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: