A Textbook of Electrical Technology Vol I PDF

FIRST MULTICOLOUR EDITION A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOLUME I BASIC ELECTRICAL ENGINEERING IN S.I. SYSTEM OF UNITS  A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOLUME I BASIC ELECTRICAL ENGINEERING...

FIRST MULTICOLOUR EDITION A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOLUME I BASIC ELECTRICAL ENGINEERING IN S.I. SYSTEM OF UNITS  A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOLUME I BASIC ELECTRICAL ENGINEERING IN S.I. SYSTEM OF UNITS (Including rationalized M.K.S.A. System) For the Examinations of B.E. (Common Course to All Branches), B.Tech., B.Sc. (Engg), Sec. A & B of AMIE(I), A.M.I.E.E. (London), I.E.R.E. (London), Grade I.E.T.E., Diploma and other Competitive Examinations B.L. THERAJA A.K. THERAJA Revised by : S.G. TARNEKAR B.E. (Hons), M.Tech., (El. Machines) Ph.D. (Electrical Power Systems) Former Professor & Head, Electrical Engineering Department Visvesvaraya National Institute of Technology, Nagpur S. CHAND AN ISO 9001 : 2000 COMPANY 2005 S. CHAND & COMPANY LTD. RAM NAGAR, NEW DELHI-110 055  S. CHAND & COMPANY LTD. 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Ph : 2226801, 2284815 z Blackie House, 103/5, Walchand Hirachand Marg , Opp. G.P.O., Mumbai-400 001. Ph : 22690881, 22610885 z 3, Gandhi Sagar East, Nagpur-440 002. Ph : 2723901 z 104, Citicentre Ashok, Govind Mitra Road, Patna-800 004. Ph : 2671366, 2302100 Marketing Offices : z 238-A, M.P. Nagar, Zone 1, Bhopal - 462 011 z A-14, Janta Store Shopping Complex, University Marg, Bapu Nagar, Jaipur - 302 015, Phone : 0141-2709153 © Copyright Reserved All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the Publisher. First Edition 1959 Subsequent Editions and Reprint 1960, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73 (Twice), 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84 (Twice), 85, 86, 87, 88 (Twice), 89, 91, 93, 94, 95, 96, 97, 98 (Twice), 99, 2000, 2001, 2002 Thoroughly Revised Twenty-third Edition 2002 Reprint 2003, 2004 First Multicolour Edition 2005 Other Parts Available Volume II : AC & DC Machines Volume III : Transmission, Distribution & Utilization Volume IV : Electronic Devices & Circuits Combined Edition Also Available ISBN : 81-219-2440-5 PRINTED IN INDIA By Rajendra Ravindra Printers (Pvt.) Ltd., 7361, Ram Nagar, New Delhi-110 055 and published by S. Chand & Company Ltd., 7361, Ram Nagar, New Delhi-110 055.  Preface to the Twenty-Third A Revised Multicoloured Edition A uthors feel happy to present to their esteemed readers this revised first multicoloured edition of Vol. I of “A Textbook of Electrical Technology”. To provide a comprehensive treatment of topics in ‘‘Basic Electrical Engineering’’ both for electrical as well as non-electrical students pursuing their studies in civil, mechanical, mining, textile, chemical, industrial, environmental, aerospace, electronic and computer engineering, information technology both at the Degree and Diploma level. Based on the suggestions received from our esteemed readers, both from India and abroad, the scope of the book has been enlarged according to their requirements. Establishment of Technological Universities have taken place in recent past. This resulted into a pool of expert manpower within a large area. Unification of syllabi has taken place and the question papers set during the last 4-5 years have a wider variety and are of more inquisitive nature. Solutions to these with brief logical reasonings have been added for the benefit of our student community. Many universities include a brief coverage on methods of “Electrical Power Generation”, in their first and basic paper on this subject. Hence, this revision includes an introductory chapter on this topic. It is earnestly hoped that with these extensive additions and revisions, this revised edition will prove even more useful to our numerous readers in developing more confidence while appearing at national competitive examinations. I would like to thank my Publishers particularly Mr. Ravindra Kumar Gupta, M.D. and Mr. Bhagirath Kaushik, Regional Manager (Western India) of S. Chand & Company Ltd., for the personal interest they look in the publishing of this revised and enlarged edition. Our student-friends, teacher-colleagues, Booksellers and University authorities have been showing immense faith and affection in our book, which is acknowledged with modesty and regards. We are sure that this revised edition will satisfy their needs to a still greater extent and serve its cause more effectively. S.G. TARNEKAR (v) T Preface to the Twenty-second Edition T he primary objective of Vol. I of A Textbook of Electrical Technology is to provide a comprehensive treatment of topics in ‘‘Basic Electrical Engineering’’ both for electrical as well as non- electrical students pursuing their studies in civil, mechanical, mining, textile, chemical, industrial, environmental, aerospace, electronic and computer engineering both at the Degree and Diploma level. Based on the suggestions received from our esteemed readers, both from India and abroad, the scope of the book has been enlarged according to their requirements. Almost half the solved examples have been deleted and replaced by latest examination papers set upto 1994 in different engineering colleges and technical institutions in India and abroad. Following major additions/changes have been made in the present edition : 1. Three new chapters entitled (a) A.C. Network Analysis (b) A.C. Filter Networks and (c) Fourier Series have been added thereby widening the scope of the book. 2. The chapter on Network Theorems has been updated with the addition of Millman’s Theorem (as applicable to voltage and current sources or both) and an article on Power Transfer Efficiency relating to Maximum Power Transfer Theorem. 3. The additions to the chapter on Capacitors include detailed articles on Transient Relations during Capacitor Charging and Discharging Cycles and also the Charging and Discharging of a Capacitor with Initial Charge. 4. Chapter on Chemical Effects of Current has been thoroughly revised with the inclusion of Electronic Battery Chargers, Static Uninterruptable Power Supply (UPS) Systems, High Temperature Batteries, Secondary Hybrid Cells, Fuel Cells and Aircraft and Submarine Batteries. 5. A detailed description of Thermocouple Ammeter has been added to the chapter on Electrical Instruments. 6. The chapter on Series A.C. Circuits has been enriched with many articles such as Determination of Upper and Lower Half-power Frequencies, Value of Edge Frequencies and Relation between Resonant Power and Off-resonance Power. It is earnestly hoped that with these extensive additions and revisions, the present edition will prove even more useful to our numerous readers than the earlier ones. As ever before, we are thankful to our publishers particularly Sh. Ravindra Kumar Gupta for the personal interest he took in the expeditious printing of this book and for the highly attractive cover design suggested by him. Our sincere thanks go to their hyperactive and result-oriented overseas manager for his globe-trotting efforts to popularise the book from one corner of the globe to the other. Lastly we would love to record our sincere thanks to two brilliant ladies; Mrs. Janaki Krishnan from ever-green fairy land of Kerala and Ms. Shweta Bhardwaj from the fast-paced city of Delhi for the secretarial support they provided us during the prepration of this book. AUTHORS (vi) CONTENTS CONTENTS 1. Electric Current and Ohm’s Law Electron Drift Velocity—Charge Velocity and Velocity of Field Propagation—The Idea of Electric Potential— Resistance—Unit of Resistance—Law of Resistance—Units of Resistivity—Conductance and Conductivity—Effect of Temperature on Resistance—Temperature Coefficient of...1—50 Resistance—Value of a at Different Temperatures— Variation of Resistivity with Temperature—Ohm’s Law— Resistance in Series—Voltage Divider Rule—Resistance in Parallel—Types of Resistors—Nonlinear Resistors— Varistor—Short and Open Circuits—‘Shorts’ in a Series Circuit—‘Opens’ in Series Circuit—‘Open’s in a Parallel Circuit—‘Shorts’ in Parallel Circuits—Division of Current in Parallel Circuits—Equivalent Resistance—Duality Between Series and Parallel Circuits—Relative Potential— Voltage Divider Circuits—Objective Tests. 2. DC Network Theorems...51—174 Electric Circuits and Network Theorems—Kirchhoff’s Laws—Determination of Voltage Sign—Assumed Direction of Current—Solving Simultaneous Equations— Determinants—Solving Equations with Two Unknowns— Solving Equations With Three Unknowns—Independent and Dependent Sources —Maxwell’s Loop Current Method—Mesh Analysis Using Matrix Form—Nodal Analysis with Voltage Sources—Nodal Analysis with Current Sources—Source Conversion—Ideal Constant- Voltage Source—Ideal Constant-Current Source— Superposition Theorem—Thevenin Theorem—How to Thevenize a Given Circuit ?