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    Electrical engineering know it all Electrical engineering know it all Document Transcript

    • Electrical Engineering
    • Newnes Know It All SeriesPIC Microcontrollers: Know It AllLucio Di Jasio, Tim Wilmshurst, Dogan Ibrahim, John Morton, Martin Bates, Jack Smith, D.W. Smith, andChuck HellebuyckISBN: 978-0-7506-8615-0Embedded Software: Know It AllJean Labrosse, Jack Ganssle, Tammy Noergaard, Robert Oshana, Colin Walls, Keith Curtis, Jason Andrews,David J. Katz, Rick Gentile, Kamal Hyder, and Bob PerrinISBN: 978-0-7506-8583-2Embedded Hardware: Know It AllJack Ganssle, Tammy Noergaard, Fred Eady, Lewin Edwards, David J. Katz, Rick Gentile, Ken Arnold,Kamal Hyder, and Bob PerrinISBN: 978-0-7506-8584-9Wireless Networking: Know It AllPraphul Chandra, Daniel M. Dobkin, Alan Bensky, Ron Olexa, David Lide, and Farid DowlaISBN: 978-0-7506-8582-5RF & Wireless Technologies: Know It AllBruce Fette, Roberto Aiello, Praphul Chandra, Daniel Dobkin, Alan Bensky, Douglas Miron, David Lide,Farid Dowla, and Ron OlexaISBN: 978-0-7506-8581-8Electrical Engineering: Know It AllClive Maxfield, Alan Bensky, John Bird, W. Bolton, Izzat Darwazeh, Walt Kester, M.A. Laughton, AndrewLeven, Luis Moura, Ron Schmitt, Keith Sueker, Mike Tooley, DF Warne, Tim WilliamsISBN: 978-1-85617-528-9Audio Engineering: Know It AllDouglas Self, Richard Brice, Don Davis, Ben Duncan, John Linsely Hood, Morgan Jones, Eugene Patronis,Ian Sinclair, Andrew Singmin, John WatkinsonISBN: 978-1-85617-526-5Circuit Design: Know It AllDarren Ashby, Bonnie Baker, Stuart Ball, John Crowe, Barrie Hayes-Gill, Ian Grout, Ian Hickman, WaltKester, Ron Mancini, Robert A. Pease, Mike Tooley, Tim Williams, Peter Wilson, Bob ZeidmanISBN: 978-1-85617-527-2Test and Measurement: Know It AllJon Wilson, Stuart Ball, GMS de Silva, Tony Fischer-Cripps, Dogan Ibrahim, Kevin James, Walt Kester,M A Laughton, Chris Nadovich, Alex Porter, Edward Ramsden, Stephen Scheiber, Mike Tooley, D. F. Warne,Tim WilliamsISBN: 978-1-85617-530-2Mobile Wireless Security: Know It AllPraphul Chandra, Alan Bensky, Tony Bradley, Chris Hurley, Steve Rackley, John Rittinghouse, JamesRansome, Timothy Stapko, George Stefanek, Frank Thornton, Chris Lanthem, John WilsonISBN: 978-1-85617-529-6For more information on these and other Newnes titles visit: www.newnespress.com
    • Electrical Engineering Clive Maxfield John Bird M. A.Laughton W. Bolton Andrew Leven Ron Schmitt Keith Sueker Tim Williams Mike Tooley Luis Moura Izzat Darwazeh Walt Kester Alan Bensky DF WarneAMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGOSAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Newnes is an imprint of Elsevier
    • Newnes is an imprint of Elsevier30 Corporate Drive, Suite 400, Burlington, MA 01803, USALinacre House, Jordan Hill, Oxford OX2 8DP, UKCopyright © 2008, Elsevier Inc. All rights reserved.No part of this publication may be reproduced, stored in a retrieval system, or transmitted inany form or by any means, electronic, mechanical, photocopying, recording, or otherwise,without the prior written permission of the publisher.Permissions may be sought directly from Elsevier’s Science & Technology RightsDepartment in Oxford, UK: phone: (ϩ44) 1865 843830, fax: (ϩ44) 1865 853333,E-mail: permissions@elsevier.com. You may also complete your request onlinevia the Elsevier homepage (http://elsevier.com), by selecting “Support & Contact”then “Copyright and Permission” and then “Obtaining Permissions.”Library of Congress Cataloging-in-Publication DataApplication submittedBritish Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library.ISBN: 978-1-85617-528-9 For information on all Newnes publications visit our Web site at www.elsevierdirect.comTypeset by Charon Tec Ltd., A Macmillan Company. (www.macmillansolutions.com)Printed in the United States of America08 09 10 10 9 8 7 6 5 4 3 2 1
    • ContentsAbout the Authors .............................................................................................................xvChapter 1: An Introduction to Electric Circuits ................................................................11.1 SI Units .......................................................................................................................11.2 Charge .........................................................................................................................21.3 Force ...........................................................................................................................21.4 Work ............................................................................................................................31.5 Power ..........................................................................................................................41.6 Electrical Potential and e.m.f. .....................................................................................51.7 Resistance and Conductance .......................................................................................51.8 Electrical Power and Energy .......................................................................................61.9 Summary of Terms, Units and Their Symbols............................................................71.10 Standard Symbols for Electrical Components ............................................................81.11 Electric Current and Quantity of Electricity ...............................................................81.12 Potential Difference and Resistance .........................................................................101.13 Basic Electrical Measuring Instruments ...................................................................111.14 Linear and Nonlinear Devices ..................................................................................111.15 Ohm’s Law ................................................................................................................121.16 Multiples and Submultiples ......................................................................................131.17 Conductors and Insulators ........................................................................................161.18 Electrical Power and Energy .....................................................................................161.19 Main Effects of Electric Current ...............................................................................20Chapter 2: Resistance and Resistivity ..............................................................................212.1 Resistance and Resistivity.........................................................................................212.2 Temperature Coefficient of Resistance .....................................................................25Chapter 3: Series and Parallel Networks .........................................................................313.1 Series Circuits ...........................................................................................................313.2 Potential Divider .......................................................................................................34 w w w.ne w nespress.com
    • vi Contents3.3 Parallel Networks ......................................................................................................373.4 Current Division........................................................................................................433.5 Relative and Absolute Voltages ................................................................................48Chapter 4: Capacitors and Inductors ...............................................................................534.1 Introduction to Capacitors ........................................................................................534.2 Electrostatic Field .....................................................................................................534.3 Electric Field Strength ..............................................................................................554.4 Capacitance ...............................................................................................................564.5 Capacitors .................................................................................................................564.6 Electric Flux Density ................................................................................................584.7 Permittivity ...............................................................................................................594.8 The Parallel Plate Capacitor......................................................................................614.9 Capacitors Connected in Parallel and Series ............................................................644.10 Dielectric Strength ....................................................................................................704.11 Energy Stored............................................................................................................714.12 Practical Types of Capacitors....................................................................................724.13 Inductance .................................................................................................................764.14 Inductors ...................................................................................................................784.15 Energy Stored............................................................................................................80Chapter 5: DC Circuit Theory ..........................................................................................815.1 Introduction ...............................................................................................................815.2 Kirchhoff’s Laws ......................................................................................................815.3 The Superposition Theorem......................................................................................895.4 General DC Circuit Theory.......................................................................................955.5 Thévenin’s Theorem .................................................................................................995.6 Constant-Current Source.........................................................................................1065.7 Norton’s Theorem ...................................................................................................1075.8 Thévenin and Norton Equivalent Networks............................................................1115.9 Maximum Power Transfer Theorem .......................................................................117Chapter 6: Alternating Voltages and Currents ..............................................................1236.1 The AC Generator ...................................................................................................1236.2 Waveforms ..............................................................................................................124w ww. n e w n e s p r e s s .c o m
    • Contents vii6.3 AC Values ...............................................................................................................1266.4 The Equation of a Sinusoidal Waveform ................................................................1336.5 Combination of Waveforms ....................................................................................1396.6 Rectification ............................................................................................................146Chapter 7: Complex Numbers ........................................................................................1497.1 Introduction .............................................................................................................1497.2 Operations involving Cartesian Complex Numbers ...............................................1527.3 Complex Equations .................................................................................................1557.4 The polar Form of a Complex Number...................................................................1577.5 Applying Complex Numbers to Series AC Circuits ...............................................1587.6 Applying Complex Numbers to Parallel AC Circuits .............................................171Chapter 8: Transients and Laplace Transforms ............................................................1858.1 Introduction .............................................................................................................1858.2 Response of R-C Series Circuit to a Step Input ......................................................1858.3 Response of R-L Series Circuit to a Step Input ......................................................1928.4 L-R-C Series Circuit Response ...............................................................................1998.5 Introduction to Laplace Transforms........................................................................2058.6 Inverse Laplace Transforms and the Solution of Differential Equations ................215Chapter 9: Frequency Domain Circuit Analysis ...........................................................2299.1 Introduction .............................................................................................................2299.2 Sinusoidal AC Electrical Analysis ..........................................................................2299.3 Generalized Frequency Domain Analysis ..............................................................257 References ...............................................................................................................315Chapter 10: Digital Electronics ......................................................................................31710.1 Semiconductors .......................................................................................................31710.2 Semiconductor Diodes ............................................................................................31810.3 Bipolar Junction Transistors ...................................................................................31910.4 Metal-oxide Semiconductor Field-effect Transistors .............................................32110.5 The transistor as a Switch .......................................................................................32210.6 Gallium Arsenide Semiconductors .........................................................................32410.7 Light-emitting Diodes .............................................................................................32410.8 BUF and NOT Functions ........................................................................................327 w w w.ne w nespress.com
    • viii Contents10.9 AND, OR, and XOR Functions ............................................................................32910.10 NAND, NOR, and XNOR Functions ....................................................................32910.11 Not a Lot ...............................................................................................................33110.12 Functions Versus Gates .........................................................................................33210.13 NOT and BUF Gates .............................................................................................33310.14 NAND and AND Gates ........................................................................................33510.15 NOR and OR Gates...............................................................................................33610.16 XNOR and XOR Gates .........................................................................................33710.17 Pass-Transistor Logic............................................................................................33910.18 Combining a Single Variable With Logic 0 or Logic 1 ........................................34210.19 The Idempotent Rules ...........................................................................................34210.20 The Complementary Rules ...................................................................................34310.21 The Involution Rules .............................................................................................34410.22 The Commutative Rules .......................................................................................34410.23 The Associative Rules...........................................................................................34410.24 Precedence of Operators .......................................................................................34510.25 The First Distributive Rule ...................................................................................34610.26 The Second Distributive Rule ...............................................................................34610.27 The Simplification Rules ......................................................................................34810.28 DeMorgan Transformations ..................................................................................34910.29 Minterms and Maxterms .......................................................................................35110.30 Sum-of-Products and Product-of-sums .................................................................35110.31 Canonical Forms ...................................................................................................35210.32 Karnaugh Maps .....................................................................................................35310.33 Minimization Using Karnaugh Maps ...................................................................35410.34 Grouping Minterms...............................................................................................35510.35 Incompletely Specified Functions .........................................................................35610.36 Populating Maps Using 0s versus 1s.....................................................................35910.37 Scalar Versus Vector Notation ..............................................................................36010.38 Equality Comparators ...........................................................................................36110.39 Multiplexers ..........................................................................................................36310.40 Decoders ...............................................................................................................36410.41 Tri-State Functions................................................................................................36510.42 Combinational Versus Sequential Functions ........................................................36710.43 RS Latches ............................................................................................................367w ww. n e w n e s p r e s s .c o m
    • Contents ix10.44 D-Type Latches .....................................................................................................37310.45 D-Type Flip-Flops.................................................................................................37410.46 JK and T Flip-Flops ..............................................................................................37710.47 Shift Registers .......................................................................................................37810.48 Counters ................................................................................................................38110.49 Setup and Hold Times ...........................................................................................38310.50 Brick by Brick .......................................................................................................38410.51 State Diagrams ......................................................................................................38610.52 State Tables ...........................................................................................................38710.53 State Machines ......................................................................................................38810.54 State Assignment ..................................................................................................38910.55 Don’t Care States, Unused States, and Latch-Up Conditions...............................392Chapter 11: Analog Electronics .....................................................................................39511.1 Operational Amplifiers Defined ............................................................................39511.2 Symbols and Connections .....................................................................................39511.3 Operational Amplifier Parameters ........................................................................39711.4 Operational Amplifier Characteristics ..................................................................40211.5 Operational Amplifier Applications......................................................................40311.6 Gain and Bandwidth .............................................................................................40511.7 Inverting Amplifier With Feedback ......................................................................40611.8 Operational Amplifier Configurations ..................................................................40811.9 Operational Amplifier Circuits .............................................................................41211.10 The Ideal Op-Amp ................................................................................................41811.11 The Practical Op-Amp ..........................................................................................42011.12 Comparators ..........................................................................................................45011.13 Voltage References................................................................................................459Chapter 12: Circuit Simulation ......................................................................................46512.1 Types of Analysis ..................................................................................................46612.2 Netlists and Component Models ...........................................................................47612.3 Logic Simulation...................................................................................................479Chapter 13: Interfacing ..................................................................................................48113.1 Mixing Analog and Digital ...................................................................................48113.2 Generating Digital Levels From Analog Inputs....................................................484 w w w.ne w nespress.com
    • x Contents13.3 Classic Data Interface Standards ..........................................................................48713.4 High Performance Data Interface Standards.........................................................493Chapter 14: Microcontrollers and Microprocessors......................................................49914.1 Microprocessor Systems .......................................................................................49914.2 Single-Chip Microcomputers ................................................................................49914.3 Microcontrollers....................................................................................................50014.4 PIC Microcontrollers ............................................................................................50014.5 Programmed Logic Devices ..................................................................................50014.6 Programmable Logic Controllers..........................................................................50114.7 Microprocessor Systems .......................................................................................50114.8 Data Representation ..............................................................................................50314.9 Data Types ............................................................................................................50514.10 Data Storage ..........................................................................................................50514.11 The Microprocessor ..............................................................................................50614.12 Microprocessor Operation ....................................................................................51214.13 A Microcontroller System ....................................................................................51814.14 Symbols Introduced in this Chapter......................................................................523Chapter 15: Power Electronics .......................................................................................52515.1 Switchgear ............................................................................................................52515.2 Surge Suppression.................................................................................................52815.3 Conductors ............................................................................................................53015.4 Capacitors .............................................................................................................53315.5 Resistors ................................................................................................................53615.6 Fuses .....................................................................................................................53815.7 Supply Voltages ....................................................................................................53915.8 Enclosures .............................................................................................................53915.9 Hipot, Corona, and BIL ........................................................................................54015.10 Spacings ................................................................................................................54115.11 Metal Oxide Varistors ...........................................................................................54215.12 Protective Relays ..................................................................................................54315.13 Symmetrical Components .....................................................................................54415.14 Per Unit Constants ................................................................................................54615.15 Circuit Simulation .................................................................................................547w ww. n e w n e s p r e s s .c o m
    • Contents xi15.16 Simulation Software .............................................................................................55115.17 Feedback Control Systems....................................................................................55215.18 Power Supplies......................................................................................................559Chapter 16: Signals and Signal Processing ...................................................................60916.1 Origins of Real-World Signals and their Units of Measurement ..........................60916.2 Reasons for Processing Real-World Signals .........................................................61016.3 Generation of Real-World Signals ........................................................................61216.4 Methods and Technologies Available for Processing Real-World Signals ...........61216.5 Analog Versus Digital Signal Processing .............................................................61316.6 A Practical Example .............................................................................................614 References .............................................................................................................617Chapter 17: Filter Design ...............................................................................................61917.1 Introduction ...........................................................................................................61917.2 Passive Filters .......................................................................................................62117.3 Active Filters .........................................................................................................62217.4 First-Order Filters .................................................................................................62817.5 Design of First-Order Filters.................................................................................63017.6 Second-Order Filters .............................................................................................63217.7 Using the Transfer Function .................................................................................63617.8 Using Normalized Tables ......................................................................................64117.9 Using Identical Components .................................................................................64117.10 Second-Order High-Pass Filters ...........................................................................64217.11 Bandpass Filters ....................................................................................................65017.12 Switched Capacitor Filter .....................................................................................65417.13 Monolithic Switched Capacitor Filter...................................................................65717.14 The Notch Filter ....................................................................................................65917.15 Choosing Components for Filters .........................................................................66317.16 Testing Filter Response .........................................................................................66517.17 Fast Fourier Transforms ........................................................................................66617.18 Digital Filters ........................................................................................................694 References .............................................................................................................732Chapter 18: Control and Instrumentation Systems .......................................................73518.1 Introduction ...........................................................................................................735 w w w.ne w nespress.com
    • xii Contents18.2 Systems .................................................................................................................73718.3 Control Systems Models .......................................................................................74118.4 Measurement Elements .........................................................................................74718.5 Signal Processing ..................................................................................................76118.6 Correction Elements .............................................................................................76918.7 Control Systems ....................................................................................................78018.8 System Models ......................................................................................................79118.9 Gain .......................................................................................................................79318.10 Dynamic Systems .................................................................................................79718.11 Differential Equations ...........................................................................................81218.12 Transfer Function ..................................................................................................81618.13 System Transfer Functions ...................................................................................82218.14 Sensitivity .............................................................................................................82618.15 Block Manipulation ..............................................................................................83018.16 Multiple Inputs ......................................................................................................835Chapter 19: Communications Systems...........................................................................83719.1 Introduction ...........................................................................................................83719.2 Analog Modulation Techniques ............................................................................83919.3 The Balanced Modulator/Demodulator ................................................................84819.4 Frequency Modulation and Demodulation ...........................................................85019.5 FM Modulators .....................................................................................................86019.6 FM Demodulators .................................................................................................86219.7 Digital Modulation Techniques.............................................................................86519.8 Information Theory ...............................................................................................87319.9 Applications and Technologies .............................................................................899 References .............................................................................................................951Chapter 20: Principles of Electromagnetics ..................................................................95320.1 The Need for Electromagnetics ............................................................................95320.2 The Electromagnetic Spectrum .............................................................................95520.3 Electrical Length ...................................................................................................96020.4 The Finite Speed of Light .....................................................................................96020.5 Electronics ............................................................................................................96120.6 Analog and Digital Signals ...................................................................................96420.7 RF Techniques ......................................................................................................964w ww. n e w n e s p r e s s .c o m
    • Contents xiii20.8 Microwave Techniques .........................................................................................96720.9 Infrared and the Electronic Speed Limit ...............................................................96820.10 Visible Light and Beyond .....................................................................................96920.11 Lasers and Photonics ............................................................................................97120.12 Summary of General Principles ............................................................................97220.13 The Electric Force Field........................................................................................97320.14 Other Types of Fields ............................................................................................97520.15 Voltage and Potential Energy ................................................................................97620.16 Charges in Metals .................................................................................................97820.17 The Definition of Resistance.................................................................................98020.18 Electrons and Holes ..............................................................................................98020.19 Electrostatic Induction and Capacitance ...............................................................98220.20 Insulators (dielectrics)...........................................................................................98620.21 Static Electricity and Lightning ............................................................................98820.22 The Battery Revisited ...........................................................................................99220.23 Electric Field Examples ........................................................................................99320.24 Conductivity and Permittivity of Common Materials...........................................994 References .............................................................................................................995Chapter 21: Magnetic Fields ........................................................................................100321.1 Moving Charges: Source of All Magnetic Fields ...............................................100321.2 Magnetic Dipoles ................................................................................................100521.3 Effects of the Magnetic Field ..............................................................................100821.4 The Vector Magnetic Potential and Potential Momentum ..................................101821.5 Magnetic Materials .............................................................................................101921.6 Magnetism and Quantum Physics.......................................................................1022 References ...........................................................................................................1024Chapter 22: Electromagnetic Transients and EMI .....................................................102722.1 Line Disturbances ...............................................................................................102722.2 Circuit Transients ................................................................................................102822.3 Electromagnetic Interference ..............................................................................1030Chapter 23: Traveling Wave Effects .............................................................................103323.1 Basics ..................................................................................................................103323.2 Transient Effects .................................................................................................103523.3 Mitigating Measures ...........................................................................................1038 w w w.ne w nespress.com
    • xiv ContentsChapter 24: Transformers ............................................................................................103924.1 Voltage and Turns Ratio ......................................................................................1040Chapter 25: Electromagnetic Compatibility (EMC) ....................................................104725.1 Introduction .........................................................................................................104725.2 Common Terms...................................................................................................104825.3 The EMC Model .................................................................................................104925.4 EMC Requirements.............................................................................................105225.5 Product design.....................................................................................................105425.6 Device Selection .................................................................................................105625.7 Printed Circuit Boards ........................................................................................105625.8 Interfaces .............................................................................................................105725.9 Power Supplies and Power-Line Filters ..............................................................105825.10 Signal Line Filters ...............................................................................................105925.11 Enclosure Design ................................................................................................106125.12 Interface Cable Connections ...............................................................................106325.13 Golden Rules for Effective Design for EMC ......................................................106525.14 System Design ....................................................................................................106625.15 Buildings .............................................................................................................106925.16 Conformity Assessment ......................................................................................107025.17 EMC Testing and Measurements ........................................................................107225.18 Management Plans ..............................................................................................1075 References ...........................................................................................................1076Appendix A: General Reference ...................................................................................1077A.1 Standard Electrical Quantities—Their Symbols and Units ................................1077Appendix B: ...................................................................................................................1081B.1 Differential Equations .........................................................................................1081Index ..............................................................................................................................1091Note from the Publisher: The authors of this book are from around the world and as suchsymbols vary between US and UK styles.ww w. n e w n e s p r e s s .c o m
    • About the AuthorsAlan Bensky MScEE (Chapter 19) is an electronics engineering consultant with over25 years of experience in analog and digital design, management, and marketing.Specializing in wireless circuits and systems, Bensky has carried out projects forvaried military and consumer applications. He is the author of Short-range WirelessCommunication, Second Edition, published by Elsevier, 2004, and has written severalarticles in international and local publications. He has taught courses and gives lectureson radio engineering topics. Bensky is a senior member of IEEE.John Bird BSc (Hons), CEng, CMath, CSci, FIET, MIEE, FIIE, FIMA, FCollT RoyalNaval School of Marine Engineering, HMS Sultan, Gosport; formerly University ofPortsmouth and Highbury College, Portsmouth, U.K., (Chapters 1, 2, 3, 4, 5, 6, 7, 8,Appendix A) is the author of Electrical Circuit Theory and Technology, and over 120textbooks on engineering and mathematical subjects, is the former Head of AppliedElectronics in the Faculty of Technology at Highbury College, Portsmouth, U.K.More recently, he has combined freelance lecturing at the University of Portsmouth,with technical writing and Chief Examiner responsibilities for City and GuildsTelecommunication Principles and Mathematics, and examining for the InternationalBaccalaureate Organisation.John Bird is currently a Senior Training Provider at the Royal Naval School of MarineEngineering in the Defence College of Marine and Air Engineering at H.M.S. Sultan,Gosport, Hampshire, U.K. The school, which serves the Royal Navy, is one of Europe’slargest engineering training establishments.Bill Bolton (Chapter 18, Appendix B.) is the author of Control Systems, and manyengineering textbooks, including the best-selling books Programmable Logic Controllers(Newnes) and Mechatronics (Pearson—Prentice-Hall), and has formerly been a seniorlecturer in a College of Technology, Head of Research, Development and Monitoringat the Business and Technician Education Council, a member of the Nuffield AdvancedPhysics Project, and a consultant on a British Government Technician Education Projectin Brazil and on Unesco projects in Argentina and Thailand. w w w.ne w nespress.com
    • xvi About the AuthorsIzzat Darwazeh (Chapter 9) is the author of Introduction to Linear Circuit Analysis andModelling. He holds the University of London Chair of Communications Engineeringin the Department of Electronic and Electrical at UCL. He obtained his first degreein Electrical Engineering from the University of Jordan in 1984 and the MSc andPhD degrees, from the University of Manchester Institute of Science and Technology(UMIST), in 1986 and 1991, respectively. He worked as a research Fellow at theUniversity of Wales-Bangor—U.K. from 1990 till 1993, researching very high speedoptical systems and circuits. He was a Senior Lecturer in Optoelectronic Circuits andSystems in the Department at Electrical Engineering and Electronics at UMIST. Hemoved to UCL in October 2001 where he is currently the Head of Communicationsand Information System (CIS) group and the Director of UCL Telecommunications forIndustry Programme. He is a Fellow of the IET and a Senior Member of the IEEE.His teaching covers aspects of wireless and optical fibre communications,telecommunication networks, electronic circuits and high speed integrated circuitsand MMICs. He lectures widely in the U.K. and overseas. His research interests aremainly in the areas of wireless system design and implementation, high speed opticalcommunication systems and networks, microwave circuits and MMICs for optical fibreapplications and in mobile and wireless communication circuits and systems. He hasauthored/co-authored more than 120 research papers. He has co-authored (with LuisMoura) a book on Linear Circuit Analysis and Modelling (Elsevier 2005) and is theco-editor of the IEE book on Analogue Optical Communications (IEE 1995). Hecollaborates with various telecommunications and electronic industries in the U.K. andoverseas and has acted as a consultant to various academic, industrial, financial andgovernment organisations.Walt Kester (Chapters 16, 17) is the author of Mixed-Signal and DSP Design Techniques.He is a corporate staff applications engineer at Analog Devices. For over 35 years atAnalog Devices, he has designed, developed, and given applications support for high-speed ADCs, DACs, SHAs, op amps, and analog multiplexers. Besides writing manypapers and articles, he prepared and edited eleven major applications books which form thebasis for the Analog Devices world-wide technical seminar series including the topics ofop amps, data conversion, power management, sensor signal conditioning, mixed-signal,and practical analog design techniques. He also is the editor of The Data ConversionHandbook, a 900ϩ page comprehensive book on data conversion published in 2005 byElsevier. Walt has a BSEE from NC State University and MSEE from Duke University.w ww. n e w n e s p r e s s .c o m
    • About the Authors xviiMichael Laughton BASc, (Toronto), PhD (London), DSc (Eng.) (London), FREng,FIEE, CEng (Chapters 25) is the editor of Electrical Engineer’s Reference Book, 16thEdition. He is the Emeritus Professor of Electrical Engineering of the University ofLondon and former Dean of Engineering of the University and Pro-Principal of QueenMary and Westfield College, and is currently the U.K. representative on the EnergyCommittee of the European National Academies of Engineering, a member of energy andenvironment policy advisory groups of the Royal Academy of Engineering, the RoyalSociety and the Institution of Electrical Engineers as well as the Power Industry DivisionBoard of the Institution of Mechanical Engineers. He has acted as Specialist Adviserto U.K. Parliamentary Committees in both upper and lower Houses on alternative andrenewable energy technologies and on energy efficiency. He was awarded The Institutionof Electrical Engineers Achievement Medal in 2002 for sustained contributions toelectrical power engineering.Andrew Leven (Chapter 17, 19) is the author of Telecommunications Circuits andTechnology. He holds a diploma in Radio Technology, HNC, BSc (Hons) Electronics,MSc Astronomy, C. Eng M.I.E.E, Teaching Diploma, M.I.P., International Education andTraining Consultant (Formerly Senior Lecturer in Telecommunications, Electronics andFibre Optics at James Watt College of Higher Education, U.K.)A. Maddocks (Chapter 25) was a contributor to Electrical Engineer’s Reference Book,16th Edition.Clive “Max” Maxfield (Chapter 10) is the author of Bebop to the Boolean Boogie. Heis six feet tall, outrageously handsome, English and proud of it. In addition to being ahero, trendsetter, and leader of fashion, he is widely regarded as an expert in all aspects ofelectronics and computing (at least by his mother).After receiving his B.Sc. in Control Engineering in 1980 from Sheffield Polytechnic (nowSheffield Hallam University), England, Max began his career as a designer of centralprocessing units for mainframe computers. During his career, he has designed everythingfrom ASICs to PCBs and has meandered his way through most aspects of ElectronicsDesign Automation (EDA). To cut a long story short, Max now finds himself Presidentof TechBites Interactive (www.techbites.com). A marketing consultancy, TechBitesspecializes in communicating the value of its clients’ technical products and servicesto non-technical audiences through a variety of media, including websites, advertising,technical documents, brochures, collaterals, books, and multimedia. w w w.ne w nespress.com
    • xviii About the AuthorsIn addition to numerous technical articles and papers appearing in magazines andat conferences around the world, Max is also the author and co-author of a numberof books, including Bebop to the Boolean Boogie (An Unconventional Guide toElectronics), Designus Maximus Unleashed (Banned in Alabama), Bebop BYTES Back(An Unconventional Guide to Computers), EDA: Where Electronics Begins, The DesignWarrior’s Guide to FPGAs, and How Computers Do Math (www.diycalculator.com).In his spare time (Ha!), Max is co-editor and co-publisher of the web-deliveredelectronics and computing hobbyist magazine EPE Online (www.epemag.com). Maxalso acts as editor for the Programmable Logic DesignLine website (www.pldesignline.com) and for the iDESIGN section of the Chip Design Magazine website (www.chipdesignmag.com).On the off-chance that you’re still not impressed, Max was once referred to as an“industry notable” and a “semiconductor design expert” by someone famous who wasn’tprompted, coerced, or remunerated in any way!Luis Moura (Chapter 9) is the author of Introduction to Linear Circuit Analysis andModelling. He received the diploma degree in electronics and telecommunications fromthe University of Aveiro, Portugal, in 1991, and the PhD degree in electronic engineeringfrom the University of North Wales, Bangor, U.K. in 1995. From 1995 to 1997 he workedas a research Fellow in the Telecommunications Research Group at University CollegeLondon, U.K. He is currently a Lecturer in Electronics at the University of Algarve,Portugal. In 2007 he took one year leave of absence to work in the company LimeMicrosystems U.K. as Senior Design Engineer. He was designing frequency synthesisersfor multi-mode/multi-standard wireless transceivers.Ron Schmitt (Chapters 20, 21) is the author of Electromagnetics Explained. He is theformer Director of Electrical Engineering, Sensor Research and Development Corp.Orono, Maine.Keith H. Sueker (Chapters 15, 22, 23) is the author of Power Electronics Design. Suekerreceived his BEE with High Distinction from the University of Minnesota, he continuedhis education at Illinois Institute of Technology where he received his MSEE, he alsocompleted his course work for his PhD. He spent many years working for WestinghouseElectric Corporation in various positions. He then moved on to Robicon Corporationas a consulting engineer, he retired in 1993. His responsibilities included analyticalw ww. n e w n e s p r e s s .c o m
    • About the Authors xixtechniques and equipment design for power factor correction and harmonic mitigation.Sueker has written a number of IEEE papers and several articles for trade publications.Also, he has prepared a monograph and 90 minute video tape on these subjects. He andMr. R. P. Stratford have presented tutorial sessions on power factor and harmonics atIEEE-IAS annual meetings, and he has presented additional tutorials in other cities. Healso presented a tutorial on transformers for the local IEEE-IAS in the spring of 1999and repeated it in the fall of 2003. Sueker delivered a tutorial on power electronics forthe local IEEE-IAS/PES in the spring of 2005. He was also pleased to serve on the IEEEcommittee for awarding the “IEEE Medal for Engineering Excellence” for four years.He is currently a Life Senior Member of the IEEE and also a registered ProfessionalEngineer in the Commonwealth of Pennsylvania.Mike Tooley (Chapters 11, 12, 14, 24) is the author of Electronics Circuits. He is theformer Director of Learning Technology at Brooklands College, Surrey, U.K.Douglas Warne (Chapters 25) is the editor of Electrical Engineers Reference book,16th Edition. Warne graduated from Imperial College London in 1967 with a 1st classhonours degree in electrical engineering, during this time he had a student apprenticeshipwith AEI Heavy Plant Division, Rugby, 1963–1968. He is currently self-employed,and has taken on such projects as Co-ordinated LINK PEDDS programme for DTI,and the electrical engineering, electrical machines and drives and ERCOS programmesfor EPSRC. Initiated and manage the NETCORDE university-industry network foridentifying and launching new R&D projects. He has acted as co-ordinator for theindustry-academic funded ESR Network, held the part-time position of Research ContractCo-ordinator for the High Voltage and Energy Systems group at University of Cardiff andmonitored several projects funded through the DTI Technology Programme.Tim Williams (Chapters 11, 13, 15) is the author of The Circuit Designer’s Companion.He is employed with Elmac Services, Chichester, U.K. w w w.ne w nespress.com
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    • CHAPTE R 1 An Introduction to Electric Circuits John Bird1.1 SI UnitsThe system of units used in engineering and science is the Système International d’Unités(International system of units), usually abbreviated to SI units, and is based on the metricsystem. This was introduced in 1960 and is now adopted by the majority of countries asthe official system of measurement.The basic units in the SI system are listed with their symbols, in Table 1.1.Derived SI units use combinations of basic units and there are many of them. Twoexamples are: ● Velocity—meters per second (m/s) ● Acceleration—meters per second squared (m/s2) Table 1.1: Basic SI units Quantity Unit length meter, m mass kilogram, kg time second, s electric current ampere, A thermodynamic temperature kelvin, K luminous intensity candela, cd amount of substance mole, mol w w w.ne w nespress.com
    • 2 Chapter 1 Table 1.2: Six most common multiples Prefix Name Meaning M mega multiply by 1,000,000 (i.e., ϫ106) k kilo multiply by 1,000 (i.e., ϫ103) m milli divide by 1,000 (i.e., ϫ10Ϫ3) μ micro divide by 1,000,000 (i.e., ϫ10Ϫ6) n nano divide by 1,000,000,000 (i.e., ϫ10Ϫ9) p pico divide by 1,000,000,000,000 (i.e., ϫ10Ϫ12)SI units may be made larger or smaller by using prefixes that denote multiplication ordivision by a particular amount. The six most common multiples, with their meaning, arelisted in Table 1.2.1.2 ChargeThe unit of charge is the coulomb (C) where one coulomb is one ampere second.(1 coulomb ϭ 6.24 ϫ 1018 electrons). The coulomb is defined as the quantity ofelectricity that flows past a given point in an electric circuit when a current of one ampereis maintained for one second. Thus,charge, in coulombs Q ϭ Itwhere I is the current in amperes and t is the time in seconds.Example 1.1If a current of 5 A flows for 2 minutes, find the quantity of electricity transferred.SolutionQuantity of electricity Q ϭ It coulombsI ϭ 5 A, t ϭ 2 ϫ 60 ϭ 120 sHence, Q ϭ 5 ϫ 120 ϭ 600 C1.3 ForceThe unit of force is the newton (N) where one newton is one kilogram meter persecond squared. The newton is defined as the force which, when applied tow ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 3a mass of one kilogram, gives it an acceleration of one meter per second squared.Thus,force, in newtons F ϭ mawhere m is the mass in kilograms and a is the acceleration in meters per second squared.Gravitational force, or weight, is mg, where g ϭ 9.81 m/s2.Example 1.2A mass of 5000 g is accelerated at 2 m/s2 by a force. Determine the force needed.SolutionForce ϭ mass ϫ acceleration kg m ϭ 5 kg ϫ 2 m/s2 ϭ 10 ϭ 10 N s2Example 1.3Find the force acting vertically downwards on a mass of 200 g attached to a wire.SolutionMass ϭ 200 g ϭ 0.2 kg and acceleration due to gravity, g ϭ 9.81 m/s2Force acting downwards ϭ weight ϭ mass ϫ acceleration ϭ 0.2 kg ϫ 9.81 m/s2 ϭ 1.962 N1.4 WorkThe unit of work or energy is the joule (J) where one joule is one Newton meter.The joule is defined as the work done or energy transferred when a force ofone newton is exerted through a distance of one meter in the direction of the force.Thus,work done on a body, in joules W ϭ Fswhere F is the force in Newtons and s is the distance in meters moved by the body in thedirection of the force. Energy is the capacity for doing work. w w w.ne w nespress.com
    • 4 Chapter 11.5 PowerThe unit of power is the watt (W) where one watt is one joule per second. Power isdefined as the rate of doing work or transferring energy. Thus, Wpower in watts, P ϭ twhere W is the work done or energy transferred in joules and t is the time in seconds. Thus,energy, in joules, W ϭ PtExample 1.4A portable machine requires a force of 200 N to move it. How much work is done if themachine is moved 20 m and what average power is utilized if the movement takes 25 s?SolutionWork done ϭ force ϫ distance ϭ 200 N ϫ 20 m ϭ 4000 Nm or 4 kJ work donePower ϭ time taken 4000 J ϭ ϭ 160 J/s ϭ 160 W 25 sExample 1.5A mass of 1000 kg is raised through a height of 10 m in 20 s. What is (a) the work doneand (b) the power developed?Solution(a) Work done ϭ force ϫ distance and force ϭ mass ϫ acceleration Hence, work done ϭ (1000 kg ϫ 9.81 m/s2 ) ϫ (10 m ) ϭ 98100 Nm ϭ 98.1 kNm or 98.1 kJ 8 work done 98100 J(b) Power ϭ ϭ ϭ 4905 J/s time taken 20 s ϭ 4905 W or 4.905 kWw ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 51.6 Electrical Potential and e.m.f.The unit of electric potential is the volt (V) where one volt is one joule per coulomb. Onevolt is defined as the difference in potential between two points in a conductor which,when carrying a current of one ampere, dissipates a power of one watt, i.e., watts joules/secondvolts ϭ ϭ amperes amperes joules joules ϭ ϭ ampere seconds coulombsA change in electric potential between two points in an electric circuit is called apotential difference. The electromotive force (e.m.f.) provided by a source of energy suchas a battery or a generator is measured in volts.1.7 Resistance and ConductanceThe unit of electric resistance is the ohm (Ω) where one ohm is one volt per ampere. Itis defined as the resistance between two points in a conductor when a constant electricpotential of one volt applied at the two points produces a current flow of one ampere inthe conductor. Thus, Vresistance, in ohms R ϭ Iwhere V is the potential difference across the two points in volts and I is the currentflowing between the two points in amperes.The reciprocal of resistance is called conductance and is measured in siemens (S). Thus, 1conductance, in siemens G ϭ Rwhere R is the resistance in ohms.Example 1.6Find the conductance of a conductor of resistance (a) 10 Ω, (b) 5 kΩ and (c) 100 mΩ.Solution 1 1(a) Conductance G ϭ ϭ siemen ϭ 0.1 S R 10 w w w.ne w nespress.com
    • 6 Chapter 1 1 1(b) G ϭ ϭ S ϭ 0.2 ϫ 10Ϫ3 S ϭ 0.2 mS R 5 ϫ 103 1 1 103(c) G ϭ ϭ Sϭ S ϭ 10 S R 100 ϫ 10Ϫ3 1001.8 Electrical Power and EnergyWhen a direct current of I amperes is flowing in an electric circuit and the voltage acrossthe circuit is V volts, then,power, in watts P ϭ VIElectrical energy ϭ Power ϫ time ϭ VIt joulesAlthough the unit of energy is the joule, when dealing with large amounts of energy, theunit used is the kilowatt hour (kWh) where1 kWh ϭ 1000 watt hour ϭ 1000 ϫ 3600 watt seconds or joules ϭ 3,600,000 JExample 1.7A source e.m.f. of 5 V supplies a current of 3 A for 10 minutes. How much energy isprovided in this time?SolutionEnergy ϭ power ϫ time and power ϭ voltage ϫ current.Hence,Energy ϭ VIt ϭ 5 ϫ 3ϫ(10 ϫ 60) ϭ 9000 Ws or J ϭ 9 kJExample 1.8An electric heater consumes 1.8 MJ when connected to a 250 V supply for 30 minutes.Find the power rating of the heater and the current taken from the supply.w ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 7SolutionEnergy ϭ power ϫ time, energypower ϭ time 1.8 ϫ 106 J ϭ ϭ 1000 J/s ϭ 1000 W 30 ϫ 60 si.e., Power rating of heater ϭ 1 kW P 1000Power P ϭ VI , thus, I ϭ ϭ ϭ 4A V 250Hence, the current taken from the supply is 4 A.1.9 Summary of Terms, Units and Their Symbols Table 1.3: Electrical terms, units, and symbolsQuantity Quantity Symbol Unit Unit symbolLength l meter mMass m kilogram kgTime t second sVelocity v meters per second m/s or m sϪ1Acceleration a meters per second squared m/s2 or m sϪ2Force F newton NElectrical charge or quantity Q coulomb CElectric current I ampere AResistance R ohm ΩConductance G siemen SElectromotive force E volt VPotential difference V volt VWork W joule JEnergy E (or W) joule JPower P watt W w w w.ne w nespress.com
    • 8 Chapter 1 Conductor Two conductors Two conductors crossing but not joined together joined Fixed resister Alternative symbol for fixed resister Variable resistor Cell Battery of 3 cells Alternative symbol for battery Switch Filament lamp Fuse A V Ammeter Voltmeter Alternative fuse symbol Figure 1.1: Common electrical component symbols1.10 Standard Symbols for Electrical ComponentsSymbols are used for components in electrical circuit diagrams and some of the morecommon ones are shown in Figure 1.1.1.11 Electric Current and Quantity of ElectricityAll atoms consist of protons, neutrons and electrons. The protons, which havepositive electrical charges, and the neutrons, which have no electrical charge, arecontained within the nucleus. Removed from the nucleus are minute negatively chargedparticles called electrons. Atoms of different materials differ from one another by havingdifferent numbers of protons, neutrons and electrons. An equal number of protons andelectrons exist within an atom and it is said to be electrically balanced, as the positive andw ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 9negative charges cancel each other out. When there are more than two electronsin an atom the electrons are arranged into shells at various distances from thenucleus.All atoms are bound together by powerful forces of attraction existing betweenthe nucleus and its electrons. Electrons in the outer shell of an atom, however, areattracted to their nucleus less powerfully than are electrons whose shells are nearer thenucleus.It is possible for an atom to lose an electron; the atom, which is now called an ion,is not now electrically balanced, but is positively charged and is thus able to attractan electron to itself from another atom. Electrons that move from one atom to anotherare called free electrons and such random motion can continue indefinitely. However,if an electric pressure or voltage is applied across any material there is a tendencyfor electrons to move in a particular direction. This movement of free electrons,known as drift, constitutes an electric current flow. Thus current is the rate of movementof charge.Conductors are materials that contain electrons that are loosely connected to the nucleusand can easily move through the material from one atom to another.Insulators are materials whose electrons are held firmly to their nucleus.The unit used to measure the quantity of electrical charge Q is called the coulomb C(where 1 coulomb ϭ 6.24 ϫ 1018 electrons).If the drift of electrons in a conductor takes place at the rate of one coulomb per secondthe resulting current is said to be a current of one ampere.Thus, 1 ampere ϭ 1 coulomb per second or 1 A ϭ 1 C/s. Hence, 1 coulomb ϭ 1 amperesecond or 1 C ϭ 1 As. Generally, if I is the current in amperes and t the time in secondsduring which the current flows, then I ϫ t represents the quantity of electrical charge incoulombs, i.e., quantity of electrical charge transferred, Q ϭ I ϫ t coulombsExample 1.9What current must flow if 0.24 coulombs is to be transferred in 15 ms? w w w.ne w nespress.com
    • 10 Chapter 1SolutionSince the quantity of electricity, Q ϭ It, then Q 0.24 0.24 ϫ 103 240Iϭ ϭ Ϫ3 ϭ ϭ ϭ 16 A t 15 ϫ 10 15 15Example 1.10If a current of 10 A flows for 4 minutes, find the quantity of electricity transferred.SolutionQuantity of electricity, Q ϭ It coulombsI ϭ 10 A; t ϭ 4 ϫ 60 ϭ 240 sHence, Q ϭ 10 ϫ 240 ϭ 2400 C1.12 Potential Difference and ResistanceFor a continuous current to flow between two points in a circuit a potential differenceor voltage, V, is required between them; a complete conducting path is necessary to andfrom the source of electrical energy. The unit of voltage is the volt, V.Figure 1.2 shows a cell connected across a filament lamp. Current flow, by convention,is considered as flowing from the positive terminal of the cell, around the circuit to thenegative terminal.The flow of electric current is subject to friction. This friction, or opposition, is calledresistance R and is the property of a conductor that limits current. The unit of resistance Figure 1.2: Current floww ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 11is the ohm; 1 ohm is defined as the resistance which will have a current of 1 ampereflowing through it when 1 volt is connected across it, i.e., potential differenceresistance R ϭ current1.13 Basic Electrical Measuring InstrumentsAn ammeter is an instrument used to measure current and must be connected in serieswith the circuit. Figure 1.2 shows an ammeter connected in series with the lamp tomeasure the current flowing through it. Since all the current in the circuit passes throughthe ammeter it must have a very low resistance.A voltmeter is an instrument used to measure voltage and must be connected in parallelwith the part of the circuit whose voltage is required. In Figure 1.2, a voltmeter isconnected in parallel with the lamp to measure the voltage across it. To avoid a significantcurrent flowing through it, a voltmeter must have a very high resistance.An ohmmeter is an instrument for measuring resistance.A multimeter, or universal instrument, may be used to measure voltage, current andresistance. The oscilloscope may be used to observe waveforms and to measure voltagesand currents. The display of an oscilloscope involves a spot of light moving across ascreen. The amount by which the spot is deflected from its initial position depends onthe voltage applied to the terminals of the oscilloscope and the range selected. Thedisplacement is calibrated in volts per cm. For example, if the spot is deflected 3 cm andthe volts/cm switch is on 10 V/cm, then the magnitude of the voltage is 3 cm ϫ 10 V/cm,i.e., 30 V.1.14 Linear and Nonlinear DevicesFigure 1.3 shows a circuit in which current I can be varied by the variable resistor R2.For various settings of R2, the current flowing in resistor R1, displayed on the ammeter,and the p.d. across R1, displayed on the voltmeter, are noted and a graph is plotted of p.d.against current. The result is shown in Figure 1.4(a) where the straight line graph passingthrough the origin indicates that current is directly proportional to the voltage. Since the w w w.ne w nespress.com
    • 12 Chapter 1 Figure 1.3: Circuit in which current can be varied Figure 1.4: Graphs of voltage vs. current: (a) linear device (b) nonlinear devicegradient, i.e., (voltage/current), is constant, resistance R1 is constant. A resistor is thus anexample of a linear device.If the resistor R1 in Figure 1.3 is replaced by a component such as a lamp, then the graphshown in Figure 1.4(b) results when values of voltage are noted for various currentreadings. Since the gradient is changing, the lamp is an example of a nonlinear device.1.15 Ohm’s LawOhm’s law states that the current I flowing in a circuit is directly proportional to theapplied voltage V and inversely proportional to the resistance R, provided the temperatureremains constant. Thus, V V Iϭ or V ϭ IR or Rϭ R Iw ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 13Example 1.11The current flowing through a resistor is 0.8 A when a voltage of 20 V is applied.Determine the value of the resistance.SolutionFrom Ohm’s law, V 20 200resistance R ϭ ϭ ϭ ϭ 25 Ω I 0.8 81.16 Multiples and SubmultiplesCurrents, voltages and resistances can often be very large or very small. Thus multiplesand submultiples of units are often used. The most common ones, with an example ofeach, are listed in Table 1.4.Example 1.12Determine the voltage which must be applied to a 2 kΩ resistor in order that a current of10 mA may flow.SolutionResistance R ϭ 2 kΩ ϭ 2 ϫ 103 ϭ 2000 Ω Table 1.4: Common multiples and submultiples of units Prefix Name Meaning Example M mega multiply by 1,000,000 (i.e., ϫ106) 2 MΩ ϭ 2,000,000 ohms k kilo multiply by 1000 (i.e., ϫ103) 10 kV ϭ 10,000 volts m milli divide by 1000 (i.e., ϫ10Ϫ3) 25 25 mA ϭ A 1000 ϭ 0.025 amperes μ micro divide by 1,000,000 (i.e., ϫ10Ϫ6) 50 50 μV ϭ V 1000 000 ϭ 0.00005 volts w w w.ne w nespress.com
    • 14 Chapter 1Current I ϭ 10 mA 10 10 ϭ 10 ϫ 10Ϫ3 A or or A 103 1000 ϭ 0.01 AFrom Ohm’s law, potential difference,V ϭ IR ϭ (0.01) (2000) ϭ 20 VExample 1.13A coil has a current of 50 mA flowing through it when the applied voltage is 12 V.What is the resistance of the coil?Solution V 12 12 ϫ 103Resistance R ϭ ϭ ϭ I 50 ϫ 10Ϫ3 50 12 000 ϭ ϭ 240 Ω 50Example 1.14A 100 V battery is connected across a resistor and causes a current of 5 mA to flow.Determine the resistance of the resistor. If the voltage is now reduced to 25 V, what willbe the new value of the current flowing?Solution V 100 100 ϫ 103Resistance R ϭ ϭ ϭ I 5 ϫ 10Ϫ3 5 ϭ 20 ϫ 103 ϭ 20 kΩCurrent when voltage is reduced to 25 V, V 25 25Iϭ ϭ ϭ ϫ 10Ϫ3 ϭ 1.25 mA R 20 ϫ 10 3 20w ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 15Example 1.15What is the resistance of a coil that draws a current of (a) 50 mA and (b) 200 μA froma 120 V supply?Solution V 120(a) Resistance R ϭ ϭ I 50 ϫ 10Ϫ3 120 12 000 ϭ ϭ ϭ 2400 Ω or 2.4 kΩ 0.05 5 120 120(b) Resistance R ϭ Ϫ6 ϭ 200 ϫ 10 0.0002 1200 000 ϭ ϭ 600 000 Ω or 600 kΩ r 2 or 0.6 MΩExample 1.16The current/voltage relationship for two resistors A and B is as shown in Figure 1.5.Determine the value of the resistance of each resistor.SolutionFor resistor A, V 20 A 20 2000Rϭ ϭ ϭ ϭ ϭ 1000 Ω or 1 kΩ I 20 mA 0.02 2 Figure 1.5: Current/voltage for two resistors A and B w w w.ne w nespress.com
    • 16 Chapter 1For resistor B, V 16 V 16 16 000Rϭ ϭ ϭ ϭ ϭ 3200 Ω or 3.2 kΩ I 5 mA 0.005 51.17 Conductors and InsulatorsA conductor is a material having a low resistance which allows electric current to flowin it. All metals are conductors and some examples include copper, aluminium, brass,platinum, silver, gold and carbon.An insulator is a material having a high resistance which does not allow electric currentto flow in it. Some examples of insulators include plastic, rubber, glass, porcelain, air,paper, cork, mica, ceramics and certain oils.1.18 Electrical Power and Energy1.18.1 Electrical PowerPower P in an electrical circuit is given by the product of potential difference V andcurrent I. The unit of power is the watt, W. Hence, P ϭ V ϫ I wattsFrom Ohm’s law, V ϭ IR.Substituting for V in equation (1.1) gives: P ϭ (IR) ϫ Ii.e., P ϭ I2R watts VAlso, from Ohm’s law, I ϭ RSubstituting for I in the equation above gives: V PϭVϫ R V2i.e., Pϭ watts RThere are three possible formulas that may be used for calculating power.w ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 17Example 1.17A 100 W electric light bulb is connected to a 250 V supply. Determine (a) the currentflowing in the bulb, and (b) the resistance of the bulb.Solution PPower P ϭ V ϫ I , from which, current I ϭ V 100 0 10 2(a) Current I ϭ ϭ ϭ ϭ 0.4 A 250 25 5 V 250 2500(b) Resistance R ϭ ϭ ϭ ϭ 625 Ω I 0.4 4Example 1.18Calculate the power dissipated when a current of 4 mA flows through a resistance of 5 kΩ.SolutionPower P ϭ I2R ϭ (4 ϫ 10Ϫ3)Ϫ2(5 ϫ 103) ϭ 16 ϫ 10Ϫ6 ϫ 5 ϫ 103 ϭ 80 ϫ 10Ϫ3 ϭ 0.08 W or 80 mWAlternatively, since I ϭ 4 ϫ 10Ϫ3 and R ϭ 5 ϫ 103 then from Ohm’s law,voltage V ϭ IR ϭ 4 ϫ 10Ϫ3 ϫ 5 ϫ 10Ϫ3 ϭ 20 VHence, power P ϭ V ϫ I ϭ 20 ϫ 4 ϫ 10Ϫ3 ϭ 80 mWExample 1.19An electric kettle has a resistance of 30 Ω. What current will flow when it is connected toa 240 V supply? Find also the power rating of the kettle.Solution V 240Current, I ϭ ϭ ϭ8A R 30Power, P ϭ VI ϭ 240 ϫ 8 ϭ 1920 W ϭ 1.95 kW ϭ power rating of kettle w w w.ne w nespress.com
    • 18 Chapter 1Example 1.20A current of 5 A flows in the winding of an electric motor, the resistance of the windingbeing 100 Ω. Determine (a) the voltage across the winding, and (b) the power dissipatedby the coil.SolutionPotential difference across winding, V ϭ IR ϭ 5 ϫ 100 ϭ 500 VPower dissipated by coil, P ϭ I2R ϭ 52 ϫ 100 ϭ 2500 W or 2.5 kW(Alternatively, P ϭ V ϫ I ϭ 500 ϫ 5 ϭ 2500 W or 2.5 kW)Example 1.21The hot resistance of a 240 V filament lamp is 960 Ω. Find the current taken by the lampand its power rating.SolutionFrom Ohm’s law, V 240 24 1current I ϭ ϭ ϭ ϭ A or 0.25 A R 960 96 4 ⎛1⎞Power rating P ϭ VI ϭ (240) ⎜ ⎟ ϭ 60 W ⎜ ⎟ ⎜4⎟ ⎝ ⎠1.18.2 Electrical Energy Electrical energy ϭ power ϫ timeIf the power is measured in watts and the time in seconds then the unit of energy iswatt-seconds or joules. If the power is measured in kilowatts and the time in hours thenthe unit of energy is kilowatt-hours, often called the unit of electricity. The electricitymeter in the home records the number of kilowatt-hours used and is thus an energy meter.Example 1.22A 12 V battery is connected across a load having a resistance of 40 Ω. Determinethe current flowing in the load, the power consumed and the energy dissipated in2 minutes.w ww. n e w n e s p r e s s .c o m
    • An Introduction to Electric Circuits 19Solution V 12Current I ϭ ϭ ϭ 0.3 A R 40Power consumed, P ϭ VI ϭ (12)(0.3) ϭ 3.6 WEnergy dissipated ϭ power ϫ time ϭ (3.6 W)(2 ϫ 60 s) ϭ 432 J (since 1 J ϭ 1 Ws)Example 1.23A source of e.m.f. of 15 V supplies a current of 2 A for 6 minutes. How much energy isprovided in this time?SolutionEnergy ϭ power ϫ time, and power ϭ voltage ϫ currentHence, energy ϭ Vt ϭ 15 ϫ 2 ϫ (6 ϫ 60) ϭ 10 800 Ws or J ϭ 10.8 kJExample 1.24An electric heater consumes 3.6 MJ when connected to a 250 V supply for 40 minutes.Find the power rating of the heater and the current taken from the supply.Solution energy 3.6 ϫ 106 JPower ϭ ϭ (or W ) ϭ 1500 W time 40 ϫ 60 si.e., power rating of heater ϭ 1.5 kW P 1500Power P ϭ VI , thus I ϭ ϭ ϭ 6A V 250Hence, the current taken from the supply ϭ 6 A w w w.ne w nespress.com
    • 20 Chapter 11.19 Main Effects of Electric CurrentThe three main effects of an electric current are: (a) magnetic effect (b) chemical effect (c) heating effectSome practical applications of the effects of an electric current include:Magnetic effect: bells, relays, motors, generators, transformers, telephones, car ignition, and lifting magnetsChemical effect: primary and secondary cells, and electroplatingHeating effect: cookers, water heaters, electric fires, irons, furnaces, kettles, and soldering ironsw ww. n e w n e s p r e s s .c o m
    • CHAPTE R 2 Resistance and Resistivity John Boyd2.1 Resistance and ResistivityThe resistance of an electrical conductor depends on four factors, these being: (a) thelength of the conductor, (b) the cross-sectional area of the conductor, (c) the type ofmaterial and (d) the temperature of the material.Resistance, R, is directly proportional to length, l, of a conductor. For example, if thelength of a piece of wire is doubled, then the resistance is doubled.Resistance, R, is inversely proportional to cross-sectional area, a, of a conductor, i.e.,R is proportional to 1/a. Thus, for example, if the cross-sectional area of a piece of wireis doubled, then the resistance is halved.Since R is proportional to l and R is proportional to 1/a, then R is proportional to l/a. Byinserting a constant of proportionality into this relationship, the type of material used maybe taken into account. The constant of proportionality is known as the resistivity of thematerial and is given the symbol ρ (Greek rho). Thus, ρlresistance R ϭ ohms aρ is measured in ohm meters (Ωm).The value of the resistivity is the resistance of a unit cube of the material measuredbetween opposite faces of the cube. w w w.ne w nespress.com
    • 22 Chapter 2Resistivity varies with temperature and some typical values of resistivities measured atabout room temperature are given in Table 2.1.Note that good conductors of electricity have a low value of resistivity and goodinsulators have a high value of resistivity.Example 2.1The resistance of a 5 m length of wire is 600 Ω. Determine (a) the resistance of an 8 mlength of the same wire, and (b) the length of the same wire when the resistance is 420 Ω.SolutionResistance, R, is directly proportional to length, l, i.e., R ∝ l. Hence, 600 Ω ∝ 5 m or600 ϭ (k)(5), where k is the coefficient of proportionality. Hence, 600kϭ ϭ 120 5When the length l is 8 m, then resistanceR ϭ kl ϭ (120)(8) ϭ 960 ΩWhen the resistance is 420 Ω, 420 ϭ kl, from which 420 420length l ϭ ϭ ϭ 3.5 m k 120Example 2.2A piece of wire of cross-sectional area 2 mm2 has a resistance of 300 Ω. Find (a) theresistance of a wire of the same length and material if the cross-sectional area is 5 mm2, and(b) the cross-sectional area of a wire of the same length and material of resistance 750 Ω. Table 2.1: Typical resistivity values Copper 1.7 ϫ 10Ϫ8 Ωm (or 0.017 μΩm) Ϫ8 Aluminum 2.6 ϫ 10 Ωm (or 0.026 μΩm) Carbon (graphite) 10 ϫ 10Ϫ8 Ωm (or 0.10 μΩm) Glass 1 ϫ 108 Ωm (or 104 μΩm) Mica 1 ϫ 10Ϫ13 Ωm (or 107 μΩm)w ww. n e w n e s p r e s s .c o m
    • Resistance and Resistivity 23SolutionResistance R is inversely proportional to cross-sectional area, a, i.e., R ∝ (1/a )So 300 Ω ∝ (1/ 2 mm 2 ) or 300 ϭ (k )(1/2) from which the coefficient of proportionality,k ϭ 300 ϫ 2 ϭ 600(a) When the cross-sectional area a ϭ 5 mm2 then R ϭ (k )(1/5) ϭ (600)(1/5) ϭ 120 Ω (Note that resistance has decreased as the cross-sectional area is increased.)(b) When the resistance is 750 Ω then 750 ϭ (k)(1/a), from which cross-sectional area, k 600 aϭ ϭ ϭ 0.8 mm 2 750 750Example 2.3A wire of length 8 m and cross-sectional area 3 mm2 has a resistance of 0.16 Ω. If the wireis drawn out until its cross-sectional area is 1 mm2, determine the resistance of the wire.SolutionResistance R is directly proportional to length l, and inversely proportional to the cross-sectional area, a, i.e., R ∝ (l/a ) or R ϭ k (l/a ) , where k is the coefficient of proportionality.Since R ϭ 0.16, l ϭ 8 and a ϭ 3, then 0.16 ϭ (k )(8 / 3) from whichk ϭ 0.16 ϫ (3/8) ϭ 0.06If the cross-sectional area is reduced to {1/3} of its original area, then the length must betripled to 3 ϫ 8, i.e., 24 m.New resistance R ϭ k (l/a ) ϭ 0.06 (24 /1) ϭ 1.44 ΩExample 2.4Calculate the resistance of a 2 km length of aluminum overhead power cable if thecross-sectional area of the cable is 100 mm2. Take the resistivity of aluminum to be0.03 ϫ 10Ϫ6 Ωm. w w w.ne w nespress.com
    • 24 Chapter 2SolutionLength l ϭ 2 km ϭ 2000 m; area, a ϭ 100 mm2 ϭ 100 ϫ 10Ϫ6 m2; resistivityρ ϭ 0.03 ϫ 10Ϫ6 Ωm ρl (0.03 ϫ 10Ϫ6 Ωm)(2000 m)Resistance R ϭ ϭ a (100 ϫ 10Ϫ6 m 2 ) 0.03 ϫ 2000 ϭ Ω 100 ϭ 0.6 ΩExample 2.5Calculate the cross-sectional area, in mm2, of a piece of copper wire, 40 m in length andhaving a resistance of 0.25 Ω. Take the resistivity of copper as 0.02 ϫ 10Ϫ6 Ωm.Solution ρl ρlResistance R ϭ so cross-sectional area a ϭ a R (0.02 ϫ 10Ϫ6 Ωm)( 40 m) ϭ 0.25 Ω ϭ 3.2 ϫ 10Ϫ6 m 2 ϭ (3.2 ϫ 10Ϫ6 ) ϫ 10Ϫ6 mm 2 ϭ 3.2 mm 2Example 2.6The resistance of 1.5 km of wire of cross-sectional area 0.17 mm2 is 150 Ω. Determine theresistivity of the wire.Solution ρlResistance R ϭ a Ra (150 Ω)(0.17 ϫ 10Ϫ6 m 2 )so resistivity ρ ϭ ϭ l (1500 m) ϭ 0.017 ؋ 10؊6 Ωm or 0.017 μΩmExample 2.7Determine the resistance of 1200 m of copper cable having a diameter of 12 mm if theresistivity of copper is 1.7 ϫ 10Ϫ8 Ωm.w ww. n e w n e s p r e s s .c o m
    • Resistance and Resistivity 25SolutionCross-sectional area of cable, a ϭ πr 2 ϭ π ( 12 ) ϭ 36π mm 2 ϭ 36π ϫ 10Ϫ6 m 2 2 2 ρl (1.7 ϫ 10Ϫ8 Ωm)(1200 m)Resistance R ϭ ϭ a (36π ϫ 10Ϫ6 m 2 ) 1.7 ϫ 1200 ϫ 106 1.7 ϫ 12 ϭ 8 ϫ 36π Ωϭ Ω 10 36π ϭ 0.180 Ω2.2 Temperature Coefficient of ResistanceIn general, as the temperature of a material increases, most conductors increase inresistance, insulators decrease in resistance, while the resistance of some special alloysremains almost constant.The temperature coefficient of resistance of a material is the increase in the resistanceof a 1Ω resistor of that material when it is subjected to a rise of temperature of 1°C.The symbol used for the temperature coefficient of resistance is α (Greek alpha).Thus, if some copper wire of resistance 1Ω is heated through 1°C and its resistanceis then measured as 1.0043 12 then α ϭ 0.0043 Ω/Ω°C for copper. The units areusually expressed only as “per °C.” So, α ϭ 0.0043/°C for copper. If the 1Ωresistor of copper is heated through 100°C then the resistance at 100°C would be1 ϩ 100 ϫ 0.0043 ϭ 1.43 Ω.Some typical values of temperature coefficient of resistance measured at 0°C are given inTable 2.2.(Note that the negative sign for carbon indicates that its resistance falls with increase oftemperature.) Table 2.2: Typical values of temperature coefficient of resistance Copper 0.0043/°C Aluminum 0.0038/°C Nickel 0.0062/°C Carbon Ϫ0.00048/°C Constantan 0 Eureka 0.00001/°C w w w.ne w nespress.com
    • 26 Chapter 2If the resistance of a material at 0°C is known, the resistance at any other temperature canbe determined from: Rθ ϭ R0 (1 ϩ α 0 θ)where Rθ ϭ resistance at 0°C Rθ ϭ resistance at temperature θ°C α0 ϭ temperature coefficient of resistance at 0°CExample 2.8A coil of copper wire has a resistance of 100 Ω when its temperature is 0°C. Determineits resistance at 70°C if the temperature coefficient of resistance of copper at 0°C is0.0043/°C.SolutionResistance Rθ ϭ R0 (1 ϩ α0θ)So resistance at 70°C, R70 ϭ 100[1 ϩ (0.0043)(70)] ϭ 100[1 ϩ 0.301] ϭ 100(1.301) ‫ 1.031 ؍‬ΩExample 2.9An aluminum cable has a resistance of 27 Ω at a temperature of 35°C. Determine itsresistance at 0°C. Take the temperature coefficient of resistance at 0°C to be 0.0038/°C.SolutionResistance at θ°C, Rθ ϭ R0(1 ϩ α 0θ) RθHence resistance at 0ЊC, R0 ϭ (1 ϩ α0 θ) 27 ϭ [1 ϩ (0.0038)(35)] 27 27 ϭ ϭ 1 ϩ 0.133 1.133 ϭ 23.83 Ωw ww. n e w n e s p r e s s .c o m
    • Resistance and Resistivity 27Example 2.10A carbon resistor has a resistance of 1 kΩ at 0°C. Determine its resistance at 80°C.Assume that the temperature coefficient of resistance for carbon at 0°C is Ϫ0.0005/°C.SolutionResistance at temperature θ°C, Rθ ϭ R0(1 ϩ α0θ)i.e., Rθ ϭ 1000[1 ϩ (Ϫ0.0005)(80)] ϭ 1000[1 Ϫ 0.040] ϭ 1000(0.96) ϭ 960 ΩIf the resistance of a material at room temperature (approximately 20°C), R20, and thetemperature coefficient of resistance at 20°C, α20, are known then the resistance Rθ attemperature θ°C is given by: Rθ ϭ R20 [1 ϩ α20 (θ Ϫ 20)]Example 2.11A coil of copper wire has a resistance of 10 Ω at 20°C. If the temperature coefficient ofresistance of copper at 20°C is 0.004/°C, determine the resistance of the coil when thetemperature rises to 100°C.SolutionResistance at temperature θ°C, R ϭ R20[1 ϩ α20(θ Ϫ 20)]Hence resistance at 100°C,R100 ϭ 10[1 ϩ (0.004)(100 Ϫ 20)] ϭ 10[1 ϩ (0.004)(80)] ϭ 10[1 ϩ 0.32] ϭ 10(1.32) ϭ 13.2 Ω w w w.ne w nespress.com
    • 28 Chapter 2Example 2.12The resistance of a coil of aluminum wire at 18°C is 200 Ω. The temperature of the wireis increased and the resistance rises to 240 Ω. If the temperature coefficient of resistanceof aluminum is 0.0039/°C at 18°C determine the temperature to which the coil has risen.SolutionLet the temperature rise to θ°Resistance at θ°C, Rθ ϭ R18[1 ϩ α18(θ Ϫ 18)]i.e., 240 ϭ 200[1 ϩ (0.0039)(θ Ϫ 18)] 240 ϭ 200 ϩ (200)(0.0039)(θ Ϫ 18) 240 Ϫ 200 ϭ 0.78(θ Ϫ 18) 40 ϭ 0.78(θ Ϫ 18) 40 ϭ θ Ϫ 18 0.78 51.28 ϭ θ Ϫ 18, from which, θ ϭ 51.28 ϩ 18 ϭ 69.28°CHence the temperature of the coil increases to 69.28°C.If the resistance at 0°C is not known, but is known at some other temperature θ1, then theresistance at any temperature can be found as follows:R1 ϭ R0 (1 ϩ α0θ1) and R2 ϭ R0(1 ϩ α0θ2)Dividing one equation by the other gives: R1 1 ϩ α0 θ1 ϭ R2 1 ϩ α0 θ 2where R2 ϭ resistance at temperature θ2.Example 2.13Some copper wire has a resistance of 200 Ω at 20°C. A current is passed through thewire and the temperature rises to 90°C. Determine the resistance of the wire at 90°C,w ww. n e w n e s p r e s s .c o m
    • Resistance and Resistivity 29correct to the nearest ohm, assuming that the temperature coefficient of resistance is0.004/°C at 0°C.SolutionR20 ϭ 200 Ω, α0 ϭ 0.004/°CR20 [1 ϩ α 0 (20)] ϭR90 [1 ϩ α 0 (90)] R20 [1 ϩ 90α 0 ]Hence, R90 ϭ [1 ϩ 20α 0 ] 200[1 ϩ 90(0.004)] ϭ [1 ϩ 20(0.004)] 200[1 ϩ 0.36] ϭ [1 ϩ 0.08] 200(1.36) ϭ ϭ 251.85 Ω (1.08)So, the resistance of the wire at 90°C is 252 Ω. w w w.ne w nespress.com
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    • CHAPTE R 3 Series and Parallel Networks John Bird3.1 Series CircuitsFigure 3.1 shows three resistors R1, R2 and R3 connected end to end, i.e., in series,with a battery source of V volts. Since the circuit is closed, a current I will flowand the voltage across each resistor may be determined from the voltmeter readings V1,V2 and V3.In a series circuit: (a) the current I is the same in all parts of the circuit; therefore, the same reading is found on each of the two ammeters shown, and, (b) the sum of the voltages V1, V2 and V3 is equal to the total applied voltage, V, i.e., V ϭ V1 ϩ V2 ϩ V3 Figure 3.1: Series circuit w w w.ne w nespress.com
    • 32 Chapter 3From Ohm’s law:V1 ϭ IR1, V2 ϭ IR2, V3 ϭ IR3 and V ϭ IRwhere R is the total circuit resistance.Since V ϭ V1 ϩ V2 ϩ V3then IR ϭ IR1 ϩ IR2 ϩ IR3Dividing throughout by I gives: R ϭ R1 ϩ R2 ϩ R3So, for a series circuit, the total resistance is obtained by adding together the values of theseparate resistances.Example 3.1For the circuit shown in Figure 3.2, determine (a) the battery voltage V, (b) the totalresistance of the circuit, and (c) the values of resistance of resistors R1, R2 and R3, giventhat the voltages across R1, R2 and R3 are 5 V, 2 V and 6 V, respectively.Solution(a) Battery voltage V ϭ V1 ϩ V2 ϩ V3 ϭ 5 ϩ 2 ϩ 6 ϭ 13 V V 13(b) Total circuit resistance R ϭ ϭ ϭ 3.25 Ω I 4 Figure 3.2: Circuit for Example 3.1w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 33 V1 5(c) Resistance R1 ϭ ϭ ϭ 1.25 Ω I 4 V2 2 Resistance R2 ϭ ϭ ϭ 0.5 Ω I 4 V 6 Resistance R3 ϭ 3 ϭ ϭ 1.5 Ω I 4 (Check: R1 ϩ R2 ϩ R3 ϭ 1.25 ϩ 0.5 ϩ 1.5 ϭ 3.25 Ω ϭ R)Example 3.2For the circuit shown in Figure 3.3, determine the voltage across resistor R3. If the totalresistance of the circuit is 100 Ω, determine the current flowing through resistor R1. Findalso the value of resistor R2.SolutionVoltage across R3, V3 ϭ 25 Ϫ 10 Ϫ 4 ϭ 11 V V 25Current I ϭ ϭ ϭ 0.25 A, which is the current flowing in each resistor R 100 V2 4Resistance R2 ϭ ϭ ϭ 16 Ω I 0.25Example 3.3A 12 V battery is connected in a circuit having three series-connected resistors havingresistances of 4 Ω, 9 Ω and 11 Ω. Determine the current flowing through, and the voltageacross the 9 Ω resistor. Find also the power dissipated in the 11 Ω resistor. Figure 3.3: Circuit for Example 3.2 w w w.ne w nespress.com
    • 34 Chapter 3 Figure 3.4: Circuit for Example 3.3SolutionThe circuit diagram is shown in Figure 3.4.Total resistance R ϭ 4 ϩ 9 ϩ 11 ϭ 24 Ω V 12Current I ϭ ϭ ϭ 0.5 A , which is the current in the 9 Ω resistor. R 24Voltage across the 9 Ω resistor, V1 ϭ I ϫ 9 ϭ 0.5 ϫ 9 ϭ 4.5 VPower dissipated in the 11 Ω resistor, P ϭ I2R ϭ 0.52(11) ϭ 0.25(11) ϭ 2.75 W3.2 Potential DividerThe voltage distribution for the circuit shown in Figure 3.5(a) is given by: ⎛ R ⎞ ⎟ V1 ϭ ⎜ ⎜ 1 ⎟ ⎜ R ϩ R ⎟V ⎜ 1 ⎝ ⎟ 2⎠ ⎛ R ⎞ ⎟ V2 ϭ ⎜ ⎜ 2 ⎟ ⎜ R ϩ R ⎟V ⎜ 1 ⎝ ⎟ 1⎠The circuit shown in Figure 3.5(b) is often referred to as a potential divider circuit. Sucha circuit can consist of a number of similar elements in series connected across a voltagesource, voltages being taken from connections between the elements. Frequently thedivider consists of two resistors as shown in Figure 3.5(b), where: ⎛ R ⎞ ⎟ VOUT ϭ ⎜ ⎜ 2 ⎟ ⎜ R ϩ R ⎟ VIN ⎜ 1 ⎝ ⎟ 2⎠w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 35 Figure 3.5: Potential divider circuitA potential divider is the simplest way of producing a source of lower e.m.f. from asource of higher e.m.f., and is the basic operating mechanism of the potentiometer, ameasuring device for accurately measuring potential differences.Example 3.4Determine the value of voltage V shown in Figure 3.6.SolutionFigure 3.6 may be redrawn as shown in Figure 3.7, and voltage ⎛ 6 ⎞ ⎟ (50) ϭ 30 VV ϭ⎜ ⎜6ϩ4⎟ ⎜ ⎝ ⎟ ⎠Example 3.5Two resistors are connected in series across a 24 V supply and a current of 3 A flows inthe circuit. If one of the resistors has a resistance of 2 Ω determine (a) the value of theother resistor, and (b) the voltage across the 2 Ω resistor. If the circuit is connected for 50hours, how much energy is used? w w w.ne w nespress.com
    • 36 Chapter 3SolutionThe circuit diagram is shown in Figure 3.8.(a) Total circuit resistance R ϭ V ϭ 24 ϭ 8 Ω I 3 Value of unknown resistance, Rx ϭ 8 Ϫ 2 ϭ 6 Ω(b) Voltage across 2 Ω resistor, V1 ϭ IR1 ϭ 3 ϫ 2 ϭ 6 V Alternatively, from above, ⎛ R ⎞ ⎟ V ϭ ⎛ 2 ⎞ (24) ϭ 6 V V1 ϭ ⎜ ⎜ 1 ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ R1 ϩ Rx ⎟ ⎜ ⎝ ⎟ ⎠ ⎝2ϩ6⎟ ⎜ ⎠ Figure 3.6: Circuit for Example 3.4 Figure 3.7: Redrawn version of Figure 3.6 Figure 3.8: Circuit for Example 3.5w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 37 Energy used ϭ power ϫ time ϭVϫIϫt ϭ (24 ϫ 3 W) (50 h) ϭ 3600 Wh ϭ 3.6 kWh3.3 Parallel NetworksFigure 3.9 shows three resistors, R1, R2 and R3 connected across each other, i.e., inparallel, across a battery source of V volts.In a parallel circuit: (a) the sum of the currents I1, I2 and I3 is equal to the total circuit current, I, i.e., I ϭ I1 ϩ I2 ϩ I3, and (b) the source voltage, V volts, is the same across each of the resistors.From Ohm’s law: V V V VI1 ϭ , I2 ϭ , I3 ϭ and I ϭ R1 R2 R3 Rwhere R is the total circuit resistance. Figure 3.9: Parallel resistors w w w.ne w nespress.com
    • 38 Chapter 3Since I ϭ I1 ϩ I2 ϩ I3 V V V Vthen ϭ ϩ ϩ R R1 R2 R3Dividing throughout by V gives: 1 1 1 1 ϭ ϩ ϩ R R1 R2 R3This equation must be used when finding the total resistance R of a parallel circuit. Forthe special case of two resistors in parallel: 1 1 1 R ϩ R1 ϭ ϩ ϭ 2 R R1 R2 R1 R2 R1 R2 ⎛ ⎞Hence, Rϭ ⎜ i.e., product ⎟ ⎜ ⎟ R1 ϩ R2 ⎜ ⎝ sum ⎟ ⎠Example 3.6For the circuit shown in Figure 3.10, determine (a) the reading on the ammeter, and(b) the value of resistor R2. Figure 3.10: Circuit for Example 3.6w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 39SolutionVoltage across R1 is the same as the supply voltage V. Hence, supply voltageV ϭ 8 ϫ 5 ϭ 40 V. V 40(a) Reading on ammeter, I ϭ ϭ ϭ2A R3 20(b) Current flowing through R2 ϭ 11 Ϫ 8 Ϫ 2 ϭ 1 A V 40 Hence, R2 ϭ ϭ ϭ 40 Ω I2 1Example 3.7Two resistors, of resistance 3 Ω and 6 Ω, are connected in parallel across a battery havinga voltage of 12 V. Determine (a) the total circuit resistance and (b) the current flowing inthe 3 Ω resistor.SolutionThe circuit diagram is shown in Figure 3.11.(a) The total circuit resistance R is given by: 1 1 1 1 1 ϭ ϩ ϭ ϩ R R1 R2 3 6 1 2 ϩ1 3 ϭ ϭ R 6 6 6 Hence, R ϭ ϭ 2Ω 3 Figure 3.11: Circuit for Example 3.7 w w w.ne w nespress.com
