EE302Lecture15 - Me C nt C sh urre ircuit Analysis I n the...

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Unformatted text preview: Me C nt C sh urre ircuit Analysis I n the le se cture you will le s, arn: How to apply Me C nt Analysis: sh urre arn ntify and assign m sh curre e nts. • Le to ide arn t ultane alge ous braic e quations. • Le to writethese of sim arn xpre s rm • Le to e ss thequantitie to calculatein te s of the m sh curre in thecircuit. e nts To ide ntify and analyzecircuits re quiring a “S rm sh” approach. upe e 1 De finitions Branch: a path that conne two node cts s. Essential branch: a branch that conne two e ntial cts sse node s. Loop: any close conne d ction of branche s. Mesh: a loop that doe not contain othe loops. s r 2 Exam ple R3 Me 3 sh R4 R2 Oute Loop is not a r Me sh + - 1A R1 Why? 2V Me 1 sh Me 2 sh 3 Me Curre Analysis sh nt How m e ntial node (ne)? any sse s 4 How many essential branches (be)? 6 3 Number of mesh curre nts? I n ge ral, num r m sh curre = be - (ne -1) ne be e nts I n our case num r m sh curre = 6 - (4 - 1) , be e nts =3 4 Me C nt Me sh urre thod Me curre m thod provide anothe syste atic tool sh nt e s r m t o solvecircuit proble s. m Only applicableto planar circuits. • Not an issuein EE302. Usem sh curre as theinde nde variable and e nts pe nt s writeKVL e quations around e m sh. ach e 5 Moreon Me C nt Me sh urre thod Be causeof thepassivesign conve ntion, thedire ction of the m sh curre flow autom e nt atically de s thepolarity of the fine branch voltage . s Me curre arenot ne ssarily thesam as branch curre sh nts ce e nts. Me curre autom sh nts atically satisfy KC sincee m sh L ach e curre both e rs and le s e nodeit e nt nte ave ach ncounte rs. 6 Me C nt Analysis Me sh urre thodology 1. I dentify all of themeshes in thecircuit. 2. Definethecurrent in each mesh as flowing clockwise. 3. Assign each mesh current a name: (I 1, I 2, I 3) or (I A, I B, I C) 4. Using KVL, sumthevoltagedrops around themesh and set them = 0. Passivesign conve ntion is use for all voltageand arede rm d in d te ine re lation to them sh curre value e nt I n som branche thecurre you will useis thediffe nceof two m sh e s, nt re e curre nts. Moreon this late r. 5. Current sources imply constraints on themesh currents. 7 Exam ple Assum them sh curre flow e e nts clockwise. Apply KVL around them sh e R1 V Me e sh quation be e com s: -V + I R1 + I R2 = 0 Thesign of e te in theKVL e ach rm quation is thesam sign as first e e ncounte d in going re around theloop as indicate by thearrow. d I R2 8 A MoreC ple Exam om x ple R1 V1 R2 V2 I1 R3 I2 De two m sh curre I 1 and I 2. fine e nts Le look at m sh 1. t’s e C nt in R1 is I 1. urre S conve ign ntion follows I 1. What is thecurre in R3? nt t nt • Ne curre in R3 = (I 1 - I 2). KVL around m sh 1: e -V1 + I 1R1 + (I 1-I 2)R3 = 0 9 A MoreC ple Exam om x ple R1 V1 R2 V2 I1 R3 I2 Now, le look at m sh 2. t’s e C nt in R2 is I 2. urre S conve ign ntion