—General Instructions for Finding Thevenin Equivalent Circuit—Reciprocity Theorem—Delta/Star Transformation—Star/Delta Transformation—Compensation Theorem—Norton’s Theorem—How to Nortanize a Given Circuit ?—General Instructions for Finding Norton Equivalent Circuit— Millman’s Theorem—Generalised Form of Millman's Theorem—Maximum Power Transfer Theorem—Power Transfer Efficiency—Objective Tests. 3. Work, Power and Energy...175—188 Effect of Electric Current—Joule’s Law of Electric Heat- ing—Thermal Efficiency—S-I. Units—Calculation of Kilo-watt Power of a Hydroelectric Station—Objective Tests. (vii) 4. Electrostatics...189—212 Static Electricity—Absolute and Relative Permittivity of a Medium—Laws of Electrostatics—Electric Field— Electrostatic Induction—Electric Flux and Faraday Tubes— —Field Strength or Field Intensity or Electric Intensity (E)— Electric Flux Density or Electric Displacement D—Gauss Law—The Equations of Poisson and Laplace—Electric Potential and Energy—Potential and Potential Difference— Potential at a Point—Potential of a Charged Sphere— Equipotential Surfaces—Potential and Electric Intensity Inside a Conducting Sphere—Potential Gradient— Breakdown Voltage and Dielectric Strength—Safety Factor of Dielectric—Boundary Conditions—Objective Tests. 5. Capacitance...213—256 Capacitor—Capacitance—Capacitance of an Isolated Sphere—Spherical Capacitor — Parallel-plate Capacitor— Special Cases of Parallel-plate Capacitor—Multiple and Variable Capacitors—Cylindrical Capacitor—Potential Gradient in Cylindrical Capacitor—Capacitance Between two Parallel Wires—Capacitors in Series—Capacitors in Parallel—Cylindrical Capacitor with Compound Dielectric—Insulation Resistance of a Cable Capacitor— Energy Stored in a Capacitor—Force of Attraction Between Oppositely-charged Plates—Current-Voltage Relationships in a Capacitor—Charging of a Capacitor—Time Constant— Discharging of a Capacitor—Transient Relations during Capacitor Charging Cycle—Transient Relations during Capacitor Discharging Cycle—Charging and Discharging of a Capacitor with Initial Charge—Objective Tests. 6. Magnetism and Electromagnetism...257—296 Absolute and Relative Permeabilities of a Medium—Laws of Magnetic Force—Magnetic Field Strength (H)—Magnetic Potential—Flux per Unit Pole—Flux Density (B)—— Absolute Parmeability (m) and Relative Permeability (mr)— Intensity of Magnetisation (I)—Susceptibility (K)—Relation Between B, H, I and K—Boundary Conditions—Weber and Ewing’s Molecular Theory—Curie Point. Force on a Current-carrying Conductor Lying in a Magnetic Field— Ampere’s Work Law or Ampere’s Circuital Law—Biot- Savart Law—Application of Biot—Savart Law—Force Between two Parallel Conductors—Magnitude of Mutual Force—Definition of Ampere—Magnetic Circuit— Definitions—Composite Series Magnetic Circuit—How to Find Ampere-turns ?—Comparison Between Magnetic and Electric Circuits—Parallel Magnetic Circuits—Series- (viii)  Parallel Magnetic Circuits—Leakage Flux and Hopkinson’s Leakage Coefficient—Magnetisation Curves— Magnetisation curves by Ballistic Galvanometer— Magnetisation Curves by Fluxmete—Objective Tests. 7. Electromagnetic Induction...297—316 Relation Between Magnetism and Electricity—Production of Induced E.M.F. and Current—Faraday’s Laws of Electromagnetic Induction—Direction of Induced E.M.F. and Current—Lenz’s Law—Induced E.M.F.—Dynamically- induced E.M.F.—Statically-induced E.M.F.—Self- Inductance—Coefficient of Self-Inductance (L)—Mutual Inductance—Coefficient of Mutual Inductance (M)— Coefficient of Coupling—Inductances in Series— Inductances in Parallel—Objective Tests. 8. Magnetic Hysteresis...317—338 Magnetic Hysteresis—Area of Hysteresis Loop—Properties and Application of Ferromagnetic Materials—Permanent Magnet Materials—Steinmetz Hysteresis Law—Energy Stored in Magnetic Field—Rate of Change of Stored Energy—Energy Stored per Unit Volume—Lifting Power of Magnet—Rise of Current in Inductive Circuit—Decay of Current in Inductive Circuit—Details of Transient Current Rise in R-L Circuit—Details of Transient Current Decay in R-L Circuit—Automobile Ignition System—Objective Tests. 9. Electrochemical Power Sources... 339—374 Faraday’s Laws of electrolysis—Polarisation or Back e.m.f.—Value of Back e.m.f.—Primary and Secondary Batteries—Classification of Secondary Batteries base on their Use—Classification of Lead Storage Batteries— Parts of a Lead-acid Battery—Active Materials of Lead- acid Cells—Chemical Changes—Formation of Plates of Lead-acid Cells—Plante Process—Structure of Plante Plates—Faure Process—Positive Pasted Plates—Negative Pasted Plates—Structure of Faure Plates—Comparison : Plante and Faure Plates—Internal Resistance and Capacity of a Cell—Two Efficiencies of the Cell— Electrical Characteristics of the Lead-acid Cell—Battery Ratings—Indications of a Fully-Charged Cell—Application of Lead-acid Batteries—Voltage Regulators—End-cell Control System—Number of End-cells—Charging Systems—Constant-current System-Constant-voltage System—Trickle Charging—Sulphation-Causes and Cure— Maintenance of Lead-acid Cells—Mains operated Battery Chargers—Car Battery Charger—Automobile Battery (ix)  Charger—Static Uninterruptable Power Systems—Alkaline Batteries—Nickel-iron or Edison Batteries—Chemical Changes—Electrical Characteristics—Nickel-Cadmium Batteries—Chemical Changes—Comparison : Lead-acid and Edison Cells—Silver-zinc Batteries—High Temperature Batteries—Secondary Hybrid Cells—Fuel Cells— Hydrogen-Oxygen Fuel Cells—Batteries for Aircraft— Batteries for Submarines—Objective Tests. 10. Electrical Instruments and Measurements...375—452 Classification of AC Motors—Induction Motor: General Principal—Construction—Squirrel-cage Rotor—Phase- wound Rotor—Production of Rotating Field—Three-phase Supply—Mathematical Proof—Why does the Rotor Rotate ?—Slip—Frequency of Rotor Current—Relation between Torque and Rotor Power Factor—Starting Torque—Starting Torque of a Squirrel-cage Motor—Starting Torque of a Slip-ring Motor—Condition for Maximum Starting Torque—Effect of Change in Supply Voltage on Starting Torque—Rotor E.M.F and Reactance under Running Conditions—Torque under Running Condition—Condition for Maximum Torque Under Running Conditions—Rotor Torque and Breakdown Torque—Relation between Torque and Slip—Effect of Change in Supply Voltage on Torque and Speed—Effect of Change in Supply Frequency Torque and Speed—Full-load Torque and Maximum Torque— Starting Torque and Maximum Torque—Torque/Speed Curve—Shape of Torque/Speed Curve—Current/Speed Curve of an Induction Motor—Torque/Speed Characteristic Under Load—Plugging of an Induction Motor—Induction Motor Operating as a Generator—Complete Torque/Speed Curve of a Three-phase Machine—Measurement of Slip— Power Stages in an Induction Motor—Torque Developed by an Induction Motor—Torque, Mechanical Power and Rotor Output—Induction Motor Torque Equation— Synchronous Watt—Variation in Rotor Current—Analogy with a Mechnical Clutch—Analogy with a D.C. Motor— Sector Induction Motor—Linear Induction Motor— Properties of a Linear Induction Motor—Magnetic Levitation—Induction Motor as a Generalized Transformer—Rotor Output—Equivalent Circuit of the Rotor—Equivalent Circuit of an Induction Motor—Power Balance Equation—Maximum Power Output— Corresponding Slip—Objective Tests. 11. A.C. Fundamentals...453—496 Generation of Alternating Voltages and Currents— Equations of the Alternating Voltages and Currents— (x)  Alternate Method for the Equations of Alternating Voltages and currents—Simple Waveforms—Complex Waveforms—Cycle—Time-Period—Frequency— Amplitude—Different Forms of E.M.F. Equation—Phase— Phase Difference—Root Mean Square (R.M.S.) Value— Mid-ordinate Method—Analytical Method—R.M.S. Value of a Complex Wave—Average Value—Form Factor— Crest or Peak Factor—R.M.S. Value of H.W. Rectified A.C.—Average Value—Form Factor of H.W. Rectified —Representation of Alternating Quantities—Vector Diagrams Using R.M.S. Values—Vector Diagrams of Sine Waves of Same Frequency—Addition of Two Alternating Quantities—Addition and Subtraction of Vectors—A.C. Through Resistance, Inductance and Capacitance—A.C. through Pure Ohmic Resistance alone—A.C. through Pure Inductance alone—Complex Voltage Applied to Pure Inductance—A.C. through Capacitance alone Objective Tests. 12. Complex Numbers... 497—506 Mathematical Representation of Vectors—Symbolic Notation—Significance of Operator j—Conjugate Complex Numbers—Trigonometrical Form of Vector—Exponential Form of Vector—Polar Form of Vector Representation— Addition and Subtraction of Vector Quantities— Multiplication and Division of Vector Quantities— Power and Root of Vectors—The 120° Operator— Objective Tests. 