Basic Electrical Engineering PDF

www.EngineeringBooksPdf.com BASIC ELECTRICAL ENGINEERING For B.E./B.Tech. and Other Engineering Examinations V.K. MEHTA ROHIT MEHTA S. CHAND & COMPANY PVT. LTD. (AN ISO 9001 : 2008 COMPANY) RAM NAGAR, NEW DELHI – 110 055 www.EngineeringBooksPdf.com S. CHAND & COMPANY LTD. (An ISO 9001 : 2008 Company) Head Office: 7361, RAM NAGAR, NEW DELHI - 110 055 Phone: 23672080-81-82, 9899107446, 9911310888 Fax: 91-11-23677446 Shop at: schandgroup.com; e-mail: [email protected] Branches : AHMEDABAD : 1st Floor, Heritage, Near Gujarat Vidhyapeeth, Ashram Road, Ahmedabad - 380 014, Ph: 27541965, 27542369, [email protected] BENGALURU : No. 6, Ahuja Chambers, 1st Cross, Kumara Krupa Road, Bengaluru - 560 001, Ph: 22268048, 22354008, [email protected] BHOPAL : Bajaj Tower, Plot No. 243, Lala Lajpat Rai Colony, Raisen Road, Bhopal - 462 011, Ph: 4274723. 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Ph. 0353-2520750 (Marketing Office) VISAKHAPATNAM : Plot No. 7, 1st Floor, Allipuram Extension, Opp. Radhakrishna Towers, Seethammadhara North Extn., Visakhapatnam - 530 013, (M) 09347580841, [email protected] (Marketing Office) © 1988, V.K. Mehta and Rohit Mehta All rights reserved. No part of this publication may be reproduced or copied in any material form (including photo copying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the copyright owner. Any breach of this will entail legal action and prosecution without further notice. Jurisdiction : All disputes with respect to this publication shall be subject to the jurisdiction of the Courts, tribunals and forums of New Delhi, India only. First Edition 1988; Subsequent Editions and Reprints 1991, 95, 97, 98, 2001, 2006, 2007, 2008 (Thrice), 2009 (Twice), 2010 (Twice); 2011; Revised Edition 2012 ISBN : 81-219-0871-X Code : 10A 113 www.EngineeringBooksPdf.com Preface to Sixth Edition The general response to the Fifth Edition of the book was very encouraging. Authors feel that their work has been amply rewarded and wish to express their deep sense of gratitude to the large number of readers who have used it and in particular to those of them who have sent helpful sugges- tions from time to time for the improvement of the book. The popularity of the book is judged from the fact that authors frequently receive feedback from many quarters including teachers, students and serving engineers. This feedback helps the authors to make the book up-to-date. In the present edition, many new topics/numericals/illustrations have been added to make the book more useful. Authors lay no claim to the original research in preparing the book. Liberal use of materials available in the works of eminent authors has been made. What they may claim, in all modesty, is that they have tried to fashion the vast amount of material available from primary and secondary sources into coherent body of description and analysis. The authors wish to thank their colleagues and friends who have contributed many valuable suggestions regarding the scope and content sequence of the book. Authors are also indebted to S. Chand & Company Ltd., New Delhi for bringing out this revised edition in a short time and pricing the book moderately inspite of heavy cost of paper and printing. Errors might have crept in despite utmost care to avoid them. Authors shall be grateful if these are pointed out along with other suggestions for the improvement of the book. V.K. MEHTA ROHIT MEHTA Disclaimer : While the authors of this book have made every effort to avoid any mistake or omission and have used their skill, expertise and knowledge to the best of their capacity to provide accurate and updated information. The authors and S. Chand does not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S.Chand and the author expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication. Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual. (iii) www.EngineeringBooksPdf.com (iv) www.EngineeringBooksPdf.com Contents Chapter Pages 1. Basic Concepts 1—35 Nature of Electricity—Unit of Charge—The Electron—Energy of an Electron—Valence Electrons— Free Electrons—Electric Current—Electric Current is a Scalar Quantity—Types of Electric Current—Mechanism of Current Conduction in Metals—Relation Between Current and Drift Velocity——Electric Potential—Potential Difference—Maintaining Potential Difference— Concept of E.M.F. and Potential Difference—Potential Rise and Potential Drop——Resistance— Factors Upon Which Resistance Depends—Specific Resistance or Resistivity—Conductance— Types of Resistors—Effect of Temperature on Resistance—Temperature Co-efficient of Resistance—Graphical Determination of a—Temperature Co-efficient at Various Temperatures— Summary of Temperature Co-efficient Relations—Variation of Resistivity With Temperature— Ohm’s Law—Non-ohmic Conductors—Electric Power—Electrical Energy—Use of Power and Energy Formulas—Power Rating of a Resistor—Non-linear Resistors—Objective Questions. 2. D.C. Circuits 36—105 D.C. Circuit—D.C. Series Circuit—D.C. Parallel Circuit—Main Features of Parallel Circuits— Two Resistances in Parallel—Advantages of Parallel Circuits—Applications of Parallel Circuits—D.C. Series-Parallel Circuits—Applications of Series-Parallel Circuits—Internal Resistance of a Supply—Equivalent Resistance—Open Circuits—Short Circuits—Duality Between Series and Parallel Circuits—Wheatstone Bridge—Complex Circuits—Kirchhoff’s Laws—Sign Convention—Illustration of Kirchhoff’s Laws—Method to Solve Circuits by Kirchhoff’s Laws—Matrix Algebra—Voltage and Current Sources—Ideal Voltage Source or Constant-Voltage Source—Real Voltage Source—Ideal Current Source—Real Current Source—Source Conversion—Independent Voltage and Current Sources—Dependent Voltage and Current Sources—Circuits With Dependent-Sources—Ground—Voltage Divider Circuit— Objective Questions. 3. D.C. Network Theorems 106—238 Network Terminology—Network Theorems and Techniques—Important Points About Network Analysis—Maxwell’s Mesh Current Method—Shortcut Procedure for Network Analysis by Mesh Currents—Nodal Analysis—Nodal Analysis with Two Independent Nodes—Shortcut Method for Nodal Analysis—Superposition Theorem—Thevenin’s Theorem—Procedure for Finding Thevenin Equivalent Circuit—Thevenin Equivalent Circuit—Advantages of Thevenin’s Theorem—Norton’s Theorem—Procedure for Finding Norton Equivalent Circuit—Norton Equivalent Circuit—Maximum Power Transfer Theorem—Proof of Maximum Power Transfer Theorem—Applications of Maximum Power Transfer Theorem— Reciprocity Theorem—Millman’s Theorem—Compensation Theorem—Delta/Star and Star/ Delta Transformation—Delta/Star Transformation—Star/Delta Transformation—Tellegen’s Theorem—Objective Questions. 4. Units—Work, Power and Energy 239—259 International System of Units—Important Physical Quantities—Units of Work or Energy— Some Cases of Mechanical Work or Energy—Electrical Energy—Thermal Energy—Units of Power—Efficiency of Electric Device—Harmful Effects of Poor Efficiency—Heating Effect of Electric Current—Heat Produced in a Conductor by Electric Current—Mechanical Equivalent of Heat (J)—Objective Questions. 5. Electrostatics 260—294 Electrostatics—Importance of Electrostatics—Methods of Charging a Capacitor—Coulomb’s Laws of Electrostatics—Absolute and Relative Permittivity—Coulomb’s Law in Vector Form—The (v) www.EngineeringBooksPdf.com (vi) Superposition Principle—Electric Field—Properties of Electric Lines of Force—Electric Intensity or Field Strength (E)—Electric Flux (ψ)—Electric Flux Density (D)—Gauss’s Theorem— Proof of Gauss’s Law—Electric Potential Energy—Electric Potential—Electric Potential Difference—Potential at a Point Due to a Point Charge—Potential at a Point Due to Group of Point Charges—Behaviour of Metallic Conductors in Electric Field—Potential of a Charged Conducting Sphere—Potential Gradient—Breakdown Voltage or Dielectric Strength—Uses of Dielectrics—Refraction of Electric Flux—Equipotential Surface—Motion of a Charged Particle in Uniform Electric Field—Objective Questions. 6. Capacitance and Capacitors 295—349 Capacitor—How does a Capacitor Store Charge ?—Capacitance—Factors Affecting Capacitance— Dielectric Constant or Relative Permittivity—Capacitance of an Isolated Conducting Sphere— Capacitance of Spherical Capacitor—Capacitance of Parallel-Plate Capacitor with Uniform Medium—Parallel-Plate Capacitor with Composite Medium—Special Cases of Parallel-Plate Capacitor—Multiplate Capacitor—Cylindrical Capacitor—Potential Gradient in a Cylindrical Capacitor—Most Economical Conductor Size in a Cable—Capacitance Between Parallel Wires—Insulation Resistance of a Cable Capacitor—Leakage Resistance of a Capacitor—Voltage Rating of a Capacitor—Capacitors in Series—Capacitors in Parallel—Joining Two Charged Capacitors—Energy Stored in a Capacitor—Energy Density of Electric Field—Force on Charged Plates—Behaviour of Capacitor in a D.C. Circuit—Charging of a Capacitor—Time Constant— Discharging of a Capacitor—Transients in D.C. Circuits—Transient Relations During Charging Discharging of Capacitor—Objective Questions. 7. Magnetism and Electromagnetism 350—385 Poles of a Magnet—Laws of Magnetic Force—Magnetic Field—Magnetic Flux—Magnetic Flux Density—Magnetic Intensity or Magnetising Force (H)—Magnetic Potential—Absolute and Relative Permeability—Relation Between B and H—Important Terms—Relation Between mr and χm—Refraction of Magnetic Flux—Molecular Theory of Magnetism—Modern View about Magnetism—Magnetic Materials—Electromagnetism—Magnetic Effect of Electric Current— Typical Electromagnetic Fields—Magnetising Force (H) Produced by Electric Current—Force on Current-Carrying Conductor Placed in a Magnetic Field—Ampere’s Work Law or Ampere’s Circuital Law—Applications of Ampere’s Work Law—Biot-Savart Law—Applications of Biot-Savart Law—Magnetic Field at the Centre of Current-Carrying Circular Coil—Magnetic Field Due to Straight Conductor Carrying Current—Magnetic Field on the Axis of Circular Coil Carrying Current—Force Between Current-Carrying Parallel Conductors—Magnitude of Mutual Force—Definition of Ampere—Objective Questions. 8. Magnetic Circuits 386—429 Magnetic Circuit—Analysis of Magnetic Circuit—Important Terms—Comparison Between Magnetic and Electric Circuits—Calculation of Ampere-Turns—Series Magnetic Circuits—Air Gaps in Magnetic Circuits—Parallel Magnetic Circuits—Magnetic Leakage and Fringing— Solenoid—B-H Curve—Magnetic Calculations From B-H Curves—Determination of B/H or Magnetisation Curve—B-H Curve by Ballistic Galvanometer—B-H Curve by Fluxmeter— Magnetic Hysteresis—Hysteresis Loss—Calculation of Hysteresis Loss—Factors Affecting the Shape and Size of Hysteresis Loop—Importance of Hysteresis Loop—Applications of Ferromagnetic Materials—Steinmetz Hysteresis Law—Comparison of Electrostatics and Electromagnetic Terms—Objective Questions. 9. Electromagnetic Induction 430—480 Electromagnetic Induction—Flux Linkages—Faraday’s Laws of Electromagnetic Induction— Direction of Induced E.M.F. and Current—Induced E.M.F.—Dynamically Induced E.M.F.— Statically Induced E.M.F.—Self-inductance (L)—Magnitude of Self-induced E.M.F.—Expressions for Self-inductance—Magnitude of Mutually Induced E.M.F.—Expressions for Mutual Inductance—Co-efficient of Coupling—Inductors in Series—Inductors in Parallel with no Mutual Inductance—Inductors in Parallel with Mutual Inductance—Energy Stored in a Magnetic Field— www.EngineeringBooksPdf.com (vii) Magnetic Energy Stored Per Unit Volume—Lifting Power of a Magnet—Closing and Breaking an Inductive Circuit—Rise of Current in an Inductive Circuit—Time Constant—Decay of Current in an Inductive Circuit—Eddy Current Loss—Formula for Eddy Current Power Loss—Objective Questions. 10. Chemical Effects of Electric Current 481—520 Electric Behaviour of Liquids—Electrolytes—Mechanism of Ionisation—Electrolysis—Back e.m.f. or Polarisation Potential—Faraday’s Laws of Electrolysis—Relation Between E and Z— Deduction of Faraday’s Laws of Electrolysis—Practical Applications of Electrolysis—Cell—Types of Cells—Lead-Acid Cell—Chemical Changes During Discharging—Chemical Changes During Recharging—Formation of Plates of Lead-acid Cells—Construction of a Lead-acid Battery— Characteristics of a Lead-acid Cell—Curves of a Lead-acid Cell—Indications of a Fully Charged Lead-acid Cell—Load Characteristics of a Lead-acid Cell—Sulphation of Plates—Methods of Charging Batteries—Important Points About Charging of Lead-Acid Batteries—Effects of Overcharging—Care of Lead-acid Batteries—Applications of Lead-acid Batteries—Voltage Control Methods—Alkaline Batteries—Nickel-Iron Cell or Edison Cell—Electrical Characteristics of Nickel-Iron Cell—Nickel-Cadmium Cell—Comparison of Lead-acid Cell and Edison Cell— Silver-Zinc Batteries—Solar Cells—Fuel Cells—Objective Questions. 11. A.C. Fundamentals 521—577 Alternating Voltage and Current—Sinusoidal Alternating Voltage and Current—Why Sine Waveform?—Generation of Alternating Voltages and Currents—Equation of Alternating Voltage and Current—Important A.C. Terminology—Important Relations—Different Forms of Alternating Voltage—Values of Alternating Voltage and Current—Peak Value—Average Value—Average Value of Sinusoidal Current—R.M.S. or Effective Value—R.M.S. Value of Sinusoidal Current—Importance of R.M.S. Values—Form Factor and Peak Factor—Complex Waveforms—R.M.S. Value of a Complex Wave—Phase—Phase Difference—Representation of Alternating Voltages and Currents—Phasor Representation of Sinusoidal Quantities—Phasor Diagram of Sine Waves of Same Frequency—Addition of Alternating Quantities—Subtraction of Alternating Quantities—Phasor Diagrams Using R.M.S. Values—Instantaneous Power—A.C. Circuit Containing Resistance Only—A.C. Circuit Containing Pure Inductance Only—A.C. Circuit Containing Capacitance Only—Complex Waves and A.C. Circuit—Fundamental Power and Harmonic Power—Objective Questions. 12. Series A.C. Circuits 578—633 R-L Series A.C. Circuit—Impedance Triangle—Apparent, True and Reactive Powers—Power Factor—Significance of Power Factor—Q-factor of a Coil—Power in an Iron-Cored Choking Coil—R-C Series A.C. Circuit—Equivalent Circuit for a Capacitor—R-L-C Series A.C. Circuit—Resonance in A.C. Circuits—Resonance in Series A.C. Circuit (Series Resonance)— Resonance Curve—Q-Factor of Series Resonant Circuit—Bandwidth of a Series Resonant Circuit—Expressions for Half-power Frequencies—To Prove : fr = f1 f 2 —Expressions for Bandwidth—Important Relations in R-L-C Series Circuit—Applications of Series Resonant Circuits—Decibels—Objective Questions. 13. Phasor Algebra 634—665 Notation of Phasors on Rectangular Co-ordinate Axes—Significance of Operator j—Mathematical Representation of Phasors—Conversion from One Form to the Other—Addition and Subtraction of Phasors—Conjugate of a Complex Number—Multiplication and Division of Phasors—Powers and Roots of Phasors—Applications of Phasor Algebra to A.C. Circuits—R-L Series A.C. Circuit—R-C Series A.C. Circuit—R-L-C Series A.C. Circuit—Power Determination Using Complex Notation—Power Determination by Conjugate Method—A.C. Voltage Divider— Objective Questions. www.EngineeringBooksPdf.com (viii) 14. Parallel A.C. Circuits 666—721 Methods of Solving Parallel A.C. Circuits—By Phasor Diagram—By Phasor Algebra—Equivalent Impedance Method—Admittance (Y)—Importance of Admittance in Parallel A.C. Circuit Analysis—Admittance Triangle—Admittance Method for Parallel Circuit Solution—Application of Admittance Method—Some Cases of Parallel Connected Elements—Series-Parallel A.C. Circuits—Series-to-Parallel Conversion and Vice-Versa—Resonance in Parallel A.C. Circuits (Parallel Resonance)—Graphical Representation of Parallel Resonance—Q-factor of a Parallel Resonant Circuit—Bandwidth of Parallel Resonant Circuit—Key Points About Parallel Resonance—General Case for Parallel Resonance—Comparison of Series and Parallel Resonant Circuits—Objective Questions. 15. Polyphase Circuits 722—829 Polyphase System—Reasons for the Use of 3-phase System—Elementary 3-Phase Alternator— Some Concepts in 3-Phase System—Interconnection of Three Phases—Star or Wye Connected System—Important 3-Phase Terminology—Voltages and Currents in Balanced Y-Connected Supply System—Checking Correct Connections for Y-connected Alternator—Delta (D) or Mesh Connected System—Correct and Incorrect D Connections of Alternator—Voltages and Currents in Balanced D Connected Supply System—Advantages of Star and Delta Connected Systems— Constancy of Total Power in Balanced 3-phase System—Effects of Phase Sequence—Phase Sequence Indicator—Y/D or D/Y Conversions for Balanced Loads—3-phase Balanced Loads in Parallel—Use of Single-Phase Wattmeter—Power Measurement in 3-phase Circuits—Three- Wattmeter Method—Two-Wattmeter Method—Proof for Two-Wattmeter Method—Determination of P.F. of Load by Two-wattmeter Method (For balanced Y or D load only)—Effect of Load p.f. on Wattmeter Readings—Leading Power Factor—How to Apply p.f. Formula ?—One-Wattmeter Method–Balanced Load—Reactive Power with Two-Wattmeter Method—Reactive Power with One Wattmeter—Unbalanced 3-Phase Loads—Four-Wire Star-Connected Unbalanced Load— Unbalanced D-Connected Load—Unbalanced 3-Wire Star-Connected Load—Methods of Solving Unbalanced 3-wire Y load—Solving Unbalanced 3-Wire Y Load by Kirchhoff’s Laws—Solving Unbalanced 3-wire Y Load By Loop Current Method—Solving Unbalanced 3-Wire Y Load by Y/D Conversion—Solving Unbalanced 3-Wire Y Load by Millman’s Theorem—Significance of Power Factor—Disadvantages of Low Power Factor—Causes of Low Power Factor—Power Factor Improvement—Power Factor Improvement Equipment—Calculations of Power Factor Correction—Objective Questions. 