follows I 2. What is thecurre in R3? nt t nt • Ne curre in R3 = (I 2 – I 1). KVL around m sh 2: e (I 2-I 1)R3 +I 2R2 + V2 = 0 10 A MoreC ple Exam om x ple R1 V1 R2 V2 I1 R3 I2 KVL around m sh 1: e -V1 + I 1R1 + (I 1-I 2)R3 = 0 (I 2-I 1)R3 +I 2R2 + V2 = 0 KVL around m sh 2: e For each mesh, thesign convention follows that m sh curre e nt. 11 A MoreC ple Exam om x ple R1 V1 R2 V2 I1 R3 I2 -V1 + I 1R1 + (I 1-I 2)R3 = 0 (I 2-I 1)R3 +I 2R2 + V2 = 0 Group te s to ge e rm t quations in standard form : I 1(R1 + R3) - I 2 R3 = V1 -I 1(R3) + I 2(R2 + R3) = -V2 12 Branch C nts vs. Me C nts urre sh urre R1 A R2 V2 V1 I1 R3 I2 I f weusean am e r to m asurethecurre in R1, wewould find m te e nt I R1 = I 1 I 1 is thesam as thebranch curre in R1 and is a physical curre e nt nt I f wem asurethecurre in R3, it would bee e nt qual to (I 1 - I 2) which e quals thebranch curre nt. Theam e r could not dire re I 1 or I 2. m te ctly ad 13 Branch C nts vs. Me C nts urre sh urre R10 R11 R4 R1 R15 R14 R5 R2 R18 S3 R3 R6 R9 R17 R16 R13 R8 R12 R7 S1 S2 Them sh curre I 5 doe e nt s not corre spond to any physical branch curre nt and cannot bem asure e d with an am e r. m te 14 Me C nts and KCL sh urre R10 R11 R4 R12 R7 S1 I1 R1 R15 I2 R14 R5 I3 R13 R8 S2 Notethat e m sh ach e curre both e rs and nt nte e e ry nodeit xits ve e ncounte rs. KCL is inherently satisfied. I4 R18 S3 R2 I5 R17 R6 I6 R16 R3 I7 I8 I9 R9 15 What about curre source nt s? S upposewehavea curre nt sourceinclude in them sh. d e R2 R1 i2 R3 + - C weuseKVL and sumthe an voltagedrops for that m sh? – 80 V e no But wealre know thevalue ady of i3 is –3A. R4 3A i3 R5 i1 16 C nt S urre ourceC ontinue d What is thevalueof m sh e curre i3? nt R -3A! Why? i2 R R + - The3A sourcete us the lls dire ction of curre in that m sh. nt e I t is oppositeto thede d fine dire ction of i3. 80 V C nt source cre a urre s ate constraint e quation in them sh e curre m thod. nt e R 3A i3 R i1 i3 = -3 am ps 17 C nt S urre ourceExam ple Me 1: sh (I 1 - I 2)R2 + (I 1 – I 3)R4 -80 = 0 Me 2: sh (I 2 – I 1)R2 + R1I 2 + (I 2 – I 3)R3 = 0 Me 3: C sh annot writeKVL, but I3 = - 3 R1 I2 R2 R3 80 V + - R4 3A S ubstitutefor I 3 and sim plify: (R2 + R4)I 1 - R2I 2 = 80 – 3(R4) -R2I 1 + (R1+R2+R3)I 2 = -3(R3) I1 I3 R5 18 Me C nt S m sh urre um ary Me e sh quations havea te for every e m nt in them sh. rm le e e Count the elements; count the terms! Think care fully about thevoltageacross e e m nt. ach le e Thevoltagepolarity corre sponds to thedire ction of them sh e curre nt. Le how to e ss thevoltagefor e ry kind of e m nt – arn xpre ve le e re sistors, curre source voltagesource com nt s, s, binations. Each te in a m sh e rm e quation is a voltage. Each curre sourceyie a constraint equation. nt lds 19 S tandard Exam ple Find thepowe dissipate by the8-ohmre r d sistor. 