    • 40 Chapter 3 ⎛ ⎞ ⎜ Alternatively, R ϭ R1 R2 ϭ 3 ϫ 6 ϭ 18 ϭ 2 Ω ⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎟ ⎝ R1 ϩ R2 3ϩ6 9 ⎠ V 12(b) Current in the 3 Ω resistance, I1 ϭ ϭ ϭ 4A R1 3Example 3.8For the circuit shown in Figure 3.12, find (a) the value of the supply voltage V and (b) thevalue of current I.Solution(a) Voltage across 20 Ω resistor ϭ I2R2 ϭ 3 ϫ 20 ϭ 60 V; hence, supply voltage V ‫ 06 ؍‬V since the circuit is connected in parallel.(b) Current I1 ϭ V ϭ 60 ϭ 6 A; I 2 ϭ 3 A R1 10 V 60 I3 ϭ ϭ ϭ 1A R3 60 Current I ϭ I1 ϩ I2 ϩ I3 and hence, I ϭ 6 ϩ 3 ϩ 1 ϭ 10 A 1 1 1 1 1 ϩ 3 ϩ 6 10 Alternatively, ϭ ϩ ϩ ϭ ϭ R 60 20 10 60 60 60 Hence, total resistance R ϭ ϭ 6Ω 10 V 60 Current I ϭ ϭ ϭ 10 A R 6 Figure 3.12: Circuit for Example 3.8w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 41Example 3.9Given four 1 Ω resistors, state how they must be connected to give an overall resistance of(a) 1/4 Ω (b) 1 Ω (c) 1 1 3 Ω (d) 2 1 2 ΩSolution(a) All four in parallel (see Figure 3.13), 1 1 1 1 1 4 1 Since ϭ ϩ ϩ ϩ ϭ , i.e., R ϭ Ω R 1 1 1 1 1 4(b) Two in series, in parallel with another two in series (see Figure 3.14), since 1 Ω and 1 Ω in series gives 2 Ω, and 2 Ω in parallel with 2 Ω gives: 2ϫ2 4 ϭ ϭ 1Ω 2ϩ2 4(c) Three in parallel, in series with one (see Figure 3.15), since for the three in parallel, 1 1 1 1 3 1 1 1 ϭ ϩ ϩ ϭ , i.e., R ϭ Ω and Ω in series with 1 Ω gives 1 Ω R 1 1 1 1 3 3 3(d) Two in parallel, in series with two in series (see Figure 3.16), since for the two in parallel Figure 3.13: Circuit for Example 3.9(a) Figure 3.14: Circuit for Example 3.9(b) w w w.ne w nespress.com
    • 42 Chapter 3 Figure 3.15: Circuit for Example 3.9(c) Figure 3.16: Circuit for Example 3.9(d) Figure 3.17: Circuit for Example 3.10 1ϫ1 1 1 1Rϭ ϭ Ω, and Ω, 1 Ω and 1 Ω in series gives 2 Ω 1ϩ1 2 2 2Example 3.10Find the equivalent resistance for the circuit shown in Figure 3.17.SolutionR3, R4 and R5 are connected in parallel and their equivalent resistance R is given by: 1 1 1 1 6 ϩ 3 ϩ 1 10 ϭ ϩ ϩ ϭ ϭ R 3 6 18 18 18 18Hence, R ϭ ϭ 1.8 Ω 10The circuit is now equivalent to four resistors in series and the equivalent circuitresistance ϭ 1 ϩ 2.2 ϩ 1.8 ϩ 4 ϭ 9 Ω.w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 43 Figure 3.18: Current division circuit3.4 Current DivisionFor the circuit shown in Figure 3.18, the total circuit resistance RT is given by: R1 R2RT ϭ R1 ϩ R2 ⎛ RR ⎞and V ϭ IRT ϭ I ⎜ 1 2 ⎟ ⎜ ⎟ ⎜R ϩR ⎟ ⎜ 1 ⎝ ⎟ 2⎠ V I ⎛ RR ⎞ ⎛ R ⎟ϭ⎜ ⎞ ⎟ (I )Current I1 ϭ ϭ ⎜ 1 2 ⎜ ⎟ ⎜ 2 ⎟ ⎜ ⎜R ϩR ⎟ ⎜ R1 ϩ R2 ⎟ ⎟ ⎜ ⎟ R1 R1 ⎝1 2⎠ ⎝ ⎠Similarly, V I ⎛ RR ⎞ ⎛ R ⎟ϭ⎜ ⎞ ⎟ (I )current I 2 ϭ ϭ ⎜ 1 2 ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎜R ϩR ⎟ ⎜ R1 ϩ R2 ⎟ ⎟ ⎜ ⎟ R2 R2 ⎝ 1 2⎠ ⎝ ⎠Summarizing, with reference to Figure 3.18: ⎛ R ⎞ ⎟ ⎛ R ⎞ ⎟ I1 ϭ ⎜ ⎜ 2 ⎟ ⎜ R ϩ R ⎟ (I ) and I2 ϭ ⎜ ⎜ 1 ⎟ ⎜ R ϩ R ⎟ (I ) ⎜ 1 ⎝ ⎟ 2⎠ ⎜ 1 ⎝ ⎟ 2⎠Example 3.11For the series-parallel arrangement shown in Figure 3.19, find (a) the supplycurrent, (b) the current flowing through each resistor and (c) the voltage across eachresistor. w w w.ne w nespress.com
    • 44 Chapter 3 Figure 3.19: Circuit for Example 3.11Solution(a) The equivalent resistance Rx of R2 and R3 in parallel is: 6 ϫ 2 12 Rx ϭ ϭ ϭ 1.5 Ω 6ϩ2 8 The equivalent resistance RT of R1, Rx and R4 in series is: RT ϭ 2.5 ϩ 1.5 ϩ 4 ϭ 8 Ω V 200 Supply current I ϭ ϭ ϭ 25 A RT 8(b) The current flowing through R1 and R4 is 25 A. The current flowing through R2 ⎛ R ⎞ ⎟ ⎛ 2 ⎞ ϭ⎜ ⎟ ⎜ ⎟ ⎜ R ϩ R ⎟ I ϭ ⎜ 6 ϩ 2 ⎟ 25 ⎜ 3 ⎟ ⎜ 2 ⎝ ⎟ ⎜ ⎝ ⎠ 3⎠ ϭ 6.25 A The current flowing through R3 ⎛ R ⎞ ⎟ ⎛ 6 ⎞ ϭ⎜ ⎟ ⎜ ⎟ ⎜ R ϩ R ⎟ I ϭ ⎜ 6 ϩ 2 ⎟ 25 ⎜ 2 ⎟ ⎜ 2 ⎝ ⎟ ⎜ ⎝ ⎠ 3⎠ ϭ 18.75 A (Note that the currents flowing through R2 and R3 must add up to the total current flowing into the parallel arrangement, i.e., 25 A.)w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 45 Figure 3.20: Equivalent circuit of Figure 3.19 Figure 3.21: Circuit for Example 3.12(c) The equivalent circuit of Figure 3.19 is shown in Figure 3.20. voltage across R1, i.e., V1 ϭ IR1 ϭ (25)(2.5) ϭ 62.5 V voltage across Rx, i.e., Vx ϭ IRx ϭ (25)(1.5) ϭ 37.5 V voltage across R4, i.e., V4 ϭ IR4 ϭ (25)(4) ϭ 100 V Hence, the voltage across R2 ϭ voltage across R3 ϭ 37.5 VExample 3.12For the circuit shown in Figure 3.21 calculate (a) the value of resistor Rx such that thetotal power dissipated in the circuit is 2.5 kW, and (b) the current flowing in each of thefour resistors.Solution(a) Power dissipated P ϭ VI watts, hence, 2500 ϭ (250)(I) 2500 i.e., I ϭ ϭ 10 A 250 w w w.new nespress.com
    • 46 Chapter 3 V 250 From Ohm’s law, RT ϭ ϭ ϭ 25 Ω, where RT is the equivalent circuit resistance. I 10 The equivalent resistance of R1 and R2 in parallel is: 15 ϫ 10 150 ϭ ϭ 6Ω 15 ϩ 10 25 The equivalent resistance of resistors R3 and Rx in parallel is equal to 25 Ω Ϫ 6 Ω, i.e., 19 Ω. There are three methods whereby Rx can be determined. Method 1 The voltage V1 ϭ IR, where R is 6 Ω, from above, i.e., V1 ϭ (10)(6) ϭ 60 V Hence, V2 ϭ 250 V Ϫ 60 V ϭ 190 V ϭ voltage across R3 ϭ voltage across Rx V2 190 I3 ϭ ϭ ϭ 5 A. Thus, I4 ϭ 5 A also, R3 38 since I ϭ 10 A V2 190 Thus, Rx ϭ ϭ ϭ 38 Ω I4 5 Method 2 Since the equivalent resistance of R3 and Rx in parallel is 19 Ω, 38 Rx ⎛ ⎞ 19 ϭ ⎜ i.e., product ⎟ ⎜ ⎟ 38 ϩ Rx ⎜ ⎝ sum ⎠ ⎟ Hence, 19(38 ϩ Rx) ϭ 38Rx 722 ϩ 19Rx ϭ 38Rx 722 ϭ 38Rx Ϫ 19Rx ϭ 19Rx 722 Thus, Rx ϭ ϭ 38 Ω 19w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 47 Method 3 When two resistors having the same value are connected in parallel, the equivalent resistance is always half the value of one of the resistors. In this case, since RT ϭ 19 Ω and R3 ϭ 38 Ω, then Rx ‫ 83 ؍‬Ω could have been deduced on sight. ⎛ R ⎞ ⎟ ⎛ 10 ⎞(b) Current I1 ϭ ⎜ ⎟ ⎜ ⎟ ⎜ R ϩ R ⎟ I ϭ ⎜ 15 ϩ 10 ⎟ (10) ⎜ 2 ⎟ ⎜ 1 ⎝ ⎟ ⎠ ⎜ ⎝ ⎠ 2 ⎛2⎞ ϭ ⎜ ⎟ (10) ϭ 4 A ⎜ ⎟ ⎜5⎟ ⎝ ⎠ ⎛ R ⎞ ⎟ ⎛ 15 ⎞ Current I 2 ϭ ⎜ ⎟ ⎜ ⎟ ⎜ R ϩ R ⎟ I ϭ ⎜ 15 ϩ 10 ⎟ (10) ⎜ 1 ⎜ ⎟ ⎜ 1 ⎝ 2 ⎟ ⎠ ⎝ ⎠ ⎛3⎞ ϭ ⎜ ⎟ (10) ϭ 6 A ⎜ ⎟ ⎜5⎟ ⎝ ⎠ From part (a), method 1, I3 ‫ ؍‬I4 ‫ 5 ؍‬AExample 3.13For the arrangement shown in Figure 3.22, find the current Ix.SolutionCommencing at the right-hand side of the arrangement shown in Figure 3.24, the circuitis gradually reduced in stages as shown in Figures 3.23(a)–(d). 17From Figure 3.23(d), I ϭ ϭ 4A 4.25 Figure 3.22: Circuit for Example 3.13 w w w.ne w nespress.com
    • 48 Chapter 3 Figure 3.23: Solution to Example 3.13, in four stages ⎛ 9 ⎞ ⎟ ( I ) ϭ ⎛ 9 ⎞ ( 4) ϭ 3 AFrom Figure 3.23(b), I1 ϭ ⎜ ⎜ ⎟ ⎜9ϩ3⎟ ⎜ ⎝ ⎟ ⎠ ⎜ ⎟ ⎜ 12 ⎟ ⎝ ⎠ ⎛ 2 ⎞⎟ ( I ) ϭ ⎛ 2 ⎞ (3) ϭ 0.6 AFrom Figure 3.22, I x ϭ ⎜ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝2 ϩ8⎟ 1 ⎜ ⎠ ⎝ 10 ⎟ ⎜ ⎠3.5 Relative and Absolute VoltagesIn an electrical circuit, the voltage at any point can be quoted as being “with reference to”(w.r.t.) any other point in the circuit. Consider the circuit shown in Figure 3.24. The totalresistance,RT ϭ 30 ϩ 50 ϩ 5 ϩ 15 ϭ 100 Ω 200and current, I ϭ ϭ 2A 100w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 49 I ϭ 2A 30 ⍀ A 50 ⍀ B 200 V 5⍀ C 15 ⍀ Figure 3.24: Relative voltageIf a voltage at point A is quoted with reference to point B then the voltage is written asVAB. This is known as a relative voltage. In the circuit shown in Figure 3.24, the voltage atA w.r.t. B is I ϫ 50, i.e., 2 ϫ 50 ϭ 100 V and is written as VAB ϭ 100 V.It must also be indicated whether the voltage at A w.r.t. B is closer to the positive terminalor the negative terminal of the supply source. Point A is nearer to the positive terminalthan B so is written as VAB ϭ 100 V or VAB ϭ ϩ100 V or VAB ϭ 100 V ϩve.If no positive or negative is included, then the voltage is always taken to be positive.If the voltage at B w.r.t. A is required, then VBA is negative and is written asVBA ϭ Ϫ100 V or VBA ϭ 100 V Ϫve.If the reference point is changed to the earth point then any voltage taken w.r.t. theearth is known as an absolute potential. If the absolute voltage of A in Figure 3.24 isrequired, then this will be the sum of the voltages across the 50 Ω and 5 Ω resistors, i.e.,100 ϩ 10 ϭ 110 V and is written as VA ϭ 110 V or VA ϭ ϩ110 V or VA ϭ 110 V ϩve,positive since moving from the earth point to point A is moving towards the positiveterminal of the source. If the voltage is negative w.r.t. earth then this must be indicated;for example, VC ϭ 30 V negative w.r.t. earth, and is written as VC ϭ Ϫ30 V orVC ϭ 30 V Ϫve.Example 3.14For the circuit shown in Figure 3.25, calculate (a) the voltage drop acrossthe 4 kΩ resistor, (b) the current through the 5 kΩ resistor, (c) the power developedin the 1.5 kΩ resistor, (d) the voltage at point X w.r.t. earth, and (e) the absolutevoltage at point X. w w w.ne w nespress.com
    • 50 Chapter 3 1 k⍀ 4 k⍀ 5 k⍀ X 1.5 k⍀ 24 V Figure 3.25: Circuit for Example 3.14Solution(a) Total circuit resistance, RT ϭ [(1 ϩ 4) kΩ in parallel with 5 kΩ] in series with 1.5 kΩ 5ϫ 5 i.e., RT ϭ ϩ 1.5 ϭ 4 kΩ 5ϩ5 V 24 Total circuit current, IT ϭ ϭ ϭ 6 mA RT 4 ϫ 103 By current division, current in top branch ⎛ 5 ⎞ ⎟ ϫ 6 ϭ 3 mA ϭ⎜ ⎜ 5 ϩ1ϩ 4 ⎟ ⎜ ⎝ ⎟ ⎠ Hence, volt drop across 4 kΩ resistor ϭ 3 ϫ 10Ϫ3 ϫ 4 ϫ 103 ϭ 12 V(b) Current through the 5 kΩ resistor ⎛ 1ϩ 4 ⎞ ⎟ ϫ 6 ϭ 3 mA ϭ⎜ ⎜ 5 ϩ1ϩ 4 ⎟ ⎜ ⎝ ⎟ ⎠(c) Power in the 1.5 kΩ resistor ϭ IT R ϭ (6 ϫ 10Ϫ3 )2 (1.5 ϫ 103 ) ϭ 54 mW 2w ww. n e w n e s p r e s s .c o m
    • Series and Parallel Networks 51(d) The voltage at the earth point is 0 volts. The volt drop across the 4 kΩ is 12 V, from part (a). Since moving from the earth point to point X is moving towards the negative terminal of the voltage source, the voltage at point X w.r.t. earth is Ϫ12 V.(e) The absolute voltage at point X means the voltage at point X w.r.t. earth; therefore, the absolute voltage at point X is Ϫ12 V. Questions (d) and (e) mean the same thing. w w w.ne w nespress.com
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    • CHAPTE R 4 Capacitors and Inductors John Bird4.1 Introduction to CapacitorsA capacitor is an electrical device that is used to store electrical energy. Next to theresistor, the capacitor is the most commonly encountered component in electricalcircuits. For example, capacitors are used to smooth rectified AC outputs, they are usedin telecommunication equipment—such as radio receivers—for tuning to the requiredfrequency, they are used in time delay circuits, in electrical filters, in oscillator circuits,and in magnetic resonance imaging (MRI) in medical body scanners, to name but a fewpractical applications.4.2 Electrostatic FieldFigure 4.1 represents two parallel metal plates, A and B, charged to different potentials.If an electron that has a negative charge is placed between the plates, a force will act onthe electron tending to push it away from the negative plate B towards the positive plate, A.Similarly, a positive charge would be acted on by a force tending to move it toward thenegative plate. Any region such as that shown between the plates in Figure 4.1, in whichan electric charge experiences a force, is called an electrostatic field. The direction ofthe field is defined as that of the force acting on a positive charge placed in the field. InFigure 4.1, the direction of the force is from the positive plate to the negative plate.Such a field may be represented in magnitude and direction by lines of electric forcedrawn between the charged surfaces. The closeness of the lines is an indication of thefield strength. Whenever a voltage is established between two points, an electric field willalways exist. Figure 4.2(a) shows a typical field pattern for an isolated point charge, and w w w.ne w nespress.com
    • 54 Chapter 4 Figure 4.1: Electrostatic field Figure 4.2: (a) Isolated point charge (b) Adjacent charges of opposite polarityFigure 4.2(b) shows the field pattern for adjacent charges of opposite polarity. Electriclines of force (often called electric flux lines) are continuous and start and finish on pointcharges. Also, the lines cannot cross each other. When a charged body is placed closeto an uncharged body, an induced charge of opposite sign appears on the surface of theuncharged body. This is because lines of force from the charged body terminate on itssurface.The concept of field lines or lines of force is used to illustrate the properties of an electricfield. However, it should be remembered that they are only aids to the imagination.The force of attraction or repulsion between two electrically charged bodies isproportional to the magnitude of their charges and inversely proportional to the square ofthe distance separating them, q1q2 q1q2i.e., force ∝ or force ϭ k d2 d2w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 55where constant k Ϸ 9 ϫ 109 in air.This is known as Coulomb’s law.Hence, the force between two charged spheres in air with their centers 16 mm apart andeach carrying a charge of ϩ1.6 μC is given by: q1q2 (1.6 ϫ 10Ϫ6 )2force ϭ k ≈ (9 ϫ 109 ) d2 (16 ϫ 10Ϫ3 )2 ϭ 90 newtons4.3 Electric Field StrengthFigure 4.3 shows two parallel conducting plates separated from each other by air. Theyare connected to opposite terminals of a battery of voltage V volts.Therefore an electric field is in the space between the plates. If the plates are closetogether, the electric lines of force will be straight and parallel and equally spaced, exceptnear the edge where fringing will occur (see Figure 4.1). Over the area in which there isnegligible fringing, VElectric field strength, E ϭ volts/meter dwhere d is the distance between the plates. Electric field strength is also called potentialgradient. Figure 4.3: Two parallel conducting plates separated by air w w w.ne w nespress.com
    • 56 Chapter 44.4 CapacitanceStatic electric fields arise from electric charges, electric field lines beginning and endingon electric charges. Thus, the presence of the field indicates the presence of equal positiveand negative electric charges on the two plates of Figure 4.3. Let the charge be ϩQcoulombs on one plate and ϪQ coulombs on the other. The property of this pair of platesthat determines how much charge corresponds to a given voltage between the plates iscalled their capacitance: Qcapacitance C ϭ VThe unit of capacitance is the farad F (or more usually μF ϭ 10Ϫ6 F or pF ϭ 10Ϫ12 F),which is defined as the capacitance when a voltage of one volt appears across the plateswhen charged with one coulomb.4.5 CapacitorsEvery system of electrical conductors possesses capacitance. For example, there iscapacitance between the conductors of overhead transmission lines and also between thewires of a telephone cable. In these examples, the capacitance is undesirable but has to beaccepted, minimized or compensated for. There are other situations where capacitance isa desirable property.Devices specially constructed to possess capacitance are called capacitors (or condensers,as they used to be called). In its simplest form, a capacitor consists of two plates that areseparated by an insulating material known as a dielectric. A capacitor has the ability tostore a quantity of static electricity.The symbols for a fixed capacitor and a variable capacitor used in electrical circuitdiagrams are shown in Figure 4.4.The charge Q stored in a capacitor is given by:Q ϭ I ϫ t coulombswhere I is the current in amperes and t the time in seconds.w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 57 Figure 4.4: Symbols for a fixed capacitor and a variable capacitorExample 4.1(a) Determine the voltage across a 4 μF capacitor when charged with 5 mC.(b) Find the charge on a 50 pF capacitor when the voltage applied to it is 2 kV.Solution(a) C ϭ 4 μF ϭ 4 ϫ 10Ϫ6 F; Q ϭ 5 mC ϭ 5 ϫ 10Ϫ3 C Q Q 5 ϫ 10Ϫ3 5 ϫ 106 Since C ϭ then V ϭ ϭ Ϫ6 ϭ V C 4 ϫ 10 4 ϫ 103 5000 ϭ 4 Hence, voltage ϭ 250 V or 1.25 kV(b) C ϭ 50 pF ϭ 50 ϫ 10Ϫ12 F; V ϭ 2 kV ϭ 2000 V 5ϫ 2 Q ϭ CV ϭ 50 ϫ 10Ϫ12 ϫ 2000 ϭ 108 ϭ 0.1 ϫ 10Ϫ6 So, charge ϭ 0.1 μCExample 4.2A direct current of 4 A flows into a previously uncharged 20 μF capacitor for 3 ms.Determine the voltage between the plates.SolutionI ϭ 4 A; C ϭ 20 μF ϭ 20 ϫ 10Ϫ6 F;t ϭ 3 ms ϭ 3 ϫ 10Ϫ3s w w w.ne w nespress.com
    • 58 Chapter 4Q ϭ It ϭ 4 ϫ 3 ϫ 10Ϫ3 C Q 4 ϫ 3 ϫ 10Ϫ3 12 ϫ 106Vϭ ϭ ϭ ϭ 0.6 ϫ 103 C 20 ϫ 10Ϫ6 20 ϫ 103 ϭ 600 VSo, the voltage between the plates is 600 V.Example 4.3A 5 μF capacitor is charged so that the voltage between its plates is 800 V. Calculate howlong the capacitor can provide an average discharge current of 2 mA.SolutionC ϭ 5 μF ϭ 5 ϫ 10Ϫ6 F; V ϭ 800 V;I ϭ 2 mA ϭ 2 ϫ 10Ϫ3 AQ ϭ CV ϭ 5 ϫ 10Ϫ6 ϫ 800 ϭ 4 ϫ 10Ϫ3 C Q 4 ϫ 10Ϫ3Also, Q ϭ It. Thus, t ϭ ϭ ϭ 2s I 2 ϫ 10Ϫ3Therefore, the capacitor can provide an average discharge current of 2 mA for 2 s.4.6 Electric Flux DensityUnit flux is defined as emanating from a positive charge of 1 coulomb. Thuselectric flux Ψ is measured in coulombs, and for a charge of Q coulombs, the fluxΨ ϭ Q coulombs.Electric flux density D is the amount of flux passing through a defined area A that isperpendicular to the direction of the flux: Qelectric flux density, D ϭ coulombs/meter 2 AElectric flux density is also called charge density, σ.w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 594.7 PermittivityAt any point in an electric field, the electric field strength E maintains the electric flux andproduces a particular value of electric flux density D at that point. For a field establishedin vacuum (or for practical purposes in air), the ratio D/E is a constant ε0, i.e.,D ϭ ε0Ewhere ε0 is called the permittivity of free space or the free space constant. The value of ε0is 8.85 ϫ 10Ϫ12 F/m.When an insulating medium, such as mica, paper, plastic, or ceramic, is introduced intothe region of an electric field the ratio of D/E is modified:D ϭ ε 0 εrEwhere εr, the relative permittivity of the insulating material, indicates its insulating powercompared with that of vacuum: flux density in materialrelative permittivity εr ϭ flux density in vacuumHere, εr has no unit. Typical values of εr include: air, 1.00; polythene, 2.3; mica, 3–7;glass, 5–10; water, 80; ceramics, 6–1000.The product ε0εr is called the absolute permittivity, ε.ε ϭ ε 0 εrThe insulating medium separating charged surfaces is called a dielectric. Compared withconductors, dielectric materials have very high resistivities. Therefore, they are used toseparate conductors at different potentials, such as capacitor plates or electric power lines.Example 4.4Two parallel rectangular plates measuring 20 cm by 40 cm carry an electric charge of0.2 μC. Calculate the electric flux density. If the plates are spaced 5 mm apart and thevoltage between them is 0.25 kV determine the electric field strength. w w w.ne w nespress.com
    • 60 Chapter 4SolutionCharge Q ϭ 0.2 μC ϭ 0.2 ϫ 10Ϫ6 C;Area A ϭ 20 cm ϫ 40 cm ϭ 800 cm2 ϭ 800 ϫ 10Ϫ4 m2 Q 0.2 ϫ 10Ϫ6 0.2 ϫ 10 4Electric flux density D ϭ ϭ ϭ A 800 ϫ 10Ϫ4 800 ϫ 106 2000 ϭ ϫ 10Ϫ6 ϭ 2.5 μC/m 2 800Voltage V ϭ 0.25 kV ϭ 250 V; Plate spacing, d ϭ 5 mm ϭ 5 ϫ 10Ϫ3 m V 250Electric field strength E ϭ ϭ ϭ 50 kV/m d 5 ϫ 10Ϫ3Example 4.5The flux density between two plates separated by mica of relative permittivity 5 is2 μC/m2. Find the voltage gradient between the plates.SolutionFlux density D ϭ 2 μC/m2 ϭ 2 ϫ 10Ϫ6 C/m2;ε0 ϭ 8.85 ϫ 10Ϫ12 F/m; εr ϭ 5.D ϭ ε0 εr ,E Dhence, voltage gradient E ϭ ε 0 εr 2 ϫ 10Ϫ6 ϭ V/m 8.85 ϫ 10Ϫ12 ϫ 5 ϭ 45.2 kV/mExample 4.6Two parallel plates having a voltage of 200 V between them are spaced 0.8 mm apart.What is the electric field strength? Find also the flux density when the dielectric betweenthe plates is (a) air, and (b) polythene of relative permittivity 2.3.w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 61Solution V 200Electric field strength E ϭ ϭ ϭ 250 kV/m D 0.8 ϫ 10Ϫ3(a) For air: εr ϭ 1 D ϭ ε 0 εr . Hence, E Electric flux density D ϭ Eε0εr ϭ (250 ϫ 103 ϫ 8.85 ϫ 10Ϫ12 ϫ 1) C/m2 ϭ 2.213 μC/m2(b) For polythene, εr ϭ 2.3 Electric flux density D ϭ Eε0εr ϭ (250 ϫ 103 ϫ 8.85 ϫ 10Ϫ12 ϫ 2.3) C/m2 ϭ 5.089 μC/m24.8 The Parallel Plate CapacitorFor a parallel plate capacitor, as shown in Figure 4.5(a), experiments show thatcapacitance C is proportional to the area A of a plate, inversely proportional tothe plate spacing d (i.e., the dielectric thickness) and depends on the nature of thedielectric: ε 0 εr ACapacitance, C ϭ farads dwhere ε0 ϭ 8.85 ϫ 10Ϫ12 F/m (constant) εr ϭ relative permittivity A ϭ area of one of the plates, in m2, and d ϭ thickness of dielectric in mAnother method used to increase the capacitance is to interleave several plates as shownin Figure 4.5(b). Ten plates are shown, forming nine capacitors with a capacitance ninetimes that of one pair of plates. w w w.ne w nespress.com
    • 62 Chapter 4 Figure 4.5: Parallel plate capacitorIf such an arrangement has n plates, then capacitance C ϰ (n Ϫ 1). ε 0 εr A(n Ϫ 1)Thus, capacitance C ϭ farads dExample 4.7(a) A ceramic capacitor has an effective plate area of 4 cm2 separated by 0.1 mm ofceramic of relative permittivity 100. Calculate the capacitance of the capacitor inpicofarads. (b) If the capacitor in part (a) is given a charge of 1.2 μC, what will be thevoltage between the plates?Solution(a) Area A ϭ 4 cm2 ϭ 4 ϫ 10Ϫ4 m2; d ϭ 0.1 mm ϭ 0.1 ϫ 10Ϫ3 m; ε0 ϭ 8.85 ϫ 10Ϫ12 F/m; εr ϭ 100 ε 0 εr Capacitance C ϭ farads d 8.85 ϫ 10Ϫ12 ϫ 100 ϫ 4 ϫ 10Ϫ4 ϭ F 0.1 ϫ 10Ϫ3 8.85 ϫ 4 8.85 ϫ 4 ϫ 1012 ϭ Fϭ pF 1010 1010 ϭ 3540 pFw ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 63 Q 1.2 ϫ 10Ϫ6(b) Q ϭ CV thus, V ϭ ϭ V ϭ 339 V C 3540 ϫ 10Ϫ12Example 4.8A waxed paper capacitor has two parallel plates, each of effective area 800 cm2. If thecapacitance of the capacitor is 4425 pF, determine the effective thickness of the paper ifits relative permittivity is 2.5.SolutionA ϭ 800 cm2 ϭ 800 ϫ 10Ϫ4 m2 ϭ 0.08 m2;C ϭ 4425 pF ϭ 4425 ϫ 10Ϫ12 F;ε0 ϭ 8.85 ϫ 10Ϫ12 F/m; εr ϭ 2.5 ε 0 εr A ε ε ASince C ϭ then d ϭ 0 r d C 8.85 ϫ 10 Ϫ12 ϫ 2.5 ϫ 0.08Hence, d ϭ ϭ 0.0004 m 4425 ϫ 10Ϫ12So, the thickness of the paper is 0.4 mm.Example 4.9A parallel plate capacitor has nineteen interleaved plates each 75 mm ϫ 75 mm separatedby mica sheets 0.2 mm thick. Assuming the relative permittivity of the mica is 5, calculatethe capacitance of the capacitor.Solutionn ϭ 19; n Ϫ 1 ϭ 18;A ϭ 75 ϫ 75 ϭ 5625 mm2 ϭ 5625 ϫ 10Ϫ6 m2;εr ϭ 5; ε0 ϭ 8.85 ϫ 10Ϫ12 F/m;d ϭ 0.2 mm ϭ 0.2 ϫ 10Ϫ3 m w w w.ne w nespress.com
    • 64 Chapter 4 ε 0 εr A(n Ϫ 1)Capacitance C ϭ d 8.85 ϫ 10Ϫ12 ϫ 5 ϫ 5625 ϫ 10Ϫ6 ϫ 18 ϭ F 0.2 ϫ 10Ϫ3 ϭ 0.0224 μF or 22.4 nF4.9 Capacitors Connected in Parallel and Series4.9.1 Capacitors Connected in ParallelFigure 4.6 shows three capacitors, C1, C2 and C3, connected in parallel with a supplyvoltage V applied across the arrangement.When the charging current I reaches point A it divides, some flowing into C1, someflowing into C2 and some into C3. Therefore, the total charge QT (ϭI ϫ t) is dividedbetween the three capacitors. The capacitors each store a charge and these are shown asQ1, Q2 and Q3, respectively. Hence:QT ϭ Q1 ϩ Q2 ϩ Q3 Figure 4.6: Three capacitors connected in parallelw ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 65But QT ϭ CV, Q1 ϭ C1V, Q2 ϭ C2V and Q3 ϭ C3V. Therefore, CV ϭ C1V ϩ C2V ϩ C3Vwhere C is the total equivalent circuit capacitance,i.e., C ϭ C1 ϩ C2 ϩ C3It follows that for n parallel-connected capacitors,C ϭ C1 ϩ C2 ϩ C3 ϩ ϩ Cnthat is, the equivalent capacitance of a group of parallel-connected capacitors is the sumof the capacitances of the individual capacitors. (Note that this formula is similar to thatused for resistors connected in series.)4.9.2 Capacitors Connected in SeriesFigure 4.7 shows three capacitors, C1, C2 and C3, connected in series across a supplyvoltage V. Let the voltage across the individual capacitors be V1, V2, and V3, respectively,as shown.Let the charge on plate “a” of capacitor C1 be ϩQ coulombs. This induces an equal butopposite charge of ϪQ coulombs on plate “b”. The conductor between plates “b” and “c”is electrically isolated from the rest of the circuit so that an equal but opposite chargeof ϩQ coulombs must appear on plate “c”, which, in turn, induces an equal and oppositecharge of ϪQ coulombs on plate “d”, and so on.When capacitors are connected in series the charge on each is the same. Figure 4.7: Three capacitors connected in series w w w.ne w nespress.com
    • 66 Chapter 4In a series circuit: V ϭ V1 ϩ V2 ϩ V3 Q Q Q Q QSince V ϭ then ϭ ϩ ϩ C C C1 C2 C3where C is the total equivalent circuit capacitance, 1 1 1 1i.e., ϭ ϩ ϩ C C1 C2 C3It follows that for n series-connected capacitors:1 1 1 1 1 ϭ ϩ ϩ ϩ ϩC C1 C2 C3 CnThat is, for series-connected capacitors, the reciprocal of the equivalent capacitance isequal to the sum of the reciprocals of the individual capacitances. (Note that this formulais similar to that used for resistors connected in parallel.)For the special case of two capacitors in series:1 1 1 C ϩ C1 ϭ ϩ ϭ 2C C1 C2 C1C2 ⎛ ⎞Hence, C ϭ C1C2 ⎜ i.e., product ⎟ ⎜ ⎟ C1 ϩ C2 ⎜ ⎝ sum ⎟ ⎠Example 4.10Calculate the equivalent capacitance of two capacitors of 6 μF and 4 μF connected(a) in parallel and (b) in series.Solution(a) In parallel, equivalent capacitance C ϭ C1 ϩ C2 ϭ 6 μF ϩ 4 μF ϭ 10 μF(b) In series, equivalent capacitance C is given by: C1C2 Cϭ C1 ϩ C2w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 67 This formula is used for the special case of two capacitors in series. 6ϫ4 24 Thus, C ϭ ϭ ϭ 2.4 μF 6ϩ4 10Example 4.11What capacitance must be connected in series with a 30 μF capacitor for the equivalentcapacitance to be 12 μF?SolutionLet C ϭ 12 μF (the equivalent capacitance), C1 ϭ 30 μF and C2 be the unknowncapacitance. 1 1 1For two capacitors in series ϭ ϩ C C1 C2 1 1 1 C ϪCHence, ϭ Ϫ ϭ 1 C2 C C1 CC1 CC1 12 ϫ 30 and C2 ϭ ϭ C1 Ϫ C 30 Ϫ 12 360 ϭ ϭ 20 μF 18Example 4.12Capacitances of 1 μF, 3 μF, 5 μF and 6 μF are connected in parallel to a direct voltagesupply of 100 V. Determine (a) the equivalent circuit capacitance, (b) the total charge and(c) the charge on each capacitor.Solution(a) The equivalent capacitance C for four capacitors in parallel is given by: C ϭ C1 ϩ C2 ϩ C3 ϩ C4 i.e., C ϭ 1 ϩ 3 ϩ 5 ϩ 6 ϭ 15 μF(b) Total charge QT ϭ CV where C is the equivalent circuit capacitance i.e., QT ϭ 15 ϫ 10Ϫ6 ϫ 100 ϭ 1.5 ϫ 10Ϫ3 C ϭ 1.5 mC w w w.ne w nespress.com
    • 68 Chapter 4(c) The charge on the 1 μF capacitor Q1 ϭ C1V ϭ 1 ϫ 10Ϫ6 ϫ 100 ϭ 0.1 mC The charge on the 3 μF capacitor Q2 ϭ C2V ϭ 3 ϫ 10Ϫ6 ϫ 100 ϭ 0.3 mC The charge on the 5 μF capacitor Q3 ϭ C3V ϭ 5 ϫ 10Ϫ6 ϫ 100 ϭ 0.5 mC The charge on the 6 μF capacitor Q4 ϭ C4V ϭ 6 ϫ 10Ϫ6 ϫ 100 ϭ 0.6 mC [Check: In a parallel circuit: QT ϭ Q1 ϩ Q2 ϩ Q3 ϩ Q4 Q1 ϩ Q2 ϩ Q3 ϩ Q4 ϭ 0.1 ϩ 0.3 ϩ 0.5 ϩ 0.6 ϭ 1.5 mC ϭ QT]Example 4.13Capacitances of 3 μF, 6 μF and 12 μF are connected in series across a 350 V supply.Calculate (a) the equivalent circuit capacitance, (b) the charge on each capacitor and(c) the voltage across each capacitor.SolutionThe circuit diagram is shown in Figure 4.8. Figure 4.8: Circuit diagram for Example 4.13w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 69(a) The equivalent circuit capacitance C for three capacitors in series is given by: 1 1 1 1 ϭ ϩ ϩ C C1 C2 C3 1 1 1 1 4 ϩ 2 ϩ1 i.e., ϭ ϩ ϩ ϭ ϭ C 3 6 12 12 So the equivalent circuit capacitance 12 5 Cϭ ϭ 1 μF 7 7(b) Total charge QT ϭ CV, hence, 12 QT ϭ ϫ 10Ϫ6 ϫ 350 ϭ 600 μC or 0.6 mC 7 Since the capacitors are connected in series, 0.6 mC is the charge on each of them.(c) The voltage across the 3 μF capacitor, Q 0.6 ϫ 10Ϫ3 V1 ϭ ϭ C1 3 ϫ 10Ϫ6 ϭ 200 V The voltage across the 6 μF capacitor, Q 0.6 ϫ 10Ϫ3 V2 ϭ ϭ C2 6 ϫ 10Ϫ6 ϭ 100 V The voltage across the 12 μF capacitor, Q 0.6 ϫ 10Ϫ3 V3 ϭ ϭ C3 12 ϫ 10Ϫ6 ϭ 50 V w w w.ne w nespress.com
    • 70 Chapter 4 [Check: In a series circuit V ϭ V1 ϩ V2 ϩ V3 V1 ϩ V2 ϩ V3 ϭ 200 ϩ 100 ϩ 50 ϭ 350 V ϭ supply voltage.]In practice, capacitors are rarely connected in series unless they are of the samecapacitance. The reason for this can be seen from the above problem where the lowestvalued capacitor (i.e., 3 μF) has the highest voltage across it (i.e., 200 V) which meansthat if all the capacitors have an identical construction they must all be rated at the highestvoltage.4.10 Dielectric StrengthThe maximum amount of field strength that a dielectric can withstand is called thedielectric strength of the material. VmDielectric strength, Em ϭ dExample 4.14A capacitor is to be constructed so that its capacitance is 0.2 μF and to take a voltage of1.25 kV across its terminals. The dielectric is to be mica which, after allowing a safetyfactor of 2, has a dielectric strength of 50 MV/m. Find (a) the thickness of the micaneeded, and (b) the area of a plate assuming a two-plate construction. (Assume εr formica to be 6.)Solution V V(a) Dielectric strength, E ϭ , i.e., d ϭ d E 1.25 ϫ 103 ϭ m 50 ϫ 106 ϭ 0.025 mmw ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 71 ε 0 εr A(b) Capacitance, C ϭ d Cd hence, area Aϭ ε 0 εr 0.2 ϫ 10Ϫ6 ϫ 0.025 ϫ 10Ϫ3 2 ϭ m 8.85 ϫ 10Ϫ12 ϫ 6 ϭ 0.09416 m2 ϭ 941.6 cm24.11 Energy StoredThe energy, W, stored by a capacitor is given by: 1Wϭ CV 2 joules 2Example 4.15(a) Determine the energy stored in a 3 μF capacitor when charged to 400 V. (b) Find also the average power developed if this energy is dissipated in a time of 10 μs.Solution 1(a) Energy stored W ϭ CV 2 joules 2 1 ϭ ϫ 3 ϫ 10Ϫ6 ϫ 4002 2 3 ϭ ϫ 16 ϫ 10Ϫ2 2 ϭ 0.24 J Energy 0.24(b) Power ϭ ϭ W ϭ 24 kW time 10 ϫ 10Ϫ6Example 4.16A 12 μF capacitor is required to store 4 J of energy. Find the voltage to which thecapacitor must be charged. w w w.ne w nespress.com
    • 72 Chapter 4Solution 1 2WEnergy stored W ϭ CV 2 hence, V 2 ϭ 2 C ⎛ 2W ⎞ ⎛ 6⎞and V ϭ √ ⎜ ⎜ ⎟ ϭ √ ⎛ 2 ϫ 4 ⎞ ϭ √ ⎜ 2 ϫ 10 ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ C ⎠⎟ ⎜ Ϫ6 ⎟ ⎝ 12 ϫ 10 ⎠ ⎝ 3 ⎟ ⎜ ⎟ ⎠ ϭ 816.5 VExample 4.17A capacitor is charged with 10 mC. If the energy stored is 1.2 J find (a) the voltage and(b) the capacitance.Solution 1 QEnergy stored W ϭ CV 2 and C ϭ 2 V 1 ⎛Q ⎞ 2 1Hence, W ϭ ⎜ ⎟ V ϭ QV ⎜ ⎟ ⎜V ⎟ 2⎝ ⎠ 2 2Wfrom which Vϭ QQ ϭ 10 mC ϭ 10 ϫ 10Ϫ3C and W ϭ 1.2 J 2W 2 ϫ 1.2(a) Voltage V ϭ ϭ ϭ 0.24 kV or 240 V Q 10 ϫ 10Ϫ3 Q 10 ϫ 10Ϫ3(b) Capacitance C ϭ ϭ F V 240 10 ϫ 106 ϭ μF ϭ 41.67 μF 240 ϫ 1034.12 Practical Types of CapacitorsPractical types of capacitors are characterized by the material used for their dielectric.The main types include: variable air, mica, paper, ceramic, plastic, titanium oxide, andelectrolytic.w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 73 Figure 4.9: End view of variable air capacitor Figure 4.10: Older construction mica capacitor1. Variable air capacitors. These usually consist of two sets of metal plates (such as aluminum) one fixed, the other variable. The set of moving plates rotate on a spindle as shown by the end view of Figure 4.9. As the moving plates are rotated through half a revolution, the meshing, and therefore the capacitance, varies from a minimum to a maximum value. Variable air capacitors are used in radio and electronic circuits where very low losses are required, or where a variable capacitance is needed. The maximum value of such capacitors is between 500 pF and 1000 pF.2. Mica capacitors. A typical older type construction is shown in Figure 4.10. Usually the whole capacitor is impregnated with wax and placed in a bakelite case. Mica is easily obtained in thin sheets and is a good insulator. However, mica is expensive and is not used in capacitors above about 0.2 μF. A modified form of mica capacitor is the silvered mica type. The mica is coated on both sides with a thin layer of silver, which forms the plates. Capacitance is stable and less likely to change with age. Such capacitors have a constant capacitance with change of temperature, a high working voltage rating and a long service life and are used in high frequency circuits with fixed values of capacitance up to about 1000 pF. w w w.ne w nespress.com
