ELECTRIC CURRENT AND RESISTANCE (Ch. 26 of Text)

Inthissection,wediscuss

electriccurrentintermsoftheflowofelectriccharge conductionmechanismsindifferentmaterials,especiallymetals therelationbetweencurrent,voltage,andresistanceusingOhmslaw electricpower

ELECTRICCURRENTTheflowofchargeconstituteselectriccurrent.Quantitatively,currentisthenetrateofchargecrossinganarea.If Q isthechargethatflowsthroughacrosssectionalarea A intime t ,thenthecurrentis QI

t

TheSIunitofcurrentistheampere(A):1A=1C/s.Byconvention,thedirectionofcurrentistakentobethedirectionofpositivechargeflow.Thus,ifelectronsmoveleftward,thecurrentisdirectedrightward.Theinstantaneouscurrentisgivenby

(instantaneous current)dQIdt

Acurrentmayconsistofonekindofmovingcharge,orboth.Ifitisboth,thenthenetcurrentisthesumofthecurrentscarriedbythepositiveandnegativecharges.Current:AMicroscopicLookCurrentdependsonthespeedofthechargecarriers,theirdensity,andtheircharge.Insomecases,likeabeamofprotonsinanaccelerator,speedmeanstheactualspeedofthecharges.Butintypicalconductors,chargesaremovingaboutathighspeedswithrandomthermalvelocitiesthatdontresultinanetflowofcharge.Whenacurrentispresent,thereisanadditionalandusuallyverysmalldriftspeed( dv )superposedonthechargesrandommotion,anditisthis dv thatdeterminesthecurrent.

SomiSticky NoteThis is the average current.

Thefigureshowsconductorthatcontains n chargesperunitvolume(units:C/m3),eachwithcharge q anddriftspeed dv .Let Abetheconductorscrosssectionalarea.Consideralength L oftheconductor:ithasvolume AL andcontainsnAL chargecarriersforatotalchargeof Q nALq .Movingat dv ,thischarge( Q )takestime / dt L v topassagivenpoint.Thenthecurrentis

/ dd

Q nALqI nAqvt L v

Example:Acopperwirehasaradiusof0.815mm.Assuminganumberdensityof 28 38.5 10 /n m ,calculatethedriftspeedforacurrentof1A.Solution:Usetheformulaabovewith q e and 2A r . 52 3.5 10 0.035 mm/sm/s .( )d

I IvneA ne r

This is an extremely low speed! Estimate how long it takes for one of these drifting electrons to move one metre. Example:Inacertainparticleaccelerator,acurrentof0.5mAiscarriedbya5MeVprotonbeamthathasaradiusof1.5mm.(a)Findthenumberdensityofprotonsinthebeam.(b)Ifthebeamhitsatarget,howmanyprotonhitin2s?Solution:(a)Use /n I qAv with 191.6 0 C1q e and 2A r . Find v from the KE. (Recall that 191eV 1.6 10 J .) 2 6 19 13 7

2( ) ) J1 5 10 (1.6 10 8 10 3.1 10J s2

m/KK m vm

v where 271.67 10m kg was used. Plugging in this value of v in /n I qAv , you'll find 131.4 19n protons /m3. (b) The total charge that strikes the target in 2 s is

3(0.5 10 C/s)(2 s) 1 mC.Q I t To find N , the number of protons hitting the target, divide Q by e :

3

1519 6.3 1010

10 C .1.6 C

QNe

SomiSticky NoteDrift speed is a really tiny speed. Even snails move faster than this.

SomiSticky NotemA means milliAmps, or one-thousandth of an amp.

CurrentDensity:Currentsarentconfinedtowires.CurrentsintheEarth,inchemicalsolutions,inyourbody,andinionizedgasesflowinilldefinedpaths,andtheirmagnitudeanddirectioncanvarywithposition.Tocharacterizesuchdiffusecurrents,itisconvenienttointroducecurrentdensity, J ,avectorwhosedirectionateachpointisthatofthelocalcurrentandwhosemagnitudeisthecurrentperunitarea(units:A/m2).

dd d

nAqv nqvI vJ qA A

J n Conduction mechanisms:

Resistance and Ohm's Law Electricfieldsexertforcesoncharges,soitisthepresenceofelectricfieldsinconductorsthatresultsinacurrent.Inaconductorinequilibrium,thereisnoelectricfield.Whenapotentialdifferenceisappliedacrossaconductor,thenanelectricfieldisproducedinsidetheconductor,anditisnolongerinequilibrium.Newtonslawsuggeststhatanelectricfieldshould

Conduction occurs differently in different types of materials: In metallic conductors, current is carried by free electrons. In ionic solutions, current is carried by positive and negative ions. Plasmas are ionized gases, with current carried by electrons and ions. Semiconductors involve current carried by

both electrons and holesabsences of electrons in a crystal structure.

Semiconductors are at the heart of modern electronics.

Their electrical properties can be altered by the controlled addition of small amounts of impurities.

Superconductors offer zero resistance to the flow of current, and thus can transmit electric power without loss of energy.