20 Ω 1A 8Ω 10 Ω + 20 Ω 10 V - 4.5 W 20 PracticeProble 1 m Find I X 1A 8Ω 100V + - Ix 2Ω 10Ω 4Ω 3Ω 6A 21 Me C nt Analysis sh urre S rm she upe e s Exam Proble -S rm sh ple m upe e S upposea circuit has a curre sourcebe e two nt twe n m she as shown. es I f wetry to useKVL to sum t hevoltage around a loop, s wee ncounte an additional r unknown: thevoltageacross t hecurre source nt . 10 Ω 3Ω ib 2Ω 100 V + - ia 6Ω 5A + ? ic 4Ω + - 50V 23 TheUnknown VoltageApproach x De unknown voltageV . fine Writem sh e e quation for m sh A: e 10 Ω -100 + 3(ia-ib) + Vx + 6ia = 0 -Vx + 2(ic-ib) +50 + 4ic = 0 100 V + - Writem sh e e quation for m sh C e: 3Ω ib 2Ω I f weadd the e se quations, Vx drops out. + x + - 50V -100 + 3(ia-ib) + Vx + 6ia = 0 2(ic-ib) + 50 -Vx + 4ic = 0 ia 6Ω 5A V ic - 4Ω -100 + 3(ia-ib) + 2(ic-ib) +50 + 4ic + 6ia = 0 24 TheS rm sh Approach upe e Me ntally re ovethe m curre source. nt Writem sh curre e nt e quations using t he original m sh curre e nts around there sulting m sh V e 100 3Ω ib 2Ω + - ia 6Ω 5A + Vx ic 4Ω + - 50V -100 + 3(ia-ib) + 2(ic-ib) +50 + 4ic + 6ia = 0 S eas be ! am fore 9ia - 5ib + 6ic = 50 ----grouping te s in standard form rm 25 TheS rm sh Approach upe e C nt in thecurre sourceis urre nt constraine d: 10 Ω (ic-ia) = 5A ⇒ ic = 5 + ia 3Ω ib 2Ω Me B: sh -3ia + (3 + 10 + 2) ib – 2ic = 0 S ubstituting for ic: -3ia + 15ib – 10 - 2ia = 0 -5ia + 15ib = 10 100 V + - ia 6Ω 5A + ? ic 4Ω + - 50V 26 TheS rm sh Approach upe e For thesupe e rm sh: 9ia - 5ib + 6ic = 50 S ubstituting for ic: 9ia - 5ib + 6(5+ia) = 50 15ia – 5ib = 20 Final Equations: 15ia - 5ib = 20 -5ia + 15ib = 10 6Ω 100 V 3Ω 10 Ω ib 2Ω + - ia 5A + ? ic 4Ω + - 50V i c = 5 + ia S olving, ia = 1.75A, ib = 1.25A, ic = 6.75A 27 TheS rm sh Approach upe e Weusetheoriginal m sh e curre to writetheKVL nts e quation! But weapply KVL around t hesupe e rm sh. Re e be wedon’t know m m r, t hevoltageacross the curre source nt . 3Ω 10 Ω ib 2Ω 100 V + - ia 6Ω 5A + ? ic 4Ω + - 50V i c = 5 + ia 28 PracticeProble m Find thepowe supplie by the2.2V source r d . 2Ω 4.5A + - 3Ω 2.2V + 5V 9Ω 3V + - 4Ω 2A 6Ω 3Ω 1Ω 29 Me C nt Analysis sh urre De nde S pe nt ource s De aling with De nde S pe nt ource s Wene d to e ss thede nde sourcevaluein te s of e xpre pe nt rm t hem sh curre value e nt s Them sh approach is e e xactly thesam as thepre e vious e ple e pt for this change xam s xce . -3Vx 2Ω 5A ia + + - 14Ω 3Ω 5Ω 10V + Vx ic x c b ib 1Ω V = 3(i – i ) 31 De nde S pe nt ourceExam ple Find thepowe de re by thede nde source r live d pe nt -3Vx 2Ω 5A ia + + - 14Ω 3Ω 5Ω 10V + Vx ic 1Ω ib 32 PracticeProble m Find thepowe be dissipate in the2Ω r ing d re sistor. 5Ω 2Ω + 60V 4A 4Ω 5A ia 3Ω 6ia + - 33 S hortcut and C ck Me he thod Add all re sistors containe in a m sh d e t oge r the This is thenum r for that m sh curre be e nt variable 30 Ω For e re ach sistor that is com on to m m ore t han onem sh e Put ne gativethat valuein front of theothe r m sh curre variable e nt + - 5Ω i2 90 Ω 26 Ω 8Ω Placethene gativesumof thevoltage source on theRHSof thee s quation 80 V i1 i3 34 S hortcut and C ck Me he thod Le look at Me 1 t’s sh Add all re sistors containe in a d m sh toge r e the um • S is 31Ω 31i1 30 Ω For e re ach sistor that is com on m t o m than onem sh, Put ore e ne gativethat valuein fromof the othe m sh curre variable re nt • 5Ω -5i2 • 26Ω -26i3 5Ω i2 90 Ω 80 V + - 26 Ω 8Ω Placethene gativesumof the voltagesource on theRHSof the s e quation um • S is -80 V 1 i1 i3 2 3 35 Hints for S ss ucce Me e sh quations havea te for every e m nt in them sh. rm le e e Count the elements; count the terms! Think care fully about thevoltageacross e e m nt. ach le e Thevoltagepolarity corre sponds to thedire ction of them sh curre e nt. Le how to e ss thevoltagefor e ry kind of e m nt – re arn xpre ve le e sistors, curre nt source voltagesource com s, s, binations. Usesupermeshes whe ne de n e d. Each te in a m sh e rm e quation is a voltage. Each curre sourceyie a constraint equation. nt lds You m not ne d all them sh voltage to answe theque ay e e s r stion. Only solvefor theone you ne d. Think ahe s e ad! 36 Me C nt Analysis sh urre Focus on writing theKVL e quations corre ctly. I f thee quations arewrong, theanswe is suret o bewrong! r Re e be that curre source re mm r nt s quiree care xtra . 37 Que stions? Me C nt Analysis sh urre Me C nt Analysis sh urre A Thre Me Exam e sh ple Me C nt Exam sh urre ple Find thepowe associate with e source r d ach . 6Ω 2Ω + 40V 8Ω Ia Ib 6Ω 4Ω + 20V - Ic Me a: sh Me b: sh Me c: sh - 40 + 2 I a + 8 (I a – I b)= 0 8 (I b – I a) + 6 I b + 6 (I b – I c) = 0 6 (I c – I b) + 4 I c + 20 = 0 40 Exam Continue ple d Original Equations: (Eq. 1) - 40 + 2 I a + 8 (I a – I b)= 0 (Eq. 2) (Eq. 3) 8 (I b – I a) + 6 I b + 6 (I b – I c) = 0 6 (I c – I b) + 4 I c + 20 = 0 Re writing in standard form : (Eq. 1) 10 I a – 8 I b + 0 I c = 40 (Eq. 2) (Eq. 3) -8 I a + 20 I b – 6 I c = 0 0 I a – 6 I b + 10 I c = -20 41 Exam Continue ple d Multiply Eq. 1 by 8/10 and add to Eq. 2: 8 I a – 6.4 I b + 0 I c = 32 -8 I a + 20 I b – 6 I c = 0 13.6I b - 6I c = 32 Multiply by 10/6 and add to Eq. 3 0 I a + 22.67 I b – 10 I c = 53.33 0Ia 6.00 I b + 10 I c 16.67 I b = 33.33 I b = 2.00 A 42 = -20 Final S olution S ubstituting I b = 2.0A into Eq. 3: 0 I a – 6 (2) + 10 I c = -20 I c = -0.800 A S ubstituting into Eq. 1: 10 I a – 8 (2) + 0 I c = 40 I a = 5.60 A 43 Now Find Powe fromS r ource s 2Ω + 40V + Vo 6Ω 6Ω Ib 4Ω + 20V - 8Ω Ia - Ic P40V = -VI a = - (40)(5.6) = -224. W (224. W de re live d) P20V = VI c = (20)(-.8A) = -16.0 W (16.0 W de re live d) Find Vo. Vo = 8(I a-I b) = 8(3.6) = 28.8V 44 Que stions? Me C nt Analysis sh urre Nodeand Me Circuit Analysis sh Whe should weusee m thod? n ach e S olving Proble s m S thefirst e , wehaveintroduce two ince xam d syste atic m thods: m e NodeVoltageAnalysis allows you to assign voltage value to e e ntial nodeoncea re re nodeis s ach sse fe nce de d. fine Me C nt Analysis allows you to assign curre sh urre nt value to all m sh curre in thecircuit. s e nts Now that wehavethe two tools which one se should weuse ? 