intermediate physics - · quickcheck 25.10 a bar magnet sits inside a coil of wire that is...
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IntermediatePhysics
PHYS102
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Dr RichardH.CyburtAssistantProfessorofPhysics
Myoffice:402cintheScienceBuilding
Myphone:(304)384-6006
Myemail:[email protected]
Mywebpage:www.concord.edu/rcyburt
Inpersonoremailisthebestwaytogetaholdofme.
PHYS102
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MyOfficeHoursTWR9:30-11:00amW4:00-5:00pm
Meetingsmayalsobearrangedatothertimes,byappointment
PHYS102
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ProblemSolvingSectionsIwouldliketohavehour-longsectionsforworkingthroughproblems.
Thiswouldbeanextracomponenttothecourseandcounttowardsextracredit
TR1-2pm
WF10-11am
S308
Ifyoucan’tmakethese,youcanstillpickuptheproblemworksheet.
PHYS102
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Midterm2Thursday,March2duringclass8-9:15
Allowedonehalfsheet(8.5x11)pieceofpaperw/notes/formuliCalculatorpencilorblue/blackpen
Review,Wednesday,March17-9pm
Bringquestions!!!
PHYS102
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IntermediatePhysics
PHYS102
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PHYS102
DouglasAdamsHitchhiker’sGuidetotheGalaxy
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Inclass!!
PHYS102
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Thislecturewillhelpyouunderstand:Faraday’sLaw
AlternatingCurrent
ACElectricityandTransformers
PHYS102
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Lenz’sLaw
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Text:p.813
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Lenz’sLaw
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Text:p.813
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Lenz’sLaw
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Text:p.813
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QuickCheck25.8Thebarmagnetispushedtowardthecenterofawireloop.Whichistrue?
Thereisaclockwiseinducedcurrentintheloop.
Thereisacounterclockwiseinducedcurrentintheloop.
Thereisnoinducedcurrentintheloop.
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QuickCheck25.8Thebarmagnetispushedtowardthecenterofawireloop.Whichistrue?
Thereisaclockwiseinducedcurrentintheloop.
Thereisacounterclockwiseinducedcurrentintheloop.
Thereisnoinducedcurrentintheloop.
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1. Upward flux from magnet is increasing.2. To oppose the increase, the field of the induced current points down.3. From the right-hand rule, a downward field needs a cw current.
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QuickCheck25.9Thebarmagnetispushedtowardthecenterofawireloop.Whichistrue?
Thereisaclockwiseinducedcurrentintheloop.
Thereisacounterclockwiseinducedcurrentintheloop.
Thereisnoinducedcurrentintheloop.
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QuickCheck25.9Thebarmagnetispushedtowardthecenterofawireloop.Whichistrue?
Thereisaclockwiseinducedcurrentintheloop.
Thereisacounterclockwiseinducedcurrentintheloop.
Thereisnoinducedcurrentintheloop.
©2015PearsonEducation,Inc.
Magnetic flux is zero, so there’s no change of flux.
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QuickCheck25.10Abarmagnetsitsinsideacoilofwirethatisconnectedtoameter.Foreachofthefollowingcircumstances1.Thebarmagnetisatrestinthecoil,2.Thebarmagnetispulledoutofthecoil,3.Thebarmagnetiscompletelyoutofthecoilandatrest,4.Thebarmagnetisreinsertedintothecoil,
Whatcanwesayaboutthecurrentinthemeter?A. Thecurrentgoesfromrighttoleft.B. Thecurrentgoesfromlefttoright.C. Thereisnocurrentinthemeter.
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QuickCheck25.10Abarmagnetsitsinsideacoilofwirethatisconnectedtoameter.Foreachofthefollowingcircumstances1.Thebarmagnetisatrestinthecoil,C2.Thebarmagnetispulledoutofthecoil,A3.Thebarmagnetiscompletelyoutofthecoilandatrest,C4.Thebarmagnetisreinsertedintothecoil,B
Whatcanwesayaboutthecurrentinthemeter?A. Thecurrentgoesfromrighttoleft.B. Thecurrentgoesfromlefttoright.C. Thereisnocurrentinthemeter.
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QuickCheck25.11Amagneticfieldgoesthroughaloopofwire,asatright.Ifthemagnitudeofthemagneticfieldis1.Increasing,2.Decreasing,3.Constant,
Whatcanwesayaboutthecurrentintheloop?Answerforeachofthestatedconditions.A. Theloophasaclockwisecurrent.B. Theloophasacounterclockwisecurrent.C. Theloophasnocurrent.
