NUR SHAHILLA - NUR ANITA HANIM F O R M E C H A N I C A L E N G I N E E R I N G ( P A R T 1 ) ELECTRICAL TECHNOLOGY quick notes:
Quick Notes: Electrical Technology NUR SHAHILLA BINTI ABD RAHIM Department of Mechanical Engineering Politeknik Kota KInabalu NUR ANITA HANIM BINTI MOHD NIZAM CRIN Department of Mechanical Engineering Politeknik Kota KInabalu for Mechanical Engineering EDITION 2023 Author: Publisher: Politeknik Kota Kinabalu 2023
Copyright Politeknik Kota Kinabalu, 2023 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any mean, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the Copyright owner. Published by: Politeknik Kota Kinabalu, No. 4, Jalan Politeknik, KKIP Barat, 55460 Kota Kinabalu, Sabah https://polikk.mypolycc.edu.my/ Tel: 088-401800 First published 2023 ©Nur Shahilla binti Abd Rahim Quick Notes: Electrical Technology for Mechanical Engineering (Part 1)
Preface Quick Notes: Electrical Technology for Mechanical Engineering (Part 1) is a special reference for lecturers and students of the Mechanical Engineering Department who take the DJJ20053 Electrical Technology course. This book contains short, concise notes and provides step-by-step solutions for every example. Practice questions with answers are provided to make it easier for users to revise.
Table of Contents
CHAPTER 1: Basic Units and Electrical Quantities
Basic Units and Electrical Quantities 1 CHAPTER 1: Basic Units and Electrical Quantities Scientific Notation Engineering Notation Metric Prefixes and SI Units Basic Electrical Quantities 1.1 Scientific Notation Scientific Notation is a special way of writing very large or very small numbers. The number is shortened to a number by multiplied by power of ten. × 10 For example: 1 5 0 0 0 0 1.5 × 105 A number A Scientific Notation Move decimal point FIVE places to the LEFT. Table 1.1 list some power of ten for both positive and negative, and the corresponding decimal numbers. The power of ten is expressed as an exponent of the base 10 in each case. Table 1.1: List Some Power of Ten 109 = 1 000 000 000 10-9 = 0.000 000 001 108 = 100 000 000 10-8 = 0.000 000 01 107 = 10 000 000 10-7 = 0. 000 000 1 106 = 1 000 000 10-6 = 0. 000 001 105 = 100 000 10-5 = 0.000 01 104 = 10 000 10-4 = 0.000 1 103 = 1 000 10-3 = 0.001 102 = 100 10-2 = 0.01 101 = 10 10-1 = 0.1 100 = 1 coefficient base exponent
Basic Units and Electrical Quantities 2 1.2 1.3 Engineering Notation Metric Prefixes The exponent indicates the number of places that the decimal point is moved to the right or left to produce the decimal number. If the power of ten is positive, the decimal point is moved to the right to get the equivalent decimal number. Engineering notation is a version of scientific notation in which the powers of ten must be multiples of three. For example: 0. 0 0 0 0 5 6 56 × 10−6 A number An Engineering Notation Move decimal point SIX places to the RIGHT. Metric Prefixes are symbols that represent each of the most used powers of ten in engineering notation. Table 1.2: Metric Prefixes Prefix Symbol Multiplier Numerical Exponential Peta P 1 000 000 000 000 000 1015 Tera T 1 000 000 000 000 1012 Giga G 1 000 000 000 109 Mega M 1 000 000 106 Kilo K 1 000 103 Hector H 100 102 Deca Da 10 101 No prefix means=1 100 deci d 0.1 10-1 centi c 0.01 10-2 milli m 0.001 10-3 micro µ 0.000 001 10-6 nano n 0.000 000 001 10-9 pico p 0.000 000 000 001 10-12 femto f 0.000 000 000 000 001 10-15 Example 1-1 Express 33, 000 in: a) Scientific notation b) Engineering notation c) Metric prefix
Basic Units and Electrical Quantities 3 Solution: a) 33 000 = 3.3 × 104 b) 33 000 = 3.3 × c) 10333 000 = 33 Example 1-2: Express 0.000 063 8 in: a) Scientific notation b) Engineering notation c) Metric prefix Solution: a) 0.000 063 8 = 6.38 × 10−5 b) 0.000 063 8 = 63.8 × 10−6 c) 0.000 063 8 = 63.8µ Example 1-3: Convert 0.25mA to µA. Solution: 0.25 = 0.25 × 10−3 = 250 × 10−6 0 . 2 5 0 Move decimal point three places to the RIGHT. 0.25 = Example 1-4: Convert 470 000 Watt into kilo Watt (kW). Solution: 470 000 = 470 × 103 4 7 0 0 0 0 Move decimal point three places to the LEFT. 470 000 = Example 1-5: Convert 22 000 picofarads(pF) into microfarads (µF). Solution: 22 000 = 22 000 × 10−12 = 0.22 × 10−6 0 2 2 0 0 0 Move decimal point six places to the LEFT. 22 000 = . µ