    • 74 Chapter 4 Figure 4.11: Typical paper capacitor Figure 4.12: Cross-section of tube of ceramic material 3. Paper capacitors. A typical paper capacitor is shown in Figure 4.11where the length of the roll corresponds to the capacitance required. The whole is usually impregnated with oil or wax to exclude moisture, and then placed in a plastic or aluminum container for protection. Paper capacitors are made in various working voltages up to about 150 kV and are used where loss is not very important. The maximum value of this type of capacitor is between 500 pF and 10 μF. Disadvantages of paper capacitors include variation in capacitance with temperature change and a shorter service life than most other types of capacitor. 4. Ceramic capacitors. These are made in various forms, each type of construction depending on the value of capacitance required. For high values, a tube of ceramic material is used as shown in the cross-section of Figure 4.12. For smaller values the cup construction is used as shown in Figure 4.13, and for still smaller values the disc construction shown in Figure 4.14 is used. Certain ceramic materials have a very high permittivity and this enables capacitors of high capacitance to be made which are of small physical size with a high working voltage rating. Ceramic capacitors are available in the range 1 pF tow ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 75 Figure 4.13: Cup construction Figure 4.14: Disc construction 0.1 μF and may be used in high frequency electronic circuits subject to a wide range of temperatures.5. Plastic capacitors. Some plastic materials such as polystyrene and Teflon can be used as dielectrics. Construction is similar to the paper capacitor but using a plastic film instead of paper. Plastic capacitors operate well under conditions of high temperature, provide a precise value of capacitance, a very long service life and high reliability.6. Titanium oxide capacitors have a very high capacitance with a small physical size when used at a low temperature.7. Electrolytic capacitors. Construction is similar to the paper capacitor with aluminum foil used for the plates and with a thick absorbent material, such as paper, impregnated with an electrolyte (ammonium borate), separating the plates. The finished capacitor is usually assembled in an aluminum container and hermetically sealed. Its operation depends on the formation of a thin aluminum oxide layer on the positive plate by electrolytic action when a suitable direct potential is maintained between the plates. This oxide layer is w w w.ne w nespress.com
    • 76 Chapter 4 very thin and forms the dielectric. (The absorbent paper between the plates is a conductor and does not act as a dielectric.) Such capacitors must always be used on DC and must be connected with the correct polarity; if this is not done the capacitor will be destroyed since the oxide layer will be destroyed. Electrolytic capacitors are manufactured with working voltage from 6 V to 600 V, although accuracy is generally not very high. These capacitors possess a much larger capacitance than other types of capacitors of similar dimensions due to the oxide film being only a few microns thick. The fact that they can be used only on DC supplies limit their usefulness.4.13 InductanceInductance is the name given to the property of a circuit whereby there is an e.m.f.induced into the circuit by the change of flux linkages produced by a current change.When the e.m.f. is induced in the same circuit as that in which the current is changing,the property is called self-inductance, L. When the e.m.f. is induced in a circuit by achange of flux due to current changing in an adjacent circuit, the property is called mutualinductance, M. The unit of inductance is the henry, H.A circuit has an inductance of one henry when an e.m.f. of one volt is induced in it by acurrent changing at the rate of one ampere per second.Induced e.m.f. in a coil of N turns, dΦE ϭ ϪN volts dtwhere dΦ is the change in flux in webers, and dt is the time taken for the flux to change inseconds (i.e., dΦ/dt is the rate of change of flux).Induced e.m.f. in a coil of inductance L henrys, dIE ϭ ϪL volts dtwhere dI is the change in current in amperes and dt is the time taken for the current tochange in seconds (i.e., dI/dt is the rate of change of current). The minus signs in each ofthe above two equations remind us of its direction (given by Lenz’s law).w ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 77Example 4.18Determine the e.m.f. induced in a coil of 200 turns when there is a change of flux of25 mWb linking with it in 50 ms.Solution dΦ ⎛ 25 ϫ 10Ϫ3 ⎞ ⎟Induced e.m.f. E ϭ ϪN ϭ Ϫ(200) ⎜ ⎜ ⎟ dt ⎜ 50 ϫ 10Ϫ3 ⎟ ⎝ ⎟ ⎠ ϭ؊100 voltsExample 4.19A flux of 400 μWb passing through a 150-turn coil is reversed in 40 ms. Find the averagee.m.f. induced.SolutionSince the flux reverses, the flux changes from ϩ400 μWb to Ϫ400 μWb, a total change offlux of 800 μWb dΦ ⎛ 800 ϫ 10Ϫ6 ⎞⎟Induced e.m.f. E ϭ ϪN ϭ Ϫ(150) ⎜⎜ ⎟ dt ⎝ 40 ϫ 10Ϫ3 ⎟ ⎜ ⎟ ⎠ ⎛ 150 ϫ 800 ϫ 103 ⎞⎟ ϭ Ϫ⎜ ⎜ ⎟ ⎟ ⎜ ⎝ 40 ϫ 106 ⎟ ⎠Hence the average e.m.f. induced E ‫ 3؊ ؍‬VExample 4.20Calculate the e.m.f. induced in a coil of inductance 12 H by a current changing at the rateof 4 A/s.Solution dIInduced e.m.f. E ϭ ϪL ϭ Ϫ(12)(4) ϭ ؊48 volts dtExample 4.21An e.m.f. of 1.5 kV is induced in a coil when a current of 4 A collapses uniformly to zeroin 8 ms. Determine the inductance of the coil. w w w.ne w nespress.com
    • 78 Chapter 4SolutionChange in current, dI ϭ (4 Ϫ 0) ϭ 4 A; dt ϭ 8 ms ϭ 8 ϫ 10Ϫ3 s;dI 4 4000 ϭ Ϫ3 ϭ ϭ 500 A/s;dt 8 ϫ 10 8E ϭ 1.5 kV ϭ 1500 V ⎛ dI ⎞Since |E | ϭ L ⎜ ⎟ ⎜ ⎟ ⎜ dt ⎟ ⎝ ⎠ |E | 1500inductance, L ϭ ϭ ϭ 3H (dI /dt ) 500(Note that |E | means the “magnitude of E,” which disregards the minus sign.)4.14 InductorsA component called an inductor is used when the property of inductance is required in acircuit. The basic form of an inductor is simply a coil of wire.Factors which affect the inductance of an inductor include: (i) the number of turns of wire—the more turns, the higher the inductance. (ii) the cross-sectional area of the coil of wire—the greater the cross-sectional area the higher the inductance. (iii) the presence of a magnetic core—when the coil is wound on an iron core, the same current sets up a more concentrated magnetic field and the inductance is increased. (iv) the way the turns are arranged—a short thick coil of wire has a higher inductance than a long thin one.Two examples of practical inductors are shown in Figure 4.15, and the standard electricalcircuit diagram symbols for air-cored and iron-cored inductors are shown in Figure 4.16.An iron-cored inductor is often called a choke since, when used in AC circuits, it has achoking effect, limiting the current flowing through it. Inductance is often undesirablein a circuit. To reduce inductance to a minimum, the wire may be bent back on itself, asshown in Figure 4.17, so that the magnetizing effect of one conductor is neutralized byw ww. n e w n e s p r e s s .c o m
    • Capacitors and Inductors 79 Figure 4.15: Two examples of practical inductorsFigure 4.16: Standard electrical symbols for air-cored and iron-cored inductors Figure 4.17: Wire coiled around an insulator to form an inductor w w w.ne w nespress.com
    • 80 Chapter 4that of the adjacent conductor. The wire may be coiled around an insulator, as shown,without increasing the inductance. Standard resistors may be non-inductively wound inthis manner.4.15 Energy StoredAn inductor possesses an ability to store energy. The energy stored, W, in the magneticfield of an inductor is given by: 1 2Wϭ LI joules 2Example 4.22An 8 H inductor has a current of 3 A flowing through it. How much energy is stored in themagnetic field of the inductor?Solution 1 2 1Energy stored, W ϭ LI ϭ (8)(3)2 ϭ 36 joules 2 2w ww. n e w n e s p r e s s .c o m
    • CHAPTE R 5 DC Circuit Theory John Bird5.1 IntroductionThe laws that determine the currents and voltage drops in DC networks are: (a) Ohm’slaw, (b) the laws for resistors in series and in parallel, and (c) Kirchhoff’s laws (seeSection 5.2). In addition, there are a number of circuit theorems that have been developedfor solving problems in electrical networks. These include: (i) the superposition theorem (see Section 5.3), (ii) Thévenin’s theorem (see Section 5.5), (iii) Norton’s theorem (see Section 5.7), and (iv) the maximum power transfer theorem (see Section 5.9).5.2 Kirchhoff’s LawsKirchhoff’s laws state: (a) Current Law. At any junction in an electric circuit the total current flowing towards that junction is equal to the total current flowing away from the junction, i.e., ΣI ϭ 0. Thus, referring to Figure 5.1: I1 ϩ I2 ϭ I3 ϩ I4 ϩ I5 or, I1 ϩ I2 Ϫ I3 Ϫ I4 Ϫ I5 ϭ 0 w w w.ne w nespress.com
    • 82 Chapter 5 Figure 5.1: Junction showing Kirchhoff’s current law Figure 5.2: Loop showing Kirchhoff’s voltage law (b) Voltage Law. In any closed loop in a network, the algebraic sum of the voltage drops (i.e., products of current and resistance) taken around the loop is equal to the resultant e.m.f. acting in that loop. Thus, referring to Figure 5.2: E1 Ϫ E2 ϭ IR1 ϩ IR2 ϩ IR3 (Note that if current flows away from the positive terminal of a source, that source is considered by convention to be positive. Thus, moving anticlockwise around the loop of Figure 5.2, E1 is positive and E2 is negative.)Example 5.1(a) Find the unknown currents marked in Figure 5.3(a). (b) Determine the value of e.m.f. E in Figure 5.3(b).Solution(a) Applying Kirchhoff’s current law: For junction B: 50 ϭ 20 ϩ I1 . I1 ‫ 03 ؍‬A For junction C: 20 ϩ 15 ϭ I2 . I2 ‫ 53 ؍‬Aw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 83 Figure 5.3: Figures for Example 5.1 For junction D: I1 ϭ I3 ϩ 120 i.e., 30 ϭ I3 ϩ 120. I3 ‫ 09؊ ؍‬A (i.e., in the opposite direction to that shown in Figure 5.3(a)) For junction E: I4 ϩ I3 ϭ 15 i.e., I4 ϭ 15 Ϫ (Ϫ90). I4 ‫ 501 ؍‬A For junction F: 120 ϭ I5 ϩ 40. I5 ‫ 08 ؍‬A(b) Applying Kirchhoff’s voltage law and moving clockwise around the loop of Figure 5.3 (b) starting at point A: 3 ϩ 6ϩE Ϫ 4 ϭ (I)(2) ϩ (I)(2.5) ϩ (I)(1.5) ϩ (I)(1) ϭ I(2 ϩ 2.5 ϩ 1.5 ϩ 1) i.e., 5 ϩ E ϭ 2(7), since I ϭ 2 A E ‫ 9 ؍ 5 ؊ 41 ؍‬VExample 5.2Use Kirchhoff’s laws to determine the currents flowing in each branch of the networkshown in Figure 5.4.SolutionProcedure1. Use Kirchhoff’s current law and label current directions on the original circuit diagram. The directions chosen are arbitrary, but it is usual, as a starting point, to w w w.ne w nespress.com
    • 84 Chapter 5 Figure 5.4: Network for Example 5.2 Figure 5.5: Labeling current directions assume that current flows from the positive terminals of the batteries. This is shown in Figure 5.5 where the three branch currents are expressed in terms of I1 and I2 only, since the current through R is I1 ϩ I2.2. Divide the circuit into two loops and apply Kirchhoff’s voltage law to each. From loop one of Figure 5.5, and moving in a clockwise direction as indicated (the direction chosen does not matter), gives: E1 ϭ I1r1 ϩ (I1 ϩ I 2 ) R, i.e., 4 ϭ 2 I1 ϩ 4 (I1 ϩ I 2 ), i.e., 6 I1 ϩ 4 I 2 ϭ 4 (5.1) From loop 2 of Figure 5.5, and moving in an anticlockwise direction as indicated (once again, the choice of direction does not matter; it does not have to be in the same direction as that chosen for the first loop), gives: E2 ϭ I 2 r2 ϩ (I1 ϩ I 2 ) R, i.e., 2 ϭ I 2 ϩ 4 (I1 ϩ I 2 ), i.e., 4 I1 ϩ 5I 2 ϭ 2 (5.2)w ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 85 Figure 5.6: Possible third loop3. Solve equations (1) and (2) for I1 and I2. 2 ϫ (1) gives: 12 I1 ϩ 8 I 2 ϭ 8 (5.3) 3 ϫ (2) gives: 12 I1 ϩ 15I 2 ϭ 6 (5.4) 2 (3) Ϫ (4) gives: Ϫ7I2 ϭ 2 hence, I 2 ϭ Ϫ ϭ ؊0.286 A 7 (i.e., I2 is flowing in the opposite direction to that shown in Figure 5.5.) From (1) 6I1 ϩ 4 (Ϫ0.286) ϭ 4 6 I1 ϭ 4 ϩ 1.144 5.144 Hence, I1 ϭ ϭ 0.857 A 6 Current flowing through resistance R is, I1 ϩ I 2 ϭ 0.857 ϩ (Ϫ0.286) ϭ 0.571 A Note that a third loop is possible, as shown in Figure 5.6, giving a third equation that can be used as a check: E1 Ϫ E2 ϭ I1r1 Ϫ I2r2 4 Ϫ 2 ϭ 2I1 Ϫ I2 2 ϭ 2I1 Ϫ I2 [Check: 2I1 Ϫ I2 ϭ 2(0.857) Ϫ (Ϫ0.286) ϭ 2] w w w.ne w nespress.com
    • 86 Chapter 5Example 5.3Determine, using Kirchhoff’s laws, each branch current for the network shown inFigure 5.7.Solution1. Currents and their directions are shown labeled in Figure 5.8 following Kirchhoff’s current law. It is usual, although not essential, to follow conventional current flow with current flowing from the positive terminal of the source.2. The network is divided into two loops as shown in Figure 5.8. Applying Kirchhoff’s voltage law gives: For loop one: E1 ϩ E2 ϭ I1 R1 ϩ I 2 R2 i.e., 16 ϭ 0.5I1 ϩ 2 I 2 (5.5) For loop two: E2 ϭ I2R2 Ϫ (I1 Ϫ I2) R3 Figure 5.7: Network for Example 5.3 Figure 5.8: Labeling current directionsw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 87 Note that since loop two is in the opposite direction to current (I1 Ϫ I2), the voltage drop across R3 (i.e., (I1 Ϫ I2) (R3)) is by convention negative. Thus, 12 ϭ 2 I 2 Ϫ 5(I1 ϩ I 2 ) i.e., 12 ϭ Ϫ5I1 ϩ 7 I 2 (5.6)3. Solving equations (5.1) and (5.2) to find I1 and I2: 10 ϫ (1) gives 160 ϭ 5I1 ϩ 20 I 2 172 (5.6) ϩ (5.7) gives 172 ϭ 27 I 2 hence, I 2 ϭ ϭ 6.37 A (5.7) 27 From (1): 16 ϭ 0.5I1 ϩ 2(6.37) 16 Ϫ 2(6.37) I1 ϭ ϭ 6.52 A 0.5 Current flowing in R3 ϭ I1 Ϫ I2 ϭ 6.52 Ϫ 6.3 ϭ 0.15 AExample 5.4For the bridge network shown in Figure 5.9 determine the currents in each of theresistors.SolutionLet the current in the 2 Ω resistor be I1; then by Kirchhoff’s current law, the current in the14 Ω resistor is (I Ϫ I1). Let the current in the 32 Ω resistor be I2 as shown in Figure 5.10.Then the current in the 11 Ω resistor is (I Ϫ I2) and that in the 3 Ω resistor is (I Ϫ I1 ϩ I2). Figure 5.9: Bridge network for Example 5.4 w w w.ne w nespress.com
    • 88 Chapter 5Applying Kirchhoff’s voltage law to loop one and moving in a clockwise direction asshown in Figure 5.10 gives: 54 ϭ 2 I1 ϩ 11 (I1 Ϫ I 2 ) i.e., 13I1 Ϫ 11I 2 ϭ 54 (5.8)Applying Kirchhoff’s voltage law to loop two and moving in an anticlockwise directionas shown in Figure 5.10 gives:0 ϭ 2I1 ϩ 32I2 Ϫ 14(I Ϫ I1)However, I ϭ 8AHence, 0 ϭ 2 I1 ϩ 32 I 2 Ϫ 14(8 Ϫ I1 )i.e., 16 I1 ϩ 32 I 2 ϭ 112 (5.9)Equations (5.8) and (5.9) are simultaneous equations with two unknowns, I1 and I2. 16 ϫ (1) gives: 208 I1 Ϫ 176 I 2 ϭ 864 (5.10) 13 ϫ (2) gives: 208 I1 ϩ 416 I 2 ϭ 1456 (5.11)(4) Ϫ (3) gives: 592I2 ϭ 592 I2 ϭ 1ASubstituting for I2 in (1) gives:13I2 Ϫ 11 ϭ 54 65 I1 ϭ ϭ 5A 13 Figure 5.10: Labeling directionsw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 89the current flowing in the 2 Ω resistor ϭ I1 ϭ 5 Athe current flowing in the 14 Ω resistor ϭ I Ϫ I1 ϭ 8 Ϫ 5 ϭ 3Athe current flowing in the 32 Ω resistor ϭ I2 ϭ 1 Athe current flowing in the 11 Ω resistor ϭ I1 Ϫ I2 ϭ 5 Ϫ 1 ϭ 4 A andthe current flowing in the 3 Ω resistor ϭ I Ϫ I1 ϩ I2 ϭ8Ϫ5ϩ1 ϭ 4A5.3 The superposition TheoremThe superposition theorem states:“In any network made up of linear resistances and containing more than one source ofe.m.f., the resultant current flowing in any branch is the algebraic sum of the currentsthat would flow in that branch if each source was considered separately, all other sourcesbeing replaced at that time by their respective internal resistances.”Example 5.5Figure 5.11 shows a circuit containing two sources of e.m.f., each with their internalresistance. Determine the current in each branch of the network by using thesuperposition theorem. Figure 5.11: Circuit for Example 5.5 w w w.ne w nespress.com
    • 90 Chapter 5SolutionProcedure:1. Redraw the original circuit with source E2 removed, being replaced by r2 only, as shown in Figure 5.12(a).2. Label the currents in each branch and their directions as shown in Figure 5.12(a) and determine their values. (Note that the choice of current directions depends on the battery polarity, which, by convention is taken as flowing from the positive battery terminal as shown.) R in parallel with r2 gives an equivalent resistance of: 4 ϫ1 ϭ 0.8 Ω 4 ϩ1 From the equivalent circuit of Figure 5.12(b), E1 4 I1 ϭ ϭ ϭ 1.429 A r1 ϩ 0.8 2 ϩ 0.8 From Figure 5.12(a), ⎛ 1 ⎞ ⎟ I ϭ 1 (1.429) ϭ 0.286 A I2 ϭ ⎜ ⎜ ⎟ ⎝ 4 ϩ1⎟ 1 5 ⎜ ⎠ And, ⎛ 4 ⎞ ⎟ I ϭ 4 (1.429) I3 ϭ ⎜ ⎜ ⎟ ⎝ 4 ϩ1⎟ 1 ⎜ ⎠ 5 ϭ 1.143A by current division Figure 5.12: (a) Redrawn circuit; (b) Equivalent circuitw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 913. Redraw the original circuit with source E1 removed, being replaced by r1 only, as shown in Figure 5.13(a).4. Label the currents in each branch and their directions as shown in Figure 5.13(a) and determine their values. r1 in parallel with R gives an equivalent resistance of: 2ϫ4 8 ϭ ϭ 1.333 Ω 2ϩ4 6 From the equivalent circuit of Figure 5.13(b) E2 2 I4 ϭ ϭ ϭ 0.857 A 1.333 ϩ r2 1.333 ϩ 1 From Figure 5.13(a) ⎛ 2 ⎞ ⎟ I ϭ 2 (0.857) ϭ 0.286 A I5 ϭ ⎜ ⎜2ϩ4⎟ 4 ⎜ ⎝ ⎟ ⎠ 6 ⎛ 4 ⎞ ⎟ I ϭ 4 (0.857) ϭ 0.571 A I6 ϭ ⎜ ⎜2ϩ4⎟ 4 ⎜ ⎝ ⎟ ⎠ 65. Superimpose Figure 5.13(a) on to Figure 5.12(a) as shown in Figure 5.14.6. Determine the algebraic sum of the currents flowing in each branch. Resultant current flowing through source 1, i.e., I1 Ϫ I6 ϭ 1.429 Ϫ 0.571 ϭ 0.858 A (discharging) Figure 5.13: (a) Redrawn circuit; (b) Equivalent circuit w w w.ne w nespress.com
    • 92 Chapter 5 Figure 5.14: Superimposed circuits Figure 5.15: Resultant currents and their directions Resultant current flowing through source 2, i.e., I4 Ϫ I3 ϭ 0.857 Ϫ 1.143 ϭ ؊0.286 A (charging) Resultant current flowing through resistor R, i.e., I2 ϩ I5 ϭ 0.286 ϩ 0.286 ϭ 0.572 AThe resultant currents with their directions are shown in Figure 5.15.Example 5.6For the circuit shown in Figure 5.16, find, using the superposition theorem, (a) the currentflowing in and the voltage across the 18 Ω resistor, (b) the current in the 8 V battery and(c) the current in the 3 V battery.w ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 93 Figure 5.16: Circuit for Example 5.6 Figure 5.17: (a) Redrawn circuit; (b) Equivalent circuitSolution1. Removing source E2 gives the circuit of Figure 5.17(a).2. The current directions are labeled as shown in Figure 5.17(a), I1 flowing from the positive terminal of E1. E1 8 From Figure 5.17(b), I1 ϭ ϭ ϭ 1.667 A 3 ϩ 1.8 4.8 ⎛ 18 ⎞ ⎟I From Figure 5.17(a), I 2 ϭ ⎜ ⎜ ⎟ ⎝ 2 ϩ 18 ⎟ 1 ⎜ ⎠ 18 ϭ (1.667) ϭ 1.500 A 5 20 ⎛ 2 ⎞ ⎟I I3 ϭ ⎜ ⎜ 2 ϩ 18 ⎟ 1 ⎜ and ⎟ ⎝ ⎠ 2 ϭ (1.667) ϭ 0.167 A 20 w w w.ne w nespress.com
    • 94 Chapter 53. Removing source E1 gives the circuit of Figure 5.18(a), (which is the same as Figure 5.18(b)).4. The current directions are labeled as shown in Figures 5.18(a) and 5.18(b), I4 flowing from the positive terminal of E2: E2 3 From Figure 5.18(c), I 4 ϭ ϭ 2 ϩ 2.571 4.571 ϭ 0.656 A ⎛ 18 ⎞ ⎟ I ϭ 18 (0.656) From Figure 5.18(b), I 5 ϭ ⎜ ⎜ ⎟ ⎝ 3 ϩ 18 ⎟ 4 ⎜ ⎠ 21 ϭ 0.562 A 5 ⎛ 3 ⎞ ⎟ I ϭ 3 (0.656) I6 ϭ ⎜ ⎜ ⎟ ⎝ 3 ϩ 18 ⎟ 4 ⎜ ⎠ 21 ϭ 0.094 A5. Superimposing Figure 5.18(a) on to Figure 5.17(a) gives the circuit in Figure 5.19. Figure 5.18: (a) Step 1; (b) Step 2; (c) Step 3 Figure 5.19: Result of superimposingw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 956. (a) Resultant current in the 18 Ω resistor: ϭ I3 Ϫ I6 ϭ 0.167 Ϫ 0.094 ϭ 0.073 A voltage across the 18 Ω resistor ϭ 0.073 ϫ 18 ϭ 1.314V (b) Resultant current in the 8V battery: ϭ I1 ϩ I5 ϭ 1.667 ϩ 0.562 ϭ 2.229 A (discharging) (c) Resultant current in the 3 V battery: ϭ I2 ϩ I4 ϭ 1.500 ϩ 0.656 ϭ 2.156 A (discharging)5.4 General DC Circuit TheoryThe following points involving DC circuit analysis need to be appreciated beforeproceeding with problems using Thévenin’s and Norton’s theorems: (i) The open-circuit voltage, E, across terminals AB in Figure 5.20 is equal to 10 V, since no current flows through the 2 Ω resistor; therefore, no voltage drop occurs. (ii) The open-circuit voltage, E, across terminals AB in Figure 5.21(a) is the same as the voltage across the 6 Ω resistor. The circuit may be redrawn as shown in Figure 5.21(b). ⎛ 6 ⎞ ⎟ (50) E ϭ⎜ ⎜ ⎟ ⎝6ϩ4⎟ ⎜ ⎠ by voltage division in a series circuit, i.e., E ‫03 ؍‬V Figure 5.20: Example circuit w w w.ne w nespress.com
    • 96 Chapter 5 Figure 5.21: (a) Example circuit; (b) Redrawn circuit Figure 5.22: (a) Example circuit; (b) Second example circuit (iii) For the circuit shown in Figure 5.22(a) representing a practical source supplying energy, V ϭ E Ϫ Ir, where E is the battery e.m.f., V is the battery terminal voltage and r is the internal resistance of the battery. For the circuit shown in Figure 5.22(b), V ϭ E Ϫ (ϪI)r, i.e., V ϭ E ϩ Ir. (iv) The resistance “looking-in” at terminals AB in Figure 5.23(a) is obtained by reducing the circuit in stages as shown in Figures 5.23(b) to (d). The equivalent resistance across AB is 7 Ω. (v) For the circuit shown in Figure 5.24(a), the 3 Ω resistor carries no current and the voltage across the 20 Ω resistor is 10 V. Redrawing the circuit gives Figure 5.24(b), from which, ⎛ 4 ⎞ ⎟ ϫ 10 ϭ 4 V E ϭ⎜ ⎜4ϩ6⎟ ⎜ ⎝ ⎟ ⎠ (vi) If the 10 V battery in Figure 5.24(a) is removed and replaced by a short-circuit, as shown in Figure 5.24(c), then the 20 Ω resistor may be removed. The reason for this is that a short-circuit has zero resistance, and 20 Ω in parallel with zero ohms gives an equivalent resistance of: (20 ϫ 0/ 20 ϩ 0), i.e., 0 Ω. The circuitw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 97 Figure 5.23: (a) Stage 1; (b) Stage 2; (c) Stage 3; (d) Stage 4, solution Figure 5.24: (a) Example circuit; (b) Step 1; (c) Step 2; (d) Step 3; (e) Step 4, equivalent resistance is then as shown in Figure 5.24(d), which is redrawn in Figure 5.24(e). From Figure 5.24(e), the equivalent resistance across AB, 6ϫ4 rϭ ϩ 3 ϭ 2.4 ϩ 3 ϭ 5.4 Ω 6ϫ4(vii) To find the voltage across AB in Figure 5.25: Since the 20 V supply is across the 5 Ω and 15 Ω resistors in series then, by voltage division, the voltage drop across AC, w w w.ne w nespress.com
    • 98 Chapter 5 Figure 5.25: Example circuit ⎛ 5 ⎞ ⎟ (20) ϭ 5 V VAC ϭ ⎜ ⎜ ⎟ ⎝ 5 ϩ 15 ⎟ ⎜ ⎠ ⎛ 12 ⎞ ⎟ (20) ϭ 16 V. Similarly, VCB ϭ ⎜ ⎜ ⎟ ⎝ 12 ϩ 3 ⎟ ⎜ ⎠ VC is at a potential of ϩ20 V. VA ϭ VC Ϫ VAC ϭ ϩ20 Ϫ 5 ϭ 15 V and VB ϭ VC Ϫ VBC ϭ ϩ20 Ϫ 16 ϭ 4 V. The voltage between AB is VA Ϫ VB ϭ 15 Ϫ 4 ϭ 11V and current would flow from A to B since A has a higher potential than B. (viii) In Figure 5.26(a), to find the equivalent resistance across AB the circuit may be redrawn as in Figures 5.26(b) and (c). From Figure 5.26(c), the equivalent resistance across AB, 5 ϫ 15 12 ϫ 3 ϭ ϩ ϭ 3.75 ϩ 2.4 ϭ 6.15 Ω 5 ϩ 15 12 ϩ 3 (ix) In the worked problems in Sections 5.5 and 5.7 following, it may be considered that Thévenin’s and Norton’s theorems have no obvious advantages compared with, say, Kirchhoff’s laws. However, these theorems can be used to analyze part of a circuit and in much more complicated networks the principle ofw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 99 Figure 5.26: (a) Example circuit; (b) Redrawn circuit; (c) Redrawn circuit replacing the supply by a constant voltage source in series with a resistance (or impedance) is very useful.5.5 Thévenin’s TheoremThévenin’s theorem states:The current in any branch of a network is that which would result if an e.m.f. equalto the voltage across a break made in the branch, were introduced into the branch,all other e.m.f.’s being removed and represented by the internal resistances of thesources.The procedure adopted when using Thévenin’s theorem is summarized below. Todetermine the current in any branch of an active network (i.e., one containing a sourceof e.m.f.): (i) remove the resistance R from that branch, (ii) determine the open-circuit voltage, E, across the break, (iii) remove each source of e.m.f. and replace them by their internal resistances and then determine the resistance, r, “looking-in” at the break, (iv) determine the value of the current from the equivalent E circuit shown in Figure 5.27, i.e., I ‫؍‬ R؉ r w w w.ne w nespress.com
    • 100 Chapter 5 Figure 5.27: Equivalent circuit Figure 5.28: Circuit for Example 5.7Example 5.7Use Thévenin’s theorem to find the current flowing in the 10 Ω resistor for the circuitshown in Figure 5.28(a).SolutionFollowing the above procedure:(i) The 10 Ω resistance is removed from the circuit as shown in Figure 5.28(b)(ii) There is no current flowing in the 5 Ω resistor and current I1 is given by: 10 10 I1 ϭ ϭ ϭ 1A R1 ϩ R2 2 ϩ8 Voltage across R2 ϭ I1R2 ϭ 1 ϫ 8 ϭ 8 V Voltage across AB, i.e., the open-circuit voltage across the break, E ϭ 8 Vw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 101 Figure 5.29: Network for Example 5.8(iii) Removing the source of e.m.f. gives the circuit of Figure 5.28(c). R1 R2 2ϫ8 Resistance, r ϭ R3 ϩ ϭ 5ϩ R1 ϩ R2 2 ϩ8 ϭ 5 ϩ 1.6 ϭ 6.6 Ω(iv) The equivalent Thévenin’s circuit is shown in Figure 5.28(d). E 8 8 Current I ϭ ϭ ϭ ϭ 0.482 A Rϩr 10 ϩ 6.6 16.6 The current flowing in the 10 Ω resistor of Figure 5.28(a) is 0.482 AExample 5.8For the network shown in Figure 5.29(a) determine the current in the 0.8 Ω resistor usingThévenin’s theorem.SolutionFollowing the procedure:(i) The 0.8 Ω resistor is removed from the circuit as shown in Figure 5.29(b). w w w.ne w nespress.com
    • 102 Chapter 5 12 12(ii) Current I1 ϭ ϭ ϭ 1.2 A 1 ϩ 5 ϩ 4 10 Voltage across 4 Ω resistor ϭ 4I1 ϭ (4) (1.2) ϭ 4.8 V. Voltage across AB, i.e., the open-circuit voltage across AB, E ϭ 4.8 V.(iii) Removing the source of e.m.f. gives the circuit shown in Figure 5.29(c). The equivalent circuit of Figure 5.29(c) is shown in Figure 5.29(d), from which, 4ϫ6 24 resistance r ϭ ϭ ϭ 2.4 Ω 4ϩ6 10(iv) The equivalent Thévenin’s circuit is shown in Figure 5.29(e), from which, E 4.8 4.8 current I ϭ ϭ ϭ rϩR 2.4 ϩ 0.8 3.2 I ϭ 1.5A ϭ current in the 0.8 Ω resistorExample 5.9Use Thévenin’s theorem to determine the current I flowing in the 4 Ω resistor shown inFigure 5.30(a). Find also the power dissipated in the 4 Ω resistor.SolutionFollowing the procedure: (i) The 4 Ω resistor is removed from the circuit as shown in Figure 5.30(b). Figure 5.30: Circuit for Example 5.9 showing steps for solutionw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 103 E1 Ϫ E2 4Ϫ2 2(ii) Current I1 ϭ ϭ ϭ A r1 ϩ r2 2 ϩ1 3 ⎛2⎞ 2 Voltage across AB, E ϭ E1 Ϫ I1r1 ϭ 4 Ϫ ⎜ ⎟ (2) ϭ 2 V ⎜ ⎟ ⎟ ⎜ ⎝3⎠ 3 Alternatively, voltage across AB, E ϭ E2 Ϫ I1r2 ⎛2⎞ 2 ϭ 2 Ϫ ⎜ ⎟ (1) ϭ 2 V ⎜ ⎟ ⎟ ⎜ ⎝3⎠ 3(iii) Removing the sources of e.m.f. gives the circuit shown in Figure 5.30(c), from which resistance, 2 ϫ1 2 rϭ ϭ Ω 2 ϩ1 3(iv) The equivalent Thévenin’s circuit is shown in Figure 5.30(d), from which, E 22 8/3 current, I ϭ ϭ 2 3 ϭ rϩR 3 ϩ 4 14 / 3 8 ϭ ϭ 0.571 A 14 ϭ current in the 4 Ω resistor Power dissipated in 4 Ω resistor, P ϭ I2R ϭ (0.571)2 (4) ϭ 1.304 WExample 5.10Use Thévenin’s theorem to determine the current flowing in the 3 Ω resistance of thenetwork shown in Figure 5.31(a). The voltage source has negligible internal resistance.Solution(Note the symbol for an ideal voltage source in Figure 5.31(a), which may be used as analternative to the battery symbol.) w w w.ne w nespress.com
    • 104 Chapter 5Following the procedure: (i) The 3 Ω resistance is removed from the circuit as shown in Figure 5.31(b). 2(ii) The 1 Ω resistance now carries no current. 3 ⎛ 10 ⎞ ⎟ (24) Voltage across 10 Ω resistor ϭ ⎜ ⎜ 10 ϩ 5 ⎟ ⎜ ⎝ ⎟ ⎠ ϭ 16 V Voltage across AB, E ϭ 16 V.(iii) Removing the source of e.m.f. and replacing it by its internal resistance means that the 20 Ω resistance is short-circuited as shown in Figure 5.31(c) since its internal resistance is zero. The 20 Ω resistance may thus be removed as shown in Figure 5.31(d). From Figure 5.31(d), resistance, 2 10 ϫ 5 r ϭ1 ϩ 3 10 ϩ 5 2 50 ϭ1 ϩ ϭ 5Ω 3 15 Figure 5.31: Circuit for Example 5.10 showing steps for solutionw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 105(iv) The equivalent Thévenin’s circuit is shown in Figure 5.31(e), from which E 16 16 current, I ϭ ϭ ϭ ϭ 2A rϩR 3ϩ5 8 ϭ current in the 3 Ω resistance.Example 5.11A Wheatstone Bridge network is shown in Figure 5.32(a). Calculate the current flowing inthe 32 Ω resistor, and its direction, using Thévenin’s theorem. Assume the source of e.m.f.to have negligible resistance.SolutionFollowing the procedure: (i) The 32 Ω resistor is removed from the circuit as shown in Figure 5.32(b).(ii) The voltage between A and C, ⎛ R ⎞ ⎟ ( E ) ϭ ⎛ 2 ⎞ (54) ϭ 8.31 V VAC ϭ ⎜ ⎜ 1 ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ R1 ϩ R4 ⎟ ⎜ ⎝ ⎟ ⎠ ⎝ 2 ϩ 11 ⎟ ⎜ ⎠ Figure 5.32: Network for Example 5.11 showing steps for solution w w w.ne w nespress.com
    • 106 Chapter 5 The voltage between B and C, ⎛ R ⎞ ⎟ ( E ) ϭ ⎛ 14 ⎞ (54) ϭ 44.47 V VBC ϭ ⎜ ⎜ 2 ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ R2 ϩ R3 ⎟ ⎜ ⎝ ⎟ ⎠ ⎝ 14 ϩ 3 ⎟ ⎜ ⎠ The voltage between A and B ϭ 44.47 Ϫ 8.31 ϭ 36.16 V Point C is at a potential of ϩ54 V. Between C and A is a voltage drop of 8.31 V. The voltage at point A is 54 Ϫ 8.31 ϭ 45.69 V. Between C and B is a voltage drop of 44.47 V. The voltage at point B is 54 Ϫ 44.47 ϭ 9.53 V. Since the voltage at A is greater than at B, current must flow in the direction A to B.(iii) Replacing the source of e.m.f. with a short-circuit (i.e., zero internal resistance) gives the circuit shown in Figure 5.32(c). The circuit is redrawn and simplified as shown in Figure 5.32(d) and (e), from which the resistance between terminals A and B, 2 ϫ 11 14 ϫ 3 22 42 rϭ ϩ ϭ ϩ 2 ϩ 11 14 ϩ 3 13 17 ϭ 1.692 ϩ 2.471 ϭ 4.163 Ω(iv) The equivalent Thévenin’s circuit is shown in Figure 5.32(f), from which, E 36.16 current, I ϭ ϭ ϭ 1A r ϩ R5 4.163 ϩ 32 The current in the 32 Ω resistor of Figure 5.32(a) is 1 A, flowing from A to B.5.6 Constant-Current SourceA source of electrical energy can be represented by a source of e.m.f. in series with aresistance. In Section 5.5, the Thévenin constant-voltage source consisted of a constante.m.f. E in series with an internal resistance r. However, this is not the only form ofrepresentation. A source of electrical energy can also be represented by a constant-currentsource in parallel with a resistance. It may be shown that the two forms are equivalent. Anideal constant-voltage generator is one with zero internal resistance so that it supplies thesame voltage to all loads. An ideal constant-current generator is one with infinite internalresistance so that it supplies the same current to all loads.Note the symbol for an ideal current source (BS 3939, 1985), shown in Figure 5.33.w ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 107 Figure 5.33: Symbol for ideal current source5.7 Norton’s TheoremNorton’s theorem states:The current that flows in any branch of a network is the same as that which would flowin the branch if it were connected across a source of electrical energy, the short-circuitcurrent of which is equal to the current that would flow in a short-circuit across thebranch, and the internal resistance of which is equal to the resistance which appearsacross the open-circuited branch terminals.The procedure adopted when using Norton’s theorem is summarized below.To determine the current flowing in a resistance R of a branch AB of an active network: (i) short-circuit branch AB,(ii) determine the short-circuit current ISC flowing in the branch,(iii) remove all sources of e.m.f. and replace them by their internal resistance (or, if a current source exists, replace with an open-circuit), then determine the resistance r, “looking-in” at a break made between A and B,(iv) determine the current I flowing in resistance R from the Norton equivalent network shown in Figure 5.33, i.e., ⎛ r ⎞ ⎟I I ϭ⎜ ⎜ r ϩ R ⎟ SC ⎜ ⎝ ⎟ ⎠Example 5.12Use Norton’s theorem to determine the current flowing in the 10 Ω resistance for thecircuit shown in Figure 5.34(a). w w w.ne w nespress.com