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QuickCheck25.11Amagneticfieldgoesthroughaloopofwire,asatright.Ifthemagnitudeofthemagneticfieldis1.Increasing,B2.Decreasing,A3.Constant,C
Whatcanwesayaboutthecurrentintheloop?Answerforeachofthestatedconditions.A. Theloophasaclockwisecurrent.B. Theloophasacounterclockwisecurrent.C. Theloophasnocurrent.
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Themagneticfieldisconfinedtotheregioninsidethedashedlines;itiszerooutside.Themetalloopisbeingpulledoutofthemagneticfield.Whichistrue?
A. Thereisaclockwiseinducedcurrentintheloop.
B. Thereisacounterclockwiseinducedcurrentintheloop.
C. Thereisnoinducedcurrentintheloop.
QuickCheck25.12
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Themagneticfieldisconfinedtotheregioninsidethedashedlines;itiszerooutside.Themetalloopisbeingpulledoutofthemagneticfield.Whichistrue?
A. Thereisaclockwiseinducedcurrentintheloop.
B. Thereisacounterclockwiseinducedcurrentintheloop.
C. Thereisnoinducedcurrentintheloop.
1. The flux through the loop is into the screen and decreasing.
2. To oppose the decrease, the field of the induced current must point into the screen.
3. From the right-hand rule, an inward field needs a cwcurrent.
QuickCheck25.12
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Section25.4Faraday’sLaw
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Faraday’sLawAninducedemf ℇ istheemf associatedwithachangingmagneticflux.
ThedirectionofthecurrentisdeterminedbyLenz’slaw.Thesizeoftheinducedemf isdeterminedbyFaraday’slaw.
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Faraday’sLawFaraday’slaw isabasiclawofelectromagneticinduction.Itsaysthatthemagnitudeoftheinducedemf istherateofchange ofthemagneticfluxthroughtheloop:
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Faraday’sLawAcoilwireconsistingofN turnsactslikeN batteriesinseries,sotheinducedemf inthecoilis
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Faraday’sLawsTherearetwofundamentallydifferentwaystochangethemagneticfluxthroughaconductingloop:1. Theloopcanmoveorexpandorrotate,creatinga
motionalemf.2. Themagneticfieldcanchange.
Theinducedemf istherateofchangeofthemagneticfluxthroughtheloop,regardlessofwhatcausesthefluxtochange.
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Faraday’sLaws
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Text:p.815
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Faraday’sLaws
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Text:p.815
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ExampleProblemThefollowingfigureshowsa10-cm-diameterloopinthreedifferentmagneticfields.Theloop’sresistanceis0.1Ohms.Foreachsituation,determinethemagnitudeanddirectionoftheinducedcurrent.
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EddyCurrentsTherearetwo“loops”lyingentirelyinametalsheetbetweentwomagnets.
Asthesheetispulled,theloopontherightisleavingthemagneticfield,andthefluxisdecreasing.
AccordingtoFaraday’slaw,thefluxchangeinducesacurrenttoflowaroundtheloop.Lenz’slawsaysthecurrentflowsclockwise.
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EddyCurrentsTheloopontheleftsideofthemetalentersthefieldandsothefluxthroughitisincreasing.Lenz’slawrequirestheinduced“whirlpool”currentontheleftlooptobecounterclockwise.
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EddyCurrentsEddycurrentsarethespread-outwhirlpoolsofaninducedcurrentinasolidconductor.Bothwhirlpoolsofcurrentaremovinginthesamedirectionastheypassthroughthemagnet.Themagneticfieldexertsaforceonthecurrent,oppositethedirectionofpull,actingasabrakingforce.
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EddyCurrentsBecauseofthebrakingforceexertedbythemagneticfield,anexternalforceisrequiredtopullametalthroughamagneticfield.Ifthepullingforceceases,themagneticbrakingforcequicklycausesthemetaltodecelerateuntilitstops.
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Section26.1AlternatingCurrent
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AlternatingCurrentAbatteryproducesaconstantemf soaflashlighthasaconstantglow.Theelectricitydistributedtohomesisdifferent.Ratherthanaconstantemf,ithasasinusoidalvariation.Alightbulb inyourhomeisonwhentheemf ispositive,butnotwhenitisnegative.Theresultingflickeristoorapidtonoticeundernormalcircumstances.