13. Series A.C. Circuits...507—556 A.C. through Resistance and Inductance—Power Factor— Active and Reactive Components of Circuit Current- I—Active, Reactive and Apparent Power—Q-factor of a Coil—Power in an Iron-cored Chocking Coil—A.C. Through Resistance and Capacitance—Dielectric Loss and Power Factor of a Capacitor—Resistance, Inductance and Capacitance in Series—Resonance in R-L-C Circuits— Graphical Representation of Resonance—Resonance Curve—Half-power Bandwidth of a Resonant Circuit— Bandwidth B at any Off-resonance Frequency— Determination of Upper and Lower Half-Power Frequencies—Values of Edge Frequencies—Q-Factor of a Resonant Series Circuit—Circuit Current at Frequencies Other than Resonant Frequencies—Relation Between Resonant Power P0 and Off-resonant Power P—Objective Test. (xi) 14. Parallel A.C. Circuits...557—598 Solving Parallel Circuits—Vector or Phasor Method— Admittance Method—Application of Admittance Method— Complex or Phasor Algebra—Series-Parallel Circuits— Series Equivalent of a Parallel Circuit—Parallel Equaivalent of a Series Circuit—Resonance in Parallel Circuits— Graphic Representation of Parallel Resonance—Points to Remember—Bandwidth of a Parallel Resonant Circuit— Q-factor of a Parallel Circuit—Objective Tests. 15. A.C. Network Analysis...599—626 Introduction—Kirchhoff's Laws—Mesh Analysis—Nodal Analysis—Superposition Theorem—Thevenin’s Theorem—Reciprocity Theorem—Norton’s Theorem— Maximum Power Transfer Theorem-Millman’s Theorem. 16. A.C. Bridges...627—640 A.C. Bridges—Maxwell’s Inductance Bridge—Maxwell- Wien Bridge—Anderson Bridge—Hay’s Bridge—The Owen Bridge—Heaviside Compbell Equal Ratio Bridge— Capacitance Bridge—De Sauty Bridge—Schering Bridge— Wien Series Bridge—Wien Parallel Bridge—Objective Tests. 17. A.C. Filter Networks...641—654 Introduction—Applications—Different Types of Filters— Octaves and Decades of frequency—Decible System— Value of 1 dB—Low-Pass RC Filter—Other Types of Low-Pass Filters—Low-Pass RL Filter—High-Pass R C Filter—High Pass R L Filter—R-C Bandpass Filter—R-C Bandstop Filter—The-3 dB Frequencies—Roll-off of the Response Curve—Bandstop and Bandpass Resonant Filter Circuits—Series-and Parallel-Resonant Bandstop Filters—Parallel-Resonant Bandstop Filter—Series- Resonant Bandpass Filter—Parallel-Resonant Bandpass Filter—Objective Test. 18. Circle Diagrams...655—664 Circle Diagram of a Series Circuit—Rigorous Mathematical Treatment—Constant Resistance but Variable Reactance—Properties of Constant Reactance But Variable Resistance Circuit—Simple Transmission Line Circuit. (xii) 19. Polyphase Circuits...665—752 Generation of Polyphase Voltages—Phase Sequence— Phases Sequence At Load—Numbering of Phases— Interconnection of Three Phases—Star or Wye (Y) Connection—Values of Phase Currents—Voltages and Currents in Y-Connection—Delta (D) or Mesh Connection—Balanced Y/D and D/Y Conversions— Star and Delta Connected Lighting Loads—Power Factor Improvement—Power Correction Equipment—Parallel Loads—Power Measurement in 3-phase Circuits—Three Wattmeter Method—Two Wattmeter Method—Balanced or Unbalanced load—Two Wattmeter Method-Balanced Load—Variations in Wattmeter Readings—Leading Power Factor—Power Factor-Balanced Load—Balanced Load- LPF—Reactive Voltamperes with One Wattmeter— One Wattmeter Method—Copper Required for Transmitting Power Under Fixed Conditions—Double Subscript Notation—Unbalanced Loads—Unbalanced D-connected Load—Four-wire Star-connected Unbalanced Load—Unbalanced Y-connected Load Without Neutral— Millman’s Thereom—Application of Kirchhoff's Laws— Delta/Star and Star/Delta Conversions—Unbalanced Star-connected Non-inductive Load—Phase Sequence Indicators—Objective Tests. 20. Harmonics...753—778 Fundamental Wave and Harmonics—Different Complex Waveforms—General Equation of a Complex Wave— R.M.S. Value of a Complex Wave—Form Factor of a Copmplex Wave—Power Supplied by a Complex Wave— Harmonics in Single-phase A.C Circuits—Selective Resonance Due to Harmonics—Effect of Harmonics on Measurement of Inductance and Capacitance— Harmonics in Different Three-phase Systems—Harmonics in Single and 3-Phase Transformers—Objective Tests. 21. Fourier Series...779—814 Harmonic Analysis—Periodic Functions—Trigonometric Fourier Series—Alternate Forms of Trigonometric Fourier Series—Certain Useful Integral Calculus Theorems— Evalulation of Fourier Constants—Different Types of Functional Symmetries—Line or Frequency Spectrum— Procedure for Finding the Fourier Series of a Given Function—Wave Analyzer—Spectrum Analyzer—Fourier Analyzer—Harmonic Synthesis—Objective Tests. (xiii) 22. Transients...815—834 Introduction—Types of Transients—Important Differ- ential Equations—Transients in R-L Circuits (D.C.),— Short Circuit Current—Time Constant—Transients in R-L Circuits (A.C.)—Transients in R-C Series Circuits (D.C.)—Transients in R-C Series Circuits (A.C)—Double Energy Transients—Objective Tests. 23. Symmetrical Components...835—854 Introduction—The Positive-sequence Components— The Negative-sequence Components—The Zero-sequence Components—Graphical Composition of Sequence Vectors—Evaluation of VA1 or V1—Evaluation of VA2 or V2—Evaluation V A0 or V0—Zero Sequence Components of Current and Voltage—Unbalanced Star Load form Unbalanced Three-phase Three-Wire System— Unbalanced Star Load Supplied from Balanced Three- phase Three-wire System—Measurement of Symmetrical Components of Circuits—Measurement of Positive and Negative-sequence Voltages—Measurement of Zero- sequence Component of Voltage—Objective Tests. 24. Introduction to Electrical Energy Generation...855—864 Preference for Electricity—Comparison of Sources of Power—Sources for Generation of Electricity—Brief Aspects of Electrical Energy Systems—Utility and Consumers—Why is the Three-phase a.c. system Most Popular?—Cost of Generation—Staggering of Loads during peak-demand Hours—Classifications of Power Transmission—Selecting A.C. Transmission Voltage for a Particular Case—Conventional Sources of Electrical Energy—Steam Power Stations (Coal-fired)—Nuclear Power Stations—Advantages of Nuclear Generation— Disadvantages—Hydroelectric Generation—Non- Conventional Energy Sources—Photo Voltaic Cells (P.V. Cells or SOLAR Cells)—Fuel Cells—Principle of Operation—Chemical Process (with Acidic Electrolyte)— Schematic Diagram—Array for Large outputs—High Lights—Wind Power—Background—Basic Scheme— Indian Scenario. Index (xiv)  VOLUME – I BASIC ELECTRICAL ENGINEERING Learning Objectives C H A P T E R 1 ➣ Electron Drift Velocity ➣ Charge Velocity and Velocity of Field Propagation ELECTRIC ➣ The Idea of Electric Potential Resistance ➣ Unit of Resistance CURRENT ➣ Law of Resistance ➣ Units of Resistivity Conductance and AND OHM’S Conductivity ➣ Temperature Coefficient of Resistance LAW ➣ Value of α at Different Temperatures ➣ Variation of Resistivity with Temperature ➣ Ohm’s Law ➣ Resistance in Series ➣ Voltage Divider Rule ➣ Resistance in Parallel ➣ Types of Resistors ➣ Nonlinear Resistors ➣ Varistor ➣ Short and Open Circuits ➣ ‘Shorts’ in a Series Circuit ➣ ‘Opens’ in Series Circuit ➣ ‘Open’s in a Parallel Circuit ➣ ‘Shorts’ in Parallel Circuits ➣ Division of Current in Parallel Circuits ➣ Equivalent Resistance ➣ Duality Between Series and Parallel Circuits ➣ Relative Potential Ohm’s law defines the relationship ➣ Voltage Divider Circuits © between voltage, resistance and current. This law is widely employed while designing electronic circuits 2 Electrical Technology 1.1. Electron Drift Velocity The electron moves at the Suppose that in a conductor, the number of free electrons Fermi speed, and has only 3 available per m of the conductor material is n and let their a tiny drift velocity superimposed axial drift velocity be ν metres/second. In time dt, distance by the applied electric field travelled would be ν × dt. If A is area of cross-section of the conductor, then the volume is νAdt and the number of elec- trons contained in this volume is νA dt. Obviously, all these electrons will cross the conductor cross-section in time dt. If drift e is the charge of each electron, then total charge which crosses Vd velocity the section in time dt is dq = nAeν dt. Electric Since current is the rate of flow of charge, it is given as Field E dq nAeν dt i= = ∴ i = nAeν dt dt 2 Current density, J = i/A = ne ν ampere/metre 6 2 29 Assuming a normal current density J = 1.55 × 10 A/m , n = 10 for a copper conductor −19 and e = 1.6 × 10 coulomb, we get 6 29 −19 −5 1.55 × 10 = 10 × 1.6 × 10 × ν ∴ν = 9.7 × 10 m/s = 0.58 cm/min It is seen that contrary to the common but mistaken