    • 108 Chapter 5SolutionFollowing the above procedure: (i) The branch containing the 10 Ω resistance is short-circuited as shown in Figure 5.34(b).(ii) Figure 5.34(c) is equivalent to Figure 5.34(b). 10 I SC ϭ ϭ 5A 2(iii) If the 10 V source of e.m.f. is removed from Figure 5.34(b), the resistance “looking- in” at a break made between A and B is given by: 2ϫ8 rϭ ϭ 1.6 Ω 2 ϩ8(iv) From the Norton equivalent network shown in Figure 5.34(d), the current in the 10 Ω resistance, by current division, is given by: ⎛ 1.6 ⎞ ⎟ (5) ϭ 0.482 A I ϭ⎜ ⎜ 1.6 ϩ 5 ϩ 10 ⎟ ⎜ ⎝ ⎟ ⎠ as obtained previously in Example 5.7 using Thévenin’s theorem.Example 5.13Use Norton’s theorem to determine the current I flowing in the 4Ω resistance shown inFigure 5.35(a). Figure 5.34: Circuit for Example 5.12 showing stepsw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 109SolutionFollowing the procedure: (i) The 4 Ω branch is short-circuited, as shown in Figure 5.35(b). 4 2(ii) From Figure 13.45(b), I SC ϭ I1 ϩ I 2 ϭ ϩ ϭ 4A 2 1(iii) If the sources of e.m.f. are removed the resistance “looking-in” at a break made between A and B is given by: 2 ϫ1 2 rϭ ϭ Ω 2 ϩ1 3(iv) From the Norton equivalent network shown in Figure 5.35(c)the current in the 4 Ω resistance is given by: ⎛ 2 /3 ⎞ ⎟ (4) ϭ 0.571 A I ϭ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎝ (2 / 3) ϩ 4 ⎠as obtained previously in problems 2, 5 and 9 using Kirchhoff’s laws and the theorems ofsuperposition and Thévenin.Example 5.14Use Norton’s theorem to determine the current flowing in the 3 Ω resistance of thenetwork shown in Figure 5.36(a). The voltage source has negligible internal resistance.SolutionFollowing the procedure: (i) The branch containing the 3 Ω resistance is short-circuited, as shown in Figure 5.36(b). Figure 5.35: Circuits for Example 5.13 w w w.ne w nespress.com
    • 110 Chapter 5 Figure 5.36: Circuits for Example 5.14(ii) From the equivalent circuit shown in Figure 5.36(c), 24 I SC ϭ ϭ 4.8 A 5(iii) If the 24 V source of e.m.f. is removed the resistance “looking-in” at a break made between A and B is obtained from Figure 5.36(d) and its equivalent circuit shown in Figure 5.36(e) and is given by: 10 ϫ 5 50 1 rϭ ϭ ϭ3 Ω 10 ϩ 5 15 3(iv) From the Norton equivalent network shown in Figure 5.36(f) the current in the 3 Ω resistance is given by: ⎛ 1 ⎞ ⎟ ⎜ ⎜ 3 ⎟ ⎜ 3 ⎟ ⎟ (4.8) ϭ 2 A, I ϭ⎜⎜ ⎟ ⎟ ⎜ 1 ⎜ 3 ϩ1 ϩ 3 ⎟ 2 ⎟ ⎜ ⎜ 3 ⎝ ⎟ ⎟ ⎠ 3 as obtained previously in Example 5.10 using Thévenin’s theorem.Example 5.15Determine the current flowing in the 2 Ω resistance in the network shown in Figure5.37(a).w ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 111 Figure 5.37: Circuits for Example 5.15SolutionFollowing the procedure: (i) The 2 Ω resistance branch is short-circuited as shown in Figure 5.37(b).(ii) Figure 5.37(c) is equivalent to Figure 5.37(b).(iii) If the 15 A current source is replaced by an open circuit then from Figure 5.37(d) the resistance “looking-in” at a break made between A and B is given by (6 ϩ 4) Ω in parallel with (8 ϩ 7) Ω, i.e., (10)(15) 150 rϭ ϭ ϭ 6Ω 10 ϩ 15 25(iv) From the Norton equivalent network shown in Figure 5.37(e)the current in the 2 Ω resistance is given by: ⎛ 6 ⎟ ⎞ I ϭ⎜ ⎟ (9) ϭ 6.75 A ⎜6ϩ2⎟ ⎜ ⎝ ⎠5.8 Thévenin and Norton Equivalent NetworksThe Thévenin and Norton networks shown in Figure 5.38 are equivalent to each other.The resistance “looking-in” at terminals AB is the same in each of the networks, i.e., r. w w w.ne w nespress.com
    • 112 Chapter 5If terminals AB in Figure 5.38(a) are short-circuited, the short-circuit current is given byE/r. If terminals AB in Figure 5.38(b) are short-circuited, the short-circuit current is ISC.For the circuit shown in Figure 5.38(a) to be equivalent to the circuit in Figure 5.38(b)the same short-circuit current must flow. Thus, ISC ϭ E/r.Figure 5.39 shows a source of e.m.f. E in series with a resistance r feeding a loadresistance R. E E/r ⎛ r ⎞E ⎟From Figure 13.50, I ϭ ϭ ϭ⎜ ⎜ ⎟ rϩR (r ϩ R ) /r ⎝ r ϩ R ⎟ r ⎜ ⎠ ⎛ r ⎞ ⎟I I ϭ⎜ ⎜ r ϩ R ⎟ SC ⎜i.e., ⎟ ⎝ ⎠From Figure 5.40, it can be seen that, when viewed from the load, the source appears as asource of current ISC, which is divided between r and R connected in parallel. Figure 5.38: Equivalent Thévenin and Norton networks Figure 5.39: Source E in series with resistance r feeding load resistance Rw ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 113Thus the two representations shown in Figure 5.38 are equivalent.Example 5.16Convert the circuit shown in Figure 5.41 to an equivalent Norton network.SolutionIf terminals AB in Figure 5.41 are short-circuited, the short-circuit current 10 I SC ϭ ϭ 5A 2The resistance looking-in at terminals AB is 2 Ω. The equivalent Norton network isshown in Figure 5.42.Example 5.17Convert the network shown in Figure 5.43 to an equivalent Thévenin circuit. Figure 5.40: Source when viewed from load Figure 5.41: Circuit for Example 5.16 w w w.ne w nespress.com
    • 114 Chapter 5 Figure 5.42: Equivalent Norton network Figure 5.43: Network for Example 5.17SolutionThe open-circuit voltage E across terminals AB in Figure 5.43 is given by:E ϭ (ISC) (r) ϭ (4) (3) ϭ 12 V.The resistance looking-in at terminals AB is 3 Ω. The equivalent Thévenin circuit is asshown in Figure 5.44.Example 5.18(a) Convert the circuit to the left of terminals AB in Figure 5.45(a) to an equivalent Thévenin circuit by initially converting to a Norton equivalent circuit. (b) Determine the current flowing in the 1.8 Ω resistor.Solution(a) For the branch containing the 12 V source, converting to a Norton equivalent circuit gives ISC ϭ 12 / 3 ϭ 4 A and r1 ϭ 3 Ω. For the branch containing the 24 V source, converting to a Norton equivalent circuit gives ISC2 ϭ 24/ 2 ϭ 12 A and r2 ϭ 2 Ω.w ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 115 Figure 5.44: Equivalent Thévenin circuit Figure 5.45: Circuits for Example 5.18 Thus Figure 5.45(b) shows a network equivalent to Figure 5.45(a). From Figure 5.45(b) the total short-circuit current is 4 ϩ 12 ϭ 16 A, 3ϫ 2 and the total resistance is given by: ϭ 1.2 Ω 3ϩ2 Thus Figure 5.45(b) simplifies to Figure 5.45(c). The open-circuit voltage across AB of Figure 5.45(c), E ϭ (16)(1.2) ϭ 19.2 V, and the resistance “looking-in” at AB is 1.2 Ω. The Thévenin equivalent circuit is as shown in Figure 5.45(d).(b) When the 1.8 Ω resistance is connected between terminals A and B of Figure 5.45(d) the current I flowing is given by: 19.2 Iϭ ϭ 6.4 A 1.2 ϩ 1.8Example 5.19Determine by successive conversions between Thévenin and Norton equivalent networksa Thévenin equivalent circuit for terminals AB of Figure 5.46(a). Determine the currentflowing in the 200 Ω resistance. w w w.ne w nespress.com
    • 116 Chapter 5 Figure 5.46: Circuits for Example 5.19SolutionFor the branch containing the 10 V source, converting to a Norton equivalent network gives: 10I SC ϭ ϭ 5 mA and r1 ϭ 2 kΩ. 2000For the branch containing the 6 V source, converting to a Norton equivalent network gives: 6I SC ϭ ϭ 2 mA and r2 ϭ 3 kΩ. 3000Thus, the network of Figure 5.46(a) converts to Figure 5.46(b).Combining the 5 mA and 2 mA current sources gives the equivalent network of Figure5.46(c) where the short-circuit current for the original two branches considered is 7 mAand the resistance is:2ϫ3 ϭ 1.2 kΩ.2ϩ3w ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 117Both of the Norton equivalent networks shown in Figure 5.46(c) may be convertedto Thévenin equivalent circuits. The open-circuit voltage across CD is: (7 ϫ 10Ϫ3)(1.2 ϫ 103) ϭ 8.4 V and the resistance looking-in at CD is 1.2 kΩ.The open-circuit voltage across EF is (1 ϫ 10Ϫ3) (600) ϭ 0.6 V and the resistance“looking-in” at EF is 0.6 kΩ. Thus, Figure 5.46(c) converts to Figure 5.46(d). Combiningthe two Thévenin circuits gives:E ϭ 8.4 Ϫ 0.6 ϭ 7.8 V, and the resistance,r ϭ (1.2 ϩ 0.6) kΩ ϭ 1.8 kΩ.Thus, the Thévenin equivalent circuit for terminals AB of Figure 5.46(a) is as shown inFigure 5.46(e).Therefore, the current I flowing in a 200 Ω resistance connected between A and B isgiven by: 7.8 7.8Iϭ ϭ ϭ 3.9 mA 1800 ϩ 200 20005.9 Maximum Power Transfer TheoremThe maximum power transfer theorem states:The power transferred from a supply source to a load is at its maximum when theresistance of the load is equal to the internal resistance of the source.In Figure 5.47, when R ϭ r the power transferred from the source to the load is amaximum.Typical practical applications of the maximum power transfer theorem are found in stereoamplifier design, seeking to maximize power delivered to speakers, and in electric vehicledesign, seeking to maximize power delivered to drive a motor.Example 5.20The circuit diagram of Figure 5.48 shows dry cells of source e.m.f. 6 V, and internalresistance 2.5 Ω. If the load resistance RL is varied from 0 to 5 Ω in 0.5 Ω steps, calculatethe power dissipated by the load in each case. Plot a graph of RL (horizontally) againstpower (vertically) and determine the maximum power dissipated. w w w.ne w nespress.com
    • 118 Chapter 5 Figure 5.47: When r ‫ ؍‬R, power transfer is maximum Figure 5.48: Circuit for Example 5.20Solution E 6When RL ϭ 0, current I ϭ ϭ ϭ 2.4 A, and r ϩ RL 2.5power dissipated in RL, P ϭ I2RL,i.e., P ϭ (2.4)2 (0) ϭ 0 W E 6When RL ϭ 0.5 Ω, current I ϭ ϭ ϭ 2A r ϩ RL 2.5 ϩ 0.5and P ϭ I2RL ϭ (2)2 (0.5) ϭ 2 W 6When RL ϭ 1.0 Ω, current I ϭ ϭ 1.714 A 2.5 ϩ 1.0and P ϭ (1.714)2 (1.0) ϭ 2.94 Ww ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 119With similar calculations the following table is produced: RL(Ω) E P ϭ I2RL(W) Iϭ r ϩ RL 0 2.4 0 0.5 2.0 2.00 1.0 1.714 2.94 1.5 1.5 3.38 2.0 1.333 3.56 2.5 1.2 3.60 3.0 1.091 3.57 3.5 1.0 3.50 4.0 0.923 3.41 4.5 0.857 3.31 5.0 0.8 3.20A graph of RL against P is shown in Figure 5.49. The maximum value of power is 3.60 W,which occurs when RL is 2.5 Ω, i.e., maximum power occurs when RL ϭ r, which is whatthe maximum power transfer theorem states.Example 5.21A DC source has an open-circuit voltage of 30 V and an internal resistance of 1.5 Ω. Statethe value of load resistance that gives maximum power dissipation and determine thevalue of this power.SolutionThe circuit diagram is shown in Figure 5.50. From the maximum power transfer theorem,for maximum power dissipation,RL ϭ r ϭ 1.5 ⍀ E 30From Figure 5.50, current I ϭ ϭ ϭ 10 A r ϩ RL 1.5 ϩ 1.5Power P ϭ I2RL ϭ (10)2(1.5) ϭ 150 W ϭ maximum power dissipated w w w.ne w nespress.com
    • 120 Chapter 5 Figure 5.49: Graph of RL vs. P Figure 5.50: Circuit diagram for Example 5.21Example 5.22Find the value of the load resistor RL shown in Figure 5.51(a) that gives maximum powerdissipation and determine the value of this power.SolutionUsing the procedure for Thévenin’s theorem: (i) Resistance RL is removed from the circuit as shown in Figure 5.51(b).(ii) The voltage across AB is the same as the voltage across the 1 Ω resistor: ⎛ 12 ⎞ ⎟ (15) ϭ 12 V Hence, E ϭ ⎜ ⎜ ⎟ ⎝ 12 ϩ 3 ⎟ ⎜ ⎠(iii) Removing the source of e.m.f. gives the circuit of Figure 5.51(c), 12 ϫ 3 36 from which resistance, r ϭ ϭ ϭ 2.4 Ω 12 ϩ 3 15w ww. n e w n e s p r e s s .c o m
    • DC Circuit Theory 121 Figure 5.51: Circuits for Example 5.22(iv) The equivalent Thévenin’s circuit supplying terminals AB is shown in Figure 5.51(d), from which current I ϭ E/(r ϩ RL). For maximum power, RL ϭ r ϭ 2.4 Ω. 12 Thus, current, I ϭ ϭ 2.5 A. 2.4 ϩ 2.4 Power, P, dissipated in load RL, P ϭ I2RL ϭ (2.5)2 (2.4) ϭ 15 W w w w.ne w nespress.com
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    • CHAPTE R 6 Alternating Voltages and Currents John Bird6.1 The AC GeneratorLet a single turn coil be free to rotate at constant angular velocity symmetrically betweenthe poles of a magnet system as shown in Figure 6.1.An e.m.f. is generated in the coil (from Faraday’s laws) which varies in magnitude andreverses its direction at regular intervals. The reason for this is shown in Figure 6.2.In positions (a), (e) and (i) the conductors of the loop are effectively moving along themagnetic field, no flux is cut and hence, no e.m.f is induced. In position (c) maximumflux is cut and maximum e.m.f is induced. In position (g), maximum flux is cut andmaximum e.m.f is again induced. However, using Fleming’s right-hand rule, the inducede.m.f is in the opposite direction to that in position (c) and is shown as ϪE. In positions(b), (d), (f) and (h) some flux is cut and some e.m.f is induced. If all such positions Figure 6.1: Coil rotates at constant angular velocity w w w.ne w nespress.com
    • 124 Chapter 6 (a) (b) (c) (d) (e) (f) (g) (h) (i) N N N N N N N N N S S S S S S S S S ϩE Induced e.m.f Revolutions of loop 0 1 1 1 3 1 3 – – – – – 8 4 8 2 4 ϪE Figure 6.2: One cycle of alternating e.m.f producedof the coil are considered, in one revolution of the coil, one cycle of alternating e.m.fis produced as shown. This is the principle of operation of the AC generator (i.e., thealternator).6.2 WaveformsIf values of quantities that vary with time t are plotted to a base of time, the resultinggraph is called a waveform. Some typical waveforms are shown in Figure 6.3. Waveforms(a) and (b) are unidirectional waveforms, for, although they vary considerably withtime, they flow in one direction only (i.e., they do not cross the time axis and becomenegative). Waveforms (c) to (g) are called alternating waveforms since their quantities arecontinually changing in direction (i.e., alternately positive and negative).A waveform of the type shown in Figure 6.3(g) is called a sine wave. It is the shape of thewaveform of e.m.f produced by an alternator and thus, the mains electricity supply is of“sinusoidal” form.One complete series of values is called a cycle (i.e., from O to P in Figure 6.3(g)).The time taken for an alternating quantity to complete one cycle is called the period orthe periodic time, T, of the waveform.w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 125 Figure 6.3: Typical waveformsThe number of cycles completed in one second is called the frequency, f, of the supplyand is measured in hertz, Hz. (The standard frequency of the electricity supply in the U.S.is 60 Hz and in Great Britain is 50 Hz.) 1 1Tϭ or f ϭ f TExample 6.1Determine the periodic time for frequencies of (a) 50 Hz and (b) 20 kHz.Solution 1 1(a) Periodic time T ϭ ϭ ϭ 0.02 s or 20 msv f 50 1 1(b) Periodic time T ϭ ϭ ϭ 0.000 05 s or 50 μs f 20 000Example 6.2Determine the frequencies for periodic times of (a) 4 ms, (b) 4 μs. w w w.ne w nespress.com
    • 126 Chapter 6Solution 1 1 1000(a) Frequency f ϭ ϭ Ϫ3 ϭ ϭ 250 Hz T 4 ϫ 10 4(b) Frequency f ϭ 1 ϭ 1 ϭ 1000 000 T 4 ϫ 10Ϫ6 4 ϭ 250, 000 Hz or 250 kHz or 0.25 MHzExample 6.3An alternating current completes 5 cycles in 8 ms. What is its frequency?SolutionTime for 1 cycle ϭ 8 ms ϭ 1.6 ms ϭ periodic time T 5 1 1 1000 10 000Frequency f ϭ ϭ Ϫ3 ϭ ϭ T 1.6 ϫ 10 1.6 16 ϭ 625 Hz6.3 AC ValuesInstantaneous values are the values of the alternating quantities at any instant of time.They are represented by small letters, i, υ, e, etc. (See Figures 6.3(f) and (g).)The largest value reached in a half cycle is called the peak value or the maximum valueor the amplitude of the waveform. Such values are represented by Vm, Im etc. (SeeFigures 6.3(f) and (g).) A peak-to-peak value of e.m.f is shown in Figure 6.3(g) and is thedifference between the maximum and minimum values in a cycle.The average or mean value of a symmetrical alternating quantity (such as a sine wave),is the average value measured over a half cycle (since over a complete cycle the averagevalue is zero). area under the curveAverage or mean value ϭ length of basew ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 127The area under the curve is found by approximate methods such as the trapezoidal rule,the mid-ordinate rule or Simpson’s rule. Average values are represented by VAV , IAV , etc.For a sine wave,average value ϭ 0.637 ϫ maximum value(i.e., 2/π ϫ maximum value)The effective value of an alternating current is that current which will produce the sameheating effect as an equivalent direct current. The effective value is called the root meansquare (rms) value and whenever an alternating quantity is given, it is assumed to bethe rms value. The symbols used for rms values are I, V, E, etc. For a nonsinusoidalwaveform as shown in Figure 6.4 the rms value is given by: ⎛ i 2 ϩ i2 ϩ 2 ϩ in ⎞ 2 ⎟Iϭ ⎜1 ⎜ ⎜ ⎟ ⎟ ⎝ n ⎟ ⎠where n is the number of intervals used.For a sine wave,rms value ϭ 0.707 ϫ maximum value(i.e., 1/ 2 ϫ maximum value) rms valueForm factor ϭ For a sine wave, form factor ϭ 1.11 average value Figure 6.4: Nonsinusoidal waveform w w w.ne w nespress.com
    • 128 Chapter 6 maximum valuePeak factor ϭ For a sine wave, peak factor ϭ 1.41 rms valueThe values of form and peak factors give an indication of the shape of waveforms.Example 6.4For the periodic waveforms shown in Figure 6.5 determine for each: (i) frequency,(ii) average value over half a cycle, (iii) rms value, (iv) form factor, and (v) peak factor.Solution(a) Triangular waveform (Figure 6.5(a)) (i) Time for 1 complete cycle ϭ 20 ms ϭ periodic time, T. Hence, frequency f ϭ 1 ϭ 1 ϭ 1000 T 20 ϫ 10Ϫ3 20 ϭ 50 Hz (ii) Area under the triangular waveform for a half cycle ϭ 1 ϫ base ϫ height ϭ 1 ϫ (10 ϫ 10Ϫ3 ) ϫ 200 2 2 ϭ 1 volt second Figure 6.5: Waveforms for Example 6.4w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 129 Average value of waveform area under curve 1 volt second ϭ ϭ length of base 10 ϫ 10Ϫ3 second 1000 ϭ ϭ 100 V 10 (iii) In Figure 6.5(a), the first 1/4 cycle is divided into 4 intervals. Thus, rms value ϭ ͱ ⎛ υ1 ϩ υ 2 ϩ υ3 ϩ υ 2 ⎞ ⎜ ⎜ ⎜ ⎝ 2 2 4 2 4⎟⎟ ⎟ ⎟ ⎠ ϭ ͱ ⎛ 252 ϩ 752 ϩ 1252 ϩ 1752 ⎞ ⎜ ⎜ ⎜ ⎝ 4 ⎟ ⎟ ⎟ ⎟ ⎠ ϭ 114.6 V (Note that the greater the number of intervals chosen, the greater the accuracy of the result. For example, if twice the number of ordinates as that chosen above are used, the rms value is found to be 115.6 V) rms value 114.6 (iv) Form factor ϭ ϭ ϭ 1.15 average value 100 maximum value 200 (v) Peak factor ϭ ϭ ϭ 1.75 rms value 114.6(b) Rectangular waveform (Figure 6.5(b)) (i) Time for 1 complete cycle ϭ 16 ms ϭ periodic time, T 1 1 1000 Hence, frequency, f ϭ ϭ Ϫ3 ϭ T 16 ϫ 10 16 ϭ 62.5 Hz w w w.ne w nespress.com
    • 130 Chapter 6 area under curve (ii) Average value over half a cycle ϭ length of base 10 ϫ (8 ϫ 10Ϫ3 ) ϭ 8 ϫ 10Ϫ3 ϭ 10 A (iii) The rms value ϭ ͱ ⎛ i1 ϩ i2 ϩ ⎜ ⎜ ⎜ ⎝ 2 2 n ϩ in ⎞ 2 ⎟ ϭ 10 A ⎟ ⎟ ⎟ ⎠ However, many intervals are chosen, since the waveform is rectangular. rms value 10 (iv) Form factor ϭ ϭ ϭ1 average value 10 maximum value 10 (v) Peak factor ϭ ϭ ϭ1 rms value 10Example 6.5The following table gives the corresponding values of current and time for a half cycle ofalternating current.time t (ms) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0current i (A) 0 7 14 23 40 56 68 76 60 5 0Assuming the negative half cycle is identical in shape to the positive half cycle, plot thewaveform and find (a) the frequency of the supply, (b) the instantaneous values of currentafter 1.25 ms and 3.8 ms, (c) the peak or maximum value, (d) the mean or average value,and (e) the rms value of the waveform.SolutionThe half cycle of alternating current is shown plotted in Figure 6.6.(a) Time for a half cycle ϭ 5 ms. The time for 1 cycle, i.e., the periodic time, T ϭ 10 ms or 0.01s. 1 1 Frequency, f ϭ ϭ ϭ 100 Hz T 0.01w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 131 Figure 6.6: Half cycle of alternating current for Example 6.5(b) Instantaneous value of current after 1.25 ms is 19 A, from Figure 6.6. Instantaneous value of current after 3.8 ms is 70 A, from Figure 6.6.(c) Peak or maximum value ϭ 76 A. area under curve(d) Mean or average value ϭ length of base Using the mid-ordinate rule with 10 intervals, each of width 0.5 ms gives: area under curve. ϭ (0.5 ϫ 10Ϫ3)[3 ϩ 10 ϩ 19 ϩ 30 ϩ 49 ϩ 63 ϩ 73 ϩ 72 ϩ 30 ϩ 2] (see Figure 6.6) ϭ (0.5 ϫ 10Ϫ3)(351) w w w.ne w nespress.com
    • 132 Chapter 6 (0.5 ϫ 10Ϫ3 )(351) Hence, mean or average value ϭ 5 ϫ 10Ϫ3 ϭ 35.1 A(e) rms value ͱ ⎛ 32 ϩ 102 ϩ 192 ϩ 302 ϩ 492 ϩ 632 ϩ ⎞ ⎟ ⎜ ⎜ ⎟ ⎜ 732 ϩ 722 ϩ 302 ϩ 22 ⎟ ⎟ ϭ ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ 10 ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎝ ⎟ ⎠ ϭ ͱ ⎛ ⎜ 10 ⎟ ⎜ ⎝ ⎞ ⎜ 19157 ⎟ ϭ 43.8 A ⎟ ⎠Example 6.6Calculate the rms value of a sinusoidal current of maximum value 20 A.SolutionFor a sine wave, rms value ϭ 0.707 ϫ maximum value ϭ 0.707 ϫ 20 ϭ 14.14 AExample 6.7Determine the peak and mean values for a 240 V mains supply.SolutionFor a sine wave, rms value of voltage V ϭ 0.707 ϫ VmA 240 V mains supply means that 240 V is the rms value, V 240Vm ϭ ϭ ϭ 339.5 V ϭ peak value 0.707 0.707Mean value VAV ϭ 0.637 Vm ϭ 0.637 ϫ 339.5 ϭ 216.3 VExample 6.8A supply voltage has a mean value of 150 V. Determine its maximum value and its rms value.w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 133SolutionFor a sine wave, mean value ϭ 0.637 ϫ maximum value mean value 150Hence, maximum value ϭ ϭ ϭ 235.5 V 0.637 0.637rms value ϭ 0.707 ϫ maximum value ϭ 0.707 ϫ 235.5 ϭ 166.5 V6.4 The Equation of a Sinusoidal WaveformIn Figure 6.7, OA represents a vector that is free to rotate anticlockwise about 0 at anangular velocity of ω rad/s. A rotating vector is known as a phasor.After time t seconds the vector OA has turned through an angle ωt. If the line BC isconstructed perpendicular to OA as shown, then, BCsin ωt ϭ i.e., BC ϭ OB sin ωt OBIf all such vertical components are projected onto a graph of y against angle ωt (inradians), a sine curve results of maximum value OA. Any quantity that varies sinusoidallycan be represented as a phasor.A sine curve may not always start at 0°. To show this, a periodic function is represented byy ϭ sin(ωt ϩ φ), where φ is the phase (or angle) difference compared with y ϭ sin ωt. InFigure 6.8(a), y2 ϭ sin(ωt ϩ φ) starts φ radians earlier than y1 ϭ sin ωt and is said to leady1 by φ radians. Phasors y1 and y2 are shown in Figure 6.8(b) at the time when t ϭ 0. Figure 6.7: Rotating vector OA and plot of rotation showing resulting sine curve w w w.ne w nespress.com
    • 134 Chapter 6 Figure 6.8: Phase angle, leading and laggingIn Figure 6.8(c), y4 ϭ sin(ωt Ϫ φ) starts φ radians later than y3 ϭ sin ωt and is said to logy3 by φ radians. Phasors y3 and y4 are shown in Figure 6.8(d) at the time when t ϭ 0.Given the general sinusoidal voltage, υ ϭ Vm sin(ωt ؎ φ), then (i) Amplitude or maximum value ϭ Vm (ii) Peak-to-peak value ϭ 2 Vm (iii) Angular velocity ϭ ω rad/s (iv) Periodic time, T ϭ 2π/ω seconds (v) Frequency, f ϭ ω/2π Hz (since ϭ 2πf ) (vi) φ ϭ angle of lag or lead (compared with v ϭ Vm sin ωtExample 6.9An alternating voltage is given by υ ϭ 282.8 sin 314t volts. Find (a) the rms voltage,(b) the frequency and (c) the instantaneous value of voltage when t ϭ 4 ms.w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 135Solution(a) The general expression for an alternating voltage is υ ϭ Vm sin(ωt Ϯ φ). Comparing υ ϭ 282.8 sin 314t with this general expression gives the peak voltage as 282.8 V The rms voltage ϭ 0.707 ϫ maximum value ϭ 0.707 ϫ 282.8 ϭ 200 V(b) Angular velocity, ω ϭ 314 rad/s, i.e. 2πf ϭ 314 314 Frequency, f ϭ ϭ 50 Hz 2π(c) When t ϭ 4 ms, υ ϭ 282.8 sin(314 ϫ 4ϫ10Ϫ3) ϭ 228.2 sin(1.256) ϭ 268.9 V ° ⎛ 180 ⎞ ⎟ Note that 1.256 radians ϭ ⎜1.256 ϫ ⎜ ⎟ ⎟ ⎜ ⎝ π ⎠ ϭ 71.96° Hence, υ ϭ 282.8 sin 71. 96° ϭ 268.9 VExample 6.10An alternating voltage is given byυ ϭ 75 sin(200πt Ϫ 0.25) volts.Find (a) the amplitude, (b) the peak-to-peak value, (c) the rms value, (d) the periodictime, (e) the frequency, and (f) the phase angle (in degrees and minutes) relative to75 sin 200πt.SolutionComparing υ ϭ 75 sin(200πt Ϫ 0.25) with the general expression υ ϭ Vm sin(ωt Ϯ φ) gives:(a) Amplitude, or peak value ϭ 75 V(b) Peak-to-peak value ϭ 2 ϫ 75 ϭ 150 V w w w.ne w nespress.com
    • 136 Chapter 6(c) The rms value ϭ 0.707 ϫ maximum value ϭ 0.707 ϫ 75 ϭ 53 V(d) Angular velocity, ω ϭ 200π rad/s 2π 2π 1 Hence, periodic time, T ϭ ϭ ϭ ω 200π 100 ϭ 0.01 s or 10 ms 1 1(e) Frequency, f ϭ ϭ ϭ 100 Hz T 0.01(f) Phase angle, φ ϭ 0.25 radians lagging 75 sin 200πt ⎛ ⎞° ⎜ 0.25 ϫ 180 ⎟ ϭ 14.32° ϭ 14°19′ 0.25 rads ϭ ⎜ ⎟ ⎜ ⎝ π ⎟⎠ Hence, phase angle ϭ 14°19Ј laggingExample 6.11An alternating voltage, υ, has a periodic time of 0.01s and a peak value of 40 V.When time t is zero, υ ϭ Ϫ20 V. Express the instantaneous voltage in the formυ ϭ Vm sin(ωt Ϯ φ).SolutionAmplitude, Vm ϭ 40 V 2πPeriodic time T ϭ hence, angular velocity, ω 2π 2π ωϭ ϭ ϭ 200π rad/s T 0.01υ ϭ Vm sin(ωt ϩ φ) becomes υ ϭ 40 sin(200πt ϩ φ)VWhen time t ϭ 0, υ ϭ Ϫ20 V i.e., Ϫ20 ϭ 40 sin φw ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 137 Ϫ20so that sin φ ϭ ϭ Ϫ0.5 40 ⎛ π ⎞⎟ radsHence, φ ϭ sinϪ1 (Ϫ0.5) ϭ Ϫ30° ϭ ⎜Ϫ30 ϫ ⎜ ⎟ ⎜ ⎝ 180 ⎟ ⎠ π ϭ Ϫ rads 6 ⎛ π⎞Thus, υ ‫ 04 ؍‬sin ⎜ 200πt ؊ ⎟ V ⎜ ⎟ ⎜ ⎝ 6⎟ ⎠Example 6.12The current in an AC circuit at any time t seconds is given by:i ϭ 120 sin(100πt ϩ 0.36) amperes. Find:(a) the peak value, the periodic time, the frequency and phase angle relative to 120 sin 100πt,(b) the value of the current when t ϭ 0,(c) the value of the current when t ϭ 8 ms,(d) the time when the current first reaches 60 A, and(e) the time when the current is first a maximum.Solution(a) Peak Value ϭ 120 A 2π 2π Periodic time T ϭ ϭ (since ω ϭ 100π) ω 100π 1 ϭ ϭ 0.02 s or 20 ms 50 1 1 Frequency, f ϭ ϭ ϭ 50 Hz T 0.02 w w w.ne w nespress.com
    • 138 Chapter 6 ° ⎛ 180 ⎞ ⎟ Phase angle ϭ 0.36 rads ϭ ⎜ 0.36 ϫ ⎜ ⎟ ⎜ ⎝ π ⎟⎠ ϭ 20°38؅ leading(b) When t ϭ 0, i ϭ 120 sin(0 ϩ 0.36) ϭ 120 sin 20°38Ј ϭ 49.3 A ⎡ ⎛ 8 ⎞ ⎤(c) When t ϭ 8, i ϭ 120 sin ⎢100π ⎜ 3 ⎟ ϩ 0.36 ⎥ ⎜ ⎟ ⎟ ⎢ ⎜ ⎝ 10 ⎠ ⎥ ⎣ ⎦ ϭ 120 sin 2.8733(ϭ120 sin 164°38Ј) ϭ 31.8 A(d) When i ϭ 60 A, 60 ϭ 120 sin(100πt ϩ 0.36) 60 thus, ϭ sin(100πt ϩ 0.36) 120 π so that (100πt ϩ 0.36) ϭ sinϪ1 0.5 ϭ 30° ϭ rads 6 ϭ 0.5236 rads 0.5236 Ϫ 0.36 ϭ 0.521 ms Hence, time t ϭ 100π(e) When the current is a maximum, i ϭ 120 A Thus, 120 ϭ 120 sin(100πt ϩ 0.36) ϭ sin(100πt ϩ 0.36) π (100πt ϩ 0.36) ϭ sinϪ1 1 ϭ 90° ϭ rads 2 ϭ 1.5708 rads 1.5708 Ϫ 0.36 Hence, time t ϭ ϭ 3.85 ms 100πw ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 1396.5 Combination of WaveformsThe resultant of the addition (or subtraction) of two sinusoidal quantities may bedetermined either: (a) by plotting the periodic functions graphically (see worked Examples 6.13 and 6.16), or (b) by resolution of phasors by drawing or calculation (see worked Examples 6.14 and 6.15).Example 6.13The instantaneous values of two alternating currents are given by i1 ϭ 20 sin ωt amperesand i2 ϭ 10 sin(ωt ϩ π/3) amperes. By plotting i1 and i2 on the same axes, using the samescale, over one cycle, and adding ordinates at intervals, obtain a sinusoidal expression fori1 ϩ i2.Solution ⎛ π⎞i1 ϭ 20 sin ωt and i2 ϭ 10 sin ⎜ ωt ϩ ⎟ are shown plotted in Figure 6.9. ⎜ ⎟ ⎜ ⎝ 3⎟ ⎠ Figure 6.9: Plots for Example 6.13 w w w.ne w nespress.com
    • 140 Chapter 6Ordinates of i1 and i2 are added at, say, 15° intervals (a pair of dividers are usefulfor this).For example, at 30°, i1 ϩ i2 ϭ 10 ϩ 10 ϭ 20 A at 60°, i1 ϩ i2 ϭ 8.7 ϩ 17.3 ϭ 26 A at 150°, i1 ϩ i2 ϭ 10 ϩ (Ϫ5) ϭ 5 A, and so on.The resultant waveform for i1 ϩ i2 is shown by the broken line in Figure 6.9. It has thesame period, and frequency, as i1 and i2. The amplitude or peak value is 26.5 A.The resultant waveform leads the curve i1 ϭ 20 sin ωt by 19°. ⎛ π ⎞⎟ rads ϭ 0.332 radsi.e. ⎜19 ϫ ⎜ ⎟ ⎜ ⎝ 180 ⎟ ⎠The sinusoidal expression for the resultant i1 ϩ i2 is given by:iR ϭ i1 ϩ i2 ϭ 26.5 sin(ωt ϩ 0.332) AExample 6.14Two alternating voltages are represented by υ1 ϭ 50 sin ωt volts and υ2 ϭ 100sin (ωt Ϫ π/6)V. Draw the phasor diagram and find, by calculation, a sinusoidalexpression to represent υ1 ϩ υ2.SolutionPhasors are usually drawn at the instant when time t ϭ 0. Thus, υ1 is drawn horizontally50 units long and υ2 is drawn 100 units long lagging υ1 by π/6 rads, i.e., 30°. This isshown in Figure 6.10(a) where 0 is the point of rotation of the phasors.Procedure to draw phasor diagram to represent υ1 ϩ υ2: (i) Draw υ1 horizontal 50 units long, i.e., Oa of Figure 6.10(b).(ii) Join υ2 to the end of υ1 at the appropriate angle, i.e., ab of Figure 6.10(b).w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 141 Figure 6.10: Phasor diagrams for Example 6.14(iii) The resultant υR ϭ υ1 ϩ υ2 is given by the length Ob and its phase angle φ may be measured with respect to υ1.Alternatively, when two phasors are being added the resultant is always the diagonal ofthe parallelogram, as shown in Figure 6.10(c).From the drawing, by measurement, υR ϭ 145 V and angle φ ϭ 20º lagging υ1.A more accurate solution is obtained by calculation, using the cosine and sine rules.Using the cosine rule on triangle Oab of Figure 6.10(b) gives:υ 2 ϭ υ1 ϩ υ 2 Ϫ 2 υ1υ 2 cos 150° R 2 2 ϭ 502 ϩ 1002 Ϫ 2(50)(100) cos 150° ϭ 2500 ϩ 10 000 Ϫ (Ϫ8660)υ R ϭ Ί (21160) ϭ 145.5 V w w w.ne w nespress.com
    • 142 Chapter 6 100 145.5Using the sine rule, ϭ sin φ sin 150° 100 sin 150°from which sin φ ϭ ϭ 0.3436 145.5and φ ϭ sinϪ1 0.3436 ϭ 20°6Ј ϭ 0.35 radians, and lags υ1Hence, υR ϭ υ1 ϩ υ2 ϭ 145.5 sin(ωt Ϫ 0.35) VExample 6.15Find a sinusoidal expression for (i1 ϩ i2) of Example 6.13, (a) by drawing phasors, (b) bycalculation.Solution(a) The relative positions of i1 and i2 at time t ϭ 0 are shown as phasors in Figure 6.11(a). The phasor diagram in Figure 6.11(b) shows the resultant iR, and iR is measured as 26A and angle φ as 19° or 0.33 rads leading i1. Hence, by drawing, iR ‫ 62 ؍‬sin(ωt ؉ 0.33) A Figure 6.11: Phasor diagrams for Example 6.15w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 143(b) From Figure 6.11(b), by the cosine rule: iR ϭ 202 ϩ 102 Ϫ 2(20)(10)(cos 120°) 2 from which iR ϭ 26.46A 10 26.46 By the sine rule: ϭ sin φ sin 120° from which φ ϭ 19.10° (i.e., 0.333 rads) By calculation iR ‫ 64.62 ؍‬sin(ωt ؉ 0.333) AAn alternative method of calculation is to use complex numbers. (See Chapter 7.) ⎛ π⎞Then i1 ϩ i2 ϭ 20 sin ωt ϩ 10 sin ⎜ ωt ϩ ⎟ ⎜ ⎟ ⎜ ⎝ 3⎟ ⎠ π ≡ 20∠0 ϩ 10∠ rad 3 or 20∠0° ϩ 10∠60° ϭ (20 ϩ j 0) ϩ (5 ϩ j8.66) ϭ (25 ϩ j8.66) ϭ 26.46∠19.106° or 26.46∠0.333 rad ≡ 26.46 sin(ωt ؉ 0.333) AExample 6.16Two alternating voltages are given by υ1 ϭ 120 sin ωt volts and υ2 ϭ 200 sin(ωt Ϫ π/4)volts. Obtain sinusoidal expressions for υ1 Ϫ υ2 (a) by plotting waveforms, and (b) byresolution of phasors.Solution(a) υ1 ϭ 120 sin ωt and υ2 ϭ 200 sin(ωt Ϫ π/4) are shown plotted in Figure 6.12. Care must be taken when subtracting values of ordinates especially when at least one of the ordinates is negative. For example: at 30°, υ1 Ϫ υ2 ϭ 60 Ϫ (Ϫ52) ϭ 112 V at 60°, υ1 Ϫ υ2 ϭ 104 Ϫ 52 ϭ 52 V at 150°, υ1 Ϫ υ2 ϭ 60 Ϫ 193 ϭ Ϫ133 V, and so on. w w w.ne w nespress.com