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AlternatingCurrentInChapter25wesawthatanelectricalgeneratorworksbyrotatingacoilofwireinamagneticfield.Thesteadyrotationofthecoilcausestheemf andtheinducedcurrentinthecoiltooscillatesinusoidally,alternatingpositiveandthennegative.Theoscillationforceschargestoflowfirstinonedirectionandthentheotherahalfcyclelater.Thisisanalternatingcurrent:AC.Iftheemf isconstantandthecurrentisalwaysinthesamedirection,theelectricityiscalleddirectcurrent: DC.
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AlternatingCurrentTheinstantaneousemf ofanACvoltagesourcecanbewrittenas
ℇ0isthepeakormaximumemf,T istheperiodofoscillation,andf =1/Tisthefrequency.
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ResistorCircuitsIncircuitswherethecurrentandvoltageareoscillating,weusei torepresenttheinstantaneous currentandv fortheinstantaneousvoltage.ThepotentialdifferenceacrossaresistorR iscalledtheresistorvoltagevR.ItisrelatedtothecurrentiR throughtheresistorbyOhm’slaw:
vR =iRR
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ResistorCircuits
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ResistorCircuits
WecananalyzeacircuitwitharesistorconnectedacrossanACemf thesamewaywedidwithaDCresistorcircuit.Kirchhoff’slooplawsaysthatthesumofallthepotentialdifferencesaroundaclosedpathiszero:
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ResistorCircuitsTheresistorvoltageisasinusoidalvoltageatfrequencyf:
vR =VR cos(2p ft)VR isthepeakormaximumvoltage,theamplitudeofthesinusoidallyvaryingvoltage.Inthesingleresistorcase,VR =ℇ0,sothecurrentthroughtheresistoris
IR=VR/Risthepeakcurrent.©2015PearsonEducation,Inc.
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ResistorCircuitsTheresistor’sinstantaneouscurrentandvoltageoscillateinphase.
Thecurrentisatitsmaximumandminimumvalueswhenthevoltageisatitsmaximumandminimumvalues.
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ACPowerinResistorsTheinstantaneouspoweriswrittenas
p =iR2R =[IR cos(2p ft)]2R =IR2R[cos(2p ft)]2
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ACPowerinResistorsThepoweroscillatestwiceduringeverycycleoftheemf.
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ACPowerinResistorsThecurrentinalightbulbreversesdirection120timespersecond,sothepowerreachesamaximumtwicepersecond.Howeverthebulbglowssteadily,sotheaveragepower ismoreuseful.TheaveragepowerPR is
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ACPowerinResistorsRecallthatthepowerinaDCcircuitisPR=I 2R =V 2/R.
Theroot-mean-squarecurrentandtheroot-mean-squarevoltagearemoreusefulexpressionsofpower:
Usingtheroot-mean-squarevalues,wewritetheaveragepowerdissipatedinanACcircuitas:
Aslongasyouworkwithrms voltagesandcurrents,alltheexpressionsforDCpowercarryovertoACpower.
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Example26.1TheresistanceandcurrentofatoasterThehotwireinatoasterdissipates580Wwhenpluggedintoa120Voutlet.a. Whatisthewire’sresistance?b. Whataretherms andpeakcurrentsthroughthewire?PREPAREWe’veseenthatthe120Voutletvoltageisanrms value.ThefilamenthasresistanceR.Itdissipates580Wwhenthere’sanrmsvoltageof120Vacrossit.WecansolveEquation26.9forR andthenuseEquations26.9and26.8tofindtherms currentandthepeakcurrent.
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Example26.1Theresistanceandcurrentofatoaster(cont.)SOLVE
a. WerearrangeEquation26.9tofindtheresistancefromthermsvoltageandtheaveragepower:
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Example26.1Theresistanceandcurrentofatoaster(cont.)b. AsecondrearrangementofEquation26.9allowsustofindthe
currentintermsofthepowerandtheresistance,bothofwhichareknown:
FromEquation26.8,thepeakcurrentis
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Example26.1Theresistanceandcurrentofatoaster(cont.)ASSESS Wecandoaquickcheckonourworkbycalculatingthepowerfortheratedvoltageandcomputedcurrent:
PR =Irms Vrms =(4.8A)(120V)=580WThisagreeswiththevaluegivenintheproblemstatement,givingusconfidenceinoursolution.
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ExampleProblemA120V(rms)ACpowersupplyisconnectedtoamotor,whichisratedat100W.A. Whatistherms currentinthecircuit?B. Now,supposethatthewiresusedtoconnectthemotortothepowersupply
havearesistanceof7.0Ω.Assumethattherms currentstaysthesame.Whatisthevoltagedropacrosstheresistanceofthewires?Whatisthevoltageatthemotornow?