view, the electron drift velocity is rather very slow and is independent of the current flowing and the area of the conductor. → N.B.Current density i.e., the current per unit area, is a vector quantity. It is denoted by the symbol J. → Therefore, in vector notation, the relationship between current I and J is : → → → I = J.a [where a is the vector notation for area ‘a’] For extending the scope of the above relationship, so that it becomes applicable for area of any shape, we write : I = J.d a The magnitude of the current density can, therefore, be written as J·α. 24 3 Example 1.1. A conductor material has a free-electron density of 10 electrons per metre. −2 When a voltage is applied, a constant drift velocity of 1.5 × 10 metre/second is attained by the 2 electrons. If the cross-sectional area of the material is 1 cm , calculate the magnitude of the current. −19 Electronic charge is 1.6 × 10 coulomb. (Electrical Engg. Aligarh Muslim University) Solution. The magnitude of the current is i = nAeν amperes Here, n = 1024 ; A = 1 cm2 = 10−4 m2 −19 −2 e = 1.6 × 10 C ; v = 1.5 × 10 m/s 24 −4 −19 −2 ∴ i = 10 × 10 × 1.6 × 10 × 1.5 × 10 = 0.24 A 1.2. Charge Velocity and Velocity of Field Propagation The speed with which charge drifts in a conductor is called the velocity of charge. As seen from above, its value is quite low, typically fraction of a metre per second. However, the speed with which the effect of e.m.f. is experienced at all parts of the conductor resulting in the flow of current is called the velocity of propagation of electrical field. It is indepen- 8 dent of current and voltage and has high but constant value of nearly 3 × 10 m/s.  Electric Current and Ohm’s Law 3 Example 1.2. 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? Solution. i = neAν or ν = i/neA 28 −19 −4 −7 ∴ ν = 1/(8.5 × 10 × 1.6 × 10 × 1 × 10 ) = 7.35 × 10 m/s = 0.735 μm/s Time taken by the charge to travel conductor length of 10 km is 3 10 × 10 t = distance = 10 = 1.36 × 10 s velocity 7.35 × 10−7 Now, 1 year = 365 × 24 × 3600 = 31,536,000 s 10 t = 1.36 × 10 /31,536,000 = 431 years 1.3. The Idea of Electric Potential In Fig. 1.1, a simple voltaic cell is shown. It consists of copper plate (known as anode) and a zinc rod (i.e. cathode) immersed in dilute sulphuric acid (H2SO4) contained in a suitable vessel. The chemical action taking place within the cell causes the electrons to be removed from copper plate and to be deposited on the zinc rod at the same time. This transfer of electrons is accomplished through the agency of the diluted H2SO4 which is known as the electrolyte. The result is that zinc rod becomes negative due to the deposition of electrons on it and the copper plate becomes positive due to the removal of electrons from it. The large number of electrons collected on the zinc rod is being attracted by anode but is prevented from returning to it by the force set up by the chemical action within the cell. Fig. 1.1. Fig. 1.2 But if the two electrodes are joined by a wire externally, then electrons rush to the anode thereby equalizing the charges of the two electrodes. However, due to the continuity of chemical action, a continuous difference in the number of electrons on the two electrodes is maintained which keeps up a continuous flow of current through the external circuit. The action of an electric cell is similar to that of a water pump which, while working, maintains a continuous flow of water i.e., water current through the pipe (Fig. 1.2). It should be particularly noted that the direction of electronic current is from zinc to copper in the external circuit. However, the direction of conventional current (which is given by the direction 4 Electrical Technology of flow of positive charge) is from copper to zinc. In the present case, there is no flow of positive charge as such from one electrode to another. But we can look upon the arrival of electrons on copper plate (with subsequent decrease in its positive charge) as equivalent to an actual departure of positive charge from it. When zinc is negatively charged, it is said to be at negative potential with respect to the electrolyte, whereas anode is said to be at positive potential relative to the electrolyte. Between themselves, copper plate is assumed to be at a higher potential than the zinc rod. The difference in potential is continuously maintained by the chemical action going on in the cell which supplies energy to establish this potential difference. 1.4. Resistance It may be defined as the property of a substance due to which it opposes (or restricts) the flow of electricity (i.e., electrons) through it. Metals (as a class), acids and salts solutions are good conductors of electricity. Amongst pure metals, silver, copper and aluminium are very good conductors in the given order.* This, as discussed earlier, is due to the presence of a large number of free or loosely-attached electrons in their atoms. These vagrant electrons assume a directed motion on the application of an electric potential difference. These electrons while flowing pass through the molecules or the atoms of the conductor, collide and other atoms and electrons, thereby producing heat. Those substances which offer relatively greater Cables are often covered with materials that difficulty or hindrance to the passage of these electrons do not carry electric current easily are said to be relatively poor conductors of electricity like bakelite, mica, glass, rubber, p.v.c. (polyvinyl chloride) and dry wood etc. Amongst good insulators can be included fibrous substances such as paper and cotton when dry, mineral oils free from acids and water, ceramics like hard porcelain and asbestos and many other plastics besides p.v.c. It is helpful to remember that electric friction is similar to friction in Mechanics. 1.5. The Unit of Resistance The practical unit of resistance is ohm.** A conductor is said to have a resistance of one ohm if it permits one ampere current to flow through it when one volt is impressed across its terminals. For insulators whose resistances are very high, a much bigger unit 6 is used i.e., mega-ohm = 10 ohm (the prefix ‘mega’ or mego meaning 3 a million) or kilo-ohm = 10 ohm (kilo means thousand). In the case of −3 very small resistances, smaller units like milli-ohm = 10 ohm or mi- −6 cro-ohm = 10 ohm are used. The symbol for ohm is Ω. George Simon Ohm * However, for the same resistance per unit length, cross-sectional area of aluminium conductor has to be 1.6 times that of the copper conductor but it weighs only half as much. Hence, it is used where economy of weight is more important than economy of space. ** After George Simon Ohm (1787-1854), a German mathematician who in about 1827 formulated the law known after his name as Ohm’s Law.  Electric Current and Ohm’s Law 5 Table 1.1. Multiples and Sub-multiples of Ohm Prefix Its meaning Abbreviation Equal to 6 Mega- One million MΩ 10 Ω Kilo- One thousand kΩ 103 Ω Centi- One hundredth – – −3 Milli- One thousandth mΩ 10 Ω Micro- One millionth μΩ 10 −6 Ω 1.6. Laws of Resistance The resistance R offered by a conductor depends on the following factors : (i) It varies directly as its length, l. (ii) It varies inversely as the cross-section A of the conductor. (iii) It depends on the nature of the material. (iv) It also depends on the temperature of the conductor. Fig. 1.3. Fig. 1.4 Neglecting the last factor for the time being, we can say that l l R ∝ or R = ρ...(i) A A where ρ is a constant depending on the nature of the material of the conductor and is known as its specific resistance or resistivity. If in Eq. (i), we put 2 l = 1 metre and A = 1 metre , then R = ρ (Fig. 1.4) Hence, specific resistance of a material may be defined as the resistance between the opposite faces of a metre cube of that material. 1.7. Units of Resistivity From Eq. (i), we have ρ = AR l 6 Electrical Technology In the S.I. system of units, A metre 2 × R ohm AR ρ = = ohm-metre l metre l Hence, the unit of resistivity is ohm-metre (Ω-m). 3 It may, however, be noted that resistivity is sometimes expressed as so many ohm per m. Although, it is incorrect to say so but it means the same thing as ohm-metre. 