01 JOHNSON-2e Electric Circuit Analysis PDF [PDF]

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Second Edition



ELECTRIC



CIRCUIT ANAI-YSIS David E. Johnson Birminghadl-southem College



Johnny R. Johnson University of North Alabama



John L. Hilburn President, lvlicrocompuler Systems Inc



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hentice Hall, Englewood Cliffs, New Jeftey 07632



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@ 1992, 1989 by Pretrtice-Ha , Itrc.



A Simon & Schust€r Company



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Edglewood Cliffs. New Jers€y 07632



All rights resereed. No pan of this book oay tre rcproduced, in any folm or by any means, without pemission in witingftorn the publisher.



10987654321 ISBN 0-I3-eq5335-7 Prcntice.Hall Intemational (tJK) Ijm;tei., L dm Prentice-Hall of Austnlia Pty. Iimited, ^SldEJ



hentice-Hall Canada Inc., ?ororro PrenticeHal l Hispanoamericatra. S.A., Mar,co Prentic€.qall of India hivaf, Llmited, New Delhi PrenticeHall of Japan. hc.- robo Simon & Schustor Asia Pt€ . LIA. , Singapore Editora Prentice-Hall do B.3sil, t,td^., Rio d. ,Iareiro Reprbted urder Authority of Prcsidential De.re€ No. 285 as anended by P-D. Nos. 400



& 1203



by



global Publishing. r,!i,!, ,biri!.i!.r tr, )6r. 30r-09]1 b 13 f! rb. !0r-09r9



ISBN 971_656-012_5



To the memory of our Mothers, Bessie Morris lohnson and Fannie Mae Page Hilburn



11 AC Steady-State Analysis 11.1 Nodal Analy$is



ll.2



345



34



Mesh Analysis 11.3 Network Theo.ema



351



11.4 Phasor Diagams 11.5 SPICE for AC Steady-Stare Circuits 11.6 Summary



3&



Problems



Conputer ApplicatioD Problemt



354 360 366



36



375



12 AC Steady-State Power 12.1 12.2 12.3 12.4 12.5 12.6 12.7



Average Power Superpositidn and Power



RMS Values Pow€r Factor Complex Power Power Measurement



376 377 384 388



390 394 398



Sllrnnury



400



Problems



,lO0



13 Three-Phase Circuits 13.1 Single-Phase, Tlree-Wile Sysreo! 13.2 Three-Phase Y-Y Systems



I3.3 The Delta ConDecriotr 13.4 Y-A Transformations



lJ.5



Powet M%surement



t3.6 SPICE for



Three-phas€ Circuit Analvsis



13.7 Summary Problems



Computer Application Problems



408 409 414 421 424 429 433 43s 436 439



IX



Contents Preface



x



l



1 lntroduction



1



1.1 Definitions and Units 1.2 Charge and Current 1.3 Voltage, Energy, atrd Power



2



1.4 Passive and Active Elements 1.5 Circuit Analysis



1.6 Summary' Problems



5 8



t2 t4 l5 15



') Resistive Circuits 2.1



Obm's Law



4,2 Kirchhoff's Laws 2.3 Series Resistance and Voltage Division 2.4 2.5 2.6 2.7



Parallel Resistance ard Current Divisiol Analysis Examples Ammeters, Voltmeters, and Ohflneters Physical Resistors



18 l9 24 31 36



2.E Summary



42 47 50 52



Problems



53



,



Jt Dependent Sources 3.1 Definitions 3.2 Ciicuits with Dependent Souices 3.3 Operational Amplifiers



3.4 Amplifier Circuits 3.5 Summary Prcblems



59 60 62 63



6



70 '11



4 Analysis .Methods 4.1



Nodal Analysis



4,2 An Example 4.3 Circuits Containing Voltage Sources 4.4 Circuits ContaininC Op AmPs .4.5 Mesh Analysis 4.6 Circuits Containilg Curent souces 4.7 Dudity 4.8 Computer-Aided Circuit Analysis Using SPICE 4.9 Summary



77 18 a2 84 89



9l 95 98 101



Problems



Computrr Application Problems



) Network Theorems 5.1 Linear Circuits



J.2



Superposition Th6vdnin's and Nonon's Theorems Practical Sources



5.3 5-4 5.5 Maximum Power Tmnsfer 5.6 sPIcE and Th6venin Equivalent Circuits



5.7



Summary



hoblems Computer Application Prcblems



.vt



Contenls



116 tt7



t2t I



135 140



t42 t44 t44



150



|



6 lndependence of Equations 6.1 6.2 6.3 6.4 6,5



6.6



Graph of a Network Trc€s and Cotre€s Independent Voltage Equations Independent Current Equations A Circuit Application Sutrlmary Problems



151 152 154



r56 160



t64 166 166



7 Energy-Storage Elements



7.1 7.2 7.3 7.4 7.5 7.5 7.7 7.E 7.9



Capacitors Energy SloIage in Capacitors Series and Parallel Capacitors



Inductors



Ene4y Storage in Inductors



169 170



174



t7't 180 184



DC Steady State



186 189



Practical Capacitori arld Induclorc



192



Duality and Linearity



193 195 199 199



Series and Parallel Inductors



7.10 Singqlar Circuits 7.11 Surnmary Problems



o () Simple RC and RL Circuits



t,l



207 208



4.4



Source-Free RC Circuit Time CoDslants Source-Free iZ Ciftuit Response to a Constant Forcing Function



t.5



The Ceneral Cas€



221 225



A Shortcut hocedur€ 8.7 The Unit Step Functiotr



228 232



4.2



t,3 E.6



212 216



vlt



8.6 The SteP Response 8-9 ApDlicarion of SuP€rposition



and $e Tmnsient Response Summary Problems ComPuter APPlication Problems



t.10 SiiCE 8.11



231 242



26 A9 249 251



o Second-Order Circuits 9.1



9.2 9.3 9,4 9.5



9.6 9.1 9.8 9.9



Circuits with Two Storage Elements Second-order Equations The Natural ResPonse Types of Natural Frequencies The Forced Response Excitatiol at a Natural FrequencY The Complete Response The Parallel RaC Circuit The Series RLC Circuit i.tt.rn"i""-ft{",ft"a" f"r Obtainiog fte Describing Equations nesponses o' Hi gber-order ci$uits



9.10 9.11 9.12 Surnflary



i"piii:i--rt-.".t Problems



Computer APplication hoblems



259



2fi 263 5 26i1



nl



n1 n1



281



286



2n' 294 29E 298 306



10 Sinusoidal Excitation and Phasors 10.1



ro.2



Properties of Sinusoids An Rl Circuit Example



308



3t3



10.4 10.5



ComPlexExcitatioN



314 318 321



Phasors



3U



l0-7



LDDedance



10.9



Phasor Chcuits



ii-i a" n-r"-u,i* i6,e



r-l"thod Using complex Numbers



Voftago-Cutt nt Retationships for Phasots



ad



Adtnrflance and lrnpedance combinatioos



i6.i r-i,liJ"ni. u*s 10.10 summary Problems



vnl



307



128



3t2 336 339 340



11 AC Steady-State Analysis 11.1 Nodal Analvsis



ll.2



Mesh Analvsis 11.3 Network Theorema 11.4 Phasor Diagrams SPICE for AC Steadv-State Circuits 11.6 Summary Problems Cooputer ApptcatioD Problems



ll.5



345



34 351



354 360



3&



36 36 375



12 AC Steady-State Power 12.1 Average Power 12.2 Superpositidn ard Power 12.3 RMS Values 12.4 Pow€r Factor 12.5 Complex Power 12.6 Power M€asurement



D.7



Slrmmarv Problems



376 377 384 388 390



394 398



400 ,to0



13 Three-Phase Circuits 13.1 Single-Phase, Tlree-Wile Sysien! 13.2 Three-Phase Y-Y Systems 13.3 The Delta ConDectiotr 13.4 Y-A Transformations 13.5 Power Measurement 13.6 SPICE for Three-Phas€ Circuit Analvsis 13.7 Surnmary Problems Computer Application Problems



408



w 414



42t 424 429 433 43s 436 439



IX



14 Complex Frequency and Network Functions 14.1 14.2 14.3 14.4



The Damped Sinusoid Complex Frequercy and Generalized Phasors Impedanc€ and Admittance Network FunctioDs



14-S



Poles



14.6 14,7



and



ros The Natural Response from the Network Function



NaturalFrequencies l4,E Two-Port Networks 14.9 Applications of Two'Port Parameters 14.10 Interconnections of Two-Port Netwo*s



l4.ll



Summary Prcblems



440 441 444 441.



451 455



45'l 460 462 471 47.6



482 442



15 Frequency Response 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9



489



Amplitude and Phase Responses Filters



4.



Resonance Bandpass Functions and Qualiry Factor Use of Pole-. zero Plots



497



Scaling lhe Network Function The Decibel Frequency Response with SPICE



505



Sunmary Problems



515 515



computer Application hoblems



522



493 499 502 509



5t2



16 Transformers 15.1 16.2 16,3 16.4 15,5



524



Mutual Indactance Energy Storage Circuits with Linear Tmnsforme$



525 534



Reflected Impcdance



541 543



The ldeal Transformer



537



16.6 Equivalelt Circuits 16.7 SPICE Analysis for TraDsfoioers 16.E Summary Problems Computer Application Prcblems



549 551



553 554 559



17 Fourier Series



560



17.1 17.2 17.3 17.4 17.5 17.6 17.7



The Trigonomenic Fouder Series Symmet y hoperties Response to Periodic Excitations Average Power and RMS Values The Expon€ntial Fourier Series



561 568



Frequency Spectra Fourier Series and SPICE



f7.q



Sunrnary



hoblems



589 592 594 594



Computer Application Problems



598



576 582 585



1B Fourier Transforms 18.1 18.2 18.3 18.4



The Fouder Integral Development of the Fourier TraDsform Fouder Trar$form Prcperties Fourjer Transform Operations



1E,5 Netwo* Functions 18,6 Parseral's Equation for Fourier Transforms



lE.7



Surnmary



hoblems



599 600 603 sn6



ffi



614



6t6 620 620



19 Lailace Transforms 19.1 19.2 19.3 19.4



Definition Some Special Results Using Linearity Translation Theorems



Convolutiotr



Contents



623 624



628 631 635



xl



19.5 19.6 19.7 19,8 19,9



The Impuls€ Futrctior the Ioverse Tramform Differeirtiation Theorcms Applicatiors to Integodiffercntial Equatiotrs



639



ffi



654 658



Swrmary



661



6t



Problems



20 Laplace Transform Applications



m.l 2O.2



665



Applicatio! to Elect ic Cilcuitr The Transformed Circuit



66 610 677



21.3 Network Functions 20.4 Step and Impulse Reslronses 20.5 Stability 20.6 Initial- alld Final-Value Thmrems 2,0.7



20.8 20.9



681



688 691 695 698 705 708 709



Steady-State Sinusoidal Response Bode Plots Quadratic Factors



Z).10 Sunmary Problems



Appendix A



Determinants and Cramer,s Rule



715



Appendix B



Gaussian Elimination



719



Appendix C



Complex Numbers



722



Appendix D



Euler's Formula



727



Appendix



E



Computer Methods



730



Appendix



F



Circuit Design Methodology



740



Appendix G lndex



xll



.



Answers to Selected Odd-Numbered



Problems



761



769



Preface This book was writteD for a two-serrester or three-quatter cou$e in Inear cirruit



analysis. This is a basic course h electrical engineerilg aDd is u;ail; the;rd;;t's firsl encounler with his or her choseD field of s;dv. ri" iextbook to coiei thorougbly tlrc futrdamo als of ttre *Uieit aaa a:t tfre as easy to,understadd as possible. These wele our objecives *[oughoot the Gtini



t ii^p"n*, t"r-f.-, fi s;;'d_";



of the book-



The book is a[ expanded teatrne of our efi\er woil., Basic Electic Circuit and is designed primarily Jor readers desiring o ,o"re ot taplace translorms aDd Fourier analysis. We hope that the earlier'book "o.of"ra wil coo_ nnue to serve the needs of students wbo are prirDarily inlercsted in nontratrsform melhods, and fhat this book will be anractive ti tho." int"r"s"a in u U-na". age of operational matbemadcs methods. "o""i Most studerts in basic circujt theory will already have sodied electricitv and magnetism iD a physics course. fnis bacigrouna is #lpfirt. prerequisil,e for reading lhe book. The maerial presented here mav be stood by a studenr who has had standard coursesio affie..otid _a ilr.era ;dcuiui. The differential equalions theory requird in circuit analy.is h tuilyA;?;p.n;'rb€ Dook and lniegmEd witb the approFiate cirujt theory opics. -p,"a"ot"a Even delerminants, Gaussian eiiminatioD. and complex nuJbber theo.V ur. io I ne operatonal anphber js i_Dtroduced immedialely after "pp"nOi""il lhe discussion of the rcsistor, ard appears, as a natter of couse, along *itn i".irmo, irducton. as a baric el€ment throughout the book: Likewise. O.p"nd""i "a*itoo,-*a ,o,r."", _a lneE constructroo uslng operational arnplifers are discussed early aDd eacountercd muhtrely jn almost every chapter. To help the rcader Mderstand the lexnra.l maGrial, examDles arc litrerallv suoplied and numerous exercises. with a[swers. ar€ given at rt" ira of Un""ffr'.".1, section. Most of $e worked-out examples are cleady marked and ou!0t .ea ior easi relereDce. tloblems. some more dimoh and som; less difEcult thaD the end_oi_ section exercises, are given at the end of every chapter, anal answeN th" numbered problems are giveo in a! appeDdix. Most bf rhe exercises anJ -.0r._. rave bEen designed !o yield easier answers. such as 6 volts rather lhan 6.12? volts. without sacrificing tbei! pracrhal aspects. We hop" in tf,i, *uy to .ini^lr. uoo" sanly tedious ari$rletic and male the srhiect of circuit theory more iiteresting atrd rewardine.



.



_



Arutlsk-



LiarJ,



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xnt



.A special effoft has be€n made to include a Nmber of probless attd exercis€,s witb r€alistic element values. Netwo* scaling, also !rcs€ntad, can be used to mrke aI he .erBaioirg Fobleos practical. In lhe chapte. orl amplitude and phas€ responses, problems arc given on such Factical cir€uits as elechic flters. Active filte$, using op€rational amplifiers, as well as passive fflten, are usad as exampl€s. Fhally, a sel€ct fow fiorcises and problems are uscd !o ertotrd tho tlrcory discussed in tie chapters. In this way, optional material is ircluded without adding to lhe t6xt. The fi$t nirc cbapteN of the book arc devoted to temiaology and time-domaitr analysis and the last elever chaptqs deal with fte4uency-domain analysis. Some material may be omitted wibout any loss in continuity. The chapoer on network lopology, an fuieresting subject, could be covered € irely. in pan, or not at aU. Chagl€i$ l?-20 on Fourier and Laplace rnethods constirut€ a detailed matrnent of these subjecb and nay bo omifted if thes€ topics arE to bo covered in a s€palate course. In deed, one could omit the classical differential equations approach to circuit theory and go direldy to the Laplace and Fourie! transfom methods. For the leader interes[ed h computer solutions of circuits, the computer-aided circuit analysis pogran SPICE is described in aq appgqdix, atrd exaoples usilg SPICE are given itr the last sectiols of selecte-d chapteN. A s€pardte section of computer-aided pmblems follow probl€m setl at the end of the cheter. Il this way the compuler-aided naterial can be easily omitted so desircd. Colcrr is used throughout the book to highlight lhe more iflportaft equatioDs, to help clarify the figure,s, and genenlly to make the book morc readable. Finaly, to 6ake the strbirct of electric circuits mor€ real-ard e joyable lo the student, we have opened each chapter with a picllle and a short biogaphy of a farnous eleakical pione€r, note.d engineer, or inveotor whose wo* ha! contributed importandy to circuit theory. ln this second edition, we have retained the basic formdt and features that have bcen well rcceived h the past, We have made chaqes tbmughout the book to better clarif] rhe material and have greatly expanded the number of end-of-chapter problems, which have be€n completely reds€d for this editiofl. We have also added a summatT sectio{ at the eid of e4ch chapter, emphasizing the coacepts introduced in ftat chapler, An appendix on design has bs.€n added for those who alo inlerosted itr an introduction to the design of cLcuits as well as the amlysis of them. The matodal is Fesented in an appendix so that it may be easily ooditted by those intercsted only



if



in analysis. For the chapler-op€ning illustratioDs we are gmteful to the Print Collertioo, Mniam and Ira D. Wallach DivisioD of Ai, ftirlts and Photographs, The New Yort Public Lrbrary. Astor. knox. and Tilden Foutdations for Chapters I, 5. 6, ?. 19. and 20; the Library of Con8rcss for Chapters 2, 3, 8, 11, and 17; lho Snithsonian Institution lor Chapters 4,9, 12. 14, afi 18; and-the marvelous book, Dictionary of Am?rican Pofiroits (Dove! Publications. Inc.. 1967, edited by Haywood aod Blarche Cirker), for Chapters 15 ad 16. Th€ photogaphs for ChapteN l0 and 13 are courtesy of lhe GercIal Electdc Company, whom we also gatefirlly acknowl" edge.



There are many people who have provided invaluable assistarce and advice concaaing this book. We are indebM to o]ll colleagues @d our students for the form the book has taken, add to Professo$ M. E. Van Valkenburg, A. P. Sage, S. R. Laxpati, and S. K. Mitra, who rcviewed the fust-editio! manuscript of Bdric



xtv



Elpctric Circuit Atatysis aod made uany helpful commeDts and suggestions. we arc also gratefrtl for the invaluable rcviews of the first e.dition of Electrk Circuit Anolyrir by Adce M. Davis. San Jose State University; John A. FleminS' Texas A & M UniveNiry: Deverl Humphreys. Brigha$ Young Universiryl Tim Jordanides' Califomia State University, Irrg Beach; K. S P. Kumar, Udve$ity of Mirmesota; Terry W, Martin, Univgrsity df Arkansas; Micha€l P. Smlth, Widener UDiversity; J. Eldon Ste€lnan, New Mexico State Universig; James Svoboda, Cla*son Uliv€r$ity; and Belh



L. Ko€stcr, Coocord,



Massachusetts.



Dovid E. Joh$on ,Iohnhy R. Johnson



Join L. Hitbun



xv



1 lntro duction



,



Elect c circuil theory had ils real beginning on March 20, 1800, when the llalian physicist Alessandro Volta announced his invention of the electic battery. This magnificent device allow6d Volta to produce culeri elect.icity,



a steady, conlinuous llow of



This endlcss circulation of the electtic fuid may appear paradoxical, but it h no less true and real, end you ntay feel it with



rent electricity was arimal eleeticily by the organisms them-



caused



solves. Volta, on lhe olher hand, maintainod that curent €lectricity \uas molallic electicity,lhe source of



which was the dissimilar



metal



probes atlached to lhe frogs' legs.



Alessandro Voha Bolh mgn wor€ right. There is an anielgctricity, as opposed to static eleclriclty, producgd in bursts by pr€vimal olectricity, and Galvani became ous elsclical machinos such as the lamous as a founder of nerve physiLeyd€n iar and Volta's own electophotus. ology. Volla's great invgntion, however, revolulionjz€d Volta was born in the ltalian city of Como, then the use of elecldcily and gave the world one ol its a parl of the Austrian Empire, and at age 18 he was greatesl benelits, the olectric currenl. Volla was showperforming electrical expadments and corresponding ered with honors durinq his lifetime. Napoleon mado with well-known Europoan elect cal investigators. ln him a senator and laler a counl in the French Empire. 1782 he became professor ol physics at the UniveF Atter Napoleon's dgleat, the Austrians allowed Volta sity of Padua, wh€ro he became involved in contro- to return lo his llalian eslate as a citizen in good versy with anolhef well-known electdcal plon€€r, Luigi stead. Volta was rowarded 54 years alter his dealh Galvani, prolessor ol analomy at Bologna. Galvani's when lhg unit ot eloctromo{ve lorce was otficially experimenls with lrogs had led him lo beli€ve lhat cur- named the voll



.



E



LJlectric



circuiL analyqis. in nearly every electrlcal engine€ring curriculum, is the first course taken in the major area by an electrical engineering student. Virtually all bmnches of electrical engineering, $rch as elechonics, power systems, communication systems, rotating machinery, aDd conool theoty, are based on circuit theory. The orly topic in elect cal elgineering more basic than circuits is elecfomagneiic field theory, and even there many problems are solved by means of equivalent electric circuits. Thus it is no exaggeration to say that the basic circuit tleory course a student firut encounters in electrical engineering is the most impo ant cou$e in his or her cu iculum. To begin oul study of electric circrlits we need to know what an electric circuit is, what we mean by its analysis, what quantities are associated with it, in what units these quantities are measuled, and the basic definitions and conventions used in circuit theory. These are the topics we consider ill this chapter.



1.1 DEFINITIONS AND UNITS An elecljic circuit, or elec*\c network, is a collection of electrical elements interconnected in some specified way. l,ater we shall define the electrical elements.in a formal manner, but for the present we shall be content to represent a general tlroterminal element as shown in Fig. 1. 1 The terminals a and D are accessible for connections with other elements. Examples with which we are all familiar, and which we shall fdrmally consider in later sections, include resistors, inductors, capacitors, batteries, and genemtors.



FtCURt



t.r



Cenemltwo terminal electricai element



More complicated circuit elements may have more than two terminals. Tmnsis_ tors and opemtional amplifiers are commor examples. Also a number of simple elements may be combined by interconnecting their terminals to form a single package having any runrb€r of accessible terminals. We shall consider some multiterminal eIements larel, but our main concern will be simple two-terminal devices. An example of an electric circuit with six elements is shown in Fig. 1.2. Some authors distinguish a circuit from a rctwork by requiring a circuit to contain at least one closed path such as path abco. We sball use the terrns interchangeably, but we Chapt€r



1



lnnoduction



flcURt 1.2 tle.ri..ir.rit little or no practical intere\t. To be more specific in defining a circuil element we need to consider certain quantitles associated with it, such as votfaS? and ctDerl. These quaDtities and otheis, vhen they adse, must be catefirlly defrcd. This can be done only if we have a standard system of units so that when a quantity is described by measuing it, we can all agree on what the mea$uement means. Fortunately, there is such a standard srstem of units that is used today by vntua y al the professional engineering societies and the authors ol most modem engineering textbooks. This system, which we shall use throughout the book, is the International System of Units (abbteviated SI), adopM in 1960 by the General Conference on Weights and Measures There are six basic units itr the SL and all other units are derived from them. Four of the basic units, the meter, kilogla$, second, and coulomb, are impoltant to ctcuit theorists. and we shall c.nsider them in some detail. The remaining two basic units are the kelvin and the candela, which are important to such people as the electron device physicist 6nd the illiminatioo engine€r. The SI units are very precisely defined in terms of permanent and reproducible quanlities. However. the defnilions are hiShly esoteric and in some cases ate comprehensible only to atomic scieDlists.r Therefore, we shall be conlent to name lhe basic units and relalq them to the very frmiliar Britisft '']stem o/ UnitJ, which includes inches, feet, pounds, aird so on. The basic unit of length in the SI is the neter, abbrevialed m, which is related to the British system by the fact that I inch is 0.0254 m. The basic unit of mass is tl:r, kiloSrarn (kg), alld the basic unit of time is the secol (s). In terms of the Biitish units, I pound-mass is e'Jictly 0.45359237 kE, ard the second is the same in both may note that without al least one closen path the circuit is of



systems.



The fourth unit in the St is tl\e coulonb (C), which is the basic unit used to measuIp elect c charge. We shall defer the definition of this unit until the next section when we consider chalge and qAretrt. The nafie coulomb was chosen to honor



the French scientist, inventor, aod army engineq Charles Augustin de Coulomb rcomot.t€ &finttiotrs of rhe basic u[r$ may be found in r oumb.r of souce\, q(h a .IEEE Rerommcnd;d PBr'e for Unib in Publish.d Sci.nd6c and Tehnical Wo*." by C H Page er al. (IEEE J, no. l. pp. 169-171, M$cn 1966) Sp€.rrM,



"ol.



Section



1.1



Definitions and Unit!



(1763-1806), who was an early pioned in the fields of friction, ele.tricity, ard magnelism. .



We might note at this point that all SI units named for famous p€ople have abheviations that are capitalized; otherwise, lowercase abbreviations are most ofteo used. It is also worth mentioning that we could choose units other than the ones we have selected to form the basic units. For example, instead of the coulomb lve cordd take the ampere (A), the unit of electdc current to be conside.ed laler. h thk case



the coulomb could then be obtained as a derived urit. There are three derived units in addition to the ampere that we shall find us€ful in circuit theory. They are the units used to measure force, work or energy, and power. The fundamental unit of force is the reuon (N), which is the force required to accelemte a l-kg mass by I meter per second per second (l m/s). Thus I N = l kg-m/s'?. The newton is named, of cou$e, for the geat Engish scientist, astronomer, and mathematician Sir Isaac Newton (1642-172?). Newton's accomplishments are too numerous to be Iisteal in a mer€ chapter, The fundamental unit of {,or* or energy is the jorre (J), named foi the British physicist James P. Joule (1818-1889), vho shared in the discovery of the larr of conservation of energy and helped establish the idea that heat is a form of energy, A joule is the work done by a constant l-N forc€ applied through a l-m distadce. Thus



lJ:1N-m.



The last derived unit we shall consider is the udtt (W), which is the fundamental unit of power, the mte at which work is done o! energy is expended. Th€ watt is defined lo be I J/s and is narned in honor ofJames Walr (1736-1819), the Scoltish engineer whose engine design fust made steam power pmcticable. Before we leave the subject of.units we should point out that one of the greatest advantages the Sl has over the Blitish syslem is its incorporation of the de.imal system to relate larger and smaller units to the basic unit. The vadous powers of 10 are denoted by standard prefixes, some of which are given, along with their abbievi-



ations, in Table



l l.



. TABI-E



l.l



Prefixes in the 5l



Mulriple



l0' ld I l0



Symbol



ziga



kilo rnilli



G



M k



p



As an example, a second was once thought to be a very short time, and fractions such as 0.1 or 0.01 of a s€cond were unimaginably short. l.{owadays in some applications, such as digital computers, the second is an impracticably large udit. As a result, times such as I ftuosecond (l !s or l0_e s) are in commi}n !6e. Another common example is I gram (g) = I0 I kg. chapter



1



lntroduction



EXAMPLE 1.1



In the 100 m



1972 Olympics one of Mart spitz's seven gold medals was for $'inming in 51.22 s. Co[vert his average sFed to miles per hour. We begio by oot-



ing that



I



lm=oJ25i4-ln



=



/ t \o-ozs+



\/t fr\/ I



i"/



mi\



l.o;/\sxo r, /



1m=0.00062137mi Therefore. the avemge speed is



#T- (#9(,*.',"lu)(,*;) =



4.37 mph



we note that the units cancel in every step, ${ich may be used to irdicate what frctors are needed in the conversion process.



EXERCISES 1.r.1



Find the number of nanoseconds in (a) 0.5 s, (b) 30 ms, and (c) 15 ps. x 103; O) 3 x l0'; (c) 15,000



inswer (a) 5 1.1.2



1.1.3



l-t.4



Find (a) the number of $econds in 22 ps, O) the number of kilometers in I milc, and (c) the work done by a constant force of 200 pN adied to a mass of l0 g for a distance of 50 m. (a\ 2.2 x 10 '; (b) 1.609; (c) 10 rJ ^nswer Sebastian Coe broke three world track records in 1979 by rundng 800 meters in I



minute 42.4 seconds, the mile in 3 minutes 49.0 secodds, and 1500 meters h 3 Ehutes 32.1 seconds. Find his avenge speed in miles per hour for each eveot. Answet l7.5, 15.'1. 15.8 Robert Hayes set a world record in 1963 by runring 100 yards in 9.1 seconds. Id tie 1988 olympics Bell Jobnson mn the lm-rneter dash in 9.79 seconds. He rras disqualified ald Carl lewis declared the winner with a time of 9.92 se4onds; atr Amcrican record. Find the average speeds in mph of erch ofthese thre€ ruorea!. Answer tjayes, 22.48; Johnson, 22.85; l*wis, 22.55.



1.2 CHARGE AND CURRENT We are hmiliar with gravitational forces of attractiod between Mies, which are tcsponsible for holding us on the earth and which cause ao apple dislodged ftom s Eec to fall to the ground mther than to soar upward into the sky, There are Mi€s' ho*.



Section



'1.2



Charge and Current



fir out of propoltion to their masses. Also, sucb forces are observed to tre repulsive as well as attractive and are clearly not gravitatioDal forces. We explain these forces by saying that they are electrical in naturo and caused by the presence of elscfical charges. We explain th€ existence of forces of both atever, that aifi"act eabh othgr by forces



traction and repulsioo by po€tulating that there are two kinds of charges, positive and negative, and that unlike charges attract aDd like charges repel. As we know, according to modern theory, natter is rinde up of atoms, which are composed of a numb6 of funda$ental particles. The most important of these particles are protons (positive charges) and neuhons (neueal, with no charge) found in the nucleus of the atom and electrons (negative charges) moving in orbit about the nucleus. Normally, the atom is €lectrically neutral, the negalive charge of the electroDs balaocing the positive charge of tne protons. Particles nray become positively charged by losing electoons to other particles and become negatively charged by gaining elecFons &om other parricles. As an erample. we may Foduce a negative charge on a balloon by rubbing it against our hai!. The balloon will then stick to a l,all or the ceiling, which are uncharged. Relative to the aegativ€ly charged balloon, the neutral wall and ceiling are opposirely charged. We now define the cortorr, (C), discussed in the previous section, by stating that the chaige of an election is a negative one of 1.6021 X 10 te coulomb, putting it another way, a coulomb is the charge of about 6.24 10'3 electlons. These are: of course, mind-boggling numbers. but their sizes enable us to use more manageable numbers, such as 2 C, in the circuit theory to follow. The symbol for charge will be taken as 0 or 4, the capital letter usually denotirg constant charges such as P = 4 C, and the lowercase letter indicating a tim€!"arying charge. In the latter case we may emphasize the time dependency by writing {(t). This practice involving capital and lowercase letters will be carried over to the other electrical qualtities as \rell. Th€ prioary purpose of an elecric circuil is to move or [ansfer charges along specified paON. This motion of charges constitutes .an electrb current, denoted by the letters i or 1, taken from the French word "intensit6.,, Formally, curre{t is the tirne rate of change of charge, given by



x



.dS



'-a



(



1.1)



I!e_ basic rltrit of cur€nt is tlrc anpere (A), mmed for Andi6 Marie Amp&e



(1?75-1836), a Rench mathematician and physicist who formulated laws of electromagnetics in the 1820s. Aa anpere is 1 coulomb per se.ond. In circuit tbeory curent is genemlly thought of as the movement of positive charges. This cotvetrtion stems from Benjamin Franklin (1706-l?90), *to guessed lhat electricity traveled tom positive ro Degativg. We now know ,that in metal colductors the curent is the moyement of elecrons that have been pulled loose ftom the orbits of the atoms of the metal. Thus we should distinguish convearional current (thc movement of positive charges), which is us€d in eleatric network theory, and Chapler



i



lntroduclion



ekctron a$reDf' Lnless otherwise stal€d, our coltcem will be with conventional cunent. As an example. suppose lhe curretrr in the wire of Fig. | .3(a) is 1 = j A. Thal is, 3 CA pass some specific poiot in the wirc. This is symbolized by the arrow ta-. beled 3 A, whose directiotr indicates that the motioa is from teft to right. This situa_ tion is equi lent to thal depicted by Fig. 1.3(b), which indicates -3 a/s or -3 A in the directiotr from right to left.



-la



rr/



-../ G)



.



(b)



FICURE 1.3 Two representations of



$e same current



Figxre |.4 represenrs a geneml circuil element wirh a currenl i flowing from . the left loward the right te.minal. The loral charge entering rhe element #tween time ,o and t is found by integnting (1.1). The result is



.



b:.so)-ean=[iat



(1.2)



We shotrid note at this point that we ale considering ihe network _ be elecrrirally neurml. _



elements to



That is. no net positive or negativJcharge can accumujate in rhe element. A posirive charge entering must be aciompanied by an equal posilive chaqe leaving (or, equivalently, an equal negative charge entering;. Tirus rtre current shown entering the left tenxdai in Fig. i.+ must leive the rifit terrninal.



RGURT



EXAMPLT 1.2



1.4 Cun€nt flowing in a Benerat



etement



Suppose that. the current ente ng a terminal of an element is i = 4t charge entering the terminal between t 0 and f 3 is given by



:



:



A, The total



n-l 'J" q,a,- Bc There are seveml types of current in common use, some of which are shown in



Fig. 1.5. A constalt cuqent, as shown in Fig, 1.5(a), will be termed a alirect current, or &. An ahernating cwrent, ot &, is a sinusoidal curetrt. such as that of Fig. 1.5(b). Figure 1.5(c, and (d) iuusEale. respctively. an expon?ntial c$rem and a rawtooth ctrtent, There ate many commerciil uses for dc, such as in flashlights and power sup_



plies for elechonic citcuits, and, of cou$e, ac is the common house current found all over the world. E\ponential currents appear quire oflen (whether we $.anl them or nor:r when a swilch is actualed lo close a path itr atr energized circuit. Sawloolh waves are useful in equipmelt, such as oscilioscopes, usdd ior displaying electrical chamcteristics on a screen Section



1.2



Charee and Curreot



rICURE



1.5



(a) Dcj (b)ac; (c) exponential current; (d) saMooth cunent



EXERCISES



1.2.1



I.2.2



How many electrons are reprcsented by a charge of 0.64084 pC? Answer 4 nillion The total charge entedng a terminal of aE element is given by



o'



. 1.2.3



Find the curent i at An\wet 4,20 mA



=



o



^u



J r=:')'



*



".



The current entering a terminal is given by



i:l+?'sin2ntA Find the totai charge eltering the terminal between



Ahs\t 2.5



t=



0 and



t=



1.5



s.



C



1.3 VOITAGE, ENERGY, AND POWER Charges in a condd,stor, exemplified by ftee electrons, may move in a mndom manDer. However, if we want some concerted motion on their part, Such as is the case with an electric cu$ent, w€ must apply an exieftal or so-called eredromotire force (EMF). Thus work is done on the chatges. We shall define ro'age "aqoss" an element as the work done in moving a unit charge (+1 C) through the element ftom one terminal to the othff. The unit of voltage, or potential ilifferehce, as itis sofie-



times called, is tle yolt (V), named ir honol of the Italian physicist Alessandro Giuseppe Antonio Anastasio Volta (1745-1827), who invented the voltaic battery. Since vohage is the number of joules of work pedormed on 1 coulomb, we may say tlat I V is I J/C. Thus the volt is a derived SI unit, exptessible in terms of other units. We shall represent a voltage by o or V and use the +, polaity convention shown in Fig. 1.6. That is, terminal A is r volts positive with lespect to terminal B. Putting it another way in terms of potential differerce, l€tninal A is dt a potential ol



-



Chdpler



I



lnlroduclion



FIGURE



1.6 Voltase polarity convention



o volts higher than teminal B. I! t€nns of work, it is clear that moving a unit charge ftom B to A requires o joules of wor*.



Some authors prefer to describe the voltage acrocs an element in terms of voltage drops aDd rr',rer. Referring m Fig. 1.6, a voltage drop of o volts occuls itr moving ftom A to B. In contrast, a voltage dse of o volts occus in moving from B to A.



As examples, Fig. 1.7(a) aDd (b) are two versions of e,\actly the siune voltage. In (a), a terminal A iS +5 V above termiml B, and in (b), terminal A is -5 V above A (or +5 V below A). A



FICURE



1,7 Two equivalent voltage representatio



We may also ]use a double-subscript notation o", to denote the potential of point a with respect to point D. In this case we have in geneml, od = -or,. Thus in Fig. 1.7(a), o,$ = 5 V and 0,r 5 V. In transferring chage through a]] element work is being done, as we have said. Or, putting it arother way, energy is being supplied. Ib know whether energy is being supplied ro the element or ,] the element to the rest of the circuit, \re must know nol only the polariry of the voltage across the element. but also the direclion of the cu.rent though the element. If a positive cudent enters the positive terminal, then an extemal force must be drivitrg the current aDd is thus supplying or dekvering e\ergy to the element. TtP- de'rent is absorbing energy in this case. If, on the other hand, a positive current leaves the positive terminal (enteN the negative terminal), then the elametrt is delivering energy to the extemal cilcuit. As examples, in Fig. 1.8(a) the elcrlelt is absorbing energy. A positive current enterc the positive terminal. This is also the case in Fig. 1.8(b). In Fig. 1.8(c)



-.



fIGURE 1.8 Various voltage-current rclatiorships



24.



-o--:= -



sv



(al Section



1.3



I HA iii I



sv ffi,"



-. (bl



voltage, Energy, and Power



24



-----\*-\



(c)



'



*"



(d)



rul



and (d) a positive curreDt eoters the negative terminal, aDd therefore tbe elemenl is delrverrng energy in both caj€s. us consider now lhe .at? at which energy is being delivered to or bv a circrrit elemenr. If lhe voltage aqoss the element iiu and a im"ff *rrg. aq i. through. the. elemeat from the positive to the negarive terminal, sorbed by the elemeDt, say A!r, is given by



kt



;".d tdth" :;dy ;:



A}r,=aAs If



the time involved is 4,/, then the mte at which the work is being done, or the ergy lr is being expended, is given by



.. Aw AFO



AI . AH



Aa AT



(1.3)



7;=D;=Di ;llc;e



en_



by definition the ftte ar which energy is expended is power, denoted



byp, we



We_might observe that (1.4) is dimensiotally conect since rhe units of



oi



are



(tC)(C/s) or J/s, which is watts (W), defined eadier. . .The gTntiri:: r, g]d i are generaly tuncdons of rime. which we may also de_ note by o(r) aDd i(t). Tberefore, p given by d.4.) is a rime_varyins ouanrirv. It rs



somelimes called the instu)ntanaous Wwet because its value stant of time at which o and i are measured.



ftcURE



r.9



i"



ti" io;., ui;" i;:



Typical etement with vottage and cu.rent



Summarizing. tbe llp;cal element of fig. 1.9 ts absorbing Dower. Eiven bv rhe polariry of u or i fbu not borhr is reversed, ihin rh. .i"rn.nt j'. detrvenng power. p oi. to the exlemal circuit, Of course. lo say that an element d€livers a negativ€ power, say -10 W, is equivalent to saying that it absoOs afosi_ tive power, in this case + l0 W.



p -_ ui. If eilher



-



EXAMPTE 1.3



In Fig. Ls(a)



and O, lhe elemeor is absorbing power of p _ (5)(2) l0 W. Lln FiB. 1..8ft, the 2 A leaves the nega!.ive rerminal. and ttrus 2 e enten *" p"r-lri". terminal.l In Fig. 1.8(c) and (d) it is delivering tO W to tfre externat circuii, since the 2 A leaves the positive termlnal, or, equilalently, -2 A enters th" terminal.



=



p;*;;



10



Chaprer



1 lntoduction



Before ending our discussion df power and energy, let us solve (1.4) for the en-



eryy rr delivered to an element between time to and t. We have, upon integrating both sides between t, and t,



w(t) EXAMPTE 1.4



-



w(a) =



f"vi



at



(1.s)



1.9 i : 2t A arld o = 6 V, the energy delivered to the element between t=0andt:2sisgivenby



If in Fig.



w(2)



-



w(0,



t'1



- JrI



\qQI) dt =



24



t



' Since the left member of (1.5) represents ihe energy delivercd to the element between ro and t, we may interpret w(/) as the drcrgy deliveted to the element betwe€n the beginning of time and / and r (to) as the energy between the beginning of o, the energy delivtime and to. In the beginning of time, which let us say is ! ered to the element was zero; that is,



=



x,( @)=o If ro :



-@ in (l.5), we shall have the energy delivered to the element ftom the be-



ginning up to



t,



gived by



*ttt



- | ."'ix



(1.6)



This is consistent with (1.5) since



wttt =



l_aidt



r



By (1.6) lhis may be written



= l-ti dt f^"'o'



w@=wA";s+ which is (1.5).



t



f^



oi dt



EXERCISES



1.3.1 Findl) if t =



E rnA and the element is (a) absorbing power of delivering to the external circuit a power p = 16 mW. 2V Answet \^) 5 V:



p



:



40



mw and (b)



lbl



+



EXERCISt 1.1.1 Section



1.1



VoltaEe, Energ/, and Power



11



1,3.2



Ifi = 5 Aando = 12 V in Exercise 1.3.1, find (a) the power absorbed by the element and (b) the energy delivered to the elemeDt betwe€n 2 ard 4 s. Arurer (a) 60 w; ft) 120 J



1.3-3



A two-terminal element



absorbs an energy



n



as showtr.



If



the curent ettering the



positive terminal is



find the element volrage Ansx,et



-50,



5



y



i = 100 cos 10007r mA at r = I ms and at t = 4 ns.



txERctsE 1.3.3



1.4 PASSIVE AND ACTIVE ETEMTNTS We may classify circuit elements into two broad categoies, p4rriye elements and dctive elemens, by conside ng the energy delivered to or by them. A circuit element is said to be pd.rriye if tle total energy delivered to it ftom the rest of the circuit is always nonnegative. That is, refering to (1.6), for all t we have



,(i = |



:



f-,,*



(1.7)



=o



-p(tyat The polarities of tr.and i are as shown in Fig. 1.9. As we shall



see



later, e/,adples



of



passive elements are resistors, capacitors, and inductors. An dctiye element is one that is not passive, ol cou$e. That is, (1.7) does not hold for all time. Examples of active elements are generators, batteries, and electronic devices that requhe power supplies. We are not ready at this stage to begin a formal disculsion of the various passive elements- This will be done in later chapters. In this section ve give a brid discussion of two very important active elements, the independent voltage souce and lhe independent currenl source. A\ depeident voltage rource is a two-terminal element, such as a battery o! a generator, that maintains a specified voltage between its terminals. The voltage is completely independent of the curent through the element. The symbol for a voltage source having o volts across its terminals is shown in Fig. 1.10. The polarit is as shown, indicating that terminal d is 1, volts above terminal ,. Thus if o > 0. terminal d is at a higher potential than terminalr. The opposite is hue, of course,



i



if



u0.



I.6



If the tunction Sraphed in prob. t.6 is the charge (mC) entering tho positive terminal of



:



16e ta W, ihe curent i is nonnegative, and the voltage is 1) : 4t (o is in volts and i is in amperes). Find rhe voltaae and the total charge delivered to the elemenr fiom 0 to



p



1.15



The cunent ent€ring the positive t€rminal of an element is i = 4e-' A. Find rhe power delivered to the element as a function of time and the energy deliver€d to the element ftom 0 to t > 0 if the voltage is (a) o = 3i, (1,)



D:2i,



and



{c, o=3lidr+6.



lThe



voltage is in volts if the currert is in amperes.)



Ch.pler



1



lnrrodu.non



1.16



If rhe current entering the positive terminal of



is D = ar, a > 0. taj Find rhe energy deliv. ered to the elemenr between 0 and f seconds, (b) Fhd a dc cunent I b > 0 rhat detivers the same energy as the ac current of (a) in



an element is



:



i:4 sin 2rA. I > 0 :0,r 0 and the chaqe delivered to the element bet$een 0 and 7r/4 s, if (a) i, = 2r, (b)



,



= 2fr,



volts



if i



*a



:



@) n



2



!;r



dr



-



atr elemenl is



i = asinfr'A,, >0



4 (D is in



:0,, 0isthe



resistance.



The slmbol used to represent the ohm is the capital Creek le]ler omela lA\. Thus by (2.1) we have R = 0/i, so rhat



1O=lV/A Seclion



2-l



Ohm's law



19



FICURI 2.1 Circuit svmbol for the resistor



In some applicadons. such



as eleclronic clcuiLs. rhe ohm is all incoDvenientlv sfirall unit and units such as kilo-ohms ot sifiply kilohms tko.t and mega-ohnls or n;BohnL, (MO) are common.



EXAMPLE



2.I



If R



:



3 O and o



=



6 V in Fig. 2.1, the curent is



. o 6V i=i=jO=2A lf R ls changed to I kO, lhe currenl



.



'



6V I kf)



is



=ffi=ox1o-3A=6;A



The process is obviously shortened by noting that



i FA.



I



V/kO



= I mA. MMO :



and so on. Since R is constant, (2.1) is the equation ofa srraight tine. For this reason, the resistor is called a llnelr/ rcrr'sror. A graph of u ve{sus i is shown in Fig. 2.2, which is a line passing through rhe origin with a slope ofR. Obviously, a stmight line is the only gmdr possible fol which thd mtio of o ro i is constani for all i.



.



FICURI 2.2 Voltage curent characreristjc for a linear .esisror Resistors whose resistances do not remain constant for different terminal currents are known as n tlineat rcsistors. For such a resistor, the rcsistance is a func_ tion of the clrrent flowing in the devic€. A simple example of a nonlineat rcsistor is an incandescent lamp. A tFical voltage-current characteristic for this device is shown in Fig. 2.3, wlEre we see that the gaph is no longer a straight line. Since R



20



.haprer



2



Re'i\ri!P arrcuit\



flCURt 2.3 Typical voltage currenl characterGiic for



a nonlinear resistor



is not a constant, the analysis of a circuit containing nonlinear resistors is more difncult. ln reality, all practical resistors are nonlinear be.cause the electrical characteristics of all conductors are affected by environmental factors such as temperature. Many matedals, however, closely approximate an ideal linear resistor over a desired opemling region. We shall conbentrate on these types of elements and simply refer to them a5 resistors. An examination of (2-1), in conjunction with Fig. 2.1, shows if i > 0 (curent entering the upper terminal), then o > 0. Thus the culrent enters the terminal of higher potential and exits from that of the lower potential. Next, suppose that < 0 (curent entering the lower terminal). Then o < 0, and the lower terminal is higher in potential than the upper one. Once again, the current enters the torminal of higher potential. Since charges are transported ftom a higher to a lower potential in passing though the resistor, the energy lost by a charge q (energy = qo) is absorbed by the resistor in the form of heat. The rate at which energy is dissipated is, by definition, the instantaneous power



i



p(r) = o(r);(r) = Rtlr) =



#



(2.2)



A graph of (2.2), shown in Fig. 2.4, reveals that p(r) is a parabolic,(and thus nonlinear) function of i(t) or o(t) which is nonnegative. (The horizontal scales.are, of course, different ir the two cases.) Thus, for a linear resistor, the instantaneous power is nonlinear even though lhe voltage-curent relationship is linear. FIGURE



Section



2.1



Ohm's Law



2.4 Craph of the innanlaneous



power for a resinor



21



Tle condition for passivity, given in (1.7), 1"0)



is



l" p(t) dt >



0



Therefore, since p(t) is nonnegative, we see that the integral above is nonnegative and lhal lhe resi\tor is. ,ndeed. a pa\srve elerirenr. The beginning student often encounters difficulty in determining the proper algebraic sign in applying Ohm's law when the voltage assignment differs from that of Fig. 2.1. By (2.1) th€ voltage is R times the cuffent entering the positive terminal. Thus, if the current i enters the negative terminal as in Fig. 2.5, the current entering the positive terminal i and therefore Ohm's law is o R( l), or



is



-



o= xi In addirion ro rrs re.i\tance. a resisrur is also charactcri/ed b) its power mting, or hratteqe rutinq, which is the maximum power lhe resistor can dissipate without being damaged by overheating. Thus if a resistor is to dissipate a power p its power rating should be at least p and preferably higher. (The power used in the power rati'rg is awruBe power, to be discussed in Chapter 12, but for direct currents the average and instantaneous bowe.s are the same.)



FIGURE



2.s



Re, stor wiih a reveKed voltage ?\signment



Another important quantity wbich is very useful in circuit analysis is known ,



as



onlrcr.r,rc". dcnn(J b) 1



G= *



(2.3)



The SI unit for conductance is siemens, symbolized by S and named for the brothers Werner and William Siemens, two noted Gerrnan engineers of the late nineteeDth century. Thus S = A/V. (Another unit of conductance, widely used in the United Statcs is tbe mho, which is "ohm" spelled backwards. The symbol for the mho is the inverted omega (U).) Combining (2.1)-(2.3), we see that alternative expressions for Ohm's law and instantaneous power are



I



I



(2.4)



.,)



Chapter



2



Resistive Circuits



and



oo):if



=c,'ttl



(2.5)



As a final note, the concept of resistance may be used to define two very com_ mon circuit theory terms , short circuit and open circutt . A short circuit is a; ideal conductor between two points and thus may be thought of as a resistance of zero ohms. It cal carry any current, depending on lhe rest of the cftcuit, but the voltage across it is zero. AralogoBly, an open circuit is a break in the cicuit through which no current can flow. Thus it may be considered to bd an infnite resistance, and it may have_any voltage, again depending on tle rest of the circuit. EXAMPTE 2-2



[,et us find the current i and the power absorbed by the l -kO resistor of Fig. 2.6. From (2.3) and (2.4), C = # = 10 I S and i = tO 3 x t2 A = 12 mA. Also, (2.5) yields p(r) = l0-3 x 12, W = 144 mW, which is the minimum power mring required for the resislor.



FICURE



2.6 Curr€nt



voltage example



The current in this example is a direct current since its talue does not change ._ with time. Suppos€ that we now replace the 12-V source by the time-varying voltale D = l0 cos t V and repeat the foregoing procedure. The current is



. l0cosrV l0cosrmA t,rc, = and the instantaneous power is



P=01cos'?rW whicb is always oonnegative. The current, in this case, is an altemating current.



EXERCISES



2.1.1 2.1.2



The terminal voltage of a 20-kO resistor is 100 V Find (a) the conductance, (b) rhe terminal cu.rent, and (c) the minimum wattage of the resistor. anrwer (a) 50 pS; (b) 5 mA; (c) 0.5 W The instantaneous power absorbed by a resistor is 4 sin, 37?t W If the curent is 40 sin 377t mA. find 1, and ,R. Answet t00 sin 37 t V, 2.5 kA 7



Seclion



2.1



Ohm's Law



23



2_1.3



Find r' and the power delivered to the resistor. Answer



-2 p,4,24 1tW



,'{i,EXtRCtSE 2.1.3



2.2 KIRCHHOFF'S IAWS Thus far we have considered Ohm's law and how it may be used to find the current, voltage, and power associated with a resistor. Hovr'€ver, Ohm's law by itself cannot be used to anallze even the simplest circuit. In addition, we must have two laws fi.st



.



'



stated by the German physicist Gustav Kirchhoff (1824 188?) in 1847. The two .laws are formally known as Kirchhof's curent law and Kirchhoff's voltage law. These laws, together with the terminal characteristics for the larious circuit elements, permit systematic methods of solution for any electrical net*o!k. We shall not attempt to prove Kirchhoff's laws here since the concepts necessary fm the proof are developed in studies of electromagnetic field theory. A circuit consists of two ot niore circuit elements connected by means of perfect conductors. Perfect conductors are zero-resistance wires which allow curent to flow fteely but accumulate no cha.ge and no energy. In this case, the energy can be considered to reside, or be lumped, entirely within each circuit element, and thus the network is called. a lufiped-parameter circuit. A point of connection of two or more circuit elements is called a node. An example of a circuit with three nodes is showo in Fig. 2.7(a). Node 1 consists of the entire connection at the top of the circuit. The beginner quite often mistakes points a and , for nodes. It should be noticed, however, that a and , are connected by a petlect conductor and qan be considered electrically as being identical points. This is readily demonshated by redrawing the circuit in the form of Fig. 2.7(b), where at node I all connections are shown at a silgle point. Similar cornments apply for node 2. Node 3 is required for the interconnection of the independent voltage source add lhe resistor. With these concepts, we are now ready to discuss the all-important la*s of Kirchhofi. Kirchhoff' s current law (KCL) states that The algebraic sum of th€ cur€nts entering atry node is zero.



For example, the currents entering the node of Fig. 2.8 are ir leaving t3 entering). Therefore, KCL for this case is



is



r,+i,+(-r3)+ta:0 24



Chapter



2



Resistive Circuits



i, i2,



ir, and



ia (since



flCURl 2.7 ,dt



eF



FICURE



rodF. .uit .b



rh.ee



roo., 1,i



.edrdwn



2.8 Cuffenrs flowing into a node



Fdr the sake of argument, let us suppose that the sum is not zero. case! we would have



In



such a



i+i,-i)+la=V+0 where V has units of C/s and hence must be the rate at which charges are accumulated in thg node. However, a node consists of perfect conductor. a"*_ mulate charges. In addition, a basic principle of physics states that ^-nd "unnot charges can nei_ ther be created nor destroyed (consenation of chirge). Therefore, our aiumption is not valid, and V must be zero, demonstrating the ptausibility of KCL_ that in our example we multiply both sides of the KCL equation by _Supqgse - l. obtaininS



_



I irr tr i:j + rir) r(-ii)-0 From Fig. 2.8 \re see thar rhe lelt-hdnd \ide is simply ihe sum of lhe currenrs leaving the node. This demonslrates an equivalent statement for KCL: The algebraic sum of the currents leaving any nod€ is zero.



Let us now rearmnge the Feceaing equation in the form



i



where ir, ir, and are entering the node and ir is leaving. This forlll of th€ equation illustrates another statement for KCL, stated as



'llre som of the currents entering any node'equalF the sum of the currents l€aving the node,



In geneml, a mathematical expression of KCL is



S; :n where



i, is



(2.6)



the nth curent entering (or leaving) the nod€ and N is the number



of



such t ode cudents.



EXAMPLE



2.3



As an example of KCL, let us find the current i in Fig. 2.9. Summing the currents



.



entering the node, we have



5+r (-3)-2=0 or



i= -6A. that *6 A entering



the node is equivalent to 6 A leaving the node. ThereWe note fore, it is not necessary to guess the correct current direction prior to solving the problem. We still arrive at the corect an$r'er in the end.



FIGURE



2.9



Example of KCL



We may find the curent i more directly by considering it as eDtering the node and thus equating it to the other tkee currents leaving the node. The result is



i=-3+2+(-5)



= -6A



which agrees with rhe prcvious answer. We now move on to Kirchhoff's voltaee law



(KVL), which



states that



The alg€braic slm of lfie volaeges around atry closed path is zero.



26



Chapter



2



Resistive Circ!its



As an illustration, application of this statement to the closed path abcda ofFig. 2.lO gives



-r! +



01



-q=0



(2.'7)



where the algebraic sign for each voltage has t€en taken as positive when going ftom + to (higher to lower potential) and negative when going from to + (lower to higher potential) in taversing the element. Using this convention we are equating the sum of the voltage drops around the loop to zero. We could use the opposite convention as well, in which case the sum of the voltage dse6 is zero.



-



-



FICURE



2.10 Voltages around a closed path



As in the case of KCL, we shall not attempt a proof of KVL. llowever, to ilof(2.7), let us assume that its right member is not zero. That



lustrate the plausibility



is,



-Dr



-



D1



- Dr=O+0



The left member of this equation is by definition the work .equired to move a unit charge aroudd the path @dcra. A lumped-parameter circuit is a conrerydtiye system, which means that lhe wor* r€quited to move a charge arou[d any closed path is zero, (This is proved in a later study of electlomagtetic theory.) Thus our assurnption is not valid, and O is indeed zero. We should point out, however, lhat all electrica.l systems are not conseffative. In fact, electrical power g€neration, mdio *"ves, and sunlight, to mention oDly a



few, are consequences of nonconservadve systems, The application of KVL is independent of the direction in which the path is traversed. Consider, for ex^mple, the path adcba in Fig. 2.10. Summing the voltages, we find



r-rr+or=0



which is equivalent to (2.7). In geneml, a mathematical representation for KVL is



),,=o



(2.8)



D" is the nth voltage in a loop of N voltages. The sign of each voltage is cho' sen as described earlier for (2.7).



where



Section



2.2



Kirchhoff's Laws



.,.f



EXAMPLE



2.4



As an example of the use of KVL, let us find o in Fig. 2.11. TraversiDg the circuit a clockwise direction, we have



ir



15+o+10+2=0 or o



=



3 V. Suppose that we now pedorm a counterclockwise traversai.



case,.



In



such a



L5 2 10 D=0



or u = 3 V, which, of course, is the same result obtained for the clockwise traversal.



- 2V +



d



FIGURI 2.11 Circuil to illustrate



Still another version of LVL for Fig. 2.11 yields



l)+10+2=15 where lhe sum of the vollageq wilh one po)aity is equated io the sum oflhe voltages with the opposite polarity. State-d another vay, the voltage rises equal the voltage drops, which is another statemrcnt of KVL. Finally, we may solve for o direcdy by noting that it is the voliage or. and thus is equal to the sum of the voltages ftom to c through the other three elements. That is, the voltage between two terminals is the same regardless of the path taken betwe€n them. Thus in Fig. 2. I I we have



,



r:=15-2-10=3V ln each ot the previous examples. KVL has been applied around conducring paths, such as arcda above. The law, however, is valid for an) closed path. Consider, for instance, the path dcdd of Fig. 2.11. We note that movement directly ftom d to c is not along a conducting path. Applying KVL to this closed path yields ,.. + 10 2 = 0, where o." is the potential of point a with respect to c. Thrs 12 V. We could also have chosen the path arca, for which a*



+



-



-15 + Therefore, o- = - 12 tain the same result.



EXAMPTE 2.5



28



v,



r)



- 0-



-- -15 + 3 - o". =



0



which demonstates the use of different closed palhs to ob-



Consider finding a, and u, in the network of Fig. 2.12. Summing the curents enter+ 2 i2 0 or ing node a gives -4 + 1 + tr = 0, or ir = 3 A. At node



iz:



r, -i' - : I A. At node c, i2+L-3= 0, or i3=4 A. Therefore, at node d,



Chaprer



?



Resistive Circuits



FICURI



2.I2



Network tor example of KCL and KVL



-5 A.



r': 0, or r, = Next, KVL about the path drcdd gives oz rr" = 0. From Ohm's ld,$ t2= 5i2 = -5 V. Therefore, -15 V. Before concluding our discussion of Kirchhoff's laws, consider the rctwork of Fig. 2.13, in which seveml elements are shown within a closed surface S. We recall that the current entering each element equals lhat leaving the device so that each element stores zero net charge. ThereforeJ the total net charge stored within the surface is zero. requiring that



-t -



1



-10 a



-



-



-



o,:



tr+i+t3+ta=0



s.



This result illustrates a generalization of KCL, which states The alg€braic sum of the currents enlering any closed sllrface is zem.r



To illustrate the plausibility of the generalized KCL, let us writer KCL equations at nodes a, b, c, and d of Fig. 2.13. The results are 'The suh.e cmot



pass rbmugh an el€m€nt, \drich is comidered



to be concentmred ar a point in



lump€d-paraneter circuils. Section



2.2



Kir.hholf's



Laws



to



i,=i5-i"



n--n-i.-,,-no ,]3=-i+rs+tro '



i4



=



tu



+ ri +



re



Adding these equarions yielG



ir



+rr -ir-ta=0



as noted previously. From the general;zed KCL. we se€ immediately in Fig. 2.12 for a surbce en. closing points a, b, c, d that 4 + 2 3 0, or i, = -5 A.



^id



-i, -



-



EXERCISES



2.2.i



Find i a\d Answer



o*.



-l A,62\



t4o



Find 1) and i. Answet l'1 V,3



A,



EXERCTSE 2.2.2



30



I



Chapl€.2



Resistive Circuits



-



2.3 SERIES RESISTANCE



,



.



AND VOLTAGE DIVISION



Now lhat the iaws of Ohm ,n.l Kirchhoff hare been introduced. we are prepared to anallze resistive circuirs. We begin with a rjmp,fe cir.ril. which we define as one rhat can be completely described by a single equation. One t'?e, which we shall consider in this section, is a circuit consisting of a single closed path, or loop, of ele.uy i. Then Ohm,s law and llents. By KCL each element has a comm6n "orren'i,in i that completely KVL. applied around the loop yield a single equation describes the circuir. Elements are said to be coonected in series when rhey all cary the same cua_ rent. Clearly, the networks of this section consisl entirely of elements connected in s€ries. An impoitant circuit of this type, consisting of two resistors and at1 independent.voltage source, provides an excellent starting point. We shall first analyzjthis special case and then develop the more geneml case. A singleloop cilcuit having two.resistots and an independent voltage source is shown in Fig. 2-14(a). The first step in the analysis procedure is the assignment of curents and voltages to all elements in the network. In this circuit, it is obvious from KCL $at all elemenls carry the same currcAt. We may arbitrarily call this cur_ rent i in the direcrion shown guo,,r," iirr. ,."r".'"ii"" "ii.ip',i is not neces_ true current direction in rMking assignments. The corect assignment sary, as we will see, and usually is nol possible. even for the e,\p€rt.) We next make the voltage aslignments ior Rr and R2 as or and o?, respectivel',. These assignments are also aibit€ry but in the figure have been chosen to satisfy Ohm's law fd a posilive algebraic sign.



ii



Ci*t*i..j.



.



i



flCUXt 2.14



(a) Sin8le,joop



circuit;



(b) equivatent circuit



The second step in the analysis is the applicatio4,of KVL, which



ields



D=U!+D) where, from Ohm's law,



tr = Rti = Rzi



az



(2.9\



Combining lhes€ equarions. we find



o=Rti+Rzi Section



2.3



series Reshtance and VolraSc Division



31



and



ol



Rt



, , P,= R,= tR=E o' respectively. The total power absorbed is



Rt+R, =,C---J



=,'



The power delivered by the source also equals oi, indicating tlat the power delivered by the source equals that absorbed by ltr and Rz. This result is known as cozser'oo -



tion of power ,sometifies also referred to oflen useful in circuit analysis. EXAMPTE 2.7



as Tellegen's



theoreml. a property which is



o = 120 sin r V and or = 48 sin r V in Fig. 2.15. IJt us Dow deter&, i, and the instantaneous powe! associa&d with each element. By KVL, l,r = 72 sin I V, and by (2.131.



Suppose that



mine Rr, u2



=



r



?2 sin



r=



-29 90+R,



t2o s;n r



- 60 O. Hence .lR": R, + n, = 150 O, and, ftom (2.10), t:(120sbt)/150:0.8sintA. The instantaneous power to Rr and R, is h: Rri'1= 38.4 sin'?tW and p, = R i2 = 51.6 sin'?rW. Thus the power delivered by the source is 96 sin'? , W. The power absorbed by lR, is n,i'1 = 96 si r W, which yields Rr



which of course is the power delivered by the soulce-



FICURE



2.15 Single-loop circuit



f,et us now extend our analysis to include the series colnection ofN resistoN and an independent voltag€ source, as shown in Fig. 2.16. This is a voltage divider with N voltages. KvL gives



l)=t)r+02+.



Seclron



2.1



Series Rerisrance and Volt.Be Division



33



FlGUnt 2,16 Single-loop circuit with N



series rcsislors



in which or



=



Rri



a, = Rzi



(2.r4)



tr'v



=



&i



Thergfore,



a-Rri*Rzi+' ..+nNt Solving this equation lor



i



.,



yields



" Rr+&+...+R,



(2.15)



i!



I2t us Do\r select n" the.ciicuit of Fig. 2.14(b) so that (2.11) is satisfred. Equivalence of (2.11) and (2.15) requires that (2.16)



Thercfore, the equivaleot r€sistance of lV series Esistors is simply the sum of the dividual resistances. Substitutiry (2.15) and (2.16) into (2.14), ve find



34



Chapter



2



Resistive Circuils



h-



R,



R. R?



(2.t7\



R" o':&o



which are the equations describing the voltage division property for y'{ series resistors. Again, we see rhar rhe vokage divjdes; direct pro;o;do;to rhe resistance. The instantaneous power delivered to the serie; c;bination, ftom (2.2) and



(2.17), is



^-Dl+oi..-."i 'hRzR*



-fr*- -*!;,'



=H,* =ftr="



This power is equal to that^dejivered by the souce, verirying conseFEtion of power for rhe series connection of /V resistors EXAMPLE



2.8



In Fig. 2.16, ler us find or and i if/V = 10, R, = 60 O, the other nine resistances are each 10 0, and o.= 75 V. The equi lent resistance is X, = 60 + 9(10) = 150 ft, and by voltage divisio& ur =



B) ohm's law we



have



m lso(75)



-



l0 v



.D'15 r-E=150=0.5A



EXERCISES



2,3.1



Find (a) the equivalent iesistanc€ seen by the soutce, (b) the curetrt i, (c) the power delivered by the source, (d) or, (e) oz, and (f) the minirnum wattage re4riiJ for the



6-(}



resistor.



Answa (a)



Section



2.3



U



O;



b)



0.5 A; (c) 6



W



(d) 5 V; (e)



Seriet Resistan.e and votiaEe Divisjon



-4



V;



(f)



1.5



w



-tt



-r-



0Ir



r0r}



8J)



EXERCtSt 2.3.1



In Fig. 2.14(a),



2-3-2



o= l6e'V,...2=4e'V, and&=24



instantaneous power delivered to Rr, and (c) the



curent i.



O. Find (a) n,, G) the



Aa\wer (a) 8 O; {b) 2e ,,W (c.) 0.5e ,A A resistive load2 requires 4 V and dissipates 2 W. A 12-V storage battery is available to opemte the load- Referring to Fig. 2.14(^), il R, represents the load and o the l2-V battery, find (a) the culrenr i, (b) the necessary resistance Rr, and (c) the minimum wattage of Rr. AA'wet lat 0.5 Ai (b) t6 Or (c) 4 W ln rhe voltage divid€r sho*n. the power delivered by lhe rource is 9 mW and D = D/4. Find R, o, r ,, and i. Answer 9 kA, t2 V, 3 V, 0.75 mA



2.3.3



.2.3.4



--i*



l ko 4r(o



EXtRCTST 2.3.4



2.4



---- PARALLEL



RESISTANCE AND CURRTNT DIVISION



Another important simple circuit is lhe single-node-pair resistive circuit. ln anahzing these networks. we shall first examine a special case and lhen develop the m;re general case, as was doDe for the single-loop network. Elements are connecled h Waltel when the same vohage is corunon lo each of them. The single-node-pair circuit shown in Fig. 2.1?(a) is a pamllel condection of two resisto$ and aD independent curent source sjnce by KVL all three elements have the same voltage o.



'zA



36



bar'



Chapre,



is an elencnt or coliecdon of el€menb connecred b€tw@n the ourpur termimts. tn rhis case lh€



?



R;sisnve



Ci(uils



(a) FIGURE



2.17



O)



(a) Single-node-pair



circuit;



(b) equivalent circuit



Applying KCL at the upper node lelds



i=it+i2 where, ftom Ohm's law, (2.18)



L=



Gza



Combining lhese equations gives



' and



i:G1o+Gzt)



solvilg for o, i{e find



i Gt+G2



"



(2.19)



In the chcuit of Fig. 2.17(b), if Gp is selected such that



i



(2.20)



D": ^ =t) ' lrp



then the network is an equivalent circuit to that of Fig. 2.17(a). Comparing (2.19) and (2.20). we see that



(2.2t)



Go=Gt+Gz



Clearly, Gp is the equivalent conductance of the two par"allel conductances. In telms



of rcsistances, (2.21) becomes GP



111 R, Rl RP:



R,



R,R,



R, + R,,...



(2.22)



Therefore, the equi\ale[t resistance of two resisto$ connected in paiallel is equal to the product of their resistalces divided by their sum. It is interesting to note that G, is geater thaD either Gr or Gr; therefore, & is less than either Rr or R2. From this re-



Section



2.4



Pahllel Resistance and Curent Division



37



sult, we see that connecting resistors in pamllel reduces the ovemll resistance. In the special case R2 fir, we see from (2.22\ that Rp = &/2. Substiruring f2.19) inro (2.18) givas



:



tt.6=



cn .Gz



Grt Q.23\



i divides between conductances Gr and G in dtect proportion to thei conductaDces, demonstmting ttrf, principle of current dir:ision. The circuit, of course, is a arrent diaider. Ilis cortuion pmctice to give the rcsistor !alues in circuit diagrams in ohms (resistance) and oot siem€ns (conductarce). In tems of resistance lues, (2.23) becomes The clrrent of the sourc€



. ' i,



?R:



+R,



(2.24\



RI



- RI+R' Therefore, the curent divides in irverse porpottion to the resistances. We see that the larger curent flows through the smaller resistance. The power abso6ed by the pamllel combination is



b+p,-Rti?+R?i7 Rit - (R, r 6zf R, t =



P) i'z



14,



-



RrRr.,



R' + R'?-I- =



^yn,



',r



which equals that delivered to the rctwork by the current source.



EXAMPLE 2.9



R



Suppose in Fig. 2.l7{a) rhat =3O,nr-6O.andi=3A.Then.from(2.22). Rp = (3)(6)/(3 + O. Fitotrl. (2.24), t, = 3(3) 2 A, and i, =;(3) = I A. The voltage a = R1L R2tr: (3)(2) = 6 V. The current of the source flows



6):2 :



tbrough an equivalent resistance



D:



-



of Rp. Hence the



voltage



is also given



by



R,i = (2)(3\ = 6V. I-€t us now consider the more geneml currett divider of N parallel conduccurent source, as shown in Fig. 2.18. KCL gives



tances and an independent



i=L+i2.+...+iN 38



Chapter



2



Resistiv€



Circlits



for which



L=Gto



?:.G* (2.25)



iN



:



GNo



Therefore, we have



i=Gp+G2aI'..iGxo ftom which



G+



FICURE



G1



+-.'+GN



(2.26)



2.18 single node pair crrcuit l^rth N pdrallel conductan(c,



Il we now select cp in Fig. 2.1?O) such that (2.20) is satisfied, then (2.26) requires that



Gp=6+G.+..



+



CN:>Gi



Q.27)



In terms of resistances, this equation becomqs



I



l



l



RP Rt R,



I {r n* *u' R'



(2.28)



Hence the reciprocal of the equivaletrt relistance is simply the sum of the reciprocals of the nsistanceq



Seclion



2.4



Parallel Resistance and Curent Division



39



Combining (2.25)-(2.28), we find



, .



=4,=\, G" Rr



G1.



R,,



(2.29)



G^ G,



R^ RV



Again the currents divide in inverse proportion to the resistances.



TXAMPLE 2.10



\rye observe



in (2.28) that for N > 2 an expression for



Rp is more complicated than



(2.22). Formulas could be obtained, of course, for 1V : 3, 4, and so on, but it is (2.28) direcdy. As an example, suppose for N = 3 that usually easier to O, and Rr = 6 f,I. Then &:4o",R2 = 12^Wly



rl1ll Ro4t262 and Rp



EXAMPTE 2.11



= 2O.



Let us now find the equirale t resisfance R* of the network of Fig. 2.19(a),



as



viewed ftom terminals x-y. Such reductions are very helpfirl in analyzing many t]?es of circuits, as we shall see in the next section. The process is carried out by successive combinations of parallel and series conn€cted resistors. In Fig. 2.19(a), the student often errs in takir4 combinations sucb as the 7- and 12-O resistors to be in sedes. we see, however, that at nod€ a, a cuffent in the 7-O resistor would divide betwe€n the l- and 12-O resistors; hence they cannot be in series. The l- and 5-O rcsistod, however, would carry the same current. Therefme they are in series, having an equivalent resistance of 6 f} as shown in Fig. 2.19(b). we now obserye tlat the same voltage would occur aoross the 6- and l2-O resistors. indicaling a parallel 4 C), as shown in, connection having an equivalent resistance of (6)(12)16 + 12) Fig. 2.19(c). It is apparent that the ?- and 4-() resistors of this network are in series,



:



yielding an equivalert resistance for the entire network of ll O [Fig- 2.19(d)]. Therefore, from terminals r-], the network could be rcplaced by a single resistor of 1l O. This is useful in determining, for inslance, the power delivered by a sourco connected to terminals.rt. Suppose that a 22-V source is applied. Then the cunent flowing ftom the source is i: i = 2 A, which gives an instantaneous power p(t) = (22\(2, - 44 W delivered to ttre resistor network.



40



c!apter2



Re5ielive ( rr.urts



FICURt 2.19 Steps in determining the equivaleni resistance of a network



EXERCISES



2.4,1



Find the equi lent resistance seen by the souce ard use the result to lind



and o. Answer 8



Q,6 A,



5



A, 30 V _i--



3f,,



2,4.2 If i = 9 A and ,; = 6 A h Fig. 2.1?(a), find the ratio RzlRr. arsr", I 2.4.3 A load requires 3 A and absorbs 48 W. If only a 5-A curent souce is available, the required resistance to place in parallel with rhe load. easqcr 8 O



2.4.4 In Fig.2.l8, if



2.4.5



,V



= 3, Rr:9O, R2=12o",



and



o:12sinrV'and



stantaneous power delivered by the curdnt souc€ is 24 sin, , and (c) rr. Answt (^) 24 A. @) 2 sin r A; (c) 0.5 sin r A



Find the equi%lent resistarce answer 10 Q,2 A



Section



2-4



seeD



find



the in_ W, find (a) Rr, (b) ,,



by the source and the curent i.



Parallel Resisrance and Crirrenr Division



41



2.5 ANATYSIS EXAMPLES



In'this section we analyze s€veral circuits to illustrate tho analysis have considered so



EXAMPII



2.12



Let us find



i, or,



techniques we



hr



and o", in the circuit of Fig.



2.20(a) KVL and Ohm'slaw give



30+mi+30t+20+50t=0 FIcURE 2.20 rarsingle-loop(ir(uitilb)equivalentcrrcu,t:



42



Chapt€r



2



Resistive Circuits



r.r anolher equi\,alcnt (ircurt



__i*



which simplifies to rhe reduced equation



-tb + rooi=o Therefore, we hav€ i = 0.1 A and or = 30i = 3 V, The circuir of Fig. 2,20(b) is equivalent to that of Fig. 2.20(a) as far as the curent j is concemed siice lottr circuits are described by tle reduced equation. ln fact, Fig. 2.20(b) rnay be obtained directly ftom Fig. 2.20(a) since in the latter all rhe elefients cary the same cudent i and thus are in series. Adding the three resistances yields the equil"lent resistance oi 100 O, and adding the source voltages algebraically gives the equiralent source of 10 V. This may be easier to see in Fig. 2.20(c), where the 20-V souce has b€en mored nexl lo rhe Jo-V soulce. To find o", we apply KVL in Fig. 2.20(a) to the loop consisting of the direct path rd, the ?0-O rcsistor, and the 30-V soulce. This yields



-xab ftom which D",



EXAMPLE



2.I3



-



20t



+



30 = 0



28 V.



l,et us show that conservation of power holds for the circuit of Fig. 2.20(a). Since by Example 2.12 we have i = 0.1 A, the powers absotbed by the resisto$ are



w



= Pro =



20(0 1)z



0.2



30(0



= 1)'? =



0.3 W



=



50(0 1)?



-



0.5



Pmo



aDd



P:oo



w



The 20-V source is also absorbing power since the current enters its positive terminal. This power is given by



Pzov=2ni=2W The l0-V source delivers power given by



P3ov=30i=3W Since 3



-



2 + 0.2 + 0.3 + 0.5, the delivered power equals the absorbed power



and thus conservation of power holds.



EXAMPLE 2.14



l,et us 6nd i and o a[d show that conservation of power holds for the ciroir of Fig. 2.21,1a *lnd\ tkee conductances and two independent curent sources ate conDected i! pamllel. Applying KCL to tle upper node yields 10 sin ?r,



-



0.01o



-



0.02o



-



5



zr



-



5)



(10 sin



Section



2.5



Analysis Exampt€s



-



-



0.07o



=



0



0.1o



-



0



43



vlo.ors



t)



FICURE



{)'n



io.o2s



2.21 Single-node-pair circuit



5) A connected apparent ftom this result that a current so&c€ of (10 sin ,., of 0.1 S (a l0-O resistor).would be an equiralent circuit as frr as i) is concerned lse€ Fig. 2.17{b)'l. Solving for o and subsequently for i, we have



-



It is to



a conductance



1)=l00siD'rt-50v i = o.o2t = 0.02(100 sin - 50) : 2 ,in tt, ''t



IA



Now let us consider coDservation of power for the circuit of Fig. 2.21. The to-



tal power absorbed by the conductances i5



px = Got:z = 0.1(100 sin zr - 50)' = 1000 sin'?zr 1000 sin 'rt + 250 W The power delivered by the lellmost source is



p, = l0 sin rt(100 sin zt



-



50) W



Similarly. the 5-A soutce delivers



P,



=



5(10J sin



7t -



50) W



Thus the total power delivered by these souces is



pd.t: h +



P2



=



1000 sin'?



tt -



1000 sin



tt + 250 W



which equals thal absorbed by the conductances.



IXAMPIE



.



2.15



us find i, o, and the power delivered by the source in the circuit of Fig. 2.22(a). We begin by obtaining successive combinations of paiallel and sedes resistor connectiom. The 4- and 8-() resistances (in series) add to give 12 O. These 12 O are in parallel with the 6-() resistor, giving atr equivalent \rdlue of (12X6)/(l? + 6) = 4 O [Fig. 2.22(b)). We now add the 12- and 4-() re,sistances, which are in pamllel with the 16 o, giving (16X16)/(16 + 16) = 8 0, as shown in Fig. 2.22(c). This is the equivalent resistance as seen looking into the circuit at terminals a-r. The equivalent resistance seen by the source, ftoh Fig. 2.22(c) is n* = 2 + 8 10 O, so that the



t€t



:



curent ir



is



ir=roa-3A Therefore. the power delivered by the source is



a = (30)(3) :



44



Chapt€r



2



Resistive Cncuits



e0



w



FICUR[ 2.22 Circuit for analvsis example u5ing voltage division



By voltage division we see in Fig. 2.22(c) that



/8\



''=(2*r-:J:o=:+v which is the voltage across points 4-, in the circuit. Pioceeding to F19 2.22(b), we see that or is the voltage across the series combination of the 12- and 4-f,) resistors; hence, again using voltage division, we fiod



/a\ o,=l ' \12'4l lt -6v which is the voltage acloss points c-d in the circuit. In Fig. 2 22(a), 02 is the voltage across the series connection of the 4- and 8-() resistors Therefore, voltaSe division yields



/B\



" l- o/"



4v



Finally, by current divisiot in Fig. 2.22(b), 'xe h^'te



i=jir=lsA



' EXAMPLE 2.15



kt



us find the cunent i in Fig 2.23(a) We note that the two 6-() resisto$ on the right of Fig. 2.23(a) are equivalett to 12 O, which is in parallel with the 4-O resistor. This parallel combination is equivalent to (4)(12)/(4 + 12) = 3 O as shown in Fig. 2.23ab). If we now leplace the two series lesistors to the right of points r-j' and Section



2-s



Analysh Examples



4s



FICURE



2.23 Circuit for analysis example usinS current division



the parallel 3- and 6-cl .e.sistors to the left of t-) by their respective series and parallel equilelents, we obtain the circuit of Fig. 2.23(c). Using current division, ry€ 6nd



/'\



r=[t-eJx12=34 A second application of current division to Fig. 2.2314) ylelds



o ,=l ' \4t6'6,/ \i,-/l)"r=lo 4 \4/



EXERCISES



2.5,1



Find o", and the power delivered by the 5-V sourc€. Aasner 5 V, 0-5 W



.



30O , EXERCISE 2



Find



n



rpsistor.



46



40O



5



1



and colstruct an equilalent circuit having orle curent solllce and a silgte



Answu



R= 20Q, t=



Chapler



2



3



sh



ResGlive Ckcuits -



tA



directed upw.rd, 10O



rooa {rrn



( l)r",,o



"



EXERCtST 2.5.2



2.5.3



Find t' and t. Aruper 0.8,0.7 A



2,5,4



Find o and the power delivered by the souce. Arsr er 4 V,864 W



txfRclst



28O



2.5,3



4(}



EXtRCtSt 2.5-4



2.6 AMMETTRS, VOTTMETERS, AND OHMMETERS A good example of the useftilness of curent add voltage division is demonstrated in the design of simple two-terminal measuring instruments, such as a$meter$, voltmete$, and obmm€te$. An idaal ainneter frrcasules the curent flowing thaough its terminals and has zero voltage across its terminals. In contrast, Nt ideal voh cter measures the voltage aooss it! terminals and has a teaminal cunent of z€ro. An tdeal ohmmeter measres the lesistaDce connected betwe€n its teminals and delivers zero power to the resistance. The Factical measuring inst uments that we shall consider only approximate tle ideal devices. The ammelers, for instance, will not have zero terminal voltages_ Similarly, tbe voltmeters will not have zero terminal currents, and the obrDmeters will not have zero power delivered from their termiDals. A popular type of artunetet consists of a mechanical movement ktown as a D'Arsolval meter. This device is constructed by suspending an electrical coil be-



Section



2.6



Ammeter!, Voltmete6, and Ohmmete6



47



tween thc poles of a permarent maglet. A & cur!€nt passing through the coil causes a rotation;f thc coil, as a result of magnetic forces, that is proportional to the cu.rent. A pohter is attached to the coil so that the lotation' m meter deflection, can b€ visually observe-d. D'Arsonral meteas arc characte zed by therJ fltll'scale current, which is the current that will cause the meter to read its gleatest ralue Meter move_ ment! are conunon having hrll_scale curretlts ftom l0 /,A to l0 ltlA. An e4uivalent circuit for th€ D'ArsoNal meter consists of an ideal atrljneter in series withl rcsistance Rir, as shoen in Fig. 2.24. In this circuit' nM reprcsents the resistance of the electrical coil. Cleady, a voltage appears across the atnmete. telminals as a result of the current i iowi!8 through ft|'. Rll is usually a few oluns, and the terminal voltage for a full-scale curent is nominally ftom 20 to 200 mV.



FICURE



2.24 Equivalent circuit for a D'Arsonval nreter



The D'Arsonral meter of Fig. 2.24 is an armeter which is suitable for measuriru & curents not greater than the full_scalc curent lFs Sup-pose, however, that we *Ish to me""ue a i*rent which exceeds IFs. It is apparent that we trIust not allow a cGrent greater than IFs to ffow through the device. A ciicuit to accomplish this is showl in Fig. 2.25, *ttere & is a pamllel resistanae that reduces the culrent iowinS throueh the meter coil.



FICUnE



2,25 Ammeter circuit



From curretrt division we se€ that



r^ =



*f ur*



(Clearly; this is where iFs is tbe current which produces lns in the D'ArsoNtsl meter' have for Re, we Solving measute) can the amlneter the maximum culrent



,



RrlFs



(2.J0)



A dc voltsneler can be constructed using the basic D'Arson\al meter by placing in series with the device. as shown in Fig 2 26 lt is obvious that the



a resistance R,



48



Chapler



2



Resistive Circuits



,'



^



FIGURE



full-scale voltage, o



=



2.26 Vok;eter circuit



oFs, occlEs when the mdter



cudent is 16. Therefore, fiom



KVL,



'



-DFs



+



R"IFS



+



R/1Fs



-



0



ftom which



n:5



- n.



(2.31)



The current sensitivity of a voltmeter, e\pressed in ohms per volt, is the value obtain€d by dividhg the resistance of the voltmerer by irs firll-scale voltage. Therefore.



o/v (I|ote.



raring



R' - L-l-4" =



(2 J2)



"-"



means apFoximately equal to.) simple ohrnmeter circuil emploing a D,Arson\,al meter lor measuinp an unlnown resistance & is shown in Fig. 2.27. ln this circuit the batrery E cauis a cuffent i to fow when R. is connected into the circuit. Applying KVL, we have



A



'



-E + (R, + R',. + R)r = 0



ftom *{rich



a,={-1n,+n; I



i I



I



section



I



t



2-6



Ammete.s, Vokmeterc, and Ohmmeters



49



We select E and R, such that for



& = 0, i = IFs. The.efore,



,E '*-i'-R" Combiniog ttre las two €quations, we find



*.= (ls -,),^. \r 'J'"n '",r '



(2.33)



A very popular genonl-purpos€ meter *ftich combine'e the thre€ pr€viously described circuits is the yOM (volhleter-ohnlrneter-millialnmeter). In the VOM, proand & so that a wide dynamic range of operation is visions are made for changing



n



provided.



EXERCISES



2.6.1 2,6,2



2.6.3



A D'ANotrval meter has lFs : I nA and R|/ : 50 O. Determine lte in Fig. 2.25 so that tFs is (a) 1.0 mA, (b) l0 mA, and (c) 100 mA A'r,er (a) I!finite; (b) 51556 O; (c) 0.505 O In Fig. 2.26, determine R" and the O/V rating for a voltmeter to have a full-scale voltage of 100 V usirg a D'ArsoN"l meter with (a) Rv : 100 O and,aFs = 50 }tA and O) nv = 50 O and 1Fs : I dA. Answer ({ 2MA,20 kO/V; (b) 100 kO, I k0/v what voltage would each meter d€sign of Exelcise 2.6.2 measure in the ckcuit shown? why are the two measuements differcnt? Answer 99.5



Y,



.9



\



EXtRCtSE 2.6.3



2.6.4



The metq movement of Exercise 2.6.1 is used to form the ohmmetq circuit of Fig. 2.27. Determine R" and so that i = 1Fs/2 lnA when & = l0 kO Allr)ral 9.95 kO, l0 V



t



2.7 PHYSICAT RESISTORS Resiston are manufacurred from a \adety of materials and are a!"ilable in dany sizes and values. Their chancteristics include a rcminal resistarce value, an accu_ racy with which the actual resislatce aPproximaie'! the nomioal value {ktrown as tor-



50



Chapter



2



Resillive Circuits



a stabilily a5 a function of temperature, humidity, and other enviroDfiental factors. Th€ oost common t)rpe of resistd found in electrical circuit! is the carbon compo6ition or carbol 6lm r€sistor. The composition t]?e is made of hot-pressed carbon granules. The carbon filln device consists of carbon powdet which is deposited on an insulathg substrate. A tt?ical resisto. of this type is shown in Fig. 2.2E. Multic.olored baads, shown as c, b, c, a\d % loleratrc'-, are paidted on the resistor lrcdy to hdicate the nominal value of the resistance. The color code for the bands is given in Table 2.1.



e/ar.e), a power dissipation, and



l



tll FICURE



2.28 Ca6on



resistor



Bands, a. D, and c give the nomi0al resislance of the resistor. and lhe tolerance band give€ the percentage by which the resistance may deviate from its nominal %lue. Referring to Fig. 2.28. the resistance is



n = (10d+



r)l0t



%



tolerance



(2.34)



by which we mean that the % tole.ance of the nominal resistarce is to be added or subtracted to give the r:anga in which the resistance lies.



ffi?



coro, coau ror carbon



Resisrors



-2



SilY€r*



-l



Gold* Bla.k



0



5 BIU€



I Red



z



Otuge



3



G",y



6



'| 8



9



% Tolennc€ Band



!5%



Gold



Silve.



.*flFF @l@ ady



EXAMPLI 2.17



ro



b.rd. oly



Suppose that we have a resistor with band colors of yellow, violet, red, and silver. The r€sistor will have a value given by



R:(4x10+j\xtO,!tO% 4700 ! 470 A Therefore the resistance value lie,s between 4230 and 5170 O.



seclion



I



I



I



2.7



Ph)5i@l Reirrors



51



Values of carbon resistors range from 2.7 to 2 2 x 107 O, with lEttages ftom w. For resistance \alues less than l0 O, we see ftom (2.34) rhat rhe-thitd badmust be gold or silver. Cafton tesisto$ are inexpensrve but have lhe disadvatrtage of a rclatively high variation ol resistance with tempemture. Another resistor type which is commonly used in adicatioos requidng a high i)wer dissipation is the wire-\dound r€sistot. This device coosisB of a metallic wire,



+



to 2



usualty a nickel-chromium alloy, wormd on a ceramic corc. Irw-tempemufe_ coefficient wire permits the hhicatioo of resistors that are very paecise and stable, having accuracy and stability of the order of 1l% to !0-0/]1 The metal film resistor is arother valuable and useful resistor type. Thes€ resistors are made by vacuum deposition of a thin layer on a low-thermal-expaNion substrate. The resistance is then adjusted by etching or grinding a p*tern though the filIn. Accuracy and stability for tbese re6istors approaches that of wire-wound types, and high resistanc€ values arc much €asier to attain. In the futwe, the reader may be more likely to encounter resistors io it teSr4ted cr'rcairs, which were developed in the late 1950s and came hto their own in the 1960s. An integrated circuit is a single monbllthi. cbip of sefiicdtductor (firte'iil with conducting properties between tho€e of a conductor and an insulator) in which active and passive elements are hbricated by difrision aDd deposition proc€sses. lnilable on chip€ about i i!. legrated circuils containing hundreds of elements arc square. An integrated-circuit resistor has the tyPical structure thowtr in Fig. 2.29.



a



FIGURE



2.29 lntegrated-ciicuit retiator



EXERCISE



2.7.1



Find the resistance mnge of catbon tesistors having color bands of (a) brown' black' red, silvei: (b) red, violet, yellow, silver: and (c) blue, gray, gold, gold. .rmwar (a) 900-1100 O; (b) 243-291 kA,, @) 6.46-7.t4 O



2.8 SUMMARY This chapter has b€en devoted to the /esistor, the simplest of the two-terminal elements. We have considercd the resistance afld conductance of a resistot, their uDits (ohms arld siemens) , some examples of rerirdye ctlcuitr (those made uP of resis^nd



52'



Chapter



2



Reshtive Circuits



tors and sources). We have covered Olrn's lltw, o = Ri, the most basic Fitrciple of circuit thmry, and the relation p = Ri'?for tbe power dissipated by a resistor. We now know aboul Klrchhofr's cwrent and ,oltage lnws ard how to apply them to anallze resistive circuits. We have considered r€n€r and pararl€, re€istq connections and the concepts of roltage a d currerd division to simplify tlrc analysis by obtaining equivalent circaitr. We have also learned of metering devicns-arntneters, vohmeters, and ohmneters-whiclt may be used to measure currents, voltages, and resistances in actual circuits. Finally, we have briefly considered physical resi$tors and the color code for reading their resistance values in the cas€ of caaboo resistors, sis



As we will see in Chaptq 4, the conc€pts of this chapter are vital to the analyof general resisrive circuits.



PROBTEMS



2.1 A l-ko



r€sistor is cormected to a battery and 6 trrA flow. What curent will flow if the bal tery is connecaed to a 30-O resistor? Wllat is the terminal voltage of the battery?



2.2. A 6-V bxttery



is connected to the eods of



of the wre?



A toaster is essentially



a rcsistor that becomes



i! calries a current. If a to€sEa is of tm V fnd



dissipqting 9@ W at a voltage its current aJd its re9islgrce.



2.4



Find tbe energy used by a rdster witb a rrsistaDce of 12 O, which is openled ar 120 V for



2.5 2.6



Find i and od.



a



1000-ft length of conducting wire and a l2-mA curent flows. wllat is the resistance per foot



2.3



hot wllen



10 s.



Find



i', r,,



and od.



PROBTEM 2.S



PtoEt-EM 2.6



Chapter



2



Problem!



53



.



Find ir,



i,,



and o.



2.4 2.9



Find i atrd R.



2,10



Find i, o"r, and an equii'atent cncuit for



Find o.



i containinS a single source and a single resistor.



2-tt A lo-V souice



in series with several resistors



carries a cufrent of 50 mA, what resistance must b€ conrccted in seri€s with the source and the resistors to timit the curr€nt to 20 mA?



?ROBLf,M 2.7



2-12



PROf,ttM



.



A 50.V



source and lwo resisrors, nr and R,, ar€ comect€d in s€ries. trfnz = 4n,. find the voltages across the two resistors.



2.A



PROATEM 2.9



2-13.



A l2-V



source in series with a r€sistive load R caries a cur:rent of 60 mA. If a resistor Rr is added in series with lhe source and load, find



Rr so thal the voltage across



PROBIIM 2.10



54



I



Chapler



2



Resistive Circuits



il



is 8



V



2.14



Design a voltag€ divider to provide 4, 10, and 20 V, all with a conrmon negative terminal, ftom a 25-V source. The source is to deliver 25 mw of power.



2-ts



Design a voltage divider to provide 2, 6, 10, 14 V, all with a common negFtiv€ U,



^nd



2.16



telminal, ftom a 50-V source, which delivers 100 mW of power. A voltage divider is to be constructed with a



A current divider consists of l0 resistors in paiallel. Nine of them have equal resistances of 60 kO and the tenth is a 20-kO resistor. Find the equil"l€nr resisrance of the divide.,



2-22



60-V source and a number of 10-kO resistors.



Find the minimum number of



2.17



Find all the possible output vohag€s less than 14 V that can be obtained by construcling vokage dividers with a l4-V source in series with tbree resistors of 2, 4, and 8 O, respectively.



2.18



Find t and the power delivered to the 4-O r€-



2.19 Find o. 2.20 A 20-O



and, if th€ total current entering the divider is 40 mA, find the currenl in the tenth resistor.



resistors



required if the output voltage is (a) 40 V and (b) 30 v.



l0-O resisror, and a resistor R are comected in parall€l 1o form an equivalenl resistance of 4 O. Find R and the cunenl il cames lf a 6-A cuffenr souce is re,sisror, a



2-23



Find n and



j1.



480



-lL 6(}



,'l 24V



),n



30r)



4EO



PROBTEM 2.23



connected to the combination.



2.21



A current divider consists of the parallel connection of a 20-, a 40-, a 60-, and a 120 kO resistor. Find the equivalent resistanc€ of the



divider, and



if the total current



entermg



LU 2,25 2.2q



the divider is 120 niA, find the cunent in the 2O-kO resistor-



F]JJd



x a'rd i.



Find t, ir, and.,. Find all the possible values of equivalent rcsis, tance tfiat can be obtained by someone baving thee 6-r) resistors.



PROSIEM 2.18



PROBIIM 2,19



.Chapter



i I I I



t



I



2



Problemt



55



l6lr'i



I' 12



to



PROBTEM 2.24



PROBTEM 2.25



' 2-27



PROBLE|i 2.28



Tbo 2 kO resistors are in series. Whcn a rcsistor lQ is connected in parallel with one 'of , tllem- the re\istance of lhe comhnalion is 2.100 O, Find R and the cu[ent it canies if the combinition is connected across the nals of a l2-V



battery.



2.2E



Find i, and



2.29.



Find /.



temi-



i I



I



N



60



v



i.



2.30 Find, and i. 2.3r Find i and R. 2-32 Find i



.56



'



PROBLTM 2.29



Chapter



2



Resistive Circuits



3f,



a



t24a



\ l,r2A ,



+t



I



PROEIIM 2.30 PROBTEM 2.33



2.35 Fird rhe power ab.orbed by the 6-Q resislor. 2i36 Find R and o using curent and voltage division.



.



. 2.39



PROBLEM 2.3.I



2.33



(a) Find the equivalent resistance looking in rerminals a-, if terminais c-d arc ope+ and tl terminals c-d are sho.ted together. (b) Find the €quivalent resistance looking in terminals c-d if terminals a-b arc open, and if terminals



d-, 2-34



are shoited together.



Find i and



,.



Find ir and t . ,Find j,, jz, and



r.



2.,10 A D'Arsonval



rneter has a full-scale qment of I rnq and a r€sistance of 4.9 O. If a 0.1-O panllei resistor is used in Fig. 2.25, what is lFs? What voltage occurs across the meter?



2.41 A n.oiD}-A/V voltmeter has a tult scale voltage of 120 V What curent ffows in the meter when measuring 90 V?



Find i, ol, and o,.



rov



2.J7 2.38



2.42 T$o lo-kf)



resistors are connecred iD sefles across a 100-V souice. What voltage will the voltmeter of Prob. 2.41 measure across one of



I



the lO-kQ resisrors? Repeat for two 1-MO re sistors in series.



0, as it is for a practical souce, rhe source can never deliver an infinite current, ds.an ideal source can.



FTCURE



5.23 Pracrical volrase source connected to a load



Rr



For a given practical vohage source tfixed values of D, and ns in Fig. 5.2J). . the load resistance Rr determines lhe current drawn from thi terminals. F6r exam_ ple, in Fig. 5.23 the load current is



.



'-R,+&



(s.2s)



Also, by voltage division we have



Rta.



&+R.



(5.26)



Therefore. as we vary Rr both i and D vary. A sketch of D versus R. is shown in Fis. 5.24. along wilh rhe ideal case. which is dashed. For large values ol R, relative io &, o is very nearly.equal to the ideal !6lue of or. (ffR! is infinite, colresponding to an open circuit. then 0 is oi., Seclion



5.4



Practical Sources



135



'0,cr FICUnE



5.24 Practical and ideal voltage source chardclerietics



We may replace the p.actical voltage souce of Fig- 5.23 by a practical current source by rcwriting (5.24) as



'R,& which



if



we deline



"R"



,:i,n described by this equation with voltage u and curent i is.shown in the shaded rectangle of Fig. 5.25. The circ{it is thus a practical current source and is seen to consist of an ideal cqrrent source in parallel with an internal resistance.



A circuit



FIGU*I 5.25 Practical current source connected to a load



RL



Figures 5.23 and 5.25 arc equivaletrt al the terminals if Rs is the same in both if (5.27) holds. This equivalence is valid, moreover, if the ideal sources are independelt or dependent sou$es. In the case of indepeDdent sotrrces the two pmctical souces are simply the Thevetritr and the Norton equivalents of the same cases and



circuit. By current division, we



ind, in Fig. 5.25,



.



Ri,



'-n+R. 136



Chapter



s



Nebvork Theorens



(5.28)



Therefore, for a given current source (fixed values of L and Rs), the load current depends on lRr, A sketch of i versus R. is shown in Fig. 5.26, along with the ideal case, whieh is dashed.



nCURE 5.26 Practical and ideal current source characterisrics



EXAMPLE 5.11



Very often network analysis can be gearly simplifi€d by changing practical voltage sources to pmctical current sources, and vice versa, by the use of Fig. 5_23 and 5.25, or equivalently. by means of Norlon's and Thevenin s theorems. For example. suppose that we wish to find the current i shown in Fig. 5.27. We could solve the problem in a number of wzys, such as replacing everytbing except the 4 O resistor by its Thevenin equitalent and rhen frnding i. However, we illushate instead the method of successive traDsformation of sourses.



FICURI 5-27 Cjrcuit with tuo practical sou.ces



Let us begin by replachg the 32V source and inlemal 3-O resistancc by a praclical cuffenl source of a 3-O internal resistance and a +-A ideal source. Then Iet us replace the 4-A source and the intemal 2-(! resistance by a voltage source of 2(4) = 8 V and a 2-O intemal reisrance. We are applying, respectively, Norton,s and Theverin's theotems, or equivaleDtly, (5.27). The results of these two source transfomations arc shown in Fig. 5.28. We may now combine the paralel 3- alld 6-(} resistances and the series l- and 2-O resistances, as shown fu Fig. 5.29(a), rcp€at the souce tansformation procedure. Contfuting the Focess, as shown in Fig. 5.29(b), (c), and (d), we finatty arrive at an equiulent circuit (itrsofa. as t is concerned) which can be analyzed by inspection. In this case, ftom Fig. 5.29(d), the answer is



a



!4,!g i ;!oi=ze Section5,4 Practiolsources



137



IICURE



'



5.28



Result of two transfomations applied



FIGURE



5.29



Step6 in



obhinins i in



to Figj 5.27



FiE. 5.28



This Focedure may seem unduly 1() g, but it should be obseryed that most of the steps may be



EXAMPTE 5.12



caried out mentally.



We often combine sources as we do resistors to obtain equivalent sources. For example, if we are interested only in i in Fig. 5.29(d), we may combine the three series resistors, as we know, but we may also combine the sedes sources, as.we did in Sec. 2.5. They leprcsent a net source of$ = 16 V with a poladty like that of the



-f



larger source- Thus an equivalenl circuit insofar as i is concened is that of Fig. 5.30. Similarly, we may combine parallel curent sources to obtain an equivalent source,



f-*-l



r6v(l)



L,l



|



FICURI 5.30 Circuit equivalent to Fjg. 5.29 tor finding i



EXERCISES



5.4.1 5.4.2



Solve Exercise 5.3.1 by using source transformations. By source transformations, replace the entire cfcuit except for the 16-0 resistor by an equivalent ciruit with a single source and a single resistance R. Using the result,



find o. A swet



R-24f),a=201,1



rzJ)



16(}



rxrRctst s.4.2 5.4-3



Convert all the souces in the frgure for Exercise 4.2.3 to voltage sources ard find or.



Alswer 20



Se.ri6n s



y



4 Pr..ri.al Soxr.er



139



5.5 MAXIMUM POWER



TRANSFER



There are many applications in circuit theory where it is desirable to obtain the mdximum possible power that a given practical source can deliver. It is very easy, usilg Thevenin's theorem, to see what fivximum power a source is capable of delive ng and how to load the source so as to obtain this maximum power. That is the subject of this section. I-et us begin with the practical voltage source shown eadier in Fig. 5.23 with a load resistance lRz . The power pr delivered to the resistor R. is given by



P. = and



\&; &/ R.



(5.29)



it is this quantity that we wish to maxirnrze.



Since the source is assumed to be given, os and Rs are fixed, and thus p. is a function ofRr. To maximize p. we can make dpL/dRz: 0 and solve forR.. From (5.29) we obtain



dp, t';L ,l(& + Rr, - 2(R- + R.)&l (R" + Rrv l



d& =



(5.30)



which results in (5.3 r )



It



may be readily shown that



a'p:l ,;l p, =_-9i.n d&-I R, 8R;



and therefore (5.31) is the condition that maximizes p.. We see, therefore, that the maximum power is delivered by a given practical source when the Ioad Rz is equal to the intemal resistance of the source. This statement is sometimes called the maximum power transfer theorem. We have developed it for a voltage source, but in view of Norton's theorem it also holds for a practical current source. The maximum power that the pmctical voltage souce is capable of delivering to the load is given by (5.29) and (5.31) to be



ol



140



Chapter



5



Network Theorems



(5.32)



In the case of the practical current source, the maximum deliverable power is



This may be seen Aom Fig. 5.25 and {5.J1) or by 15.32, and Norgn's theorem. We may extend the maximum power tlansfer. theorem to a linear ciicuit mther than a single source by means of thevenia's theorem: That is, the maximum power is obtained ftom a lirear ciruit at a given pair of terminals when the teminals are Ioaded by the Thevenin re,.istance of the circuit. This is obviously hue since by Thevenin's theorem the cftqrit is equi\,alent to a practical voltage souce with internal rcsistarce iR,i . EXAMPLE 5.13



Wq rnay draw the maximum power ftom the circuit of Fig. 5.18(a) nals d-, with the Thevenin resistaoce.



if



we load termi-



n.=R,i=6O :



Since L 5 A by (5.22), we may draw the Nortdn equil"lent circuit with the required R, as shown in Fig. 5.31. The power supplied to lbe load is given by



which for &-



:



6 vields



p*



=



37.5



w



Any other lalue of R, will result in a lowgr value of p. For then we have



p = 37.19 W



aad for Rr



=



7 (1, we havep



e"xample,



if



& = 5 O,



=



37,28 W.



-



4 O. and (c) when R



EXERCISES



5.5.1



Find the power delivercd Io R when (a) lt 12 O, (b) R receives the maximum power. An'wet tal 4.32rvi (6) 4 Wi (c) 4.5 W (when R = 8 O)



=



S€ction



5.5 Maimum PNe.



Transter



141



FXtRCtSE 5-5.1



5.5.2



Show that the two networks are equivalent at terminals pated in the 4-f,) resistor in each case. easwer (a) 9 w; (b) 1 w



d-,



and find the power dissi-



txtRclsE 5.5.2 Find the maximum power delivered to the load R, in Fig. 5.23 fixed and ?R, is uriable. A s$,er a?/RLWhenRs = O



if



os and Rr.



)



0 are



5.6 SPICE AN D THEVENIN CQUIVALENT CIRCUITS SPICE can be used direcdy for determining the Thevenin equi leDt for conplex circuits using the .TF command. These equivalent circirits are very useful in determinin8 load conditioos for maximum power tmnsfer, as discussed in the preceding sec-



tion. EXAMPLE 5.14



As an example of the utility of SPICE, consider finding the Thevenin equivalent circuit to the left of R, in Fig. 5.32(a). Since the open-circuit voltage at terminals a-, is equal to the voltage across the 6-() resistor (no current flows in the 4-O resistor), we need only to apply a SPICE simulation to the circuit of Fig. 5.32(b) and add the 4 ,f) to the output resistance found for the open-circuited terminals. lt should be noted that a dummy voltage source rd = 0 has been insert€d to provide the required current i, definirion for the CCVS rr. A circuit file for this circuit is I'IE\,ENIN EQUIVAIEIIT CIRCUIT FOR



r2 0tDc10



142



I



Chdpre'



5 \elworl



lheorems



FIC. 5.32(b)



lI



I



t



lr I



E2 236 \lD 30DCO .lF v(2) . r2 .



i



END



The solulion for thjs program is



NODE VOL?AGE NODE ( 1) 20.0000 ( 2J



i.



VOLTACE



30.0000



NODE



( 3)



VOLTAGE



b. oooo



VOLTACE $OURCE CT'RRENIS



.



NAIIE r,'D



I



C'IMRENA



5 0008



00



TOTAI Porr^jliti DISSIPAI]ION 0. O0E+00 **** snrlLLrSICNA! CIIARAC1ERISTICS



v(2)



/r2 =



3: O00E+0O



INPUT RESIS'IANCE AT



ot{Ptrt



$AATS



I2 = 2,OOOE+OO v(2) = 3.000E+o0.



RESISTANCS A1



FIGURE



5.32 Circuits for obtainin8 the Theven'n equjvalent using SPICI li, v



i=



lo



G) Section



5.6



SalCE and



Th6v;nin tquiv:lent Circuits



143



:



The open-circuit voltage is v(2) 30 V, and the output resistance is 3 f,). Therefore, the total Thevenin resistance is 3 + 4 ='7 A. The resulting Thevenin equiwlent circuit is shown in Fig. 5.32(c). Thus maximum power fiansfer occurs when R = 7 O and the maximum power is (30 / 14\'z x 1 32.14 W.



:



EXERCISE



5.6.1



Using SPICE, determine the Theveniir equivalent circuit to the left of terminals a-r. Find the wlue of n for maximum power tmnsfer and the vaiue of the maximum power. ,answer 12 V, 4 kO, 4 kO, 9 m'rV



EXERCTSE 5.6.1



5.,/ SUMMARY We have considered in this chapter what liredl circuits and linedr elements are, and ho\r't superposition may be applied to analyze such circuits with more than one source by analyzing single-source circuits and combining the tesults. The concept of superposition leads to frevenin' s and Norton's theorems, which allow us to replgce complicated circuits by simpleThevenin and Norton equi)aleht circuits containing only a resistance (the Thevenin resistance) and a single sou,tce. Pracical so /ces, made up of id?dl souces and an intenwl resistance, were considercd and shown to delivet a tnximum power wlrcn loaded with a resistance equal to their intemal resistance. Finally, SPICE was shown to be an exhemely useful tool for the topics of this chapter because of the ease rrith which it can be used to calculate the open-citclit yoltages and short-circuit currertr needed in the Thevenin and Notton circuits, respectively.



PROBTEMS



5.1



Solve Prob. 2.34 using the property of propoF



tionality.



5.2



Solve PIob. 2.38 using the property ofproporrion lity. (Suggestion: Let mA, and work tol'"Id the source.)



i:



144



Chapter5



I



Nei.lvorkTheorems



Solve Exercise 4.4.1 using the piop€ y of f,.oportionality. Solye Prob. 4.32 using the property of proportionality. Use the property of proportionality to find



l'



.



PROALIM 5.5



5,6 5.7 5.t 5.9 5.10 5.ll



5.12 5.13



Solve Prob. 4.34 using the pioperty gfpropor, tionaliiy. (SuSgertou: trt i = cos 2r A.) Solve Ererclse 4.b.1 usrng superposilion.



5.14 5.15



Solve Prob.4.S using superposition. Solve



Hob.4.l0



using superposition..



Find o using superposition if R



:



2



O.



Using superposition, find the power delivered to the 4 f,) resistor in the circuit ofprob. 4.30. Find r using supeposirion Find t using superposition. (Sr8g"rrion: Find t,



Solle Prob.4.26 using superposition. Find o using



sup€rposition.



12



v



PROATEM



4fl



l)



6(,,



,l'"" (l),^ PROBTIM 5.14 Chapter



s



s-r2



Problems



( PROBLTM 5-I5



145



5,16



s.20



Replace the network to lhe lefl of lerminals d b by its Thevenin equivalent and use lhe result



Find tbe Norton equivalent of lhe circuit to the lefi of teminais d-r. and use the result to :0c}



a i -



4!)



PROBttM



PROBLTM 5.16



5.17



Find the Thevenin equivalent ofeverything except the 4-() res;stor in the clrcuil of Prob. 5.16 and use the result to lind the power deliv, ered to the 4 () rcs:stor.



5,18



Find r b) replacinS everyrhing rn rhe cir(uir e{cefl rhe 4 Q resistor by itq Thelenin equi!-



l)o^ |



5.21- Find i by rcpiacing the network to the left of terminals d-, by its Norlon equivaleot. 2()



6



8Q



)rs n



.^ft



PROBTIM 5.21



2c,



In Prob 5,21, replace the network to the right of terminals c-d by its Thevenin equil"lent



PROBLEM 5.T8



5.19



and use lhe r€sult to find o.



5.23



Find the Norton equivalent of rhe circuit to tbe left of terminah d-r, and use the result to



Replace the circuit to the left of terminals a-, by its Thevenin equivalent. and use the result



to find o.



find ,. s,24



Replaae'the circuit to the left of terminals a-, by its Thevenin equivalent and use the result to find t).



5-25



Find the Thevenin equivalent of the circuit external.to the 4-O resistor and use the resuk to



20()



l'



find t. 5.26



Find the Noron equivalent of the circuit to the left of terminah a and use the result to find the power delivered to th€ 6-kO resistor.



s,21



Replace everything except the resistor R = 2 kO by its Thevenin equivalent, and use lhe resuh ro find i.



4Q



PROBIIM 5.19



't46



5_20



Chaprer



5



NeMork Theoreme



,



'1, PROBttM 5-23



PROE|-[M 5.25



PROBIIM 5.24



PXOBIIM 5,26



PROaIM 5.27 Chapter



s



Problems



147



5.28



5.29



5.30



Replace everything except the 4-O resistor by its Thevenin equivalenr circuit and use rhe resuft to find i. 'In the circuit of Prob. 5.16, find the power delivered to the 4-O resistor by using successive source lmrlsformations to obtain the Thevenin equivalent of everything etcept the 4 O resis,



5.3t



Find i by using source transformations to ob tain an equivalent circuit insofar as rhe 2-ko resistor is concerned, containing only one source and one resistor. in addition lo the 2 kO resistor.



5.32



Solve Prob.5.16 using source lransforma-



1or.



5.33



Solve Prob.5.17 usirg source Fansforma:



Use successive source transfo nations to obtain the Thevenin equinlent of the circuit to the left of terminals d-r. From th€ resultfind r.



5.34



iions.



Find



ahe maximum power that can be deliv-



ered to resistor R iD the circuit of Prob.



5



12



32r)



* PROBTEM 5.28



RROOLTM 5.30



PROAGM 5.31



144



5.35 5.36



Find the maximum power that can be deliv' ered ro resisror R and rhe value of R for ma\i. mum power in lhe circuit of Prob. 5.27. Find the value of R thal will draw rhe rna\i. mum power ffom the rest of the circuit. Also



5:38



Find the maximum power thar can b€ delive.edtoR if (a)R, = 12Oand(b)Rr = 30O



6nd the maximum power.



0



i



0v



PROSIEM 5.36



5.37



Find a resistaoce 4 to be placed between termirals d-, to dlaw the maximum power. AIso find the maximum power.



PnoBtEM 5.38



5.39



Find the l"lue of R that will dmw the maximum power ftom the rest of the chcuit. Also find the rnaximum po\rer.



5.40



Find the value of n that wil drare tle rnaximum power ftom the rest of the circuit. Also find tho maximtm power drawn by n.



PROBTEM 5.37



j-*



oo



PNOBLTM 5.39 Chapter s



119



6 lndependence of Equations



,t



The appljcation of Kirchhoff's laws to Euler calculated w ithout a clergyman: He graduated fom the a circuit ol many nodes and loops apparent effofi, as me Universily of Basel in 1724 and joined lhe Russian Academy of Scican bg oxtremoly ditticuli, unless we brcathe, or as eaglbs uso a branch ol malhemallcs known ences rn satnt PercrsDurg tn I /2/ on sustain themselves in the as grcph theory, which we introduce lhe invitation of Cathe ne He vtind. in this chapter. (A circuil with only 10 seNed in a similar capacity at the Domi ique Arago German Academy of Sciences at the nodes and no parallel elements, lor example, could have as many as 103 request ol Frededck lhe Great in loops.) The falher of graph lheory 1741. He ',tas pefiaps lhe most was lhe great Swiss mathematician Leonhard Eulor, prolific mathematician ol all lime, even continuing lo whose farious 1736 paper, "The Seven Bridges ot dictate books and papers After he became blind in Kdnigsberg," was the lirsl treatise on lhe subjeet. He 1766. He slilltound lime lor 13 children aM 2 wives, also made original important conkibutions to every ihe second of whom he took when he was 69 years branch of the mathematics ol his day, and Eule/s loF' old. Swiss mathematicians are slill publishing his pamula is lhe basis of lhe phasor molhod of solving ac pers, and it is estimaled that his works will eventually circuits discussed in Chapler 10. lill 60 ro 80 large volumes.. Euler was born in Basel, Switzerland, lhe son ot



'



l.



151



,( ln



elecfxic network is determined by the tlpe of elements it contains and the manner in which the elements arc connected. We have spent considemble time in the pievious chapters considering the elements themselve! and thek volt-ampere characteristics. ln this chapter we consider the manner in which the network elements are connected, or, as it is sometimes called, the network topology. As we shall see, a study of the lopology of the network ptovides us with a systematic way of determining how many equations are required in the aoalysis, which on€s are independent, and rhe best set of equations to select for the most straightfor$ard analysis.



6.1 GRAPH OF A NETWORK



'



To illustrate the problems involved in the analysis of more comptcated oetworks, let us consider the circuit of Fig. 6.1. The resistors are numbered . . . ,9, vtith values of resistance, say Rr, Rr, . . . , Re. Suppose that we arc required to perform a loop analysis, in which case we ireed to w te a set of independetrt KVL equations. (We note that the circuit is nonplanar, and thus we cannot perform a mesh analysis_ Anyone doubting this is welcome to tly redmwing the circuit in a planar frshion.) Th6.e a.e 15 loops in the circuit, as may be verified by sevedl means, one of



l,2,



which is ftial atd efior (much trial and more error). For the curious leader the 15 loops are (1, 3, 4, s), (t,3,'7,9), (2,3, 5, 6), (t,2,8,9\. (1,2, 4,6), (4, 5,'t , g), FICURt



152



Chapter



6



lndependence ol fquations



6.1



Nonplanar circuit



(2,3, 4, s,8,9), (1,2, 5, 6, 7, 9), (1, 3, 5, 6,8,9\, (2,3,7,8), (2,3, 4,6,7, e),



and (l, 2, 4, 5, 7, 8). That is, rcsistors I, 3, 7, and 9 (with the source Ds) form a loop; and so on. To undertake a loop analysis of Fig. 6. I we need to know which of these loops are independent and how many are required. To ansrwer these questions we need consider only how the elements are connected; it is unimportant what kird of elements are involved. To facilitate matters, then, we may retain the nodes of the nel work and, replace its elements by lines. The network topology thus is preserved in a



(4, 6, 8, 9), (5, 6, 7, 8\,



3,4,



and 5 form a loop;



(l,3, 4,6,7,8),



l,



much simpler configumtion. The configuration of lines and nodes obtained by replacing the elements of a network by lines is called the 8/apft of the netwod(. The lines of the graph are called its bmnches, aad the nodes of the Sraph are, of course, the nodes of the network. As an exrmple, the gmph of the network of Fig. 6.1 is shown in Fig. 6.2. lt has nine branches and six nodes. (We could consider a node between resistor 9 and os, but irasmuch as this will not affect the number of loqps, we have chosen to consider these two seri€s elements as one element, namely, a nonideal voltage source.)



FIGURf



6.2



Graph of a network



We say that a graph is connected i,f tlrcte is a path of one or more branches between any two nodes. The graph of Fig. 6.2 is eviden y connected. An examplg of a graph that is not connected is shown in Fig. 6.3, There is, for instance, no path between nodes r and d. For the present we consider only connected graphs. FICURE



6.3



Unconnected graph



A,,"\



d



()



/ \"Y /



section



6.1



C.aph ot a Network



\



153



EXERCISES



6.1,1



Show that the gaph is planar.



.IXERC|SE 6,1.r



txrRcrst 5.r.2



6.2 TREES



AND COTREES te? of a gmph as a connected portion, or subgiaph, of the gaph that contains all the trodes but no loops. As an example, Fig. 6.4(b) is a tree of the graph oflig. 6.4(a). The ftee is connecred, has no loops, and co[tains all the nodes of the We define a



graph.



' 154 I



I



Generally, a graph has many trees. The configuation of Fig. 6.4(c) is evidently another tree of the graph of Fig. 6.4(a), silce it satisfies all the requirements. I Chapter6



lndependenceof Equations



(b)



G) FICURE



(c)



6.4 Graph and two of ih



trees



This particular graph has 24 tIees, which the reader may wish to try to discover. lt will help in enumerating the trees to notice that each one has exactly tkee bmnches, since it takes at Ieast three lines to connect four nodes and more tha! three lines will form a loop. There are 35 ways to select seven branches three at a time, but 11 of these combinations are not trees. The bmnches of the graph which are not in the chosen tree arc called lirLi, and together rvith their nodes form the cotle€ coresponding to the tree. Thus the tree of Fig. 6.4(b) has the cotree of links 4, 5, 6, and 7. In the geneml case, Iet B be the number of branches and N be the number of I nodes in a given graph. Then any tree of the graph contains N nodes and N branches. The number of nodes follows fiom the dennidon of a tree. and the number us of tree branches may be established by the following constructio[ argument. build tle tlee starting with one branch and the two nodes to which it is connected. Each additional bGnch connected to build the tree adds one additional node. The number of nodes is, therefore, one more than the numter of b€nches, and since 1 branches- The number of links in any cotre€ there are N nodes, there mDst be N (1{ is tlereforeB l) orA N + l.



lrt



-



EXAMPTE 6.1



-



-



-



As ar example, the graph of Fig. 6.4(a) has /V = 4, and thus the number of tree I = 3- The selectioq of the tree, of blaoches in both Figs. 6.4(b) and (c) is N coulse, determines the cotree. The tree of Fig. 6.4(b) is redrawn in Fig- 6.5, with the tree bmnches shown as solid lines and the links of the cotlee as dashed lines. 4 + = 4. The number of links in this case, since B = 7, is 7



-



-



FIGURE



Se.rion



tI



6.2



6.5



Trees and Cotrees



|



Tree branches and links of a graph



155



EXAMPLE



5,2



The graph of Fig. 6.6 has tree branch voltages or , 02, and q, irdicated by the solid If the link and lines. Tle dashed lines are the links whose voltages are 01, 's, labeted oa is added to the tree, the circuit of 02, or, oa is formed By KVL a'ound this circuit, we may obtain



'6



aa=Dt-t2 ln like inaDner, adding link



os



to the tree yields



as=Dt-D2 and adding



link



o6



to the tree yields



1,6=1)r-u3 Thus the link voltages may be found fiom the tree bmnch voltages.



FIGURE



6.6



Tr€e and lihk voltages



A systematic way of writing equations involving tree branch voltages is to imagine opening a tree branch, noting that this sepamtes the tree itrto two parts. Cu ents flow beiween the two parts through the tree braDch imagined to be open and through links. Thus by KCL the aigebraic sum of these currents in a given direction is zero. This procedue may then be repeated for the other hee b€nches. EXAMPLE 6.3



us consider the circuit in Fig. 6.7(a). A gaph of the circuit is shown in Fig. 6.7(b) with the tree branches shown as solid lines and the lints as dashed lines Since the tree bmnch voltages are independent, we have included in the tree the 20-V source. Thus the number of unknowns is rcduced by one. By KVL we may find the link voltages in terms of the ftee voltages, with the results shown in Fig. 6.7(b): If we imagine the tree bianch (d, ,) labeled ul as open, then the tree is separated into two parts. These two parts are connected by branch (d, ,) and links (a, i), (b, d), and (d, c), as indicated by the line marked I summing the curents across the line in the direction of the arorv, we have



l€t



ot+b+at



n



1=o -_2t z



Repeating the procedure for tree bmnch (b, equation



-(or + Se.tion



6.3



oJ



independenl voltage Equalions



c), labeled or, leads to line



2a2+



lI



and the



Il=0 157



I I



.'



20



.a



-) !



..,i. ,



G)-(b) tlcuRt 6,7



CircLrir and i|\ Sraph



:



:



Solving these equations-, we have or 8 V and ou I V A[ the link voltages, and' coDsequently all the link and trer curr€nts. may now be found.



A cut set o{ a graph is a minimum set of elements which when l./r, or removed. sepamtes th€ graph into two parti. Tte set of elemens we have been discussing - a ree branch. which when cut separates the tree inlo two pafls, and the lioks bgtween thgse two parts-is an example of a cut se1, The.two parts oia gr4ph . detemined by a cut set will either be nodes or suDernodes and lhus bv KCL the algebraic suln of the currents leaving eitherpar! is zero. That is. the aljebraic sum of the currents i! a cui set is zero. .



EXAMPLE 6.4



kt



us sup;rose that the gaph -of Fig. 6.8ia) has cunens ih the directio{ of the arrows shown oq th9 el€rneirts, Selecting the Eee i, i;, i6, i7, shown by the sold branches of Fig. 6.8(b). we have the links ii, r:, rj, shdwtr dashed. itriting tree braDch i ftom rlrc g.aph curs rhe tree into two.lqrts (braqch i, is one part and ir, in is the other part). Therefore, 6nrch 4 and li;.i; i", ii", i5 (betweerthd rro t e". parts) conslirute the cut sel CS as shown. KCL for this cul sel is



frcURE 6.8



.



Chapter



6



(a) Craph.of a



circuiti 6) die of its cut



lndepCndence of Equatiors



sets; (c) otber cut sets



which is also KCL for the supeffode co[tdining i6. Cut sets coffesponding to tree branches ir, ia, and i6 are shown in Fig. 6.8(c), with KCL given.by



i4+



L-iz=O i5- L+ i2:O t2- h+ i6:o



respectNely. A cut set nor based on the given tree is ir. i,. i.. L. consisling of the elements connected to, ot incident to, node a in Fig. 6.8(a). Such a cut set is called afi incidence cut set (e)emetrts incident to a node), and its KCL equation,



tr-t,+t5+t:0 is simply the nodal equation al node d.



In any circuit analysis procedure where the unknowns are voltages, we need to find only the,ry - I tree bmnch voltages which constitute aD independent set. This means thal only N I independent voltage equations are rcquired in the analysis. Since dnJ independent set of equations will suffice, then ary independent set of lV - I rohages consrirutes a solution.



- I



Anolher independent sel of N vohages. othel than lhe tlee branch voltages, is the set of nondatum node voltages, considered itr the nodal method of Chapter 4. Ib see this, we note that any nondatum node is in the tree and is con, nected tbrough tree branches to the datum node. Thus evcly nondatum node voltage is an algebraic sum.of tree bmnch voltages (the tree bmnches between the nondatum and datum nodes). On the other hand, evety tree bratch voltage is the difference between its two node voltages. In summary, the node voltages may be determined from the tree voltages and vice versa. Thus the noodatum node voltages me also an independent set. (Of course, writing KCL at a node is the same as equating to zerc the aigebmic sum of the currents in the incidellt cut set for that node.) EXAMPLT 6.5



I



ln lhe example of FiB. 6.7 we see thal if i( rhe datum node. lhen the nondatum node voltages o", or, and o. are related to the tree voltages or, or, and 20 by



u" 2O ab=20-]l! n.: m - q



a2



Conversely, we have



'



02: ob-



20-



ac



D,



The nodal merhod, in many cases, is easier to apply than the loop method beto find. In the loop method, as exemplified by the example



cause the nodes arc €asy



Se.tion



6.3



lndependent Voltage Eguations



159



of Fig. 6.1, the appropriate loops may be difficult to identify. In the next section we consider a method based on gpph theory of finding sets of independent loop equa-



EXERCISES



6.3.1



In the daph of the cicuit of Prob. 4.8, select the tlee of the voltage souces and the l2-O rcsistor. Using the method of this section, write one KCL equation and determine D,



6.3.2



Select the hee of the voltage source, and the 6- and l2-f,1 resistors, and use the merhod of rhis seclion ro find r'. AnJq pr 18 V



Answ bV



EXtRCISE 6.3,2



6.3.3



Using an apprcpriate tree and the methods of this section, find o in PIob. 4.27. (Not?. The tree should contain the voltage o alrd the three sources as well as one other branch. )



6.4 INDEPENDENT CURRENT EQUATIONS As we have seen in the example ofFig. 6.1, it is not always eary to ideDtify the independent loops for a loop analysis of a cLcuir. To develop a systematic means of wdtilg loop equations, let us consider a geneml network with B branches and N nodes. Corresponding to a given tree there are B - N + I links. suppose that all the link currents are made zero by open-circuiting the links (replacing the links by open ckcuits). Since the tree contains no loops, then all the tlee bratrch curents ille zero also. The tree cunents thus depend on the link currerts; thlat is, they may be expressed in terms of the link curf;nts, for if a tree current were independent of the link currents it could not be forced to zerc by open: cicuiting thc links. Moreoven, if one link is not open-circuited, a loop is left in the gaph, and a cuftent will flow in the link. A link curreDt thus is not dependent on



150



F



chapler



6



lndependence oi Equations



the other link culrents. In summary, the B N + I link curents are an independent set, and the loop analysis of the cfucuit requires B N + I independent equations. One systematic way to find B N + t independent loops is to start with the tlee and add one of the links. This determines the loop containing that link, since adding the link to the tree closes a loop. Remove this link and add another link to the tree, determining a second loop. ContiDue the process until the B N + I loops are found. The set is independent because each loop contains a different link.



-



-



-



,



EXAMPLE 6.5



I€t



Fig. 6.1. One tree consists of branches 1, 5, 7, 8, 9, with conesponding links 2, 3,4, and 6, as shown in Fig. 6.9. Closing the links one at a time results in the four independent loops I, II, III, and IV shown. Loop I contfis ltuk 2 and ree branches 8, 9, and l: loop II contains 3, 'l, g, and,l: loop III contains 4, 5, 7, and 9; and loop IV contaiis 6, 5, 7, and 8. These four us reconsider the circuit of



and



loops are sufncient fm performing a loop analysis.



i,\



h__,



-!



FICUXI 6.9 toops of Fig. EXAMPLE 6.7



6.1



To illustraE lhe use of link currents in circuil analysis, let us retum to the examDle of Fi8. 6.7(a,. The graph is redrawn m Fig. 6. t0. showinS the link turrents i,,l.r, and 11 A. We have.chosen the current soruce as a liak beciuse the link currents are an independent set. This reduces the number of unknowns by one. Genemlly, for this reason, one should place voltage souces in the tree anal current sources in the



links.



The tree bmnch currents, as in the geneml case, may be found fiom the liok currents, as shown in Fig. 6.10. Closing the links labeted jr and ,:, forms loop6 I aDd 2, as indicated. Applying KVL to these loops yields, from Figs. 6.7(a) andt.l0.



2i' 20+, rr+11 =0 1l - i, +i,t ----i, -0 the solution of which is 5eclron



I



I



6,4



ir = 6 A



and i?



lndependenr Curen( Equatrons



= 9 A. 161



FICURT



6.10 Craph of Fis. 6.7(a)



Since the link currenr I I A is known. we needed only t!.ro loops involving link culrents ir and ir. Incidentally, in this simple example the links were chosen so that the link curredts are also mesh culrents. This is, of cowse, not the case in geneml.



The results obtained thus far in this chapter ate valid for general netwol*s, which may be either planar or nonplanar. ln the special case of planar networks, as we saw in Chapter 4, a mesh analysis is possible. In the circuils of that chapter the mesh cunents were indepefident and were sufficient in number to perform the analysis. We shall now show that this is the case in geneml for planar;etwo.ks. Irt us begin by taking apart the planar circuit with M meshes and reconstructing it one mesh at a time. The first mesh in the recoNtruction has.the same numbr, say kr, of nodes and bmnches, for the first branch has two nodes, each additional branch adds one n€w iode, and the last branch iidds lro nodes since it is tied back to a node of the first bmnch. This is illustrated by the graph of four meshes in Fig. 6.11(a). The first mesh constructed, shown in Fig. 6.11(b); has the same number of branches and nodes, namely four in this caseAfter the Iist mesh, each subsequent mesh is formed by connecting brarches and nodes to previous meshes. Each time the number of nodes added is one less than the number of bmnches, trecause each added branch adds one node, except for the last branch, which is connected to a node of a previously added mesh. This process is illustmted in Fig. 6.11(b), (d), ard (e). Thus, if the second mesh adds tr branches, it adds only &, 1 nodes. Similarl). the third mesh adds k. branches and k, I nodes. and so on. The lasr mesh. the Mth, adds kr bnnches and &M I nodes. If in the completed gaph the number of branches is and the number of nodes is N. we have



-



I



-



-



k,+k1+...+k .



k, + (t,



- t)+...+(t,-



=B



(6.1)



1)=N



(6.2)



The latter equation may be written



k'+k2+ ..+k"-(M-t} which by (6.1) becomes



R



162



chapte.6



lndepencjehceof tquations



- \M l) = N



=



1'1



{



\



f



1



,/9 )



)



(d)



. FICURI 6.'l



I



G) Planar circuit and its meshes



Solving for the number of meshes, we have



M=B



-



N



+I



(6.31



which is also the +umber of links in &e graph. The rn€sh currenls lberefore constirute an appropriale set of curreirts to completely descdbe a planar network. They are the same in number as the independent set of link currents, and they are independent since each new mesh contains at least one bGnch not in the previous meshes.



Select an applopriate tre€ and use the method of this section to find



i in tho cicuit of



Exercise 6.3.2. Answet 2 A



Section



6.4



lndependent Current [quations



163



t6.5 A CIRCUIT APPTICATION EXAMPI-E



6.8 .



As a final illustmtion in this chapter we analyze a modemtely complicated circuit, shorrn in Fig. 6.12(a). Its graph is shown in Fig. 6.12(b), where we have selected a tree shown by the solid lines- Note that voltage sources and voltages driving depen, dent sources are placed in the tr€e and cutent sources are ilaced in the cotree. Currents driving dependent sources would also be placed in the cotree, if possible.



(bl



FICURI



6,12 Network and



its graph



From the graph we see ihat lhere are lour ree branches labeled ur. o,. J1,,. and 10. Thus if the tree bmnch voltage method is used, there will be only two unknowns, or and o?, requking two equations, There are five link currents in the graph, one of which is knowri (the 6-,4 source) and another, the 2or source, which may be expressed in terms of the trce bratrch curent ie, and subsequently io terms of other link curents. Thus if the link curr€nt method is used, we must have three equations. Accordingly, we shall aralyze the circuit using bmnch voltages.



164



Chapter



6



lndependence of aqualions



sets I and II shown in Fig The two necessary equations are (CL fot "ut(a, d), respectively. These are 6.12(b), based on tree branch (r, e) and tree branch



| +\



+ zw, + 6 +



30".



=



o



Zat-2ar+3az:0



(6.4)



(6.5)



Inspection of the graph shows that



a, 3r1 Dtu=at-2oz-lO l),.=1)'-10 ae =



Substituting these values voltages, we have



into (6.4) and (6.5) and solving for the tree branch



o'=-11V,



az=-2nV



\e



ma] note that nodal analysis is equally eary to apply. The node vollages may be expressed in terms of the two unknowns or and o?, and thus.only two nodal equations are required. We leave the details to the problems (Prob. 6.4).



EXERCISES



6.5.1



Write one KVL equation and find An,wer I A



i,



using the m€thod of Sec. 6.4.



EXERCTSE 6.5.1



6.5.2



Using an appropriate tree for the graph of Prob. 4.26, find the current i flowing to the right in the 4-fl rqsistor. (An appropriate tree should n t contain the current sources or the curent i. Thus, using the methods of Sec. 6.4, only one K\rL equation is required.) Answer 6.5 A



Section



6.5



A Circuit Appli.atior!



165



6.5.3



Using the method of Sec. 6.4, {ind the power delivered to the 8'f,) resislor Aruwer 8 W



EXERCTSE 6.5.3



6.6 SUMMARY In this chapter we have defined a planar circuit and lls graph, corsisting of the ,o/es of the circuit and lines, ot branclks, which replace the elements of the circuit. We have considered a tree of lhe graph (a set of connected bnnches containing all lhe nodes but no loops) and a coar€? (the rest of the graph after the tree is removed). The branches of a tree are, of course, tree bmnches, and those of the cotree aie called link. We have seen that the tree branch voltages form an independent set of voltages and the link currents form an independent set of currents- Other indepen dent currents arc those of a cut set, which is a minimum number of graph branches whose removal cuts the gnph into two parts: These facts enable us to write KVL or KCL for an independent set of voltages or of currents, and thus readily find a set of describing equations, no matter how complicated the circuit.



PROBTEMS 6.1



Find a tree, if possible, that contains all ahe voltage sources and lhe branches whos€ voltages control dependenl souces but does not contain orrent sources or ttrnnches whose currents control dependent sources. use this tree with an appropriate graph theory mothod toJind 0r.



6.2



Select a tree as described in Prob. 6.1 and use an appropriate.gmph theory method to find o, .



6.3



Solve Prcb. 4.16 selecting an appropdate tree and using gaph theory methods.



6.4



Find



01 and



o, in Fig. 6.12 using nodal



ana-



PROBLEM



lysis.



6.5



Select the tree



ofthe 8-O resistor



resistor with voltage method to find 0.



166



I



Chapier



o,, :ind



and ahe



4-O



6'6



4.14.



use lhe cut set



6.7



6



l.dependence of Equations



6.I



Use tlrc cut set method to solve for t in ProbSolve Prob. 4.15 using cut seis.



PROELEM 6.?



PROBLEM 6.8



6.10 b



that the number



of tr



PROaUM chapter



6



Problems



trees



in a



ladder graph



Fibona\i nunbet a".



- t, a) - aa t at - 2. - a2 + a. : 5. and so on. That i:. excepr for do and each Fi. ",, rhe prebonacci number is obrained by addjng :



All



the resisrances are 1 O, €lement l, is also a O resttor. et€menls r and ) are indepfndenl l-A curfent source\ drrected upward, and ele_ men( - L an rnd€pendent 3.A current sourcc directed ro rhe left Setecting an appropriare rree dnd using Brrph rheory merhods, 6nd i. lNore rhat lhA ci,cuiL is simitar to lhat ot Fig. 6.1 and i' lhus nonptdnar Howe"er. onlr onc loop equation is required in this case.) frnd , in Prob. o I u\ing graph rheor) methods il elernent r is d 5 V sou'ce w h positive lermirulal rhe top. v is a J-V source $ilh po,s rlrre rerminal at the bo om. : is a l7-V source wilh poerive terminal al the lefl. and D is a 7-V source with positive aerminal ar the lefi.



of



branches is the defrned by do a. a! at + a2 = 3, a4



I



6.9



(a)



and rb) shown here aJe e),amptes of graph\ ot ladder nerworks. There is a theolem that srdte\



PROSLTM 6,5 6.8



The figure for Exerc'(e 6.2.2 and fisures



-



vious two. V€rify that the rheorem holds for the ladder gBphs shoMr and also for ihe graph in the figur€ for Exercise 6.2.2.



6-lr



The gmphs shown in rar and (b) are two basic nonplanar graphs tthe one rn (b) i5 Lhar in rhe



figure for Prob. 6.8 redmwnl. Branch a,b in



(a) is an ideil



6V



voltage source wrrh its Dosi-



live lerminal at lhe rop and in (b) is a 4-A



ideal currenr source djrecled upward. AIt lhe olher bmnche\ in borh hgures are t.O reJislors. Find i shown rn each figure osing graph theory methods.



5_rO



167



b



'-t (a)



(b)



PROBI FM 6.11



6.12 6.13 6.14 6:15



158



metiod.



6.16



Find o in the circuit of Prob. 4.17 using the Iint currenr merhod.



6,17



Solve Prob. 4.26 nsing graph tieory methods.



6.18



Solve



6.19



Solve Prob. 4-2 using graph thaory methods.



6.20



Solve Prob. 4.3 using graph themy methods.



Solvb Prob. 4.25 using the cut set



Find rr and o, in Fig. 6.12(a) using link renls as the unlnowns. Find o and t)r inProb.4.15 using renr melhod.



chapte.5



method.



clrr-



the'link€ur-



Solve Prob. 4.25 using the link current



lodependenaeof Equaiions



ftob- 4.29 using graph theory neihods.



Energy-storage Elements



On August 29, 1831, Michael Faraday, the greal English chemist and physicisi, discovered electromagnelic induclion, when he lound that movinq a magnel through a coil of copper



My greatest discovery was Michael Faraday. Sir H mphry



Dary



dream by becoming assistanl al the Royal lnstiiution to his jdol, the greal chemist Sir Humphry DaW. He remajned at the lnstitution for 54 yearc,



taking over Davy's position when wire caused an ele:tric currenl lo Davy .elired. Faraday was perhaps flow in lhe wire. Since the eleclic motor and genera- lhe greatest experimentalist who ever lived, with tor are based on lhis principle. Faraday s discovery achievements lo hjs credit in nearly all the areas ol profoundly changed the course of world hlslory. When



physical science under investiqation in his tim€. To asked by lhe British prime minister years later what describe the phenofiena he investigated, he and a use could be made of his discoveries, Faraday science-philosopher friend invented new words, such quipped, "Some day il miqht be possible io tax ihem." as eleclrolysis, eleclroMe, ion, anode, and calhode. Faraday, one ot 10 children ot a blacksmilh, To honor him, the unit ot capacitance is named lhe was bern near London. He was first apprentic€d to a larad. bookbinde., but age 22 he realized his boyhood



d



t



169



*"



UO



huu" considered otly resistive circuits, that is, circuits containing ro no* .esistorc and sources. The terminal chamcteristics of these elements are simple alSe_ braic equations which lead to circuit equations that are algebmic ln this chapter we shall inffoduce two important dynamic circuit elerrents, the capacitor and the inductor, whose termiial equations me differential mther than algebraic equations. These elements are rcferred to as dJnamic L'''cavse, in the ideal case, they store energy which can be retrieved at some later time. Another term which is us9d, foi this rea_ son; is storaSe elements. We first describe the property of capacitance and discuss the mathematical model of an ideal device. The terminal characteristics and energy relations will then be given, followed by derivations for parallel and sedes connections of two or more capacitors. We then repeat this procedu.e for the inductor. The chapter concludes with a drscussion of pnctical capacitors and inductors and their equivalent circuits



7.1 CAPACITORS



-



alevice tbat consists of two conducting bodies that are sepamted by a nonconducting material. Such a nonconducting material is known as an insulator or a dielearnc. Because of the dielecffic, charges cannot move ftom on€ conducting body to the other within the device. They must therefore b€ transported between the conducling bodies via extemal circuitry connected to the teminals of the capacitor. One very simple t'?e, called a parallel_plate capacitor, is shown Fig. 7. l. The conducting bodies are flat, rectangtlar conductors that are sepamted by the dielectric material.



A capacitor is atwo-terminal



it



To describe the charge-voltage .elationship for the device, let us transfer charge from one plate to the other. Suppose, for instance, that by means of some ex_



FIGURE



\ 170



t-



Chapter



7



Energy Stoege Elemenls



7.1 Parallel'plate capacitor



te al circuit, we take a small charge, say 44, from the lower plate to the upper plate. This. of course. deposits a char8e Ag on the lop plale and leares a charge of -A4 on the bottom plate. Since moving these charges requires the separation of unlike charges (recall that unlike charges atffact one another), a small amount of work is performed, and the top plate is mised to a potential of say Ao with respect to lhe bofiom plate. Each increment of charge Aq that we transfer inqeases the potential difference between the plates by Ao. Therefore, the potential difference between the plates is proportional to the charge being tmnsferred. This suggests that a change in the terminal vollage by an amount Ao causes a corresponding change in the charge on the upper plate by an amount &. Thus the charge is proportional to the potential difference. That is, if a terminal voltage o corresponds to a charge 4 on the capacitor (+q on the top plate and -4 on the bottom plate), then the capacitor has been charged to the voltage o, which is proportional to the charge 4. We thus may wtite



of



q=CD



(7.l)



where C is the constant of proportionaiity, knoli'n as the capacitance of the device, in coulombs per volt. The unit of capacitance is known as the /d/dd (F), named for the famous Bdtish physicist Michael Faraday (1791-1867). Capacitors that satisfy ('l.I) arc ca1led ltueat carycitorursince thek charge-voltage relationship is the equation of a straight line having a slope of C. It is interesting to note in the example above that the net charge within the capacitor is always zero. Charges removed from one plate always appear on the other so that the total charge remains zero. We should also observe that charges leaving one terminal enter the other. This fact is consistent with the requirement that current entering one terminal must exit the other in a two-terminal devic€. Since the current is defined as the mte of change of charge, differentiating



(?.1), we find that



i: C:dt



(1 .2)



which is the curent-voltage telation for a capaciror. The circuit qimbol for the capacitor and the current-voltage convention which satisfies (7.2) are shown in FrE. '7.2. lt is apparent tlat moving a charge of A4 in Fig. 7- 1 ftom the lower to the ul4)er corducto. represents a cuffent flowing into the upper terminal. The movement of this charg€ causes the uppq terminal to become more positive than the lowet one by an amount Ar. Hence the currert voltage convention of Fig. 7.2 is satisfed. If either the voltage polarity or the curent direction is reversed, then the current enteling the positive terminal is and (7.2) becomes



-i



, .



^dt) aa



We recall that for this case a minus sign was also required in Ohm's law.



Section



7.1



Capacitors



171



7.2 Ci.cuir



FICURE



EXAMPLE 7.1



Suppose thar the voh4ge on a



I



symbol for a capacitor



-,,F capaciror t)



:



is



6 cos 2000,



V



Then the cur.ent is dD



i = ci =-



lo 6r- l2.ooo sin 2ooor)



A



t2 sin 2000, mA



In (7.2) we see that if o is constant, the curent i is zero. Therefore, a capacitor acts like an open circuit to a dc voltage. One the other hand, the more mpially o chalges, the larger is the current flowing though its teminals. Considet, for example, a voltage that increases linearly from 0 to d t s, given by



o=0, = at. - l,



Mn t



c'



(?.8)



Thus the equivalent capacitance of N pamllel capacitors is simply the sum of the individual capacitances- An initial voltage, of course, would be equal to that which is presenr acro\s rhe parallel combinatron. It is interesting to notice that the equivalent capacitance of series and pamllel capacitors is analogous !o the equi\?lent conductance of sedes and parallel conduc-



EXERCISES



7.3.1



.



7,J.2



Find lhe maximum and minimum values of capacitance that can be obtained ftom ten l-lrF capacitors. A,rwsl l0 rrF, 0.1 pF Find the equiralent capacitance.



Anr'€/ l0 !



F 60



EXERCISt 7.3.2



7.3,3



Derive an equation for current division between two paftllel capacito$ by finding ir and



t. A,,**



C' ,



c,rc,



C'



,



c, +c,'



fxtRcrsE 7.3.3



Seclion



z-3



Series and Parallel Capacitors



179



?.3.4



Derive an equation for voltage division between two uncharged series capacitors by



finding



,



1lr and 0z-



C:



a,



+



C,



c,--c,



cr-



txERctsE 7.3.4



7.4 INDUCTORS In the p.evious sections we folmd that the electdcal chamcteristics of the capilcitor ale the result of forces that exist between electric charges. Just as static charges exert forces upon one another, it is fourd that moving charges, or. currents, also influence one another. The foice which is experienced by two neighboring ctment-carrying wiies was experimentally determined by Ampdre in the early nineteenth century. These forces can be chamcterized by the existeice of a nagnetic feld. The magnetic field, in turn, can be thought of in terms ol a magnetic fux that forms closed loops about electric currents. The origin of the flux, of course, is the electric culrents. The study of magnetic fields, like that of electric fields, comes io a later course on elecftomagnetic theory. An inductor is a two-terminal device that consists of a coiled conducting wire. A current flowing tfuough the device produces a magnetic flux d which forms closed loops encircling the coils making up the indrctor, as shown by the simple model of Fig. 7.7. Suppose that the coil contains N tums aDd that the flux d passes through each tuft. In this case, the total flux linked by the N tums ofthe coil, denoted by,{, is



^-N4



This total flux is commonly refered to as theJfur liriage. The unit of magnetic ffux is the werer (Wb), named for the German physicist Wilhelm Weber (1804,1891). In a linear inductor, the ffux linkage is directly proportional to the current flowing though the device. Therefore, we may \r'rite



I= Li t,



(i.9)



where the constant of proportionality, is the ifiductance in webers per ampere. The unit of I Wb,A rs known as the h"nry (H). named for the American physicist Joseph Henry (l?97- 1878).



180



ChapterT



tner8y-StorageEleinents



FIGURE



7,7



Simple model of an inductor



l.



In (7.9) we see that an inqease in i produces a corresponding increase in This increase in ,\ Foduces a voltage in the l{-tuln coil. The fact that voltages occur with changiq magaetic flux was fi$t discovered by Henry. Heffy, however, repeating the mistake of Cavendish with the resistor, failed to publish his findings. As a result, Fa.aday is credited with discovedng the law of elecromagnetic induction. This law states that the voltage is equal to fhe dme mte of change of the total magnetic flux. h rnathematical form. the law is dA



which with (7.9) yields



.di



(7.10)



Clearly, as i inqeases, a voltage is develop€d actoss the terminals of the inductor; the polarity of which is show! in Fig. 7.?. This voltage apposes an itrcte.ise in i, for if this were not the case, that is, if the polarity were reversed, the hduced voltage would "aid" the current. Physically, this cannot be tnte, because the culrcnt would increase indef nitely. The circuit syEbol and the curent-voltage convention for the inductor are shown io Fig. 7.8. Just as iD the cases of the rcsistor artd the capacitor, if either the curent directiotr or the voltage assignment, but not both, ate tevelsed, a negative sigd must be us€d in the right-hand side of (7.10). From (7.10), if i is consta0t, the voltage o is zero. Thercfore, an inductor acts like a sho.t circuit to a alc cu$ent. On the other hand, the more rapidly i changes, the greafer is the vok4ge that appears across its terminals.



Section



7,4



lnductoE



,181



FICURE



EXAMPLE 7.5



7.8 Circuir symbol for an inducror



Consider a current that decreases linearly from



=l-bt, A l-H



I



to 0 A



h ,-t



s, defined by



ah\



inductor having this terminal curent has a iermfual voltage given by



o=0, --b, =0.



r 0, as we will see in Chapter 8, we need certain initial conditions (values of certain cuments and voltages, as well as their derilatives, at r = 0*), which may be found ftom the dc sleady state condirion!.



r-_r IXAMPI-E



7.9



I,,et us consider the RaC circuit switch is opened at t = 0. At a



of Fig. 7.13(a), which is in dc steady state when the , just prior to the switching action, the circuit is shown in Fig. 7.13(b), where the srritch is closed, the capacitor is an open circuit,



Section



2,7



OC Steadv State



=0



189



-#dxff')":.



I



FICUIE 7.13



(a) RtC circuir,



and the jdductor



ii



ibl circuit at t = 0-;



(c) circuir ar



t = 01



:



a short circuit. From this circuit we have r(0-) l0/5 = 2A l0r 6 V The circuir at r = 0* (jusr after rhe swilch is is-opengit) op€n€d) is 0 if o(0) - 0. Find lhe maximum value



7.9



ofp



and the smallest value of t for which



Th€ current 4 sin 2r



A.



in a



0.25-F capacitor is



The initial voltage is 4



v. Find



the maxinum and rninimum Elues of the stored energy and the smallest time at which each oc-



1-tal



The current ttuough a 0.01-F capacitor is as shown. Find the voltage o and the power al



t = l0 ms and at t =



40



ns ifo(O)



:



PAOBLEM



i=



0.



7



.13/



?.r4 Ii o(0-) : 72 Y find t', i,, k, t,., and , at / 1=0-andatt=0+. 7y'5 lI itto r= + A. 6nd i,. n. i.. and '. al t:0*. 7.16 Find the equivalent capacitance seen at telmi nals a , if the capacitances are all in ttF. 7.17 The capacitances shown ale all in pF Find the equivalent capacitance seen at terminals a-b.



pRoBttM 7.10



200



Chapter



T



Ene€y'Storage tlements '



pF



7.18



The capacitances shown are all in



7.19



Find the voltage across a 10-mH inductor



Find



ihe equivalent capacitance seen at terminals



t : l0 ms, 30 ms, and 60 ms if as shown.



at



the current is



I



I



PROSLIM 7.14



PROBIEM 7.15



PROBLEM 7,16



i



tI



A



7.20



Find the terminal vohage of a 10-nH inductor



if



A, O) 20t A, (c)' l0 srn lm/ A. and (d) l0(l ?-') A. ?.21 The cu[ent through a 0.1 H itrductor is i: 10 cos 10t mA. Find (a) the termtuat the current is (a) 2



voltage, (b) the power, (c) the stored €nergy, and (d) the rnaximum lalue of the power being abdorbed.



7.22 lf i(O\ = 0 is tle initial cu.retrt in a l-n*I inductor, find the cunent for I > 0 lor the two cases of inductor voltag€s shown.



7.24



tr



a(o-):



7zI5



tf t (0 )



If a(0 ) :



50 V, frnd



i(0.)



2.5 V, find



i{0') afld .



and o(0+).



,1,(0 I



0, find i(0.) and o(0+). 7. 1.21 lf n(O ) = 9 V, i(0 ) = 1 A, and the switch



r = 0, find i(0*) and di(0.)/dr. The inductances shown are aI in mH. Find is opened at



7.28



lhe equi!alenL rnduclance seen ar lerminal\



1.29 fi



the inductances are



all in n*I. find



equivalent inductance seen at terminals



viv) 5



0



t0



the



a-r.



7.30 7,31



Find aq if aI the induclances are in ml{.



7,32



The circuit is in dc steady stare at



30



*5



.



(a) Find the raximum and minimum values of inductance that can be obtained fioln ten 5-mH inducron. (b) Find a connecrion us'ns a1l ten inductors that lelds an equivalent inductance of 20 rnH-



r, i,. r,.



di, -.dl



and



di,



at



,=0



-dt the circuit is in dc steady state at t find r, and r, at, = 0 andatr=0+.



7.33 ff



. Find



/ : 0'.



=0



,



7.34 Find o(0-1, t'(0.), and di(o+)/dt if t(0 ) = 2 A and the switch is op€ned at



r-0.



'



(b)



PROBLEM 7.22



7.23



For tbe cncuit shown in (a), the source voltage



is siven in (b)- Find the qtrrent t if t(0) = -t -q for (a) 0



a.5



circuit;



(a) More general RC



(b) its equ;v;lent at



t=



0



; ic) its equivalent



0 l



Therefore, by (8.7) the voltage is



o(r) = 46e-'rro Y



I



If ve $,ant q in Fig. 8.5(a), we may fnd it ftom o using voltage divisioD. since or is acroos the equialeot resistatrcr of 6(3)^6 + 3) = 2 O, theo r.ll



2



- :_i_;o 2+a = 8?-'llo V



l



EXERCISES



8.2.1 In ,R



a series



= l0 kO



XC circuit, determioe (a) tfoJR = 5 kO ad C =2lLF, (b) C fq and r = 20 ps, and (o) n foi o(t) on a 2-rrF capacitor to halve evcry



20 ms.



An$rer (a) 10 Ds; (b) 2 nF; (c) 14.43 kO



'



S€ction



8.2



Time constants



215



I



4.2.2



A series RC circuit consists of a 20-kO resistor and a 0.05-pF capacitd. It is desired to decrease the curent in the netiork by a factor of 5 without changing the capaci-



toi voltage. Find the necessary lues ofR and C.



,,rwe. 4.2.3



100 kO, 0.01 AF The circuit is in steady state al r = position 2 Find o for t >0.



0



att:0.



and the switch is moved ftom position



I



ro



29



EXERCtSt 8.2.3



8.2,4



Findifort > 0 ifthe Answer O.25e-"



circuit is in steady state at



A



t=0



.



EXERCTSt 8.2.4



_I



8.3 SOURCE.FREE



RI CIRCUIT



In this section we study the se es connection of a! inductor and a resistor, as shown in Fig. 8.6. We assume that the iDductor is canying a curent Ic at time t : 0. As in the case of the source-free RC cicuit, there are no cunent oi voltage sources in the network, and the current and voltage lesponses are due entirely to the energy stored in the inductor. The stored energy at t = 0 is given by wt@)



=



iua



(8.8)



Surnning the voltages arouDd the circuit, we have



.ti



Ldt+ Ri=0



&R A+ Li=t) 216



Chapter



B



simple RC and Rr Crrcurts



(8.e)



FICURE



8,6 Soutcejree Rl circuit



This equation is of the same form as that of (8.2) for the nC ctcuit. We may therefore solve it by separating the \ariables. Instead, however, let us introduce a s€cond, very powerful method, which we generalize in Chapter 9. The method consists of assuminS or guessing (a perfectly legitimate mathematical technique) a general form of the solution based on an inspection of the equation to be solved. In guessing e solution, we shall include s€veral unknown constants and determine iheir values so that our assumed solution satisfies the differential equation and the iDitial conditiors for the network. As seen ftom (8.9), i must be a function that does not change its form upon differentiation; that is, dt/dt is a muhiple of The only function that satisfies this requirement is an exponential function of t, such as



i



(8.1o)



i(t) = Ae"



will take this function as our guess, whele A and r ale constants to be deter_ mined. Substituting the solution into (8.9), we obtain We thus



(,*f)o""=o ftom which we se€ that our solution is valid if Ae" = 0 or if s = -iR/,. The first case is disregaded since, by (8.10), it results ir i = 0 foi all t and cannot satisry i(0) : Io. Thus we take the case s = -n/r, and (E.10) becomes



i(t)



:



Ae



R'/L



The constant A can now be determined ftom the initial condition t(0)



=



10. This



condition requAes that



i(0)=70=^ Therefore, the solution becomes (8.



1



l)



Sinc€ the solution is an exponential function, as in the RC case, it also has a time constant r. In terms of 7 we may *.rite the currCnt in the geneml form



s€ction



0.3



Source-fr€e Rt Circuit



217



,where by comparison with



ItO:



(V-S/A)/(V/A)



(8.1l) we



s€e



that. = L/R. Evidendy,6€ u aof rare incressitrg C h 6e lC cilsuit, itr-



= s. Increasing t, like



crcases th€ time constaDt. However, an incl€ale in R, iD coDtrait to lhe iC circuit, lowers the value of the time consta . A graph of a q?ical curtctt r€spoffe is shoivn



in Fis. 8.7.



The instantaneous power delivered to the resisror



i!



Fig. 8.6 is



p(t\=Ri1(r)=N2oc1t/L Therefor€, the energy absorbed by the .esistor



wl-\



as



time be.omes



itrflite



is giveo by



= J"I pTdt = |t-



=



NAe-'1RtL d!



4Lr6



Comparing this result with (8.8), we s€e that the energy initially stored ir the hductor is dissipated by the resistor, as expect€d. Suppose that \re had chosen to'6nd the inductor voltage o in thg circuit itrsrosd of the cwrent i. Applying KCL. we fnd



t[' *=-; j" odt+i@):o which is an irt?8?uJ equation. Difrereotiati.og this e-quatiotr witb respoct to ser that



ldo Rdr



I L



daR -+-D=ll dtL 218



chapter



I



simple RC and Rt circuits



timr,



we



This equation is a ditrerential €quation that we can solve using one of the methods discussed previously. It is also interesting to note that if we replace o by iR, we have



diR dtL =+-i=0 which is (8.9), obtained using KVL. Ftom these last two results satisfy the same e4uation and thus have the same form.



EXAMPIE 8.3



rii/e see



that o and



i



[,er us now determine i and o in lhe more general RL circuir of Fig. 8.8. which we assume is ir a dc steady-state condition at t = 0-. Therefore, recalling that an inductor is 4 short circuit to dc, we have



i(o-):g=2A Since the current in the inductor is continuous at



,=



0, we have



r(0")=r(o):2A r=O



8.8 More seneral Rl circuit



FICURE



The time constant for the network for t > 0 is clearly the ratio of the inductance and the equivalent lesistance as se€n ftom the termitrals of the inductor. The equi lent rcsistarce is



Re



=



5o



+



(Q1!!0-



-



roo



o



and hence the time constant is



"=!=6.1" Therefore,



since 10 :



i(0*) =



2



A, we have



i(t\ = 2-rw o Sununidg the voltages given by



of the ilductor and the 50-O resistor, the volrage o(r) is u(r)



= l0;+



50i



= -100e.te V se.rion



8.3



Source-free Rl Circuit



219



TXAMPLE



8.4



Consider the network of Fig. 8.9, which contain$ a dependent voltage source. The initial curent is t(0) = 10. Summing the vollages around rhe loop, w€ find that ,.li Z- tRi+ki=0



or



di /R+k\



a*\ t )i=o Comparing this equation to (8.9), we see thar the equations arc identical is replaced byR + t. Thus from (8.11) we have



i\t)



=



Ioe t''r''



ifR in (9.9)



L



The time constant in this case, which is modified by the presence of the dependent soulce, is given by



-L '



n{t



This is not surprising since in this case the dependent soarce b€haves like a rt-O resistor.



i()



FICURI



8.9



RL circujt contajning a dependent voltage source



EXERCISES the series Ra circuit of Fig. 8.6, determine (a) the inductor voltage for R : 50 rrlH, and 10 = 16 mA, 0) f ifR = l0kOand r: 10ps, and (c) R for the inductor curent in a 0.01-H in&rctu to halve every l0O ps. answer @) 4e '* Y O) 0.1 H; (c) 69.3 O A series RL ciJcuit has a l-H inductor. Deterftine the lue of lR for the stored energy to halve every 10 ms. tnswer 50 ln 2 = 34.(6 Q The circuit is in steady state at r = 0-. Fitd i and o for r > 0. 2, Answer 3e-a A. -6e y



8,3.1 In



25OA,L =



8.3.2 8.3.3



220



Chapter



S



Simple RCand Rl Circuirs



EXERCtSt



t.3,4



The circuit is in Answet -be "''! Y



steady state at



t = 0-.



ai.3



Find o for



t > 0.



EXtRCtSt 8.3.4



8.4 RESPONSE TO A CONSTANT FORCING FUNCTION



In the preceding sections we have considered soulc€_ftee circuits whode responses have been the risult of initial energies stored i! caiacitors and inductors. AI independ€nt eullent or voltage sources were removed or srtitched ou1 of the circuits pdor to finding the natural lespons€s. It was shown that these lesponses, when arising in circuits containilrg a single capacitor or inductor aDd an equivalent resistor, die out with increasing time. In this section we examine circuits which, in addition to havitrg initial stoted energies, arc d ven by constant hdependent curent or voltage sources' orjforcinS funcn:o.rs. For these circuits we shall obtain solulions which are the result of ioserting or switching sources into the nelworks We shall find lhat the responses in lhese caies, ualike those of soufce_ftee circuits, consist of two parts, one of which is always a constanl, L€t us begin by considering the circuit of Fig 8.10. The network comists of the parallel connection of a constant current source and a resistor which is switched



at time



t



-



0 asoss a capacitor having a voltage o(0-)



:



% v. Fot



t > 0, the



swirch is closed and a nodal equation at the upper lode is given by



r4i+!=n -dt R Section



8.4



Response to a Constant Forcing Fundion



221



'



FIGURE



8.lO Driven



RC



dallo dt RC'



netwo*



(8.12)



C



Equations of this tt?e that have constant forcing tunctions (1o in this case) can be solved by the method of sepamtiot of ttriables. We may first write (8.12) in the



form



da=_o-RIo dt RC Mdtiplyirg both



sid€s by



dt/(o



-



N6) ar]d forming itrd€finite integrals, we have



Iau



J



t.



tt -rtJ* Rto



or ln (o



-



Rr6)



=



-#* *



wherc tr is a constant of iltegration- This result can be *,ritten



as



a - RIo= ?-r/R't'r or, solving for t,, we find



o= where we hav€ takenA



:



ex,



Ae-4Rc



+



RIo



a constant to be determined



the circuit.



(8.1.1)



bv the initial condition



of



kom (8.13) we set that the geneml solution for the voltagg response consists of two parts, atr exponential frrnctiol and a cobstant functioD. The exponential function is of the identical form as that of the natuml tesponse in a source-free circuit composed ofi and C. Since this part of rhe solution is chamctedzed entirely by the nC tiDe constant, we shall refer to it as the r4tural response a, of the driven circuit. As h the case of $e source-&ee circuit. this response approaches zero as time incteas€s.



The second part of the solution, given by R/o, beam a close resemblance to the forcing function 10. In fact, as time increases, the natural response disappears, and the solution is simply Rio. This compon€nt is due entitely to the forcinC fuoction, and we shall call it the/orcel response al of the dtiven circuit. A reader who has had a course in diferential equations will, of course, recognize the natural restrnnse o,



222



Chapter



8



Simple fiC and Rt Circuils



and the forced respons€ ol as the homogeneous response o,, and the particular response



De.



IJt



us now eyaluate tlrc constant A in (8.13). As in the case of the souc€-ftee circuit, its value must be selected so that the initial voltage condition is satisfied. At



t = 0*,



we se€ that



t,(0*)-o(0)=Yo Th€rcforc, at



t = 0*,



(8.13) r€quires that



Vo=.4+RIo or



A=Vo-'Jo



.



Subsdruling this value of A back into our solution yields



a(t) = PJo+ (Vo-



(8.14)



RIo\e"rRc



We should observe in this solution that the constant A is now determined aot only by the initial voltage (o. energy) on the capacitor but also by the forcing function 10.



Craphs of



o,, or,



aDd



rcsponse D" for yo RJo plete response is shown.



-



FICURE



8.ll



>



Ii



(a) the natual u a.e showll in Fig. 8.11(a) and (b). 0 and the forced r€sponse o, are. shown. In (b) the com-



Craphs of voltaSe response for the driven RC network of Fig. 8.10:



(a) natural and forced respons€s; (b) complete respons€



vo



Section



8.4



-



Rto



Respons€ to a Constant Forcint Function



223



The current



ir



t > 0 is Vo - RIo ,.,, = L^drt= _ __T_? ,t,.



the capaciior for



,tt



whereas the cu.rlent in the resistor is



iR=k- i(=6+b--E!r"," RIt is interesting to note that the resisto! voltage has changed abrupdy ftom RIo at , = 0 ro vo at t : O* . The capacitor voltage, as pohted out previously, is conrinuous,



The solutions that we have encountered so



fir



ro this chapter are often teferred



to in othe! more descdptiv€ tems. Two such ternrs that are very popular are the transient response alfr tbe steady-state r.sponse. T'trc tmnsient respolse is the tradsi_ tory portion of the complete response which approaches zero as time increises. The steady-state lespons€, otr the other hand, is that part of the complete response ,rdich rcmains after the tmnsient response has become rcro. In the case of dc sotnces, the steady-state response is constant atrd is the dc steady state discussed itr Sec. 7.7. In our example we see that the transient response and the natural response are identical, as arc those fff the source-ftee circuits of the previous se-ctions. The steadylstate response is therefore identical to the forced respoos€. In our example 0, and tx = 10, & lalues that constitute the dc these responses Ne a = RIo, ic steady-state condition. We should not conclude from the discussion above that the natutal ard folced responses arc always the tra{si€nt and steady-state responses. If the forcing functioD is a transitory firnction, for instance, the steady-state rcsponse is zero, as we shall see in Chapter 9. In this case th€ complete rcsponse is tbe traDsient response.



:



EXERCISES



E.4,1



Find



D



for



tntwer I0



t > 0 if the ctcuit is in steady state at r : - 6e 5u \l



l0



v -:-



txtRclst 8.4,2



224



The circuit is in steady state at Answer



3



Chapter



S



-



0-.



e a'A



t = 0-.



Simple RCand Ra Cicuits



Find



8,4.1



i for



t > 0.



I' 2H



txERctsE 8.4.2



8.4.3



Find u for Answer 24



t > 0 if the circuit is in steady state at t = 0 . - 8e-3' \



txERctst 8.4.3



8.5 THE GENERAL CASE The equations describing the networks of the previous sections are all speaial cases of a general expression given by



flt,,:a



(8.15)



ll.here is the uDknown variable, such as o or i, and P and C are constants. If, for instanc€, we compare this equation to that of (8.2) for the source-free circuit of Sec. 8.1, we see that 1 = a, P = I/RC, ar]d Q = 0. The same relations are valid for the forced ltc circuit of Sec. 8.4 except thal Q = Io/C. A solution of (8.15) for P and C constant can be found by separation of !"riables. However, let us introduce another method which also is applicable when O is a firlction of time, an important case in later chapters. This method, known as the integraling faclor method, consists of multiplying the equation by a hctor that makes its left+and side a perfect derivative and simply integating both sides.