Basic Units and Electrical Quantities 4 1.4 Standard International, SI Unit The system of units used in engineering and science is International System of units, usually abbreviated to SI units, and is based on the metric system. The SEVEN base quantities of SI units: Table 1.3: SI Base Units SI Base Units Base quantity Base Unit Name Symbol Name Symbol Time t second s Length l meter m Mass m kilogram kg Electric current I, i ampere A Thermodynamic temperature T kelvin K Amount of substance n mole mol Luminous intensity lv candela cd SI Derived Units is derived from the basic seven base SI units as shown in Table 1.4. Table 1.4: SI Derived Units SI Derived Units Derived quantity Equivalent SI Unit Base Units Quantity Name Symbol Frequency, f Hertz Hz 1/s @ s-1 Force. F Newton N kg ms-2 Pressure, p Pascal Pa kg m-1 s -2 Energy, E Joule J N-m Power, P Watt W J/s Electric charge, Q Coulomb C sA Electric potential, V Volt V W/A Electric resistance, R ohm Ω V/A Celcius temperature,T Degree celcius C K
Basic Units and Electrical Quantities 5 1.5 Electrical Quantities The base unit of charge (Q) are the coulomb (C). This base unit of charge, also used in defining the base units of other electrical quantities such as current, I and voltage, V. So, these electrical units are derived from SI Base Units. Table 1.5: Electrical Quantities Electrical parameter Symbol Description Measuring unit Electromotive force e.m.f EMF E Energy transfer to an electric circuit per unit electric charge by a generator or a battery. Volt (V) Electrical charge Q A property of matter that causes itself to move when exposed to an electromagnetic field. There are two type of charge which is positive and negative charge. Coulomb (C) Current I The rate of flow of electric charge. Ampere (A) Voltage V The potential difference between two electronic points or an electric circuit. Volt (V) Resistance R The ability of a conductor in an electrical circuit to limit or oppose the flow of current in the circuit. Ohm (Ω) Resistivity (rho) It is the characteristic of conductive material to opposition or decrease the current flows in it. Ohm meter (Ωm) Conductor - A material that can conduct electricity and -
Basic Units and Electrical Quantities 6 heat (for example copper and iron). Insulator - A material that cannot conduct electricity and heat (for example plastic, rubber, wood and glass). - Semiconductor - A material that is between and insulator and a conductor (for example silicon, germanium and gallium arsenide). Widely used in electronics segments, transistor and microchip. -
Basic Units and Electrical Quantities 7 Problem 1-1 1. Convert the following number into scientific notation. a) 45000 b) 0.000482 c) 325600000 2. Express the following number into metric prefixes: a) 56000000 b) 0.00024 c) 0.000000000040 3. Convert 47 000pF into µF. 4. Convert 5000 nanoamperes to microamperes. 5. Convert 56 000Ω to MΩ. 6. Convert 0.05MV to kV. 7. Which of the following is not electrical quantities? (a) Current (b) Power (c) Resistance (d) Mass 8. The quantity of 2.2 × 103 is same as: (a) 220 (b) 2200 (c) 0.022 (d) 22 000 9. The unit for charge is: (a) Q (b) V (c) C (d) W 10. Hertz is a unit of: (a) Energy (b) Capacitance (c) Frequency (d) Current
CHAPTER 2: Introduction to Electrical Circuit
Introduction to Electrical Circuit 8 CHAPTER 2: Introduction to Electrical Circuit Resistor Resistance and Resistivity Ohms Law Electrical Power and Energy Electrical Circuit Basic Meters 2.1 Resistor 2.1.1 Resistor is a passive electrical component that creates resistance to the flow of current and produces a voltage drop across it. The resistor only consumes electrical energy and does not deliver electrical energy. The resistance is represented by the symbol R and the unit for resistor is Ohm (Ω). Resistor used: To limit/reduce the current. To divide the voltages. Resistors are used extensively throughout electrical and electronic circuits. Types of resistors: Linear Resistor: Values change with change in applied temperature and voltage are known as linear resistors. There are two types of linear resistors: Fixed: A fixed resistor is one in which the value of its resistance cannot change. For example, thin film resistors, carbon composition resistors, wire wound resistors and thick film resistors. Variable: A variable resistor is one which the values can be change through a dial, knob, and screw or manually by a proper method. For example, potentiometers, rheostats and trimmers.
Introduction to Electrical Circuit 9 2.1.2 Non-Linear Resistor: values change according to the temperature and voltage applied and is not dependent on Ohm’s law. Types of non-linear resistors are: Thermistors Varistors Photo resistors Figure 2.1: Symbol of Resistors Figure 2.2: Types of Resistors Resistor Color Codes Resistor values are often indicated with color codes. Practically all leaded resistors with a power rating up to one watt are marked with color bands. The coding is defined in the international standard IEC 60062. This standard describes the marking codes for resistors and capacitors.
