电路基础英文版Chapter 11

Chapter11ACPowerAnalysis要求深刻理解与熟练掌握的重点内容有：正弦稳态电路的有功功率、无功功率、视在功率和功率因数的概念及计算，复功率的概念及最大功率传输。难点：提高功率因数、功率分析•11-1Introduction•11-2InstantaneousandAveragePower•11-3MaximumAveragePowerTransfer•11-4EffectiveorRMSValue•11-5ApparentPowerandPowerTransfer•11-6ComplexPower•11-8PowerFactorCorrection•11-10SummaryandReview•11-7ConservationofACPower11.1IntroductionInthischapter,ourgoalsandobjectivesinclude•Determiningtheinstantaneouspowerdeliveredtoanelement•Definingtheaveragepowersuppliedbyasinusoidalsource•Usingcomplexpowertoidentifyaverageandreactivepower•Identifyingthepowerfactorofagivenload,andlearningmeansofimprovingitintegration积分法integrand被积函数theloadparameter负载参数thecomplexconjugate共轭复数trigonometricidentity三角恒等式Domesticload家用电器quadrant象限pressingiron压力熨斗inertia惯性11-2InstantaneousandAveragePower1.Instantaneouspowerisp(t)=v(t)i(t))cos()cos()()()(imvmtItVtitvtp)cos()()cos()(imvmtItitVtv)]cos()[cos(21coscosBABABA)2cos(21)cos(21)(ivmmivmmtIVIVtp)cos(21ivmmIVTheaveragepoweristheaverageoftheinstantaneouspoweroveroneperiod.TdttpTP0)(1Substitutinginstantaneouspowerp(t)gives)(ivcoscos21)cos(21)2cos(1211)cos(21)2cos(211)cos(2110000VIIVIVdttTIVdtTIVdttIVTdtIVTPmmivmmTivmmTivmmTivmmTivmm2.AveragePower.Thus,averagepoweriscoscos21VIIVPmm3.AveragePowerabsorbedbyR,L,C.When0ivVoltageandcurrentisinphase022111cos00222RmmmmmPVIVIIRVIIRWhen090ivWehaveapurelyreactivecircuit0)90cos(210mmCLIVPPInSummary:Aresistiveload(R)absorbspoweratalltimes,whileareactiveload(LorC)absorbszeroaveragepower.电抗性负载11.3MaximumAveragePowerTransferjXRZjXRZthththLLL;Thecurrentthroughtheloadis)()(jXRjXRVZZVILLthththLththZLZththVIAveragepowerdeliveredtoloadis:OurobjectiveistoadjusttheloadparametersRLandXLsothatPismaximum.TodothiswesetP/RLandP/XLequaltozero.Weobtain)1(222)()(2LXXRRVLLRRIPLthLththP/RL=0P/XL=0jXRZjXRZthththLLL*thLXX)2()(22ththLthLRXXRRFormaximumaveragepowertransfer,theloadimpedanceZLmustbeequaltothecomplexconjugateoftheTheveninimpedanceZth.SettingRL=RthandXL=-XthinEq.(1)givesusthemaximumaveragepowerasththLXXRRVLLRVRRIPLthLthth42)()(2222thththLZXRR22)0(Inasituationinwhichtheloadispurelyreal,theconditionformaximumpowertransferisobtainedfromEq.(2)bysettingXL=0;thatis,Example11.1：Giventhatv(t)=120cos(377t+450）Vandi(t)=10cos(377t-100)AFindtheinstantaneouspowerandtheaveragepowerabsorbedBythepassivelinearnetworkofFig.11.1.Solution:Theinstantaneouspowerisgivenbyp=vi=1200cos(377t+450)cos(377t-100)ApplyingthetrigonometricidentityWIVPiumm2.34455cos600)]10(45cos[1012021)cos(21000givesOrp(t)=344.2+600cos(754t+350)WTheaveragepoweris)]cos()[cos(21coscosBABABAp(t)=600[cos(754t+350)+cos550]Example11.2：ForthecircuitshowninFig.11.2,findtheaveragepowersuppliedbythesourceandtheaveragepowerabsorbedbytheresistor.ISolution:243050jIAIIR057.56118.1ThecurrentthroughtheresistorisWP5)57.5630cos(118.1500VIVRR057.56472.44TheaveragepowersuppliedbythevoltagesourceisAndthevoltageacrossitisWhichisthesameastheaveragepowersupplied.Zeroaveragepowerisabsorbedbythecapacitor.TheaveragepowerabsorbedbytheresistorisWP5118.1472.4Example11.3:DeterminethepowergeneratedbyeachsourceandtheaveragepowerabsorbedbyeachpassiveelementinthecircuitFig11.3.Solution:WeapplymeshanalysisasshowninFig11.3.formesh1,Formesh2,AIjIjorIjIjj002020121.7958.10860124030605306010)510(AI0104Forthevoltagesource,thecurrentflowingfromitisI2=10.58∠79.10A.Andthevoltageacrossitis60∠300V,sothattheaveragepowerisWP6.415)1.7930cos(58.1060005V1=20I1+j10(I1-I2)=80+j10(4-4-j10.39)=183.9+j20=184.984∠6.210VWP6.735)021.6cos(4984.18401TheaveragepowersuppliedbythecurrentsourceisWRIP32020)4(2220;090cos)9.52()8.105(403PPP1+P2+P3+P4+P5=-735.6+320+0+0+415.6=0V3=-j5I2=(5∠-900)(10.58∠79.10)=52.9∠(79.10-900)Example11.4Giventhetime-domainvoltage,)6/cos(4VtvFindboththeaveragepowerandanexpressionfortheinstantanouspowerthatresultwhenthecorrespondingphasorvoltageisappliedacrossanimpedanceZ=2∠600Ω.Solution:ThephasorcurrentisV/Z=2∠-600A,andaveragepoweris:WP260cos)2)(4(210Thetime-domainvoltage,Vttv6cos4)(time-domaincurrent,Atti)606cos(2)(0Andinstantaneouspower,WttttP)603cos(42)606cos(6cos8)(0011.4EffectiveorRMSValueTheeffectivevalueofaperiodiccurrentisthedccurrentthatdeliversthesameaveragepowertoaresistorastheperiodiccurrent.Theeffectivevalueofaperiodicsignalisitsrootmeansquare(rms)value.TeffdtiTI021TeffdtuTU02111.5ApparentPowerandPowerFactor1.ReactivePower)(sinsin21VarVIIVQmm2.ApparentPowerSPpfiucos)cos(3.PowerFactor)(VAIVS乏ZZIVIVIVZiuiuiu)()(Thepowerfactoristhecosineofthephasedifferentbetweenvoltageandcurrent.Itisalsothecosineoftheangleoftheloadimpedance.11.6ComplexPower1.DefinejIVjIVIVQjPSiuiuiuiu)sin()cos()sin()cos()(1)sin()cos()(iuiuiujjeiuSIVIVIVIVSiuiu*)()(SothatQjPBjVGVBjGVYVVYVIVSQjPXjIRIXjRIZIIZIIVSS