2 If l is in centimetres and A in cm , then ρ is in ohm-centimetre (Ω-cm). Values of resistivity and temperature coefficients for various materials are given in Table 1.2. The resistivities of commercial materials may differ by several per cent due to impurities etc. Table 1.2. Resistivities and Temperature Coefficients Material Resistivity in ohm-metre Temperature coefficient at at 20ºC (× 10− ) 20ºC (× 10− ) 8 4 Aluminium, commercial 2.8 40.3 Brass 6–8 20 Carbon 3000 – 7000 −5 Constantan or Eureka 49 +0.1 to −0.4 Copper (annealed) 1.72 39.3 German Silver 20.2 2.7 (84% Cu; 12% Ni; 4% Zn) Gold 2.44 36.5 Iron 9.8 65 Manganin 44 – 48 0.15 (84% Cu ; 12% Mn ; 4% Ni) Mercury 95.8 8.9 Nichrome 108.5 1.5 (60% Cu ; 25% Fe ; 15% Cr) Nickel 7.8 54 Platinum 9 – 15.5 36.7 Silver 1.64 38 Tungsten 5.5 47 14 Amber 5 × 10 10 Bakelite 10 10 12 Glass 10 – 10 15 Mica 10 16 Rubber 10 14 Shellac 10 Sulphur 1015  Electric Current and Ohm’s Law 7 Example 1.3. A coil consists of 2000 turns of copper wire hav- 2 ing a cross-sectional area of 0.8 mm. The mean length per turn is 80 cm and the resistivity of copper is 0.02 μΩ–m. Find the resistance of the coil and power absorbed by the coil when connected across 110 V d.c. supply. (F.Y. Engg. Pune Univ. May 1990) Solution. Length of the coil, l = 0.8 × 2000 = 1600 m ; 2 −6 2 A = 0.8 mm = 0.8 × 10 m. l −6 −6 R = ρ = 0.02 × 10 × 1600/0.8 × 10 = 40 Ω A 2 2 Power absorbed = V / R = 110 /40 = 302.5 W Example 1.4. An aluminium wire 7.5 m long is connected in a 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 μΩ-m ; that of the aluminium is 0.028 μΩ-m. (F.Y. Engg. Pune Univ. May 1991) Solution. Let the subscript 1 represent aluminium and sub- script 2 represent copper. l1 l R ρ l a R1 = ρ and R2 = ρ2 2 ∴ 2= 2. 2. 1 a1 a2 R1 ρ1 l1 a2 R ρ l ∴ a2 = a1. 1. 2. 2...(i) R2 ρ1 l1 Now I1 = 3 A ; I2 = 5 −3 = 2 A. If V is the common voltage across the parallel combination of aluminium and copper wires, then V = I1 R1 = I2 R2 ∴ R1/R2 = I2/I1 = 2/3 2 2 π ×1 a1 = πd = = π mm 2 4 4 4 Substituting the given values in Eq. (i), we get π × 2 × 0.017 × 6 = 0.2544 m 2 a2 = 4 3 0.028 7.5 2 ∴ π × d2 /4 = 0.2544 or d2 = 0.569 mm Example 1.5. (a) A rectangular carbon block has dimensions 1.0 cm × 1.0 cm × 50 cm. (i) What is the resistance measured between the two square ends ? (ii) between two opposing rectan- −5 gular faces / Resistivity of carbon at 20°C is 3.5 × 10 Ω-m. (b) A current of 5 A exists in a 10-Ω resistance for 4 minutes (i) how many coulombs and (ii) how many electrons pass through any section of the resistor in this time ? Charge of the electron = 1.6 × 10−19 C. (M.S. Univ. Baroda) Solution. (a) (i) R = ρ l/A 2 −4 2 Here, A = 1 × 1 = 1 cm = 10 m ; l = 0.5 m −5 −4 ∴ R = 3.5 × 10 × 0.5/10 = 0.175 Ω 2 −3 2 (ii) Here, l = 1 cm; A = 1 × 50 = 50 cm = 5 × 10 m −5 −2 −3 −5 R = 3.5 × 10 × 10 /5 × 10 = 7 × 10 Ω 8 Electrical Technology (b) (i) Q = It = 5 × (4 × 60) = 1200 C Q 1200 = n = e = 20 (ii) −19 75 × 10 1.6 × 10 Example 1.6. Calculate the resistance of 1 km long cable com- posed of 19 stands of similar copper conductors, each strand being 1.32 mm in diameter. Allow 5% increase in length for the ‘lay’ (twist) of each strand in completed cable. Resistivity of copper may be taken as 1.72 × 10−8 Ω-m. Cross section of a packed bundle of strands Solution. Allowing for twist, the length of the stands. = 1000 m + 5% of 1000 m = 1050 m Area of cross-section of 19 strands of copper conductors is 19 × π × d2/4 = 19 π × (1.32 × 10−3)2/4 m2 −8 l 1.72 × 10 × 1050 × 4 = Now, R = ρ = 2 −6 0.694 Ω A 19π × 1.32 × 10 Example 1.7. 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 percent more current than the latter and the latter is 47 percent longer than the former. Determine the ratio of their cross sectional areas. (Elect. Engg. Nagpur Univ. 1993) Solution. Let suffix 1 represent lead and suffix 2 represent iron. We are given that ρ1/ρ2 = 49/24; if i2 = 1, i1 = 1.8; if l1 = 1, l2 = 1.47 l1 l2 Now, R1 = ρ1 A and R2 = ρ 2 A 1 2 Since the two wires are in parallel, i1 = V/R1 and i2 = V/R2 i2 R ρl A ∴ = 1 = 11 × 2 i1 R2 A1 ρ2l2 A2 i ρl 1 × 24 × 1.47 = 0.4 ∴ = 2× 22 = A1 i1 ρ1l1 1.8 49 Example 1.8. A piece of silver wire has a resistance of 1 Ω. What will be the resistance of manganin wire of one-third the length and one-third the diameter, if the specific resistance of manganin is 30 times that of silver. (Electrical Engineering-I, Delhi Univ.) l l2 Solution. For silver wire, R1 = 1 ; For manganin wire, R = ρ2 A A 1 2 R2 ρ l A ∴ = 2× 2× 1 R1 ρ1 l1 A2 Now A1 = πd12/4 and A2 = π d22/4 ∴ A1/A2 = d12/d22 2 R2 ρ l ⎛d ⎞ ∴ = 2 × 2 ×⎜ 1 ⎟ R1 ρ1 l1 ⎝ d 2 ⎠ 2 2 R1 = 1 Ω; l2/l1 = 1/3, (d1/d2) = (3/1) = 9; ρ 2/ρ 1 = 30 ∴ R2 = 1 × 30 × (1/3) × 9 = 90 Ω  Electric Current and Ohm’s Law 9 −8 Example 1.9. The resistivity of a ferric-chromium-aluminium alloy is 51 × 10 Ω-m. A sheet of the material is 15 cm long, 6 cm wide and 0.014 cm thick. Determine resistance between (a) opposite ends and (b) opposite sides. (Electric Circuits, Allahabad Univ.) Solution. (a) As seen from Fig. 1.5 (a) in this case, l = 15 cm = 0.15 cm A = 6 × 0.014 = 0.084 cm2 = 0.084 × 10−4 m2 −8 l 51 × 10 × 0.15 R = ρ A= −4 0.084 × 10 − = 9.1 × 10 3 Ω (b) As seen from Fig. 1.5 (b) here l = 0.014 cm = 14 × 10−5 m A = 15 × 6 = 90 cm2 = 9 × 10−3 m2 Fig. 1.5 − ∴ R = 51 × 10−8 × 14 × 10−5/9 × 10−3 = 79.3 × 10 10 Ω Example 1.10. The resistance of the wire used for telephone is 35 Ω per kilometre when the −8 weight of the wire is 5 kg per kilometre. If the specific resistance of the material is 1.95 × 10 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. Here R = 35 Ω; l = 1 km = 1000 m; ρ = 1.95 × 10−8 Ω-m −8 l or A = ρl ∴ A = 1.95 × 10 × 1000 = –8 2 Now, R = ρ 55.7 × 10 m A R 35 If the second case, if the wire is of the material but weighs 20 kg/km, then its cross-section must be greater than that in the first case. 20 × 55.7 × 10−8 = 222.8 × 10−8 m 2 Cross-section in the second case = 5 −8 l 1.95 × 10 × 16000 = Length of wire = 2 × 8 = 16 km = 16000 m ∴ R = ρ = −8 140.1 Ω A 222.8 × 10 Tutorial Problems No. 1.1 1. Calculate the resistance of 100 m length of a wire having a uniform cross-sectional area of 0.1 mm2 if the wire is made of manganin having a resistivity of 50 × 10−8 Ω-m. If the wire is drawn out to three times its original length, by how many times would you expect its resistance to be increased ? [500 Ω; 9 times] 2. A cube of a material of side 1 cm has a resistance of 0.001 Ω between its opposite faces. If the same volume of the material has a length of 8 cm and a uniform cross-section, what will be the resistance of this length ? [0.064 Ω] 3. 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 per cent more current than the latter and the latter is 47 per cent longer than the former. Determine the ratio of their cross-sectional area. [2.5 : 1] 4. A rectangular metal strip has the following dimensions : x = 10 cm, y = 0.5 cm, z = 0.2 cm Determine the ratio of resistances Rx, Ry, and Rz between the respective pairs of opposite faces. [Rx : Ry : Rz : 10,000 : 25 : 4] (Elect. Engg. A.M.Ae. S.I.) 2 5. The resistance of a conductor 1 mm in cross-section and 20 m long is 0.346 Ω. Determine the specific −8 resistance of the conducting material. [1.73 × 10 Ω-m] (Elect. Circuits-1, Bangalore Univ. 1991) 6. When a current of 2 A flows for 3 micro-seconds in a coper wire, estimate the number of electrons crossing the cross-section of the wire. (Bombay University, 2000) Hint : With 2 A for 3 μ Sec, charge transferred = 6 μ-coulombs −6 −19 + 13 Number of electrons crossed = 6 × 10 /(1.6 × 10 ) = 3.75 × 10 10 Electrical Technology 1.8. Conductance and Conductivity Conductance (G) is reciprocal of resistance*. Whereas resistance of a conductor measures the opposition which it offers to the flow of current, the conductance measures the inducement which it offers to its flow. l 1 A σA From Eq. (i) of Art. 1.6, R = ρ A or G = ρ. l = l where σ is called the conductivity or specific conductance of a conductor. The unit of conductance is siemens (S). Earlier, this unit was called mho. It is seen from the above equation that the conductivity of a material is given by G siemens × l metre σ = G l = 2 = G l siemens/metre A A metre A Hence, the unit of conductivity is siemens/metre (S/m). 1.9. Effect of Temperature on Resistance The effect of rise in temperature is : (i) to increase the resistance of pure metals. The increase is large and fairly regular for normal ranges of temperature. The temperature/resistance graph is a straight line (Fig. 1.6). As would be presently clarified, metals have a positive temperature co-efficient of resistance. (ii) to increase the resistance of alloys, though in their case, the increase is relatively small and irregular. For some high-resistance alloys like Eureka (60% Cu and 40% Ni) and manganin, the increase in resistance is (or can be made) negligible over a considerable range of temperature. (iii) to decrease the resistance of electrolytes, insulators (such as paper, rubber, glass, mica etc.) and partial conductors such as carbon. Hence, insulators are said to possess a negative temperature-coefficient of resistance. 