16. Electrical Instruments and Electrical Measurements 830—935 Classification of Electrical Measuring Instruments—Types of Secondary Instruments—Principles of Operation of Electrical Instruments—Essentials of Indicating Instruments—Deflecting Torque—Controlling Torque—Damping Torque—Ammeters and Voltmeters—Permanent-Magnet Moving Coil (PMMC) Instruments (Ammeters and Voltmeters) —Extension of Range of PMMC Instruments—Extension of Range of PMMC Ammeter—Extension of Range of PMMC Voltmeter— Voltmeter Sensitivity—Dynamometer Type Instruments (Ammeters and Voltmeters)—Deflectiing Torque (Td) of Dynamometer Type Instruments in Terms of Mutual Inductance—Range Extension of Dynamometer Type Instruments—Moving-Iron (M.I.) Ammeters and Voltmeters—Attraction Type M.I. Instruments—Repulsion Type M.I. Instruments—Td of M.I. Instruments in Terms of Self-Inductance—Sources of Errors in Moving Iron Instruments—Characteristics of Moving- Iron Instruments—Extending Range of Moving-Iron Instruments—Comparison of Moving Coil, Dynamometer type and Moving Iron Voltmeters and Ammeters—Hot-Wire Ammeters and Voltmeters—Thermocouple Instruments—Electrostatic Voltmeters—Attracted Disc Type Voltmeter—Quadrant Type Voltmeter—Multicellular Electrostatic Voltmeter—Characteristics of Electrostatic Voltmeters—Range Extension of Electrostatic Voltmeters—Induction Type Instruments—Induction Ammeters and Voltmeters—Characteristics of Induction Ammeters and Voltmeters—Wattmeters—Dynamometer Wattmeter—Characteristics of Dynamometer Wattmeters—Wattmeter Errors—Induction Wattmeters—Three-phase Wattmeter—Watthour www.EngineeringBooksPdf.com (ix) Meters or Energy Meters—Commutator Motor Meter—Mercury Motor Watthour Meter— Induction Watthour Meters or Energy Meters—Single-Phase Induction Watthour Meters or Energy Meters—Errors in Induction Watthour Meters—Three-Phase Watthour Meter—D.C. Potentiometer—Direct Reading Potentiometers—Modern D.C. Potentiometers—Crompton D.C. Potentiometers—Volt Ratio Box—Applications of D.C. Potentiometers—A.C. Potentiometer— Drysdale A.C. Potentiometer—Ballistic Galvanometer—Vibration Galvanometer—Frequency Meters—Vibrating-Reed Frequency meter—Electrodynamic Frequency Meter—Moving-Iron Frequency Meter—Power Factor Meters—Single-Phase Electrodynamic Power Factor Meter—3- Phase Electrodynamic Power Factor Meter—Moving-Iron Power Factor Meter—3-Voltmeter Method of Determining Phase Angle—Ohmmeter—Megger—Instrument Transformers—Current Transformer (C.T.)—Potential Transformer (P.T.)—Advantages of Instrument Transformers— Objective Questions. 17. A.C. Network Analysis 936—983 A.C. Network Analysis—Kirchhoff’s Laws for A.C. Circuits—A.C. Mesh Current Analysis—A.C. Nodal Analysis—Superposition Theorem for A.C. Circuits—Thevenin’s Theorem for A.C. Circuits—Norton’s Theorem for A.C. Circuits—Thevenin and Norton Equivalent Circuits— Millman’s Theorem for A.C. Circuits—Reciprocity Theorem—Maximum Power Transfer Theorem for A.C. Circuits—A.C. Network Transformations—Objective Questions. n Index 985—989 www.EngineeringBooksPdf.com (x) www.EngineeringBooksPdf.com 1 Basic Concepts Introduction Everybody is familiar with the functions that electricity can perform. It can be used for lighting, heating, traction and countless other purposes. The question always arises, “What is electricity” ? Several theories about electricity were developed through experiments and by observation of its behaviour. The only theory that has survived over the years to explain the nature of electricity is the Modern Electron theory of matter. This theory has been the result of research work conducted by scientists like Sir William Crooks, J.J. Thomson, Robert A. Millikan, Sir Earnest Rutherford and Neils Bohr. In this chapter, we shall deal with some basic concepts concerning electricity. 1.1. Nature of Electricity We know that matter is electrical in nature i.e. it contains particles of electricity viz. protons and electrons. The positive charge on a proton is equal to the negative charge on an electron. Whether a given body exhibits electricity (i.e. charge) or not depends upon the relative number of these par- ticles of electricity. (i) If the number of protons is equal to the number of electrons in a body, the resultant charge is zero and the body will be electrically neutral. Thus, the paper of this book is electrically neutral (i.e. paper exhibits no charge) because it has the same number of protons and electrons. (ii) If from a neutral body, some *electrons are removed, there occurs a deficit of electrons in the body. Consequently, the body attains a positive charge. (iii) If a neutral body is supplied with electrons, there occurs an excess of electrons. Conse- quently, the body attains a negative charge. 1.2. Unit of Charge The charge on an electron is so small that it is not convenient to select it as the unit of charge. In practice, coulomb is used as the unit of charge i.e. SI unit of charge is coulomb abbreviated as C. One coulomb of charge is equal to the charge on 625 × 1016 electrons, i.e. 1 coulomb = Charge on 625 × 1016 electrons Thus when we say that a body has a positive charge of one coulomb (i.e. +1 C), it means that the body has a deficit of 625 × 1016 electrons from normal due share. The charge on one electron is given by ; 1 Charge on electron = − = – 1.6 × 10–19 C 625 × 1016 1.3. The Electron Since electrical engineering generally deals with tiny particles called electrons, these small particles require detailed study. We know that an electron is a negatively charged particle having negligible mass. Some of the important properties of an electron are : (i) Charge on an electron, e = 1.602 × 10–19 coulomb (ii) Mass of an electron, m = 9.0 × 10–31 kg (iii) Radius of an electron, r = 1.9 × 10–15 metre * Electrons have very small mass and, therefore, are much more mobile than protons. On the other hand, protons are powerfully held in the nucleus and cannot be removed or detached. www.EngineeringBooksPdf.com 2 Basic Electrical Engineering The ratio e/m of an electron is 1.77 × 1011 coulombs/kg. This means that mass of an electron is very small as compared to its charge. It is due to this property of an electron that it is very mobile and is greatly influenced by electric or magnetic fields. 1.4. Energy of an Electron An electron moving around the nucleus possesses two types of energies viz. kinetic energy due to its motion and potential energy due to the charge on the nucleus. The total energy of the electron is the sum of these two energies. The energy of an electron increases as its distance from the nucleus increases. Thus, an electron in the second orbit possesses more energy than the electron in the first orbit ; electron in the third orbit has higher energy than in the second orbit. It is clear that electrons in the last orbit possess very high energy as compared to the electrons in the inner orbits. These last orbit electrons play an important role in determining the physical, chemical and electrical properties of a material. 1.5. Valence Electrons The electrons in the outermost orbit of an atom are known as valence electrons. The outermost orbit can have a maximum of 8 electrons i.e. the maximum number of valence electrons can be 8. The valence electrons determine the physical and chemical properties of a mate- rial. These electrons determine whether or not the material is chemically active; metal or non-metal or, a gas or solid. These electrons also determine the electrical properties of a material. On the basis of electrical conductivity, materials are generally classified into conductors, insu- lators and semi-conductors. As a rough rule, one can determine the electrical behaviour of a mate- rial from the number of valence electrons as under : (i) When the number of valence electrons of an atom is less than 4 (i.e. half of the maximum eight electrons), the material is usually a metal and a conductor. Examples are sodium, magnesium and aluminium which have 1, 2 and 3 valence electrons respectively. (ii) When the number of valence electrons of an atom is more than 4, the material is usually a non-metal and an insulator. Examples are nitrogen, sulphur and neon which have 5, 6 and 8 valence electrons respectively. (iii) When the number of valence electrons of an atom is 4 (i.e. exactly one-half of the maximum 8 electrons), the material has both metal and non-metal properties and is usually a semi-conductor. Examples are carbon, silicon and germanium. 