Introduction to Electrical Circuit 10 Figure 2.3: Resistor Color Code Table How to read resistor codes? The resistance value also can be measured using ohmeter or multimeter. Example 2-1: What is the color code of 2.2kΩ 5% four band resistor? Solution: 2- Red 2- Red X100 - Orange 5%- Gold Red, Red, Orange and Gold Grey-8 Red-2 Orange-x1k Gold-5% 82kΩ5%
Introduction to Electrical Circuit 11 Example 2-2: What is the value of a four band resistor which has the first band-Green, second band-Blue, third band-Orange and the fourth band-Gold? Solution: Green – 5 Blue -6 Orange x 1000 Gold - 5% 56kΩ5% Example 2-3: What is the value of a six-band resistor, which has band color Green, Blue, Black, Orange, Violet, Brown? Solution: Green - 5 Blue - 6 Black -0 Orange - 3 Violet – 0.1 Brown – Temp. co-efficient 100ppm/C 560KΩ 0.1% Temp. co-eff. 100ppm/C Example 2-4: Determine the value of the four band resistor in Figure 2-a below. Figure 2-a. Solution: 1 st band – Yellow (4) 2 nd band – Violet (7) 3 rd band – Brown (x10) 4 th band – Silver (10%) R = 470Ω 10%
Introduction to Electrical Circuit 12 Problem 2-1 1. What is the value of a resistor which has first three color bands are Red, Black, and Orange? 2. What is the color code for the following four band resistors? a) 500kΩ, tolerance ± 20% b) 2.7MΩ, tolerance ± 5% c) 330Ω, tolerance ± 10% d) 1MΩ, tolerance ± 5% e) 3.21kΩ, tolerance ± 10% 3. Find the value of the following resistors below: a) b) c) d) e)
Introduction to Electrical Circuit 13 2.2 Resistance and Resistivity Resistance and resistivity are closely related, and they refer to different measurements and resistance factors: Electrical Resistance, R – measures the overall resistance of conductor. This includes resistivity, length, temperature, material, and gauge or thickness. Resistance is measured in Ohms. Resistivity, (rho) - refers to the natural resistance of a material or substance only. Resistivity is represented by the symbol ρ, and this measurement will not change as length or thickness changes. Resistivity is measured in Ohms-metres. Figure 2.4: Resistance vs Resistivity The resistance of a conductor or rod to the flow of electric currents is depends on its geometry and its composition. There are FOUR factors effect that affect the resistance of a conductor: Table 2.1: Factors Effect the Resistance of Conductor Length, l Longer materials have greater resistance. Cross-sectional area, A Larger areas offer LESS resistance. Temperature, T The higher temperature results in higher resistances. Resistivity, Higher resistance, higher resistivity The resistance of a wire can be expressed by following mathematical expression: =
Introduction to Electrical Circuit 14 Where, R = Resistance (Ω) = Resistivity (Ωm) l = length of conductor (m) A = Cross-sectional area (m2 ) Resistivity varies with temperature and some typical values of resistivity measured at about room temperature are given below: Copper: 1.7 x 10-8 Ωm (or 0.017 µΩm) Aluminium: 2.6 x 10-8Ωm (or 0.026 µΩm) Carbon (graphite): 10 x 10-8 Ωm (or 0.10 µΩm) Example 2-5: A wire of length 2m, cross sectional area 0.05m2 and resistance 20Ω. What is the Resistivity of the wire? Solution: Given: l=2m; A=0.05 m 2 ; R=20Ω. = = = (20)(0.05) (2) = . Example 2-6: The heating element has 100Ω resistance, 1000cm length and 0.7mm diameter. Calculate the resistivity. Solution: Given: R=100Ω; l=1000cm=10m; d=0.7mm. = 2 = ()(0.35 × 10−3 ) 2 = 3.85 × 10−72 = = (100)(3.85 × 10−7 ) 10 = . × −
Introduction to Electrical Circuit 15 Example 2-7 Calculate the total resistance of 100m of 2.5mm2 copper wire if the resistivity of copper at 20C is 1.72X10-8Ωm. Solution: Given: l=100m; A=2.5mm2 =(2.5x10-3 ) 2 = = (1.72 × 10−8 )(100) (2.5 × 10−6) = . Ω Example 2-8 The resistivity of an aluminium coil is 0.017µΩm, 1.5mm radius and 25Ω resistance. Calculate the length of the coil. Solution: Given: =0.017µΩm; r=1.5mm; R=25Ω. Example 2-9 Determine the diameter of the tungsten filament of a light bulb in Figure 2-b below. Given: l=6cm, R=0.35Ω and =5.6x10-8Ωm. Figure 2-b Solution: = = (5.6 × 10−8 )(0.06) 0.35 = 9.6 × 10−92 = 2 = ()(1.5 × 10−3 ) 2 = 7.07 × 10−62 = = (25)(7.07 × 10−6 ) 0.017µ = = ( 2 ) 2 = 2 4 = ( 4 ) 1/2 = ( 4(9.6 × 10−9 ) ) 1/2 = .