]



Section



a.5



The General Case



225



IJt



us begin by considering the



deri\ttive of



a



Foduct, given by



d ye^t__dtd]"", * r,""



;tt



=



(! * ,r\"" sides of (8.15) by eP', we have



if we multiply both



Frcm this result we see that



d.



j,r



r*t-.



= q'e"



Iniegrating both sldes of the equation, we find



je,,=[q".**o 'J where A is a constant of integratiotr. Solving for



),



we have



v="-olQ"*at+*o



(8..16)



which is valid, of coulse, if C is a function of time or a constant. If 0 is dot a constant, we must carry out the integration to find y. Examples of this twe are given in Exercises 8.5.2 and 8.5.4. In the important dc case wheie q is a constant, (8.16) becomes



n



) = Ae-P1 l=



=t,+vr



(8



17)



), = Ae P' and, y1:0/P are the natural and forced responses. We obsefle y, has th9 same mathematical form as the soulce-ftee natural rcsponse and that yl is always a constant which is Foportional to 0. In addition, l/P is the time co!where that



'



stant in the natuml rcsponse.



EXAMPTE



8.5



[.et us find i, for t > 0 in the ckcuit of Fig. 8.12, given that i(0) = 1 A. Althougb the circuit is a somewhat complex combinatioD of elements, the solutio! of (8.17) is valid since the network contains a constant forcing function and d single enetgystorage element (the inductor). The loop e-quations for the circuit are



Str-4L:10 -ai,+L2iz't*=o tlt Elimimring lr ftom these equations, we find that



**t0,,=t dt 226



chapier



s



simple Rc and Ri (tr(u'tc



I a



t



. rTCUXE 8.12 Driven



RL



Comparing this e4uation with (8.15), we se€ that yields



circuir



P = l0



and



t



O = 5. Henco (8.17)



i,=Aetb+l Adying



the initial coudiriotr, we hove



i,(o)=A+.+=r Therefore. ,4



=I



atrd rhe solution is giveD by



.



h =+€_,0,++A



We nay also obtain (8.17) for the case C constatrt by obeerviry (8.15) and (8.16). Making Q = 0 in (8.16) yields the latrrral reqlonse.) = y" = Ae-P', which then must be the solutio! of. (8.15) when O = 0. Thst is, the natutal response sarisfies



ff*et=o and may be found as in Sec. 8.3by



tryin|



r. = Af" wbich results in



s+p=0 Thus we have r = -P, which yields y, = Ae-"', as befote. The forccd respon$e ), may be'found by ,iling in (8.15) a furction like O. Sioce O is cotrstant in this cas€, Lhe al solution is thed a constalt. That is. we try



t



'



Yt= K



which subsliruled inlo {8.15) eives



0



+pK=e.



or



o h= K=-P



I



as belme.



l



Sedion



8.5



The CenehlCas€



227



l



I



I



EXERCISES



8.5.1 Findofor, >0if t(0) = l Aandos = 50V. Awwer 2O



*



l5e-3' Y



i



tso



I



10c,



t8



rxERctst 8.s.1



: 30e ' V 8.5.3 Findifort >0if t(0) = 4AandL Ancwer 2 't 2e 3' A 8,5.2



Solve Exercise 8.5.1 if o, Ahrwer 33e-3' l&e s'v



8A. 6ct



l' ril



,t;



1)



rxERctsE 8.5.3



8.5.4



Solve Exercise 8.5.3 Answer 0.4(8 cos



if i, =



t+



sin



13 cos t



A.



t + 2-&) A



8.5 A SHORTCUT PROCTDURE



kt



us now introduce a shortcut procedure that is very useful for finding the currents and voltages iD circuits with dc sources. The technique irvolves fomulatirg the so-



lution by melely inspecting the circuit.



EXAMPIE



8,5



Consider Example 8.5, especialy Fig. 8.12. We know that



i,=i,+i,, .



where i2" and irr are the natuml and forced responses, resp€ctively. Since iz" has the, same form as the source-fiee response, we can look at tie network in the abserce of the forcing function (i.e., make the lo-V source zero by replacing it by a short circuit), as shown in Fig. 8.13(a). The natural response is then



iz,



228



chapter



8



simple Rcand Rr cncuirs



= Ae



td



{t bo



""*



"."611;-l tr"l



D



(r) G)



(b)



FICURI a.l3 Circuits for finding the response of Fig. 8.12: (a) circuit for finding (b) circuit for findins i,i



i.;



The forced response is constant; therefore,. insohr as the forced response is concefied, it does not matter at what time we look at the circuit. We may choose then to look at the circuit in the stE dy state when i?" is zelo. At this time the inductor is a short circuit, as shown in Fig. 8.13(b), from which it follows that



4=i



Therefore,



i2= Ae-N'+l The constant A is now determined as before from the initial condition, rr(0)



= t.



A word of caution is appropriate at this point. When evaluating the constant A, the initial condition should always be applied to the complete response-never to the natuml rcsponse alone-because the initial condition is always given for the current. not for a part of it. EXAMPLE 8.7



[,et us fi nd



i for



r.



0 in Fig. 8. 14. given u{0,



=



24 V. The currenl is given by



i:i"+v To obtain i, we note that it has the same/orm as o", the natuml response of the capacitor voltage. In fact, the natural response of erery cunent or voltage iII the cir. cuit has tle same form as p,. This is true because all the other curents and voltages in the soulce-ftee circuit may be obtained ftom o. by applying one or more of the opeiations of addition, subtraction (in KCL and KVL), differentiation, and integration, none of which changes the nature of the exponeotial e-'l'. Examining the souce-free circuit (the current souce open circuited), we se-e that thp time constant 0.2 s. Therefore, for the capacitor voltage is r



:



i': FTGURE



4"-s'



8.14 Driven



RC



circuit



rto



se.rion



i I



F



8.6



A Shoncut Proc€dure



229



3 In the steady state the capacitor is an open circuit, and the forced response is, by inspecrion



tt=



1 A'



Therefore,



i(t\=Ae-3'+l To €rduate A, we must find the value of t(0*). Since D(0) = o(0*) ming the voltages around the right-hand mesh, for t = 0*, we have



4i(0)+6[l -



,(0'J] +



24



=



:



24



V,



sum-



0



OI



'(0.)



=



3



Substituting this initial curent into our solution, we find that



3=A+I Thercfore,A=2and



i=l+2e-5'A. EXAMPLT 8.8



Before concluding this section, let us d€termine i aDd o in the circuit of Fig. 8.15(a). The network is in a dc steady-state condition at , = 0- with the switch open; therefore, the iriductor atrd capacitor aie a short circuit ard atr open circuit, resFctively, at this time. The capacitor voltage is equal to the voltage that appea$ acroes the 20-() resistor, and the indrctor curreDt is €qual to the curent in the l5-O .esistor. By cudent divisiotr, the currents in the 15- aod m-O lesistors are easily shown to be 2 and I A. resp€ctively. Thus



i(o-) =



2



A



and



o(0)=60V When the siwitch closes al t = 0, we observe thal nodes a and, are shortcircuited together, and we can redmw the network as shown ia Fig. 8.15(b). It should be noticed that the 30-f,) !€sistor ne€d not b€ included in this circuit because the s,titch is a short circuit across ia terminals. The combination is equivalent to a 30-.Q resistor in paiallel with a 0-() resistor, which, of course, is 0 f,I or a short circuit. IJt us next consider the current i leaving d in Fig. 8.150) through thc t5-O resistor. Frcm KCL, this sa$e current must enter a through the l-H irductor; hence no curent flowing in the circuit to the left of d can enter the other part of the circuit to the right of a, and vice versa. Thus after the switch is clo6ed, the network redtrces to two independent circuits, each of which can be solved individually. The first circlit, consisting of the l-H inducto and the 15-() resistor, is simply a source-free rRL net'rvork having



i(0*)



i-



230



Chapter



8



Simple RCand



ft



Circuits



:



i(0-) = k-ts' A



2



A. Therefore,



ib) FICURI 8.'15 (al Circuit containinC an inductance and a capacjtance; (b) equivalent



circuitfo.t>O



lhe second ctcuit. composed of all the elemenls lo the righr of d. is simply a : x16 ) = 6O V From our shortcut p.ocedue, we



driven RC network with o(0*) 6nd



.



a= 40+me-'V



The shorkui procedure presented in this secrion is appticable also to circuirs containinS dependent sources. However, no savings in time or effort usually re.sult because the circuit equations still have to be written fol the source-fie€ and dc steady-state cases



.



EXERCISES



8.6.1



t.6.2



Solve Exercise 8.5.1 using the shortcut procedure of this section. Find o and i for t > 0 if the circuit is in dc steadv stare O Answer



2(l



-



€-ld) v,4(1 +



atr:



"-'d)



rnA



txERct$ a-6.2



231



8.6.3



Find i for Ans-e, le



t > 0 if the circuit is in dc steady state at t = 0 '*' - 3e-,* + 5 mA :tr)



:H



-kO



,'A



txERcrsE 8.6.3



8.6.4 Findifort > 0 inthe circuitof Ex. 8 6.2 if both switches areclosedatt:0 i(0) = 0. (SrSSestion. Find i, as before and note that +is due to two sources ) 5oo0r Answer



L '



12



and



mA



12e



THE UNIT STEP FUNCTION In the previous sections we have anallzed clrcuits in which energy sources have been suddehly inserted into the networks- At the instant these souces are applied the voltages or curents, at the points of application, change abruptly. Forcing functions whose values change in this manner are called ritgularity ftuctions Therc arcm ny singularity functions that are useful in circuit analysis. One of the most important is ihe unit step tunction, so named by the English engineer Oliver Heaviside (1850- 1925). The nit step function is the function equal to zero for all negative values of its argument and equal to I for all positive ralues of its aryument. If we denote the unit step tunction by the Embol , (r) , a mathematical descliption is



u(r) =



=



o, ,



(s.ls)



0



From a graph of (8.18), shown in Fig, 8.16, we se€ that at t = 0' !(t) changes abruptlyiom 0 to l. Some authors define (0) to be 1, bulwe are leaving !(t)



undefinedatr=0. FIGURE



232



Chapter



S



8.r6



G'aph of Ihe unir \rep fun' rion uh)



Simple RC and Rl Circuils



The unit step fuDction rnay be used to rcpreseot voltages or curcnts with finite djscontinuilies. For example, a voltage step of y volts is represeoled by the product Vu(t). Cleaiy, this voltage is 0 for t < 0 and y volts for, > 0. A voltage step souce of l/ volts is shown in Fig. 8.17(a), A circuit that is equivalent to this solrc; is shown in Fig. 8.17(b). A shott circuit exists for < 0, and the voltage is, of course, zero. For > 0, a voltage y appears at the terminals. We have assumed in our model that the switching action occurs in zero time.



I



/



FIGURE



Ll7



(a) Voltage step source of V



volb; (b)equivatenr chcuit



Equivalent circuits for a cudetrt step souce of 1 am!'eres arc shown in Fig. open circuit exists for , < 0. and rhe current is zero. For r > 0, tie swilching aclion causes a terminal curreD! of 1 arnperes lo flow.



^ - An 8.i8. _



FICURI 8.18 (a) Currcnt step source of / amperes; (b)equjvalent circuit



The switching action shown in Fig. 8.17 can only be apFoximated i! actual circuits. However. in maDy cases, it is not necessary lo rcquLe rhat the voltage souce be a short circuit for t < 0, as we slEll s€e i4 the next section. If the terminals of a network to c4tich the source is to be connected rernain at 0 V for t < 0, a series conrcction of a source y aDd a switch is equivalent to the voltage step genera-



tor, as shown itr Fig. 8.19. Equivalent circuits fqr



FICURI 8.19 {a) Network with V applied ar r



Section



I



L



8.7 fhe



Unit step Frlnction



a current step generator



:



in a



net_



O; (b) equivalent circujt



233



work arc shown in Fig. 8.20. In each case the cutrent in the netwolk teminals must be zero for r < 0.



Iu(t\



(a)



. FIGURE



S.20



(a) Network



with i applied at t



:



o) 0; (b) equivalent circuit



I-et us now retum to our definition of the unit step furctlon Siven in (8.18). We /o in the three places that it ocmay generaliz€ this dennitio! by rcplacing , by t



-



curs, which resulE ilr



u(t



The iulction 8.21.



u



(,



*



-td= 0, =1,



,o) is the fuaction



tro



(8.Ie)



l,l(t\ delnted by to se.o ds, as shown in Fig



HCUnf 0.21 Graph of l}le unt



step function



u(t



t')



Multiplyiog (E.19) by V or l give's us a voltage step rource or a curent step source who"" ialue cha4es abruptly at time to. Equi\'"alent netrvorks for thes€ to sources are obtained in Figs. 8.17-8.20 by taking aI actions rclaled to $tirching



.Eflrratt=ln. EXAMPIE 8.9



SreD functioDs are very usefirl



in formuiatilg more complex funclions Thke' for inFig 8 22a) From this fr8ure we se€ thal



stance, lhe rectangular voltage pulse of



orG)=0' : v' =0, 234



Chapler



8



Simple RC and



fl



Ckcuils



toand -r(r -



a(t):



vlu(t)



Tq check this result, we see that, for



,>



(b) square wave



ro) becomes



u(!



-l



for



r>



- k\l



ro. we may (8.20)



t < 0,



"r(,)=v(0-0r=0



For0l



as



in.



r)l



*



12(t



-



e-:t)e1,-,,u(t



- t\



(8.22)



Fig. 8.2?(b).



EXERCISES



E.8.1



'.



Find the step responses Ansv',et



(l -



i



and



, [i, = u(t) A),. - e-'la(r) V



0.5e to)r(r) A, 5(l



'



'5c)



EXERC|ST 8,8.1



8.E.2



8.8.J



Find the response i to rrs



= 42u(t)



Answ"r 2(t - e ?t)u\) A Find i in Exercise 8,8.1 if Answet



Section



tol(l



8.8



-



V.



i, = i0['/(r)



0.5e-i&)!(r)



.Th6 Step Response



- (l -



-



t1(, 0.5e-,0r,



- l)l A. ")r(, - l)l A 241



l, l2a



txERctsE 8.8.2



8.8,4



Find o if a" = 2"-'*rrr, V and there is no initial slored energy. (This circuit i! lik€ the intedrator of Fig. 8:24, except thal the capacitor has a resistot in parallel with it, making it a practical, or lrssy, capacitor. Thus the circuit is.alled a,rss)



iuegmtot.l lnswer



2(e-M'



e-rmJll(t) V



APPLICATION OF SUPERPOSITION In this section we consjder the use of superposition for obtaining solutions of RC ard RL circuits containin€ two or more independent sources. EXAMPLE



'



8.14



Let us consider th€ circuit ot Fig. 8.26. The value of the independent currenl source is given bv



rs=10!(t)-l0r(r-l) This source is equi lent to a pair of independent current sources connected in parallel. Thus if we let



L=i+i, where



ir = l0l](r)



aDd



6=



lou{r



-



l).thecircuitofFig.8.26canberedrawnas



shown in Fig. 8.28. From the principle of suprposition, 1'e rnay write the outFn voltage



242



chapter



8



Simple Rc and Rl circuits



as



5H



3c,
t



O



is in steady state



if the circuit is in



steady state



>oifr(0:sA.



9c)



I



PROBTEM 8.'I



O



I



I



:



PROBTEM A.I1



r'/



I0mf



30s,



chapter



S



Problems



251



PROBEMa.I3.



PROILIM 8.14 0.5irv



PROELEM



t>



8.16



Find i for



8.11



Find o and i for



0



stateatt=0. E.lE



if the circuit



r>



is



in



st€ady state



0 if the circuit is iD steady



The circuit shown is in a dc steady-state condition at t = 0 . Find D for > 0.



i



lqo



S.1y



8.19



Find i for



E.20



Find t



a.2t



Find o for



8.22.



Find



t=0



PROBTTM 0.16



2s2



I



t



Chapter



8



Simple RCand Rr eircuals



the circuit is ia steady srare



fort > 0 if i(0) = 2A.



at{=0. ,



, > 0if



for



t > 0 if the circuit ls in steady ,>



0



if the circuit



is



in



state



stoady state



'i io



2.5.0



30v



t2()



4H



PROETEM 8.18



PROALEM 8.22



253



I



I



t > 0 if the circuit is id steady



8.2.!



Find o for



8.24 a-25



Find t and o for



att:0



state



: -6 v Reo€at Bob. 8.24 if the l6v source is repl;ed by a lo" r'-V source *ilh lhe same po,>



0 if 0(0)



IaritY



t,26



Fiod o for



alr = 0



t.2?



Find i for



t.28



Find



,>



0 if the circuit is in steady state



r>



0



Find i for



8.30



Find o and i for



is



in



steady state



at, = 0-. ol for , > 0 if there is no initial



stored



0



if th€ circuit is in



I>



0



steady state



if the circuit



is in sleady



stateatr=0-.



8.31



Find o for all t if (a) os - l2t/il,r 4(I



8.32



Find o for



.



if the crcuit



I>



t.29



/>0 if



o":



124(t)



V and (b)



I)l v o(0)



= 2 v and



!,s- 6V, O) a*:6e-' v,



and (c)



6e-3' Y.



8.33 Find o if t" = 2? 3'!(t) V 8.34 FindlJfort > 0if os = 3!(r)V



energy.



6(I



u



PROBLTM 8.24



PROELEM 8.23



PROBLTM 8,26



PROBTEM 8.27



254



chapter



S



s'mple



RC and Rt Circuits



(a)



t)'=



12



PROBTEM 8.29



PIOBLEM 8.30



3s'



\ $r PROa[tM e31



PROBLIM A32 14f,r



PROBLEM



Chapter



8.!3_/,--



8



Probl€ms



PROBLTM 8.34



255



8.35 Find , for i>0 t.36 8.37



if D,-2e-3' V



o.(0) = 0 Find l, if D,



and



2u



Find for capacitor



= 2e x'u(t, V Find o for r > O if Dt(O) : 0 and ,, = t



6.36



a.



.



8,39



(t) v.



8.40



r>



0 the curenr downward



= 2e ' V,



znd (.1 Ds



Find o for all r



if



(a)



8.y./



PROBIIM 8.38



256



Chapter



S



Simple RC and Rr Circuitt



=



2 cos 2tV.



tbe



O)



,, :2l(,) V, and (b)



q : 2tu(t) u(t - t)lv. Find , for all time if os = 2u (r) V



Ii'



PRoBLEM



ir



if r(0) = 0, and (a) r, :4 V,



PROALEM



AI9



PROELIM 8.40



COMPUTER APPLICATION PROBLEMS t.41 Use SPICE to solve (a) Prob.8.l9for0< t < I s, (b) Prob.8.28 for 0 < I < (c) Prob. 8.36 for 0 < t < 1.5 s. 8.42 Use SPICE to plot the output voltage D for 0 < r < 25 ms. 100



0.5 s,



and



ko



l0 0



l0



PROBLTM 8.42 ; I



Chapler



8



Computer Applicatio; Problems



257 ,l {



E.43



a.+4



"



2sS



Chapter



E



Simple RC and Rl Circuits



PRdBLE ,!



8.i4



9 Second-O rder Circuits



Electromagnetism was discovered in



lhe spring oi 1820 by the Danish physicisl Hans Christian Oersted



An aaempt should be nade to see whether ekctricity, in i|s nnst latent stage, has afiy ejlect on the nagnet as



cation, were accepted by lhe UniveF



sity of Copenhagen, where



Hans



studied astronomy, ahemislry, rnathewhen he demonslraled lhat lhe neematics, physics, and phaamacy, He dle oJ a compass moved when placed completed his training in phamacy in near a curreni-carrying wjre, By JuJy 1797 and lwo years lalsr received his Hans Christian Oersted of lhal year, he was certain that an doclorate in philosophy. Afler a baiel electrjc current produced about il a stint as a pharmacist, he was alcircular magnelic field, and he pubtracled to the world of science, which lished his results in a short paper, written in Latin, and was in ferment al lhe time over Volta's discovery of carried by the major scientific journals of Eurcpe. the eleckic batlery. Between 1800 and 1820, he was Oersted was born in the town of Rudkobing, on a university teacher, researcher, publisher, and one of the Danish island of Langeland, the elder son of an the mosl sought-afler lecturers ol his day. apoihecary, Soren Chrislian Oerstod. Because of Oersted's great discovery had an enormous imlamily problems, Hans and his younger brolher were pacl on the scienlific world, and he was showered placed wilh a Cerman wigmaker.while lhey were still with honors and awards. The Royal Sociely of London young boys, The brothers' intellectual abilities and ex- gave him the Copley Medal, and lhe Frgnch Academy traordinary thirst for knowledge were soon apparent to awarded him a prize of 3000 gold francs. ln his honor the townspeople, who did what they muld to educate the oersted was chosen as lhe standard cgs unit of them. ln 1794 lhe brothers, wilh no pior lormal edu- magnetic lield intensity..



259



L



T n the case I of Iinear circuits with energy-storage elemen!s the describing equations



(those relaling rhe outputs lo rhe inpurs) may be expressed a. linear d iffer.-" iili .""r_ llons. because the lerminal relations of the elemenrs are sucb that the terms in rtre loop or nodal equations arc derivatives, int€rals, oi multiples of th" lhe source variables. Evidently, a single differenriation ol an .quu,ion *iff any inlegrals lhal h may conlain. so that in general the Ioop or nodal equations tor a given circuit may be considered to be differential equations. The descridirg"qoution then may be obtain€d fiom these equations. The circuits containing stomge elements that we hav€ considered so far were first-order circuits. That is, they wete described by first_order diffe.""riui This is always the case when thcre is only one siomge element "q."io".. switching action convsts the circuit into two or more-independenicircuits tuu, ing no more than one stomge element. "uctr In this chapter we consitler second-order circuits, which, as we shall see, con_ tain t\ro stomge elements and hav€ describing equations that are second-order iiffer_ . ential_equations. In general, nth-order circuits, containing n storage elements, are describ€d by rrth-order differential eqlarions. The results fL first_ ind second_order circuits (r, = I and n = 2) may be readily extended to the general case, but we shali not do so here. However, a solution of a third-order differJntial equation is outlined in.Prob. 9.39, which may be us€d to solve a third-order circuit given in prob. 9.40. Higher-order citcuits are he3ted in more detail in Chapter 14. Another, very elegant, method of solving higher-order circuits. as well as firs! and second-order ones. is the Laplace rransform method gi\en inChapler lg. An in_ lerested reader may go directly to this chapter withour rhi need for reading the inter_ vening chapters,



unkn;;;;;; ,.ro".



^



pr"."rt;.l;il;;



.



9.1 CIRCUITS WITH TWO STORAGE ELEMTNTS To introduce the subjectof second-ordq citcuits. let us begrn with lhe circuit of Fia_ 9.l. where the output to be found is the mesh currenr ir. The circuir contains rvr'o stomge elements, the inductors, and as we shall see, i, satisfies a second_order dif_ ferential equation. Methods of solving such equations will be considered in later sec_ tions of this chapler.



260



\ Chapter



9



s€cond-Order Cncuits



I?) FIGURE



9.1 Circuit with two



inductors



The mesh equations of Fig. 9.1 are given by



.li,



2; + t2it



4i,



ax



(e.



r)



-'ai,-+rai,-o 'dt From the second of these we have t ldi i,-i\;t4t,)



\ (e.2)



which diferentiated results in



di,



_t(d,i? + 4di,\ dt I



(e.3)



dt 4\dt'



Substituting (9.2) add (9.3) into the fiist equation of (9.1) to eliminate afler mulriplying the resultirg equation through by 2.



ff



* rcff *



rci.= 2.,



i,,



we have,



(e.4)



The describing equation for the output ,; $ thus a seconl-order differcntial equation. That is, it is a differential equation in which the highest derivative is second-order. For this rcason we refer to Fig. 9.1 as a recorl-ord?4 circuit and note that, typically, second-order circuits contaitr two stomge elements. There are exceplions, however. to the rule lhat two-storage-elemenl circuits have second-order describitrg equations. For example, let us consider the cLcuit of Fig. 9.2, which has two capacitors. With tle refereoce node taken as indicated, FIGURI



Sedion I



I I L



I



I



9-1



9.2 Circuit with two



Circuits with Two Stohge Units



capacitors



261



nodal equalions at the nodes labeled or and



02 are



given by



---' + rr (e.5)



h, -: + 2a1



The choice of the oode voltages ur and o? as the unknowns has resulted in two first-order differential equations, eact containing only one of the uDkno\rns. When this happens, we say that the equatiois te uncoupled, and thus no elimination procedure is required to sepa&te the variables. It was the elimination procedure which, applied to (9.1), gave the second-order equation of (9.4). The eqlations bf (9.5) may be solved separat-ely by the methods ol the Fevious chapter. Eyidently, Fig. 9.2, although it contains two siomge elements, is not a secondorder circuit. The same voltage o, is across each RC combination, aDd thus the circuit may be redmwn as two 6$t-order circuits. If the souce were a practical souce rather than at ideal source, then the circuit would be a second-older circuit (see Prob. 9.1).



EXERCISES



9,1.1



Find the equation satisfied by th€ mesh current



Ansi?r



i, + _di, .. d., tA t 6t,= dt2 d;



i,.



d,



3r)



2A



IH



I



EXIRCISE 9,1.1



9.1,2 Let Ds= 8e 2'y, n(0*) = 2 A anO r:(0*)=9 A in Exercise 9.1.1, and find diz(0*)/dt (the value of dirldt d t = o*\. Ans\9et -23 Als 9.1.3 For the values of iz(o.), dn@.)ldt, and os given in Exercise 9.1.2, show that t, in Exercise 9. t.I is given by i. = 3e-' + 4e 2t +



2e-61



A



(Sug8ertbr. Substitute the answer into the differential equation, etc.)



262



Chapter 9



9.2 -E



SECOND.ORDER EQUATIONS ln Chapter 8 se considered first-order crrcuils in some delail and saw thar rherr describing equations werc lint-order difierential €quations of the form



dr



(e.6)



In Sec. 9.1. we defined second-order circuits as those having two storage elements nirh de\cribints equarions that were second-mder dlflerenrial equation\. given generally by



dlx



dx



E+o'A+dor=/(rl In (9.6) and (9.7) the a's are real constants,



and/(r)



i



(e.?)



may be either a voltage or a current,



ofthe independent sources. As an example, for the circuit of Fig. 9.1, the desoibing equation was (9.4). Comparing this equation with is a known function



(9.7), we see that d,



:



:



10, ao 16, /(r) = 2Ds, and x = iz. From Chapter 8 we know that the complete response satisfying (9.6) is given



by



(e.8) where r, is the natual response obtained when = 0 and _rr. is the forced response, which satisfies (9.6). The forced response, in cont$st to the natuml response, contains no arbitmry constants. l,et us see if this same procedure will apply to the second-order equation (9.7). By a solution to (9.7) we shall mean a function which sarisfies (9.7) identically. That is, when.t is substituted into (9.7), the left member becomes idenrically / (r). We shall also require that r contain two arbitrary constants since we must be able to satisfy the two conditions imposed by the initial energy stored in the two srorage



/(l)



i



elements.



I{.r" is the natural resporse [i.e., the response the equation



d2x, dx" --:-i+dr---+(hx"=l)



when/(r) = 0], ir must



satisfy



(e.e)



Since each term contains.r, to the same degr€e, namely I (the right member may be thought of as 0 0&), this equation is sometimes called the l/rmogeneous eg]d.ation. If.rr" is to satisfy the original equation, as it did in tie firct-order case, then by (9.7) we must have



:



d|t,



dx,



i'^;''ao.'fk\ \er non



9.1



se, ond-Ordpr I



qLDlroh



(9.10)



263



Adding (9.9) and (9.10) and reananging the terms, we may wdte



d,. ;(r,



d. - x) - at:(\



+ t) - an6, + x!\ = Ihl



(9.1 1)



The rearnngement is possible, of couse, because the equations involved are linear. Comparing (9.7) ard (9.11) we see that (9.8) is our solution, as it \ras in the first-order case. That is, r satisfying (9,7) is made up of two components, a nafuml response r, satisfying the homogeneous equation (9.9) and a.forced restrronse x,. satisfying the original equation (9.10) or (9.7). As we shall see, the natural response will contain two arbitrary constarts and, as ilr the first-order case, the forced response will have no arbitmry constants. We considd methods of finding the natuml and forced reponses in the next three sections. Of cotrlse, if the driving, or forcing, tunctions are such that/(t) = 0 in (9.7), then the forced response is zqo, and the solution of the difrerential e4uation is simply the natural response. A reader ryho has had a course in diferential equations will note that the natuml response and the forced response are also called respectively the complementary solution and the particular solution. The complementary solution contains the arbirary conslants and the particular solution, as its name implies. contains no arbiffary constants-



EXERCISES



9.2.1



Show that



and



are each solutions



of



dzt



dx



Et5E* regaidless of the values of the coDstaDts



6x=O



A! and,4r.



9.2.2



Show that



9.2.3



is also a solution of the differcntial equation ol Exercise 9.2.1. Show that if the dght member of the differertial equation of-Exercise 9.2.1 is changed frorp 0 to 12, then



x=xt+t2=Ate'}'-A-e"



x:



A,e-, + Aze-1,+



2



is a solution. Thus the natural respotrse is Are a + Are r'and the forced tesponse is 2.



264



Chapter



9



Second-Orde. Circuils



9.3 THT NATURAT RTSPONSE The natuml response



r, of



the geneml solution



t=x,+x, of (9.7) must satisfy the bomogeneous equation, which we repeat



as



d'?x dr -,+at+!anr-O dt' d r :,r" must be a furction which does not change



(e.r2)



Evidently, the solutioD its form when it is differentiated. That is, the function, irs first deri!"tive, and its secodd derivative must all have the same form, for otheiwise the combination in the left member of the equafion could not become identically zerc for all t. We are therefore led to trv (e.



r3)



since this is the only function which retains its folm when it is repeatedly differentiated. This is, of couse, tbe same fimction that worked so well for us in the fustorder case of Chapter 8- Also, as in the first-order case, A and r aie constants to be



determined. Substituting (9.13) for



r



in (9.12), we have



As2e" + Asate"'+ Aaoe"':



O



or



,4e"(J'?+drr+do)=0 Since Ae" cannot be zero [for then by (9.13) tial energy-stomge conditionsl, we have



r, = 0, aod we cannot satisfy any ini-



12+drr+do=0



(e.14)



This equation is called tbg characteristic equar-on ind is simply the result of replacing de vatives in (9.12) by powers of r. That is, -r, tle zeroth derivarive, is replaced by ro, the first derivative by st, and the second dedvative by rr. Since (9. 14) is a quairatic equation, we have not one'solution, as in the firstorder case, but two solutioDs, say sr and rr, given by the quadratic formula as



-a,+!a"-qa"



(e.l5)



Therefore, we have two Datural components of the folm (9.13), which we denote by (e.16)



Section



9.1



The Natural Response



265



The coefficients Ar and A, are, of cou$e, arbitmry. Either of the two solutions (9.16) wilt satisfy the homogeneous equation, because substiruting either into (9.12) reduces it to (9.14). As a matter of fact, because (9.12) is a linear equation, the r m of the solutions



(9.l6)



is also a solution. That is,



(e.t7)



x.=til+\42 is a solulion of



t9.l2). To



see



into (9. t2). This results in



rhis. \ e have only lo subsrirute the expression for &



d, d. t x"z\ ' atj,(xa' r', r do(r,, - r'.,.r ),,(x"' r td t" - /d'*, * a'\dx"t a"x.,)-



li



d;:



\i



=0+0=0



'



d.x",



''-i; ' ^,')



sioce both .r"r and -r,, satisfy (9.12). By (9.16) and (q.17) we have



,"= Ale\'+



(e.18)



A1e"21



which is a more gengral solution (unless J! = s2) than either equation of (9.16). In fuct, (9.18) is ca,lled the eeneral sol rrb, of the homogeneous equation if sr and r, are d[stihct (i.e., not equal) roots of the characterisric equation (9.14). EXAMPLE



9.1



The homogeteous equation corresponding to (9.4) is given by



dlt, _-



-E



d(



I



and lhus the characreflsitic equalion



di,



t0_.: dt



+ l6t?.



0



(e.1e)



is



r.+lor+16=o The root! are's : -2



r: -8, so that the general solution is given by ,t + A2e-st i2 = ^e (9.19), The reader may verify by direct substitution that (9.20) and



satisfies



O.2o)



regardless



of



the value of the arbitrary constants. Because (9.18) is the natual response, the numbers rr and 12 me sometimes calle.d the natural ftequ?rcr€r of the circuit. Evidently, they play the same role as the negative reciprmal of the time corstants considered in Chapter 8. There me, of course, two time constants in the second-order case as compared to one in the first-



ordea case. For example, the'nalual ftequencies of the iircuit of Fig. 9.1 are r = - 2, and r = - 8, as displayed in (9.20); the time constants of the two terms are then



266



j



and



Chapter



9



j.



Second-Order Carcuils



Thb unit of natuml ftequency, which is the inverse of that of the time constant, is the reciprocal of seconds. Thal is. it is a dimen\ionless quan ty drrided b) \econds. Therefore, st is dimensionless, as it must be in e"'.



EXERCISES



9.3.1



Given the linear differential equation



u



o



te-', xz : I, and -tj + .rr2 are all solulions. Civen the nonlir€al differential equation show that -rr



9,3.2



- tt#, tt 2t#



=



,dtl-ta=0 rr :



show that



9.3.3



t'? and



r: = I



are both solutions but that -rl



+



-r, is nol a solution.



Civen



d'1t dx .dt(a) -+6-+8,r=0 dt d'1t dt (b\E+6A+9r=t) find the chamcteristic equation and the natural frequencies in each case. Answet \a\..*2, -4; (b) -3, -3



9.4 TYPES OF NATURAT FREQUENCIES Since the natural frequencies of a second-order circuit are lhe roots of a quadratic chamcteristic equation, they may be real. imaginary. or complex numbers. The oa4ao of(9.15), which may be tule ofthe roots is determined by the discriminantai positive (corresponding to real, distinct roots), negative (complex roots), or zero (real. e4ual roots).



-



EXAMPLE



9.2



Consider the circuit of Fig. 9.3, where the response to be found is the voltage o. Fdr variety, rather than writing two loop equations or two nodal equations, let us mix the two. The nodal equation at node t? is



4 and



the



- + i + lda -4dt =



{)



ght mesh equation is



di



Ril--l



rU



Seclion 9"j.



.



Types of Naloral Frequencies



\



267



,l



:) I



|r



Hi



FIGUtE 9-3 S€ond-order ckcuil



(We have thus avoided terms with integrals. The reader is asked to use a strictly nodal analysis in Exercise. 9.4.4.) Substituting for i from the first equation into the second, we have



lt ,,-\l "/ t - al-



,



*



L4\dt ^lJ(ou



t'a' r'l t\a ' - *)l=



Drllerenralng and simplifying the resuh. we



have



r,t, n,



+r"



*4



*,o -



=



nu"



"



*ff



The natural component o, satisnes the homogeneous equalion



4a"



'^ - r,4" '' dt + rn r +r," _ o dt' -,o



from which lhe chamcteristic equation is



.



,r'?



+



(R



+ 1)r + R + 4 =



0



Using the quadratic formula, we'have the natuml frequencies,



,, IfR =



,_-rn.lr+VF'



x-ts



r9.:t)



6 O in (9.21), the natural ftequencies arc real and distinct, gived by



sr.r:



_2,



_5



6.22)



If R = 5 O. lhe natural hequencies are real and equal. given by s,., = FinaUy,



ifR =



1



(9.23)



f,), the natural ftequencies are complex nurntrers, given by



s\2 =



j



-3, -3



!'



-l t j2



(9.24\



1. (In elect ical engineerilg we cannot use i, as the rnathematiciatrs = do, for the imaginary number unit, since this would result in confusion with the current. Complex numbers arc considered in Appendix C for the reader who needs to review the subject.) where



268



chapte.



e



second-order circuits



Distincl Real Roots: Overdamped



Case



If the natural



frequencies sr.2 are real and distinct, the natural response is given by (9.18). This case is cal]'d the oretdampel cdse, because for a real circuit sr and r, are negative so that the response decays, ot is damped o,rt, with time. As an exampie, in the case of (9.22) we have



6 - A,e '1' + A,e-5' Complex Roots: Underdamped Case If



the natural frequencies are complex, then in geieral we have



st?=a ! where (t and



p



iB



are real numbers. By (9.18) the natural response in the geneml case



is



x^



=



Aieb+iB' + A,eG iF'



(9.2sJ



This appears to be a complex quantity and not a suitable answer for a real current or voltage. However, because Ar and Ar are complex numbers, it is rnathematically correct, although somewhat inconvenient, To put the natural response (9.25) in a better form, let us consider l€r'slormuld, given by



''



ed=cosd+jsin0 and its ahernative form, obtained by replacing d by



(9.26)



-d



er'=cosd-jsind



(9.27)



These results are derived in Appendix D. They are named for the great Swiss matbematician konhard Euler (pronounced "oiler"), who lived from 1707 to 1783. Euler's greatness is attested to by the fact that the symbol € for the base of the natural logarithmic system was chosen in his honor. Using (9.26) and (9.27), we may write (9.25) as



xn= eot(he)Bt + Are-jg!)



= e"'l^kos pt + j sin pr) + A?(cos B. j sin Fr)l = e"'[(Ar + At cos pr + (7e1 - 7Az) sin F4 Since Ar and,4, are arbitrary, let us rename the constants as



+ A2= Bl iA, jAz: Bz At



so that



t" = e"t(&



Section



9.4



Types of Natural r.€quencies



cos



Bt + 82 sin Bt)



(e.28)



269



The case of complex roots is called lhe undetdamped case- For a real circuit, a is negative so that the response (9.28) is damped out with time. Because of the sinusoidal terms, however, the damping is accompanied by oscillations. which distinguishes this case from the overdamped case.



'



TXAMPLE



9.3



lnlhecaseof{a.24rwehaveo



- -landB - 2. $lhar | , - c '(Bt cos 21 . B) sin 2/)



where



8r and 8, are, of course, arbitmry.



Real Equal Roots: Critically Damped Case The last type of natural liequencies we may have are those that are real and equal, say



(e.2e)



'



These chamcterize the cfitically damped case, \thich is the dividing line between the overdamped and underdamped cases. In the crirically damped case, (9. 18) is not the geneml solution since both.r,r and are of the form Aet', and thus there is only



t,



one indep€ndent arbitrary constant. For (9.29) characteristic equation must be



(r



-



k)'?:



s2



-



2ks



to be the natural frequencies, the



+



't1



:0



and therelore lhe homogeneous equalion mu(t be



d'x^ -zkT+k'x,=\) - d^^ E



(9.30)



Since we know that,4ee is a solution for A arbitrary, let us try



.



x^



=



h(t)b,,



Srbstituting this expression into (9.30) and simplifying, we have



Ee"=o Therefore, ll(l) must be such that its second derivative is zero for all t. This is true ,i? (t) is a tr)lynominal of degree I, or



.



if



.h(t) = At + Azt



where A, and A, are arbitmry constantJ. The geneml solution case, sr., = k, is thus



x^- (h + A,t)ek



in the repeated-root



.



(9.31)



which may be verilied by direct substitution into the homogeneous equation (9.30). EXAMPLE



9.4



In the case of (9.23) we tlave rr,2



I



p,



270



Chapter



9



Second'Order Circnits



= -3, -3,



:



(A1



+



and thus



tut)e t'



EXERCISES



9.4.1



Flnd the natuml frequencies of a circuil described by



d:xt atA d\ I dt'1 if (a) ar



:



5, do



Answer (a)



9.4.2 Find.r in Ex. d\(O)/dt



-



Find



r if



1:



13; and (c)



j3,(c\ -4, -4



a, = 8,



ao



=



16.



9.4.1 with the arbitrary constants determined so that r(0)



6.



Aniwet (a) 6e-'



9,4.3



= 4; (b) at = 4, a! =



-t, -4; (b) -2



aox=tJ



-



3e a'; (b) e 2(3 cos 3r + 4 sin 3r); (c) (3



-



3 alld



+ 18r)e-i



djt



A'z-25r=O 9.4.4



x:



Ar cos 5t + A, sin 5/ If the node voltage at node Z' in Fig. 9.3. is or, show that the two nodal equaiions



Answer are



LDr *, u-Dr



4



R



lda



iA=u



and



D,-D t' ur dt t-i(o) =o i. --" Differentiate the second of these and substitute into the rcsult the \alue of the first equation to obtain the describing equation for Fig. 9.3.



,r



from



9.5 THE FORCED RESPONSE The forced response rr of the general second-otder circuit must satisfy (9.10) and contain no arbitlaty constants. There arc a number of methods for finding .rr, but for our puryoses we shall use the proced[e of guessing the solution, which has worked so well for us in the past. We know fiom out experience with first-order ctcuits that the forced response has the form of rhe drivitg function. A constatt source results io a constant fotced tesponse, and so on. However, the rcspons€ must satisfy (9.10) identically, which means that fust and secotrd derilatives of rr., as well as r, itself, will appear in the left member of (9.10). Thus we are led to try as rr. a combination of the right rnember of (9.10) and its derilatives. EXAMPLE



9,5 kt



us consider the case



o, =



16



V in Fig. 9.1. Then by (9.4), for i? =



: dP+l0nt+ l6t Section



9.s



The forced Response



32



r,



we lrave



(9.32)



271



The natuml response was given earlier in (9.20) by



r,:



A)e 2' +



A*-e



Since the right member of (9.32) is a constant and (namely zero), let us try



(9.33)



all its derivatives are



consranr



/ is a constant to be determined. We note that A is not arbifary but is a particular value that hopefully rdakes -!. a solution of (9.32). Substituting ,r. inro (9.32) where



yields 16A



:



32



of



xr=A=2 Therefore, the genetal solution of (9.32) is



x\t)



=



e'r'



t A,e s'-



2



A knowledge of the initial energy stored in the inductors can now be used to evaluate Ar and Ar. In the case of constant forcing functions we tuay often obtain -q from the circuit itself. In the example just considered, .!. is the steady-state value of i in Fig. 9.1 when os = 16 V. At steady state the inductors are short circuits, as shown in Fig. 9.4, so that, from the figure, we have



go



f""l ) FICURI



EXAMPTE



9.6



9.4 Circuir of



Fig. 9.1 in the sready stare



Suppose that in Fig. 9.1 we have



1),=20cos4,V Then by (9.4), again for t, =



i,



we have



*+10*+16,r=,+0cos4r



(e.34)



The natuml response.r, is given by (9.33), as before. To find the forced response.v .J,le need to seek a solution which contains all lhe terms, and their possible derivatives, in the .ight member of (9.34). The coefficients of these terms will then be de-



272



Chapter 9



termined by requiing r/to satisfy the differential equation. In the case under considemtion, the only term is a cos 4t term and the al.



=



.q



A cos 4r +



t I sin 4r



(9.35)



contains this term;nd all its possible deri\atives (which are cos 4t and sin 4t times constants).



From (9.35.) we have



dt,



I 48co\4t



4As''1.4t



id, t, 7/ -



-16A



4t



cos



-



168 sin 4r



Substituting these values and (9.35) into (9.34) and collecting terms, we have 408 cos



-



4t



,10A sin



4r = 40 cos 4r



Si-Dce this must be an identity, the coeffrcients of like terms must be the same on both sides of the e4uation. ln the case of the cos 4t terms we have



408



-



40



and for the sin 4t terms we have



.



Thus



40A=0 A=



0and8-



l.sothar xt



=



sjJt



4t



(9.36)



The general solution of (9.34), from (9.33) and (9.36), is given by



i,=



7=



tr" z''A,e 3'+ sin 4t



(q.31)



This may be readily verified by direct substitution. Some of the more comnon lorcin8 functions/rttwhich occur in (g.7) are listed in the fusr column of Thble 9. 1. The general form of the coFespoDding forced response is given in the second column, which may be useful for formulating the rial solution



\.



TABtt 9.1 lrial Forced



Responses



(0



KA I t1 sin



,r,



AI+R Atz+Bt+C cos



,t



e"' sia bt, e'' .os



Section



9-5



The Forced Response



bt



,,{ sin



"-(A



r, + , cos rt sin rt + a cos ,,



273



EXERCISES



9.5.1



Find the forced response



if



d".*ttu+i'=rrrt 'dt '^



a'



where f(t) is giver by (a) 6, (b) 8e Answer (a) 2; (b) -Be';(c)2t + 2



',



and (c)



6t +



14.



9.5,2 If r(0) : q and dt (O) / dt : 2, find the complete solution in Exercise 9.5.1. 3' Answer (a) 2e' + 2; (b) 9e-' - 8e-'z' + 3e-'' : (.) e-t + e + 2t + 2 9.6 EXCITATION AT A NATURAL FREQUENCY Suppose that the



circuit equation to be solved is given by



d'x . tu fi-h-b\A+abx= where



f(t\



a and b + d are known constants- In this



(9 38)



case the chamcteristic equatioD is



s'?-(a+b)s+ab=O from which tle natural frequencies are Jr =



a,



S..



=



b



Therefore, we have the natuml response



n:



Are''



i Azeb



(9.39)



where Ar and A, are aibitrary. us suppose now that the excitation functioo contaiN a natuml frequency,



Irt



say



f \fi =



e'



(9.40)



The usual procedure is to seek a forced response,



x!= Ae' and determine A so that



4



satisfies (9.38), which in this case is



d'zt ta t blV dx + abt ' e' * However, substituting



x,.



(9.41)



(9.42)



into (9.42) yields



0=e^ which is an impolsible situation. This difficulty could bave been fores€er by observing that rr" in (9.41) bas the form of one of the components of .t" in (9.39), and thus .t will satisfy the homoge-



274



chaprer



e



second-order ctrcuirr



neous equatioi coiraspording to (9.38). That is, ,/ substituted into (9.38) makes its l€ft member identically zefo. There is no poidt theo in fiying such a forcad response as (9.41). i



in



r".



I.€t us corsider what happeDs if we multiply by t the pafl of rl that is duplicated That is, let us try



xt



= Ate'



instead of (9.41). We lhen have



I



'==AGr - t\e. dt dx,



"+-Az't+2n)e^ dr' Substitlting these 1dues, along with (9.43), into (9.42), we have



Ae-la2t + 24



-



(a



+ b\{at + l) + abtf =



ed



whicb, upon simplification, b€comes



A(a-b)e^=e^ since this must be atr idenrily for all r. we musl have 1



a-



h



The general solution of (9.38), using (9.39), and (9.43), is then



.



\- Ated+Are"+



te,



Suppos€ that the excitation in Fig, 9,1 is given by



EXAMPTE 9.7



ozi Ther



ifiz



I



6e-" +



32



.{, we bave, by (9.4),



ilzx -dx dt' dt -+tU-+



l6x=12e-'+g



(e.44)



The natuml response, as before, is



&=



Are'z + Aze-3'



Noting that the right member of the differcntial equation has the telm e-z tlr commo! rtith .rr, we try



tr=A?-2'+8 Tbe hctor r has been inserled inlo the naurml trial solutioo of x, to remove rhe duplication of the term e-'. Substituting r/ into (9.44) and simpliffng, ve have



S€crion



9.5



Exciration at a Natu€l Fr€quency



I



i



,Ae-a+168:12e-'+& .



I



I



275



I



I



I



I



Thereforc, we have



A = 2 and B



:



4, so thar



xf= 2te-! +



4



The general solution is now



tz:t-:aa+t Finally, let us considet the case of (9.38) whpte b = a and/(r) is given by (9-40). That is, both natural fiequencies and the ftequency of excitarion are all th; same. In this case we h?ive tl'x



df



2tt: + a'zx:



-



ed



(e.4s)



The characte stic equdtion is



s2-2as+a?=O and thus the naruml fiequencies are



Jr=J2=d The natuml response is rhen



x"=(At+A2t)e.l



'



We know it is ftuitless to ffy as the forced response .xr given in (9.41) becaus€ it is duplicated in the traoml response. In this case, (9.43) will Dot work eitber because it, too, is duplicated. The lowest E)wer of / that is nor duplicated is 2; thus we are led to tly



tt:



Atzed



Substituting this exFession into (9.45) we have



so thal



A=



:.



The forced and complele responrs follow as before.



A geneml rule of thumb is that if a natural fiequency term of ,r, is duplicated in xr, tbe term in .V is multiplied by the lowest power of t required to remove the dupl! cation.



EXTRCISES



9.6,1



Find the forced respoDse



if dzx dx -.: | 4=.r jx - IQt d( dt



I



wheref(/) is given by (a) 2e 1' + 6e-4t (b) 1e ' + 2e-', ^nd 2e 11.- ft-3t' (b\ t(2ea - e-3t)



I



Answer (a) I



|



I I



I



276



Chapter



9



Second'Oider Circuirs



9.6.2



Find the forced response



if



dlx dx -A'z+4A+4t= lut where



9.6.3



f



(,.)



NSivenby (a)6e



,tnswer la) 3t')e-t; lbt rte 'z' Find the complete lespoDse if



'z' and



(b\ 6te-2' [Saa8?rtion'In (b), try



d)t



.;+9i -



aod



x(o)



:



Answer sin



dx(o)ldt



3l



-



:



o.



3t cos 3t



18



sin 3t



'ky xt = r(A



lsusSestion:



4- At'e '']



cos



3t + B sin 3t l



-



9.7 THE COMPTETE RESPONSE In the prcvious sections we have noted that the complete response of a circuit is the



of a natural and a forced lesponse and that the naturdl response' and thus the complete response, contains albitrary constants Th€se coffitants' as in Jhe firstChapter 8, are aletermined so that the complete lesponse satisfies olde; cases specified initial eneryy-stomge conditions' sum



;f



EXAMPLE



9.8



l,et us find t (/), for



t > 0, which



satisfies the system of equations



fr+x+sl"''a,=rc"" x(o) : 2



(e.45)



the htegral; this To begin, Iet us differentiate the fust of these equations to eliminate results in



d'1x ^dt dt' dt



5r



48e r'



=



The ;hamcteristic equation is



J"+2s+5=0 with roots



sr'z=



-l:



j2



Therefore, lhe natuEl response is x^



= e '(At



cos



2l + A,



sin 2t) i



Tiying as the forced resPonse



'J Section



9.7



The Complete Response



= Ae-" l



277



I



l



I



we see that



t = _48e-tt



EAe so that



A = -6.



The complete response is therefore



= c-'tAt



^ul



t



2t



ccrs



A,



2tl



sin



- 6" "



\9.47 )



To determine the arbitrary constants we D€€d two initial conditions. One, r(0) - 2. is grven in 19.46). To obtain the other we may elaluare the first equation of (9.46) at t = 0, resulti$g in



dx(01 b(ol -t- -



Noting the value of



r(0)



f ' 5 Joxdt = t6



aDd that the integral term is z€ro, we have



dr



(O)



,i, = 12



(9.48)



Applying the second equation of (9.46) to (9.47), we have



or



Ar = 8. To dx



;



'(0)



=



Ar 6=2



apply (9.48) we may differentiate (9.4?), obtainiog



- e-'(-2At



sin



2r + 24? cos 2r)



-



e-'(A,



+



cns 2r



A2 sin



2r)



+ l8e !



from which da (o\



;



=



2A2



-



A:



(9.49\



+ t8 = t2



From this, knowing A!, we find A, = l. At this point let us digess for a moment to note a very easy rDay to get (9.49). We may differeDtiate r(r) and immediately replace r by 0 beforc we write down the result. That is, in (9.47), the derivative of Aise-'eft = 0 (which is l)



x t:



times the dedvative of (Ar cos 2r + A2 sin 2t) at r - 0 (which is 2Ar) plus (Ar cos 2, + Az S'll. 2r) t = 0 (rrhich isA') times the derivative of e I att = 0 (which is -1) plus the de vative of -6e n at t : 0 (which is l8). Th€se steps are written down in (9.49) and can be done mentally, avoiditrg the iotermediate prior step.



Returring to oru problem, we now have the arbitrary corFtaots, so that by (9.47), the final ans,wer is



r: EXAMPLE



9,9



+



2t)



-



6e



L€t us find o, r > 0, in the circuit of Fig. 9.5



if



o,(0)



€-'(8 cos 2r



sin



5 cos 2{Xnr V. The nodal equatioD at node or is



I .



2/10)q



I



278 I I



I



chaprcr



e



5econd-o,dcr cin u,l'



l0-ror



'' = o(0)



r ,-odo, -0 -1x l0'u+ l0-'-



:0



ard o! =



flCUlE 9.5



I



I



"' of



L



4ot



D



+



2



x



l0



Example



'ff -



zr, = !o



(e.50).



cos 2000t



and the nodal equation at the inverting input of the op amp is



x to-r,, I)8dt



'1, ro"*=o



or



| .^ ,da ot=__xlo,dt



(e.51)



Substituting (9.51) into (9.50) and simplirying, we have



d2r -du 4tlt' + 2 x l0ra



+2



\



i06o



= - 2 r l0'cos2000r



Ttle characlerislic equatlon is



s'?+2xlCPr+2x106-0 sj., - l00O( - I t jl).



so that rhe natural frequencies are tlrerefore



.



'u,



: e-'*(,4'



"o"



10001



+ A, sin



The narural response is



1000,



For the forced rcsponse we shall try



ot= A cos 2fffit +



I



sin 20001



which subctituted into the differential equation yields



(-2A + 48) co6 2000t + (-4A



-



28) sin 2000,



= -20



cos 2000t



Therefore, equating coeffiiients of like terms, we have



-2A+48 = -m -4A - 28.: O fiom which A



o = e-'o'n(A, section



9.7



=2a\d,B : -4. The complete response is then cos lmor + A, sin 1000r) + 2 cos 2m0t - 4 sin 2000t



The Cohplere R€sponse



(9.52)



279



l



/ - 0-.



From (9.51). for



we see that



Dr(o')--]'lo'rd"(o-) 4dt and since or(0+)



= or(0-) =



0 we have



dulo').o dtLike 01, o is also



a capacitor voltage (acrcss the



From (9.52) and (9.54) we have



At = -2,



*-pF capacitor),



)=0



1,(0+)=t,(0



ot



rq5lr so that



(9 54)



At+2=0



and lrom (9.52) and (9.53) we have



l0O0A,



l000Ar



-



8000



=



0



fiom which A, = 6. Thus the complete response is o=a



'*(



2cos 1000r + 6sin 1000r) + 2 cos 2000r,- 4 sin 2000r V



r, t >



0. where



EXERCISES



9.7.1



Find



V'tx-+lxa-1ut



. and



9.7.2



dtI



r(0)



(a)/(t) = I



and (b)



/(t) = 21.



,+rcwo (a) (2 3t)e '1'; (b\ 3(t - t\e-" Find i, a, di/dt, and dx/dt at t -O*. An:rer 0.0,2 A/s.40 V/s



=2 +



t



a



I



txtRctst



9.7.3



I



9,7.2



For / > 0 in Exercise 9.7.2, frad (a) o and (b) t. (Sugeestion: Be.ause of Kirchhoff's laws ard the terminal relationships of the elements, i has the same nahral ftequencies as o. Thus o, is easily obtained afte! i" is found: its forc€d respon!€ is evident by inspection of the circuit.) Answer (a) e-2'( 6 cos 4, + 7 sin 4t) + 6 V; (b\ e-'z'(-2 cos 4t I sin 4r) + 2 A



-



280



chapter



g



second-o.der circuits



'



9.8 THT PARALLEL NIC CIRCUIT One of the most important second-order circuits is the parallel RLC circuit of Fig 9.6(a). We shall assume that at r = 0 there is an initial irductor current,



:



/"



(9.5s)



(0)



:vo



(9.s6)



and analyze the circuit by finding o for r



>0.



i(0) and



a[



initial capacitor voltage, 1r



FIGURE



9.6



(a) Parallel RIC circuit,



with



(b) the source killed



The single nodal equation that is nec€ssary is given by



r: tf' t/tt+ Io-('dt=i, dD



r + tJ.t



{9.57)



which is an inlegrodifferential equarion that becomes. upon differentiation.



d't 'dtt



ldD I di" R&'L"dt ght To find the natural response we make the member d'1a lb ( dt,+ RA+LD={J



zero, resulting in



1



(e.



This result follows also from killing the cuffelt source, as in Fig. 9.6(b), and irg the nodal equation. From (9.58) the charactedstic equation is



s8)



wrir



c"'+]"+j=o Section



9.8



The Parallel RrC Cir.uil



281



from which the natuml fiequencies are



/'\2RCl \' LC'



rsse)



As in the geneEl second-order case already discussed, there are three types of responses, depending on tbe nature of the discriminant, l/Rz 4C/L, in (9.59\. We shall now look briefly at these thfee cases. For simplicity we *ill take i! = 0 and considei th; source-ftee case of Fig. 9:6(b). The forced response is theo zero and the natuml response is the complete response,



-



Overdamped Case If the discriminant



is positive, that is,



t4c -



->{l



or, equildently,



L>



(9.60)



4RZC



then the natumi frequencies of (9-59) are real and distinct nCgative numbers, and we have the overdamped case,



x=AGrt+A*t2' at, = 0*, %+Rlo



From the initial conditions and (9.5?) evaluated



d0(0.)



dt



_



whrch together with {9.561 can be used



TXAMPLT 9.10



we obtain



(9.62)



RC Lo



determine the atbitmry constants.



O,,



Suppose thar R = 1 (9.59) we have Jr., =



(9.61)



= : H, C = i F, -1, -3, and hence



Vo



= 2 V,



and



10



= -3 A.



Then by



a-Ar'''+Azet Also, by (9.56) and (9.62) we have



o(0) = du(9*) dt which may be used to obtain Ar I



I



I I



t



= c v/.



: -3, and thus D.5e-'-3c''



=



5 and



A,



This overdamped case is easily sketched, as shown by the solid line of Fig. 9.7, by sketching the two components and aCding them graphicaly.



I I



2v



282



Chaprer



q



Second-Order



Crcuil'



flCURt 9,7 Sketch of an overdamped



respons€



The reason for the term overd.omryd fiay be s?*n ftom the absence of oscillations (fluctuations in sign). The element values are such as to "dadp ouf'any oscillatory tendencies. [t is, of course, possible for the response to change sign ortc?, depending on the initial conditions



Underdamped Case



If



the discrimimnt in (9.59) is negative, that is,



L
0if (a)C: jFtudO)C=;F. Answer (a\ -25e-, + e-' + 24 Vt O) U - (24 + 36t)e"'



9.9.4



V



6cl



c IH cxtRctst



9-9_4



---9lL



ALTERNATIVE METHODS fOR OETAINING



THE DESCRIBINC EQUATIONS Iir this section we consider two methods of expediting the process of obtaioing the describing equatio! of the circuit. In the cas€ of the para{etind series RIC cirJuits, a single equatiotr is requtued, which afte! difrerenriation, is the describing equatit n.



However, in many second-order circuifs there are two simultaneous circuii equa_ tions ftom which the describing e{uation is obtained after a tedious eliminarlion process.



EXAMPLE 9.1



5



I.et us consider the circuit of Fig. 9. 12 foi , > 0. Taking node qridDg nodal equatiors at qodes a and or we have



o-a' 4



--*J"



o-ot ldt) 6 4dt ur dr+i(0)=0



,



as reference and



(e. E4)



If



we arc inter€-stedin finding o we must elimitrate or aad obfain the describing equa_ tion itr terms of o. The result, as the reader may verifu, is



*4*r*+rca:*+61)"



(e.85)



In this-case the process is not ovedy complicated but it can be shortercd by the melhods we shall consider in this chapter.



290



Chapte.9



Second-Ord€rCircuits



r I I



FICURE



9.12 Cjrcuit



wath two storage ele:r€nts



The first method we shall discuss is a systernatic way of obraining r describing equation, such as (9.85), from the circuit equations, stlch as (9.84). To devrlop the method, Ier us filst introduce the diffefenfltrltio[ operatt D, which is defined by



=!dt



o That is,



Dr



:



:



dx/dt, D(Dx\



Dzt = d2x/dt2,



and so on. Also, we have, for ex-



ample,



dx ai+bx-



aDx*bx=GD t



b)x



It



is importani here to note that r is hctored out of the middle member and placed the opemtor exFession, indicating thal the operation is to be pedormed on r. Otherwise the meaning is changed radically. With these ideas in mitrd, let us rewrite (9.84) io op€raror form. This resulr, after first differenlialing the second equation. is dJ&e/



/t



s\ I la, * ,2r, - 6u,= iu, I



t



/t



\



-loa+llo+llo,=o b\b/ which when cleared of fi"actions becomes



(3D+5)o-2ar:3x, -Da + (D + 6)0, :0.



(e.86)



To eliminate or, let us "opemte" on the tust equation with D + 6, by which we mean multiply it through by D + 6 on the lefi of each term. Then let us multiply the rcond e4uation by 2, resulting in



(D + 6X3D + 5)t,



-



2(D + 6)q = 3(D + 6)os



-2Do+ 2(D +



6)o1



:0



[We notc. that colstants such as 2 commute with op€rators, i.e., (D + 6)U = 2(D + 6)r, but riabl€s do not.l Adding tbese last two equalions elimirates or and rcsults in



(D + Section



9.10



6X3D



+ s\



-



2Dlo = 3(D +



6)oa



Alt€rnative Methods for Obtaining the Describing fquataons



(9.87)



291



Multiplying the operators as if they were polynomials, collecting lerms, and dividing out the common factor 3, we have



(Dt



+'lD +



l0)o = (D + 6)os



which is the same as (9.85). The procedure may be carried out in a more direct manner by using determinants. For example, we may use Cramer's rule to obtain the expression for o from (9.86), given by



,:A,A



(e.88)



where A is the coefficient determinant



l;n



A= I



+-s )



| -D



I I



D+61



(e.8et



and Ar is given by



a'



lr,- -z :3{a + o)u' :lo" o+61 I



(e.eo)



We note that in this last expression we must be careful to write os a/tel the opemtor.



Writing (9.88)



as



{o=4, we see ftom (9.89) and (9.90) that we have the describinS equation (9.8?). The second method we consider is a mixture of the loop and nodal methods in which we seiect the inductor cureDts and the capacitor voltages as the unknowns, rather than the loop curreDts or the node voltages. We then write KVL around loops which conrain only a single inductor and KCL at nodes, or geDemlized nodes, to which only a single capacitor is cormected. In this .Ilanner each equation cont4irs only one derivative, that of an inductor current or a capacitor voltage, and no inte-. grals. The equations are then relalively easy to manipulate to find the describing equation. EXAMPLE 9.16



Usrng Flg. 9.12 again. let i. lhe inductor currenl. and r. the capacitor vohage. be the unknowDs. (These unknowns are sometimes called the rtdle variables of lhe cin. cuit-) Then for t > 0 the nodal e4uation at node a is



r+i+__=0lda 4 4dt



(e.e1)



and the loop equation around the dght mesh is



o=6i+4 It



(e.e2)



is a relatively simple rnatter to solve for i in (9.91), substitute its value into (9.92), and simplify the result to obtain (9.85). The reader may note that we have applied



292



this method, without saying so, in obtaining the describing equation of the circuit Fig. 9.3 in Sec. 9.4.



of



Some advantages of this method are that no integmls appear (thus no second derivatives occur as a result of differentiation), one unknow! is easily found in terms of the others, and the initial conditions on the first derivatives are easily obtained for use in determining the arbitrary constants in the general solution. For example, it may be se€n from Fig. 9.12 that i(0) = 1 A and D(0) = 6 V. Thus from (9.91) and



lhe ralue ol r'rl0'). we have



do (0. )



dt



:



0,(0-)



-



l0



This last method can be greatly facilitated, particularly in the case of complex circuits, by using graph theory- Since we are looking for inductor currents and capacitor voltages (the state variables), we put the inductors in the links, whose currents constitute an independent set, and the capacitors in the tree, whose bmnch voltages constitute an independent set, as we recall ftom Chapter 6. (The fiee should also contain voltage sources, the links should contain current sources, and so on, if possible.) Each inductor is then a link with current i, which forms a loop whose only other elements are tree bmnches. Therefore, Kvl.around this loop will contain only one derivative term, L (di/dt), and no integrals. This loop can easily be found since it is the only loop in the graph if the only link added to the tree is L- For example, the gaph of Fig. 9.12 is shown in Fig. 9.13 with tree branches shown as solid lines and Iinks as dashed lines. The loop containing the l-H inductor is a, ar, b, a through the branch labeled o, ard KVL around it is (9.92).



,



flCURt 9.I3 Craph of Fig. 9.12 Each capacitor is a tree branch whose current, together with



link currents,



constitutes a set of currents flowing out of a node or a generalized node, because if the capacitor is cut ftom the circuit, the ffee is separated into two parts connected only by lirlks. In the eMmple of Fig- 9- 13 the thrce curents, shown crcssing the dashed line ttuough the capacitor, labeled o, and two linl 0, in Exercise 9.9.4 by (a) the 6rst mpthod of this section using.nodal equations and (b) the second method of this section. Answer C(d'za/dt'?) + 6c (da/dr) + a:24 Solve Exercise 9.1.1 using (a) the tust method of this secrion applied to lhe me6h e4uations and (b) the second method of this section. Solve Exercise 9.7.3(a) using the second method of this secrion.



1



SPICE FOR TRANSIENT RESPONSES



OF HIGHER-ORDER CIRCUITS



'



SPICE, or any similar compiter program, be€omes vety useful in solving circuits as the order of the describing differential gquatiixs increases. The appliaation of SPICE to higher-order circuits involves essentially the same procedure as thai of the fiNt-order networks of Chapter 8.



EXAMPLE



9.17



i and o in Fig. 9.6(b) for R = 2N A, L = 10 mH, C = I pF, = lV, andi(0):0Aintheintenral0 < r < 1ms. A circuit file for a plot of



Consider finding



o(0)



fICURE 9.14 Transient response tor circuit of Fig. 9.6(bl IEOAND:



+: I([) TUIE (.)---------(+)----------



V{ r )



-1.00008+00



-5.0000t-01 o.o000a+00 5,0o0oE-ol



0.0008+00



l,0o0E+00:_----



L0008-04



2. ?3r E-0r



6.0008-04 6.5008-04 ?.000E-0!t ?.500E-0a



,.r6tB-01



5.0008-05 6.750!-01



1.500!-04 -9.1538-02. ?.0008-04 -3.602t-01 . ,.5008-04 -4.9r6E-01 . 3.000!-0t -4.392E-01 3.5001-0,1 -3.3168-01 4.000I-0.t -2.1498-01 . 4.500r-0,r -3,3658-02 . 5.000!-04 1,0?38-01 . 5.6008-04 1.932E-01 ,



3.000E-04



1.9958-01 1,355E-0r



. . -



5.639!-02. -1.?s,E-02.



8.t00E-ort -7.149!-02 . 9.0008-04 -9.3108-0?. 9.500!-0.1 -9.?838-0t . 1.000E-03 -?.€30t-02 -



294



1.0000!+00



-5.00008-01 o.0oo0a+00 5.OOO03-03 1,ooo0!-02 t.5000!-O!



Chapler



9



Second-Orde. Circuits



-:------.--



-;



ianduis PABALLEL RLC CIBCT



*



IT



OF



FIG 9.6{8)



lunderalahped case!



DATA STA'TnI'NTS



RI O



2o00HM



L 1o loiiH c1o 1uF * sET V(1) : I V and r(C) =0 ATT=0 .IC V (1) t .IBAN .



O.



OsMS



USrNC . rC STAIEI,ENT



1lls UtC



V(1) I(L)



PLC'T TRAN



. END



In this program the .IC (itritial cotrdition) comrnand sets node voltage V (l) = o, the capacitor voltage, to atr initial value of I V. The inductor current is zero sfuce i! is not specifred by an IC s{ateoent in L. The Pspice solurion is shown ir Fig. 9.14-



EXAMPLE



9.18



t€t us frndr,foro < r < 15 rns in the thtd-oder circuit of Fig. g.ts(a). hior to time t 0, os = 10 V and the s?itch is closed. making = 0, which requires that the CCCS have 7.ro cuneot (an open circuit). The circuit is redrawn Fig. 9.15(b) illustrating these cooditions. A circuit file for finding the initial ralues



:



oa



(0 )



and



r.(0-)



INIIIAL



*



i



,



h



is



CONDITIONSi FOR CIRCT'IT OF.



FIG.



9. 15 (b]



DATA STATBm.IS



vcrooocro R1 10 1lK c1 201UF L 1 4 0.1H



R3401K



*



sollrTtoN coNltoL



.DC CC



r0 10



* oulPut .I,ElNa ,



sT^lqnl|r



1



coNIBoL sTAtstEtr



DC



V(C1) r (L)



F,ID



The resulting initial Yalues are



vc



1.000E+1



v(c1)



I (L)



5- OOOE+OO



We can now write a circuit fle for t > 0 for Fig. 9.15(c) usirg thes€ lblues to find y(4) iq the desned htelval. Note that a dummy voltage souce or has be€n ino" serted to allow ,; for the CCCS between nodes 0 and 3.



:



Section



9.11



SPICE for T.ansient



Rerponss of Hither-Order Circuits



295



@ (b)



FIGURE



296



Chapter



g



9.15



(a) Third-order



Second-Order Cn(uitr



circult, (b) redrawn



at



I = 0-; {c) redrawn for t >



O



TI],ANSIEIYT RESPoNSE FoR CIRCUIT



l DATA STATEIENTS Rt0l1K R2t22K .l 2 o rlt- lc-sv \,Dt20mo C2 20



3 2UF



' .



9. 15 (c)



1C=O



L3401HlC=5M R4OlK Frx03v.t0 * SOLUIION CONIROL .



FIc.



oF



STAT'EMENA



TRAN 1MS TSMS UIC OUIPIJT CONTROL STATEMEM



PLoT iRAN v(R3) END



The plot of o, for rhis program is shown in Fig. 9.16.



Tl[lI (.)----------



V(R3)



-2.00008+oo 0.00008+00 2.0000E+00 .l.o00oE+o0



6,00ooE+0u



1,0008-01 1.035!+00 2.OOOA-03 ?.1r1!-Or



3.0001-03 4.493t-01



it.000E-0, 2.355x-01 5.000t-03 r.3?38-01



.



3,0008-03 4.932!-0t. ?.0008-03 -1.119!-02 3.0001-03 -5.2€0!-02



s.0oot-03 -?.934t-02



"



1.0001-01 -9.?,178-02 1.100t-02 -1.0038-01



t.t00E-0, -t.l43E-01 1.3008-02 -1.1?08-01 1.400!-0t -1.1?5!-01 1,5008-02 -1.1641-0r



FICURI 9.16 Transienr response for cjrcuir of Fig. 9.1s(a)



EXERCISES



9.11.1



Use SPIC!: to_plot_ ; in Fig. 9.6(a) for 0 < , < 500 ps wirh ,; = 0. I ! (r) A, L = l0 mH, C = I pF, i(0) = -t0 InA, o(0) =0.and(a)R=25O,(b)R:



9.11.2



Use SPICE to plot



R



50 O. and (c)



by l0[r(r)



9.11.3



-



!,(r



75 O.



i for



O




0 if the circuit is in sterdy state a{t=0-, Find i for I > 0 if l(0) = 4 A and o(0) =



Insert a 1-O resistor in se.ies rvith o, in Fig9.2. thereby making the souce a praclical rather than an ideal one. Show lhat in this crse



Find i for



02 satisfies the second-order €quatiotr,



8V.



sff* nff,u=+ff,u Find



3A. tt



i



for



t >0 if



4H



t,(0) = 9 A and i(0)



It



:



i.Ai'



2\1 PROBTTM 9.4



[*l*f!"



9,5 9.6 9.1



PROBLTM 9.2



49



2H



PROBLEM



294



i



Chapler



9



Second-Od€r Circuits



9.3



Find o in Prob. 9.4 Find



,



for



'Fitrd, for



r>



0



if t(0)



i > 0 if o,(0) : 2H



is changed to



ifthe circoil



I A.



is in steady state



az(O)



= r2v.



PROAEM 9.6



PROBTEM 9.7



9.t



The aircuit is in steady stat€ at and i for > 0.



9.9



Find o for



t>



0



ilthe circuit



in



steady state



9.10



Find i for



t>



0



if the circuit is in



steady state



,



att=0-.



att=0-. 9.11 Find i for a > 0 lf the circuit . att:0-.



t = 0-.



is



Fitrd o



9.f2



r = 0 . Find is (a) 8 H, O) 6 H, and



The ciicuit is in steady state at



o for r



>0 if I



(c) 4.8 H.



9.13



Find i for



t>



0



if the circuit is in



9.14



Find o for



,>



0



if the circuit



is in steady state



att=0_. att=O



steady state



is in st€ady state



PROBTIM 9.8



l



'ro3l.tM



L



Chapter9



9.9



299



Problems



I



l



I



'I I



t-



0.0r F



PROAI€M 9.10



PROBTEM 9.11



PROBttM 9.r2



40|



t=b



t5V



lo



)tlt-----Y-



l2 sI



4H PROBLIM 9.I3



PROBTEM 9.14



9.15 Findifor, >0if ,(0) = 2 V, i(0) =1A, and(a) I : 1 HandR = I O, (b)L = I H andR : 3 O, and(c)a = 2Handi = sO. 9.ld Find i for . > 0 if tr(o) = 3 A, t,(0) = -1 A, and'(a) o" - 15 V, (b) o" : 10€-'' V, and(c)o,=5e'V



9.1? 300



t>



Find I for



atr=0



O



steadv stare



_



9



Second-Order Circ!its



Find t



for, > 0if



the circuit is in steady stale



*hen the switeh is oponed 9.1 9.20



g.2t



if the circuit is in



Chapter



9.18



at



,=



0.-



t > 0if os:12"(itV Find o for t > 0 if the circuit is in stpdy slate att=0. Find o, and 1,, for , > 0 if the circuit is in Find t for



steady state at



, = 0-



PROBTEM 9.16



PROSLEM 9,17



Ir PnoBLtM 9.18



301



20v



PROSIIM 9-20



PROaIM 9.21



state



9.25



en-



9-26



9,22



Fird i, fot



9.23



Find i, t >'0, if tbere is no initial6tored cryY and {a) R = 2 n, P = 2i 6\ R = 2



r>



O



ifthe ctrcuit is in sleady



p = l;and (c)i = | a, p = 2, 9.24 Fi.d i, r > 0, if there is no initial



.



elgy and (a)



c=+F.



c = + F, (b) c = *



stored



fi,



en-



9.11



F, 8nd (c)



FR()BTIM 923



9



Second-Order Circuits



critically



(



I



Chapler



'Find lhe maximum value of fte



damp€d rcsFtrse i of Prob. 9.25 and the al which it occurs if is = l0& r) A. ,



PROB|tM 9.22



302



Find i for r > 0 if is : 10 A and the circuit,is in steedy state at a - 0-.



Find



6V



i



for r > 0



if



i(0)



: 2 A and



ti;€



'(0) =



PROBLTM 9.24



PROBLTM 9.27



PROBLTM 9.25



9.2t



Fbd o for



',



t



= 2,''n(t)



>



O



t.



if



(a)



-



t,



2zO) A. atrd lb)



9.30



Find o for



t >0.



PROSLIM 9.28



9.4



titrd t,



t>



U,



tp(U) =



O



v aldr((,) = z A.



12



ul.t) Y



PROBTIM 9.30



--*lir257 i



9.31 2H



_-Jv\r-:rtrf\6fl ;",



I > O. (b) Replace th€ c!rrent and voltage sourc€s by 2 eoo 2t A alrd 6 co6 2t V, r€spec'tively, with the same polariti;s, and fnd o for i > 0 if therc is no initial (a) Find o for



stored energy,



9.32



at



PROStC,ll 9.29



Chapt€r9



Problems



t > O, if lhe circuit is in $eady state : 0 . (Not€ that this is the circuit of Fig.



Find o,



r



9.12,)



303



I PROBTEM 9.31



9.33



Find o for 3A_



=1*.3



q



6f}



J,..



li.



rHa PNOBLEM 9.32



t >0 if o(0) =4 V and t(0):



9.34 (a) Find o for t > 0, il ol0) = 0 and o5l(t) = 2 y. (b) Repeat part (a) if th€ 4-V soulce is replaced by on€ of 26 co6 2r V (c) R€p€at part (a)



In



one of



9,35 i\



r-0



if the 4-V



source is replacld



by



2" ' v.



Find o for



r > 0 if



therc is no initial s1oled



energy.



,/l6c.s8rv



lo



tiF



9.36 Ffud the 'value, m mnge of lalu€s of p : I + R > 1 so that the circuit is (a) ove!O) underdarnped, and (c) criticatly damped. (,Vor?. The ouFut is i) and p < 3 to damped.



PROE|-CM 9.33



insue that the natural reEloBe. than grows with time.)



PROELEM 9.34



PROBLTM 9.35



304



Chapter



9



Second-Oder Cncuits



decays



ntber



.



PROBLEM 9.36



PROBLEM 9.37



9.31



Find o for



/>



eneryY and 1)j



9.38



9,39



:



PROBLEM 9.38



0 if there is no initial srored



has ihe natural fiequencies



5 V.



s-



Find o, t > 0, if (a) ,,(0) = 4 V. o(0) : 0: (b) o, (0) = 0, o(0) : 2 v; and (c) o1(0): 4 V, ,(0) : 2 V. (Note that thb rcsponse js an unforced sinusoidal response. Such a circuit is .alled a hamonic oscillatot .)



x,= A\e'+



13+6r,+llr+6=o chaprer



9



Problems



A1e



'1'



+



he



1'



Show also that the forced response is



equations- There are more natural ftequencies and thus more terms in the flatural rcsponse, For example, if



show that the chamcterisric equation



3



as its roots. Thus show that the natural re-



Higher-order differential equations may be solved in the same manner as second-order



d1t* r|'1r dx a, 6E * 1l dt+ 6x =



1, 2,



,=2 and that the geneml solution is



12



tt:t4+tJ 9.,m



Using the results of Prob. 9.39, find if there is no initial stored energy.



l,



t>



0,



305



fin



iir PROALEM 9.40



COMPUTER APPLICATION. PROBTEMS



9.41 Use SPICE to plol I in Prob. Ll I in lbe interval0 < t < I s. 9.42 Use SPICE ro plot i in Prob. 9.24 in the intervalo < I < I s. 9.43 Use SPICE in Prob. 9.37 tb plo. l) in the interval 9.44 9.45



- l,tr,



for



PROBTEM 9.45



306



0



O



fr, -



* '(*)]



112.6.)



because



A



EXERCISES



l0.t.l



Find rhe period of rhe following situsoids:



- 33'). - 1) r :.; \- 4/ - *'\-/r,



tar 4 cos (3r (bt cos '"' *" /2r



(c) 8 sin 2zr. antwt ta) 2n/3r



10.1.2



.



_ . 10.1.3



ftt t;



lc)



1\



6)'



I



Find the amplitude and phase of the following silusoids: (a) 6 cos 2r + 8 sin 2r. (b) (4V3 3) cos (2r + 30) + (3V5 4) cos (2r + 60"). [Suggestion: In (b), expand both functions and use (10.9).] Aniwqr (a).10, -53.10; (b) 5, 36.9. Find tbe frequency of rhe following sinusoids: (a) 3 cos (6ar 10'). (b) 4 sin 377r. Ansyer (a) 3i b) 60 Hz



-



-



-



312



Chapler



l0



Sinusoidal Excitation and Phaso.s



= -5 < 0



and B --



AN f,T CIRCUIT EXAMPLE As an example of a circuit with a sinusoidal excitation, let us 6Dd the forced compo_ netrt iof the curent i in Fig 10.4. The describing equation is



,d!



,r, =v."*.,



(10.11)



and followirtg the method of Chapter 9, let us assume the



trial solution



'i=Acos@r+Bsinort L



rrcURE 10.4 Rl circuit Substituting the trial solution into (10.11), \{e have



L(-@A



sin @t



+



@B cos



ot) + lt (A



cos (.)t



+



I



sin (.)t)



: v. c6



@t



Therefore, equating coefficients of like ternrs, we must have



M*aLB=V^ -oLA+RB:O from which



B= R,



uLV"



+;Li



The forced response is then



Rv-



aLV^



ot



R' + u'L'



i'' = R'+



L' cos.dt -



-:--



which by (10.9) and (10.10) may be witteD



srn



(,l



as



vta[\ - tan-'^J n= \"'r ;1, * -,1cos -



(lo.l2)



The forced response is, therefore, a sinusoid like the excitation, as we pledicted when we chose the trial solution. We may write it in the form ir



Secrion



10.2



An Rr Circuit Example



= I-



cos (@t



+ 6\



(10.13)



313



\rAere



,v^



\./R, +.,'21.2



@L



R



(10.14)



Since the natuml response is



i": it is clear that after



a



sho.t time i.



+



Ate



Rt/L



0, aDd the current seltles down to its ac steady



slate value, given by (10.12).



The method we have used is straightforward and conventional but, as the reader might agrce, is rather laborious for such a simple problem. For a second-ordef circuit, the method is more tedious, as was illustrated by the example of (9.34). For



very high-order circuits the procedure is, of course, even more complicated. Evidently, we need a bett$ method. One such method is developed in the remainder of this chapter, and irs use allows us to tlea! circuits with stomge elements in the same way we uealed reSistive circuits in Chapters 2, 4, and 5.



EXERCISES



10.2,1



Find the forced response



=



100,000 rad,'s. Al'rl'er 0.4 cos (100,000r @



10,2.2



iin



-



Fig.



10.4



if L = 60 rnH, R = 8 kO, V, = 4 V,



36.9') mA



Find the forced component of o.



. tu,c, tRl-/ll



-



RC



J cos



(.rr



- tan



[xtRctst



1



and



(,RCt V



10.2.2



0.3



AN AITERNATIVE METHOD USINC COMPLEX NUMBERS The alternative method of analyzing citcuits with sinusoidal excitations, which we consider in the remainder of the chapter, relies heavily on the coocept of complex numbers. The reader who is unfamiliar with complex numbers, or who needs to review the subject, should consuit Appendices C and D, where conplex numbers and



314



Chapter



l0



Sinusoidal Excitation and Phaso6



are discussed in some detail. For convenience, we list a few of these properties in this section b€fore considering the alternative method of analysis. The complex number A is written in rectangular form as



thei properties



A=a+



jb



(10.15)



where j = \EJ



and the real numbers d and b are the real and imaginary parts A, respectively. Equivalendy, we may say that



of



a=ReA, b=lmA where Re and Im denotr lhe real pa of and lhe imagimry parr ol The number A may be written also in the polar form,



A=lAler"=lAlld where iA I is the magnitude, given by L



and



a



el =



(10.16)



t/7'l-F



is the an8le or argument. given by



o = ,^n



'



!



These relations betiveen rectangular and polar forms are illustrated in



Fig l0.5



flCURt 10.5 Ceometrical represenlation of a complex number A EXAMPT-I 10.3



Suppose that



tan '



EXAMPLE 10.4



I



we have



A=4+ j3.



Then



.1b.9'. Therefore. the polar form is



lAl=



\/4'1



+3'?=5 and d=



A slllg ,



j 12. Since both d and are negative, the line segment repreConsider A = -5 senting A lies in the third quadrant, as shown in Fig. 10.6, ftom which we see that



-



lAl= V5'- 12:= ll



and



d= Thus we have A Section



10.3



=



l8o'-



ran-'



l



z+t.o'



131247.4'.



An Altemativ€ MeIhod Using ComPlex Numbe6



315



FICUf,E 10.6 Complex number with neSatjve real and imagjnary parts



Other usefi, results are



j=\/w j'= 't



= t/tEo"



By Euler's forrDula. wtuch is discussed in Appendix D aud which we have used



in Chaptf,r 9, we know Illat



V-



cos



tDt



+ jV^ sB at =



V^eN



Therefore, we trlEr say that



c6



@t



= Re(V-etu)



Y. sin



ot



:



V6



(r0.17)



and



'



MV^etu)



Retuning to the Rt circuir example of Fig. 10.4. we know rhat exronentials are matnernallcall, easier to handle aB excitarions &an sinusoids. Therefore. let us see what happens



if we apply the



complex excitation



,



at:



V^er''



(10.r8)



instead of rhe feal excitatioD os



= y-



cos



or =



We cannot duplicate such a complex excitation



i!



Re



r)r



(10.19)



the laboratory. but there is no rea-



considq it abstracdy. In this case the forced compoDent of the cl|r_ rent, which we call ir, satisfies soD we calurot



' 376



Chapter



10



rfr



+



nr = ar =



Sinusoidal Excitat'on and Pha5o6



v-etu,



(ro.2o)



To solvc this equation, we try



i = AeP' which, substituted into (10.20), yietds I



j@L



+ RtA?' =



V-e



-



from which V^



^- n+ Pt v-



,,,"



,..., *



VR' + d')L' using complex number division. Therefore, we have



.vVR'



- d'L'



Let us now observe that



o.,, =



*"1-4 + h"Lz""-"--'"''"]l



'



v. I t?-Zcosltor



-



I



\/R'1



which, by (10.12), is the corect folced response of rr



=



Re



,rt-\ tan'RJ



Fig l0.4.



That is'



l'



(10.21)



we have established for this erample lhe interesting resuh thal if ir is the complex response to the complex forcing function or, then = Re is the response to The reasoi for u, = ne o'. That is, o' yields ir and Re or = os yields Re ir = this is that the describing equation (10 20) contains only real coefficients. Thus, fiom (10.20), we have



i



""(rf;*",,)



i



i.



=*",,



OI



r4-tn. r't dt



and therefore. by {10.1



R{Re



i,, = y. cos.rr



l).



r=i=Rer, Ttus we



(10.22)



it is easier lo use the complex forcing functioD Lr to find the ir. Then since the real forcing functioD is Re tJ1, the real respons€



see that



complex response



is Re ir. This principle holds for all our cLcuit amlyses, since the describing equations are lincar with r€al coefficients, as may be seen in a developmelnt analogous to thar leading to r I0.22). Secrion



10.3



An Alternative Method Using complex



Numbe6



317



EXERCISES



10.3,1



Replace the real forcing function /, cos {rt in Exercise 10.2.2 by the complex forc_ ing function 1.?/.', find the resulting complex response o,, and show that the real respons€ is o = Re or,



10.3.2



Show that, for a real.



n"/,d') o3,n. ,, \ drl - dt and use this result ro esrablish (10.22). (Susgestian:



10.3.3



llJt:t = J8, where/and8 are real. ) In Exercise 10.2.2, replace the curent souce by ir = 1-€,.'A and show that the response o, has the propqty that Re or = o, where o is the original response.



/+



10.4 COMPTEX EXCITATIONS [,et us now generalize lhe results using complex excitalion functions in the preceding section- The excitation, as well as the forced response. may be a sinusoidal voltage or curent. However, to be specific, let us consider the input to be a voltage source



-



,



and.the outputto be a curent though some element. The other cases may



ered in an analogous way. In general, we know that



i



consid-



if o,



=



V_ cos



e)t +



0)



(r0.23)



the forced response is of the form



i = I-



cos



k'rt + 4)



(10.24)



in the general circuit of Fig. 10.7. Therefore, if by some means we can we have olll answer, since .r, d, and y, are known. To solve for i in Fig. 10.7, let us apply the complex excitation



as indicated



find 1. and



{,



at =



vder@t+o)



flCURt 10.7 Ceneral circuit with input and output



318



Chapter



10



Sinusoidal txcitation and Phasors



(to.zs)



and find lhe complex response ir, as shown in Fig. 10.8. Then we know from the results of the preceding section that the real response of Fig. 10.7 is



t = Rei,



(10.26)



This is a consequence oi the fact that the coefficients in the describing equation are real. as pointed out previously.



l4 FICURE



10.0 Ceneral circuit with a complex excitation



The deit;ribing equation may be solved fo! the forced response by the method



of Chapler 9. That is, since we rnay wtite the excitation



as



ar = Vneroejd which is a constalt times



€t-, then the trial solution



(t0.27\ is



i' = AeP' Comparing (10-24) and (10.26), we must have



I^.os



(@t



+



{) = Re[Aefl



which requires that



A = I.eb and hence



L=



Ihei6



er"



(10.28)



Taking rhe real pad. we have the solution (10.24r, of course. EXAMPTE 10.5



I-et us find the forced respoise



i



of



! , z r 8r = l2\4 cos (2r + 1s.., ,4



First we replace the real excitation by the complex excitation,



t),



- l2\5



e'n'-ts"



where for coovenience lhe phase is written in dege€s. (This is. of couse, ao incon-



sistent mathematical explession, but as long as w€ interpret present no difficulty.) The complex respons€ ir satisfes



9! Section



l0.4



+ s;, = + zo! tll



Compler Excilations



121f2



it



correctly



it



should



"n'r" 319



atrd



it rnust have tha g€neral form



il = Ael' Thercfore, we must have



l-4 r j4 +



or,



StAeL



=



l2leE eatens"



A= which gives



i = (3/-3cP)el :



3ei@-toPl



Thus the real answer is



=



rr



Re



i, =



3 cos (21



-



30')



.EXERCISES



10.4.1



(a) From the time-domain e{uations find the forced response l) if os = 10€i3, V. (b) Usirg the result in (a), frnd the forc€d response o if os : l0 cos 8t V. Answet (n) 2ei\3ts3' e V (b) 2 cos (8r 53.1.) V



EXERCISI 10.4.r



10.4.2



Find the forced response o in Exercise 10.4.1 tf as sin 8r : Im eJr'.) Answer 2 sm



10.4.3 Using the



(8t



-



method



10.4.4



as (2t +



of



320



Chapter



l0



y.



(Suggestion.



complex axcitation, find the forced response



36.9")



Repeat Exercise 10.4.3



Ausk?. 2 cos 4, A



sin 8r



53.1) V



20 cos 2t Y. Answer 2



= l0



A



if o,



: t.:':;:t""t



Sinusoidal frcitation and Ph6o6



i if



Ds



=



r-- 10.s PHASORS The results obtained in the preceding section may be put in much more compact form by the use of quantities called p/rdrorr, which we shall introduce in this section. The phasor method of analyzing circuits is credited generally to Charles Proteus Steinmetz (1865-1923), a famous electrical engineet with the Ceneral Electric Company in lhe early part ol this cenlur]. To b€gin, let us recall the gen€ral sinusoidal voltage,



Y. cos (irr + 0)



(10.29)



which, of course, is the source voltage Ds of the preceding section. If the frequency o is known, then D is completely specified by its amplitude y. and its phase 0. These quantities ale displayed in a related complex number,



v=vnet6=vd/g



(10.30)



which is defined as a phasor, or a phasot represandtion. To distinguish them from other complex numbe$, phaso6 are printed in boldface type, as indicated. The motivation lor the phasor dehnrrion ma) be seen from the equivalence. by Euler's fomula,



'



of V-cos (4,' + d) = Re(Vneiaejut)



(



ro.3l )



Thereforer in view of (10.29) and (10.30), we have tJ



.



EXAMPTE



10.6



= Re(ver'')



(10.32)



Suppose that we have



r'



:



10 cos



(4r



+



30') V



The Phasor rePresentation is rhen



v = 10l]q v since y- = l0 and 0 = 30'. known,



,



Conversely, since (d is readily obtained from V.



: 4 rad/s



is



assumed



to



be



In an identical fashion we define the phasor reFesentation of the time-domain current



i=1-cos(@r+d)



(r0.33)



l=t.ejo-t-14



(10.34)



to be



Section



10,5



Phaso6



321



Thus



if



:6



we knon, for example, that ar



rad/s and that



I = 21!f A, the! we



have



i=2cos(6r+15")A we have chosen lo represent sinusoids aod theL relared phason on fte basis of cosine functiolls using the hct that cos art : Re(e"). We could have cho6€n sine fimctions just as easily, using sin ort = Irn (er') (see Exercise 10.4.2). Thus if a function such



as



o=8sin(3r+3f) is given. we may change



il to



rl=8cos(3t+30'-90) = 8 cos (3r - 60') Then the phasor representation is



v = 8L60' Had we chosen to base the phrso$ on sine firnctioDs, then we would keep o a6 is ard write its phasor as 8f3E, which, of course, wodd represent 8 sin (3t + 3f) in the time domain. An illustration using sine-basdd phasors is given in Example t0. t6.



EXAMPLE



1O.7



To se€ how the use of phasors can greatly shorten the work, let us reconsidgr Fig. 10.4 and its describing equation (l0.ll), rewitte& as



t-:cll t



=



Ri



6* .,,,



Y.



0.35)



Following oul me$od, we replace the excilarion y. cos ot by the complex forcirg function



at = V^eN which may be writen



ot:Ieb 0 = 0, and therefole (10.35), we have since



V = v^19: v^. di, L-:+



.dt whose solutjon



ir b



Ri1



S8bstituting



=VeN



related to lhe real solution



i-Reij Next, hying



=w 322



chapte.10



sinusoidal Excilation and Phasors



i by



this;lue



aDd



i = ir into



as a solution, we have



.



ioLlei't +



= Vett'



Rleia'



Dividing out the factor ed, we have the phasor equation



j@LI



+ Rl:y



(10.36)



Therefore,



vv^



R Substituting



this



+



ioL



VR, +



/ .aL ' t',L' f -tan R



lue into the exptession fot



ir, we have



.vvR1 + u'L, Taking the real part, we have



i = i,



obtained earter in (10.12).



It is importart to note that if we can go directly from (10.35) to (10.36), there is a vast saving of time and effort. Also, in the process we have conveated the differential equation itrto an algebraic equation, somewhat like those encountered in resistive circuits. Inde€d, the only difference is that the Dumbers here are complex, whereas in resistive circuits they were real. With tle hand calculatot as cornmonly a€ilable as it is today, even the complexity of the numbeN presents little difficulty. In the remainder of the chapter we shall see how to bypass all the steps betwe€n (10.35) and (10.36) by studying the phasor relationships of the circuit ele ments and conside rg Kirchhoff's laws as they pertain to phasors. Indeed, as we shall see, we may go directly from the circuit to (10.36), bypassing even the step of writing down the differetrtial equation. In general, the real solutions are time-domain firnctions, and their phasofi are frcquenct-domain fitnctions (i.e.i they are functions of the ftequency o). This is il: Iustrated by the phasor I of Examplg 10.7. Thus to solve the time-domain Foblems, we may convert to phasors and solve the corresponding freque cy-domain problems, which are generally much easier. Fioally, we convert back to the time domain by fiDding the time fuDction ftom its phasor representation.



EXERCISES



10,5.1 10.5,2



FiDd the phasor representation of (a) 4 cos (2t srn (Jr b)_). Answer (a) 4 / -6r.9"i @) 2 l2s' @)



(c) I



lgt



+



45"), (b) 8 cos



-j6.



In all cases a'



An$'€r (a) 10 cos (3,



Section



10.5



Phasors



15 sin



2r, and



ti



Find the time-domain tunction represented by the phasors (a) and (c)



2t +



-



:



101:!U, O) 6 + j8,



3.



l7'); (b)



10 cos (3r



+



53.1'); (c) 6 cos (3r



-



90')



323



10.6 VOLTACE-CURRENT RELATIONSHIPS TOR PHASORS In this section we show that relationships between phasor voltage and phasor current



for resistors. inductors, and capacitors are very similar to Ohm's law fff resistors. In fact, the phasor voliage is proportional to the phasor cwrent' as in Olun's law' with the proportionality factor being a constant or a function of the tequency (, We begin by considering the voltage-curent lelation for the resisfor,



o=Ri



(10.3?)



where



'



a =V.cos(rdt+0)



i=I^cos(d+A\



(10.38)



If we apply the complex loIt^ge V-ejt't+o), the complex current which results is 1-?,1-+c), which substituted into (10.3?) yields Vdel-t+o)



: N-ei\.1+6t



Dividing out the factd ej'' results in V^eio



which, since V.er" and



/-?'



=



Rl^ejo



are the phasors



(10.39)



V and I, respectively, !€duces to



V=RI



(10.40)



Thus lhe phasor or ftequency-domain relalion lor lhe resislor is exacll] like lhe timedomain rilation. The voltage-current relations for the resistor are illuslrated in



Fis.



10.9.



FICURE



10.9 Voltage-current relations for a resistor Rjn the



(a) time and (b) freqLrency



From (10.39) we hav€ y. = RI- and 0 = 0. Thus the sinusoidal voltage and ent for a resislor have the same phase angle, iil which case they are said to be irl prdre. This phase relationship is shown in Fig. 10 10, *tere the voltage is represented by the solid line and the current by the dashed line.



'



cu



324



Chapter



10



Sinusoidal Excilation and Phaso6



IICURE 10.10 Voltage and current waveforms for a resistor



EXAMPLE IO.8



Suppose lhar the vohage 1)



=



I0 cos (t00r



+ l0') V



is applied across a 5-O resistor, with the polarity indicated in phasor voltage is



(lo.4l) Fig. 10.9(a). Then the



v = l0ll0" v



and the phasor current is



r=Y-rofo'-t,:o"e R5 Therefore.



ir



the iime domain we have



i=



2 cos (100r



+



30') A



(r0.42)



This is. of course. simply lhe result we would have obrained using Ohm's law.



In rhe case of the inductor, substituting the complex curent and voltage into the time-domain relationl



,L4 gives the complex relalion



v-"'-' n = yd 11-",- '1 = ioLI^et''+6) Again, dividing out the factor er- and identifying the phasors, we obtain the phasor relation



v = jaLt



I',



section



10.6



Vohage-Curent Relationships for



Phasors



(10.43)



325



dnsor voltag€ V. as in Ohm's law. is Foportiooal to the phasor clllrent I, with dr€ pioportionality hctor irz, Th€ voltage-current relations for the indEtor are shov/n in Fis. 10.11,



Thus the



I



y+



(



j@L\(r^/A



9f.



atr inductor ttle cuJrent ,a8s tlie voltage by Another expie3siorl that is u6ed is that the cufien! atrd voltage are 90o oxt of pbrse. This i! $hown graphically in FU.



10.12.



Finaly. let us consider the capacitor. Substituting the complex cunent aDd voltage into the dme-domain relatiotr,



i=ci-do FtcuRf 10.12 VoltaSe and cuqent waveforms ior an ihductor v,i I



L



i



I



326



Chapter



10



Sinusoidal Exciiation and Phaiots



t



give\ the complex relation I4etd' at



Again



. dividing by e!



= c 4tlv-?rd+s'l = i@CV-sit**ot



and idenlirying lhe phasors. we obtain rhe phasor relation



I : jaCV



(10.44)



I v= juc



(



10.45)



Thus the phasor voltage V is proportional to the phasor current I, with the proportionality factor given by I / jac . The voltnge-current relations for a capacitoi in the time and ftequeoq/ domains are shown in Fig. 10.13.



G)



(b)



flCURt 10.13 VoltaSe-current relations for



a capacitor in the (a) time and (b) frequenc)



domains



In the genenl case, if the capacitor voltage is given by the first equation of (10.38), then by (10.,14), the phasor curent is



t = (jac)(v^/!) = @cv^/o +



90"



Therefore, in the time domain we have i



i



I



l I



t



i:



@CVn cos



t.tt t 0 + gDo)



which, by comparison with the fu$ equation of (10.38), irdicates that in the case of a capacitor the current and voltage are out of phase with the current leading tbe voltage by 9O'. This is shown graphically in Fig. 10.14. Section10.6 voliaSe-cur€nr



Relationships for



Phasore



327



FICURt 10.14 Voltage dnd cutrent waveforms



EXAMPTE 10.9



Il rhc voltaBe phar,,I currcnl



of (lo.4l' r\ i\



applred acros\



t



= Jr l00r



l0



= lAzq



mA



lor:



capacitor



a l-/]f capacitor. (hen b] {10.44t lhe



"r{



101jff} A



The time-domain current is then



I=



coc



{t00r



-



t20"r mA



and therefore the current leads the voltage by 90'.



EXERCISES



10.6.1



10.6.2



U.ing phasors. tind the ac \te.rdy state currenl r ifo - 12costlOOOt + 30') V in (a) Fig. 10.9(a) for n - 4 kO, (b) Fig. lO.ll(a) for L = 15 mH, and (c) Fi8. l0.lJ(a)lora = . /rF. Azswer (a) 3 cos (1000t + 30') mA; (b) 0.8 cos (1000t - 60') A; (c) 6 cos (1000t + 120) mA ln Exercise 10.6.1, find i in each case at t = 2 ms. Anster (a) 2.445 mA; (b) 0.4 A; (c) 3.476 mA



10.7 IMPEDANCE AND ADMITTANCE Let us now consider a Seneral circuit of phasor quantities with two accessible terminals, as shown in Fig. 10.15. If the time-domain voltage and current at the terminais are given by (10.38), the phasor quantities at the terminals are



v = v^1! | = r.l:! 328



chapter



l0



sinusoidal Excitation and Phasors



(10.46)



I



FICURI 10.15 General phasor circuit



We define the ratio of the phasor voltage to the phasor current as lhe tmp?dakr of lhe cl..cuit. which we denore by Z. That ts.



z:7 which by (10.46) is



(10.47)



z=lzl&=Yrc-



where I Z I is the magnitude and dz the angle of



o



(10.48)



Z. Evidently,



e":0-o



zl _v_



Impedance, as is seen f()m (10.47), plays the role, in a general circuit, of resistance in resistive circuits- Indeed, (10.47) looks very much like Ohm's law; also like resistance, impedance is measured in ohms, being a mtio of volts to amperes.



It is important to



stress that igrpedance is a complex number, being the ratio



of



two complex numbers. but it i\ not a phasor. That is. il haq no corrJsponding \inusoidal time-domain funcrion ol any phy\ical meaninB. as curre'nt and voltage phasor: have. The impedance Z is written in polar form in (10.48); in rectangular form genemlly denoted by



Z=R+



jX



(



it



is



10.49)



R = Re Z is the resistfue conponent, or simply r?sistanc?, and X = In\ Z is lhe reactive conponent, ot reactance. fn general, Z = Z(ja) is a complex function ofJ.r, but R = R (o) and X = X((') are real tunctions of o. Both R and X, like Z, are measured in ohms. Evidently, comparing (10.48) and (10.49) we may write where



El _ \fN + x, .x ez R



and



R= Z y=lZ



a



i



These.relations are shown graphically in Seclion



10.7



lmpedance and Admittance



co\ 0, sin 0z



Fig. 10.16.



329



FICUIE't0.t6 Craphicat EXAMPIE



t0.lo



Suppose in



Fig. 10.15 thar V



=



representation of impedance



10156.9"



V and



7=lL^,^E$ 2/20"



I=



2120" A,. Then we have



h.g"a



In rectangular form this is



Z = 5(co\ J6.9' = 4 , j3A



Ij



sin 16.9"1



The impedances of resistors, inductors, and capacitors are readily found from relations of (10.40), (10.43), and (10.45). Disringuishing their impedances with sutrscripts R, a, and C, respeatively, we have, from iiese e{uations and (10.47),



their



V-l



ZN:R



zL=jaL:@LN:



z.: -L = -;1 = uC- j@C " aC -L7



(r0.50) q6.



In the case of a rcsistor, &e impedarce is purely resistive, its reactance being zero. Impedances of inductors and capacitors ale purely reactive, having zero rJsistive componeats. The inductiae rcactence ir deDoted by XL



= ."L



(10.51)



so that



L= and the capacitive reacance



ii



jx"



demted by



__l -aC



(10.s2)



and thus



k- jx. 330



Chapter



10



Sinusoidal fx€itation and Phaso6



,{



(10.s3)



I



Since (,, 1-, and C are positive, we see that inductive reactance is positive and that capacitiv€ rqrctance is negative. ln the general case of (10.49), we may have X = 0, in which cas€ the circuit appears to be resistive; X > 0, in which case its reactance appeaN to be inductive; and X < 0, in which case its reactance appellrs to be capacitive. These cas€s are possible when resistance, inductanc€, and capacitance are all present in the circuit, as we shall see. As an example, the cicuit with impedance given by Z = 4 + j3, \rhich we have just considered, ha! reactance X = 3 , which is of the inductive t)?€. In all cases of passive circuits, as we shall se€ in Chapter 12, the resistance lt is nonnegative. The reciFocal of impedance, denoied by



v=+



(r0.54)



is called admittan e a$d is analogous to conductance (the reciprocal of resistance) in resistive cirpuits. Evidently, since Z is a complex number, then so is Y, the standard representation being



Y=G+jB



(10.55)



The quantilies G = Re Y and B = Im Y are called conductance and susceptance, t} respectively. and are relaled to the inpedance componen6



by



Y=Gr+i8



=;="tj



(r0.s6)



The units of Y, G, and B are all siemens, sinca in general Y is the mtio of a current to a voltage phasor. To obtain the relation between components of Y and Z we may ralionalize lhe last member of (10.56), which results in



1 R- iX c+ jB= R +ji R- ix



F4uating real and imaginary parts results in



-R (10.57)



^X "--R\x, Therefore. we nole that R aDd C are rrt reciprocals excepl in the purely resistive case (X = 0). Similarly, X and ,8 are novff reciprocals, but the purely reactive case (.1R = 0) rhey are negadve reciprocals.



i!



I



Secrion



10.2



lmpedance and Admattance



331



EXAMPLE 10.11 If we have



Z=4+ j3



then



j3



l 44 t '-4+ i3- 4'+3' E rE



., f



Therefore, G = andB = Furlher examples are



--f



-



Yr=G



Y.=+ yc _ which are the admittances of



a



joc



resistor, with R



:



l/G,



an inductor, and a capacitor.



EXERCISES



10.7.1



.1S7.2



Find the impedance seen at the termrnals of the souce in Fig. 10.4 in both re.tangular and polar lorm Ans\|et R + jaL, \/R1 + a'1Ll len-t aL/R Find the admittance seen at the terminals of the source in Fig. 10.4 in both rectangular and polar form.



10.7.3 Fild



:-:



the conductaoce and susceptance



if Z is



(a) 6



- j8, (b) 0.2 +



j0.15, and (c)



/ :.35"



A"sw"r (a) 0.06,0.08; (b) 3.2. -2.4; (c)



-4,



4



'10.8 KIRCHHOFF'S LAWS AND IMPEDANCE COMBINATIONS Kirchhoff's laws hold fol phaso.s as well as for thei corresponding time-domain voltages or curents. We may see this by obsewing that



if



a



complex excitation, say



applied to a circuit, lhen complex voltages, such as Vrer(''+dr), V2ejt't+or, etc., appear across the elements in the ci.cuit. Since Kirchhoff's laws hold in the time domain, KVL applied around a typical loop results in an equation such a\ Vteit.t+\) + V1e1.t+02\ +. . ..+ yren.doN):0 Vnej\@t+o\,



is



Di!iding out the common lactor p/ . we



where 332



chaorerio



have



v,+v,+...+vd:0 v" = snusoidar



v,&,



n: r,2,.



i



. . ,N



Excitationaodphasors



I



are the phasor voltages around the loop. Thus KVL holds for phasors. A similar development will also establish KCL. ln circuits having sinusoidal excitations with a common ftequency (r, if we are interested only h the forced, or ac steady-state response, we may firld the phasor voltages or curents of every element and use Kfuchhoff's laws to complete the a.lalysis. The ac steady-state analysis is therefore identical to the resistive circuit analysis of Chapters 2, 4, and 5, with impedances replacing resistances and phasors replacing time-domain quantities. Once we have found the phasors, we can convert immediately to the time-domain sinusoidal answers.



EXAMPTE



10.12



Consider the circuit ofFig. 10.17, which consists of N impedances connected in series. By KCL for phasors, the single phasor curent I flows in each element. Therefore, the voltag€s shown actoss each element ille



\:



ZJ Vz =. Z,l



-



.



V, = Z,l and by



KVL around the circuit,



v=Vr-V:'.. =12,-2._.. Since we must also have, ftom



z,*ls



. + zN)l



=



Z,.tl



the equfualent impedance se€n at the terminals, it follows that



Za=Z'+22+...+ZN as



t



Fig. 10.17,



\ \\here



+v,



(10.58)



in thp case of series rcsistors.



frcURt 10,17



Seclion



10.8



Kirchhoff's



lmpedances connected in series



Las and lmpedanc€ Combinations



333



Similarly. as was rhe case for parallel conductlnces in Chapt€r 2, the equivalent admittance Yq of N parallel admittances is



Yq = Y, + Y, + In the case of two paEllel glements (-lf



=



.''



Y{



+



(10.J9)



2), .xe have



_ltz,z, z* = Y* Yn Y,= zn z,



(10.60)



In like manner, voltage and current division rules hold for phasor circuits, with impedances and fte$rency-domain quantities, in exactly the same way that they held for resistive circuils. with resistances and time-dornain quantitie.s. The reader is asked to establish these rules in Exercise 10.8.2. EXAMPTE



I0.13 [-€! us retuin to



the RZ cbcuit considered in Sec. 10.2. The circuit and its phasor counterpart are shown in Fig. 10.18(a) and (b), respe.tively. By KVL in the phasor circuit we have



zLt+ Rt=v^l! (jaL + R)r - v^1! ftom whilh the phasor current is



v"Ly ,'-R+Jtz



v^



/



,on



vR, + azLtl



R



Therefore, in the tine domain we have, as before,



v.



.=



' vF-;1L, flCURt 10.18



(a)



cos



/ - tat-,rr\ 'n/ \ot



Timeiomain circuit;



(b) equivalent phasor circuit



L +



iAL



v- IA



* .



334



Chapter l0



(a)



Sinusoidal Excitation and Phaso6



(b)



An alternative method of solution is to observe that the impedance Z seeo at the souce terminals is the impedance of the inductor, joL, and the resistor, R, connected in s€ries. Ther;fore,



I



z= jaL+R



and



, v



i;



v^Is



'- Z- R+ t,,L



I



as obtaineal earlier.



i



I I



EXERCISES



10,8.1



Derive (10.59).



10.E,2



Show in (a) that the voltage division rule,



v



=urttv,



ard in O) that the current division rule,



r= are



ralid, where Z,



= l/Yr



v,-2,-



v--;t'



and



Zz



z, . z,r'



= l/Yz.



t



l{



Y2



EXERCISI 10.8.2



10,E.3



Find the steady-state current i using phasors. ,qnswer,4 cos (4t 36.91 A



-



1H



+



r0.E.4



Find the



steady-;



sioD.



Answer 2



h Section



I



i I



sos



10.8



(4t



-



""t "r" 126.9"\



, t';::::"';:;,



;F



*.,



phasors and vorrage divi-



Y



Kirchhoff's Laws and lDpedanc€ Combinataons



335



10.9 PHASOR CIRCUITS



'



. . EXAMPLE



As rhe discus5ion in the preceding 5eclion suggesE. we may omit the steps of finding the describing equation in the time domain, replacing ilrc excjtations and responses by their complex forcing tunctions and then dividing the equation rbrough by ?r,.to obtain the phasor equation. We may simply start with the phasor circuit, which we will now formalli define as rhe time-domain circuit wilh the voltages and curents replaced by their piasors and the elenents identified by their impedatces, as illustmted Feviously in Fig. 10.18(b r. The describing equation obtained ftom this circuit is then the phasor equation. Solving this equation yields the phasor of thg answer, which then may be cgnverted to the lime-domain answer. The procedure from starting with the phasor circuit to obtainhg the phasor answer is identical to that used earlier in resistive circuits. The only differetce is that the numbers are. complex.



10,14 Let us find the



steady-state curlent i in Fig. 10.19(a). The phasor circuit, shown in Fig. 10.19(b), is obtained by r€placing tle voltage sou&e and the currents by their phasors and laHing rhe elements with their impedances. In the phasor circuit the impedance seen ftom the source terminals is



_ ,,(3 3



+ j3x-j3) + lJ _ It



=a-j3O FICURE



10.19 RIC time-domain and phasor circuits



IrI +lO 3f,,



5lQ'v 13f}



336



Chapter



l0



Sinusoidal Excitation and Phasors



Therefore, we have



.,54,. =,sq4 _ j3- sLl!.e"



l,



= r/16.s.



and by current division. / >':a t=1.''." \3+j3-j3i ,lr,=\A/



.s" A



In the time domain, the anslwer is



i = li



cos



(3r



+



81.9")



A



In the case of a dependent source, such as a source io, volts cotrtrolled,by a voltage D., it will appear in the phasor circuit as a sourc€ tV,, wherc V, is the phasor representation of o. , becausie D, = y. cos ((l)r + d) in the time domain rrill becomc V-e^d*o) wl\en a complex excitation is applied. Then dividing eF out of the equations leaves o. represented by its dEsor y-€d. In the samc way. to, = kV- cos (ar d) is represented by irs phasor ,tV-e&. which is t tiftes rhe phasor of o..



r



EXAMPLE



I0.15 As an example of a circuit containing a dependent souce, let us consider Fig. 10.20(a), in which it is required to find the steady-state \"alue of i. The corre$poDding phasor circuit is shown in Fig. 10,20(b). Sinc€ dlasor circuits are analyzed. exactly like resistive circuits, we may apply KCL at node a in Fig. lo.mo), req ting



v, - 1v. t+ -:= 3l0. By Ohm's law we have Vr



= 4I, '



or



r0.6r)



which subsrituted into (10_61) yields I



t4t\ - -i6 6/-qr 1 -+ _ -:-/-4s.



j2t +



i6 l= 2- += 2 J



(



2V2l



4s V2-



^



FIGURE 10.20 {di Crcurr Lontdining a dependint source; rbJ correspondinS phasor circuir



:"



Section



10.9



Phasor circuits



3te"



a



337



Therefore, we have 3



EXAMPLE



-



4s') A



10.16 I€t us find the steady-statc qrrlent ir in Example 10.14 if the source voltage 3t v. sirce o, = 5 cos (3r - 90) v, the plBsor voltage is vs: is o, = 5 O Thus V,"itr and, as before, the i$pedance se€n by the source is 5 5



t]ql



/-![



we have



.:ffi



=!-53.1'A



alld



ir =



cos



(3t_



53.1")



=



sin (3,



+



36-9') A



If



we had based the phasors on the sine instead of the cosine, we would save the filst converting the sine to the cosine and then conve ing the cosine back to the sine, or equiyalently, of first subtracting 90' and then adding 90' to the phase. Based on the sine. V" = 518 V. and st€p6 of



= l/36.9"



A,



OT



i, = sir (3t + 36.9) A In the case of an op alnp, the phasor circuit is the same as the time-domain circuit. That is, an ideal op amp in the time-domain circuit appears as an ideal op amp in the plEsor circuit, because the time-domain equations



which char-acteriz€ the current into and the voltage aqoss the input terminals retain the identical form,



I:0,



V=0



in the dmsor equations. As a final note vre observe that the phasor method of 6rding, say i, by fi$t finding I = V/Z and theD converting I to i, frils to wo* if Z(Jar) = 0. This is the case when the circuit is excited at a Datural ftequency jo, and we must then use the nethod of Sec. 9.6. An example of this case is considered in hob. 10.38.



EXERCISES



10.9.1 10.9.2



Solve Exercise 10.2.2 by dears of the phasor circuit. Find rhe steady-state voltage o using the phasor circuit.



Aruper 8 cos (8r



338



Chapter



l0



- 53.f) V



Sinusoida! Excitation and Phasors



20 >



40



EXtRCtSt 10.9.2



10,9.3 10.9.4



Find the steady-state voltage o in Exercise 10.4.1(b) using the phasor circuit. Find the steady-state voltage D using the phasor circuit, given that os = 4 cos lor V'



An',c, V2



cos



(l0r I ll5') v



txERctsE 10.9,4



....g



SUMMARY In this chapter we bave considered in some det ail the sinusoidal fit'.cfion, the dominant signai in the elecfiical Power industry. We have defined its amplitude, ph4se,



and lr;quencJ, and have considered \ts phosor representation we have seen that steady+iate sinusoidal rcsponses may be obtained using p/usors, a shorrcut method based ot conplcx exciutbrs. The phasor voltage-curent relations for resistors, in_ ductms, and capacitors are identical in form to Ohm's law, with imPeddnce, the fitio of the'ebm;nt's phasor voltage to its phasor current, playing the role of resistance. The reciprocal of ifip"fLance is admittance, the analogy to conductance y lre obtained by replacing the voltages and currents in the A phasor circutt given time-domain circuit by their phaso6 and labeling each passive element with its impedance. sioce KvL and KCL hold for phasors, and impedances behave like resisttnces, the pbasor circuit is solved exactly like a lesistive (Acuit The phasor solutions are then converted to their time-domain counterpalts, completing the



tl



't 1



analysis. I



Section



1!.10



Summary



339



I I I



i I



PROBTEMS l0.l



Civen the voltage o = 50 cqs (20OTr + 60") V, find (a) its amplitude, (b) its phase angle in degrees, (c) its phase angle in radians, (d) irs period in Im, (e) its fr€quency in rad/s, (f) its fiequency in tiz, and (g) by how many de,



cunent i : A. Convert the following functions to cosine tunctions wilh positive amplitudes: (a) 6 sin (2t + l5'), (b) -2 cos (4r + 10"), (c)



Irees



it



leads



2 cos (21l0it 10.2



8 cos



10.3



10.8



5r



-



or lags the



17")



A yoltage y- sitr (,rt V, a ftsistor R, and an inductor are a[ in s€rie.s. Show that the forced response i js identical to (10.12) except that the cosine is replaced by the sine in lhe follow ing two \lays. (a) l.{oting that V. sin r.rt = y. cos ((1,t 90'), use the compl€x excitation y-end-ee) and take the rcal part of the cufent response. (b) Noting that y- sin ('/ Irn(y-el'), usc the complex excitation y,,?Fr



,



:



and take the imaginary part of the cuffent re-



15 sin 5t.



Determine



if



or leads or lags o, and by whai



r0,9



,, = 5 cos (4t - 60'), 0, = 5 sin 4r, : 10 cos 4r, ',0, = 5 cos 4r + 12 sin 4r. (c) t' = 16 1-. Ot + \6 sin 4r, (a)



(b.)



Find o! ftom the differential equation and use the rcsult to 6nd tbe forced response o to an irput voltage of (a) 34 cos 4t V, and (b) I7 sin 4t V-



20'l



0r=4cos4t+3sin4t.



r0.4



Find r'., using only the propffties of sirusoials, if (a) t, = 6.cos 3r4, i - 4 co6 (3r - 30') A, and rr = -4V3 cos (3t + 60") A, (b) t, = 5 cos (3t + 30") A, 12 = 5 sin 3/ A, and



i, =



ir =



+



5 cos (3t



25 cos 13



(3t



150")



A, and (c) i, :



A,



= cos (3r 22.6')';A. 53.1")



ruoa



2 sin 3a A, and (Hint: cos 22.6'



PROALEM I 0.9



0.02F



.



10.10 Find lhe rcsponse ,, to the souce 4es A and use the result to fird the response 4 cos 8r A, and (b) 4 sin 8r A.



o to



(a)



fr PROELEM I O.4



10.5



4 mH, R= 6kO, Vd= rad/s.



1Cl6



10.6



13 cos



f.)



10.11 A complex voltage input



10"p*a V pro, curent output of5l2 ,o1A. Find the output current the input voltage is (a)



10.4 if the sourc€ and lhe response is i67.4') mA. (Take tan 67.4" :



Z in Fig.



0000,



2 cos (6000t



-



4 cos 400Or 24 cos (4000t



-



mA and lhe outpul is 53.1) V, nnd R



chapter



10



and C. 'J



-



V,



if



(b) 4oejt?'+af) 4 sb 151v.



V



In the figure of Exercise 10.2.2, if dle source



rs



340



PROBLEM 1O.10



duces a



Find R and



is



10.7



l0.4ifr: 5Y.N\drD- 2x



Find the forced response i in Fig.



(2r



20c.x2t V, ard



(c)



10.12 Find the phasor rcpreseltations of the tirnedomain turcrions (a) l0 cos (5r + 18'), (b) -8 cos 5r + 6 sin 5r, (c) 18 sin 5r, and (d)



Sinusoidal txcitation and Phasors



-2



sin



(5r



l0').



10.13 II (, = 20 lad/s, find tbe time-domain f,rnctions r€present€d by th€ phasors (a)



(b\ -4



- js.



- i3, (c) 5 - jl2'



10.1{ solv€ Prob- 10.4 using 10.15



-5 + i5'



(d) 10' and



(e)



phasor circuit shown, ffnd Z@ 3nd u5$ the result to find Lhe phasor current l. !t (.) ? rad/s, find the forc€d resPons€ t co.resPond_ ing to I-



10.21 For the



-



PhasPrs



Fitrd the impedance of the circuit of Fi8. 10.15



if ihe time-domain functions



repre-



sented by the Phaso.s Y ard I are (a) o 30 cos 2t + 16 sin 2t v, 1.7 cos (2t + 20') A.



t:



=



rb)D=Ret,ietrlv.



Ret(l + /)"ii'?"rdl mA : aY. cos (@, + d) V. i:Y-co6ldt+6-a) A Show that if Ro Z = i is Positive, thon



i:



(c, o 10.16



Re



Y = Rell/Zl = 6



l0.l? A circuit



PROBLIM lO.2r



10.22 Find ihe reactance X so that th€ imp€dance real For this case. find seen b) lhe sourcc



^ lhe sreadv-state currenl r\r) cofiesponding to



is also Positive



I



ifo:1-0rad/s.



has an imPedarce



5rl + ,id)(3 + ,ir) Z=+ll ju\z + Jat



.



)



;'5o



i,;ron



Find the aesistance, reactance' condutance'



and sosceptance at {, =



I rad/s If



the



time-domain voltago applied to lhe circuit is 20 cos r V, 6nd lhe steady state cutrenl



t0.1t



A eircuit



-



has an impedanco



16(2



+ rirx8



-



-



15n'r



@'1



+



64



j}or\



PROALEM 1O'22



10.23 Find



thE steady-state values 10



nH



of i



and o-



20



nH



^



aid reactance at (l, = 1, 2, and 3 nd/s. If tbe lime-domain voltag€ applied to the citcuit is 64 cG c,t V, frnd the steady-state current in esch cas€. Final the resistance



PROBI-IM I 0.23



10.19 In Prob. 10.9, use phasors, irnp€dance, and



lo.al



volkge division to 6nd o Find lhe steady-state ralue



of i if (a)



&'



=



2 rad/s. l'Iote that in the latter case the impe-drnce sefl al th€ terminals of the souce is Pulely resistive.



I



rad/s and O) a,



10.24 Find C so that the imp€ilance s€en by the source is re3l. Find the pow€r absorb€d by the 1 2-O r€sistor in this case.



=



L- zu



8rr



10.25 Find th€ steady-state current i. 10.26 Find the st€ady-slate voltage o. 10.27 Find the steady-state value of i.



10.2t



Fitrd the steady-state voltage o.



10.29 Find lhe steady-stale fllrrenis



i



and



ir



usitrg



phasors.



10.30 Find the steady-state voltage



o if



o,



=



2-5 cos 8t V.



PtoatEM 10.20



10.31 Find the



st€ady-state



ldue of o.



341



t2



',



!u ao



PROn



_L-



o.z



ttt



n



10,2a



o.2 E



2A



0-05 F



PROBT-fM 10.25



IH



I



',



1lI



PROALTM



10:6



PROELEM



IOJ'



I



342



t.-



Chapter



10



Sinusoidal E\citation and Phasors



I



I.



!F 5;



+F



.



3i "



+



4rl



I



PIQEIIM t0.30



PRO9IEM t0.29



I I i



lka _r- -_ loo nH



PROBLEM



ofo



10.3-2



Find the steady-state \alues



10,33



Find the steady-sra& \alue of i when (a) a, = f rad/s, (b) ar = 2 Ed/s, and (c) r, = 4 nd/s. [.lote tbat (b) js the r€sonant crse.l



_



and or.



IOJl



10.35 Fitrd ttle steady-state currc i. 10.36 Fitrd ttle steady-statc vollage l0 co6 10001 V10.37 Find the stesdy-stde cur€nt



Fihd the stady-stste voltage o.



2 cos 200b



v



o if



o' =



i if



o,



-



PROTLIM 10.32



!-



4a



rH



jr Pto8tE$ roJS



343



+



t)"



|'oo



#A



fo.ore



PROBLEM 10.34



*F



EO



+



iu



5('



PROELEM I0.35



iur



PROBTEM'I0.37



PROBLIM 10.36



t0-38 Find the forced response i. [Sl/SSesdbtrr Note the/ | = v tz hils to wo* sinc€ z(i 1) = 0. Solve the describing equation by the method of Ser. 9.6.1



lO.9



Fitldthe conplet respons€ i ift(o) = 2Aand o(0) = 6 V. (Srsserrion: Use phaso$ to 8et and the differential equalion to get



i,.)



i



1..



PROBTEM I 0.39 and o(0) iD Prob. 10.39 so that the natuml component vanishes and i is simply dl€ torc€d compodent



10.,f0 Determin€ i(0)



?foBttM



344



10.38



Chapter



l0



Sinusoidal Excitation and Phasore



11 AC Steady-State Analysis



The lirst prabtical appllcation ol elec: lricity is said by many to be the telegraph, developed by Samuel F. B, Morse, an American porlrait painter and invenlor, Morse built on the ideas of the famous American



What hath God wrought! [The famow message tapped out on the rtrst



Design. But the previous ye€ir his



wile had died, in 1826 his lalher died, and in 1828 his mother died. The lollowing year the diskessed lllorse teleeftphl Samuel F. B. Morse weni to Europe to recover and study lurlher. ln 1832, while retuming home physicist Joseph Henry, using the on board the passenger ship Sr//y, he met an eccenlric inventor and beopening and closing of r€lays to produce the dots and dashes (or Morse code) that repre- came int gued wilh developing a lelegraph, the principld ol which had already been considered by Henry. sent lelters and numbers. Morse was bom in Charleslown, Massachu- By 1836 Mo6e had a wolking model, and in 1837 he setts, the son ol a minister and author. He studied to acquired a parlner, Allred Vail, who ,inanced the probe an artist at Yale and the Royal Academy of Arts in jecl. Their etlorts were rewarded with a patent and the London, and by 1815 he was considered to be moder- financing by Congress of a telegraph in 1844, over ately successlui. ln 1826 he helped found and be- which Morse-on May 24, 1844 sen! his newcame lhe lirst president ol the National Academy ol tamous message, 'Whai halh God wroughtl"r



345



In



.



. .



Cfrupr", l0 we have seen that in lhe case of circuits with smusoidal inputs we may obtain the ac steady-state response by analyzing the corresponding phasor circuits. The circuits etcountered in most cases were relatively simple ones that could be anallzed by the use of voltage-current relations and current and voltage division rules . It should be clea., because of the close k inship between phasor circuits and resistive circuits, that we may extend the methods of Chapter l0 to more ge dral circuits using nodal analysis, loop analysis, Thevenin's and Norton's theorems, superposition, and so on. In this chapter we formally consider these more geneml anaisis procedures, limiting ourselves to obtaining the forced, or ac steady-state response.



11.1 NODAT ANATYSIS As we have seen, the voltage-curent relalion



V



=Zl



for passive ele rents is id;ntical in form to Ohm's law, and KVL and KCL hold in phasor circuits exactly as they did in resistive circuits. Therefore, the only difference in anallzing phasor circuits and resistive circuits is that the excitations and responses are complex quantities in the former case and real quantities in the latter case. Thus we may anallze phasor circuits in exactly the same manner in which we anallzed resistive circuits. Specifically, nodal and mesh, or loop, analysis methods apply. We illuslrate nodal analysis in this section and loop analysis in the following section. EXAMPLE 11.1



To illustrate the nodal method, let us find the ac steady-state voltages 01 and o, of Fig. I 1. l. Fi$t we obtain the phasor circuir by replacing the element values by their impedances for (1) = 2 rad/s and the sourc€s and node voltages by their phasors. This results in rhe circuit of Fig. I1.2(a). Since we are interested in finding Vi and V,, the node voitage phasors, we may replace the two sets of pamllel impedances by their equivalent impedances, resulting in the simpler, equivalent circuit of Fig.



l1.2(b).



The nodal equations, from Fig.



2(v



ll.2(b),



are



- 5/o') ' 1 JrJt 1 h.v'jt V,-V' V' -



it ' ,t *



346



( hdpler



1l



AC Srea4-\rare And\.is



iz,ts=



s'



5&"



FICURT



'll;1



Circuit to be analyzed by the phasor method



5ld 4



FIGURE



ll.2



Ib) Two versions of the phasor (ircuit corresponding to Fig.



ll.1



which in simpllfied form are



- jlv' l0 '- ,-,'' jl)V,=s -jlv, l {l Solving these equations by &terminants, \te have



Iro - jt



v,.i|r_t-4n=:l-_rl -2 itv I'i,



lz+ iz I ._



v^,1-Jl '5 Secrion 11.1



i



NodalAnalysG



t-itl to



_l )l



=g+jm:2+



j4v



'



347



In polar form these quantities are



:



-



v, = .v. -



tfs24gv . 2\/=5163.4"



.,



v



Therefore, the time-domain solutions are



\6 cos (2r - 26.6') V o" = 2\/i cos Qr + 63.41 V br =



EXAMPTE 11.2



As an exanple involving a alependent source, let us consider Fig. i 1.3, in which it is requir€d to find the forced response i.i. Taking the gound node as shown, iwe fuve the two unknowir node voltages o and o + 3000i, a6 indicated. The phasor phasorcircuit circuit in its simplest form is shown jn Fig. 11..4. from which we may observe that only one nodal equalion is needed. \lliting KCL at the generalized node, shown dashed. we have



v-4



v



v 3000I j 0'r ' :{t - j2)(t0r) {2 jt){tdr - " +



Also. from lhe phasor circuit we have



4-V - j(ld) rl6uRt



r



1



1.3 Circuit containili8 a dependent sburce



|to



,



3000i



-L' FI6URE 11.4 Phasor circuit of Fig..1.1.3



,t



on



(2



-il)ka



i



348



ehapter



11



AC Sready-State Analysis



I



Eliminating V betwe€n these rwo equations and solving for



l, ve have



t=24xro3f!):L =



24153.1: mA



Therefore. in the lime domain, we have



i=



. EXAMPLE I



1.3



24 cos (5000t



Ler us find lhe lorced response



r



+



in Fig. I L5



53.1") mA



if



rs=Y.cosa)tV We note first rhat rhe op amp and lhe two 2-kO resisrors consdtute a VCVS with gain 1 +' = 2 (see Sec. 3.4). Therefore, D = 2D2, or oz o/2, as indicated by the phasor V/2 in the phasor circuit of Fig. 11.6.



:



m



tlGURt I1.5 ( ,rcuir (ontdrnrn8 dn op amp I ,rF



FICURE i 1.6 Phasor circuit of Fig. 11.5



-t1000/d



Seclion



11.1



Nodal Analysis



tfl



349



Writing nodal equations at the nodes labeled V1 and Y/2, rre have'



.



v, (v/2) v, - v u ' Airr'o) \4(lo) ' - irctY-= (\ /2) .v, v/2 -y5*, ' - twt,,= o



v (f



v.10: .



Elimindting V' and solving lor V results in



--



2V_



V=



= lr - (,,:/Itr)l + j(v2ul1o1)



In polar form this is



_-



2V^/0



Vl o=



(l l. r)



+ (o/looo).



.l



V1-/wn I



-tan-'[l -



(r 1.2)



(,,,/rooo),1



ln the time domain we have a=



-=:!!: Vt + (r.r/1000).



cos



lirr +



o)



(l r.3)



We might note in Example I1.3 that for low frequencies, say 0 o t



0:



rle 'dt



J(t)=3- J"[2Y\t



[Sr8gertorl The iltegral is the convolution of 2y(t') ALrw{ 3 a 6t 19.4.5



Solve for



r,



for



r>



0,



Ansvret 2e-' cas



Show



t



e-z.l



if



-r'lr) +



19-4.6



all,d



t'



r(t) t I t\rr?' 'dt = 0. Jo



tiat



elf_^l,t



: luu,^, f-^rt t *



rg



+



,r



(0)



-



2



r"r1?) d'



"]= * . l"' rot f__tr,



"]



19.5 rHE IMPUI.SE FUNCTION frr we have obtaine-d a number of Laplace hansforms, ranging in complexity ftom 14, the transform of the unit step futrction, to some very complicated lirnc_



Thus



tions arising from the tmnslation thoorems atrd the convolution theorem. The reader t, is missing. In may have noticed, horaever, that one very simple transform, F(s) this section we will consider this tmnsform, and we will subsequently see that its coresponding time-domain function is orc of the most useful functions in circuit and system theory. The function whose transform is I is called the impube function a\d is denoled by 5 (r). That is,



:



9[6 (/)]



=l



(re.21)



The impulse function is not a fimction in the conventional sense, and indeed it was origina y called an "improper function' by the gteat Bdtish physicist Patrl A. M Section'19.5 The lmpulse runction



i I



I



639



Dirac (1902- ), who is noted for his prediction of the existence of the pocitive elec_ tron. or positron. but who was also a pioneer in lhe development of *rc ioDr se lirnction. The impulse funcrion catr be defned and its propertes established rigorously using generalized or distributior function theory. We give a plaulibiliLv ar8u_ ment for its existence arld use it quite successfully ar an otdinary frinction.



' -



EXAMPLE



19.'5



Irt



us_ begin by considering the finite pulse of Fig. 19.6, centered about the origitr, of width d and height 1/a, defined by



els€where



al? o FICURE



al2



19.6 Finite pulse



The area under the pulse is t and rcmaiDs fixed regardless of the \"alue of a. If a is made smaller, theD the base of the pulse shrinks and its height incr€as€6, m4intainittg the comtant area of l. In the limit as d tends toward zero, the puls€ approaches aD infinite pulse occurring over zeto time but still associated with an area of l_ This is the impulse firnction, soEdetimes called the llrut impuls€ function to iddicatc its relationship to the unit arca. Crdphically, it is represenM by an arrow erEct€d at t = 0, as shown iD Fig. 19.7(a). A more general function is 5Q r), which is an ihpulse occurring at t r, as indicated in Fig. 19.7(b).



-



:



FIGURE



G)



640



Chapter19 LaplaceTransfoms



I9.7 lmpulses



(a) 6(t) and (b)



6(t



- ')



Itr Fig. 19.7, th€ impulse llmctions are of,rtrarrSrr, 1 because of their association with the area of I in Fig. 19.6. ,{n impulse of strength,t is denoted by l6(t) and is .epresented by an arrow marked (t). Mathematically, the impulse function is defined by



t+0



6(r):0,



(1e.28)



f-uato,=, or more genemlly by



- t) = O' I at,-,tat=t 6('



t+



r (1929)



J_^



Thus the impulse is zero everywhere except at its point of discontinuity, at which it has an area of I concentrated. Physically, the impulse descdbes very well a force of very larye magnitude exerted for an extremely short time, like that of a harmer suddenly striking a mass on a spring to put it into motion. An importaDt property associated with the impulse function is the rdmprirs, or sifting, prorf,,tty, desc bed by



,)



l.'tu,ur,



dt:



(19.30)



fG)



:



r. This propqty may be made plauwhere a < ? < r and/(t) is cortinuous at t r) is zero except at t = we may write sible by noting that since 5(r



-



'.,



fb t" I fk\6k ' r\ dr = | lk)50 - t\ dt Thus



if e is sufficiently small,



mately/(r)



betwe€n



r-



/(t) is cotrtinuous at 7, then /(t) is approxir + €. Thus we maX factq/(t) out of the integral,



since



€ and



and by the nature of the impulse function we have



fb



I .f(r)5 (r J"'



fA.



fll) J.|



r) dt = =



av



,



-



11 ar



fG)



The sampling property is very us€fu1 in determining the Laplace transform the irnpulse fuDction, giv€n by



9[6(r)]



- |



c-"d(r)



of



dr



JN



Section



t I



19-5



The lmpulse Function



641



(Recall that in the event of an infinite discontiuLity of the integand ar 0, we are us, as the lower limit.) By the samptng property, with 0, we have



ing



0



"



g[6{t)l



j



=P-"I =l



ll9.]lr



which is (19.27). The shifted impulse has the more general fiansform,



f I



Yl6{r r)l which rcduces to EXAMPLE 19.-I6



Ib



I



for



h



e



"5lr r)dt



i = 0.



illustrate how an impulse function may adse, let us invert the transform



I(r) = --L s+2 This is an improper fraction (the numerator is of the same degree as the denominator), which has not been encountered thus hr. However, by long division it may be



wnllen F(s) and b5 lineariry ils inverse



= t --ls+2



is



t(t) = D(,)



2e "n(t)



At this point let us relate the impulse function to another well-known funciion by recalling the expression for the tmnsform of a deri\ative,



Jll'{ l 0



rF{r)



- /(01



where 0 has been replaced at for the case of a discontinuity at r(rr, lhis relarion becomes



f(t)



'



(1q.32)



t = 0. If



we let



/l\



llu'lt;l =r{:l -0=l \r/



Thus we would like to conclude that



du(t\



-a- = o\tt



(le.l3)



since they have the same hansform. Unfortunately, this does not follow from dgorous mathematics, treating r(r) and 6(t) as conventiooal functions. However, (19.33) may be established using generalized functior rheory and is a good plactical tool used extensively in systems and circuit analysis. A plausibility argument for its validity is given in Exercise 19.5.2.



642



r



chapter



le



Laplace rransforms



If



we formally replace/(r) in (19.32) by



6(l), we have



l8l6'(t)l = s9t6(t)l



-



0=



s



(19.34)



Thus 6 '0), called a dolrlet, has Laplace transform s, and thus is a us€fuI function, though not a conventional one. The functions 6(t) = u'(t\. 6'(t) = u'(t\ "(t),and may be shown ^nd formally, but are members of the family of singular functiow, not conventionally, by means of (19.32) to have the transforms



'l+P]=sn,



n:0,1.2,...



(r9.35)



The step and impulse are the most often us€d members of the family, but the doublet may aise occasionally and, in any case, it is useful in our later discussion of stability. Like the step and impulse functions, the doublet has a graph that cao be established by the following plausibility argument. Consider the triangular pulse p (r) of Fig. 19.8(a). Its arca is l, as was the cas€ for the rectangular pulse of Fig. 19.6. ff n --t 0 in Fig. 19.8(a), the area of 1 is preseNed, and the legs of the triangle shdnk toward zero while the altitude tends toward infinity. Thus the shape in the limiting position of Fig. 19.8(a) is precisely 6 (t), a pulse of in€nite magnitude occurring over atr itrteryal of zero time. It is possible then to take



att):



|gi a(t)



and thus



6'(r)



:*[m"r]



If



we could interchalge the.deri live and limit operations in this last expression, we would hai,e



6'(t) = lim p'(r) FIGURE



Section



19.5



19.8



(a) A triangular pulse and (b) its derivative



The lmpuhe funclion



643



*hich by Fig. 19.8(b) is seeD to



be two pulses tenditrg towatd zero bases and infnite heights in opposite directiols. Thus we are led to lepresetrt the doublet by the double anows of Fig. 19.9.



FIGURI 19.9 Craph of the doublet



Anolher usefij resull concerDhg the impulse function



is



(1e.36) /(r)6(r) = /(0)5(r) where /O is cotrtinuous at r = 0. This is plausible sioce D(r) = 0 for r + 0. Ir aaises,_for exaEple, when we differeDtiate functions that rlry as a facto!,



formally using the rule for differentiation of a Foducr. EXAMPLE



19.17



"ont^irr



bt us fiod the cur.ent i (r) in the inida[y relaxed circui! [i (0-) = 0] of Fig. 19.10 if the input is o,(r) 6(r, V. Tbe current in this case is the r.nrpulse respols;(response to a unit impulse), which will b€ considered in detail i! Sec. 20.4. The loop equation



:



is



ditt\



i*zt\tt=aut which trnnsforms to



sr(r)+2r(s)=l The curent tmnsfolm is



Irs) '' = and the



I



-----



r+2



curent is



iQ)=e'L(t)A TIGURE 19.10 Rl ci.cuit



i(r)



's(\



644



I



Chapter



19



Laplace Transforns



(t9.37\



To check the answer, we note that the initial condition



t(0-) =



0 holds. To se€



if the differential e4udion is satisfied, we need



di ;= which by { 19.36)



,,.



e



2e )'ubt



'6\tl



is



di ;= Substituting this expression is formally satisfied.



ad



^. 5k\



-



22



'uu)



(19.37) into the



diffqeltial e4uation, ."" ,"" thut it



EXERCISES



19.5.1



E%luate the integral



| (r'-



3 cos 2tA



(.)



J"



where (a) a .tnswer



19.5.2



(a)



: -l



and (b)



d



:



It



1.



3; (b) 0 Show that the pulse of Fig. 19.6 is given by



-



ub + \o/2)l so



ub



-



tu/2tl



thal ormioJ



\rt



=



du(t)



-



This is a plausibility aigument for establishing



duk\ ot = -a-



19.5.3



Use the results of Exercis€ 19.5.2 to obtain formally



)



lldt' f lttu\r)l '- f'\t)u(, t Iul6\tt :f (tl,(r) + f(o)D (r) [If"f (0) = 0, this result becomes



!f(,)o(,): dl



19.5.4



Use the result



we wrile



f,(,),(r)l



,



of Exercise 19.5.3 and (19.9), which, because of the discontinuity,



as



S.f'@l:s%f(il- f@) lo derive 9[6V)]



Section



I I



I



19.5



-



I, where/(r)



The lmpuhe Function



-



e--rl(r).



645



19.6 THE INVERSE TRANSFORM From the first sections of this chaprer, it is cl€r that we may compile a lengthy table of functions and their Laplace tlansfolms by repeated adications of the defrnition of the transform and the theorems. We may then obtain a wide }rriety of inverse tmnsforms by matching entries in the table. This is our objective in this sectiotr, and for this purpose w€ use Table 19.1, which contains most of the common futrctions /(t) and their transforms F(s), which we have derived Feviously. (Enry 10 follows directly fiom Exercise 19.3.4 with ,i replaced by 1.) For ready reference, a more complete table of tmnsforms is given inside the back cover of the book. Since we are interested only in tunctions defined for r > 0, we will omit the factor a(l) except where it is n€cessary, as in the time tmnslation formula (19.21). We have retained &(t) in intry 2 of Table 19.1, which for r >0willbe used as



l|



'



gUJ:



!



we have omitted the factor n1r; in tfre other esntries of the table. TABTE



19.1 Short Table of laplace Transforrhs



JO



FG)



1. 6(r)



I



I



2. ultt



I



3.e"'



k



'17



;,+i,-



Cfril



+



t,



Gi.f+t' 8.r



1



F I



(r +



to''(, EXAMPTE



19.18



4),-



I



I)!



(r + a)'



n:1,2,3..



To illustrate the use of the table, sup!'ose thar we are requted to find the inverse



the tmnsform



Frs)



646



chaprer



le



Laplace rransfo.ms



6 +_l l =r+4 s'+9



of



Since the invers€ transform is a linear operation, we have



2 -l1_-13 I "1111 = 659'l--L111;s-, lr+41 3 [r'+3'l



JJ-



lrl| ,rlsl



which by enfti€s 3, 4, and 2 of tlle table yields



IUI = 6c-a' I 3sin3, EXAMPTE



19.i9



3



Let us invert the tmnsform



trt.rl=-



2tr



+ l0)



(r+lxs+4)



There is no direct entry in Table l9.l that we can use to obtain /(t) in this case; however, we may obtain a partial ftaaion expansioD of tr'(r) and apply the table to each term in the expansion. Obtaining a partial fraction expansion i3 the opposite operation of getting a common denominator. That is, we ask ourselv€s what simple fitactions add together to yield F(r). Since in this case the transform isa prcper fractron (the numemtor is of lower de$ee than the detrominator), the paftial fractions will be proper fractions and mu$ lherelore be of the form



2lr



+10)



(s+ l)lr-4)



A s+l



R



rl4



The constants ,4 and B are determined so as to mate the s€cond and thfud members an identity in r. The simplest meads of determidry A a4d B is to note that



rr



F



r)F(r)=i++.=o ,'l';*;"



and



{r r



i++



-



4rF{s)



=



ol'**,ot * u



r. let us evaluate the 6rsr at J - l. = -4. t\.hich etiminates A. The results are



Since rhese musl be ideDtities for all



elininales B. and the second al.r



,4-(r i



l)F(r)



I



4)F(r)



(r



r



which



II ==':6 ztst lr,



I| =--+= ztat t,--. -l



4



Tterefore, we have



f,'(.l '



6 4 = r+l - s+4



and,.by Table 19.1,



f(t\ = 6e' Section



19.6



The lnverse Transform



a€



'' 647



we should note that the formal computation notation for what we are actually doing, which is



A



=



fim,



ofA



and B above is a shorthand



fr + l)Ffr)



and so on.



As lhe reader may recall from partial fraction expansioDs in calculus, Example p in the de19.19 is a relatively simple case of distinct linear factors, such as nominator of F(s). The pole s = p is in this case a rirpie pole, a a pole of order one, and in geneml for each such simple pole the partial fi:action oxpansion contains atermAf(s p). In the case of a denominator factor (r p)',vhercn:2,3,4, thepoles: p is a multipla pole, or a pole of order D. The factors comprising p)" are thus no! distinct, and the partial fiaction expansion, of cause, must be modified. We will consider multiple poles later.



J



-



..., (s



EXAMPLE



19.20 As



-



another e-xample involving only simple poles, the expansion



of



t"



F("J=(,+rx;2x"+) is given by



(, + lX"-- 2)r,



I



3rL



\rhere



=



;+ t, -



-



"j'



3-l



2r 1"*zrtr.rr l,_,--, zel B=t"*rx"*trl-:-,=o



A=



(: =-l The inverse rmnsform



*



*."rii"



r



2sl



= l)(r + 2) l"--3



I



3e'



.f(t):-e'+4ext-



Complex poles occur in coojugare pairs and their corresponding coetficients in if F(s) has simple poles t iB and an expans;on



the expansion are complex cotrjugates. For exagple,



a-



F(s)=t rrA-JP .-* s-a-JP :B then



e: (s a



I



ia)n(s)



I |



648



chapler



1e



taprace



rhnsfcrms



":"+n



:



and



I =k-



a+ Tprrr"r



I



l,-"-ia



I



From this we see lhat = A*, because F(s) is a ratio of polynomials in coeffrcients. The inverse transform is



I (t)



=



Aets-'tu



t



A'eta



r with



real



rqtt



which is the sum of a complex number and its conjugate. 'I'llerefore,



f (t) = 2ReIAea*iF\ $A-



lAl?n. then we have



f(t) :2Re\lAle*enq"'tl = 2lAle- cos tg + el



. EXAMPLE 19.21 The transfom



Flsl: (s



s



(re.38)



t l(r?+tu+2)



has the partial fiaction expansiod



Fhl' = (s+



r



l)(s+ I



- jlxr+ I + jl)



ABB+ I r + I - jl ' s + I I jl



r+ where



.t a=",***rl,_ =-t and



B=



I ='-;':!r*r... =----==-L(s'lXs+l+ jl)1,. 2 \/rI-,



Thus we have



!u)



An altemative form is



= Ae



'+



2



RelB'r-'-rr)rl



[r ltr | - -e + 2 Rel'-c-' LV2



: -e , + \4,e-'cos (t -I



f(t) : -e-' + a4e-(cos r cos 45' + :-p'+e-,(co€r+sinr) Section



19.6



The lnverse



Thnsform



45"



I



I



45"\



sir r sitr 45")



:



(le.3e)



649



EXAMPI-E



19.22 If ir is desimble to keep the quadnitic factor intact, Frr)



+ D is



where Cr



- l,



A=



r =-tr+ rft"'zr+



zf



the expansion of ( 19.38) is



A cs+r' -"+ I +"'+2r+2



the most general numeratoi for the quadratic term. As before



and to find C and O we ctear 119.+O) of fractions, obtaidng



s= -(s,*



+2) +



(Cs



+



D)(s



which is an identity in s. Equating coefficients of



s'?



gives



- -l



0 from which or D



{19'loj



c=



I



Eq'ati'c



: 2 The rrrssfo'm



F



+



1)



c



"*tr;el,_Ii;r.:



is tberefore



('):*



-?;h



s+l -l =ril,(r+ll_t-f"tll+t



1



The inverse tansform is the tunction of (19.39).



Multiple poles from denominaror hcrors {r T a)'requle a partjal ftaction expansion



+ A-' , *...* F(s)=--1{J+ a)' {s + af-r



1 ,.. 1-IL.. 's+a



--4+ atr l,r



p,1r1 (19_41)



where E(s) coresponds to the remaining poles of F(s).



# EXAMPLE



I9.23



From



F(s) =



(r'l)'z(s+2) ---i-



(r9.42)



.A



B (r+ l)' r-l



. coef€cients



4



C



s,2



aod C may be obtained easily. as



A



-(r



+ l)':F(s)



lt)



|l" = '^l I l*'l--,



=+



and



c.=



650



Chapter



19



.l4l



(s + 2)F(s)



lapla(e Transtorms



l,_,



=



u



m



1., =,



i



I



0=B+4



I



so that



I = -4.



Tbe transform of (19.42) is therefore



444 l)' - s+l



*



F(J) -(s-



t



-



2



so that



f(t\: $te' -



4e-t + 4e-2t



The frrst term on the right is obtained using entry 9 in Tatile 19.1. .We may also find B in (19.42) by giving s any vdlue other than s -. 0, for example. { 19.42) becomes



4



4.B



r(r)=i*



-l



or



-2. If



4



i-t



fromwhichS-'-4.



,



Wb consider next a general method of finding all the coefficients corresponding to mrltiple poles. we begin by multiplying (19.41) through by (r + d)', resulting in



(s + a)'F(s) = Differentiating



r-



A"+ A" t(s + d) + . ' + A(s + a)'-i + . . . + A(r + a)' I + (r + a)"F1(s)



.



,t times gives



IafF(r)l - (n-*)!Ar - {s -



-l{r



(19.43)



arG(s)



where



(r + arc(r,



)^L



-;-lAr-,(s as-



E\aluating (19.43) al



r = -a



A'= Section



19.6



+



a)"tt+.



yields. for



I



.



* = l.



A'(s + 2. 3. . .



ar'r+



.,



(r



-



a)'F,fr\J



n.



d"r



rn_^^"-tLls+ ,)'o(t,l,__,



The lnvers€ TRnsfoftn



'651



EXAMPTE 19.24 To illusaate this r€sult, we find B in (19.42) to be



tl



I dz"t ' __t__1 "!: - t2 - tttar''ls + zllt1.., tl



'.1 = 15,2),



1,,



=-4



whicb checks lhe previous answer. EXAMPTE T9.25 Multiple complex poles are handled exactly like multiple real poles, the only difference being that complex numbe6 are involved. For example, the transform 4



tr + I,(r')-



:



2s



+



2)l



A . B B+ C _ f , . s+l lrrl- jl)': {r+l jl)2 srl-jl'r+l+jl



.



has coefficients given by



,l I



"-(r'+2r+2)'l 4



(r+lxr+1+j1)'z



-4



|



=;,=,100'



and



. - r,1*1,, -' - r',,utTtr+ -



-



'4



4l ll asl{s + trt., i r i jlll l,_



il



rtl 1,,,,



,,,,



,f + tir2, ' r r l - ill l 'L t, I l)?ls - I + jlP I l.-



:_,



rs



,.,,



Therefore, lhe time-domain funclion is



I\tl=kt I 2Rel{119ff)te' r-rrrl + 2Rel 2?r-r*'rr'l - 4e-' + zte ' cos (r * 90') - zle-' cos t = 4e-t - 2te-t sin t - 4e-' cns t EXAMPTE



I9.26 l,et



us rnverl



F(s)



652



Chapter



19



taplace Transforms



C* (19.45)



We note rhat Lhe d€gree of rhe numemtor i\ lhe same as that of the denominator. 50 that long division is required. It is not necessary to actually perform the division since it is evident that it will result in 9 plus a proper fraction. Theaefere, let us write 9s1



F/.\



(s



---+ lxr'+ 2r + 10) A Bs+T""



-g+ r'l



|



(19.46)



r'+2r+10



We find A as before, from



a=ts I l)Fts) _J_2-q r_I0 _ To obtain B and



9\l



c,



Iet us multiply (19.46) by the denominator of F(s), obtaining



li{s) r



-9{s



The coefficients of



|



t)



2r- l0r-(s'?-2r-



lot



-{BrrC)rs+lJ



are



0=18+9-1+B hom \hich



I



26. The coelficients of ro are



yielding C



-



Therefore. we have



0:90-10+c -80.



26r + 80 I Fir)=q_"ll_r,+Zr+10 which we rearmDge



Frr)



as



r I ., t I t 3 I -q .--: *l;-, ,y,r,l-l8l r,_ly+y-l



Therefore, by Table 19.1, we have



/(rl



= 96r/)



- e-'tI '



26 cos



l, T



18



sin



lr)



As observed earlier. an explrcrt relation for lhe inverse lmnsform is gilen in purposes we may generally use partial fractions and Table to obtain the inverse tmnsforms.



Proh 19.18, but for our



l9.l EXERCISES



19.6.1



Find



tle inverse Fansform of



3r+7



.



ru/ (b)



("



+ltr



+ 2X, +



,



.



.J



(s' + lxJ: + 4)'



Selion



19.6



The lnveue Transform



553



| 2lrr | 41s t '78 and lc)r4 i5J1 gxr, (r- - 2r - 5) ..r'*5r:+4rF4 'o' t(t''4' Answet (^) 2e ' - e 2t - e r'; (b) l(cos t - cos 2t); (cr 5(t) I cos 3, lrn Jl I 2P 'cos 2r: (d)r+l+2sin2r 19.6.2 Find the inver"" t*n"for- of ?@ s(s + l) Answer 26'(t) - 36(t) Q- e-')u(t) 19.6.3



Find B and C in (19.46) by subtracting the first two terms in the right member from the left member. (Note that A = - L) An'w?t 80 Given the transform 2r'+ 5s + 6 P11 (s + l)'z(s + 2) find f (t).



B



19.6.4



26,C



" =



Anlwa



19,6,5



3te' k'+ k



''



Find rhe inverse transform of



rrrr Ans,tel 2



-



2cost +



i(sin,



- a',1";i



lcost)



19.7 DIFFERENTIATION THEOREMS In this section we complete our list of Laplace ffansiorms and properties and summaize the latter in a table for easy reference. In particular, we derive two results, using in one case differentiation in the time domain and, in the other, differentiation in rhe llequency domain. In Sec. 19.1 we derived the transform of a dedvative, which we repeat as



st f '@l = If



we formally



replace



/



by



/



'



sF(s)



- / (0)



0e



47)



, we have



ll f'k\ = sgl J tttl



/'(0)



or, by (19.47),



gL f'G\l = s"F(s) We may replace/ by/' again general result being



9t f{,t(t\] = s'F(s)



654



Chapter



19



Laplace



-



T6nsfoms



- sf(o) "r'(0)



in 19.48, to obtain gt /''(l



s"



(



Y(0) r'



7'(0)



(le 48)



)1, and so forth, with the



- . .. - /c-1)(0)



(19.49)



t€ ;r" $i



,i'



:



where/(") is the ,th derivative. The tunctions f, , . . . , f6-t) ue usumed to be continuous on (0, @), and/(') is continuous except possibly for a finite numb€r of finite discontinuities. We note that the sum of the exponent on s and the order of the derivative of/is atways, I in the right member.



f'



EXAMPLE 19.27



If in (19.49) we let /(t) = t', fo a nonnegative integer, ,f(0) =.f'(0) =... = fa ')(0) = 0. Therefore, we have 9lntl = or



Sfl,"l a result we obtained



EXAMPLE



19.28



[,er us inverr the



then



/(')(r) = zl



and



""91t"'



lnl



n.0. 1.2....



- --yln:l - -=:



(19.50)



in Exercise 19.1.4 by another method-



lmnsform F(r)



:



f)



r.l. +



I)..-



which has the partial flaction expansion



F(s)-4,+



B,*\ 2 J. J'



J



J '



I



The co€mcients may be found by the method of the preceding section, but instead, we will illustratq an altefiative method that is very effective for high-order real poles and is based simply on long division we carry out a long division of I by I + r until a remainder is obtained of degee equal to the order of the multiple pole (in this case 4). The result is



r,,,



=91-]-l s'L' s'lt+sl =6.1,



s-,'-,'+ 1".l ' s-ll



6t'666 s's+



1



ot



1t



Fr.rr 1 J'



tl



tr



A



6



l-:l- '6+--+--:J' r .t+



I



The first lour lerms in the righl member are of lhe form ( lq 50). and thus the inverse tmnsform is



f(t\:t'



3t'+6t -6+6e-'



To obtair the ftequency-domain differentiatioD formulas, we the Laplace ffansform with re.spect to r. That is,



/F(r, (t



1i Section19.7 Difier€ntiationTheorcms



will differcntiate



("



=a1"t"'"'o' 655



where F(sl



EXAMPU 19.29



=



If/G) - z(r),



g[/rrll. then F(r)



:



1/r, and



vr,,r,rr= f(l)=11 lf/(r) = cos *,. tben F{r) - r/(rr + 11,. and w€ have :ft'



cos



tul =



)/ J -:l- r", r' t,/t _ _(", _ k1, ",



We may rcpea(edly differentiate the transform lo obtain the geneml case



d"F(-rl - :-::: = f* t



_t).JQ)l?-",dt



from which we conclude that



sttf(t)): EXAMPTE



19.30



If/(r) = si! ,, then F(r) =



t-rrff9;



l/(s.'z



+



n=0,1,2,...



1), and we have (for n



,), gtt' sin ,l = (-Il:l--:l l \ ' dJ'\s' + l/



-



Table 19.2.



656



Chapterlg



taplace.TEnstorms



:



2)



(19.51)



TABLT



19.2 Short



List of Laplace Transform Prcperties



llt)



t



cf\t



2.



l,O +



F{r)



.r{r)



f,(t)



1{!



I,(s) + L(r)



,



rrG)



4 {!.!)



- /i0)



s,FG)



-r"-7(0)



s. I J\n



r(,)



d.



9.



/(c,),.



rc.



t"f(tl,



R



€-"rG)



= | J\t)s( - ndr



J,



110)



F(r + 4)



6. e-"'f(t) 7. f(t t)ult tJ 8.



y(o) r' -.



s"



F(r)c(r)



l.1r)



>0



'=0,



1,2,...



(-r)"Fd(r)



EXERCISES



19.7.1



Find the I-aplace transform of (a) t4, O) t5e-3', (c) t sin,tt, and (d) t'zsin



24 120 2k 2(ls: Answer (a)-.. (b) I *;(c) ,-;(d)-:-(s' _ (r J)" (r' r" - t /t'.,' .4-r\_4rr_6r_2 19.7.2 Find the inverse transform of L s'?(s + 1)3 Answer e-'(t2 t+l)-2t 19,7.3 De ve, lot r = 1, 2, 3, . . ., entry l0 of Table 19.1:



I



t.



)



1.,



I='''"'o' *-,1 L(, +' ,)"1 (, - l)l 19.7.4



lsuagerfibr. Use (19.50) and Foperty 6 of Table 19.2.] Use the method of long division considered in this section to invert the hansform F{sr



19.7.5



=llr



! l).



-L



(Suggestion: Change the rariable fiom r to p by letting r + 1 = p, perform the long division, and change back to .r.) Answet f(t) = 6 tte t 6e-'3t2e ' (19.51) n by integating the transform, wdtten in the Extend the formula to = -1 forrn



-



6te-t



-



FG\-ff(.n".at 657 I I



I



I



t-



between the limits -r



:



r



= 6 tq obtain



to .Y



r[tt]= L,I [ ,,,,* J,



(Suggestion: Change the order of integation iD the double inregral.)



19.8 AppUcATtoNs To tNTECRODTFFERENTtAt EQUA ONS



-



Like the Fouder transform, the Laplace ransform may be used to solve differertial equations. However, the Laplace tratsform method has the advantage of yielding the complere solution with rhe iritial conditions accounted fo; ;uromatically, as we shall see in this section. This use of rhe Laplace transform is one of it. .ori applications. "f"g;t



.



Evidently, if we tnnsform bolh merDbers of a linear differential equdtion constant coefficients, the_result will be an algeb@ic equation



with



in the transformed vari-



we may \ee-this from lhe generat tJrrnrta r t6.+sj foi ,f,. *^1rr. of ,m 1bte. oeflvarrves. lhts tirrmula also shows lhal the inillal condirions are aulomatically taken into account. we may then solve for tf," runrfo.,n oi*r"_unfnown and invert



it



ro obtain the time_doma.in ansiw_er. We have, in facr, atreaay consiaerea simpte lustralions of this t)pe earlier in the chapler.



EXAMPLE



Ie.3I



ter us nnd rhe sorution,,,,. rl1, _ro,o;



r(0) = t,



Transforming, we have



I lT],;:; " x'(o) =



r,Xrr, r-2I4lrx(s)_ 1+ fiom



which Xl'r



ii-



*"",,"",



2



3X{r)



_-a



r+2



8r t 13 = rs rr'1 lXs t2)(s_J)



The partial.ilacrion expansion is 1151



=a-l-l r+l s+2 .rrl



t > 0. is x(t)=3"'-e2t-e-3'



so that the time-domain atswer,



fot



As we have^also seen. \re may handle certain inregrodifferential ecuations di_ _, rccuy wlrnour orflerentatrng to remove lhe integlals. we need only transform the integuls by means of



sl[rc'*1-:_l LJn I s



658



Chapter



lc



lapldce rran{orm.



EXAMPI.E 19.32



we will illustrate ar integodifferential eqration with the circuit of Fig. 19.11, where we will find the current i(t) if there is no initial stored eneryy. The equations' are



di t' -dt+ 2i + 5 tI idt i(0)



:



u\t) 0



lha[sforming, we obtain



rl(r) + 2I(r)



r



sl -:l(r) -JJ



l[



4r'-r'/I2J.5-2l(r+



2



]



l)r +



41



Therefore, the resPonse is



i(fl =:?'sin2rA 2



'(') I I



FICURE



EXAMPTE T9.33



Ir.t



19,11 RIC circuit



us solve the simultaneous equations



'dr



;+r+4J=lO r



d.t'



/r' =-1,=0



r(t) and y (r), given that r(0) = 4 and f (0) = 3- we could filst eliminate y and solve the resulting second-order equation for r. \ye would then ne-ed a second coDdition otr r, which we would have to find ftom the differential equations. trt is easier with rhe Laplace tmnsform method lo lransform bolh e4utions firsl and lhen solve for



tor X{s). the ransform of.rlt,. The ransformed equations are



rxtr) 4 r x(r) -



4Y(r)



5eclion19.8 ApplicariorstoldtegrodifferentialEquations



lo



=:-



659



ftom which we have



t0



4



-+4 ,t



After some 5implification. lhis yields



!r"r=The partial fraction expansion



4c)+2rl t0 rl(r-l)z-41



be sho*n to be 'rray



2 2Is - l) X(sr=:+ s (sl l)?-4



4



(s tI)1



-4



from which the time-domain solution is



, In



=2+



2e-'(cos



2t



-



sir.



a similar manner wo may show that th€ transform



.. = 2+ s+l s (r+l)'+4



Yls) '



2t\



oly 2



+.............._ (r+l)'+4



and thus that



J



:2



+



e-'(cos



2t



+



sia



ztl



EXERCISES



19.8.1



Solve



(a)



for,r(l), for r > 0, by.using Inplace tansforms:



I' t 4x' I 3, - 25(t) r(0 ): x'(0-) = 0.



(br{+4x'+3x=4c''



r(0)=x'(0)=4, (e-' - e-t)a(t); 6) e-t(2t + 7) - 3e t' 19.8.2 Find D for t > 0 using Laplace tmnsforms if o(0) = 8 V. Answ 4 | 4e-'V Answer (a)



IXERCISE I9.A.2



660



Chapter



19



Laplace Transforms



is



19.8.3



Use Laplace transforrns to find



x(t)



and



)(t):



x'=x*2y J':2x+J where r(0) Answer



1=



=



0, y(0)



"t' -



e



=



2.



't! = et + e '



't9.9 SUMMARY ln this chapter we have delned the Laplace traksfom, which,like the Fourier transform, is an integnl opelation. The Laplace transform is thus a linear opemtion and can be used to tranform time functions into funalions of the cotuplac Irequenc) s.ll is a genemlization of bolh the phasor representation and the Fourier tmnsform, and is especially well adapted to anal)zing electric circuits- Translatkm theorems arid scale-chanSing propertier simplify the calculations and enable us to find an entire table of Laplace transfotus. Convolurion m y & lused to obtain more transforms or to solve directly for the output of a circuit whose output tmnsform is a product of two hansforms. The exttemely nsef]cl ittpuke functb, is easily defned and its properties determined by means of Laplace tmnsformsThe iwerse Laplace translrrm, needed to obtain the time-domain answer ftom the tmnsfom answer, may be obtained directly fiom the table of transforms or ftom the table and partial fraction expansion of the transform. The Laplace transform is ^ for solving both a powerful tool dffferchrial equations afud integrodifferential equatior?.r, and tlus is almost tailor-made for elect c cfucuit analysis.



PROBTEMS D.r



Find the lnplace transform oi the functions



19.2



Evaluate the integral J te"'dt and use the result and the definition of the Laplace tmnsform to find the tmnsforms of.the tunclions



(a)f(t\=1, 0