47 I ssue#1: What circuit amI analyzing? I n te s of thenum r of e rm be quations you ne d to writeout and e solve e r nodevoltageanalysis or m sh curre analysis could be , ithe e nt sim r to apply. ple How should I e valuatethis? Do a Quick I nspe ction be you start. fore S nd 60-90 se pe conds on e circuit and ask thefollowing: ach any quations do I haveto solve ? • I f I usenodevoltageanalysis, how m e e nt any quations do I haveto solve ? • I f I usem sh curre analysis, how m e s ithe e e asy • Doe e r m thod se mparticularly difficult or particularly e for this circuit? r e • What is theanswe I ne d? 48 Nodevs. Me Exam sh ple R3 R2 R4 1A R1 + - R5 2A 2V How many unknown node voltages? How many unknown mesh currents? 2 3 49 Exam Circuit #2 ple 4Ω 3Ω 20 Ω 5A 6Ω 12 A 13 Ω 7Ω iy ix I f I usenodevoltageanalysis, how m e any quations do I haveto solve ? I amle with 3 node for which I ne d to writea KC e ft s e L quations. This m ans I have3 e e quations and 3 unknowns 50 Exam Circuit #2 C ple on’t 4Ω 3Ω 20 Ω 5A 6Ω 12 A 13 Ω 7Ω iy ix I f I usem sh curre analysis, how m e e nt any quations do I haveto solve ? I amle with 4 m she for which I ne d to writea KVL e ft es e quations This means I have4 equations and 4 unknowns Although onepair is a super mesh So this reduces to threemeaningful unknowns. 51 Exam Circuit #3 ple Which m thod would you usefor this circuit? e 52 Points to Re e be mm r Quick I nspe ction is de signe to m your lifee r. d ake asie S stions for NodeVoltage ugge ntify thee ntial node sse s • I de • Pick an appropriateground!! S stions for Me Curre ugge sh nt ntify them she es • I de te ine s • De rm which one areknown!! I n both case s: You do not ne d to nam thenodevoltage or m sh curre until you pick e e s e nts your m thod. e Le to do this by “looking” at thecircuit and not writing anything down arn until you de on a m thod… cide e 53 Exam Circuit #4 ple I n te s of thenum r of e rm be quations I would ne d to e write which m thod is be m sh curre or node , e st: e nt voltage ? y 8 kΩ 0.2 kΩ 20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x 54 Equal Num r of Equations be This will happe for m circuits you will analyze n any . Now which m thod should you choose e ? Things to conside r: Do I haveto usea “supe rnode or “supe e ” rm sh”? ne se r t • I n ge ral, the areharde to se up… Which m thod is m com e ost fortablefor m ? e What amI solving for? ne e nt, e nt asie e • I n ge ral, if I ne d a curre m sh curre analysis will bee r; if I ne d a voltage nodevoltageanalysis will bee r. , asie 55 Exam Circuit # 3 ple Will I re quiree r a “supe e or “supe ithe rm sh” rnode in ” t his circuit? y 8 kΩ 0.2 kΩ 20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ No!!! x 56 Exam Circuit #3 C ple on’t y 8 kΩ 0.2 kΩ 20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x What if I amaske to solvefor thepowe be de re by the d r ing live d 25-V source ? Ne d a curre to calculate e nt Usem sh curre