Introduction to Electrical Circuit 16 Problem 2-2 1. Calculate the resistance of a 3km length of aluminum overhead power cable if the cross-sectional area of the cable is 10 mm2 . Take the resistivity of aluminum to be 0.03 x 10-6Ωm. 2. Determine the resistance of 1200m of copper cable having a diameter of 12 mm if the resistivity of copper is 1.7 x 10-8Ωm. 3. The resistance of 1.5 km of wire of cross-sectional area 0.17 mm2 is 150Ω. Determine the resistivity of the wire. 4. Calculate the diameter of a 1km copper wire with 8.1Ω resistance. Take the resistivity of copper is 1.623x10-8 Ωm. 5. A copper wire with resistivity 1.6x10-8 Ωm has a cross sectional area of 20x10-4 cm2 . Calculate the length of the wire required to make a 10Ω coil. 6. The resistance of a 2m length of cable is 2.5Ω. Determine: a. The resistance of a 7m length of the same cable. b. The length of the same wire when the resistance is 6.25Ω. 7. Express the value of resistance through the aluminum wire with a length of 100km and a diameter of 25cm if the supply is 100V. The resistivity of the wire is 25.5μΩm. 8. Find the resistance of the copper coil in Figure 2-2(a) below. What would be the resistance of a similar coil of aluminum? Given resistivity of the copper =1.7x10-8Ωm and aluminium=2.6x10-8Ωm. Figure 2-2(a) 9. The diameter of 0-gauge copper wire is 8.252 mm. Find the resistance of a 1km length of such wire used for power transmission. Resistivity of the copper wire is 1.7x10-8Ωm. 10. Determine the length of the conductor having 2.5mm diameter, 0.04Ω resistance. Take the resistivity of the conductor is 5.6x10-8 Ωm. 10mm2 10m
Introduction to Electrical Circuit 17 2.3 Ohm’s Law 2.3.1 2.3.2 Ohm’s Law states that the current, I flowing in a circuit is directly proportional to the applied voltage, V and inversely proportional to the resistance, R provided the temperature remains constant. For a constant resistance, if the voltage across a resistor is increased, the current flows through the resistor will increase; and likewise. If the voltage is kept constant, less resistance results in more current and more resistance results in less current. Ohm’s Law relationship is shown below: = By knowing two values of Voltage, Current or Resistance, we can find the third missing value by using Ohm’s Law: To find the Voltage, V; = × To find the Resistance; = Where; V = Voltage (V) I = Current (A) R = Resistance (Ω) Relationship Between Voltage, Current and Resistance The concept of voltage, current and resistance is shown in Figure 2.5. Source: https://www.eeweb.com Figure 2.5: Concepts of Voltage, Current and Resistance.
Introduction to Electrical Circuit 18 From this figure, we could understand that the harder Volt exerts, the more Ampere will get through. While the more Ohm resists (by pulling the rope), the less Ampere will get through. A simple electrical circuit including voltage, current, and resistance is illustrated in Figure 2.6. Figure 2.6: Relationship Between Voltage, Current, Resistance and Power.
Introduction to Electrical Circuit 19 2.3.3 According to Ohm’s Law for a constant value of resistance and temperature, the currents flow through a conductor of wire is directly proportional to the voltage across it as shown in Figure 2.7. Figure 2.7: Current (I) versus Voltage (V) a Constant Resistor Due to changing resistor values, the graph for a nonconstant/non-ohmic resistance is not straight like in Figure 2.7. Graph for non-constant resistor is displayed in Figure 2.8. Figure 2.8: Current (I) versus Voltage V for a Non-Constant Resistor Ohm’s Law Triangle It is easy to understand and remember this Ohm’s Law relationship by using pictures. The three parameter of V, I and R have been superimposed to form a triangle known as, Ohm’s Law Triangle as shown in Figure 2.9. Figure 2.9: Ohm’s Law Triangle Current Voltage R is constant Current Voltage R is not constant V I R
Introduction to Electrical Circuit 20 The following variations of the same equation can be created by transposing the above basic Ohms Law equation as shown in Figure 2.10: Figure 2.10: Transposing of Ohm’s Law Example 2-10: What is the voltage if a resistance of 25Ω produces a current of 0.6A? Solution: = × = (0.6)(25) = Example 2-11: What is the current produce by a 9V battery flowing through a resistance of 470Ω? Solution: = = 9 470 = . Example 2-12: A resistor is supply from a 10V dc source has 1.5mA current flowing through it. Find the value of the resistance? Solution: = = 10 1.5 = . Ω V I R V I R V I R = = × =