1.10. Temperature Coefficient of Resistance Let a metallic conductor having a resistance of R0 at 0°C be heated of t°C and let its resistance at this temperature be Rt. Then, considering normal ranges of temperature, it is found that the increase in resistance Δ R = Rt − R0 depends (i) directly on its initial resistance (ii) directly on the rise in temperature (iii) on the nature of the material of the conductor. or Rt − R0 ∝ R × t or Rt −R0 = α R0 t...(i) where α (alpha) is a constant and is known as the temperature coefficient of resistance of the conduc- tor. Rt − R0 ΔR Rearranging Eq. (i), we get α = R × t = R × t 0 0 If R0 = 1 Ω, t = 1°C, then α = Δ R = Rt − R0 Hence, the temperature-coefficient of a material may be defined as : the increase in resistance per ohm original resistance per °C rise in temperature. From Eq. (i), we find that Rt = R0 (1 + α t)...(ii) * In a.c. circuits, it has a slightly different meaning.  Electric Current and Ohm’s Law 11 It should be remembered that the above equation holds good for both rise as well as fall in temperature. As tem- perature of a conductor is decreased, its resistance is also decreased. In Fig. 1.6 is shown the temperature/resistance graph for copper and is practically a straight line. If this line is extended backwards, it would cut the tempera- ture axis at a point where temperature is − 234.5°C (a number quite easy to remember). It means that theoretically, the resistance of copper conductor will Fig. 1.6 become zero at this point though as shown by solid line, in practice, the curve departs from a straight line at very low temperatures. From the two similar triangles of Fig. 1.6 it is seen that : Rt t + 234.5 R0 = 234.5 ( = 1+ t 234.5 ) ∴ ( t ) Rt = R0 1 + 234.5 or Rt = R0 (1 + α t) where α = 1/234.5 for copper. 1.11. Value of α at Different Temperatures So far we did not make any distinction between values of α at different temperatures. But it is found that value of α itself is not constant but depends on the initial temperature on which the increment in resistance is based. When the increment is based on the resistance measured at 0°C, then α has the value of α0. At any other initial temperature t°C, value of α is αt and so on. It should be remembered that, for any conductor, α0 has the maximum value. Suppose a conductor of resistance R0 at 0°C (point A in Fig. 1.7) is heated to t°C (point B). Its resistance Rt after heating is given by Rt = R0 (1 + α0 t)...(i) where α0 is the temperature-coefficient at 0°C. Now, suppose that we have a conductor of resistance Rt at temperature t°C. Let this conductor be cooled from t°C to 0°C. Obviously, now the initial point is B and the final point is A. The final resistance R0 is given in terms of the initial resistance by the following equation R0 = Rt [1 + αt (− t)] = Rt (1 −αt. t)...(ii) Rt − R0 From Eq. (ii) above, we have αt = Rt × t Substituting the value of Rt from Eq. (i), we get R0 (1 + α 0t ) − R0 α0 α0 αt = = ∴ αt =...(iii) R0 (1 + α 0 t ) × t 1 + α0 t 1 + α0 t In general, let α1= tempt. coeff. at t1°C ; α2 = tempt. coeff. at t2°C. Then from Eq. (iii) above, we get Fig. 1.7 12 Electrical Technology α0 1 + α0 t1 α1 = or 1 = 1 + α0 t1 α1 α0 1 1 + α 0 t2 Similarly, = α2 α0 Subtracting one from the other, we get 1 − 1 1 1 1 = (t2 −t1) or = + (t2 −t1) or α2 = α 2 α1 α2 α1 1/α1 + (t2 − t1) Values of α for copper at different temperatures are given in Table No. 1.3. Table 1.3. Different values of α for copper Tempt. in °C 0 5 10 20 30 40 50 α 0.00427 0.00418 0.00409 0.00393 0.00378 0.00364 0.00352 In view of the dependence of α on the initial temperature, we may define the temperature coefficient of resistance at a given temperature as the charge in resistance per ohm per degree centigrade change in temperature from the given temperature. In case R0 is not given, the relation between the known resistance R1 at t1°C and the unknown resistance R2 at t2°C can be found as follows : R2 = R0 (1 + α0 t2) and R1 = R0 (1 + α0 t1) R2 1 + α 0t2 ∴ =...(iv) R1 1 + α0t1 The above expression can be simplified by a little approximation as follows : R2 −1 = (1 + α0 t2) (1 + α0 t1) R1 = (1 + α0 t2) (1 − α0 t1) [Using Binomial Theorem for expansion and = 1 + α0 (t2 − t1) neglecting squares and higher powers of (α0 t1)] 2 ∴ R2 = R1 [1 + α0 (t2 − t1)] [Neglecting product (α0 t1t2)] For more accurate calculations, Eq. (iv) should, however, be used. 1.12. Variations of Resistivity with Temperature Not only resistance but specific resistance or resistivity of metallic conductors also increases with rise in temperature and vice-versa. As seen from Fig. 1.8 the resistivities of metals vary linearly with temperature over a significant range of temperature-the variation becoming non-linear both at very high and at very low temperatures. Let, for any metallic conductor, ρ1 = resistivity at t1°C Fig. 1.8  Electric Current and Ohm’s Law 13 ρ2 = resistivity at t2°C m = Slope of the linear part of the curve Then, it is seen that ρ − ρ1 m = 2 t2 − t1 1 m (t t1) or ρ2 = ρ 1 + m (t2 −t1) or 2 1 2 1 The ratio of m/ρ 1 is called the temperature coefficient of resistivity at temperature t1°C. It may be defined as numerically equal to the fractional change in ρ 1 per °C change in the temperature from t1°C. It is almost equal to the temperature-coefficient of resistance α1. Hence, putting α1 = m/ρ 1, we get ρ2 = ρ 1 [1 + α1 (t2 −t1)] or simply as ρ t = ρ 0 (1 + α0 t) Note. It has been found that although temperature is the most significant factor influencing the resistivity of metals, other factors like pressure and tension also affect resistivity to some extent. For most metals except lithium and calcium, increase in pressure leads to decrease in resistivity. However, resistivity increases with increase in tension. −6 Example 1.11. A copper conductor has its specific resistance of 1.6 × 10 ohm-cm at 0°C and a resistance temperature coefficient of 1/254.5 per °C at 20°C. Find (i) the specific resistance and (ii) the resistance - temperature coefficient at 60°C. (F.Y. Engg. Pune Univ. Nov.) α0 α0 Solution. α20 = or 1 = ∴ α0 = 1 per °C 1 + α 0 × 20 254.5 1 + α0 × 20 234.5 − (i) ρ60 = ρ 0 (1 + α0 × 60) = 1.6 × 10−6 (1 + 60/234.5) = 2.01 × 10−6 Ω-cm α0 1/ 234.5 1 (ii) α60 = 1 + α × 60 = 1 + (60 / 234.5) = 294.5 per°C 0 Example 1.12. A platinum coil has a resistance of 3.146 Ω at 40°C and 3.767 Ω at 100°C. Find the resistance at 0°C and the temperature-coefficient of resistance at 40°C. (Electrical Science-II, Allahabad Univ.) Solution. R100 = R0 (1 + 100 α0)...(i) R40 = R0 (1 + 40 α0)...(ii) 3.767 1 + 100 α 0 ∴ = 1 + 40 α or α0 = 0.00379 or 1/264 per°C 3.146 0 From (i), we have 3.767 = R0 (1 + 100 × 0.00379) ∴ R0 = 2.732 Ω α0 0.00379 1 Now, α40 = 1 + 40 α = 1 + 40 × 0.00379 = 304 per°C 0 Example 1.13. A potential difference of 250 V is applied to a field winding at 15°C and the current is 5 A. What will be the mean temperature of the winding when current has fallen to 3.91 A, applied voltage being constant. Assume α15 = 1/254.5. (Elect. Engg. Pune Univ.) Solution. Let R1 = winding resistance at 15°C; R2 = winding resistance at unknown mean tem- perature t2°C. ∴ R1 = 250/5 = 50 Ω; R2 = 250/3.91 = 63.94 Ω. Now R2 = R1 [1 + α15 (t2 − t1)] ∴ 63.94 = 50 ⎡1 + 1 (t2 − 15) ⎤ ⎢⎣ 254.5 ⎥⎦ ∴ t2 = 86°C 14 Electrical Technology Example 1.14. Two coils connected in series have resistances of 600 Ω and 300 Ω with tempt. coeff. of 0.1% and 0.4% respectively at 20°C. Find the resistance of the combination at a tempt. of 50°C. What is the effective tempt. coeff. of combination ? Solution. Resistance of 600 Ω resistor at 50°C is = 600 [1 + 0.001 (50 − 20)] = 618 Ω Similarly, resistance of 300 Ω resistor at 50°C is = 300 [1 + 0.004 (50 − 20)] = 336 Ω Hence, total resistance of combination at 50°C is = 618 + 336 = 954 Ω Let β = resistance-temperature coefficient at 20°C Now, combination resistance at 20°C = 900 Ω Combination resistance at 50°C = 954 Ω ∴ 954 = 900 [ 1 + β (50 − 20)] ∴ β = 0.002 Example 1.15. Two wires A and B are connected in series at 0°C and resistance of B is 3.5 times that of A. The resistance temperature coefficient of A is 0.4% and that of the combination is 0.1%. Find the resistance temperature coefficient of B. (Elect. Technology, Hyderabad Univ.) Solution. A simple technique which gives quick results in such questions is illustrated by the diagram of Fig. 1.9. It is seen that RB/RA = 0.003/(0.001 −α) or 3.5 = 0.003/(0.001 − α) or α = 0.000143°C−1 or 0.0143 % Example 1.16. Two materials A and B have resistance temperature coefficients of 0.004 and 0.0004 respectively at a given temperature. In what proportion must A and B be joined in series to pro- duce a circuit having a temperature coefficient of 0.001 ? Fig. 1.9 (Elect. Technology, Indore Univ.) Solution. Let RA and RB be the resistances of the two wires of materials A and B which are to be connected in series. Their ratio may be found by the