1.6. Free Electrons We know that electrons move around the nucleus of an atom in different orbits. The electrons in the inner orbits (i.e., orbits close to the nucleus) are tightly bound to the nucleus. As we move away from the nucleus, this binding goes on decreasing so that electrons in the last orbit (called valence electrons) are quite loosely bound to the nucleus. In certain substances, especially metals (e.g. copper, aluminium etc.), the valence electrons are so weakly attached to their nuclei that they can be easily removed or detached. Such electrons are called free electrons. Those valence electrons which are very loosely attached to the nucleus of an atom are called free electrons. The free electrons move at random from one atom to another in the material. Infact, they are so loosely attached that they do not know the atom to which they belong. It may be noted here that all valence electrons in a metal are not free electrons. It has been found that one atom of a metal can www.EngineeringBooksPdf.com Basic Concepts 3 provide at the most one free electron. Since a small piece of metal has billions of atoms, one can expecta very large number of free electrons in metals. For instance, one cubic centimetre of copper has about 8.5 × 1022 free electrons at room temperature. (i) A substance which has a large number of free electrons at room temperature is called a conductor of electricity e.g. all metals. If a voltage source (e.g. a cell) is applied across the wire of a conductor material, free electrons readily flow through the wire, thus constituting electric current. The best conductors are silver, copper and gold in that order. Since copper is the least expensive out of these materials, it is widely used in electrical and electronic industries. (ii) A substance which has very few free electrons is called an insulator of electricity. If a voltage source is applied across the wire of insulator material, practically no current flows through the wire. Most substances including plastics, ceramics, rubber, paper and most liquids and gases fall in this category. Of course, there are many practical uses for insulators in the electrical and electronic industries including wire coatings, safety enclosures and power-line insulators. (iii) There is a third class of substances, called semi-conductors. As their name implies, they are neither conductors nor insulators. These substances have crystalline structure and contain very few free electrons at room temperature. Therefore, at room temperature, a semiconductor practically behaves as an insulator. However, if suitable controlled impurity is imparted to a semi-conductor, it is possible to provide controlled conductivity. Most common semi-conductors are silicon, germa- nium, carbon etc. However, silicon is the principal material and is widely used in the manufacture of electronic devices (e.g. crystal diodes, transistors etc.) and integrated circuits. 1.7. Electric Current The directed flow of free electrons (or charge) is called electric current. The flow of electric current can be beautifully explained by referring to Fig. 1.1. The copper strip has a large number of free electrons. When electric pressure or voltage is applied, then free electrons, being negatively charged, will start moving towards the positive terminal around the circuit as shown in Fig. 1.1. This directed flow of electrons is called electric current. Fig. 1.1 The reader may note the following points : (i) Current is flow of electrons and electrons are the constituents of matter. Therefore, electric current is matter (i.e. free electrons) in motion. (ii) The actual direction of current (i.e. flow of electrons) is from negative terminal to the positive terminal through that part of the circuit external to the cell. However, prior to Electron theory, it was assumed that current flowed from positive terminal to the negative terminal of the cell www.EngineeringBooksPdf.com 4 Basic Electrical Engineering via the circuit. This convention is so firmly established that it is still in use. This assumed direction of current is now called conventional current. Unit of Current. The strength of electric current I is the rate of flow of electrons i.e. charge flowing per second. Q \ Current, I = t The charge Q is measured in coulombs and time t in seconds. Therefore, the unit of electric current will be coulombs/sec or ampere. If Q = 1 coulomb, t = 1 sec, then I = 1/1 = 1 ampere. One ampere of current is said to flow through a wire if at any cross-section one coulomb of charge flows in one second. Thus, if 5 amperes current is flowing through a wire, it means that 5 coulombs per second flow past any cross-section of the wire. Note. 1 C = charge on 625 × 1016 electrons. Thus when we say that current through a wire is 1 A, it means that 625 × 1016 electrons per second flow past any cross-section of the wire. Q ne \ I = = where e = – 1.6 × 10–19 C ; n = number of electrons t t 1.8. Electric Current is a Scalar Quantity Q (i) Electric current, I = t As both charge and time are scalars, electric current is a scalar quantity. (ii) We show electric current in a wire by an arrow to indicate the direction of flow of positive charge. But such arrows are not vectors because they do not obey the laws of vector algebra. This point can be explained by referring to Fig. 1.2. The wires OA and OB carry currents of 3 A and 4 A respectively. The total current in the wire CO is 3 + 4 = 7 A ir- Fig. 1.2 respective of the angle between the wires OA and OB. This is not surprising because the charge is conserved so that the magnitudes of currents in wires OA and OB must add to give the magnitude of current in the wire CO. 1.9. Types of Electric Current The electric current may be classified into three main classes: (i) steady current (ii) varying current and (iii) alternating current. (i) Steady current. When the magnitude of current does not change with time, it is called a steady current. Fig. 1.3 (i) shows the graph between steady current and time. Note that value of current remains the same as the time changes. The current provided by a battery is almost a steady current (d.c.). Fig. 1.3 www.EngineeringBooksPdf.com Basic Concepts 5 (ii) Varying current. When the magnitude of current changes with time, it is called a varying current. Fig. 1.3 (ii) shows the graph between varying current and time. Note that value of current varies with time. (iii) Alternating current. An alternating current is one whose magnitude changes contin- uously with time and direction changes periodically. Due to technical and economical reasons, we produce alternating currents that have sine waveform (or cosine waveform) as shown in Fig. 1.3 (iii). It is called alternating current because current flows in alternate directions in the circuit, i.e., from 0 to T/2 second (T is the time period of the wave) in one direction and from T/2 to T second in the opposite direction. The current provided by an a.c. generator is alternating current that has sine (or cosine) waveform. 1.10. Mechanism of Current Conduction in Metals Every metal has a large number of free electrons which wander randomly within the body of the conductor somewhat like the molecules in a gas. The average speed of free electrons is sufficiently high ( 105 ms–1) at room temperature. During random motion, the free electrons collide with posi- tive ions (positive atoms of metal) again and again and after each collision, their direction of motion changes. When we consider all the free electrons, their random motions average to zero. In other words, there is no net flow of charge (electrons) in any particular direction. Consequently, no current is established in the conductor. When potential difference is applied across the ends of a conductor (say copper wire) as shown in Fig. 1.4, electric field is applied at every point of the copper wire. The electric field exerts force on the free electrons which start accelerating towards the positive terminal (i.e., opposite to the direction of the field). As the free electrons move, they *collide again and again with positive ions of the metal. Each collision destroys the extra velocity gained by the free electrons. Fig. 1.4 The average time that an electron spends between two collisions is called the relaxation time (t). Its value is of the order of 10–14 second. Although the free electrons are continuously accelerated by the electric field, collisions prevent their velocity from becoming large. The result is that electric field provides a small constant velocity towards positive terminal which is superimposed on the random motion of the electrons. This con- stant velocity is called the drift velocity. The average velocity with which free electrons get drifted in a metallic conductor under the → influence of electric field is called drift velocity ( v d ). Thedrift velocity of free electrons is of the order of 10–5 ms–1. Thus when a metallic conductor is subjected to electric field (or potential difference), free elec- → trons move towards the positive terminal of the source with drift velocity v d. Small though it is, the drift velocity is entirely responsible for electric current in the metal. Note. The reader may wonder that if electrons drift so slowly, how room light turns on quickly when switch is closed ? The answer is that propagation of electric field takes place with the speed of light. When we apply electric field (i.e., potential difference) to a wire, the free electrons everywhere in the wire begin drifting almost at once. * What happens to an electron after collision with an ion ? It moves off in some new and quite random direction. However, it still experiences the applied electric field, so it continues to accelerate again, gaining a velocity in the direction of the positive terminal. It again encounters an ion and loses its directed motion. This situation is repeated again and again for every free electron in a metal. www.EngineeringBooksPdf.com 6 Basic Electrical Engineering 1.11. Relation Between Current and Drift Velocity Consider a portion of a copper wire through which current I is flowing as shown in Fig. 1.5. Clearly, copper wire is under the influence of electric field. Let A = area of X-section of the wire n = electron density, i.e., number of free electrons per unit volume e = charge on each electron vd = drift velocity of