Introduction to Electrical Circuit 21 Example 2-13: Calculate the currents flow in the circuit of Figure 2-c below. Figure 2-c Solution: = = 10 5 = Example 2-14: If the current, 1.5A flow through a 40Ω resistor, determine the voltage supply? Solution: = × = (1.5)(40) = 2.4 Electrical Power Electric Power, P is defined as the rate of electrical energy is transferred by an electric circuit. Power, P is given by the product of voltage, V and current, I. = × Also: = 2 And: = 2 Where; P = Power (Watt) V = Voltage (V) I = Current (A) R = Resistance (Ω) Vdc 10V R 5Ω I
Introduction to Electrical Circuit 22 Example 2-15 A 100 W light bulb operates on 120 V. How much current does it require? Solution: = = 100 120 = 0.833 Example 2-16 An electric kettle has a resistance of 30Ω. What current will flow when it is connected to a 240 V supply? Find also the power rating of the kettle. Solution: Example 2-17 If a hairdryer is connected to the 230V mains supply and there is a current of 1.1A passing through it, what is its electrical power consumption? Solution: = = (1.1)(230) = 2.5 Electrical Energy Electrical Energy, E is the ability to do work; and power is the rate at which energy is used. = Thus; = × Where; P = Power (watt) E = Electrical energy (joules) t = time (second) = = 240 30 = = = (8)(240) =
Introduction to Electrical Circuit 23 Although the unit of energy is the joules, when dealing with large amounts of energy, a kilowatt-hour is used to measure the total amount of electricity used. The unit used is the kilowatt-hour (kWh) where; = − = × − = Example 2-18 A source e.m.f. of 5 V supplies a current of 3 A for 10 minutes. How much energy is provided in this time? Solution: Example 2-19 An electric heater consumes 1.8MJ when connected to a 250 V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply. Solution: Example 2-20 How many kilowatt-hours are used by a 250 W bulb burning for 8 hours? Solution: = = (250)(8) = = = (3)(5) = 15 = = (15)(10 × 60) = = = = 1 250 = = = = 1.8 (30 × 60) =
Introduction to Electrical Circuit 24 Example 2-21 Find the energy in kWh consumed in 10 hours by four devices of power 500 W each. Solution: = = (500 × 4)(10) = 20ℎ Example 2-22 A 120 W electric blanket is left on for 8 hours. a. How many joules electrical energy is transferred? b. How many kilowatt hours of electrical energy is transferred? Solution: a. Energy in joules = = (120)(8 × 60 × 60) = . b. Energy in kWh = = (120)(8) = = .
Introduction to Electrical Circuit 25 Problem 2-3 1. What is the resistance of a 60W bulb that takes 0.3A? 2. The current flowing through a heating element is 1.5A when a potential different of 15V is applied across it. Find the resistance of the element. 3. A 50V battery is connected across a resistor and causes a current of 5mA to flow. Determine the resistance of the resistor. If the voltage is now reduced to 25 V, what will be the new value of the current flowing? 4. An e.m.f. of 240V is connected across a resistance and the current flowing through the resistance is 4A. Determine: a. The resistance b. Power developed 5. A 60 W electric light bulb is connected to a 240 V supply. Determine: a. The current flowing in the bulb b. The resistance of the bulb 6. If a 75-watt lamp is turned on for one hour, how many joules of electrical energy have been converted by the lamp? 7. Calculate the heat produced by an electric heater which has a resistance of 35Ω and takes current pf 3A when it is switched on for 15 seconds. 8. A battery of e.m.f. 12 V supplies a current of 5 A for 2 minutes. How much energy is supplied in this time? 9. A dc electric motor consumes 36MJ when connected to a 250V supply for 1 hour. Find: a. The power rating of the motor b. The current taken from the supply c. The energy consumes in kWh 10. A 4kW cooker is used for 20 minutes. a. How many joules electrical energy is transferred? b. How many kilowatt-hours of electrical energy is transferred?