C. Now,supposethepowersupplyis1200V,andthemotorisratedat100Watthishighervoltage.Whatisthecurrentinthecircuitassumingnoresistanceinthewires?
D. Ifthewireshavearesistanceof7.0Ω,whatisthevoltagedropacrossthewires?Thevoltageatthemotor?
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Section26.2ACElectricityandTransformers
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ACElectricityandTransformersAtransformerisadevicethattakesanalternating voltageasaninputandproduceseitherahigherorlowervoltageasoutput.Theoperationofatransformerisbasedontheemf producedbychangingmagneticfields,sotheinputmustbeACelectricity.
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TransformerOperationAsimplifiedversionofatransformerconsistsoftwocoilswrappedonasingleironcore.
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TransformerOperationInasimpletransformer,theprimarycoil isconnectedtoanACvoltage.TheACvoltagecreatesanalternatingcurrent.Thecurrentinthecoilcreatesamagneticfieldthatmagnetizestheironcoretoproduceamuchstrongernetfield.Thestrongfluxfollowstheironcoreandentersthesecondarycoil.
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TransformerOperationThecurrentintheprimarycoilofatransformerisanalternatingcurrent;itcreatesanoscillatingmagneticfieldintheironcore.
Thechangingmagneticfieldmeansthereisachangingmagneticfluxthroughthesecondarycoil,whichinducesanemf,anACvoltageV2 inthecoil.
Aresistor,orload, isconnectedtodissipatepower.
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TransformerOperation
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TransformerOperationThepurposeofatransformeristochangethevoltage,soweneedtounderstandhowthevoltageintheprimarycoilofatransformerrelatestothatinthesecondarycoil.AccordingtoFaraday’slaw,aninstantaneousvoltageacrossN1 turnsduetothemagneticfluxF intheprimarycoilis
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TransformerOperationInanidealtransformer, allthefluxis“guided”bytheironcorethroughthesecondarycoil.Thechangingfluxinducesanemf acrossthesecondarycoilgivenby
BecauseΔF/Δt isthesameineachcoil,wecanwritearatioofinstantaneousvoltages:
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TransformerOperationWecanalsorelatethepeakandrms voltagesoftheprimaryandsecondarycoils:
Astep-uptransformer, withN2> N1,increasesthevoltage,whileastep-down transformer, withN2 < N1,lowersthevoltage.
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Wecanrelatethepeakcurrentsbetweenthecoilsinatransformerwithanexpressionthatmirrorstheexpressionofpeakvoltagesbetweenthecoils:
A step-uptransformerraisesvoltagebutlowerscurrent;astep-downtransformerlowersvoltagebutraisescurrent.
TransformerOperation
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QuickCheck26.1Iftheprimarycoilofwireonatransformeriskeptthesameandthenumberofturnsofwireonthesecondaryisincreased,howwillthisaffectthevoltageobservedatthesecondary?
A. Thevoltagewillincrease.B. Thevoltagewillstaythesame.C. Thevoltagewilldecrease.
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QuickCheck26.1Iftheprimarycoilofwireonatransformeriskeptthesameandthenumberofturnsofwireonthesecondaryisincreased,howwillthisaffectthevoltageobservedatthesecondary?
A. Thevoltagewillincrease.B. Thevoltagewillstaythesame.C. Thevoltagewilldecrease.
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QuickCheck26.2Supposethatanidealtransformerhas400turnsinitsprimarycoiland100turnsinitssecondarycoil.Theprimarycoilisconnectedtoa120V(rms)electricoutletandcarriesanrms currentof10mA.Whataretherms valuesofthevoltageandcurrentforthesecondary?
A. 480V,40mAB. 480V,2.5mAC. 30V,40mAD. 30V,2.5mA
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QuickCheck26.2Supposethatanidealtransformerhas400turnsinitsprimarycoiland100turnsinitssecondarycoil.Theprimarycoilisconnectedtoa120V(rms)electricoutletandcarriesanrms currentof10mA.Whataretherms valuesofthevoltageandcurrentforthesecondary?
A. 480V,40mAB. 480V,2.5mAC. 30V,40mAD. 30V,2.5mA
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Example26.2Analyzingastep-downtransformerAbooklighthasa1.4W,4.8Vbulbthatispoweredbyatransformerconnectedtoa120Velectricoutlet.Thesecondarycoilofthetransformerhas20turnsofwire.Howmanyturnsdoestheprimarycoilhave?Whatisthecurrentintheprimarycoil?
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Example26.2Analyzingastep-downtransformer(cont.)
