电路原理课件讲义英文版-Chapter-3

Scientistsstudytheworldasitis,engineercreatetheworldthatneverhasbeen.---TheodorevonKarmanChapter3MethodsofAnalysis3.1Introduction3.2NodalAnalysis3.3NodalAnalysiswithVoltageSources3.4MeshAnalysis3.5MeshAnalysiswithCurrentSources3.6NodalandMeshAnalysisbyInspection3.7NodalVersusMeshAnalysis3.10Summary3.1IntroductionTwopowerfultechniquesforcircuitanalysis:Nodalanalysis(basedonasystematicapplicationofKCL)Meshanalysis(basedonasystematicapplicationofKVL)3.2NodalAnalysisMethods:ChoosingnodevoltagesinsteadofelementvoltagesFindingthenodevoltagesMerits:ConvenientThenumberofequationsisreducedStepstoDetermineNodeVoltages:1.Selectanodeasthereferencenode.Assignvoltagesv1,v2,…,vn-1totheremainingn-1nodes.Thevoltagesarereferencedwithrespecttothereferencenode.2.ApplyKCLtoeachofthen-1nonreferencenodes.UseOhm’slawtoexpressthebranchcurrentsintermsofnodevoltages.3.Solvetheresultingsimultaneousequationstoobtaintheunknownnodevoltages.TheFirstStep:Selectinganodeasthereferenceordatumnode(calledthegroundwithzeropotential):.().().()ThreesymbolschassisgroundshowninFigbshowninFigaearthgroundshowninFigc()a()b()cTypicalcircuitfornodalanalysis:1201I2I1v2v1R2R3RTheSecondStep:ApplyingKCLtoeachnonreferencenodeinthecircuitAtnode1:1I2I1v2v1R2R3R1i3i2i2i12120IIiiAtnode2:2230IiiNotice:Currentflowsfromahigherpotentialtoalowerpotentialinaresistor.Then:11212120vvvIIRR1222230vvvIRRIntermsoftheconductance,thereare1211212()0IIGvGvv221232()0IGvvGvSoweobtain:12222122()vvioriGvvR1111110vioriGvR2333230vioriGvR1I2I1v2v1R2R3R1i3i2i2iTheThirdStep:Tosolveforthenodevoltages1I2I1v2v1R2R3R1i3i2i2iStandardMethods:SubstitutionmethodEliminationmethodCramer’sruleMatrixinversion12211222322GGGvIIGGGvIThelasttwoequationscanbecastinmatrixformaswhichcanbesolvedtogetv1andv2.Example110A5A1v2v6421i3i2i2iSimplifyingtheequations,thereisAtnode1:121050iiAtnode2:2350ii1120105064vvv12205042vvv12125360320vvvv122013.33vVvV1233.331.676.67iAiAiAExample2Simplifyingtheequations,thereisAtnode1:Atnode2:1v2v3v3A2xi2844xi1i2i3i130xii230xiiiAtnode3:1220xiii13123042vvvv2312200284vvvvv1323122()0482vvvvvv1231231233212470230vvvvvvvvv3.3NodalAnalysiswithVoltageSourcesAvoltagesourceisconnectedbetweenthereferencenodeandanonreferencenodeCase11v2v3v10V2481i4i2i3i5V6110(1)vVCase2Avoltagesource(dependentorindependent)isconnectedbetweentwononreferencenodesAsupernode(orgeneralizednode)isformedbyenclosinga(dependentorindependent)voltagebetweentwononreferencenodesandanyelementsconnectedinparallelwithit1v2v3v10V2481i4i2i3i5V6SupernodeAtsupernode:12340iiii133122000(2)2486vvvvvvSelectingaloopwithsupernodeByKVL,thereis2350vv235(3)vv1.Thevoltagesourceinsidethesupernodeprovidesaconstraintequationneededtosolveforthenodevoltages.2.Asupernodehasnovoltageofitsown.3.AsupernoderequirestheapplicationofbothKCLandKVL.Notice:Example3FromEqs.(1)and(2),weobtain:Atsupernode:Aroundtheloopwithsupernode:12720ii12005024vv127.335.33vVvV2V2A24107A1v2v1i2i1220(2)vvExpressingv1andv2intermsofthenodevoltages,thereis122200(1)vvExample4Atsupernode1-2:123100iii321410100326vvvvv13450iiii123452600(1)orvvvv3xvxv20V10A241631v2v3v4v1L2L3L1i2i3i4i5iAtsupernode3-4:32314403614vvvvvv1234425160(2)orvvvvForloop1:14xvvv12200(3)vv3xvxv20V10A241631v2v3v4v1L2L3L1i2i3i4i5iInaddition:Forloop2:3430(4)xvvvForloop3:336200(5)xxvvi3326ivvFromEqs.(1),(2),(3)and(4),weobtain:123426.67,6.67173.33,46.67vVvVvVvVAssignmentSolvetheproblems3.3and3.7inpage1093.4MeshAnalysisAPlanarCircuitisonethatcanbedrawninaplanewithnobranchescrossingoneanother;otherwiseitisNonplanar.AMeshisaloopwhichdoesnotcontainanyotherloopswithinit.01233A2xi2844xiQuestions:1.Howmanymeshesintherightfigure?Whatarethey?2.Ispath12301amesh?Why?StepstoDetermineMeshCurrents:1.Assignmeshcurrentsi1,i2,…,in-1tothenmeshes.2.ApplyKVLtoeachofthenmeshes.UseOhm’slawtoexpressthevoltagesintermsofthemeshcurrents.3.Solvetheresultingnsimultaneousequationstogetthemeshcurrents.Methods:ChoosingmeshcurrentinsteadofelementcurrentsandthenfindingthemeshcurrentsMerits:OnlyapplicabletoacircuitthatisplanarTheFirstStep:Assignmeshcurrentsi1,i2,…,intothenmeshesabc1V1R4R1I4I2I3I2V3Rd2R1i2i3iTheSecondStep:ApplyingKVLtoeachmeshesFormesh1:1113212()()0VRiiRii12122131()orRRiRiRiVFormesh2:212232()0RiiVRi212322()orRiRRiVabc1V1R4R1I4I2I3I2V3Rd2R1i2i3iTheThirdStep:SolvingsimultaneousequationsforthemeshcurrentsFormesh3:113432()0RiiRiV111432()orRiRRiVNotice:Thebranchcurrentsaredifferentfromthemeshcurrentsunlessthemeshisisolated..Example512132410()12()0iiiiUsemeshanalysistofindthecurrenti0inthefollowingcircuit.24V121024404i0i1i2i1i2i3iSolution:Formesh1:123115612(1)oriiiFormesh2:22321244()10()0iiiii12351920(2)oriii24V121024404i0i1i2i1i2i3iFormesh3:03132412()4()0ii