rated: 10 ohms, 12 ohms, and 15 ohms connected in delta. What would be the resistances of the equivalent wye connected load?

Solution.

𝑋𝑋 = 𝐴𝐴𝐶𝐶

𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶=

10𝑥𝑥1210 + 15 + 12

= 𝟑𝟑.𝟐𝟐𝟐𝟐 𝜴𝜴

𝑌𝑌 = 𝐴𝐴𝐵𝐵


10𝑥𝑥1510 + 15 + 12

= 𝟐𝟐.𝟏𝟏𝟓𝟓 𝜴𝜴

𝑍𝑍 = 𝐵𝐵𝐶𝐶


15𝑥𝑥1210 + 15 + 12

= 𝟐𝟐.𝟐𝟐𝟔𝟔 𝜴𝜴

Delta to Wye:

𝑋𝑋 = 𝐴𝐴𝐶𝐶

𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶 𝑌𝑌 =

𝐴𝐴𝐵𝐵𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶

𝑍𝑍 = 𝐵𝐵𝐶𝐶

𝐴𝐴 + 𝐵𝐵 + 𝐶𝐶

Wye to Delta:

𝐴𝐴 = 𝑋𝑋𝑌𝑌 + 𝑌𝑌𝑍𝑍 + 𝑍𝑍𝑋𝑋

𝑍𝑍 𝐵𝐵 =

𝑋𝑋𝑌𝑌 + 𝑌𝑌𝑍𝑍 + 𝑍𝑍𝑋𝑋𝑋𝑋

𝐶𝐶 = 𝑋𝑋𝑌𝑌 + 𝑌𝑌𝑍𝑍 + 𝑍𝑍𝑋𝑋

𝑌𝑌





1. A circuit consist of three resistors rated: 12 ohms, 22 ohms, and 35 ohms connected in delta. What would be the resistances of the equivalent wye connected load?


3. A circuit consist of three resistors rated: 6 ohms, 12 ohms, and 18 ohms connected in wye. What would be the resistances of the equivalent delta connected load?


5. Three resistors of 11-ohm resistance are connected in delta. Inside of it, there is a wye connected resistors rated of 11 ohms each. Find the resistance between any two corners.

Activity 15

On a clean sheet of paper, create your own circuit and answer it. Send your work to your designated GDrive






11.159 Ω, 6.087 Ω, 3.83 Ω


2.1Ω, 3.14 Ω, 1.14 Ω


66 Ω, 33 Ω, 22 Ω


55.5 Ω, 44.4 Ω, 37 Ω

5. Three resistors of 11-ohm resistance are connected in delta. Inside of it, there is a wye connected resistors rated of 11 ohms each. Find the resistance between any two corners. 8.25 Ω, 8.25 Ω, 8.25 Ω



1

End of Module 4



1

Module 5: Batteries Table of Contents Lesson 1: Life of battery

Lesson 2: Series and Parallel Batteries



1 Lesson 1: Life of battery Cells and Batteries

Cells and batteries are components composed of two different metals and a chemical solution such as acids or alkaline. The lifetime of a cell or battery depends on the amount of current it supplies. The greater the current, the shorter battery life. Mathematically, to compute for the life of a battery or cell, the following formula shall be used.

𝐿𝐿𝑀𝑀𝑜𝑜𝑁𝑁(𝐻𝐻𝐿𝐿𝐻𝐻𝑈𝑈𝐻𝐻) =𝐴𝐴𝑁𝑁𝑝𝑝𝑁𝑁𝑁𝑁𝑁𝑁 − 𝐻𝐻𝑜𝑜𝑁𝑁𝑁𝑁 𝑅𝑅𝐶𝐶𝐸𝐸𝑀𝑀𝐸𝐸𝑅𝑅

𝐴𝐴𝑁𝑁𝑝𝑝𝑁𝑁𝑁𝑁𝑁𝑁𝐸𝐸 𝑜𝑜𝑜𝑜 𝐸𝐸𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝐸𝐸𝐸𝐸 𝑝𝑝𝑁𝑁𝑜𝑜𝑑𝑑𝑁𝑁𝐸𝐸𝑁𝑁𝑑𝑑

Example 1. A device is connected to a battery rated 4500mAh. It is found that 250 mA is being consumed by the device. How long will the battery supply the device?