free electrons In one second, all those free electrons within a Fig. 1.5 distance vd to the right of cross-section at P (i.e., in a volume Avd) will flow through the cross-section at P as shown in Fig. 1.5. This volume contains n Avd electrons and, hence, a charge (nAvd)e. Therefore, a charge of neAvd per second passes the cross- section at P. \ I = n e A vd Since A, n and e are constant, I ∝ vd. Hence, current flowing through a conductor is directly proportional to the drift velocity of free electrons. (i) The drift velocity of free electrons is very small. Since the number of free electrons in a metallic conductor is very large, even small drift velocity of free electrons gives rise to sufficient current. (ii) The current density J is defined as current per unit area and is given by ; I n e Avd Current density, J = = = n e vd A A The SI unit of current density is amperes/m2. → Note. Current density is a vector quantity and is denoted by the symbol J. Therefore, in vector notation, → → → the relation between I and J is I = J. A → where A = Area vector Example 1.1. A 60 W light bulb has a current of 0.5 A flowing through it. Calculate (i) the number of electrons passing through a cross-section of the filament (ii) the number of electrons that pass the cross-section in one hour. Q ne Solution. (i) I = = t t It 0.5 × 1 \ n = = = 3.1 × 1018 electrons/s e 1.6 × 10 —19 (ii) Charge passing the cross-section in one hour is Q = I t = (0.5) × (60 × 60) = 1800 C Now, Q = ne Q 1800 \ n = = = 1.1 × 1022 electrons/hour e 1.6 × 10 —19 Example 1.2. A copper wire of area of X-section 4 mm2 is 4 m long and carries a current of 10 A. The number density of free electrons is 8 × 1028 m–3. How much time is required by an electron to travel the length of wire ? Solution. I = n A e vd www.EngineeringBooksPdf.com Basic Concepts 7 Here I = 10 A ; A = 4 mm2 = 4 × 10–6 m2 ; e = 1.6 × 10–19 C ; n = 8 × 1028 m–3 I 10 \ Drift velocity, vd = = = 1.95 × 10 —4 ms —1 n A e 8 × 1028 × (4 × 10 —6 ) × 1.6 × 10 —19 \ Time taken by the electron to travel the length of the wire is l 4 t = = = 2.05 × 104 s = 5.7 hours vd 1.95 × 10 —4 Example 1.3. The area of X-section of copper wire is 3 × 10–6 m2. It carries a current of 4.2 A. Calculate (i) current density in the wire and (ii) the drift velocity of electrons. The number density of conduction electrons is 8.4 × 1028 m–3. I 4.2 Solution. (i) Current density, J = = = 1.4 × 106 A/m2 A 3 × 10 —6 (ii) I = n e A vd I 4.2 \ Drift velocity, vd = = = 1.04 × 10–4 ms–1 n A e (8.4 × 10 ) × (1.6 × 10 —19 ) × 3 × 10 —6 28 Tutorial Problems 1. How much current is flowing in a circuit where 1.27 × 1015 electrons move past a given point in 100 ms ? [2.03 A] 2. The current in a certain conductor is 40 mA. (i) Find the total charge in coulombs that passes through the conductor in 1.5 s. (ii) Find the total number of electrons that pass through the conductor in that time. [(i) 60 mC (ii) 3.745 × 1017 electrons] 22 –3 3. The density of conduction electrons in a wire is 10 m. If the radius of the wire is 0.6 mm and it is carrying a current of 2 A, what will be the average drift velocity ? [1.1 × 10–3 ms–1] 4. Find the velocity of charge leading to 1 A current which flows in a copper conductor of cross-section 1 cm2 and length 10 km. Free electron density of copper = 8.5 × 1028 per m3. How long will it take the electric charge to travel from one end of the conductor to the other ? [0.735 mm/s; 431 years] 1.12. Electric Potential When a body is charged, work is done in charging it. This work done is stored in the body in the form of potential energy. The charged body has the capacity to do work by moving other charges either by attraction or repulsion. The ability of the charged body to do work is called electric potential. The capacity of a charged body to do work is called its electric potential. The greater the capacity of a charged body to do work, the greater is its electric potential. Obviously, the work done to charge a body to 1 coulomb will be a measure of its electric potential i.e. Work done W Electric potential, V = = Charge Q The work done is measured in joules and charge in coulombs. Therefore, the unit of electric potential will be joules/coulomb or volt. If W = 1 joule, Q = 1 coulomb, then V = 1/1 = 1 volt. Hence a body is said to have an electric potential of 1 volt if 1 joule of work is done to give it a charge of 1 coulomb. Thus, when we say that a body has an electric potential of 5 volts, it means that 5 joules of work has been done to charge the body to 1 coulomb. In other words, every coulomb of charge possesses an energy of 5 joules. The greater the joules/coulomb on a charged body, the greater is its electric potential. www.EngineeringBooksPdf.com 8 Basic Electrical Engineering 1.13. Potential Difference The difference in the potentials of two charged bodies is called potential difference. If two bodies have different electric potentials, a potential difference exists between the bodies. Consider two bodies A and B having potentials of 5 volts and 3 volts respectively as shown in Fig. 1.6 (i). Each coulomb of charge on body A has an energy of 5 joules while each coulomb of charge on body B has an energy of 3 joules. Clearly, body A is at higher potential than the body B. Fig. 1.6 If the two bodies are joined through a conductor [See Fig. 1.6 (ii)], then electrons will *flow from body B to body A. When the two bodies attain the same potential, the flow of current stops. Therefore, we arrive at a very important conclusion that current will flow in a circuit if potential difference exists. No potential difference, no current flow. It may be noted that potential difference is sometimes called voltage. Unit. Since the unit of electric potential is volt, one can expect that unit of potential difference will also be volt. It is defined as under : The potential difference between two points is 1 volt if one joule of work is **done or released in transferring 1 coulomb of charge from one point to the other. 1.14. Maintaining Potential Difference A device that maintains potential difference between two points is said to develop electromotive force (e.m.f.). A simple example is that of a cell. Fig. 1.7 shows the familiar voltaic cell. It consists of a copper plate (called anode) and a zinc rod (called cathode) immersed in dilute H2SO4. The chemical action taking place in the cell removes electrons from copper plate and transfers them to the zinc rod. This transference of electrons takes place through the agency of dil. H2SO4 (called electrolyte). Consequently, the copper plate attains a positive charge of +Q coulombs and zinc rod a charge of –Q coulombs. The chemical action of the cell has done a certain amount of work (say W joules) to do so. Clearly, the potential difference between the two plates will be W/Q volts. If the two plates are joined through a wire, some electrons from zinc rod will be attracted through the wire to copper plate. The chemical action of the cell now transfers an equal amount of electrons from copper plate to zinc rod internally through the cell to maintain original potential difference (i.e. W/Q). This process continues so long as the Fig. 1.7 * The conventional current flow will be in the opposite direction i.e. from body A to body B. ** 1 joule of work will be done in the case if 1 coulomb is transferred from point of lower potential to that of higher potential. However, 1 joule of work will be released (as heat) if 1 coulomb of charge moves from a point of higher potential to a point of lower potential. www.EngineeringBooksPdf.com Basic Concepts 9 circuit is complete or so long as there is chemical energy. The flow of electrons through the external wire from zinc rod to copper plate is the electric current. Thus potential difference causes current to flow while an e.m.f. maintains the potential differ- ence. Although both e.m.f. and p.d. are measured in volts, they do not mean exactly the same thing. 