电路原理课件讲义英文版-Chapter-2汇总

Chapter2BasicLaws2.1Introduction2.2Ohm’sLaws2.3Nodes,Branches,andLoops2.4Kirchhoff’sLaws2.5SeriesResistersandVoltageDivision2.6ParallelResistersandCurrentDivision2.7Wye-DeltaTransformations2.8Summary2.1IntroductionToactuallydeterminethevaluesofthesevariablesinagivencircuitrequiresthatweunderstandsomefundamentallawsthatgovernelectriccircuits.BasicLaws:Ohm’sLawKirchhoff’sLaw2.2Ohm’sLawsResistance:thephysicalproperty(orability)toresistcurrentlRAResistor:thecircuitelementusedtomodelthecurrent-resistingbehaviorofamaterial(Simplestpassiveelement)::sec:resistivitywhereAcrosstionalllengthOhmdefinedtheconstantofproportionalityforaresistortobetheresistance,ROhm’sLaw:Ohm’sLawstatesthatthevoltagevacrossaresistorisdirectlyproportionaltothecurrentiflowingthroughtheresistorviviRTheresistanceRofanelementdenotesitsabilitytoresisttheflowofelectriccurrent,measuredinohms()Twoextremepossiblecase:ShortCircuit:Ashortcircuitisacircuitelementwithresistanceapproachingzero.OpenCircuit:AnopencircuitisacircuitelementwithresistanceapproachinginfinityPassivesignconvention:viR0viRIfnotconformwithit:viRlim0RviRviRFixed:itsresistanceremainsconstant.(wire-woundtype;carbonfilmtype)Variable:variableresistorshaveadjustableresistance.(compositiontype;sliderpot)Resistor:PotentiometerIngeneralLinearresistor:obeyOhm’sLaw.Nonlinearresistor:doesnotobeyOhm’sLaw.SlopeRSlopeRConductanceistheabilityofanelementtoconductelectriccurrent,measuredinsiemens(S).Powerrepresentationofaresistor:1iGRv22vpviiRR22ipvivGG☺Thepowerdissipatedinaresistorisanonlinearfunctionofeithercurrentorvoltage☺SinceRandGarepositivequantities,thepowerdissipatedinaresistorisalwayspositive.Thus,aresistoralwaysabsorbspowerfromthecircuit.Thisconfirmstheideathataresistorisapassiveelement,incapableofgeneratingenergy.Notice:2.3Nodes,Branches,andLoopsANetworkisaninterconnectedofelementsordevices.Acircuitisanetworkprovidingoneormoreclosedpaths.Innetworktopology,westudythepropertiesrelatingtotheplacementofelementsinthenetworkandthegeometricconfigurationofthenetwork.NodesBranchesLoopsABranchrepresentsasingleelementsuchasavoltagesourceoraresistor.ANodeisthepointofconnectionbetweentwoormorebranches.ALoopisanyclosedpathinacircuit.(Aloopissaidtobeindependentifitcontainsabranchwhichisnotinanyotherloop.)1bln5215V2AabcAssignment1,Read1.7.2(ElectricityBills)inpage17,and1.8(ProblemSolving)inpage182,Solvetheproblems2.5and2.7inpage632.4Kirchhoff’sLawsKCL:Kirchhoff’sCurrentLaw(basedonthelawofconservationofcharge)KVL:Kirchhoff’sVoltageLaw(basedontheprincipleofconservationofenergy)KCLstatesthatthealgebraicsumofcurrentsenteringanode(oraclosedboundary)iszero.10NnniKCL:Thesumofthecurrentsenteringanodeisequaltothesumofthecurrentsleavingthenode.1i2i3i4i5i12345()()0iiiiiCase1(node)13425oriiiiiAlternativeformofKCL:Case2(closedboundary)Generalized:anodemayberegardedasaclosedsurfaceshrunktoapoint.Twodimension:aclosedboundaryisthesameasaclosedpath.ClosedBoundary1i2i3i4i5i12345()()()0iiiii12345oriiiiiApplicationofKCLSeriesofcurrentsources:acircuitcannotcontaintwodifferentcurrents,I1andI2,inseries,unlessI1=I2;otherwise,KCLwillbeviolatedParallelofcurrentsources:thecombinedcurrentisthealgebraicsumofthecurrentsuppliedbytheindividualsources.213IIII123orIIII1I2I3IabIabIKVLstatesthatthealgebraicsumofallvoltagesaroundaclosedpath(orloop)iszero.10MmmuKVL:AlternativeformofKVL:Illustration123450vvvvv1v2v3v4v5v12435orvvvvvSumofvoltagedrops=SumofvoltagerisesApplicationofKVLSeriesofvoltagesources:thecombinedvoltageisthealgebraicsumofthevoltagesoftheindividualsources.Parallelofvoltagesources:acircuitcannotcontaintwodifferentvoltages,V1andV2,inparallel,unlessV1=V2;otherwise,KVLwillbeviolated.1230abvvvv123aborvvvvababv1v2v3vababvExample1Forthefollowingcircuit,findvoltagesv1andv2.12200(2)vv122,3(1)vivi20V2v1v23()bi20V2v1v23()aSolution:FromOhm’sLaw,ApplyingKVLaroundtheloopgivesSubstitutingEq.(1)intoEq.(2),thereis202304iiiAThen128,12vVvVExample2Determinev0andiinthefollowingcircuit.01242460(1)ivi06(2)viFromOhm’sLaw,ApplyingKVLaroundtheloopSubstitutingEq.(1)intoEq.(2),thereis16101208iiiAand048vV12V4V02v46()a0vi12V4V02v46()b0viiSolution:Example3Findthecurrentsandvoltagesinthefollowingcircuit.1230(2)iii1122338,3,6(1)viviviByOhm’sLaw,Then2121(303)30830(3)8iiiori30V2v1v833v6()a1i3i2i1l30V2v1v833v61i3i2i2la()bAtnodea,KCLgivesApplyingKVLtoloop112300vvSolution:131233,1,24,6,6iAiAvVvVvVThenweobtain2222(303)0282iiiiA1l30V2v1v833v61i3i2i2la()bApplyingKVLtoloop223320(4)vvvvSo232363(5)2iiiiSubstitutingEq.(3)and(5)intoEq.(2),givesAssignment1,Solvetheproblems2.5and2.7inpage632,Solvetheproblems2.10,2.12,and2.17inpages64-652.5SeriesResistorsandVoltageDivisionSeriesResistorv1v2viba1R2RvvibaeqRByOhm’sLaw,weobtain1122,viRviRFromKVL,wehave120vvvThen1212()vvviRR12voriRReqviRSo12eqRRRTheequ