Solution.

𝐿𝐿𝑀𝑀𝑜𝑜𝑁𝑁(𝐻𝐻𝐿𝐿𝐻𝐻𝑈𝑈𝐻𝐻) =𝐴𝐴𝑁𝑁𝑝𝑝𝑁𝑁𝑁𝑁𝑁𝑁 − 𝐻𝐻𝑜𝑜𝑁𝑁𝑁𝑁 𝑅𝑅𝐶𝐶𝐸𝐸𝑀𝑀𝐸𝐸𝑅𝑅

𝐴𝐴𝑁𝑁𝑝𝑝𝑁𝑁𝑁𝑁𝑁𝑁𝐸𝐸 𝑜𝑜𝑜𝑜 𝐸𝐸𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝐸𝐸𝐸𝐸 𝑝𝑝𝑁𝑁𝑜𝑜𝑑𝑑𝑁𝑁𝐸𝐸𝑁𝑁𝑑𝑑

𝐿𝐿𝑀𝑀𝑜𝑜𝑁𝑁(𝐻𝐻𝐿𝐿𝐻𝐻𝑈𝑈𝐻𝐻) =4500𝑥𝑥10−3

250𝑥𝑥10−3

𝐿𝐿𝑀𝑀𝑜𝑜𝑁𝑁(𝐻𝐻𝐿𝐿𝐻𝐻𝑈𝑈𝐻𝐻) = 𝟏𝟏𝟐𝟐 𝑯𝑯𝒐𝒐𝑵𝑵𝑵𝑵𝒆𝒆





1. A device is connected to a battery rated 4000mAh. It is found that 400 mA is being consumed by the device. How long will the battery can supply the device?

2. An LED module is connected to a battery rated 4400mAh. It if the battery can supply the device for 8 hours, how much current is the device drawn?

3. A certain building is suddenly de energized. An emergency light was energized by its internal power source that can light the building for 3 hours. the emergency light has two lamps are drawing 0.625A each. What is the Ampere-Hour rating of the internal power source?

Activity 16

On a clean sheet of paper, create your own battery problem and answer it. Send your work to your designated GDrive





1. A device is connected to a battery rated 4000mAh. It is found that 400 mA is being consumed by the device. How long will the battery can supply the device?

𝟏𝟏𝟏𝟏 𝑯𝑯𝒐𝒐𝑵𝑵𝑵𝑵𝒆𝒆 2. An LED module is connected to a battery rated 4400mAh. It if the battery can

supply the device for 8 hours, how much current is the device drawn? 𝟓𝟓𝟓𝟓 𝑵𝑵𝟖𝟖

3. A certain building is suddenly de energized. An emergency light was energized by its internal power source that can light the building for 3 hours. the emergency light has two lamps are drawing 0.625A each. What is the Ampere-Hour rating of the internal power source?

𝟑𝟑𝟎𝟎𝟓𝟓𝟏𝟏 𝑵𝑵𝟖𝟖𝒎𝒎



1 Lesson 2: Series and Parallel Batteries

Series and Parallel Batteries When the batteries are connected in series, their voltages are added up and their

ampere-hour remains the same.

𝑉𝑉𝑇𝑇 = 𝑉𝑉1 + 𝑉𝑉2 + ⋯+ 𝑉𝑉𝑛𝑛 𝐴𝐴ℎ𝑇𝑇 = 𝐴𝐴ℎ1 = 𝐴𝐴ℎ2 = ⋯ = 𝐴𝐴ℎ𝑛𝑛

When the batteries are connected in parallel, their ampere-hour is added up and their voltages remain the same.

𝑉𝑉𝑇𝑇 = 𝑉𝑉1 = 𝑉𝑉2 = ⋯ = 𝑉𝑉𝑛𝑛 𝐴𝐴ℎ𝑇𝑇 = 𝐴𝐴ℎ1 + 𝐴𝐴ℎ2 + ⋯+ 𝐴𝐴ℎ𝑛𝑛

When the batteries are connected in combination, take consideration of its connection with respect whether it is a series combined with parallel or parallel combined with series.

Example 1. An electric bike has a motor rated 48 volts. How many 12 volts battery is needed for the electric bike to run where the number of 12 volt battery to be used is at minimum?