Weknowthevoltagesoftheprimaryandthesecondary,sowecancomputetheturnsintheprimarycoilusingEquation26.13.Weknowthevoltageandthepowerofthebulb,sowecanfindthecurrentinthebulb.Thisisthecurrentinthesecondary,whichwecanuseinEquation26.14tofindthecurrentintheprimary,thecurrentprovidedbytheoutlet.
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Example26.2Analyzingastep-downtransformer(cont.)SOLVE Thebulbisratedat4.8V;thisistherms voltageatthesecondary,so(V2)rms =4.8V.Thepoweroutlethastheusual(V1)rms =120V,sowecanrearrangeEquation26.13tofind
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Example26.2Analyzingastep-downtransformer(cont.)Thebulbconnectedtothesecondarydissipates1.4Wat4.8V;thisisanrms voltage,sotherms currentinthesecondaryis
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Example26.2Analyzingastep-downtransformer(cont.)WecanthenrearrangeEquation26.14tofindtherms currentintheprimary:
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Example26.2Analyzingastep-downtransformer(cont.)ASSESS Wecancheckourresultsbylookingatthepowersuppliedbythewalloutlet.ThisisP1 =(120V)(0.012A)=1.4W,thesameasthepowerdissipatedbythebulb,asmustbethecasebecausewe’veassumedthetransformerisideal.
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PowerTransmission
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Example26.3PracticalpowertransmissionToprovidepowertoasmallcity,apowerplantgenerates40MWofACelectricity.Thepowerplantis50kmfromthecity(atypicaldistance),andthe100kmofwireusedinthetransmissionline(tothecityandback)hasaresistanceof7.0Ω.
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Example26.3Practicalpowertransmission(cont.)a. Toprovide40MWofpoweratthegeneratorvoltageof25,000V,
whatcurrentisrequired?b. Whatisthepowerdissipatedintheresistanceofthetransmission
lineforthiscurrent?c. Toprovide40MWofpowerat500,000V,whatcurrentis
required?d. Whatisthepowerdissipatedintheresistanceofthetransmission
lineforthishighervoltage?
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Example26.3Practicalpowertransmission(cont.)PREPAREWecantreatthecity—andthewiresthattransmitpowertoit—asaload.Allofthevoltagesarerms valuesandthepowerisanaveragepower,sowecanfindthecurrenttoprovideagivenpowerandthepowerdissipatedusingtherelationshipsinEquation26.9.
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Example26.3Practicalpowertransmission(cont.)SOLVE
a. Toprovide40MWatthegeneratorvoltageof25,000V,thecurrentis
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Example26.3Practicalpowertransmission(cont.)
b. Passingthiscurrentthroughthetransmissionlineswillresultinpowerdissipationinthe7.0Ωresistanceofthewires.Wedon’tknowthevoltagedropacrossthewires,butwedoknowthecurrentandresistance,sowecancompute
Pdissipated inwires =(Irms)2R =(1600A)2 (7.0Ω)=18MW
Thisisnearlyhalfthepowergenerated,clearlyanunacceptableloss.
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Example26.3Practicalpowertransmission(cont.)c. Increasingthetransmissionvoltageto500kVreducesthe
necessarycurrent:
Thisisaremarkablysmallcurrenttosupplyacity.Ifyouuseseveralhigh-powerappliancesatonetime,youcouldeasilyusethismuchcurrentinyourhouse.Butthenecessarycurrentforthecitycanbesosmallbecausethevoltageissolarge.
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Example26.3Practicalpowertransmission(cont.)d. Therelativelysmallcurrentmeansthatthepowerdissipatedin
theresistanceofthewireswillbesmallaswell:Pdissipated inwires =(Irms)2R =(80A)2 (7.0Ω)=0.045MW
Thisisonlyabout0.1%ofthepowergenerated,whichisquitereasonable.
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Example26.3Practicalpowertransmission(cont.)ASSESS Thefinalresult—thepowerdissipatedinthewiresisdramaticallyreducedforanincreasedtransmissionvoltage—isjustwhattheexamplewasdesignedtoillustrate.
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PowerTransmissionTransmittingelectricityathighvoltagesmeansasmallercurrentisrequired,andthereforetheresultingpowerlossismoremanageablethanforlowvoltages(andlargercurrents).Thisiswhyelectricaltransmissionlinesrunathighvoltages.Inordertotransmitelectricityathighvoltages,weneedtousetransformerstoincreasethevoltage,whichrequiresACelectricity.ThisiswhyweuseACpowereventhoughitisslightlymoredangerousthanDCpower.
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