analysis e nt 57 Exam Circuit #3 C ple on’t y 8 kΩ 0.2 kΩ 20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x What if I amaske to solvefor thepowe be de re by the d r ing live d 20 m curre source A nt ? Ne d a voltageto calculate e Usenodevoltageanalysis 58 Exam Circuit #3 C ple on’t y 8 kΩ 0.2 kΩ 20 mA 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x What if I amaske to solvefor thepowe be absorbe by the d r ing d 4kΩ re sistor? C ould usee r a voltageor a curre ithe nt Usem thod I amm com e ost fortablewith… 59 Additional C m nts: Nodevs. Me om e sh C ircuits with se s e m nts te to bee r using m sh analysis. rie le e nd asie e C ircuits with paralle e m nts arealm always e r using nodal l le e ost asie analysis. S estude pre r m sh and useit e n whe nodal analysis is om nts fe e ve n MUC e r. H asie The alm always ge thewrong answe y ost t r! Examproble s areofte de m n signe to bee r with onem thod d asie e t han with theothe r. Always do theQuick I nspe ction be you start. fore All of m EE302 e proble s that re st a num ric answe can be y xam m que e r solve using two sim d ultane e ous quations (plus constraints). 60 PracticeC ircuits 61 PracticeC ircuit 1 20 V 100 Ω 10 V 200 Ω 1A 0.2 A 62 PracticeC ircuit 1 20 V 20V 100 Ω 10 V 200 Ω 1A A 0.2 A 63 PracticeC ircuit 1 20 V 100 Ω 10 V 200 Ω 1A 0.2 A 64 PracticeC ircuit 2 How m nodeand m sh e any e quations arene de ed t o solvethis circuit? 10i1 R1 R3 + R2 + R5 + R4 vx 8.5 V - R7 R6 Vx/5 i1 65 PracticeC ircuit 2 10i1 R1 R3 + R2 + R5 + R4 vx 8.5 V - R7 R6 Vx/5 i1 66 PracticeC ircuit 2 10i1 R1 R3 + R2 + R5 + R4 vx 8.5 V - R7 R6 Vx/5 i1 67 Quick I nspe ction Quick I nspe ction will m your lifee r. ake asie S stions for NodeVoltage ugge ntify thee ntial node sse s • I de • Pick an appropriateground!! S stions for Me Curre ugge sh nt ntify them she es • I de te ine s • De rm which one areknown!! I n both case s: Do not nam thenodevoltage or m sh curre until you pick your m thod. e s e nts e Le to do this by “looking” at thecircuit and not writing anything down arn until you de on a m thod… cide e 68 Equal Num r of Equations be This will happe for m of thecircuits you analyze n any . Now which m thod should you choose e ? Things to conside r: Do I haveto usea “supe rnode or “supe e ” rm sh”? ne se r t • I n ge ral, the areharde to se up… Which m thod is m com e ost fortablefor m ? e What amI solving for? ne e nt, e nt asie e , • I n ge ral if I ne d a curre m sh curre analysis will bee r; if I ne d a voltage nodevoltageanalysis will bee r. asie 69 Que stions? NodeVoltageAnalysis vs. Me C nt Analysis sh urre Me thod S le e ction Practice For e of the21 circuits you will begive ach n: I de ntify thenum r of e be quations you would ne d to writedown e and solvetheproble using nodevoltageanalysis. This m ans that m e t henodevoltagevalueis unknown and you ne d to solvefor it. e I de ntify thenum r of e be quations you would ne d to writedown e and solvetheproble