Known superconducting materials all require temperatures far below typical ambient temperatures.

acceleratefreechargesinaconductor,resultinginaneverincreasingcurrent.Butinmostconductors,chargescollidewiththingsusuallyionsandlosetheenergytheyvegainedfromthefield,asshownontheright.Thesecollisionsprovideaneffectiveforcethatcounterstheelectricforce,andtheendresultisthatittakesanelectricfieldtosustainasteadycurrent.Inmanymaterials,thefieldandthecurrentdensityarerelatedbyOhmsLaw:

(Ohm s law, microscopic version) J E where isknownastheconductivityofthematerial.Wellrestrictourattentiontoohmicmaterialsinwhich isindependentoftheappliedfield.(YoumaybemorefamiliarwiththemacroscopicversionofOhmslaw,V IR ,whichwelldiscussshortly.)Theconductivity, ,tellsushowlargeacurrentdensitywillresultfromagivenfield:itisameasureofhoweasilychargesinthematerialcanmove.Aperfectconductorwouldhave ;aperfectinsulator, 0 .Arelatedquantityistheresistivity, ,definedastheinverseofconductivity: 1/ .Then,Ohmslaw, J E ,canbewrittenas

EJ

.Resistance:Thefigureshowsawiresegmentoflength L andcrosssectionalarea A carryingacurrent I .Thefieldpointsinthedirectionofdecreasingpotential,so a bV V .Thepotentialdifferenceis a bV V V E L .Theratio /V I isdefinedtobetheresistanceofthesegment: Resistan )e( cVR

I

TheSIunitofresistanceistheohm( ):1 1 Volt/Ampere.Formanymaterials, R doesntdependonthevoltagedroporthecurrent.Suchmaterials,whichincludemostmetals,arecalledohmicmaterials(mentionedearlier): con t, ( )s antIRV R ThisisthemacroscopicversionofOhmslaw.Themicroscopicversionis

/J E .Toseetheirequivalence,set /V EL JL IL A inV IR :Then, .IL LIRV A RA Thus,thetwoversionsareequivalentif /R L A .

Example:ANichromewire( 610 m)hasaradiusof0.65mm.Whatlengthofwireisneededtoobtainaresistanceof 2.0 ? Solution:Solve /R L A for L .Since 2A r ,

2 )( 2.65 m. r RL

Example, A material has resistivity, a block of the material has a resistance:

Problem 26.65 (from Ch 26 of Text): A potential difference V is applied to a wire of length L . cross-sectional area A , and resistivity . You want to change the applied potential difference and stretch the wire so that the energy dissipation rate is multiplied by 30.0 and the current is multiplied by 4.00. Assuming the wire's density does not change, what are (a) the ratio of the new length to L and (b) the ratio of the new cross-sectional area to A ? Solution: We use P = i2 R = i2L/A, or L/A = P/i2. (a) The new values of L and A satisfy

LA

Pi

Pi

LA

FHGIKJ FHGIKJ

FHGIKJ

FHGIKJnew new old old2 2 2

304

3016 .

SomiSticky NoteThis problem requires knowledge of electric power. Study it only after "power" has been discussed in class. See page 9 if you wish to study it on your own.

Consequently, (L/A)new = 1.875(L/A)old, and

newnew old oldold

1.875 1.37 1.37LL L LL

. (b) Similarly, we note that (LA)new = (LA)old, and

newnew old old

old

1/1.875 0.730 0.730AA A AA

. Ohms law: Comparing microscopic and macroscopic versions

26.4: Resistance and Resistivity, Variation with Temperature:

Resistivityandresistancevarywithtemperature.Experimentally,itisfoundthattherelationbetweentemperatureandresistivityforcopperandformetalsingeneralisfairlylinearoveraratherbroadtemperaturerange.Forsuchlinearrelationswecanwriteanempiricalapproximationthatisgoodenoughformostengineeringpurposes: 0 01( ) ( )T T T Here, 0 istheresistivityvalueatareferencetemperature 0T ,usuallytakentobe293K(roomtemperature). isknownasthetemperaturecoefficientofresistance.Thetableonlastpageliststhevaluesof and forsomematerials.Multiplyingbothsidesoftheaboveequationby /L A andrecognizing

/R L A , we find a similar relation is obeyed by the resistance: 0 01( ) ( )R T R T T

Formetals, and R increase with temperature, but for semiconductors, and R decrease. (The reason is the conduction mechanism in the two systems are quite different.) Problem 26.71 (from Ch 26 of Text): (a) At what temperature would the resistance of a copper conductor be double its resistance at 20.0C? (Use 20.0C as the reference point in Eq. 26.17; compare your answer with Fig. 26.10.) (b) Does this same doubling temperature hold for all copper conductors, regardless of shape or size? Solution: Solve 0 0 0( ) ( )1 2R T R T T R for T . We find 0( ) 1T T . 3 10 1 250101 4.3 KT T T C (b) Yes, it holds for all Cu conductors, since they have the same .