    • 144 Chapter 6 Figure 6.12: Voltage plots for Example 6.16 The resultant waveform, υR ϭ υ1 Ϫ υ2, is shown by the broken line in Figure 6.12. The maximum value of υR is 143 V and the waveform is seen to lead υ1 by 99° (i.e., 1.73 radians). By drawing, υR ϭ υ1 Ϫ υ2 ϭ 143 sin(ωt ϩ 1.73) volts(b) The relative positions of υ1 and υ2 are shown at time t ϭ 0 as phasors in Figure 6.13(a). Since the resultant of υ1 Ϫ υ2 is required, Ϫυ2 is drawn in the opposite direction to ϩυ2 and is shown by the broken line in Figure 6.13(a). The phasor diagram with the resultant is shown in Figure 6.13(b) where Ϫυ2 is added phasorially to υ1. By resolution: Sum of horizontal components of υ1 and υ2 ϭ 120 cos 0° ϩ 200 cos 135° ϭ Ϫ21.42 Sum of vertical components of υ1 and υ2 ϭ 120 sin 0° ϩ 200 sin 135° ϭ 141.4w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 145 Figure 6.13: Phasor diagrams for Example 6.16From Figure 6.13(c), resultant υ R ϭ Ί [(Ϫ21.42)2 ϩ (141.4)2 ] ϭ 143.0,and tan φЈ ϭ 141.4 ϭ tan 6.6013, from which 21.42 φЈ ϭ tanϪ1 6.6013 ϭ 81°23Ј and φ ϭ 98°37Ј or 1.721 radians w w w.ne w nespress.com
    • 146 Chapter 6 Figure 6.14: Half-wave rectification Figure 6.15: Full-wave rectification By resolution of phasors, υR ‫ ؍‬υ1 ؊ υ2 ‫ 0.341 ؍‬sin(ωt ϩ 1.721) volts6.6 RectificationThe process of obtaining unidirectional currents and voltages from alternating currentsand voltages is called rectification. Automatic switching in circuits is carried out bydiodes.Using a single diode, as shown in Figure 6.14, half-wave rectification is obtained. WhenP is sufficiently positive with respect to Q, diode D is switched on and current i flows.When P is negative with respect to Q, diode D is switched off. Transformer T isolates theequipment from direct connection with the mains supply and enables the mains voltage tobe changed.w ww. n e w n e s p r e s s .c o m
    • Alternating Voltages and Currents 147 Figure 6.16: Bridge rectifier Figure 6.17: Smoothing output using capacitorsTwo diodes may be used as shown in Figure 6.15 to obtain full wave rectification.A center-tapped transformer T is used. When P is sufficiently positive with respect to Q,diode D1 conducts and current flows (shown by the broken line in Figure 6.15). When S ispositive with respect to Q, diode D2 conducts and current flows (shown by the continuousline in Figure 6.15). The current flowing in R is in the same direction for both half cyclesof the input. The output waveform is shown in Figure 6.15.Four diodes may be used in a bridge rectifier circuit, as shown in Figure 6.16 to obtainfull wave rectification. As for the rectifier shown in Figure 6.15, the current flowing inR is in the same direction for both half cycles of the input giving the output waveformshown.To smooth the output of the rectifiers described above, capacitors having a largecapacitance may be connected across the load resistor R. The effect of this is shown onthe output in Figure 6.17. w w w.ne w nespress.com
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    • CHAPTE R 7 Complex Numbers John Bird7.1 IntroductionA complex number is of the form (a ϩ jb) where a is a real number and jb is animaginary number. Therefore, (1 ϩ j2) and (5 Ϫ j7) are examples of complex numbers.By definition, j ϭ ͱ Ϫ 1 and j 2 ϭ Ϫ1(Note: In electrical engineering, the letter j is used to represent Ϫ1 instead of the letter i,as commonly used in pure mathematics, because i is reserved for current.)Complex numbers are widely used in the analysis of series, parallel and series-parallelelectrical networks supplied by alternating voltages, in deriving balance equations withAC bridges, in analyzing AC circuits using Kirchhoff’s laws, mesh and nodal analysis,the superposition theorem, with Thévenin’s and Norton’s theorems, and with delta-starand star-delta transforms, and in many other aspects of higher electrical engineering.The advantage of the use of complex numbers is that the manipulative processes becomesimply algebraic processes.A complex number can be represented pictorially on an Argand diagram. In Figure 7.1,the line 0A represents the complex number (2 ϩ j3), 0B represents (3 Ϫ j), 0C represents(Ϫ2 ϩ j2) and 0D represents (Ϫ4 Ϫ j3).A complex number of the form a ϩ jb is called a Cartesian or rectangular complexnumber. The significance of the j operator is shown in Figure 7.2. In Figure 7.2(a) thenumber 4 (i.e., 4 ϩ j0) is shown drawn as a phasor horizontally to the right of the originon the real axis. (Such a phasor could represent, for example, an alternating current, i ϭ 4sin ωt amperes, when time t is zero.) w w w.ne w nespress.com
    • 150 Chapter 7 Figure 7.1: The Argand diagramThe number j4 (that is, 0 ϩ j4) is shown in Figure 7.2(b) drawn vertically upwards fromthe origin on the imaginary axis. Multiplying the number 4 by the operator j results in ananticlockwise phase-shift of 90° without altering its magnitude.Multiplying j4 by j gives j24, i.e., Ϫ4, and is shown in Figure 7.2(c) as a phasor four unitslong on the horizontal real axis to the left of the origin—an anticlockwise phase-shift of90° compared with the position shown in Figure 7.2(b). Thus, multiplying by j2 reversesthe original direction of a phasor.Multiplying j24 by j gives j34, i.e., Ϫj4, and is shown in Figure 7.2(d) as a phasor fourunits long on the vertical, imaginary axis downward from the origin—an anticlockwisephase-shift of 90° compared with the position shown in Figure 7.2(c).Multiplying j34 by j gives j44, i.e., 4, which is the original position of the phasor shown inFigure 7.2(a).Summarizing, application of the operator j to any number rotates it 90° anticlockwiseon the Argand diagram, multiplying a number by j2 rotates it 180° anticlockwise,multiplying a number by j3 rotates it 270° anticlockwise and multiplication by j4 rotatesw ww. n e w n e s p r e s s .c o m
    • Complex Numbers 151 Figure 7.2: Significance of the j operatorit 360° anticlockwise, i.e., back to its original position. In each case, the phasor isunchanged in its magnitude.By similar reasoning, if a phasor is operated on by Ϫj then a phase shift of Ϫ90° (i.e.,clockwise direction) occurs, again without change of magnitude.In electrical circuits, 90° phase shifts occur between voltage and current with purecapacitors and inductors; this is the key as to why j notation is used so much in theanalysis of electrical networks. This is explained later in this chapter. w w w.ne w nespress.com
    • 152 Chapter 77.2 Operations Involving Cartesian Complex Numbers(a) Addition and subtraction (a ϩ jb) ϩ (c ϩ jd ) ϭ (a ϩ c) ϩ j (b ϩ d )and (a ϩ jb) Ϫ (c ϩ jd ) ϭ (a Ϫ c) ϩ j (b Ϫ d )Thus, (3 ϩ j 2) ϩ (2 Ϫ j 4) ϭ 3 ϩ j 2 ϩ 2 Ϫ j 4 ϭ 5 ؊ j 2and (3 ϩ j 2) Ϫ (2 Ϫ j 4) ϭ 3 ϩ j 2 Ϫ 2 ϩ j 4 ϭ 1 ؉ j 6(b) Multiplication(a ϩ jb)(c ϩ jd ) ϭ ac ϩ a(jd ) ϩ (jb)c ϩ (jb)(jd ) ϭ ac ϩ jad ϩ jbc ϩ j 2 bdBut j2 ϭ Ϫ1, thus,(a ϩ jb)(c ϩ jd ) ϭ (ac Ϫ bd ) ϩ j (ad ϩ bc)For example,(3 ϩ j 2)(2 Ϫ j 4) ϭ 6 Ϫ j12 ϩ j 4 Ϫ j 2 8 ϭ (6 Ϫ (Ϫ1)8) ϩ j (Ϫ12 ϩ 4) ϭ 14 ϩ j(Ϫ 8) ϭ 14 ؊ j8 8(c) Complex conjugateThe complex conjugate of (a ϩ jb) is (a Ϫ jb). For example, the conjugate of (3 Ϫ j2) is(3 ϩ j2). The product of a complex number and its complex conjugate is always a realnumber, and this is an important property used when dividing complex numbers. Thus,(a ϩ jb)(a Ϫ jb) ϭ a 2 Ϫ jab ϩ jab Ϫ j 2 b2 ϭ a 2 Ϫ (Ϫb2 ) ϭ a 2 ϩ b2 (i.e., a real number) uFor example, (1 ϩ j 2)(1 Ϫ j 2) ϭ 12 ϩ 22 ϭ 5and (3 Ϫ j 4)(3 ϩ j 4) ϭ 32 ϩ 42 ϭ 25w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 153(d) DivisionThe expression of one complex number divided by another, in the form a ϩ jb, isaccomplished by multiplying the numerator and denominator by the complex conjugateof the denominator. This has the effect of making the denominator a real number. Forexample,2 ϩ j4 2 ϩ j4 3 ϩ j4 6 ϩ j8 ϩ j12 ϩ j 2 16 ϭ ϫ ϭ3 Ϫ j4 3 Ϫ j4 3 ϩ j4 32 ϩ 42 6 ϩ j8 ϩ j12 Ϫ 16 ϭ 25 Ϫ10 ϩ j 20 ϭ 25 ؊10 20 ϭ ؉j or ؊ 0.4 ؉ j 0.8 25 25The elimination of the imaginary part of the denominator by multiplying both thenumerator and denominator by the conjugate of the denominator is often termedrationalizing.Example 7.1In an electrical circuit the total impedance ZT is given by: Z1Z 2ZT ϭ ϩ Z3 Z1 ϩ Z 2Determine ZT in (a ϩ jb) form, correct to two decimal places, when Z1 ϭ 5 Ϫ j3,Z2 ϭ 4 ϩ j7 and Z3 ϭ 3.9 Ϫ j6.7.Solution Z1Z 2 ϭ (5 Ϫ j 3)(4 ϩ j 7) ϭ 20 ϩ j 35 Ϫ j12 Ϫ j 2 21 ϭ 20 ϩ j 35 Ϫ j12 ϩ 21 ϭ 41 ϩ j 23Z1 ϩ Z 2 ϭ (5 Ϫ j 3) ϩ (4 ϩ j 7) ϭ 9 ϩ j 4 w w w.ne w nespress.com
    • 154 Chapter 7 Z1Z 2 41 ϩ j 23 (41 ϩ j 23)(9 Ϫ j 4)Hence, ϭ ϭ Z1 ϩ Z 2 9 ϩ j4 (9 ϩ j 4)(9 Ϫ j 4) 369 Ϫ j164 ϩ j 207 Ϫ j 2 92 ϭ 92 ϩ 4 2 369 Ϫ j164 ϩ j 207 ϩ 92 ϭ 97 461 ϩ j 43 ϭ ϭ 4.753 ϩ j 0.443 97 Z1Z 2Thus, ϩ Z 3 ϭ (4.753 ϩ j 0.443) ϩ (3.9 Ϫ j 6.7) Z1 ϩ Z 2 ϭ 8.65 ؊ j6.26, correct to two decimal places.Example 7.2Given Z1 ϭ 3 ϩ j4 and Z2 ϭ 2 Ϫ j5 determine in Cartesian form correct to three decimalplaces: 1 1 1 1 1(a ) ( b) (c ) ϩ (d ) Z1 Z2 Z1 Z 2 (1/Z1 ) ϩ (1/Z 2 )Solution 1 1 3 Ϫ j4 3 Ϫ j4(a) ϭ ϭ ϭ 2 Z1 3 ϩ j4 (3 ϩ j 4)(3 Ϫ j 4) 3 ϩ 42 3 Ϫ j4 3 4 ϭ ϭ Ϫj ϭ 0.120 ؊ j0.160 25 25 25 1 1 2 ϩ j5 2 ϩ j5 2 ϩ j5(b) ϭ ϭ ϭ 2 ϭ Z2 2 Ϫ j 5 (2 Ϫ j 5)(2 ϩ j 5) 2 ϩ5 2 29 2 5 ϭ ϩj ϭ 0.069 ؉ j0.172 29 29 1 1(c) ϩ ϭ (0.120 Ϫ j 0.160) ϩ (0.069 ϩ j 0.172) Z1 Z 2 ϭ 0.189 ؉ j 0.012w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 155 1 1(d) ϭ (1/Z1 ) ϩ (1/Z 2 ) 0.189 ϩ j 0.012 0.189 Ϫ j 0.012 ϭ (0.189 ϩ j 0.012)(0.189 Ϫ j 0.012) 0.189 Ϫ j 0.012 ϭ 0.1892 ϩ 0.0122 0.189 Ϫ j 0.012 ϭ 0.03587 0.189 j 0.012 ϭ Ϫ 0.03587 0.03587 ϭ 5.269 ؊ j 0.3357.3 Complex EquationsIf two complex numbers are equal, then their real parts are equal and their imaginaryparts are equal. Hence, if a ϩ jb ϭ c ϩ jd, then a ϭ c and b ϭ d. This is a usefulproperty, since equations having two unknown quantities can be solved from oneequation. Complex equations are used when deriving balance equations with AC bridges.Example 7.3Solve the following complex equations:(a) 3(a ϩ jb) ϭ 9 Ϫ j2(b) (2 ϩ j)(Ϫ2 ϩ j) ϭ x ϩ jy(c) (a Ϫ j2b) ϩ (b Ϫ j3a) ϭ 5 ϩ j2Solution(a) 3(a ϩ jb) ϭ 9 Ϫ j2. Thus, 3a ϩ j3b ϭ 9 Ϫ j2 Equating real parts gives: 3a ϭ 9, i.e., a ‫3 ؍‬ Equating imaginary parts gives: 3b ϭ Ϫ2, i.e., b ϭ ؊2 / 3 w w w.ne w nespress.com
    • 156 Chapter 7(b) (2 ϩ j)(Ϫ2 ϩ j) ϭ x ϩ jy Thus, Ϫ4 ϩ j2 Ϫ j2 ϩ j2 ϭ x ϩ jy Ϫ5 ϩ j 0 ϭ x ϩ jy Equating real and imaginary parts gives: x ‫ ,5؊ ؍‬y ‫0 ؍‬(c) (a Ϫ j2b) ϩ (b Ϫ j3a) ϭ 5 ϩ j2 Thus, (a ϩ b) ϩ j(Ϫ2b Ϫ 3a) ϭ 5 ϩ j2 Hence, aϩb ϭ 5 (7.1) and, Ϫ2b Ϫ 3a ϭ 2 (7.2) We have two simultaneous equations to solve. Multiplying equation (7.1) by 2 gives: 2 a ϩ 2b ϭ 10 (7.3) Adding equations (7.2) and (7.3) gives Ϫa ϭ 12, i.e., a ‫21؊ ؍‬ From equation (7.1), b ‫71 ؍‬Example 7.4An equation derived from an AC bridge network is given by: ⎡ 1 ⎤R1 R3 ϭ ( R2 ϩ j ωL2 ) ⎢ ⎥ ⎢ (1/R ) ϩ ( j ωC ) ⎥ ⎣ 4 ⎦R1, R3, R4 and C4 are known values. Determine expressions for R2 and L2 in terms of theknown components.SolutionMultiplying both sides of the equation by (1/R4 ϩ jωC4) gives: ( R1 R3 )(1/R4 ϩ j ωC4 ) ϭ R2 ϩ j ωL2i.e., R1 R3 /R4 ϩ jR1 R3ωC4 ϭ R2 ϩ j ωL2Equating the real parts gives: R2 ‫ ؍‬R1R3/R4w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 157Equating the imaginary parts gives:ωL2 ϭ R1R3ωC4, from which, L2 ‫ ؍‬R1R3C47.4 The Polar Form of a Complex NumberIn Figure 7.3(a), Z ϭ x ϩ jy ϭ r cos θ ϩ jr sin θ from trigonometry, ϭ r (cos θ ϩ j sin θ)This latter form is usually abbreviated to Z ‫ ؍‬r∠θ, and is called the polar form of acomplex number.r is called the modulus (or magnitude of Z) and is written as mod Z or ԽZ Խ. r is determinedfrom Pythagoras’s theorem on triangle OAZ:ԽZԽ ϭ r ϭ ͱ ( x 2 ϩ y 2 )The modulus is represented on the Argand diagram by the distance OZ. θ is called theargument (or amplitude) of Z and is written as arg Z. θ is also deduced from triangle OAZ:arg Z ϭ θ ϭ tanϪ1y/x.For example, the cartesian complex number (3 ϩ j4) is equal to r∠θ in polar form, wherer ϭ (32 ϩ 42 ) ϭ 5 and, 4θ ϭ tanϪ1 ϭ 53.13° 3 Figure 7.3: Polar form of complex numbers w w w.ne w nespress.com
    • 158 Chapter 7Hence, (3 ؉ j 4) ‫°31.35∠5 ؍‬Similarly, (Ϫ3 ϩ j4) is shown in Figure 7.3(b), 4where, r ϭ (32 ϩ 42 ) ϭ 5, θЈ ϭ tanϪ1 ϭ 53.13° 3and, θ ϭ 180° Ϫ 53.13° ϭ 126.87°Hence, (؊3 ؉ j4) ‫°78.621∠5 ؍‬7.5 Applying Complex Numbers to Series AC CircuitsSimple AC circuits may be analyzed by using phasor diagrams. However, when circuitsbecome more complicated, analysis is considerably simplified by using complex numbers.It is essential that the basic operations used with complex numbers, as outlined in thischapter thus far, are thoroughly understood before proceeding with AC circuit analysis.7.5.1 Series AC Circuits7.5.1.1 Pure ResistanceIn an AC circuit containing resistance R only (see Figure 7.4(a)), the current IR is in phasewith the applied voltage VR as shown in the phasor diagram of Figure 7.4(b). The phasordiagram may be superimposed on the Argand diagram as shown in Figure 7.4(c). Theimpedance Z of the circuit is given by: VR∠0°Zϭ ϭR I R∠0°7.5.1.2 Pure InductanceIn an AC circuit containing pure inductance L only (see Figure 7.5(a)), the current ILlags the applied voltage VL by 90° as shown in the phasor diagram of Figure 7.5(b). Thephasor diagram may be superimposed on the Argand diagram as shown in Figure 7.5(c).The impedance Z of the circuit is given by: VL ∠90° VLZϭ ϭ ∠90° ϭ X L ∠90° or jX L I L ∠0° ILw ww. n e w n e s p r e s s .c o m
    • Complex Numbers 159 Figure 7.4: (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagramwhere XL is the inductive reactance given by:X L ϭ ωL ϭ 2π fL ohmswhere f is the frequency in hertz and L is the inductance in henrys.7.5.1.3 Pure CapacitanceIn an AC circuit containing pure capacitance only (see Figure 7.5(a)), the current ICleads the applied voltage VC by 90° as shown in the phasor diagram of Figure 7.5(b). Thephasor diagram may be superimposed on the Argand diagram as shown in Figure 7.5(c).The impedance Z of the circuit is given by: VC ∠Ϫ90° VCZϭ ϭ ∠Ϫ90° ϭ XC ∠Ϫ90° or Ϫ jXC IC ∠0° IC w w w.ne w nespress.com
    • 160 Chapter 7 Figure 7.5: (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagramwhere XC is the capacitive reactance given by: 1 1XC ϭ ϭ ohms ωC 2πfCwhere C is the capacitance in farads.⎡ ⎤⎢ Note:ϪjXC ϭ Ϫ j ϭ Ϫ j ( j ) ϭ Ϫ j ϭ Ϫ(Ϫ1) ϭ 1 ⎥ 2⎢ ωC ωC ( j ) j ωC j ωC j ωC ⎥⎦⎣7.5.1.4 R–L Series CircuitIn an AC circuit containing resistance R and inductance L in series (see Figure 7.7(a)),the applied voltage V is the phasor sum of VR and VL as shown in the phasor diagramof Figure 7.7(b). The current I lags the applied voltage V by an angle lying between0° and 90°—the actual value depending on the values of VR and VL, which depend on thevalues of R and L. The circuit phase angle, that is, the angle between the current andthe applied voltage, is shown as angle φ in the phasor diagram. In any series circuit thecurrent is common to all components and is taken as the reference phasor inFigure 7.7(b). The phasor diagram may be superimposed on the Argand diagram asw ww. n e w n e s p r e s s .c o m
    • Complex Numbers 161 Figure 7.6: (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram Figure 7.7: (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagramshown in Figure 7.7(c), where it may be seen that in complex form the supply voltage Vis given by:V ϭ VR ϩ jVLFigure 7.8(a) shows the voltage triangle that is derived from the phasor diagram ofFigure 7.8(b) (triangle Oab). If each side of the voltage triangle is divided by current I, w w w.ne w nespress.com
    • 162 Chapter 7 Figure 7.8: (a) Voltage triangle; (b) Impedance triangle; (c) Argand diagramthen the impedance triangle of Figure 7.8(b) is derived. The impedance triangle may besuperimposed on the Argand diagram, as shown in Figure 7.8(c), where it may be seenthat in complex form the impedance Z is given by:Z ϭ R ϩ jX LFor example, an impedance expressed as (3 ϩ j4) Ω means that the resistance is 3 Ω andthe inductive reactance is 4 Ω.In polar form, Z ϭ ԽZ Խ ∠φ where, from the impedance triangle, the modulus ofimpedance ԽZԽ ϭ √( R 2 ϩ X L ) and the circuit phase angle φ ϭ tanϪ1 (XL /R) lagging. 27.5.1.5 R-C Series CircuitIn an AC circuit containing resistance R and capacitance C in series (see Figure 7.9(a)),the applied voltage V is the phasor sum of VR and VC as shown in the phasor diagramof Figure 7.9(b). The current I leads the applied voltage V by an angle lying between 0°and 90°—the actual value depending on the values of VR and VC, which depend on thevalues of R and C. The circuit phase angle is shown as angle φ in the phasor diagram. Thephasor diagram may be superimposed on the Argand diagram as shown in Figure 7.9(c),where it may be seen that in complex form the supply voltage V is given by:V ϭ VR Ϫ jVCw ww. n e w n e s p r e s s .c o m
    • Complex Numbers 163 Figure 7.9: (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagram Figure 7.10: (a) Voltage triangle; (b) Impedance triangle; (c) Argand diagramFigure 7.10(a) shows the voltage triangle that is derived from the phasor diagram ofFigure 7.10(b). If each side of the voltage triangle is divided by current I, the impedancetriangle is derived as shown in Figure 7.10(b). The impedance triangle may besuperimposed on the Argand diagram as shown in Figure 7.10(c), where it may be seenthat in complex form the impedance Z is given by:Z ϭ R Ϫ jXCThus, for example, an impedance expressed as (9 Ϫ j14) Ω means that the resistance is9 Ω and the capacitive reactance XC is 14 Ω. w w w.ne w nespress.com
    • 164 Chapter 7In polar form, Z ϭ ԽZ Խ∠φ where, from the impedance triangle, angle, ԽZԽ ϭ √( R 2 ϩ XC ) 2and φ ϭ tanϪ1 (XC /R) leading.7.5.1.6 R-L-C Series CircuitIn an AC circuit containing resistance R, inductance L and capacitance C in series (seeFigure 7.10(a)), the applied voltage V is the phasor sum of VR, VL and VC as shown in thephasor diagram of Figure 7.10(b) (where the condition VL Ͼ VC is shown). The phasordiagram may be superimposed on the Argand diagram as shown in Figure 7.10(c), whereit may be seen that in complex form the supply voltage V is given by:V ϭ VR ϩ j (VL Ϫ VC )From the voltage triangle the impedance triangle is derived and superimposing this on theArgand diagram gives, in complex form,Impedance Z ϭ R ϩ j ( X L Ϫ XC ) or Z ϭԽZԽ∠φwhere,ԽZԽ ϭ ͱ [ R 2 ϩ ( X L Ϫ XC )2 ] and φ ϭ tanϪ1 ( X L Ϫ XC ) /RWhen VL ϭ VC, XL ϭ XC and the applied voltage V and the current I are in phase. Thiseffect is called series resonance. Figure 7.11: (a) Circuit diagram; (b) Phasor diagram; (c) Argand diagramw ww. n e w n e s p r e s s .c o m
    • Complex Numbers 1657.5.1.7 General Series CircuitIn an AC circuit containing several impedances connected in series, say, Z1, Z2, Z3, … ,Zn, then the total equivalent impedance ZT is given by:ZT ϭ Z1 ϩ Z 2 ϩ Z 3 ϩ ϩ ZnExample 7.5Determine the values of the resistance and the series-connected inductance or capacitancefor each of the following impedances: (a) (12 ϩ j5) Ω; (b) Ϫj40 Ω; (c) 30∠60° Ω;(d) 2.20 ϫ 106∠Ϫ30° Ω. Assume for each a frequency of 50 Hz.Solution(a) From Section 24.2(d), for an R–L series circuit, impedance Z ϭ R ϩ jXL. Thus, Z ϭ (12 ϩ j5) Ω represents a resistance of 12 Ω and an inductive reactance of 5 Ω in series. Since inductive reactance XL ϭ 2πfL, XL 5 Inductance L ϭ ϭ ϭ 0.0159 H 2πf 2π(50) So, the inductance is 15.9 mH. Thus, an impedance (12 ؉ j5) Ω represents a resistance of 12 Ω in series with an inductance of 15.9 mH.(b) For a purely capacitive circuit, impedance Z ϭ ϪjXC. Thus, Z ϭ Ϫj40 Ω represents zero resistance and a capacitive reactance of 40 Ω. Since capacitive reactance XC ϭ 1/(2πfC), 1 1 Capacitance C ϭ ϭ F 2πfXC 2π(50)(40) 106 ϭ μF ϭ 79.6 μF 2π(50)(40) Thus, an impedance ؊j40 Ω represents a pure capacitor of capacitance 79.6 μF. w w w.ne w nespress.com
    • 166 Chapter 7(c) 30∠60° ϭ 30(cos 60° ϩ j sin 60°) ϭ 15 ϩ j25.98 Thus, Z ϭ 30∠60° Ω ϭ (15 ϩ j25.98) Ω represents a resistance of 15 Ω and an inductive reactance of 25.98 Ω in series. Since XL ϭ 2πfL, XL 25.98 Inductance L ϭ ϭ 2πf 2π(50) ϭ 0.0827 H or 82.7 mH Thus, an impedance 30∠60° Ω represents a resistance of 15 Ω in series with an inductance of 82.7 mH.(d) 2.20 ϫ 106 ∠Ϫ30° ϭ 2.20 ϫ 106 [cos(Ϫ30°) ϩ j sin(Ϫ30°)] ϭ 1.905 ϫ 106 Ϫ j1.10 ϫ 106 Thus, Z ϭ 2.20 ϫ 106 ∠Ϫ30° Ω ϭ (1.905 ϫ 106 Ϫ j1.10 ϫ 106 ) Ω represents a resistance of 1.905 ϫ 106 Ω (i.e., 1.905 MΩ) and a capacitive reactance of 1.10 ϫ 106 Ω in series. Since capacitive reactance XC ϭ 1/(2πfC), 1 1 Capacitance C ϭ ϭ F 2πfXC 2π(50)(1.10 ϫ 106 ) ϭ 2.894 ϫ 10Ϫ9 F or 2.894 nF Thus, an impedance 2.2 ؋ 106∠؊30° Ω represents a resistance of 1.905 MΩ in series with a 2.894 nF capacitor.Example 7.6Determine, in polar and rectangular forms, the current flowing in an inductor ofnegligible resistance and inductance 159.2 mH when it is connected to a 250 V, 50 Hzsupply.w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 167SolutionInductive reactanceX L ϭ 2π fL ϭ 2π(50)(159.2 ϫ 10Ϫ3 ) ϭ 50 ΩThus, circuit impedance Z ϭ (0 ϩ j50) Ω ϭ 50∠90° ΩSupply voltage, V ϭ 250∠0° V (or (250 ϩ j 0)V)(Note that since the voltage is given as 250 V, this is assumed to mean 250∠0° V or(250 ϩ j0)V.) V 250∠0° 250Hence, current I ϭ ϭ ϭ ∠(0° Ϫ 90°) Z 50∠90° 50 ‫°09؊∠5؍‬A V (250 ϩ j 0) 250(Ϫ j 50)Alternatively, I ϭ ϭ ϭ Z (0 ϩ j 50) j 50(Ϫ j 50) Ϫ j (50)(250) ϭ ϭ ؊ j5A A 502which is the same as 5∠؊90°AExample 7.7A 3-μF capacitor is connected to a supply of frequency 1 kHz and a current of2.83∠90°A flows. Determine the value of the supply voltage.Solution 1 1Capacitive reactance XC ϭ ϭ 2πfC 2π(1000)(3 ϫ 10Ϫ6 ) ϭ 53.05 ΩHence, circuit impedance Z ϭ (0 Ϫ j 53.05) Ω ϭ 53.05∠Ϫ90° Ω Current I ϭ 2.83∠90° A (or (0 ϩ j2.83)A)Supply voltage, V ϭ IZ ϭ (2.83∠90°)(53.05∠Ϫ90°)i.e., voltage ‫ °0∠051 ؍‬V w w w.ne w nespress.com
    • 168 Chapter 7Alternatively, V ϭ IZ ϭ (0 ϩ j 2.83)(0 Ϫ j 53.05) ϭ Ϫ j 2 (2.83)(53.05) ϭ 150 VExample 7.8The impedance of an electrical circuit is (30 Ϫ j50) ohms. Determine (a) the resistance,(b) the capacitance, (c) the modulus of the impedance, and (d) the current flowing and itsphase angle, when the circuit is connected to a 240 V, 50 Hz supply.Solution(a) Since impedance Z ϭ (30 Ϫ j50) Ω, the resistance is 30 ohms and the capacitive reactance is 50 Ω.(b) Since XC ϭ 1/(2πfC), capacitance, 1 1 Cϭ ϭ ϭ 63.66 μF 2πfX c 2π(50)(50)(c) The modulus of impedance, |Z | ϭ ͱ ( R 2 ϩ XC ) ϭ ͱ (302 ϩ 502 ) 2 ϭ 58.31 Ω XC(d) Impedance (30 Ϫ j 50) Ω ϭ 58.31∠tanϪ1 R ϭ 58.31∠Ϫ59.04° Ω V 240∠0° Hence, current I ϭ ϭ Z 58.31∠Ϫ59.04° ϭ 4.12∠59.04° AExample 7.9A 200 V, 50 Hz supply is connected across a coil of negligible resistance and inductance0.15 H connected in series with a 32 Ω resistor. Determine (a) the impedance of thecircuit, (b) the current and circuit phase angle, (c) the voltage across the 32 Ω resistor, and(d) the voltage across the coil.w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 169Solution(a) Inductive reactance XL ϭ 2πfL ϭ 2π(50)(0.15) ϭ 47.1 Ω Impedance Z ϭ R ϩ jXL ϭ (32 ϩ j 47.1) Ω or 57.0∠55.81° Ω The circuit diagram is shown in Figure 7.12. V 200∠0°(b) Current I ϭ ϭ ϭ 3.51∠Ϫ55.81° A Z 57.0∠55.81° i.e., the current is 3.51A lagging the voltage by 55.81°(c) Voltage across the 32 Ω resistor, VR ϭ IR ϭ (3.51∠Ϫ55.81°)(32∠0°) i.e., VR ‫ °18.55؊∠3.211 ؍‬V(d) Voltage across the coil, VL ϭ IX L ϭ (3.51∠Ϫ55.81°)(47.1∠90°) i.e., VL ‫ °91.43∠3.561 ؍‬VThe phasor sum of VR and VL is the supply voltage V as shown in the phasor diagram ofFigure 7.13. Figure 7.12: Circuit diagram for Example 7.9 w w w.ne w nespress.com
    • 170 Chapter 7 Figure 7.13: Phasor diagram for Example 7.9VR ϭ 112.3∠Ϫ55.81° ϭ (63.11 Ϫ j 92.89) VVL ϭ 165.3∠34.19° V ϭ (136.73 ϩ j 92.89) VHence,V ϭ VR ϩ VL ϭ (63.11 Ϫ j 92.89) ϩ (136.73 ϩ j 92.89) ϭ (200 ϩ j 0) V or 200∠0° V, correct to three significant figures.Example 7.10Determine the value of impedance if a current of (7 ϩ j16)A flows in a circuit when thesupply voltage is (120 ϩ j200)V. If the frequency of the supply is 5 MHz, determine thevalue of the components forming the series circuit.Solution V (120 ϩ j 200) 233.24∠59.04°Impedance Z ϭ ϭ ϭ I (7 ϩ j16) 17.464∠66.37° ϭ 13.36∠Ϫ7.33 Ω or (13.25 Ϫ j1.705) ΩThe series circuit consists of a 13.25 Ω resistor and a capacitor of capacitive reactance1.705 Ω. 1Since XC ϭ 2πfCw ww. n e w n e s p r e s s .c o m
    • Complex Numbers 171 1Capacitance C ϭ 2πfXC 1 ϭ 2π(5 ϫ 106 )(1.705) ϭ 1.867 ϫ 10Ϫ8 F ϭ 18.67 nF7.6 Applying Complex Numbers to Parallel AC CircuitsAs with series circuits, parallel networks may be analyzed by using phasor diagrams. However,with parallel networks containing more than two branches, this can become very complicated.It is with parallel AC network analysis in particular that the full benefit of using complexnumbers may be appreciated. The theory for parallel AC networks introduced previously isrelevant; more advanced networks will be analyzed in this chapter using j notation. Beforeanalyzing such networks admittance, conductance and susceptance are defined.7.6.1 Admittance, Conductance and SusceptanceAdmittance is defined as the current I flowing in an AC circuit divided by the supplyvoltage V (i.e., it is the reciprocal of impedance Z). The symbol for admittance is Y.Thus, I 1Yϭ ϭ V ZThe unit of admittance is the siemen, S.An impedance may be resolved into a real part R and an imaginary part X, givingZ ϭ R Ϯ jX. Similarly, an admittance may be resolved into two parts—the real part beingcalled the conductance G, and the imaginary part being called the susceptance B—andexpressed in complex form. Thus, admittance,Y ϭ G Ϯ jBWhen an AC circuit contains:(a) pure resistance, then, 1 1 Z ϭ R and Y ϭ ϭ ϭG Z R w w w.ne w nespress.com
    • 172 Chapter 7(b) pure inductance, then, 1 1 Ϫj Z ϭ jX L and Y ϭ ϭ ϭ Z jX L ( jX L )(Ϫ j ) Ϫj ϭ ϭ Ϫ jBL XL thus, a negative sign is associated with inductive susceptance, BL.(c) pure capacitance, then, 1 1 j Z ϭ Ϫ jXC and Y ϭ ϭ ϭ Z Ϫ jXC (Ϫ jXC )( j ) j ϭ ϭ ϩ jBC XC thus, a positive sign is associated with capacitive susceptance, BC(d) resistance and inductance in series, then, 1 1 Z ϭ R ϩ jX L and Y ϭ ϭ Z R ϩ jX L ( R Ϫ jX L ) ϭ 2 R ϩ XL 2 R X R X i.e., Y ϭ Ϫj 2 L 2 or Y ϭ Ϫj L R2 ϩ XL 2 R ϩ XL ԽZԽ2 ԽZԽ2 Thus, conductance, G ϭ R/ԽZ Խ2 and inductive susceptance, BL ϭ −XL/ԽZ Խ2 (Note that in an inductive circuit, the imaginary term of the impedance, XL, is positive, whereas the imaginary term of the admittance, BL, is negative.)(e) resistance and capacitance in series, then, 1 1 R ϩ jXC Z ϭ R Ϫ jXC and Y ϭ ϭ ϭ 2 Z R Ϫ jXC R ϩ XC2w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 173 R X i.e., Y ϭ ϩj 2 C 2 or R 2 ϩ XC 2 R ϩ XC R X Yϭ 2ϩj C ԽZԽ ԽZԽ2 Thus, conductance, G ϭ R/ԽZ Խ2 and capacitive susceptance, BC ϭ XC/ԽZ Խ2 (Note that in a capacitive circuit, the imaginary term of the impedance, XC, is negative, whereas the imaginary term of the admittance, BC, is positive.)(f) resistance and inductance in parallel, then, 1 1 1 jX L ϩ R ϭ ϩ ϭ Z R jX L ( R )( jX L ) ( R )( jX L ) ⎛ ⎞ ⎜ i.e., product ⎟ from which, Z ϭ ⎜ ⎟ R ϩ jX L ⎜ ⎝ sum ⎟ ⎠ 1 R ϩ jX L R jX L and, Y ϭ ϭ ϭ ϩ Z jRX L jRX L jRX L 1 1 (Ϫ j ) 1 i.e., Y ϭ ϩ ϭ ϩ jX L R ( jX L )(Ϫ j ) R 1 j or, Y ϭ Ϫ R XL Thus, conductance, G ‫/1 ؍‬R and inductive susceptance, BL ‫/1؊ ؍‬XL.(g) resistance and capacitance in parallel, then, ( R )(Ϫ jXC ) ⎛ ⎞ ⎜ i.e., product ⎟ Zϭ ⎜ ⎟ ⎟ R Ϫ jXC ⎝ ⎜ sum ⎠ 1 R Ϫ jXC R jXC and Y ϭ ϭ ϭ Ϫ Z Ϫ jRXC Ϫ jRXC Ϫ jRXC w w w.ne w nespress.com
    • 174 Chapter 7 1 1 ( j) 1 i.e., Y ϭ ϩ ϭ ϩ Ϫ jXC R (Ϫ jXC )( j ) R 1 j or, Y ϭ ϩ (7.1) R XC Thus, conductance, G ‫/1 ؍‬R and capacitive susceptance, BC ‫ ؍‬l/XC The conclusions that may be drawn from sections (d) to (g) above are: (i) that a series circuit is more easily represented by an impedance, (ii) that a parallel circuit is often more easily represented by an admittance especially when more than two parallel impedances are involved.Example 7.11Determine the admittance, conductance and susceptance of the following impedances:(a) Ϫj5 Ω (b) (25 ϩ j40) Ω (c) (3 Ϫ j2) Ω (d) 50∠40° Ω.Solution(a) If impedance Z ϭ Ϫj5 Ω, then, 1 1 j j admittance Y ϭ ϭ ϭ ϭ Z Ϫ j 5 (Ϫ j 5)( j ) 5 ϭ j 0.2 S or 0.2∠90° S Since there is no real part, conductance, G ‫ ,0 ؍‬and capacitive susceptance, BC ‫ 2.0 ؍‬S.(b) If impedance Z ϭ (25 ϩ j40) Ω then, 1 1 25 Ϫ j 40 Admittance Y ϭ ϭ ϭ 2 Z (25 ϩ j 40) 25 ϩ 402 25 j 40 ϭ Ϫ ϭ (0.0112 ؊ j 0.0180) S 2225 2225 Thus, conductance, G ‫ 2110.0 ؍‬S and inductive susceptance, BL ‫ 0810.0 ؍‬S.w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 175(c) If impedance Z ϭ (3 Ϫ j2) Ω, then, 1 1 3 ϩ j2 admittance Y ϭ ϭ ϭ 2 Z (3 Ϫ j 2) 3 ϩ 22 ⎛3 2⎞ ϭ ⎜ ϩ j ⎟ S or ⎜ ⎟ (0.231 ؉ j0.154) S ⎜ 13 ⎝ 13 ⎟ ⎠ Thus, conductance, G ‫ 132.0 ؍‬S and capacitive susceptance, BC ‫ 451.0 ؍‬S(d) If impedance Z ϭ 50∠40° Ω, then, 1 1 1∠0° admittance Y ϭ ϭ ϭ Z 50∠40° 50∠40° 1 ϭ ∠Ϫ40° ϭ 0.02∠؊40° S or 50 (0.0153 ؊ j0.0129) S Thus, conductance, G ‫ 3510.0 ؍‬S and inductive susceptance, BL ‫ 9210.0 ؍‬S.Example 7.12Determine expressions for the impedance of the following admittances: (a) 0.004∠30° S(b) (0.001 Ϫ j0.002) S (c) (0.05 ϩ j 0.08) S.Solution(a) Since admittance Y ϭ 1/Z, impedance Z ϭ 1/Y. 1 1∠0° Hence, impedance Z ϭ ϭ 0.004∠30° 0.004∠30° ϭ 250∠؊30° Ω or (216.5 ؊ j125) Ω 1(b) Impedance Z ϭ (0.001 Ϫ j 0.002) 0.001 ϩ j 0.002 ϭ (0.001)2 ϩ (0.002)2 0.001 ϩ j 0.002 ϭ 0.000 005 ϭ (200 ؉ j400) Ω or 447.2 ∠63.43° Ω w w w.ne w nespress.com
    • 176 Chapter 7(c) Admittance Y ϭ (0.05 ϩ j0.08) S ϭ 0.094∠57.99° S 1 Hence, impedance Z ϭ 0.0094∠57.99° ϭ 10.64∠؊57.99° Ω or (5.64 Ϫ j9.02) ΩExample 7.13The admittance of a circuit is (0.040 ϩ j0.025) S. Determine the values of the resistanceand the capacitive reactance of the circuit if they are connected (a) in parallel, (b) inseries. Draw the phasor diagram for each of the circuits.Solution(a) Parallel connection Admittance Y ϭ (0.040 ϩ j0.025) S, therefore conductance, G ϭ 0.040 S and capacitive susceptance, BC ϭ 0.025 S. From equation (7.1) when a circuit consists of resistance R and capacitive reactance in parallel, then Y ϭ (1/R) ϩ (j/XC). 1 1 Hence, resistance R ϭ ϭ ϭ 25 Ω G 0.040 1 1 and capacitive reactance XC ϭ ϭ ϭ 40 Ω BC 0.025 The circuit and phasor diagrams are shown in Figure 7.14.(b) Series connection Admittance Y ϭ (0.040 ϩ j0.025) S, therefore, 1 1 Impedance Z ϭ ϭ Y 0.040 ϩ j 0.025 0.040 Ϫ j 0.025 ϭ (0.040)2 ϩ (0.025)2 ϭ (17.98 Ϫ j11.24) ΩThus, the resistance, R ‫ 89.71 ؍‬Ω and capacitive reactance, XC ‫ 42.11 ؍‬Ω.w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 177 Figure 7.14: (a) Circuit diagram; (b) Phasor diagram Figure 7.15: (a) Circuit diagram; (b) Phasor diagramThe circuit and phasor diagrams are shown in Figure 7.15.The circuits shown in Figures 7.14(a) and 7.15(a) are equivalent in that they take the samesupply current I for a given supply voltage V; the phase angle φ between the current andvoltage is the same in each of the phasor diagrams shown in Figures 7.14(b) and 7.15(b).7.6.2 Parallel AC NetworksFigure 7.16 shows a circuit diagram containing three impedances, Z1, Z2 and Z3connected in parallel. The potential difference across each impedance is the same, i.e.,the supply voltage V. Current I1 ϭ V/ Z1, I2 ϭ V/ Z2 and I3 ϭ V/ Z3. If ZT is the totalequivalent impedance of the circuit then I ϭ V/ ZT . The supply current, I ϭ I1 ϩ I2 ϩ I3(phasorially). w w w.ne w nespress.com