using m sh curre analysis. This m ans that m e nt e t hem sh curre valueis unknown and you ne d to solvefor it. e nt e C onside othe difficulty factors. r r 71 Figure1 72 Figure2 73 Figure3 74 Figure4 1 mA R3 R1 10 10 V R2 R4 R5 R6 5 mA 20 V 75 Figure5 1.8 A 7Ω 2.8 Ω 8V 3V 4Ω 1Ω 8Ω 5Ω 2Ω 4Ω -1 A 9Ω 7Ω 3A 76 Figure6 20 Ω 2A 10 Ω 80 V 120 V 40 Ω 4A 77 Figure7 V1 + + - R2 + - V3 R4 V2 R3 R1 78 Figure8 4Ω 8Ω 10 Ω 2Ω 6Ω 5A 12 12 V 79 Figure9 075 A y 20 Ω 10 Ω 8V x -2 V 4Ω 16 Ω 3A 8Ω 80 Figure10 6Ω R4 I RL 20Ω 2.5 A 15Ω 10 V 81 Figure11 y 8 kΩ 20 20 mA 4 kΩ 0.2 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x 82 Figure12 2V 5Ω 20 Ω 5V 20 Ω 9Ω 15 Ω 1A 83 Figure13 R4 R2 2A R3 7V R1 1A 84 Figure14 I2 R3 R2 I1 I1 R1 V1 R4 R6 R5 RL I3 85 Figure15 R1 R3 R4 I1 I1 R2 V2 V1 RL 86 Figure16 + 3.5 mA 12 mA V1 + vx 10 kΩ 1 kΩ 1 mA -3 mA 0.03vx - - 87 Figure17 ix 6A 3Ω 9Ω 0.9Ix 6Ω 15 Ω 4A 6Ω 88 Figure18 io R3 R1 R1 R2 V1 + - R4 3io 89 Figure19 R3 V2 + + + + - io R1 R2 V1 + 4vx + + 2io R4 vx - 90 Figure20 + 4io vo R2 2vo R5 + + + + + + V1 - I1 R1 R3 io R4 91 Figure21 ix R1 I1 Vx/4 R3 V1 + - + + vx R2 - 4ix - 92 Que stions? NodeVoltageAnalysis Me C nt Analysis sh urre S le e cting theBe r Me tte thod PracticeFigure with S s olutions (Not poste with Handouts) d Exam Circuit #3 ple Which m thod would you usefor this circuit? e Node: 1 Eqn Mesh: 2 or 3 Eqn 95 Exam Circuit #3: Node ple VA VA − V 1 − I1 + =0 R1 + R 2 + R3 R5 + R 6 V1 A 96 Exam Circuit #3: Me ple sh I must write equations for each mesh!! And deal with a supermesh!! 97 Exam Circuit #4 ple I n te s of thenum r of e rm be quations I would ne d to e write which m thod is be m sh curre or node , e st: e nt voltage ? y 8 kΩ 0.2 kΩ 20 mA Mesh = 2 Eqns. & Node = 2 Eqns. 4 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x 98 PracticeC ircuits Figure 1 Figure 2 Figure 3 Figure 4 Figure 5 Figure 6 Figure 7 Figure 8 Figure 9 Figure 10 Figure 11 Figure 12 Figure 13 Figure 14 Figure 15 Figure 16 Figure 17 Figure 18 Figure 19 Figure 20 Figure 21 99 Figure1 Node: 2 Mesh: 3 100 Figure1: Node 101 Figure1: Me sh 102 Figure2 Node: 2 Mesh: 1 103 Figure2: Node 104 Figure2: Me sh 105 Figure3 Node: 3 Mesh: 3 (supermesh) 106 Figure3: Node 107 Figure3: Me sh 108 Figure4 1 mA R3 R1 10 10 V R2 R4 R5 R6 5 mA Node: 2 Mesh: 3 20 V 109 Figure4: Node 1 mA R3 R1 10 10 V R2 R4 R5 R6 5 mA 20 V 110 Figure4: Me sh 1 mA R3 R1 10 10 V R2 R4 R5 R6 5 mA 20 V 111 Figure5 1.8 A 7Ω 2.8 Ω Node: 3 8V 3V 4Ω 1Ω 8Ω 5Ω 2Ω 4Ω -1 A 9Ω 7Ω Mesh: 2 (supermesh) 3A 112 Figure5: Node 1.8 A 7Ω 2.8 Ω 8V 3V 4Ω 1Ω 8Ω 5Ω 2Ω 4Ω -1 A 9Ω 7Ω 3A 113 Figure5: Me sh 1.8 A 7Ω 2.8 Ω 8V 3V 4Ω 1Ω 8Ω 5Ω 2Ω 4Ω -1 A 9Ω 7Ω 3A 114 Figure6 20 Ω 2A Node: 2 80 V 10 Ω Mesh: 2 (supermesh) 120 V 40 Ω 4A 115 Figure6: Node 20 Ω 2A 10 Ω 80 V 120 V 40 Ω 4A 116 Figure6: Me sh 20 Ω 2A 10 Ω 80 V 120 V 40 Ω 4A 117 Figure7 V1 + + - R2 Node: 1 + V3 R4 Mesh: 3 V2 R3 R1 118 Figure7: Node V1 + + - R2 + - V3 R4 V2 R3 R1 119 Figure7: Me sh V1 + + - R2 + - V3 R4 V2 