Introduction to Electrical Circuit 26 2.5 Electrical Circuit 2.5.1 2.5.2 Electrical circuits show how components are connected in a circuit. It consists of electric components like resistor, capacitor, inductor, power supply that are connected through a wire. There are SIX types of electric circuit: Close circuit Open circuit Short circuit Series circuit Parallel circuit Series-parallel circuit Close Circuit A close circuit is when there is current flow continuously from the voltage source to load as shown in Figure 2.11 below. Figure 2.11: Close Circuit This circuit represent the ON state of the circuit. Open Circuit Open circuit is when there is a break or gap in an electric circuit which current cannot flow from source to load. This may be due to switch in OFF state or the failure in the circuit as shown in Figure 2.12. Figure 2.12: Open Circuit
Introduction to Electrical Circuit 27 2.5.3 2.5.4 Short Circuit Short circuit is when a current bypass through least resistance and eliminating the light bulb as shown in Figure 2.13. Figure 2.13: Short Circuit Most electricity will take the path of least resistance and creating a large amount of current flow through the wires. Series Circuit A series circuit is when all components are connected along a single path, as shown in Figure 2.14. Figure 2.14 Total resistance: = 1+2 + 3 + ⋯ … … . . + The current through each component is the same. = 1 = 2 = 3 = ⋯ … . . = According to Kirchoff’s Voltage Law, “the total voltage around a loop equal to zero”. ∑ = 0 Resulting KVL equation gives: + 1 + 2 + 3 + ⋯ … . . + = 0 VT R2 RN R3 I V1 VN V3 V2 R1
Introduction to Electrical Circuit 28 So, for series circuit the voltage across the circuit is the sum of the voltages across each component. = 1 + 2 + 3 + ⋯ … . . + Voltage drop at each resistor that connected in series can be find by using Voltage Divider Rule (VDR). For example, by referring to Figure 2.18; 1 = 1 1 + 2+3 × 2 = 2 1 + 2+3 × 3 = 3 1 + 2+3 × Example 2-22 Find the current and voltage drop at each resistor in Figure 2-d using: a. Ohms law method b. Voltage divider method Figure 2-d Solution: a. Ohms law method = 1 + 2 = 15 + 35 = 50Ω = = 50 50 = b. Voltage divider method 1 = 1 1 = (1)(15) = 2 = 2 1 = (1)(35) = 2 = 1 1 + 2 × 2 = 15 15 + 35 × (50) = R1 R2 V 50V R2 15kΩ 35kΩ 1 = 1 1 + 2 × 1 = 15 15 + 35 × (50) =
Introduction to Electrical Circuit 29 Example 2-23 Referring to Figure 2-e, determine the value of R3. Figure 2-e Solution: 2.5.5 Parallel Circuit When two or more resistors are individually connected between the same two junctions, they are in parallel with each other. A parallel circuit provides more than one junction for current as shown in Figure 2.15. Figure 2.15: Parallel Circuit In a parallel circuit, the voltage across each of the components is the same. = 1 = 2 = 3 = ⋯ … . . = R1 = 2.2kΩ R2 = 100Ω R3=? I=20mA Vs=100V = = 100 20 = 5Ω = 1 +2 + 3 3 = −1 − 2 3 = 5 −2.2 −100 = . Ω R1 R2 R3 I1 I2 I3 IN RN VT IT V1 V2 V3 VN
Introduction to Electrical Circuit 30 Kirchhoff’s Current Law states that “the total current flowing towards the junction is equal to the total current flowing away from the junction”. ∑ = ∑ So, total current is the sum of the currents through each component. = 1 + 2 + 3 + ⋯ … . . + The total resistance, RT: 1 = 1 1 + 1 2 + 1 3 + ⋯ … . . + 1 = 1 1 1 + 1 2 + 1 3 + ⋯ … . . + 1 For two resistors connected in parallel: Total resistance: = 1 1 1 + 1 2 = 12 1 + 2 According to Current Divider Rule (CDR), if two resistors R1 and R2 are parallel: 1 = 2 1 + 2 × 2 = 1 1 + 2 × Example 2-24 Find the total resistance and total current for the circuit in Figure 2-f below. Given V=15V R1=560Ω, R2=220Ω and R3=1.5kΩ. Figure 2-f VT R1 R2 I1 I2 V R1 R2 R3
Introduction to Electrical Circuit 31 Solution: = 1 1 1 + 1 2 + 1 3 = 1 1 560 + 1 220 + 1 1.5 = 1 1 560 + 1 220 + 1 1.5 = . Ω Example 2-25 Referring to Figure 2-g, determine: a. Total resistance, RT b. Current I1 and I2 Figure 2-g Solution: = 12 1 +2 = (4.7)(10) (4.7 + 10) = . = = 100 3.2 = . 2.5.6 Series-Parallel Circuit Series-parallel circuit is a combination of series and parallel circuit. To analyse this type of combination circuit, we need to identify which parts are series and which parts are parallel. = = 15 142.9 = . R2 R1 V 100V 4.7kΩ 10kΩ 1 = 2 1 + 2 × 1 = (10) (4.7 +10) × (31.25) = . 2 = 1 1 + 2 × 2 = (4.7) (4.7 + 10) × (31.25) = . I1 I2
Introduction to Electrical Circuit 32 Referring to Figure 2.16: Resistor R2 is parallel with R3. Resistor R1 is series with resistor R2-3. The current flow is divided into TWO branches I1 and I2. Figure 2.16: Series-Parallel Circuit Total resistance, RT: = 1 + ( 23 2+3 ) Total current, IT: Omh’s Law: = KCL: = 1 + 2 Current I1 and I2 Using CDR: 1 = 3 2 + 3 × 2 = 2 2 + 3 × Voltage drop at each resistor: 1 = 1 2 = 12 3 = 23 Example 2-26 Given that R1 = 3kΩ and R2 = 2.2kΩ. Both resistors are connected in parallel and supplied with 30Vdc. Calculate: a. The total resistance of the circuit b. The total current of the circuit c. The voltage drops at R1 d. Current flowing through R2 by using current division method. Solution: V R1 R2 R3 IT I1 I2 b. Total current, IT: = = 30 1.27 = . a. Total resistance, RT: = 12 1 +2 = (3)(2.2) (3) + (2.2) = .