    • 178 Chapter 7 Figure 7.16: Circuit with three impedances in parallel V V V VThus, ϭ ϩ ϩ and, ZT Z1 Z 2 Z 3 1 1 1 1 ϭ ϩ ϩZT Z1 Z 2 Z 3or total admittance, YT ϭ Y1 ϩ Y2 ϩ Y3In general, for n impedances connected in parallel,YT ϭ Y1 ϩ Y2 ϩ Y3 ϩ ϩ Yn (phasorially)It is in parallel circuit analysis that the use of admittance has its greatest advantage.7.6.2.1 Current Division in AC CircuitsFor the special case of two impedances, Z1 and Z2, connected in parallel (see Figure 7.17), 1 1 1 Z ϩ Z1 ϭ ϩ ϭ 2ZT Z1 Z 2 Z1Z 2The total impedance, ZT ‫ ؍‬Z1Z2/(Z1 ϩ Z2) (i.e., product/sum).From Figure 7.17, ⎛ ZZ ⎞supply voltage, V ϭ IZT ϭ I ⎜ 1 2 ⎟ ⎜ ⎟ ⎜Z ϩZ ⎟ ⎜ 1 ⎝ ⎟ 2⎠Also, V ϭ I1Z1 (and V ϭ I2Z2)w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 179 Figure 7.17: Two impedances connected in parallel ⎛ ZZ ⎞Thus, I1Z1 ϭ I ⎜ 1 2 ⎟ ⎜ ⎟ ⎜Z ϩZ ⎟ ⎜ 1 ⎝ ⎟ 2⎠ ⎛ Z ⎞ ⎟i.e., current I1 ϭ I ⎜ ⎜ 2 ⎜Z ϩZ ⎟ ⎟ ⎜ 1 ⎝ ⎟ 2⎠ ⎛ Z ⎞ ⎟Similarly, current I 2 ϭ I ⎜ ⎜ 1 ⎜Z ϩZ ⎟ ⎟ ⎜ 1 ⎝ ⎟ 2⎠Note that all of the above circuit symbols infer complex quantities either in Cartesian orpolar form.The following problems show how complex numbers are used to analyze parallel AC networks.Example 7.14Determine the values of currents I, I1 and I2 shown in the network of Figure 7.18.SolutionTotal circuit impedance, (8)( j 6) ( j 48)(8 Ϫ j 6)ZT ϭ 5 ϩ ϭ 5ϩ 8 ϩ j6 82 ϩ 6 2 j 384 ϩ 288 ϭ 5ϩ 100 ϭ (7.88 ϩ j 3.84) Ω or 8.77∠25.98° Ω w w w.ne w nespress.com
    • 180 Chapter 7 Figure 7.18: Network for Example 7.14 V 50∠0°Current I ϭ ϭ ϭ 5.70∠؊25.98° A ZT 8.77∠25.98° ⎛ j6 ⎞ ⎟Current I1 ϭ I ⎜ ⎜ ⎟ ⎜ 8 ϩ j6 ⎟ ⎝ ⎟ ⎠ ⎛ 6∠90° ⎞ ⎟ ϭ (5.70∠Ϫ25.98°) ⎜ ⎜ ⎟ ⎝ 10∠36.87° ⎟ ⎜ ⎠ ϭ 3.42∠27.15° A ⎛ 8 ⎞ ⎟Current I 2 ϭ I ⎜ ⎜ ⎟ ⎜ 8 ϩ j6 ⎟ ⎝ ⎟ ⎠ ⎛ 8∠0° ⎞ ⎟ ϭ (5.70∠Ϫ25.98°) ⎜ ⎜ ⎟ ⎜ 10∠36.87° ⎟ ⎝ ⎠ ϭ 4.56∠؊62.85° A[ Note: I ϭ I1 ϩ I 2 ϭ 3.42∠27.15° ϩ 4.56 ∠Ϫ62.85° ϭ (3.043 ϩ j1.561) ϩ (2.081 Ϫ j 4.058) ϭ (5.124 Ϫ j 2.497)A ϭ 5.70 ∠Ϫ25.98° A]Example 7.15For the parallel network shown in Figure 7.19, determine the value of supply current Iand its phase relative to the 40 V supply.w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 181 Figure 7.19: Parallel network for Example 7.15SolutionImpedance Z1 ϭ (5 ϩ j12) Ω, Z2 ϭ (3 Ϫ j4) Ω and Z3 ϭ 8 Ω Supply current V Iϭ ϭ VYT where ZT ϭ total circuit impedance, and YT ϭ total circuit admittance. ZT YT ϭ Y1 ϩ Y2 ϩ Y3 1 1 1 1 1 1 ϭ ϩ ϩ ϭ ϩ ϩ Z1 Z 2 Z 3 (5 ϩ j12) (3 Ϫ j 4) 8 5 Ϫ j12 3 ϩ j4 1 ϭ 2 ϩ 12 2 ϩ 2 ϩ 5 3 ϩ4 2 8 ϭ (0.0296 Ϫ j 0.0710) ϩ (0.1200 ϩ j 0.1600) ϩ (0.1250)i.e., YT ϭ (0.2746 ϩ j0.0890) S or 0.2887∠17.96° SCurrent I ϭ VYT ϭ (40∠0°)(0.2887∠17.96°) ϭ 11.55∠17.96° AHence, the current I is 11.55A and is leading the 40 V supply by 17.96°.Alternatively, current I ϭ I1 ϩ I2 ϩ I3 40∠0° 40∠0°Current I1 ϭ ϭ 5 ϩ j12 13∠67.38° ϭ 3.077∠Ϫ67.38° A or (1.183 Ϫ j 2.840) A w w w.ne w nespress.com
    • 182 Chapter 7 40∠0° 40∠0°Current I 2 ϭ ϭ ϭ 8∠53.13°A 3 Ϫ j4 5∠Ϫ53.13° or (4.80 ϩ j 6.40) A 40∠0°Current I 3 ϭ ϭ 5∠0° A or (5 ϩ j 0) A 8∠0°Thus, current I ϭ I1 ϩ I2 ϩ I3 ϭ (1.183 Ϫ j 2.840) ϩ (4.80 ϩ j 6.40) ϩ (5 ϩ j 0) ϭ 10.983 ϩ j 3.560 ϭ 11.55∠17.96° A, as previously obtained.Example 7.16An AC network consists of a coil, of inductance 79.58 mH and resistance 18 Ω, in parallelwith a capacitor of capacitance 64.96 μF. If the supply voltage is 250∠0°V at 50 Hz,determine (a) the total equivalent circuit impedance, (b) the supply current, (c) the circuitphase angle, (d) the current in the coil, and (e) the current in the capacitor.SolutionThe circuit diagram is shown in Figure 7.20.Inductive reactance, X L ϭ 2πfL ϭ 2π(50)(79.58 ϫ 10Ϫ3 ) ϭ 25 ΩHence, the impedance of the coil,Z COIL ϭ ( R ϩ jX L ) ϭ (18 ϩ j 25) Ω or 30.81∠54.25° Ω Figure 7.20: Circuit diagram for Example 7.16w ww. n e w n e s p r e s s .c o m
    • Complex Numbers 183 1Capacitive reactance, XC ϭ 2πfC 1 ϭ 2π(50)(64.96 ϫ 10Ϫ6 ) ϭ 49 ΩIn complex form, the impedance presented by the capacitor ZC is ϪjXC, i.e., Ϫj49 Ω or49∠Ϫ90° Ω.(a) Total equivalent circuit impedance, Z COIL XC ⎛ ⎞ ZT ϭ ⎜ i.e., product ⎟ ⎜ ⎟ Z COIL ϩ ZC ⎜ ⎝ sum ⎠ ⎟ (30.81∠54.25°)(49∠ Ϫ90°) ϭ (18 ϩ j 25) ϩ (Ϫ j 49) (30.81∠54.25°)(49∠ Ϫ90°) ϭ 18 Ϫ j 24 (30.81∠54.25°)(49∠ Ϫ90°) ϭ 30∠ Ϫ 53.13° ϭ 50.32∠(54.25° Ϫ 90° Ϫ (Ϫ53.13°)) ϭ 50.32∠17.38° or (48.02 ؉ j15.03) Ω V 250∠0°(b) Supply current I ϭ ϭ ZT 50.32∠17.38° ϭ 4.97∠؊17.38° A(c) Circuit phase angle ϭ 17.38° lagging, i.e., the current I lags the voltage V by 17.38°. V 250∠0°(d) Current in the coil, I COIL ϭ ϭ Z COIL 30.81∠54.25° ϭ 8.11∠؊54.25° A V 250∠0°(e) Current in the capacitor, IC ϭ ϭ ZC 49∠Ϫ90° ϭ 5.10∠90° A w w w.ne w nespress.com
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    • CHAPTE R 8 Transients and Laplace Transforms John Bird8.1 IntroductionA transient state will exist in a circuit containing one or more energy storage elements (i.e.,capacitors and inductors) whenever the energy conditions in the circuit change, until the newsteady state condition is reached. Transients are caused by changing the applied voltage orcurrent, or by changing any of the circuit elements; such changes occur due to opening andclosing switches. In this chapter, such equations are developed analytically by using bothdifferential equations and Laplace transforms for different waveform supply voltages.8.2 Response of R-C Series Circuit to a Step Input8.2.1 Charging a CapacitorA series R-C circuit is shown in Figure 8.1(a).A step voltage of magnitude V is shown in Figure 8.1(b). The capacitor in Figure 8.1(a) isassumed to be initially uncharged.From Kirchhoff’s voltage law, supply voltage,V ϭ vC ϩ vR (8.1) dvc dvCVoltage vR ϭ iR and current i ϭ C , so, vR ϭ CR dt dtTherefore, from equation (8.1) dvCV ϭ vC ϩ CR (8.2) dt w w w.ne w nespress.com
    • 186 Chapter 8 C R V VC VR i Switch 0 t V (a) (b) Figure 8.1: (a) Series R-C circuit; (b) Step voltage of magnitude VThis is a linear, constant coefficient, first order differential equation. Such a differentialequation may be solved (find an expression for voltage vC) by separating the variables.Rearranging equation (8.2) gives: dvC V Ϫ vC ϭ CR dt dvC V Ϫ vCand ϭ dt CR dvC dtfrom which, ϭ V Ϫ vc CR dvC dtand integrating both sides gives ∫ V Ϫ vC ϭ∫ CR tHence, Ϫln(V Ϫ vC ) ϭ ϩk (8.3) CRwhere k is the arbitrary constant of integration. dvC(To integrate ∫ V Ϫ vC make an algebraic substitution, u ϭ V Ϫ vC —see EngineeringMathematics or Higher Engineering Mathematics, J.O. Bird, 2004, 4th edition, Elsevier.)When time t ϭ 0, ␯C ϭ 0, hence, Ϫln V ϭ k. tThus, from equation (8.3), Ϫln(V Ϫ vC ) ϭ Ϫ lnV CRw ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 187 vC V vc ϭ V (1ϪeϪt/CR) 0 t Figure 8.2: Exponential growth curve of Equation 8.4Rearranging gives: tln V Ϫ ln(V Ϫ vC ) ϭ CR V t ln ϭ by the laws of logarithms V Ϫ vC CR Vi.e., ϭ e t/CR V Ϫ vC V Ϫ vC 1and ϭ t/CR ϭ eϪt/CR V e V Ϫ vC ϭ VeϪt/CR V Ϫ VeϪt/CR ϭ vCi.e., capacitor voltage, vc ϭ V (1 Ϫ eϪt/CR ) (8.4)This is an exponential growth curve, as shown in Figure 8.2.From equation (8.1),vR ϭ V Ϫ vC ϭ V Ϫ [V (1 Ϫ eϪt/CR )] from equation (8.4) ϭ V Ϫ V ϩ VeϪt/CRi.e., resistor voltage, vR ϭ Ve؊t/CR (8.5) w w w.ne w nespress.com
    • 188 Chapter 8 vR V vR ϭVeϪt/CR 0 t Figure 8.3: Exponential decay curve of Equation 8.5This is an exponential decay curve, as shown in Figure 8.3. dvCIn the circuit of Figure 8.1(a), current i ϭ C dt dHence, i ϭ C [V (1 Ϫ eϪt/CR )] from equation (8.4) dt di.e., iϭC [V Ϫ VeϪt/CR ] dt ⎡ ⎛ Ϫ1 ⎞ Ϫt/CR ⎤ ϭ C ⎢ 0 Ϫ (V ) ⎜ ⎟e ⎥ ⎢ ⎜ CR ⎟ ⎜ ⎝ ⎟ ⎠ ⎥ ⎣ ⎦ ⎡ V Ϫt/CR ⎤ ϭC⎢ e ⎥ ⎢⎣ CR ⎥⎦ V Ϫt/CRSo, current, i ϭ e (8.6) R Vwhere is the steady state current, I. RThis is an exponential decay curve as shown in Figure 8.4.After a period of time, it can be determined from equations (8.4) to (8.6) that the voltageacross the capacitor, vC, attains the value V, the supply voltage, while the resistor voltage,vR, and current i both decay to zero.Example 8.1A 500 nF capacitor is connected in series with a 100 kΩ resistor and the circuit isconnected to a 50 V, DC supply. Calculate (a) the initial value of current flowing,w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 189 i V — V R i ϭ — eϪt/CR R 0 t Figure 8.4: Exponential decay curve of Equation 8.6(b) the value of current 150 ms after connection, (c) the value of capacitor voltage 80 msafter connection, and (d) the time after connection when the resistor voltage is 35 V.Solution V(a) From equation (8.6), current, i ϭ eϪt/CR R Initial current, i.e., when t ϭ 0, V 0 V 50 i0 ϭ e ϭ ϭ ‫ 5.0 ؍‬mA R R 100 ϫ 103 V Ϫt/CR(b) Current, i ϭ e so, when time t ϭ 150 ms or 0.15 s, R 50 Ϫ9 iϭ eϪ0.5 / ( 500ϫ10 )(100ϫ10 ) 3 100 ϫ 10 3 ϭ (0.5 ϫ 10Ϫ3 )eϪ3 ϭ (0.5 ϫ 10Ϫ3 )(0.049787) ϭ 0.0249 mA or 24.9 μA(c) From equation (8.4), capacitor voltage, vC ϭ V(1 Ϫ eϪt/CR) When time t ϭ 80 ms, Ϫ3 / ( 500ϫ10Ϫ3 ϫ100ϫ103 ) vC ϭ 50(1 Ϫ eϪ80ϫ10 ) ϭ 50(1 Ϫ eϪ1.6 ) ϭ 50(0.7981) ϭ 39.91 V w w w.ne w nespress.com
    • 190 Chapter 8(d) From equation (8.5), resistor voltage, ␯R ϭ VeϪt/CR When ␯R ϭ 35V, Ϫ9 35 ϭ 50eϪt/( 500ϫ10 ϫ100ϫ10 3) then 35 i.e., ϭ eϪt/ 0.05 50 35 Ϫt and ln ϭ 50 0.05 from which, time t ϭ Ϫ0.05 ln 0.7 ϭ 0.0178s or 17.8 ms8.2.2 Discharging a CapacitorIf after a period of time the step input voltage V applied to the circuit of Figure 8.1 issuddenly removed, by opening the switch, thenfrom equation (8.1), vR ϩ vC ϭ 0 dvCor, from equation (8.2), CR ϩ vC ϭ 0 dt dvC Ϫ1Rearranging gives: ϭ vC dt CRand separating the variables gives: dvC ϭ Ϫ dt vC CR dvC dtand integrating both sides gives: ∫ vC ϭ∫Ϫ CR tfrom which, ln vC ϭ Ϫ ϩk (8.7) CRwhere k is a constant.At time t ϭ 0 (i.e., at the instant of opening the switch), vC ϭ VSubstituting t ϭ 0 and vC ϭ V in equation (8.7) gives:ln V ϭ 0 ϩ kw ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 191Substituting k ϭ ln V into equation (8.7) gives: t ln vC ϭ Ϫ ϩ lnV CR tand ln vC Ϫ lnV ϭ Ϫ CR v t ln C ϭ Ϫ V CR vCand ϭ eϪt/CR Vfrom which, vC ϭ VeϪt/CR (8.8)That is, the capacitor voltage, vC , decays to zero after a period of time, the rate of decaydepending on CR, which is the time constant, τ. Since vR ϩ vC ϭ 0 then the magnitude ofthe resistor voltage, vR, is given by:vR ϭ VeϪt/CR (8.9) dvC dand since i ϭ C ϭ C (VeϪt/CR ) dt dt ⎛ 1 ⎞ Ϫt/CR ⎟e ϭ (CV ) ⎜Ϫ ⎜ CR ⎟ ⎜ ⎝ ⎟ ⎠i.e., the magnitude of the current, V Ϫt/CRiϭ e (8.10) RExample 8.2A DC voltage supply of 200 V is connected across a 5 μF capacitor as shown inFigure 8.5. When the supply is suddenly cut by opening switch S, the capacitor is leftisolated except for a parallel resistor of 2 MΩ. Calculate the voltage across the capacitorafter 20 s. w w w.ne w nespress.com
    • 192 Chapter 8 S ϩ 2 M⍀ 200 V 5 μF Ϫ Figure 8.5: Circuit for Example 8.2 L R vL vR i Switch V Figure 8.6: Series R-L circuitSolutionFrom equation (8.8), vC ϭ VeϪt/CR Ϫ6 ϫ2ϫ106 )After 20 s, vC ϭ 200eϪ20 /( 5ϫ10 ϭ 200eϪ2 ϭ 200(0.13534) ϭ 27.07 V8.3 Response of R-L Series Circuit to a Step Input8.3.1 Current GrowthA series R-L circuit is shown in Figure 8.6. When the switch is closed and a step voltageV is applied, it is assumed that L carries no current.From Kirchhoff’s voltage law, V ϭ vL ϩ vRw ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 193 diVoltage vL ϭ L and voltage vR ϭ iR dt diHence, V ϭ L ϩ iR (8.11) dtThis is a linear, constant coefficient, first order differential equation.Again, such a differential equation may be solved by separating the variables. di V Ϫ iRRearranging equation (8.11) gives: ϭ dt L di dtfrom which, ϭ V Ϫ iR L di dtand ∫ V Ϫ iR ϭ∫ L 1 tHence, Ϫ ln(V Ϫ iR ) ϭ ϩ k (8.12) R Lwhere k is a constant. di(Use the algebraic substitution u ϭ V Ϫ iR to integrate ∫ ) V Ϫ iR 1At time t ϭ 0, i ϭ 0, thus, Ϫ ln V ϭ 0 ϩ k R 1Substituting k ϭ Ϫ ln V in equation (8.12) gives: R 1 t 1Ϫ ln(V Ϫ iR ) ϭ Ϫ lnV R L R 1 tRearranging gives: [ ln V Ϫ ln(V Ϫ iR )] ϭ R L ⎛ V ⎞ Rt ⎟ϭand ln ⎜ ⎜ ⎟ ⎝ V Ϫ iR ⎟ ⎜ ⎠ L w w w.ne w nespress.com
    • 194 Chapter 8 i V — R V ( i ϭ — 1Ϫ eϪRt/L R ) 0 t Figure 8.7: Exponential growth curve of Equation 8.13 VHence, ϭ e Rt/L V Ϫ iR V Ϫ iR 1and ϭ Rt/L ϭ eϪRt/L V e V Ϫ iR ϭ VeϪRt/L V Ϫ VeϪRt/L ϭ iR Vand current, iϭ (1 Ϫ eϪRt/L ) (8.13) RThis is an exponential growth curve as shown in Figure 8.7.The voltage across the resistor in Figure 8.6, vR ϭ iR ⎡V ⎤Hence, vR ϭ R ⎢ (1 Ϫ eϪRt/L ) ⎥ from equation (8.13) ⎢⎣ R ⎥⎦i.e., VR ϭ V (1 Ϫ eϪRt/L ) (8.14)which again represents an exponential growth curve. diThe voltage across the inductor in Figure 8.6, vL ϭ L dtw ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 195 ⎡ ⎤i.e., vL ϭ L d ⎢ V (1 Ϫ eϪRt/L ) ⎥ ϭ LV d [1 Ϫ eϪRt/L ] dt ⎢⎣ R ⎥⎦ R dt LV ⎡⎢ ⎛ R⎞ ⎤ LV ⎛ R ϪRt/L ⎞ ϭ 0 Ϫ ⎜Ϫ ⎟ eϪRt/L ⎥ ϭ ⎜ ⎟ ⎟ ⎜ e ⎜ ⎟ ⎟ ⎟ R ⎢⎣ ⎜ ⎝ L⎠ ⎥ ⎜ R ⎝L ⎠ ⎦i.e., vL ϭ VeϪRt/L (8.15)Example 8.3A coil of inductance 50 mH and resistance 5 Ω is connected to a 110 V, DC supply.Determine (a) the final value of current, (b) the value of current after 4 ms, (c) the value ofthe voltage across the resistor after 6 ms, (d) the value of the voltage across the inductanceafter 6 ms, and (e) the time when the current reaches 15 A.Solution(a) From equation (8.13), when t is large, the final, or steady state current i is given by: V 110 iϭ ϭ ϭ 22A R 5 V(b) From equation (8.13), current, i ϭ (1 Ϫ eϪRt/L ) R 110 Ϫ3 Ϫ3 When t ϭ 4 ms, i ϭ (1 Ϫ e(Ϫ( 5)( 4ϫ10 ) / 50ϫ10 ) ) 5 ϭ 22(1 Ϫ eϪ0.40 ) ϭ 22(0.32968) ϭ 7.25 V(c) From equation (8.14), the voltage across the resistor, vR ϭ V (1 Ϫ eϪRt/L ) When t ϭ 6 ms, vR ϭ 110(1 Ϫ e(Ϫ( 5)(6ϫ10Ϫ3 ) / 50ϫ10Ϫ3 ) ) ϭ 110(1 Ϫ eϪ0.60 ) ϭ 110(0.45119) ϭ 49.63 V w w w.ne w nespress.com
    • 196 Chapter 8(d) From equation (8.15), the voltage across the inductance, vL ϭ VeϪRt/L When t ϭ 6 ms, Ϫ3 Ϫ3 vL ϭ 110e(Ϫ( 5)(6ϫ10 ) / 50ϫ10 ) ϭ 110eϪ0.60 ϭ 60.37 V (Note that at t ϭ 6 ms, vL ϩ vR ϭ 60.37 ϩ 49.63 ϭ 110V ϭ supply voltage, V.)(e) When current i reaches 15A, V 15 ϭ (1 Ϫ eϪRt/L ) from equation (8.13) R 110 Ϫ3 i.e., 15 ϭ (1 Ϫ eϪ5t/ ( 50ϫ10 ) ) 5 ⎛ 5 ⎞ ⎟ ϭ 1 Ϫ eϪ100 t 15 ⎜ ⎜ ⎟ ⎝ 110 ⎟ ⎜ ⎠ 75 and eϪ100 t ϭ 1 Ϫ 110 ⎛ 75 ⎞ ⎟ Hence, Ϫ100t ϭ ln ⎜1 Ϫ ⎜ ⎟ ⎜ ⎝ 110 ⎟ ⎠ 1 ⎛ 75 ⎞ ⎟ and time, t ϭ ln ⎜1 Ϫ ⎜ ⎟ Ϫ100 ⎝ ⎜ 100 ⎟ ⎠ ϭ 0.01145 s or 11.45 ms8.3.2 Current DecayIf after a period of time the step voltage V applied to the circuit of Figure 8.6 is suddenlyremoved by opening the switch, then from equation (8.11), di 0ϭL ϩ iR dt di di iRRearranging gives: L ϭ ϪiR or ϭ Ϫ dt dt Lw ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 197 di RSeparating the variables gives: ϭ Ϫ dt i Land integrating both sides gives: di R∫ i ϭ∫ Ϫ dt L R ln i ϭ Ϫ tϩk (8.16) LAt t ϭ 0 (i.e., when the switch is opened), ⎛ V ⎞i ϭ I ⎜ϭ , the steady state current ⎟ ⎜ ⎟ ⎟ ⎜ R ⎝ ⎠then ln I ϭ 0 ϩ kSubstituting k ϭ ln I into equation (8.16) gives: R ln i ϭ Ϫ t ϩ lnI L RRearranging gives: ln i Ϫ lnI ϭ Ϫ t L i R ln ϭ Ϫ t I L i ϭ eϪRt/L I V ϪRt/Land current, i ϭ IeϪRt/L or e (8.17) Ri.e., the current i decays exponentially to zero. ⎛V ⎞From Figure 8.6, vR ϭ iR ϭ R ⎜ eϪRt/L ⎟ from equation (8.17) ⎜ ⎟ ⎟ ⎜R ⎝ ⎠So, vR ϭ VeϪRt/L (8.18) w w w.ne w nespress.com
    • 198 Chapter 8 5A S 10 ⍀ 2H V Figure 8.8: Circuit for Example 8.4 di d ⎛V ⎞The voltage across the coil, vL ϭ L ϭ L ⎜ eϪRt/L ⎟ from equation (8.17) ⎜ ⎟ ⎟ dt dt ⎜ R ⎝ ⎠ ⎛ V ⎞⎛ R ⎞ ϭ L ⎜ ⎟ ⎜Ϫ ⎟ eϪRt/L ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ R ⎠⎝ L ⎠The magnitude of vT is given by:vL ϭ VeϪRt/L (8.19)Both vR and vL decay exponentially to zero.Example 8.4In the circuit shown in Figure. 8.8, a current of 5 A flows from the supply source. SwitchS is then opened. Determine (a) the time for the current in the 2 H inductor to fall to200 mA and (b) the maximum voltage appearing across the resistor.Solution(a) When the supply is cut off, the circuit consists of just the 10 Ω resistor and the 2 H coil in parallel. This is effectively the same circuit as Figure 8.6 with the supply voltage zero. V ϪRt/L From equation (8.17), current i ϭ e R V In this case ϭ 5A, the initial value of current. R When i ϭ 200 mA or 0.2 A,w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 199 i L R vL vR V vC C Figure 8.9: L-R-C circuit 0.2 ϭ 5eϪ10 t/ 2 0.2 i.e., ϭ eϪ5t 5 0.2 thus, ln ϭ Ϫ5t 5 1 0.2 and time, t ϭ Ϫ ln ϭ 0.644 s or 644 ms 5 5(b) Since the current through the coil can only return through the 10 Ω resistance, the voltage across the resistor is a maximum at the moment of disconnection, i.e., vRm ϭ IR ϭ (5)(10) ϭ 50V8.4 L-R-C Series Circuit ResponseL-R-C circuits are widely used in a variety of applications, such as in filters incommunication systems, ignition systems in automobiles, and defibrillator circuits inbiomedical applications (where an electric shock is used to stop the heart, in the hope thatthe heart will restart with rhythmic contractions).For the circuit shown in Figure 8.9, from Kirchhoff’s voltage law,Vϭ vL ϩ vR ϩ vC (8.20) di dvvL ϭ L and i ϭ C C , hence, dt dt d ⎛ dvC ⎞ 2 ⎟ ϭ LC d vCvL ϭ L ⎜ C⎜ ⎟ dt ⎜ dt ⎟ ⎝ ⎠ dt 2 ⎛ dv ⎞ dvvR ϭ iR ϭ ⎜C C ⎟ R ϭ RC C ⎜ ⎟ ⎟ ⎜ ⎝ dt ⎠ dt w w w.ne w nespress.com
    • 200 Chapter 8Hence, from equation (8.20): d 2 vC dvV ϭ LC 2 ϩ RC C ϩ vC (8.21) dt dtThis is a linear, constant coefficient, second order differential equation. (For the solutionof second order differential equations, see Higher Engineering Mathematics).To determine the transient response, the supply voltage, V, is made equal to zero, d 2 vC dvi.e., LC 2 ϩ RC C ϩ vC ϭ 0 (8.22) dt dtA solution can be found by letting vC ϭ Aemt, from which,dvC dv ϭ Ame mt and C ϭ Am 2 e mt dt dt 2Substituting these expressions into equation (8.22) gives:LC ( Am 2 e mt ) ϩ RC ( Ame mt ) ϩ Ae mt ϭ 0i.e., Ae mt (m 2 LC ϩ mRC ϩ 1) ϭ 0Thus, vC ϭ Aemt is a solution of the given equation provided thatm 2 LC ϩ mRC ϩ 1 ϭ 0 (8.23)This is called the auxiliary equation.Using the quadratic formula on equation (8.23) gives: ϪRC Ϯͱ [( RC )2 Ϫ 4( LC )(1)] mϭ 2 LC ϪRC Ϯͱ ( R 2C 2 Ϫ 4 LC ) ϭ 2 LCw ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 201 ϪRC R 2C 2 Ϫ 4 LCi.e., mϭ 2 LC Ϯ Ί (2 LC )2 ⎛ R 2C 2 4 LC ⎞ Ί ⎜ 2 2 Ϫ 2 2⎟ R ⎜ ϭϪ Ϯ ⎟ 2L ⎜ 4L C ⎝ 4L C ⎟⎟ ⎠ ⎡⎛ R ⎞2 ⎤ Ί ⎢⎜ ⎟ Ϫ 1 ⎥ R ϭϪ Ϯ ⎟ ⎢⎜ 2 L ⎟ (8.24) 2L ⎜ ⎢⎣⎝ ⎠ LC ⎥⎥ ⎦This equation may have either: (i) two different real roots, when (R/2L)2 Ͼ (1/LC), when the circuit is said to be overdamped since the transient voltage decays very slowly with time, or,(ii) two real equal roots, when (R/2L)2 ϭ (1/LC), when the circuit is said to be critically damped since the transient voltage decays in the minimum amount of time without oscillations occurring, or,(iii) two complex roots, when (R/2L)2 Ͻ (1/LC), when the circuit is said to be underdamped since the transient voltage oscillates about the final steady state value, the oscillations eventually dying away to give the steady state value, or,(iv) if R ϭ 0 in equation (8.24), the oscillations would continue indefinitely without any reduction in amplitude—this is the undamped condition.Damping is discussed again in Section 8.8.Example 8.5A series L-R-C circuit has inductance L ϭ 2 mH, resistance R ϭ 1 k Ω and capacitance,C ϭ 5 μF. (a) Determine whether the circuit is over, critical or underdamped. (b) If C ϭ 5 nF,determine the state of damping.Solution ⎛ R ⎞2 ⎡ 103 ⎤ 2 1012 ⎜ ⎟ ϭ⎢(a) ⎜ ⎟ ⎥ ϭ ϭ 6.25 ϫ 1010 ⎝ 2L ⎟ ⎜ ⎠ ⎢ 2(2 ϫ 10Ϫ3 ) ⎥ ⎣ ⎦ 16 1 1 109 ϭ ϭ ϭ 108 LC (2 ϫ 10Ϫ3 )(5 ϫ 106 ) 10 w w w.ne w nespress.com
    • 202 Chapter 8 ⎛ R ⎞2 1 Since, ⎜ ⎟ Ͼ ⎜ ⎟ the circuit is overdamped. ⎝ 2L ⎟ ⎜ ⎠ LC 1 1(b) When C ϭ 5 nF, ϭ Ϫ3 )(5 ϫ 10Ϫ9 ) ϭ 1011 LC (2 ϫ 10 ⎛ R ⎞2 1 Since, ⎜ ⎟ Ͻ ⎜ ⎟ ⎟ the circuit is underdamped. ⎜ ⎝ 2L ⎠ LCExample 8.6In the circuit of Example 8.5, what value of capacitance will give critical damping?Solution ⎛ ⎞2For critical damping: ⎜ R ⎟ ϭ 1 ⎜ ⎟ ⎜ 2L ⎟ ⎝ ⎠ LCfrom which, capacitance, 1 1 4 L2 4LCϭ ϭ ϭ ϭ 2 ⎛ R ⎞2 ⎟ R2 LR 2 R L⎜ L 2 ⎜ 2L ⎟ ⎜ ⎝ ⎟ ⎠ 4L 4(2 ϫ 10Ϫ3 ) ϭ ϭ 8 ϫ 10Ϫ9 F or 8 nF (103 )28.4.1 Roots of the Auxiliary EquationWith reference to equation (8.24): (i) when the roots are real and different, say m ϭ α and m ϭ β, the general solution is: vC ϭ Aeαt ϩ Beβ t (8.25) ⎡⎛ R ⎞2 ⎤ Ί ⎢⎜ ⎟ Ϫ 1 ⎥ R where, α ϭ Ϫ ϩ ⎢⎜ 2 L ⎟ 2L ⎝ ⎟ ⎢⎣⎜ ⎠ LC ⎥⎥ ⎦ ⎡⎛ R ⎞2 ⎤ Ί ⎢⎜ ⎟ Ϫ 1 ⎥ R β ϭϪ Ϫ ⎢⎜ 2 L ⎟⎟ 2L ⎜ ⎢⎣⎝ ⎠ LC ⎥⎥ ⎦w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 203(ii) when the roots are real and equal, say m ϭ α twice, the general solution is vC ϭ ( At ϩ B)eα t (8.26) R where α ϭ Ϫ 2L(iii) when the roots are complex, say m ϭ α Ϯ jβ, the general solution is vC ϭ eαt {A cos βt ϩ B sin βt} (8.27) ⎡ 1 ⎛ R ⎞2 ⎤⎥ Ί ⎢ R ⎜ ⎟ where α ϭ Ϫ and β ϭ ⎢ LC Ϫ ⎜ 2 L ⎟ ⎥ (8.28) 2L ⎢⎣ ⎜ ⎟ ⎥ ⎝ ⎠ ⎦To determine the actual expression for the voltage under any given initial condition, itis necessary to evaluate constants A and B in terms of vC and current i. The procedure isthe same for each of the above three cases. Assuming in, say, case (iii) that at time t ϭ 0,vC ϭ v0 and i(ϭC(dvC /dt)) ϭ i0 then substituting in equation (8.27): v0 ϭ e0 {A cos 0 ϩ B sin 0}i.e., v0 ϭ A (8.29)Also, from equation (8.27),dvC ϭ eαt [ϪAβ sin βt ϩ Bβ cos βt ] ϩ [ A cos βt ϩ B sin βt ](αeαt ) (8.30) dtby the product rule of differentiation. dvCWhen t ϭ 0, ϭ e0 [0 ϩ Bβ] ϩ [ A](αe0 ) ϭ Bβ ϩ αA dt dvCHence, at t ϭ 0, i0 ϭ C ϭ C ( Bβ ϩ αA) dtFrom equation (8.29), A ϭ v0 hence i0 ϭ C(Bβ ϩ αv0) ϭ CBβ ϩ Cαv0 i0 ؊ C αv 0from which, B ‫؍‬ (8.31) Cβ w w w.ne w nespress.com
    • 204 Chapter 8Example 8.7A coil has an equivalent circuit of inductance 1.5 H in series with resistance 90 Ω. It isconnected across a charged 5 μF capacitor at the moment when the capacitor voltage is10 V. Determine the nature of the response and obtain an expression for the current inthe coil.Solution 2⎛ R ⎞2 ⎡ 90 ⎤ 1 1⎜ ⎟ ϭ⎢⎜ ⎟ ⎥ ϭ 900 and ϭ⎜ 2L ⎟⎝ ⎠ ⎢⎣ 2(1.5) ⎥⎦ LC (1.5)(5 ϫ 10Ϫ6 ) ϭ 1.333 ϫ 105 ⎛ R ⎞2 1Since ⎜ ⎟ Ͻ ⎜ ⎟ the circuit is underdamped. ⎜ ⎟ ⎝ 2L ⎠ LCFrom equation (8.28), R 90 α ϭϪ ϭϪ ϭ Ϫ30 2L 2(1.5) Ί ⎡ 1 ⎛ R ⎞2 ⎤⎥and β ϭ ⎢ ⎜ ⎟ ⎢ LC Ϫ ⎜ 2 L ⎟ ⎥ ⎜ ⎟ ⎥ ⎝ ⎠ ⎢⎣ ⎦ ϭͱ [1.333 ϫ 105 Ϫ 900] ϭ 363.9With v0 ϭ 10 V and i0 ϭ 0, from equation (8.29), v0 ϭ A ϭ 10and from equation (8.31), i0 Ϫ Cαv0 0 Ϫ (5 ϫ 10Ϫ6 )(Ϫ30)(10)Bϭ ϭ Cβ (5 ϫ 10Ϫ6 )(363.9) 300 ϭ ϭ 0.8244 363.9 9 dvCCurrent, i ϭ C , and from equation (8.30), dt i ϭ C{eϪ30 t [Ϫ10(363.9) sin βt ϩ (0.8244)(363.9) cos βt ]w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 205 ϩ (10 cos βt ϩ 0.8244 sin βt )(Ϫ30eϪ30 t )} ϭ C{eϪ30 t [Ϫ3639 sin βt ϩ 300 cos βt Ϫ 300 cos βt Ϫ 24.732 sin βt ]} ϭ CeϪ30 t [Ϫ3663.732 sin βt ] ϭ Ϫ (5 ϫ 10Ϫ6 )(3663.732)eϪ30 t sin βti.e., current, i ‫ 810.0 ؊ ؍‬e؊30t sin 363.9t amperes8.5 Introduction to Laplace TransformsThe solution of most electrical problems can be reduced ultimately to the solutionof differential equations and the use of Laplace transforms provides an alternativemethod to those used previously. Laplace transforms provide a convenient method forthe calculation of the complete response of a circuit. In this section and in Section 8.6,the technique of Laplace transforms is developed and then used to solve differentialequations. In Section 8.7, Laplace transforms are used to analyze transient responsesdirectly from circuit diagrams.8.5.1 Definition of a Laplace TransformThe Laplace transform of the function of time f (t) is defined by the integral ∞∫0 eϪst f (t ) dt where s is a parameterThere are various commonly used notations for the Laplace transform of f (t) and theseinclude ᏸ{ f (t)} or L{ f (t)} or ᏸ( f ) or Lf or f (s).Also the letter p is sometimes used instead of s as the parameter. The notation used in thischapter will be f (t) for the original function and ᏸ{f(t)}for its Laplace transform, ∞i.e., ᏸ{ f (t )} ϭ ∫ eϪst f (t ) dt (8.32) 08.5.2 Laplace Transforms of Elementary FunctionsUsing equation (8.32): ∞ ⎡ eϪst ⎤∞ (i) when f ( t ) ϭ 1, ᏸ{1} ϭ ∫0 eϪst (1) dt ϭ⎢ ⎥ ⎢ Ϫs ⎥ ⎣ ⎦0 w w w.ne w nespress.com
    • 206 Chapter 8 1 ϭ Ϫ [eϪs (∞) Ϫ e0 ] s 1 ϭ Ϫ [0 Ϫ 1] s 1 ϭ (provided s Ͼ 0) s ⎛1⎞ k(ii) when f ( t ) ϭ k, ᏸ{k} ϭ k ᏸ{1} ϭ k ⎜ ⎟ ϭ ⎜ ⎟ from (i) above ⎝s⎟ ⎜ ⎠ s ∞(iii) when f ( t ) ϭ e at , ᏸ{e at } ϭ ∫ eϪst (e at )dt 0 ϭ ∫ eϪ( sϪa )t dt from the laws of indices ⎡ eϪ( sϪa )t ⎤ ∞ ϭ⎢ ⎥ ⎢ Ϫ(s Ϫ a ) ⎥ ⎣ ⎦0 1 ϭ (0 Ϫ 1) Ϫ(s Ϫ a ) 1 ϭ (provided s Ͼ a) s؊ a ∞(iv) when f ( t ) ϭ t, ᏸ{t} ϭ ∫ eϪst t dt 0 ∞ ⎡ teϪst eϪst ⎤⎥ ϭ ⎢ Ϫ∫ dt ⎢ Ϫs Ϫs ⎥⎦ 0 ⎣ ⎡ teϪst eϪst ⎤ ∞ ϭ⎢ Ϫ 2⎥ by integration by parts ⎢ Ϫs Ϫs ⎥⎦ 0 ⎣ ⎡ ∞eϪs (∞) eϪs (∞) ⎤ ⎡ e0 ⎤ ϭ⎢ Ϫ ⎥ Ϫ ⎢0 Ϫ ⎥ ⎢ Ϫs s 2 ⎥⎦ ⎢⎣ s 2 ⎥⎦ ⎣w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 207 ⎛ 1⎞ ϭ (0 Ϫ 0) Ϫ ⎜ 0 Ϫ 2 ⎟ since (ϱ ϫ 0) ϭ 0 ⎜ ⎟ ⎜ ⎝ s ⎟⎠ 1 ϭ (provided s Ͼ 0) s2(v) when f(t) ‫ ؍‬cos ωt, ∞ ᏸ{cos ωt} ϭ ∫ eϪst cos ωt dt 0 ⎡ eϪst ⎤∞ ϭ⎢ 2 (ω sin ωt Ϫ s cos ωt ) ⎥ by integration by parts twice ⎢ s ϩ ω2 ⎥ ⎣ ⎦0 s (provided s Ͼ 0) ϭ 2 s ؉ ω2A list of standard Laplace transforms is summarized in Table 8.1 below. It will notusually be necessary to derive the transforms as above—but merely to use them.The following worked problems only require using the standard list of Table 8.1.Example 8.8Find the Laplace transforms of:(a) 1 ϩ 2t Ϫ 1 t 4 3(b) 5e2 t Ϫ 3eϪtSolution ⎧ ⎪ 1 ⎪ ⎫ 1(a) ᏸ ⎪1 ϩ 2t Ϫ t 4 ⎪ ϭ ᏸ{1} ϩ 2 ᏸ{t} Ϫ ᏸ{t 4 } ⎨ ⎬ ⎪ ⎪ ⎩ 3 ⎪ ⎪ ⎭ 3 1 ⎛ 1 ⎞ 1 ⎛ 4! ⎞ ϭ ϩ 2 ⎜ 2 ⎟ Ϫ ⎜ 4ϩ1 ⎟ from 2, 7 and 9 of Table 8.1 ⎜ ⎟ ⎜ ⎟ s ⎜ ⎟ ⎜ ⎝s ⎠ 3⎝s ⎠ ⎟ 1 2 1 ⎛ 4 ϫ 3 ϫ 2 ϫ1⎞ ⎟ ϭ ϩ 2Ϫ ⎜⎜ ⎟ ⎟ s s 3⎜⎝ s5 ⎠ 1 2 8 ϭ ؉ 2؊ 5 s s s w w w.ne w nespress.com
    • 208 Chapter 8 Table 8.1: Standard Laplace transforms ∞ Time function f(t) Laplace transform ᏸ{f(t)} ‫؍‬ ∫0 e؊st f ( t ) dt 1. δ (unit impulse) 1 2. 1 (unit step function) 1 s 3. eat (exponential function) 1 sϪa 4. unit step delayed by T eϪsT s 5. Sin ωt (sine wave) ω s2 ϩ ω2 s 6. cos ωt (cosine wave) s2 ϩ ω2 1 7. t (unit ramp function) s2 2! 8. t2 s3 n! 9. tn (n ϭ 1, 2, 3…) s nϩ1 s 10. cosh ωt s2 Ϫ ω2 ω 11. sinh ωt s2 Ϫ ω2 n! 12. eat tn (s Ϫ a)nϩ1 ω 13. eϪat sin ωt (damped sine wave) (s ϩ a)2 ϩ ω2 sϩa 14. eϪat cos ωt (damped cosine wave) (s ϩ a)2 ϩ ω2w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 209 Table 8.1: (Continued) ∞Time function f(t) Laplace transform ᏸ{f(t)} ‫؍‬ ∫0 e؊st f ( t ) dt ω15. eϪat sinh ωt (s ϩ a)2 Ϫ ω2 sϩa16. eϪat cosh ωt (s ϩ a)2 Ϫ ω2(b) ᏸ{5e2t Ϫ 3eϪt} ϭ 5ᏸ{e2t} Ϫ 3ᏸ{eϪt} ⎛ 1 ⎞ ⎛ 1 ⎞ ⎟ ϭ 5⎜ ⎟Ϫ 3⎜ ⎟ ⎜sϪ2⎟ ⎜ ⎝ ⎟ ⎠ ⎜ ⎜ s Ϫ (Ϫ1) ⎟ from 3 of Table 8.1 ⎝ ⎟ ⎠ 5 3 ϭ Ϫ s Ϫ 2 s ϩ1 5(s ϩ 1) Ϫ 3(s Ϫ 2) ϭ (s Ϫ 2)(s ϩ 1) 2 s ؉ 11 ϭ 2 s ؊ s؊ 2Example 8.9Find the Laplace transform of 6 sin 3t Ϫ 4 cos 5t.Solutionᏸ{6 sin 3t Ϫ 4 cos 5t} ϭ 6ᏸ{sin 3t} Ϫ 4ᏸ{cos 5t} ⎛ 3 ⎞ ⎟ Ϫ 4 ⎛ s ⎞ from 5 and 6 of Table 8.1 ⎟ ϭ 6⎜ 2 ⎜ ⎜ s ϩ 32 ⎟ ⎜ ⎝ ⎟ ⎠ ⎜ s ϩ 52 ⎟ ⎜ 2 ⎝ ⎟ ⎠ 18 4s ϭ ؊ 2 s2؉ 9 s ؉ 25Example 8.10Use Table 8.1 to determine the Laplace transforms of the following waveforms:(a) a step voltage of 10 V which starts at time t ϭ 0,(b) a step voltage of 10 V which starts at time t ϭ 5s, w w w.ne w nespress.com
    • 210 Chapter 8(c) a ramp voltage which starts at zero and increases at 4 V/s,(d) a ramp voltage which starts at time t ϭ 1 s and increases at 4 V/s.Solution(a) From 2 of Table 8.1, ⎛ 1 ⎞ 10 ᏸ{10} ϭ 10 ᏸ{ } ϭ 10 ⎜ ⎟ ϭ 1 ⎜ ⎟ ⎜s⎟ ⎝ ⎠ sThe waveform is shown in Figure 8.10(a).(b) From 4 of Table 8.1, a step function of 10 V which is delayed by t ϭ 5 s is given by: ⎛ eϪsT ⎞ ⎟ ϭ 10 ⎛ e ⎞ ϭ 10 e؊5 s Ϫ5 s ⎟ 10 ⎜ ⎜ ⎟ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ ⎝ s ⎟ ⎠ ⎜ ⎝ s ⎠ ⎟ s This is, in fact, the function starting at t ϭ 0 given in part (a), i.e., (10/s) multiplied by eϪsT, where T is the delay in seconds. The waveform is shown in Figure 8.10(b).(c) From 7 of Table 8.1, the Laplace transform of the unit ramp, ᏸ{t} ϭ (1/s2) Hence, the Laplace transform of a ramp voltage increasing at 4 V/s is given by: 4 4ᏸ{t} ϭ s2 The waveform is shown in Figure 8.10(c).(d) As with part (b), for a delayed function, the Laplace transform is the undelayed function, in this case (4/s2) from part (c), multiplied by eϪsT where T in this case is 1 s. The Laplace transform is given by: ⎛ 4 ⎞ ؊s ⎜ ⎟e ⎜ 2⎟ ⎜s ⎟ ⎝ ⎠ The waveform is shown in Figure 8.10(d).w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 211 V V 10 10 0 t 0 5 t (a) (b) V V 4 4 0 1 t 0 1 2 t (c) (d) Figure 8.10: Waveforms for Example 8.10Example 8.11Determine the Laplace transforms of the following waveforms:(a) an impulse voltage of 8 V, which starts at time t ϭ 0,(b) an impulse voltage of 8 V, which starts at time t ϭ 2 s,(c) a sinusoidal current of 4 A and angular frequency 5 rad/s which starts at time t ϭ 0.Solution(a) An impulse is an intense signal of very short duration. This function is often known as the Dirac function. From 1 of Table 8.1, the Laplace transform of an impulse starting at time t ϭ 0 is given by ᏸ{δ} ϭ 1, hence, an impulse of 8 V is given by: 8ᏸ{δ} ϭ 8. This is shown in Figure 8.11(a).(b) From part (a) the Laplace transform of an impulse of 8 V is 8. Delaying the impulse by 2 s involves multiplying the undelayed function by eϪsT where T ϭ 2 s. Hence, the Laplace transform of the function is given by: 8eϪ2s This is shown in Fig. 8.11(b). w w w.ne w nespress.com
    • 212 Chapter 8 V V 8 8 0 t 0 2 t (a) (b) i 4 0 ␲ 2␲ t — — 5 5 Ϫ4 (c) Figure 8.11: Graphs for Example 8.11 ω(c) From 5 of Table 8.1, ᏸ{sin ωt} ϭ s 2 ϩ ω2 When the amplitude is 4 A and ω ϭ 5, then ⎛ 5 ⎞ ⎟ ϭ 20 ᏸ{4 sin ωt} ϭ 4 ⎜ 2 ⎜ ⎟ 2⎟ ⎜ ⎝ s ϩ 5 ⎠ s2 ؉ 25 The waveform is shown in Figure 8.11(c).Example 8.12Find the Laplace transforms of:(a) 2t4e3t(b) 4e3t cos 5t.w ww. n e w n e s p r e s s .c o m
    • Transients and Laplace Transforms 213Solution(a) From 12 of Table 8.1, ᏸ{2t 4 e3t } ϭ 2 ᏸ{t 4 e3t } ⎡ 4! ⎤ ϭ2⎢ ⎥ ⎢⎣ (s Ϫ 3)4ϩ1 ⎥⎦ 2(4 ϫ 3 ϫ 2 ϫ 1) 48 ϭ ϭ (s Ϫ 3) 5 (s Ϫ 3)5(b) From 14 of Table 8.1, ᏸ{4e3t cos 5t} ϭ 4 ᏸ{e3t cos 5t} ⎡ sϪ3 ⎤ ϭ 4⎢ ⎥ ⎢⎣ (s Ϫ 3)2 ϩ 52 ⎥⎦ 4(s Ϫ 3) 4( s ؊ 3) ϭ ϭ 2 s2 Ϫ 6 s ϩ 9 ϩ 25) s ؊ 6 s ؉ 34Example 8.13Determine the Laplace transforms of:(a) 2 cosh 3t,(b) eϪ2t sin 3t.Solution(b) From 10 of Table 8.1, ⎡ s ⎤ 2s ᏸ{2 cosh 3t} ϭ 2 ᏸ cosh 3t ϭ 2 ⎢ 2 ⎥ϭ ⎢⎣ s Ϫ 32 ⎥⎦ s 2 ؊9(c) From 13 of Table 8.1, 3 3 ᏸ{eϪ2 t sin 3t} ϭ ϭ 2 ( s ϩ 2) 2 ϩ 32 s ϩ 4s ϩ 4 ϩ 9 3 ϭ 2 s ؉ 4 s ؉ 13 w w w.ne w nespress.com
    • 214 Chapter 88.5.3 Laplace Transforms of DerivativesUsing integration by parts, it may be shown that:(a) for the first derivative: ᏸ{ f ′(t )} ϭ s ᏸ{ f (t )} Ϫ f (0) ⎧ dy ⎫ ⎪ ⎪ or ᏸ⎪ ⎪ ‫ ؍‬sᏸ { y} ؊ y(0) ⎨ ⎬ (8.33) ⎪ dx ⎪ ⎪ ⎪ ⎩ ⎭ where y(0) is the value of y at x ϭ 0(b) for the second derivative: ᏸ{ f Љ(t )} ϭ s 2 ᏸ{ f (t )} Ϫ sf (0) Ϫ f Ј(0) ⎧ d2 y⎫ ⎪ ⎪ or ᏸ ⎪ 2 ⎪ ‫ ؍‬s2 ᏸ { y} ؊ sy(0) ؊ yЈ(0) ⎨ ⎬ (8.34) ⎪ dx ⎪ ⎪ ⎩ ⎪ ⎭where yЈ(0) is the value of (dy/dx) at x ϭ 0Equations (8.33) and (8.34) are used in the solution of differential equations inSection 8.6.8.5.4 The Initial and Final Value TheoremsThe initial and final value theorems can often considerably reduce the work of solvingelectrical circuits.(a) The initial value theorem states: limit [ f (t )] ϭ limit [ s ᏸ{ f (t )}] t →0 s→ϱ Thus, for example, if f (t) ϭ ␯ ϭ VeϪt/CR and if, say, V ϭ 10 and CR ϭ 0.5, then f (t) ϭ ␯ ϭ 10eϪ2t ⎛ 1 ⎞⎟ from 3 of Table 8.1 ᏸ{ f (t )} ϭ 10 ⎜ ⎜