1 Solution.

𝑉𝑉𝑇𝑇 = 𝑉𝑉1 + 𝑉𝑉2 + ⋯+ 𝑉𝑉𝑛𝑛

𝑉𝑉𝑇𝑇 = 𝐸𝐸 (𝑉𝑉1)

48 = 𝐸𝐸 (𝑉𝑉1)

𝐸𝐸 = 𝟐𝟐 𝑩𝑩𝒂𝒂𝒆𝒆𝒆𝒆𝑵𝑵𝑵𝑵𝒊𝒊𝑵𝑵𝒆𝒆 Example 2. An 12 volt motor rated 12 volts needed to run for 8 hours. If the 12 volt motor drawn 0.5A, how many 12 volt, 500 mAh battery is needed for the motor to run where the number of 12 volt battery to be used is at minimum?

Solution.

𝐴𝐴ℎ𝑇𝑇 = 𝐿𝐿𝑀𝑀𝑜𝑜𝑁𝑁(𝐻𝐻𝐿𝐿𝐻𝐻𝑈𝑈𝐻𝐻)[𝐴𝐴𝑁𝑁𝑝𝑝𝑁𝑁𝑁𝑁𝑁𝑁𝐸𝐸 𝑜𝑜𝑜𝑜 𝐸𝐸𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝐸𝐸𝐸𝐸 𝑝𝑝𝑁𝑁𝑜𝑜𝑑𝑑𝑁𝑁𝐸𝐸𝑁𝑁𝑑𝑑]

𝐴𝐴ℎ𝑇𝑇 = [8ℎ][0.5A]

𝐴𝐴ℎ𝑇𝑇 = 4 𝐴𝐴ℎ

𝐴𝐴ℎ𝑇𝑇 = 𝐴𝐴ℎ1 + 𝐴𝐴ℎ2 + ⋯+ 𝐴𝐴ℎ𝑛𝑛

𝐴𝐴ℎ𝑇𝑇 = 𝐸𝐸 (𝐴𝐴ℎ1)

4 = 𝐸𝐸 (500 mAh)

𝐸𝐸 = 𝟐𝟐 𝑩𝑩𝒂𝒂𝒆𝒆𝒆𝒆𝑵𝑵𝑵𝑵𝒊𝒊𝑵𝑵𝒆𝒆

Example 3. What will be the output for the battery connection given below.

Solution.

Finding for the Voltage Rating,

𝑉𝑉𝑇𝑇 = 𝑉𝑉1 + 𝑉𝑉2

𝑉𝑉𝑇𝑇 = 12 + 12



1 𝐕𝐕𝐓𝐓 = 𝟐𝟐𝟐𝟐𝐕𝐕

Finding for the Ah Rating,

𝐴𝐴ℎ𝑇𝑇 = 𝐴𝐴ℎ1 + 𝐴𝐴ℎ2 + 𝐴𝐴ℎ3

𝐴𝐴ℎ𝑇𝑇 = 60 + 60 + 60

𝐴𝐴ℎ𝑇𝑇 = 𝟏𝟏𝟐𝟐𝟏𝟏 𝟖𝟖𝒎𝒎





1. A remote controller needed a two AAA battery for it to energize. If the AAA battery is rated 1.5 volts, what is the voltage needed by the remote controller to be energized?

2. An amplifier is connected to a 12 volt source. It draws 8.33 Amperes and needed to be run for 8 hours. how many 12 volt, 3500 mAh battery is needed for the amplifier to run where the number of 12 volt battery to be used is at minimum?

3.

Activity 17

On a clean sheet of paper, create your own battery circuits and answer it. Send your work to your designated GDrive





1. A remote controller needed a two AAA battery for it to energize. If the AAA battery is rated 1.5 volts, what is the voltage needed by the remote controller to be energized?

3 Volts 2. An amplifier is connected to a 12 volt source. It draws 8.33 Amperes and needed

to be run for 8 hours. how many 12 volt, 3500 mAh battery is needed for the amplifier to run where the number of 12 volt battery to be used is at minimum?

19 Batteries

3.

36V, 180A-hr



1

End of Module 5



1

End of Unit 1 Basics of Electrical

Engineering Circuits

REFERENCE BOOKS: Available at University of Rizal System Morong

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EE 8 BASIC ELECTRICAL ENGINEERING 1

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