Introduction to Electrical Circuit 33 Example 2-27 Calculate the total resistance between terminal A and B for the circuit in Figure 2-h below. Figure 2-h Solution: Viewing from terminal A-B: R1 is parallel with R2+R3 R7 is parallel with R5+R6 R1 R2 R3 R6 R5 R7 R4 A B 6Ω 6Ω 6Ω 5Ω 12Ω 10Ω 30Ω RA RB = (1)(2 + 3) (1 )+(2 + 3) = (6)(6 + 6) (6) + (6 + 6) = 4Ω = (7)(5 +6) (7 )+ (5 + 6) = (10)(12+ 5) (10) + (12 + 5) = 6.3Ω Total Resistance is: = +4+ = 4 + 30 +6.3 = . R4 RA RB A B c. Voltage drops at R1: Since the circuit is parallel connection, the voltage drop at VR1 equal to supply voltage: = d. Current flowing through R2: Using CDR, 2 = 1 1 + 2 × 2 = 3 3 +2.2 × 23.62 = .
Introduction to Electrical Circuit 34 Example 2-28 Referring to Figure 2-i below, calculate: a. Total resistance. b. Current flow through resistor R2 and R4. c. Voltage drop at R1, R2, R3 and R4. Figure 2-i Solution: a. Total resistance. = 1 + ( (2 +3)(4) (2 + 3)+ (4 ) ) = 25+ ( (15+ 75)(50) (15 +75)+ (50) ) = 25+ (32.14) = 57.14Ω b. Current flow through resistor R2 and R4. R1 R2 R3 R4 V 9V 25Ω 15Ω 75Ω 50Ω I1 I2 Find IT: IT = = 9 57.14 = 157.5 Using CDR, find I1 and I2: 1 = 4 (2 +3)+ 4 × 1 = 50 (15 +75)+ 50 × (0.158) = . 2 = (2 +3) (2 + 3)+ 4 × 2 = (15+ 75) 15+ 75)+ 50 × (0.158) = . I1 flow through R2 and R3 I2 flow through R4
Introduction to Electrical Circuit 35 c. Voltage drop at R1, R2, R3 and R4. Example 2-29 For the circuit in Figure 2-j, calculate: a. The value of resistor Rx such the total power dissipated in the circuit is 2.5kW b. Current Ia and Ib Figure 2-j Solution: Using Ohm’s Law: 1 = × 1 1 = (157.5) × (25) = . 2 = 1 × 2 2 = (56.43) × (15) = . 3 = 1 × 3 3 = (56.43) × (75) = . 4 = 1 × 4 4 = (101.57) × (50) = . 250V V1 V2 R1=15Ω R2=10Ω R3=38Ω Rx Ia Ib Ic Id I a. Resistor Rx Find total current using power formula P=IV: = = 2.5 250 = 10 From Ohm’s Law: = = 250 10 = 25Ω = ( 12 1+2 ) +( 3 3+ ) 25 = ( (15 × 10) (15 × 10) ) + ( (38) (38+) ) 25 = 6 + 38 (38+) 19(38+) = 38 722 = 38− 19 = Ω
Introduction to Electrical Circuit 36 Example 2-30 Refer to Figure 2-k, total current in the circuit is, IT=0.8mA. calculate: a. Value of resistor R4 b. Total resistance, RT c. Voltage supply, Vs d. Value of current at R2 and R3 e. Power dissipated in the circuit Figure 2-k Solution: = 1 (1 + 2) × = 15 (10 + 15) × 10 = b. Current Ia and Ib = 2 (1 + 2) × = 10 (10+ 15) × 10 = R1 330Ω R2 1.5kΩ R3 560Ω R4 VR4 2.4V a. Current flow through R4 is equal to IT: Using Ohm’s Law: 4 = 4 4 = 2.4 0.8 = b. Total resistance, RT: = 1 + ( 23 2 + 3 )+4 = 330 + ( (1.5)(560) (1.5 + 560) ) + 3 = . c. Voltage supply, Vs: = = (0.8)(3.74) = d. Current at R2 and R3: 2 = 3 (2 + 3) × 2 = 560 (1.5 + 560) × 0.8 = . 3 = 2 (2 + 3) × 3 = 1.5 (1.5 + 560) × 0.8 = . e. Power dissipated, P: = = (0.8)(3) = 2.4
Introduction to Electrical Circuit 37 Problem 2-4 1. Determine the total resistance for the series-parallel relationship in Figure 2-4 (a). Figure 2-4(a) 2. Determine the total resistance at point A-B in Figure 2-4(b) below. Figure 2-4(b) 3. Referring to Figure 2-4(c), find the value of resistor R3 and power dissipated in the circuit. Figure 2-4(c) 4. By using Kirchhoff’s Voltage Law, find the voltage drop at each bulb in Figure 2- 4(d). Figure 2-4(d) 25Ω + 3V - Vs 10V 10Ω R1 R2 R3 0.2A 3Ω 6Ω 15Ω 9V R1 5ΩR2 10Ω R3 8Ω R4 4Ω A B R1 3kΩR2 2kΩ R3 1.5kΩ