simple technique shown in Fig. 1.10. RB 0.003 = 5 Fig. 1.10 = RA 0.0006 Hence, RB must be 5 times RA. Example 1.17. A resistor of 80 Ω resistance, having a temperature coefficient of 0.0021 per degree C is to be constructed. Wires of two materials of suitable cross-sectional area are available. For material A, the resistance is 80 ohm per 100 metres and the temperature coefficient is 0.003 per degree C. For material B, the corresponding figures are 60 ohm per metre and 0.0015 per degree C. Calculate suitable lengths of wires of materials A and B to be connected in series to construct the required resistor. All data are referred to the same temperature. Solution. Let Ra and Rb be the resistances of suitable lengths of materials A and B respectively which when joined in series will have a combined temperature coeff. of 0.0021. Hence, combination resistance at any given temperature is (Ra + Rb). Suppose we heat these materials through t°C. When heated, resistance of A increases from Ra to Ra (1 + 0.003 t). Similarly, resistance of B increases from Rb to Rb (1 + 0.0015 t). ∴ combination resistance after being heated through t°C = Ra (1 + 0.003 t) + Rb (1 + 0.0015 t) The combination α being given, value of combination resistance can be also found directly as  Electric Current and Ohm’s Law 15 = (Ra + Rb) (1 + 0.0021 t) ∴ (Ra + Rb) (1 + 0.0021 t) = Ra (1 + 0.003 t) + Rb (1 + 0.0015 t) Rb 3 Simplifying the above, we get R =...(i) a 2 Now Ra + Rb = 80 Ω...(ii) Substituting the value of Rb from (i) into (ii) we get Ra + 3 Ra = 80 or Ra = 32 Ω and Rb = 48 Ω 2 If La and Lb are the required lengths in metres, then La = (100/80) × 32 = 40 m and Lb = (100/60) × 48 = 80 m Example 1.18. A coil has a resistance of 18 Ω when its mean temperature is 20°C and of 20 Ω when its mean temperature is 50°C. Find its mean temperature rise when its resistance is 21 Ω and the surrounding temperature is 15° C. (Elect. Technology, Allahabad Univ.) Solution. Let R0 be the resistance of the coil and α0 its tempt. coefficient at 0°C. Then, 18 = R0 (1 + α0 × 20) and 20 = R0 (1 + 50 α0) Dividing one by the other, we get 1 + 50 α 0 20 = ∴ α 0 = 1 per°C 18 1 + 20 α0 250 If t°C is the temperature of the coil when its resistance is 21 Ω, then, 21 = R0 (1 + t/250) Dividing this equation by the above equation, we have 21 R0 (1 + t/250) = ; t = 65°C; temp. rise = 65 − 15 = 50°C 18 R0 (1 + 20 α0 ) Example 1.19. The coil of a relay takes a current of 0.12 A when it is at the room temperature of 15°C and connected across a 60-V supply. If the minimum operating current of the relay is 0.1 A, calculate the temperature above which the relay will fail to operate when connected to the same supply. Resistance-temperature coefficient of the coil material is 0.0043 per°C at 6°C. Solution. Resistance of the relay coil at 15°C is R15 = 60/0.12 = 500 Ω. Let t°C be the temperature at which the minimum operating current of 0.1 A flows in the relay coil. Then, R1 = 60/0.1 = 600 Ω. Now R15 = R0 (1 + 15 α0) = R0 (1 + 15 × 0.0043) and Rt = R0 (1 + 0.0043 t) Rt 1 + 0.0043 t 1 + 0.0043 t ∴ or 600 = ∴ t = 65.4°C R15 = 1.0654 500 1.0645 If the temperature rises above this value, then due to increase in resistance, the relay coil will draw a current less than 0.1 A and, therefore, will fail to operate. Example 1.20. Two conductors, one of copper and the other of iron, are connected in parallel and carry equal currents at 25°C. What proportion of current will pass through each if the tempera- ture is raised to 100°C ? The temperature coefficients of resistance at 0°C are 0.0043/°C and 0.0063/ °C for copper and iron respectively. (Principles of Elect. Engg. Delhi Univ.) Solution. Since the copper and iron conductors carry equal currents at 25°C, their resistances are the same at that temperature. Let each be R ohm. For copper, R100 = R1 = R [1 + 0.0043 (100 − 25)] = 1.3225 R For iron, R100 = R2 = R [1 + 0.0063 (100 − 25)] = 1.4725 R If I is the current at 100°C, then as per current divider rule, current in the copper conductor is 16 Electrical Technology R2 1.4725 R I1 = I =I = 0.5268 I R1 + R2 1.3225 R + 1.4725 R R2 1.3225 R I2 = I =I = 0.4732 I R1 + R2 2.795 R Hence, copper conductor will carry 52.68% of the total current and iron conductor will carry the balance i.e. 47.32%. Example 1.21. The filament of a 240 V metal-filament lamp is to be constructed from a wire having a diameter of 0.02 mm and a resistivity at 20°C of 4.3 μΩ-cm. If α = 0.005/°C, what length of filament is necessary if the lamp is to dissipate 60 watts at a filament tempt. of 2420°C ? Solution. Electric power generated = I2 R watts = V2/R watts 2 2 ∴ V /R = 60 or 240 /R = 60 240 × 240 Resistance at 2420°C R2420 = = 960 Ω 60 Now R2420 = R20 [1 + (2420 − 20) × 0.005] or 960 = R20 (1 + 12) ∴ R20 = 960/13 Ω π(0.002) 2 Now ρ20 = 4.3 × 10−6 Ω-cm and A= cm 2 4 2 A × R20 π (0.002) × 960 ∴ l = = −6 = 54 cm ρ20 4 × 13 × 4.3 × 10 Example 1.22. A semi-circular ring of copper has an inner radius 6 cm, radial thickness 3 cm and an axial thickness 4 cm. Find the resistance of the ring at 50°C between its two end-faces. −6 Assume specific resistance of Cu at 20°C = 1.724 × 10 ohm-cm and resistance tempt. coeff. of Cu at 0°C = 0.0043/°C. Solution. The semi-circular ring is shown in Fig. 1.11. Mean radius of ring = (6 + 9)/2 = 7.5 cm Mean length between end faces = 7.5 π cm = 23.56 cm Cross-section of the ring = 3 × 4 = 12 cm2 0.0043 = 0.00396 Now α0 = 0.0043/°C; α20 = 1 + 20 × 0.0043 ρ50 = ρ 20 [1 + α0 (50 − 20)] = 1.724 × 10 −6 (1 + 30 × 0.00396) = 1.93 × 10−6 Ω-cm ρ × l 1.93 × 10−6 × 23.56 R50 = 50 = = 3.79 × 10−6 Ω A 12 Fig.1.11 Tutorial Problems No. 1.2 1. It is found that the resistance of a coil of wire increases from 40 ohm at 15°C to 50 ohm at 60°C. Calculate the resistance temperature coefficient at 0°C of the conductor material. [1/165 per °C] (Elect. Technology, Indore Univ.) 2. A tungsten lamp filament has a temperature of 2,050°C and a resistance of 500 Ω when taking normal working current. Calculate the resistance of the filament when it has a temperature of 25°C. Tem- perature coefficient at 0°C is 0.005/°C. [50 Ω] (Elect. Technology, Indore Univ.)  Electric Current and Ohm’s Law 17 3. An armature has a resistance of 0.2 Ω at 150°C and the armature Cu loss is to be limited to 600 watts with a temperature rise of 55°C. If α0 for Cu is 0.0043/°C, what is the maximum current that can be passed through the armature ? [50.8 A] 4. A d.c. shunt motor after running for several hours on constant voltage mains of 400 V takes a field current of 1.6 A. If the temperature rise is known to be 40°C, what value of extra circuit resistance is required to adjust the field current to 1.6 A when starting from cold at 20°C ? Temperature coefficient = 0.0043/°C at 20°C. [36.69 Ω] 5. In a test to determine the resistance of a single-core cable, an applied voltage of 2.5 V was necessary to produce a current of 2 A in it at 15°C. (a) Calculate the cable resistance at 55°C if the temperature coefficient of resistance of copper at 0°C is 1/235 per°C. (b) If the cable under working conditions carries a current of 10 A at this temperature, calculate the power dissipated in the cable. [(a) 1.45 Ω (b) 145 W] 6. An electric radiator is required to dissipate 1 kW when connected to a 230 V supply. If the coils of the radiator are of wire 0.5 mm in diameter having resistivity of 60 μ Ω-cm, calculate the necessary length of the wire. [1732 cm] 7. An electric heating element to dissipate 450 watts on 250 V mains is to be made from nichrome ribbon of width 1 mm and thickness 0.05 mm. Calculate the length of the ribbon required (the resistivity of −8 nichrome is 110 × 10 Ω-m). [631 m] 8. When burning normally, the temperature of the filament in a 230 V, 150 W gas-filled tungsten lamp is 2,750°C. Assuming a room temperature of 16°C, calculate (a) the normal current taken by the lamp (b) the current taken at the moment of switching on. Temperature coefficient of tungsten is 0.0047 Ω/Ω°C at 0°C. [(a) 0.652 A (b) 8.45 A] (Elect. Engg. Madras Univ.) 9. An aluminium wire 5 m long and 2 mm diameter is connected in parallel with a wire 3 m long. The total current is 4 A and that in the aluminium wire is 2.5 A. Find the diameter of the copper wire. The respective resistivities of copper and aluminium are 1.7 and 2.6 μΩ-m. [0.97 mm] 10. The field winding of d.c. motor connected across 230 V supply takes 1.15 A at room temp. of 20°C. After working for some hours the current falls to 0.26 A, the supply voltage remaining constant. Calculate the final working temperature of field winding. Resistance temperature coefficient of cop- per at 20°C is 1/254.5. [70.4°C] (Elect. Engg. Pune Univ.) 11. It is required to construct a resistance of 100 Ω having