R3 R1 120 Figure8 4Ω 8Ω Node: 2 10 Ω 2Ω 6Ω 5A Mesh: 2 12 12 V 121 Figure8: Node 4Ω 8Ω 10 Ω 2Ω 6Ω 5A 12 12 V 122 Figure8: Me sh 4Ω 8Ω 10 Ω 2Ω 6Ω 5A 12 12 V 123 Figure9 075 A y 20 Ω 10 Ω 8V x -2 V 4Ω 16 Ω 3A 8Ω Node: 2 Mesh: 2 124 Figure9: Node 075 A y 20 Ω 10 Ω 8V x -2 V 4Ω 16 Ω 3A 8Ω 125 Figure9: Me sh 075 A y 20 Ω 10 Ω 8V x -2 V 4Ω 16 Ω 3A 8Ω 126 Figure10 6Ω R4 I RL 20Ω 2.5 A Node: 1 Mesh: 2 15Ω 10 V 127 Figure10: Node 6Ω R4 I RL 20Ω 2.5 A 15Ω 10 V 128 Figure10: Me sh 6Ω R4 I RL 20Ω 2.5 A 15Ω 10 V 129 Figure11 y 8 kΩ 20 20 mA 4 kΩ 0.2 kΩ Node: 2 Mesh: 2 0.5 kΩ 1 kΩ 25 V 2 kΩ x 130 Figure11: Node y 8 kΩ 20 20 mA 4 kΩ 0.2 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x 131 Figure11: Me sh y 8 kΩ 20 20 mA 4 kΩ 0.2 kΩ 0.5 kΩ 1 kΩ 25 V 2 kΩ x 132 Figure12 2V 5Ω 20 Ω Node: 2 (supernode) Mesh: 3 5V 20 Ω 9Ω 15 Ω 1A 133 Figure12: Node 2V 5Ω 20 Ω 5V 20 Ω 9Ω 15 Ω 1A 134 Figure12: Me sh 2V 5Ω 20 Ω 5V 20 Ω 9Ω 15 Ω 1A 135 Figure13 R4 R2 2A R3 7V R1 1A Node: 1 Mesh: 1 136 Figure13: Node R4 R2 2A R3 7V R1 1A 137 Figure13: Me sh R4 R2 2A R3 7V R1 1A 138 Figure14 I2 Node: 3 Mesh: 5 R5 R4 R6 I3 RL R3 R2 I1 I1 R1 V1 (supermesh) 139 Figure14: Node I2 R3 R2 I1 I1 R1 V1 R4 R6 R5 RL I3 140 Figure14: m sh e I2 R3 R2 I1 I1 R1 V1 R4 R6 R5 RL I3 141 Figure15 R1 R3 R4 Node: 1 Mesh: 2 I1 I1 R2 V2 V1 RL 142 Figure15: Node R1 R3 R4 I1 I1 R2 V2 V1 RL 143 Figure15: Me sh R1 R3 R4 I1 I1 R2 V2 V1 RL 144 Figure16 + 3.5 mA 12 mA V1 + vx 10 kΩ 1 kΩ 1 mA -3 mA 0.03vx - - Node: 2 Mesh: 0!!! 145 Figure16: Node + 3.5 mA 12 mA V1 + vx 10 kΩ 1 kΩ 1 mA -3 mA 0.03vx - - 146 Figure16: Me sh + 3.5 mA 12 mA V1 + vx 10 kΩ 1 kΩ 1 mA -3 mA 0.03vx - - 147 Figure17 ix 6A 3Ω 9Ω 0.9Ix 6Ω 15 Ω 4A 6Ω Node: 2 Mesh: 4 (supermesh) 148 Figure17: Node ix 6A 3Ω 9Ω 0.9Ix 6Ω 15 Ω 4A 6Ω 149 Figure17: Me sh ix 6A 3Ω 9Ω 0.9Ix 6Ω 15 Ω 4A 6Ω 150 Figure18 io R3 R1 R1 R2 Node: 2 R4 3io V1 + - Mesh: 3 (supermesh) 151 Figure18: Node io R3 R1 R1 R2 V1 + - R4 3io 152 Figure18: Me sh io R3 R1 R1 R2 V1 + - R4 3io 153 Figure19 R3 V2 + + + + - io Node: 2 + R1 R2 V1 + 4vx + - 2io R4 Mesh: 3 (supermesh) vx - 154 Figure19: Node R3 V2 + + + + - io R1 R2 V1 + 4vx + + 2io R4 vx - 155 Figure19: Me sh R3 V2 + + + + - io R1 R2 V1 + 4vx + + 2io R4 vx - 156 Figure20 Node: 2 + 4io vo R2 2vo R5 - Mesh: 4 (supermesh) + + + + + + V1 - I1 R1 R3 io R4 157 Figure20: Node + 4io vo R2 2vo R5 + + + + + + V1 - I1 R1 R3 io R4 158 Figure20: Me sh + 4io vo R2 2vo R5 + + + + + + V1 - I1 R1 R3 io R4 159 Figure21 Node: 2 ix R1 I1 V1 + Vx/4 R3 Mesh: 3 (supermesh) 4ix + + vx R2 - - 160 Figure21: Node ix R1 I1 Vx/4 R3 V1 + - + + vx R2 - 4ix - 161 Figure21: Me sh ix R1 I1 Vx/4 R3 V1 + - + + vx R2 - 4ix - 162 ...
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This note was uploaded on 11/02/2009 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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EE302Lecture15 - Me C nt C sh urre ircuit Analysis I n the...

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