PowerinElectricCircuits Whenthereisanelectricfieldinaconductor,thefreeelectronsareacceleratedforashorttime,givingtheelectrongasincreasedKE,butthisadditionalenergyisquicklydissipatedasthermalenergybecauseofcollisionswithlatticeions.ThisincreaseinthermalenergyiscalledJouleheat. The figure shows a segment of length L and area A carrying a current I . In time

t , a charge Q enters from the left (at potential 1V ) and exits from the right (at potential 2V ). Let 1 2V V V . The change (decrease) in the PE of Q is

2 1)( ( ) .U Q V Q QV V V The rate of energy loss is the time derivative:

Power .dU VIdQVdt

P VIdt

P VI is the rate at which Joule heat is produced in the conductor. Using V IR (Ohm's law), we may write this in other forms:

22Power dissipated in a resistor VP VI I R

R

Example:

The units of power are WATTS. Note that 1 W = 1 Volt.Ampere.

Problem 26.45 (from Ch 26 of Text): A 1250 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in 1.0 h?

Worked Solution

THINK

Let P be the power dissipated, i the current in the heater, and V the potential difference across the heater. The three quantities are related by P iV .

EXPRESS

The current is given by /i P V . Using Ohm's law V iR , the resistance of the heater can be written as

2

./

V V VRi P V P

ANALYZE a. Substituting the values given, we have

b. Similarly, the resistance is

c. The thermal energy E generated by the heater in time 1.0 h 3600 st is

LEARN

Current in the heater produces a transfer of mechanical energy to thermal energy, with a rate of the transfer equal to 2 /P Vi V R .

EMFandBatteriesTomaintainacurrentinaconductor,weneedasupplyofelectricalenergy.Adevicethatsupplieselectricalenergyiscalledasourceofemf.(Thelettersemfstandforelectromotiveforce,atermthatisnowrarelyused.)Examplesofemfsourcesareabattery(whichconvertschemicalenergyintoelectricalenergy)andagenerator(whichconvertsmechanicalenergyintoelectricalenergy).Asourceofemfdoesworkonapositivechargepassingthroughit,raisingitsPE.Theworkdoneonunitchargeiscalledtheemf, ,ofthesource.Theunitof isthevolt.Anidealbatteryisasourcethatmaintainsaconstantpotentialdifference(PD)betweenitsterminals,independentofthecurrentflow.Thefigureshowsasimplecircuitconsistingofanidealemfandaresistor.Assumingconnectingwirestohavenegligibleresistance,wefind a b c dV V V V .Thecurrentisclockwiseandhasthevalue /I R by Ohm's law. Consideracharge

Q movingthroughtheemf(fromto+)inatime t .ItsgaininPEis Q .TherateatwhichPEisgainedisthepoweroutputoftheemf:

QP It

Asthechargeflowsthrough R ,thisPEisconvertedintothermalenergy:thepoweroutputofthesourceequalsthepowerdissipatedin R (energyconservation).Inarealbattery,thePDacrossthebatteryterminals(calledtheterminalvoltage)dependsonthecurrentflowing,asthegraphshows.Inthecircuitontheright,thebatteryisrealandhasaninternalresistance r .Theterminalvoltageisnot anymore;itisnow a bV V Ir . Thisobviouslydecreaseslinearlywiththecurrent.Sincetheconnectingwireshavenoresistance,

a b I RV IV r ,whichcanbesolvedforthecurrent: ( )I R r .

SomiSticky NoteAn ideal battery has no internal resistance.

SomiSticky NoteThe battery in the figure is a REAL battery in that it has an internal resistance: r.

Inthepreviouscase,weassumed 0r ,andso I R andtheterminal

voltagewas a b IRV V .Realbatteriessuchasagoodcarbatteryusuallyhaveaninternalresistanceoftheorderofafewhundredthsofanohm,sotheeffectofthe Ir termisnegligibleunlessthecurrentisverylarge.Onesignofabadbatteryislargeofvalueof r .Ifyoususpectyourcarbatteryisbad,checkingtheterminalvoltagewithavoltmeter(whichdrawsonlyasmallcurrent)isnotsufficient:youneedtocheckitwhilealargecurrentisdrawnfromthebattery,suchaswhileyouarestartingyourcar.Carbatteriesareoftenratedinamperehours(A.h),whichisthetotalchargetheycandeliver:1A.h=(1C/s)(3600s)=3600C.ThetotalenergystoredinthebatteryisW Q .Example:An11 resistorisconnectedacrossabatteryofemf6Vandinternalresistance1 .Find(a)thecurrent,(b)theterminalvoltage,(c)thepowerdeliveredbythebattery,(d)thepowerdeliveredtothe11 resistor,and(e)thepowerdissipatedbythebatterysinternalresistance.(f)Ifthebatteryisratedat150A.h,howmuchenergydoesitstore?

Solution:(a) 6 A( ) (11 1) 0.5 .I R r

(b) 6 (0.5)(1) 5. V5 .a bV V Ir (c) 3 W.batP I

(d) 2 2. W75 .RP I R

(e) Powerdissipatedasheatinsidethebatteryis 2 0.25I r W.Thisisequalto

bat RP P ,whichisnotsurprising(energyconservation). (f) Thetotalenergystoredistheemftimesthetotalcharge:

3600C1h(150 ) 3.2A. .h M46 V JW Q