a temperature coefficient of 0.001 per°C. Wires of two materials of suitable cross-sectional area are available. For material A, the resistance is 97 Ω per 100 metres and for material B, the resistance is 40 Ω per 100 metres. The temperature coefficient of resistance for material A is 0.003 per °C and for material B is 0.0005 per °C. Deter- mine suitable lengths of wires of materials A and B. [A : 19.4 m, B : 200 m] 12. The resistance of the shunt winding of a d.c. machine is measured before and after a run of several hours. The average values are 55 ohms and 63 ohms. Calculate the rise in temperature of the winding. (Temperature coefficient of resistance of copper is 0.00428 ohm per ohm per °C). [36°C] (London Univ.) 13. A piece of resistance wire, 15.6 m long and of cross-sectional area 12 mm2 at a temperature of 0°C, passes a current of 7.9 A when connected to d.c. supply at 240 V. Calculate (a) resistivity of the wire (b) the current which will flow when the temperature rises to 55°C. The temperature coefficient of the resistance wire is 0.00029 Ω/Ω/°C. Ω-m (b) 7.78 A] (London Univ.) [(a) 23.37 μΩ 14. A coil is connected to a constant d.c. supply of 100 V. At start, when it was at the room temperature of 25°C, it drew a current of 13 A. After sometime, its temperature was 70°C and the current reduced to 8.5 A. Find the current it will draw when its temperature increases further to 80°C. Also, find the temperature coefficient of resistance of the coil material at 25°C. −1 [7.9 A; 0.01176°C ] (F.Y. Engg. Univ.) 15. The resistance of the filed coils with copper conductors of a dynamo is 120 Ω at 25°C. After working for 6 hours on full load, the resistance of the coil increases to 140 Ω. Calculate the mean temperature rise of the field coil. Take the temperature coefficient of the conductor material as 0.0042 at 0°C. [43.8°C] (Elements of Elec. Engg. Banglore Univ.) 18 Electrical Technology 1.13. Ohm’s Law This law applies to electric to electric conduction through good conductors and may be stated as follows : The ratio of potential difference (V) between any two points on a conductor to the current (I) flowing between them, is constant, provided the temperature of the conductor does not change. V V In other words, = constant or =R I I where R is the resistance of the conductor between the two points considered. Put in another way, it simply means that provided R is kept constant, current is directly propor- tional to the potential difference across the ends of a conductor. However, this linear relationship between V and I does not apply to all non-metallic conductors. For example, for silicon carbide, the m relationship is given by V = KI where K and m are constants and m is less than unity. It also does not apply to non-linear devices such as Zener diodes and voltage-regulator (VR) tubes. Example 1.23. A coil of copper wire has resistance of 90 Ω at 20°C and is connected to a 230- V supply. By how much must the voltage be increased in order to maintain the current consant if the temperature of the coil rises to 60°C ? Take the temperature coefficient of resistance of copper as 0.00428 from 0°C. Solution. As seen from Art. 1.10 R60 1 + 60 × 0.00428 = ∴R60 = 90 × 1.2568/1.0856 = 104.2 Ω R20 1 + 20 × 0.00428 Now, current at 20°C = 230/90 = 23/9 A Since the wire resistance has become 104.2 Ω at 60°C, the new voltage required for keeping the current constant at its previous value = 104.2 × 23/9 = 266.3 V ∴ increase in voltage required = 266.3 − 230 = 36.3 V Example 1.24. Three resistors are connected in series across a 12-V battery. The first resistor has a value of 1 Ω, second has a voltage drop of 4 V and the third has a power dissipation of 12 W. Calculate the value of the circuit current. Solution. Let the two unknown resistors be R2 and R3 and I the circuit current 3 2 4 IR3 = 4 ∴ R3 = 4 R2. Also, I = R 2 ∴ I R3 =12 and 2 Now, I (1 + R2 + R3) = 12 Substituting the values of I and R3, we get R2( 4 1 + R + 3 R2 2 ) 2 4 2 = 12 or 3R2 − 8 R2 + 4 = 0 8 ± 64 − 48 ∴ R2 = ∴ R2 = 2 Ω or 2 Ω 6 3 () 2 3 R 2 = 3 × 22 = 3 Ω or 3 2 = 1 Ω ∴ R3 = 4 2 4 4 3 3 12 = 2 A or I = 12 = 6A ∴ I = 1+ 2 + 3 1 + (2 / 3) + (1/ 3) 1.14. Resistance in Series When some conductors having resistances R1, R2 and R3 etc. are joined end-on-end as in Fig. 1.12, they are said to be connected in series. It can be proved that the equivalent resistance or total resistance between points A and D is equal to the sum of the three individual resistances. Being a series circuit, it should be remembered that (i) current is the same through all the three conductors  Electric Current and Ohm’s Law 19 (ii) but voltage drop across each is different due to its different resistance and is given by Ohm’s Law and (iii) sum of the three voltage drops is equal to the voltage applied across the three conductors. There is a progressive fall in potential as we go from point A to D as shown in Fig. 1.13. Fig. 1.12 Fig. 1.13 ∴ V = V1 + V2 + V3 = IR1 + IR2 + IR3 —Ohm’s Law But V = IR where R is the equivalent resistance of the series combination. ∴ IR = IR1 + IR2 + IR3 or R = R1 + R2 + R3 1 1 + 1 + 1 Also = G G1 G2 G3 As seen from above, the main characteristics of a series circuit are : 1. same current flows through all parts of the circuit. 2. different resistors have their individual voltage drops. 3. voltage drops are additive. 4. applied voltage equals the sum of different voltage drops. 5. resistances are additive. 6. powers are additive. 1.15. Voltage Divider Rule Since in a series circuit, same current flows through each of the given resistors, voltage drop varies directly with its resistance. In Fig. 1.14 is shown a 24-V battery connected across a series combina- tion of three resistors. Total resistance R = R1 + R2 + R3 = 12 Ω According to Voltage Divider Rule, various voltage drops are : R 2 V1 = V. 1 = 24 × = 4 V R 12 R 4 V2 = V. 2 = 24 × = 8 V R 12 Fig.1.14 R 6 V3 = V. 3 = 24 × = 12 V R 12 1.16. Resistances in Parallel Three resistances, as joined in Fig. 1.15 are said to be connected in parallel. In this case (i) p.d. across all resistances is the same (ii) current in each resistor is different and is given by Ohm’s Law and (iii) the total current is the sum of the three separate currents. Fig.1.15 20 Electrical Technology V + V + V I = I1 + I2 + I3 = R1 R2 R3 V Now, I = where V is the applied voltage. R R = equivalent resistance of the parallel combination. V V + V + V 1 1 1 1 ∴ = or R = R + R + R R R1 R2 R3 1 2 3 Also G = G1 + G2 + G3 The main characteristics of a parallel circuit are : 1. same voltage acts across all parts of the circuit 2. different resistors have their individual current. 3. branch currents are additive. 4. conductances are additive. 5. powers are additive. Example 1.25. What is the value of the unknown resistor R in Fig. 1.16 if the voltage drop across the 500 Ω resistor is 2.5 volts ? All resistances are in ohm. (Elect. Technology, Indore Univ.) Fig. 1.16 Solution. By direct proportion, drop on 50 Ω resistance = 2.5 × 50/500 = 0.25 V Drop across CMD or CD = 2.5 + 0.25 = 2.75 V Drop across 550 Ω resistance = 12 − 2.75 = 9.25 V I = 9.25/550 = 0.0168 A, I2 = 2.5/500 = 0.005 A I1 = 0.0168 − 0.005 = 0.0118 A ∴ 0.0118 = 2.75/R; R = 233 Ω Example 1.26. Calculate the effective resistance of the following combination of resistances and the voltage drop across each resistance when a P.D. of 60 V is applied between points A and B. Solution. Resistance between A and C (Fig. 1.17). = 6 || 3 = 2 Ω Resistance of branch ACD = 18 + 2 = 20 Ω Now, there are two parallel paths between points A and D of resistances 20 Ω and 5 Ω. Hence, resistance between A and D = 20 || 5 = 4 Ω ∴Resistance between A and B = 4 + 8 = 12 Ω Total circuit current = 60/12 = 5 A Fig. 1.17 20 = 4 A Current through 5 Ω resistance = 5 × —Art. 1.25 25  Electric Current and Ohm’s Law 21 5 =1A Current in branch ACD = 5× 25 ∴ P.D. across 3 Ω and 6 Ω resistors = 1 × 2 = 2 V P.D. across 18 Ω resistors = 1 × 18 = 18 V P.D. across 5 Ω resistors = 4 × 5 = 20 V P.D. across 8 Ω resistors = 5 × 8 = 40 V Example 1.27. A circuit consists of four 100-W lamps connected in parallel across a 230-V supply. Inadvertently, a voltmeter has been connected in series with the lamps. The resistance of the voltmeter is 1500 Ω and that of the lamps under the conditions stated is six times their value then burning normally. What will be the reading of the voltmeter ? Solution. The circuit is shown in Fig. 1.18. The wattage of a lamp is given by : 2 2 W = I R = V /R 2 ∴ 100 = 230 /R ∴ R = 529 Ω Fig.1.18 Resistance of each lamp under stated condition is = 6 × 529 = 3174 Ω Equivalent resistance of these four lamps connected in parallel = 3174/4 = 793.5 Ω This resistance is connected in series with