1.15. Concept of E.M.F. and Potential Difference There is a distinct difference between e.m.f. and potential difference. The e.m.f. of a device, say a battery, is a measure of the energy the battery gives to each coulomb of charge. Thus if a battery supplies 4 joules of energy per coulomb, we say that it has an e.m.f. of 4 volts. The energy given to each coulomb in a battery is due to the chemical action. The potential difference between two points, say A and B, is a measure of the energy used by one coulomb in moving from A to B. Thus if potential difference between points A and B is 2 volts, it means that each coulomb will give up an energy of 2 joules in moving from A to B. Illustration. The difference between e.m.f. and p.d. can be made more illustrative by referring to Fig. 1.8. Here battery has an e.m.f. of 4 volts. It means battery supplies 4 joules of energy to each coulomb continuously. As each coulomb travels from the positive terminal of the battery, it gives up its most of energy to resistances (2 W and 2 W in this case) and remaining to connecting wires. When it returns to the negative terminal, it has lost all its energy originally supplied by the battery. The battery now supplies fresh energy to each coulomb (4 joules in the present Fig. 1.8 case) to start the journey once again. The p.d. between any two points in the circuit is the energy used by one coulomb in moving from one point to another. Thus in Fig. 1.8, p.d. between A and B is 2 volts. It means that 1 coulomb will give up an energy of 2 joules in moving from A to B. This energy will be released as heat from the part AB of the circuit. The following points may be noted carefully : (i) The name e.m.f. at first sight implies that it is a force that causes current to flow. This is not correct because it is not a force but energy supplied to charge by some active device such as a battery. (ii) Electromotive force (e.m.f.) maintains potential difference while p.d. causes current to flow. 1.16. Potential Rise and Potential Drop Fig. 1.9 shows a circuit with a cell and a resistor. The cell provides a potential difference of 1.5 V. Since it is an energy source, there is a rise in potential associated with a cell. The cell’s potential difference represents an e.m.f. so that symbol E could be used. The resistor is also associated with a potential difference. Since it is a consumer (converter) + of energy, there is a drop in potential across the resistor. E = 1.5 V 1.5 V We can combine the idea of potential rise or drop with (RISE) (DROP) – the popular term “voltage”. It is customary to refer to the potential difference across the cell as a voltage rise and to the potential difference across the resistor as a voltage drop. Fig. 1.9 www.EngineeringBooksPdf.com 10 Basic Electrical Engineering Note. The term voltage refers to a potential difference across two points. There is no such thing as a voltage at one point. In cases where a single point is specified, some reference must be used as the other point. Unless stated otherwise, the ground or common point in any circuit is the reference when specifying a voltage at some other point. Example 1.4. A charge of 4 coulombs is flowing between points A and B of a circuit. If the potential difference between A and B is 2 volts, how many joules will be released by part AB of the circuit ? Solution. The p.d. of 2 volts between points A and B means that each coulomb of charge will give up an energy of 2 joules in moving from A to B. As the charge flowing is 4 coulombs, therefore, total energy released by part AB of the circuit is = 4 × 2 = 8 joules. Example 1.5. How much work will be done by an electric energy source with a potential differ- ence of 3 kV that delivers a current of 1 A for 1 minute ? Solution. We know that 1 A of current represents a charge transfer rate of 1 C/s. Therefore, the total charge for a period of 1 minute is Q = It = 1 × 60 = 60 C. Total work done, W = Q × V = 60 × (3 × 103) = 180 × 103 J = 180 kJ Tutorial Problems 1. Calculate the potential difference of an energy source that provides 6.8 J for every milli-coulomb of charge that it delivers. [6.8 kV] 2. The potential difference across a battery is 9 V. How much charge must it deliver to do 50 J of work ? [5.56 C] 3. A 300 V energy source delivers 500 mA for 1 hour. How much energy does this represent ? [540 kJ] 1.17. Resistance The opposition offered by a substance to the flow of electric current is called its resistance. Since current is the flow of free electrons, resistance is the opposition offered by the substance to the flow of free electrons. This opposition occurs because atoms and molecules of the substance obstruct the flow of these electrons. Certain substances (e.g. metals such as silver, copper, aluminium etc.) offer very little opposition to the flow of electric current and are called conductors. On the other hand, those substances which offer high opposition to the flow of electric current (i.e. flow of free electrons) are called insulators e.g. glass, rubber, mica, dry wood etc. It may be noted here that resistance is the electric friction offered by the substance and causes production of heat with the flow of electric current. The moving electrons collide with atoms or molecules of the substance ; each collision resulting in the liberation of minute quantity of heat. Unit of resistance. The practical unit of resistance is ohm and is represented by the symbol W. It is defined as under : A wire is said to have a resistance of 1 ohm if a p.d. of 1 volt across its ends causes 1 ampere to flow through it (See Fig. 1.10). There is another way of defining ohm. A wire is said to have a resistance of 1 ohm if it Fig. 1.10 releases 1 joule (or develops 0.24 calorie of heat) when a current of 1 A flows through it for 1 second. A little reflection shows that second definition leads to the first definition. Thus 1 A current flowing for 1 second means that total charge flowing is Q = I × t = 1 × 1 = 1 coulomb. Now the charge flowing between A and B (See Fig. 1.10) is 1 coulomb and energy released is 1 joule (or 0.24 calorie). Obviously, by definition, p.d. between A and B should be 1 volt. www.EngineeringBooksPdf.com Basic Concepts 11 1.18. Factors Upon Which Resistance Depends The resistance R of a conductor (i) is directly proportional to its length i.e. R∝ l (ii) is inversely proportional to its area of X-section i.e. 1 R∝ a (iii) depends upon the nature of material. (iv) depends upon temperature. From the first three points (leaving temperature for the time being), we have, l l R ∝ or R = ρ a a where ρ (Greek letter ‘Rho’) is a constant and is known as resistivity or specific resistance of the material. Its value depends upon the nature of the material. 1.19. Specific Resistance or Resistivity l We have seen above that R = ρ a If l = 1 m, a = 1 m2, then, R = ρ Hence specific resistance of a material is the resistance offered by 1 m length of wire of material having an area of cross-section of 1 m2 [See Fig. 1.11 (i)]. 2 1m Current 1m 1m 1m 1m (i) (ii) Fig. 1.11 Specific resistance can also be defined in another way. Take a cube of the material having each side 1 m. Considering any two opposite faces, the area of cross-section is 1 m2 and length is 1 m [See Fig. 1.11 (ii)] i.e. l = 1 m, a = 1 m2. Hence specific resistance of a material may be defined as the resistance between the opposite faces of a metre cube of the material. ρl Ra Unit of resistivity. We know R = or ρ = a l Hence the unit of resistivity will depend upon the units of area of cross-section (a) and length (l). (i) If the length is measured in metres and area of cross-section in m2, then unit of resistivity will be ohm-metre (Ω m). ohm × m 2 ρ = = ohm-m m www.EngineeringBooksPdf.com 12 Basic Electrical Engineering (ii) If length is measured in cm and area of cross-section in cm2, then unit of resistivity will be ohm-cm (Ω cm). ohm × cm 2 ρ= = ohm-cm cm The resistivity of substances varies over a wide range. To give an idea to the reader, the follow- ing table may be referred : Resistivity (Ω-m) at S.No. Material Nature room temperature 1 Copper metal 1.7 ×10–8 2 Iron metal 9.68 × 10–8 3 Manganin alloy 48 × 10–8 4 Nichrome alloy 100 × 10–8 5 Pure silicon semiconductor 2.5 ×103 6 Pure germanium semiconductor 0.6 10 7 Glass insulator 10 to 1014 8 Mica insulator 1011 to 1015 The reader may note that resistivity of metals and alloys is very small. Therefore, these materials are good conductors of electric current. On the other hand, resistivity of insulators is extremely large. As a result, these materials hardly conduct any current. There is also an intermediate class of semiconductors. The resistivity of these substances lies between conductors and insulators. 1.20. Conductance The reciprocal of resistance of a conductor is called its conductance (G). If a conductor has resistance R, then its conductance G is given by ; G = 1/R Whereas resistance of a conductor is the opposition to current flow, the conductance of a conductor is the inducement to current flow. The SI unit of conductance is mho (i.e., ohm spelt backward). These days, it is a usual practice to use siemen as the unit of conductance. It is denoted by the symbol S. Conductivity. The reciprocal of resistivity of a conductor is called its conductivity. It is denoted by the symbol σ. If a conductor has resistivity ρ, then its conductivity is given by ; 1 Conductivity, σ = ρ 1 a a We know that G = = = σ. Clearly, the SI unit of conductivity is Siemen metre−1 (S m−1). R ρl l Example 1.6. A coil consists of 2000 turns of copper wire having a cross-sectional area of 0.8 mm2. The mean length per turn is 80 cm and the resistivity of copper is 0.02 mW m. Find the resistance of the coil and power absorbed by the coil when connected across 110 V d.c. supply. Solution. Length of coil, l = 0.8 × 2000 = 1600 m; cross-sectional area of coil, a = 0.8 mm2 = 0.8 × 10–6 m2; Resistivity of copper, r = 0.02 × 10–6 Wm l 1600 \ Resistance of coil, R = ρ = 0.02 × 10–6 = 40 W a 0.8 × 10 —6 V2 (110) 2 Power absorbed, P = = = 302.5 W R 40 www.EngineeringBooksPdf.com Basic Concepts 13 Example 1.7. Find the resistance of 1000 metres of a copper wire 25 sq. mm in cross-section. The resistance of copper is 1/58 ohm per metre length and 1 