Introduction to Electrical Circuit 38 5. Refer to Figure 2-4(e), determine: a. The voltage source, Vs b. Resistor R1 and R3 Figure 2-4(e) 6. For the circuit in Figure 2-4(f), find the value of resistor R1, R2 and R3. Figure 2-4(f) 7. Referring to Figure 2-4(g), determine: a. Total voltage, VT b. Voltage drops at each resistor VR1 c. Value of R1, R2, R3 and RT d. Power dissipated by R2 and R3 Figure 2-4(g) Vs R1 R2 330Ω R3 0.1A 0.03A R1 30W R2 15W R3 30W 5A R1 R2 R3 2mA 3mA 15V 45V 0.15A 1A
Introduction to Electrical Circuit 39 8. Based on Figure 2-4(h), determine: a. Total resistance, RT b. Total current, IT c. Currents flow through R2 and R3 Figure 2-4(h) 9. Find the total power dissipated in the circuit of Figure 2-4(i) below. Figure 2-4(i) 10. Refer to Figure 2-4(j), determine the voltage drops at resistor R4. Next, calculate the total power dissipated by the circuit. Figure 2-24(j) R1 2.5kΩ R2 5kΩ R3 5kΩ V 50V Vs=30V R1 47Ω R2 100Ω R3 68Ω R4 75Ω R1 560Ω R2 330Ω R3 330Ω R4 560Ω Vs 50V
Introduction to Electrical Circuit 40 2.6 Basic Meters 2.6.1 The basic measurements for and electrical/electronic circuit usually are: Resistance, Ω Voltage, V Current, I Electrical energy, W There are different types of meter available for measuring purposes such as mention above: Ohmmeter for measuring resistance Voltmeter for measuring voltage Ammeter for measuring current Wattmeter for measuring electrical power Ohmmeter Ohmmeter is an instrument for measuring electrical resistance and continuity of an electrical circuit. Figure 2.17 shows the symbol of Ohmmeter. Figure 2.17: Ohmmeter To measure the resistance, it must be connected in parallel with the meter probe and the power source must be disconnected. Figure 2.18 shows an analog Ohmmeter. Figure 2.18: Analog Ohmmeter Ω
Introduction to Electrical Circuit 41 2.6.2 Multimeter also can be used to measure the electrical resistance when in resistance-measuring mode as shown in Figure 2.19. Figure 2.19: Multimeter with Ohm Selector Figure 2.20 shows how to measure the resistance using a multimeter. Source: https://www.matsusada.com/ Figure 2.20: Measuring Resistance Voltmeter A voltmeter is used to measure potential difference between two points in an electric circuit. Figure 2.21 shown the basic symbol of voltmeter. Figure 2.21: Voltmeter Symbol To measure the voltage/potential difference, the voltmeter is connected in parallel with the load. V
Introduction to Electrical Circuit 42 Figure 2.22 shows the AC and DC analog voltmeter. Depending on the voltmeter type and range, this type of voltmeter can measure voltages up to 1000V. (a) (b) Figure 2.22 (a): AC Analog Voltmeter (b): DC Analog Voltmeter A multimeter also can be used to measure either DC or AC voltage as shown in Figure 2.23. This type of multimeter can measure AC and DC voltage from 0 up to 500V. Figure 2.23: Multimeter with DCV and ACV Range Selector Figure 2.24(a) and Figure 2.24(b) shows how to measure AC and DC Voltage using a multimeter. Source: https://www.matsusada.com/ Figure 2.24(a): Measuring AC Voltage
Introduction to Electrical Circuit 43 2.6.3 Source: https://www.matsusada.com/ Figure 2.24(b): Measuring Dc Voltage Ammeter An Ammeter is a device used to measure an electric current in Amperes (A). An ammeter is connected in series with the circuit so that the currents flow is directly flows through the meter. The symbol of ammeter is shown in Figure 2.25. Figure 2.25: Ammeter Figure 2.26 shows an AC and DC Ammeter. (a) (b) Figure 2.26 (a): AC Analog Ammeter (b): DC Analog Ammeter A multimeter also can be used for measuring an AC and DC current as shown in Figure 2.27. This type of multimeter can measure AC current from 0 up to 1000A and DC current from 0 up to 500mA. A