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Transcript
EE215ClassProblems,Week5 Solutions Allquestionsareinitiallylistedinblack.Afterclass,thosethatwerecoveredinclassusingTopHat (asclassactivities)willbehighlightedinblue.Questionsremaininginblack(afterclass)maybe usedforadditionalreview.Solutionsarenotedinitalics. Week5,Q1 FindtheNortonEquivalentwithrespecttotheloadresistanceinthefollowingcircuit: Solution: SourcetransformationswillworktoreducethiscircuittotheTheveninEquivalent.Convertingallthe voltagesourcesinserieswithresistorstocurrentsourcesinparallelwiththoseresistorsgives: CombiningcurrentsourcesinparallelandresistorsinparallelgivestheNortonEquivalent: Week5,Q2 PartA FindtheTheveninEquivalentofthefollowingcircuitwithrespecttotheterminalsaandb. Solution: Itispossibletodoonesourcetransformationinthiscircuitonthe20Vsource(convertingitfroma 20Vsourceinserieswitha100ohmresistortoa20/100or200mAsourceinparallelwitha100ohm resistor).Afterthetransformationthe100ohmresistorisinparallelwithanother100ohmresistor andcanbecombinedintoa50ohmresistor. NomoresourcetransformationsareproductiveforreducingthecircuittoitsTheveninEquivalent. Thus,weneedtochooseanothermethodforfindingtheTheveninEquivalent.Duetothenumberof sourcesinthecircuit,wechoosethehybridmethod(Method3fromtheLectureNotes)wherefirstall thesourcesarede-activatedandtheTheveninresistancefound: TheTheveninresistancewithrespecttotheterminalsaandbis: (100+50)||100=60Ohms Now,wecaneitherfindtheopencircuitvoltageortheshortcircuitcurrentassociatedwiththe terminalsaandb.Wechoosetofindtheopencircuitvoltage(equaltotheTheveninvoltage): Thereisonejunctionpointvoltagetobefoundinthiscircuit(thatontopofthe50ohmsource).The equationis: (60-Va)/200+20mA=Va/50+200mA SolvingforVagives: Va=4.8V andtheopencircuitvoltageis:(60-Va)*0.5+Va=32.4V Thus,theTheveninEquivalentcircuitis: PartB Whatvalueofloadresistancecanweattachtothiscircuittoachievemaximumpowerdeliveredto thatloadresistance? Solution: Maximumpowertransferoccurswhentheloadresistance=theveninresistance RL=60ohms PartC Whatisthemaximumpowerthatcanbetransferredtotheload? Solution: Power=VLoad2/RL=(0.5*32.4)2/60=4.37Watts PartD Whatarethevaluesofloadresistancethatresultinhalfthemaximumpowerbeingdeliveredtothis circuit? Solution: 1/2Power=2.185W=VLoad2/RL 2.185=[32.4*(RL/(60+RL)]2/RL R=351ohmsor10ohms Week5,Q3 PartA Whatisthemaximumpowerthatcanbeprovidedbya12.6Vcarbattery? Theinternalresistanceofaleadacidcarbatteryis.01ohms. Solution: Maximumpoweristransferredtotheloadonthecarbatterywhentheloadresistanceisequaltothe internalresistanceofthebattery.Inthiscase,thevoltageacrosstheloadis0.5*12.6=6.3V Andthemaximumpoweris:3,969orabout4,000Watts PartB Whatisthemaximumpowerthatcanbeprovidedbya9Valkalinebattery? Theinternalresistanceofanalkalinebatteryis.15ohms. Solution: Maximumpowerprovidedbythealkalinebattery=4.52/0.15=135W Week5,Q4 Uponfindinganunknowncircuitonthelabbench,youmeasureitsoutputvoltagewithyour multimeter(withnoloadattached)andfindthevoltagetobe20.5V.Concernedthatyoumight blowthecircuitup,youthenconnecta150ohmresistoracrosstheoutputofthecircuitand measurethevoltageagain.Itreads18V. PartA WhatistheTheveninVoltageofthiscircuitwithrespecttotheoutputterminals? Solution: Vthevenin=Voc=20.5V PartB WhatistheNortonCurrentofthiscircuitwithrespecttotheoutputterminals? Solution Inyoursecondmeasurement,youhavecreatedthefollowingsituation/circuit: Themeasuredvoltageinthissituationwas18V(measuredacrossthe150ohmresistor).Byvoltage divider: 18=20.5*(150/(150+Rth)) SolvingforRthgives: Rth=20.8Ohms AndtheNortonCurrentis:20.5/20.8=0.99A Week5,Q5 FindtheTheveninequivalentofthefollowingcircuitwithrespecttothe400Ohmresistor Solution: Sincethiscircuitcontainsdependentsources,wecannotusesourcetransformationstofindthe TheveninEquivalent.WewilluseMethod#2(findtheopencircuitvoltageandtheshortcircuit current). TheOpenCircuitVoltage: Nodalanalysisontherighthandcircuitgives: (5-Va)/100+0.1v=Va/200 Byobservation:v=5-Va Substituting: (5-Va)/100+0.1(5-Va)=Va/200 Solving: Va=4.78Vandv=0.22Vandix=4.78/200=0.024A Byobservation: Voc=82.6ix=1.98V TheShortCircuitCurrent: Nodalanalysisontherighthandcircuitgives: (5-Va)/100+0.1v=Va/200 Byobservation:v=5-Va Substituting: (5-Va)/100+0.1(5-Va)=Va/200 Solving: Va=4.78Vandv=0.22Vandix=4.78/200=0.024A Onthelefthandsideofthecircuit 82.6ix=100isc=1.98V isc=19.8mA AndtheTheveninResistanceis: Voc/isc=100Ohms TheTheveninEquivalentis:
