dc resistance.docx
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dc resistance.docx
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dcresistance
Directcurrent:
Ohm'slaw,resistivity,resistance—seriesandparallelcombinations,
Jouleheatingeffect,electricpower,Kirchoffslaws,electricmeters.
EMFandTerminalvoltage.
Tohavecurrentinanelectriccircuit,needadevicesuchasabatteryoranelectricgenerator,thatcantransformsonetypeofenergy(chemical,mechanicalenergy)intoelectricenergy.Suchdeviceiscalledsourceorelectromotiveforceoremf.Thevoltagedifferencebetweentheterminals,whennocurrentflowtoexternalcircuitiscalledemfofthesource.thesymbolεisusuallyusedforemf.
Electriccurrent
thedirectionofthecurrentisthedirectionofpositivechargesflow.
currentisdefineastherateofchargeflowsthroughthissurfaceofareaA.
fromthedefinitioncurrentIistherateofpositivechargesflowor
TheSIunitofcurrentisampere(A)
Itisconventionaltoassigntothecurrenttothesamedirectionastheflowofpositivecharges.
Inconductorsuchcopper,aluminiumthecurrentisduetomotionofnegativechargesofelectronflows.Thenthedirectionofcurrentflowisoppositetothedirectionofflowofelectrons.
example.acurrentof2.5Aexistsinawirefor4.0minute.(a)Howmuchcharge(b)howmanymayelectronswouldthisbe.(thechargeofoneelectronis1.6x10-19C)
ΔQ=IΔt=2.5x4x60=600C
(b)noofelectron
electrons.
Resistance.
Consideraconductorofcross-sectionareaAcarryingacurrentI.ThecurrentdensityJiscurrentperunitareaAor
theunitisA/m2.
AcurrentdensityJandelectricfieldEareestablishedinaconductorwheneverapotentialdifferenceismaintainedacrossaconductor.Thecurrentdensityisproportionaltotheelectricfield.or
J=σE
theconstantproportionalityσiscalledconductivityoftheconductor.
apotentialdifferencerelatedtotheelectricfieldis
ΔV=El
becauseJ=σE,thenthepotentialdifferencecawriteas
aconductorlengthlandcross-sectionA.ApotentialdifferenceΔV=Vb–VaismaintainedacrossaconductorsetupanelectricfieldE.
ohmlaw
ItisdefinethattheresistanceastheratioofpotentialdifferenceΔVtothecurrentI
example.
Theresistanceofawire20mlongis0.10Ω.Iftheresistivityofthewireis1.68x10-8Ω.mwhatisthediameterofthewire.(b)ifthecurrentis4.0Awhatispotentialdifferenceorvoltagedrop,acrossawire.
ElectricPower
theenergytransformwhenachargeQmovesthroughpotentialdifferenceVisQV,
thenpower,therateofenergyis
example.
(a)Whatistherequiredresistanceofanimmersionheaterthatwillincreasethetemperatureof1.5kgofwaterfr0m10.00Cto50.00Cin10.0minwhileoperatingat110V.
(b)Estimatethecostofheatingthewater.(estimatethepriceof10centperkilowatt-hour.)
energytoheatthewaterisQ=mcΔT
therateofenergydeliveredbyresistorisequaltotherateofenergyenteringthewater
Resistanceandtemperature.
Thechangeofresistivityisdirectlyproportionaltothechangeoftemperatureor
ΔραΔT
Δρ=(αρ0)ΔTwhereαcalledtemperaturecoefficientofresistivity
ρ-ρ0=(αρ0)ΔTorρ=ρ0[1+αΔT]
andbecauseresistanceisproportionaltoresistivity
R=R0[1+αΔT]
example.
theresistancemadeupfromplatinum(αplatinum=3.92x10-3C-1)hasvalueof50.0Ωat20.00C.Whenimmersedinavesselcontainingmeltingindium,itsresistanceincreaseto76.8Ω.Calculatethemektingpointoftheindium.
Directcurrentcircuit.
1.thebatteryhasresistancecalledinternalresistancer.
2.Thecurrentflowfromterminal+vetoterminal–veofthebattery.
thedashedboxinthediagramrepresentthebatteryhavingemfεandinternalresistancer.
3.theterminalvoltageofthebattery
ΔV=Vb–Vais
4.theterminalvoltageofthebattery
ΔV=ε–Ir
5.theterminalvoltageacrosstheresistorRis
ΔV=IR
6.ε=IR+Ir
7.thepowerofthecircuit
Resistorsinseriesandparallel
(a)Resistorinseries.
whereReq=R1+R2
ΔV=IR1+IR2=I(R1+R2)=IReq
(b)Resistorinparallel
(i)ThecurrentIthatentertopointamustbeequaltothecurrentleavingthat
point
I=I1+I2
(ii)thepotentialdifferenceΔVisthesame.
VR2=VR1=ΔV
(iii)sinceI=I1+I2or
example
(i)Findtheequivalentresistancebetweenpointaandc.GiventhatR1=8Ω,R2=4Ω,R3=6Ω,andR4=3Ω,
(ii)IfVacis42Vfindthecurrentineachresistor.
KirchoofRulesonR-Icircuit
(i)Thesumofthecurrententeringanyjunctioninacircuitisequaltothesumof
thecurrentleavingthejunction
ΣIin=ΣIoutorI=I1+I2
+I1
(iii)Inclosedcircuitloopthesumofemfεisequaltothesumofpotential
differenceacrosstheresistance.
Σε=ΣIR
+I1
signofemfε.Whenthesourceemfεistraversedfrom–veto+vethenemfεispositive.
Whenthesourceemfεistraversedfrom+veto-vethenemfεisnegative.
example.Singleloop
findthecurrentofsingleloopcircuit.
Σε=ε1–ε2=6-12=-6
ΣIR=I(10)+I(8)
Σε=ΣIR
6-12=I(10)+I(8)
I=-3A
the-3Aindicatethedirectionofthecurrentisoppositefromtheassumeddirection
example.2loopofcurrentcircuit.
I1=I2+I3…………….
(1)
loop1bcfeb
Σε=ΣIR
10+14=I2(4)+I1(6)………..
(2)
loop11bcdab
Σε=ΣIR
10=I1(6)+I3
(2)…………..(3)
finallywehavethreeequationswiththreeunknownparameter
I1,I2andI3
ifwesolvedthreeequation
I1=2A
I2=3A
andI3=-1A
sinceI3=-1Anegativevalue,thenthedirectionofI3isopposite.thecorrectdirectionofI1,I2andI3asshowninthediagrambelow
IftheloopofclosecircuitweassumefollowclockwisethentheequationfromKirchhoffruleshownasbelow.
thesolutionofI1,I2andI3
-24=4I2–6I1……………………
(1)
10=6I1+2(I1+I2)=8I1+2I2
10=8I1+2I2multiplyby2
20=16I1+4I2…………………..
(2)
subtractequation
(2)from
(1)toeliminatesI2
44=22I1I1=2A
20=16
(2)+4I2=32+4I2
I2=-3A
I3=I1+I2=2-3=-1A
loopclockwisebefcb
-14-10=4I2–6I1
or-24=4I2–6I1
loopclockwisebcdab
10=6I1+2I3
SincetheanswerofI2andI3arenegativethenthecurrentareoppositeofdirectionthatwechooseforthem.Thevaluearecorrect.
IfR=2.8kΩwhatistheresistanceinequivalentandcurrentindiagramA
\
Findthepotentialdifferencebetweenpointaandbandthecurrentinthe20Ω
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