sq. mm cross-section. What will be the resistance of another wire of the same material, three times as long and one-half area of cross- section ? Solution. For the first case, R1 = ? ; a1 = 25 mm2 ; l1 = 1000 m For the second case, R2 = 1/58 Ω ; a2 = 1 mm2 ; l2 = 1 m R1 = ρ (l1/a1) ; R2 = ρ (l2/a2) R1 l a  1000   1  \ = 1 × 2 =  ×   = 40 R2 l2 a1  1   25  1 20 or R1 = 40 R2 = 40 × = W 58 29 For the third case, R3 = ? ; a3 = a1/2 ; l3 = 3l1 R3 l   a  \ =  3  ×  1  = (3) × (2) = 6 R1  l1   a3  20 120 or R3 = 6R1 = 6 × = W 29 29 Example 1.8. A copper wire of diameter 1 cm had a resistance of 0.15 Ω. It was drawn under pressure so that its diameter was reduced to 50%. What is the new resistance of the wire ? π Solution. Area of wire before drawing, a1 = (1)2 = 0.785 cm2 4 π Area of wire after drawing, a2 = (0.5)2 = 0.196 cm2 4 As the volume of wire remains the same before and after drawing, \ a1l1 = a2l2 or l2/l1 = a1/a2 = 0.785/0.196 = 4 For the first case, R1 = 0.15 Ω ; a1 = 0.785 cm2 ; l1 = l For the second case, R2 = ? ; a2 = 0.196 cm2 ; l2 = 4l l1 l Now R1 = ρ ; R2 = ρ 2 a1 a2 R2 l   a  \ =  2  ×  1  = (4) × (4) = 16 R1  l1   a2  or R2 = 16R1 = 16 × 0.15 = 2.4 Ω Example 1.9. Two wires of aluminium and copper have the same resistance and same length. Which of the two is lighter? Density of copper is 8.9 × 103 kg/m3 and that of aluminium is 2.7 × 103 kg/m3. The resistivity of copper is 1.72 × 10−8 Ω m and that of aluminium is 2.6 × 10−8 Ω m. Solution. That wire will be lighter which has less mass. Let suffix 1 represent aluminium and suffix 2 represent copper. l l R1 = R2 or ρ1 1 = ρ2 2 A1 A2 ρ1 ρ2 or = ( l1 = l2) A1 A2 A1 ρ 2.6 × 10 —8 or = 1 = = 1.5 A2 ρ2 1.72 × 10 —8 www.EngineeringBooksPdf.com 14 Basic Electrical Engineering m1 ( A1 l1 )d1 Ad Now = = 1 1 ( l1 = l2) m2 ( A2 l2 )d 2 A2 d 2 m1  A1   d1  2.7 × 103 or =   ×   = 1.5 × = 0.46 m2  A2   d 2  8.9 × 103 or m1/m2 = 0.46 It is clear that for the same length and same resistance, aluminium wire is lighter than copper wire. For this reason, aluminium wires are used for overhead power transmission lines. Example 1.10. A rectangular metal strip has the dimensions x = 10 cm, y = 0.5 cm and z = 0.2 cm. Determine the ratio of the resistances Rx, Ry and Rz between the respective pairs of opposite faces. ρx ρy ρz 10 0.5 0.2 Solution. Rx : Ry : Rz = : : = : : yz zx xy 0.5 × 0.2 0.2 × 10 10 × 0.5 10 1 = : : 0.04 = 2500 : 6.25 : 1 0.1 4 Example 1.11. Calculate the resistance of a copper tube 0.5 cm thick and 2 m long. The external diameter is 10 cm. Given that resistance of copper wire 1 m long and 1 mm2 in cross-section is 1/58 Ω. Solution. External diameter, D = 10 cm Internal diameter, d = 10 – 2 × 0.5 = 9 cm π 2 π Area of cross-section, a =( D − d 2 ) = (10) 2 − (9) 2  cm 2 4 4 π = (10) 2 − (9) 2  × 100 mm 2 4 ρl 1 length in metres \ Resistance of copper tube = = × a 58 area of X-section in mm 2 1 2 = × = 23.14 × 10–6 Ω = 23.14 µΩ 58 π (10) 2 − (9) 2  × 100 4  Example 1.12. A copper wire is stretched so that its length is increased by 0.1%. What is the percentage change in its resistance ? l l′ Solution. R = ρ ; R′ = ρ a a′ 0.1 Now l′ = l + × l = 1.001 l 100 As the volume remains the same, al = a′l′. l a \ a′ = a = l ′ 1.001 R′  l′   a  \ =   ×   = (1.001) × (1.001) = 1.002 R  l   a′  R′ − R or = 0.002 R R′ − R \ Percentage increase = × 100 = 0.002 × 100 = 0.2% R www.EngineeringBooksPdf.com Basic Concepts 15 Example 1.13. A lead wire and an iron wire are connected in parallel. Their respective specific resistances are in the ratio 49 : 24. The former carries 80% more current than the latter and the latter 47% longer than the former. Determine the ratio of their cross-sectional areas. Solution. Let us represent lead and iron by suffixes 1 and 2 respectively. Then as per the condi- tions of the problem, we have, ρ1 49 = ; I1 = 1.8 I2 ; l2 = 1.47 l1 ρ2 24 l1 l2 Now R1 = ρ1 ; R2 = ρ2 a1 a2 V V I1 = and I2 = R1 R2 I2 R ρl a ρ  l  a  \ = 1 = 1 1 × 2 =  1 × 1 × 2  I1 R2 a1 ρ2 l2  ρ2   l2   a1  1 49 1 a or = × × 2 1.8 24 1.47 a1 a2 1 24 \ = × × 1.47 = 0.4 a1 1.8 49 Example 1.14. An aluminium wire 7.5 m long is connected in parallel with a copper wire 6 m long. When a current of 5 A is passed through the combination, it is found that the current in the aluminium wire is 3 A. The diameter of the aluminium wire is 1 mm. Determine the diameter of the copper wire. Resistivity of copper is 0.017 µΩm ; that of the aluminium is 0.028 µΩ m. Solution. Let us assign subscripts a and c to aluminium and copper respectively. Current through Al wire, Ia = 3 A \ Current through Cu wire, Ic = 5 – 3 = 2 A Since Ra and Rc are in parallel, the voltage across them is the same [See Fig. 1.12] i.e. Ra I c 2 Ia Ra = Ic Rc or = = Rc I a 3 ρ l ρl Now Ra = a a ; Rc = c c aa ac Rc ρc lc aa \ = × × Ra ρa la ac Rc 3 ρc 0.017 lc 6 Fig. 1.12 Here = ; = ; = ; Ra 2 ρa 0.028 la 7.5 π 2 π× (1) 2 π aa = d = = mm 2 4 4 4 3 0.017 6 π / 4 \ = × × 2 0.028 7.5 ac 2 0.017 6 π or ac = × × × = 0.2544 mm2 3 0.028 7.5 4 π 2 or d c = 0.2544 \ dc = 0.569 mm 4 www.EngineeringBooksPdf.com 16 Basic Electrical Engineering Example 1.15. A transmission line cable consists of 19 strands of identi- cal copper conductors, each 1.5 mm in diameter. The length of the cable is 2 km but because of the twist of the strands, the actual length of each conductor is increased by 5 percent. What is resistance of the cable ? Take the resistivity of the copper to be 1.78 × 10–8 Ω m. Solution. Fig. 1.13 shows the general shape of a stranded conductor. Fig. 1.13 Allowing for twist, the length of the strands is l = 2000 m + 5% of 2000 m = 2100 m π —3 2 —6 2 Area of X-section of 19 strands, a = (19)   × (1.5 × 10 ) = 33.576 × 10 m   4 l —8 2100 \ Resistance of line, R = ρ = 1.72 × 10 × = 1.076 Ω a 33.576 × 10 —6 Example 1.16. The resistance of the wire used for telephone is 35 Ω per kilometre when the weight of the wire is 5 kg per kilometre. If the specific resistance of the material is 1.95 × 10–8 Ω m, what is the cross-sectional area of the wire ? What will be the resistance of a loop to a subscriber 8 km from the exchange if wire of the same material but weighing 20 kg per kilometre is used? Solution. For the first case, R = 35 Ω ; l = 1000 m ; ρ = 1.95 × 10–8 Ω m l ρl 1.95 × 10 —8 × 1000 Now R = ρ \ a = = = 55.7 × 10–8 m2 a R 35 Since weight of conductor is directly proportional to the area of cross-section, for the second case, we have, 20 a = × 55.7 × 10–8 = 222.8 × 10–8 m2 ; l = 2 × 8 = 16 km = 16000 m 5 l 16000 \ R = ρ = 1.95 × 10–8 × = 140.1 Ω a 222.8 × 10 —8 Example 1.17. Find the resistance of a cubic centimetre of copper (i) when it is drawn into a wire of diameter 0.32 mm and (ii) when it is hammered into a flat sheet of 1.2 mm thickness, the current flowing through the sheet from one face to another, specific resistance of copper is 1.6 × 10–8 W-m. Solution. Volume of copper wire, v = 1 cm3 = 1 × 10–6 m3 (i) Resistance when drawn into wire. π 2 π Area of X-section, a = d = (0.32 × 10 ) = 0.804 × 10–7 m2 —3 2 4 4 v 1× 10 —6 Length of wire, l = = = 12.43m a 0.804 × 10 —7 l 12.43 \ Resistance of wire, R = ρ = 1.6 × 10–8 = 2.473 W a 0.804 × 10 —7 (ii) Resistance when hammered into flat sheet. Length of flat sheet, l = 1.2 × 10–3 m ; Area of cross-section of flat sheet is v 1× 10 —6 10 —3 2 =a = = m l 1.2 × 10 —3 1.2 l —8 1.2 × 10 —3 \ Resistance of copper flat sheet is R = ρ = 1.6 × 10 = 2.3 × 10–8 a 10 —3 / 1.2 Ω www.EngineeringBooksPdf.com Basic Concepts 17 Tutorial Problems 1. Calculate the resistance of 915 metres length of a wire having a uniform cross-sectional area of 0.77 cm2 if the wire is made of copper having a resistivity of 1.7 × 10–6 Ω cm. [0.08 Ω] 2. A wire of length 1 m has a resistance of 2 ohms. What is the resistance of second wire, whose specific resistance is double the first, if the length of wire is 3 metres and the diameter is double of the first? [3 Ω] 3. A rectangular copper strip is 20 cm long, 0.1 cm wide and 0.4 cm thick. Determine the resistance between (i) opposite ends (ii) opposite sides. The resistivity of copper is 1.7 × 10–6 Ω cm. [(i) 0.85 × 10–4 Ω (ii) 0.212 × 10–6 Ω] 4. A cube of a material of side 1 cm has a resistance of 0.001 Ω between its opposite faces. If the same material has a length of 9 cm and a uniform cross-sectional area 1 cm2, what will be the resistance of this length ? [0.009 Ω] 5. An aluminium wire 10 metres long and 2 mm in diameter is connected in parallel with a copper wire 6 metres long. A total current of 2 A is passed through the combination and it is found that current through the aluminium wire is 1.25 A. Calculate the diameter of copper wire. Specific resistance of copper is 1.6 × 10–6 Ω cm and that of aluminium is 2.6 × 10–6 Ω cm. [0.94 mm] 6. A copper wire is stretched so that its length is increased by 0.1%. What is the percentage change in its resistance ? [0.2%] 1.21. Types of Resistors A component whose function in a circuit is to provide a specified value of resistance is called a resistor. The principal applications of resistors are to limit current, divide voltage and in certain cases, generate heat. Although there are a variety of different types of resistors, the following are the commonly used resistors in electrical and electronic circuits : (i) Carbon composition types (ii) Film resistors (iii) Wire-wound resistors (iv) Cermet resistors (i) Carbon composition type. This type of resistor is made with a mixture of finely ground carbon, insulating filler and a resin binder. The ratio of carbon and insulating filler decides the resistance value [See Fig. 1.14]. The mixture is formed into a rod and lead connections are made. The entire resistor is then enclosed in a plastic case to prevent the entry of moisture and other harmful elements from outside. Fig. 1.14 Carbon resistors are relatively inexpensive to build. However, they are highly sensitive to temperature variations. The carbon resistors are available in power ratings ranging from 1/8 to 2 W. (ii) Film resistors. In a film resistor, a resistive material is deposited uniformly onto a high- grade ceramic rod. The resistive film may be carbon (carbon film resistor) or nickel-chromium (metal film resistor). In these types of resistors, the desired resistance value is obtained by removing a part of the resistive material in a helical pattern along the rod as shown in Fig. 1.15. www.EngineeringBooksPdf.com 18 Basic Electrical Engineering Metal film resistors have better characteristics as compared to carbon film resistors. (iii) Wire-wound resistors. A wire- wound resistor is constructed by winding a resistive wire of some alloy around an insulating rod. It is then enclosed in an insulating cover. Generally, nickle- chromium alloy is used because of its very small temperature coefficient of resistance. Wire-wound resistors can safely operate Fig. 1.15 at higher temperatures than carbon types. These resistors have high power ratings ranging from 12 to 225 W. (iv) Cermet resistors. A cermet resistor is made by depositing a thin film of metal such as nichrome or chromium cobalt on a ceramic substrate. They are cermet which is a contraction for ceramic and metal. These resistors have very accurate values. 1.22. Effect of Temperature on Resistance In general, the resistance of a material changes with the change in temperature. The effect of temperature upon resistance varies according to the type of material as discussed below : (i) The resistance of pure metals (e.g. copper, aluminium) increases with the increase of temperature. The change in resistance is fairly regular for normal range of temperatures so that temperature/ resistance graph is a straight line as shown in Fig. 1.16 (for copper). Since the resistance of metals increases with the rise in temperature, they have positive temperature co-efficient of resistance. (ii) The resistance of electrolytes, insulators (e.g. glass, mica, rubber etc.) and semiconductors (e.g. germanium, Fig. 1.16 silicon etc.) decreases with the increase in temperature. Hence these materials have negative temperature co-efficient of resistance. (iii) The resistance of alloys increases with the rise in temperature but this increase is very small and irregular. For some high resistance alloys (e.g. Eureka, manganin, constantan etc.), the change in resistance is practically negligible over a wide range of temperatures. Fig. 1.16 shows temperature/resistance graph for copper which is a straight line. If this line is extended backward, it would cut the temperature axis at −234.5°C. It means that theoretically, the resistance of copper wire is zero at −234.5°C. However, in actual practice, the curve departs (point A) from the straight line path at very low temperatures. 1.23. Temperature Co-efficient of Resistance Consider a conductor having resistance R0 at 0°C and Rt at t °C. It has been found that in the normal range of temperatures, the increase in resistance (i.e. Rt − R0) (i) is directly proportional to the initial resistance i.e. Rt − R0 ∝ R0 www.EngineeringBooksPdf.com Basic Concepts 19 (ii) is directly proportional to the rise in temperature i.e. Rt − R0 ∝ t (iii) depends upon the nature of material. Combining the first two, we get, Rt − R0 ∝ R0 t or Rt − R0 = *a0 R0 t...(i) where α0 is a constant and is called temperature co-efficient of resistance at 0°C. Its value depends upon the nature of material and temperature. Rearranging eq. (i), we get, Rt = R0 (1 + α0 t)...(ii) Definition of α0. From eq. (i), we get, Rt − R0 α0 = R0 × t = Increase in resistance/ohm original resistance/°C rise in tem- perature Hence temperature co-efficient of resistance of a conductor is the increase in resistance per ohm original resistance per °C rise in temperature. A little reflection shows that unit of α will be ohm/ohm°C i.e./°C. Thus, copper has a temperature co-efficient of resistance of 0.00426/°C. It means that if a copper wire has a resistance of 1 Ω at 0°C, then it will increase by 0.00426 Ω for 1°C rise in temperature i.e. it will become 1.00426 Ω at 1°C. Similarly, if temperature is raised to 10°C, then resistance will become 1 + 10 × 0.00426 = 1.0426 ohms. The following points may be noted carefully : (i) Those substances (e.g. pure metals) whose resistance increases with rise in temperature are said to have positive temperature co-efficient of resistance. On the other hand, those substances whose resistance decreases with increase in temperature are said to have negative temperature co- efficient of resistance. (ii) If a conductor has a resistance R0, R1 and R2 at 0oC, t1oC and t2oC respectively, then, R1 = R0 (1 + a0 t1) R2 = R0 (1 + a0 t2) R2 1+ α 0 t2 ∴ =...(iii) R1 1 + α 0 t1 This relation is often utilised in determining the rise of temperature of the winding of an electrical machine. The resistance of the winding is measured both before and after the test run. Let R1 and t1 be the resistance and temperature before the commencement of the test. After the operation of the machine for a given period, let these values be R2 and t2. Since R1 and R2 can be measured and t1 (ambient temperature) and a0 are known, the value of t2 can be calculated from eq. (iii). The average rise in temperature of the winding will be (t2 − t1)°C. Note. The life expectancy of electrical apparatus is limited by the temperature of its insulation; the higher the temperature, the shorter the life. The useful life of electrical apparatus reduces approximately by half every time the temperature increases by 10°C. This means that if a motor has a normal life expectancy of eight years * It will be shown in Art. 1.25 that value of a depends upon temperature. Therefore, it is referred to the original temperature i.e. 0°C in this case. Hence the symbol a0. www.EngineeringBooksPdf.com 20 Basic Electrical Engineering at a temperature of 100°C, it will have a life expectancy of only four years at a temperature of 110°C, of two years at a temperature of 120°C and of only one year at 130°C. 1.24. Graphical Determination of a The value of temperature co-efficient of resistance can also be determined graphically from temperature/resistance graph of the material. Fig. 1.17 shows the temperature/resistance graph for a conductor. The graph is a straight line AX as is the case with all conductors. The resistance of the conductor is R0 (represented by OA) at 0°C and it becomes Rt at t°C. By definition, Rt − R0 α0 = R0 × t Fig. 1.17 But Rt − R0 = BC and t = rise in temperature = AB BC \ a0 = R0 × AB But BC/AB is the slope of temperature/resistance graph. Slope of temp./resistance graph \ α0 =...(i) Original resistance Hence, temperature co-efficient of resistance of a conductor at 0°C is the slope of temp./ resistance graph divided by resistance at 0°C (i.e. R0). The following points may be particularly noted : (i) The value of α depends upon temperature. At any temperature, a can be calculated by using eq. (i). Slope* of temperature/resistance graph Thus, α0 = R0 Slope of temperature/resistance graph and αt = Rt (ii) The value of α0 is maximum and it decreases as the temperature is increased. This is clear from the fact that the slope of temperature/resistance graph is constant and R0 has the minimum value. 1.25. Temperature Co-efficient at Various Temperatures Consider a conductor having resistances R0 and R1 at temperatures 0°C and t1°C respectively. Let a0 and a1 be the temperature co-efficients of resistance of the conductor at 0°C and t1°C respectively. It is desired to establish the relationship between a1 and a0. Fig. 1.18 shows the temperature/resistance graph of the conductor. As proved in Art. 1.24, Slope of graph a0 = R0 \ Slope of graph = α0 R0 * The slope of temp./resistance graph of a conductor is always constant (being a straight line). www.EngineeringBooksPdf.com Basic Concepts 21 Slope of graph Similarly, a1 = R1 or Slope of graph = a1 R1 Since the slope of temperature/resistance graph is constant, \ a0 R0 = a1 R1 α 0 R0 α 0 R0 or a1 = = R1 R0 (1 + α 0 t1 ) [ R1 = R0 (1 + α0 t1)] α0 \ α1 =...(i) 1 + α 0 t1 α0 Similarly,* α2 =...(ii) 1 + α 0 t2 Subtracting the reciprocal of eq. (i) from the reciprocal of eq. (ii), Fig. 1.18 1 1 1 + α 0 t 2 1 + α 0 t1 − = − = t2 − t1 α 2 α1 α0 α0 1 \ α2 =...(iii)

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