electricity magnetism lecture 9: electric current lecture 09...electric current electricity &...

Electricity & Magnetism Lecture 9: Electric Current

Today’sConcept:

ElectricCurrent

Electricity&Magne8smLecture9,Slide1

Midterm

! B9201:6:30pmto8:00pm! Coversuptotoday’slecturematerial(Unit9)andnextweek’stutorials

How do you feel about circuits

A. Icompletelyunderstandthemfromhighschool

B. NeedReview

C. S8llhopeless

D. Circuit?What’sthat?

Joke of the day: I could not resist coming to class today

(b) Measure the current through the circuit when the switch is closed and the light is lit. Enter the value in the table below in Activity 22-6d

Now suppose you connect a second bulb, as shown in Figure 22-9.

A

+-

+-

A+ -

V

+-

V

+-

V+

–

+

A

-

+ –A

Figure 22-9: Two bulbs connected in series with a voltmeter and an ammeter.

✍ Activity 22-6: Current and Voltage Measurements with Two BulbsThe predictions below should be completed before class.(a) How do you think the voltage across the battery will compare to that with only one bulb? (More, less or the same within measurement error?)

(b) What do you think will happen to the brightness of the first bulb when you add a second bulb? Explain.

(c) What will happen to the current drawn from the battery? Explain.

(d) Connect a second bulb as shown, and test your predictions. Measure the voltage across both the bulbs and the current entering both bulbs with the switch closed and record in the table.

Measurements 22-5 and 22-6

1 bulb 2 bulbs

voltage

current

Workshop Physics II: Unit 22 – Batteries, Bulbs, & Current Flow Page 22–13Authors: Priscilla Laws, John Luetzelschwab, David Sokoloff, & Ron Thornton

© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by N. Alberding & S. Johnson, 2007, 2014.

Whentheswitchisclosed,andbulbsareiden8cal

A. Topbulbisbrighter

B. BoWombulbisbrighter

C. Bothareequallybright

Whentheswitchisclosed

A. VbaWery=Vbulb

B. VbaWery<Vbulb

C. VbaWery>Vbulb

Figure 22-7 shows a simple circuit with a battery, a bulb, and two voltmeters connected to measure the voltage across the battery and the voltage across the bulb. The circuit is drawn again symbolically on the right. Note that the word across is very descriptive of how the voltmeters are connected to measure voltage.

V

+-

V+ -

V

+-

V

+

-

V+

–

V

+

–

Figure 22-7: Two voltmeters connected to measure the voltage across the battery and the bulb.

To do the next few activities, you will need:

• a D-cell alkaline battery and holder • 2 #14 light bulbs • a SPST switch • a digital voltmeter • an ammeter, 0.25 A

✍ Activity 22-4: Voltage Measurements in a Simple Circuit(a) First connect both the + and the - clips of the voltmeter to the same point in the circuit. Observe the reading. What do you conclude about the voltage when the leads are connected to each other (i.e. not across anything else)?

(b) In the circuit in Figure 22-7, predict the voltage across the battery compared to the voltage across the bulb. Explain your predictions.



Howdothecurrentsmeasuredcompare?A. Ile^<IrightB. Ile^=IrightC. Ile^>Iright

(c) Now test your prediction. Connect the circuit in Figure 22-7. Use the voltmeter to measure the voltage across the battery and then use it to measure the voltage across the bulb.

Voltage across the Battery

Voltage across the bulb

What do you conclude about the voltage across the battery and the voltage across the bulb?

Now let's measure voltage and current in your circuit at the same time. To do this, connect a voltmeter and an ammeter so that you are measuring the voltage across the battery and the current entering the bulb at the same time. (See Figure 22-8.)

V

+-

A

+-

V

+ –

+

–

A

Figure 22-8: Meters connected to measure the voltage across the battery and the current through it. (The positive terminal of the battery is at the bottom.)

✍ Activity 22-5: Current and Voltage Measurements(a) Measure the voltage across the battery when the switch is closed and the

light is lit. Enter the value in the table below in Activity 22-6d

Page 22-12 Workshop Physics II Activity Guide SFU

© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by N. Alberding & S. Johnson, 2007,2014.



A

+-

+-

A+ -

V

+-

V

+-

V+

–

+

A

-

+ –A







1 bulb 2 bulbs

voltage

current



HowdowtheVoltagesmeasuredcompare?A. Vle^<Vright

B. Vle^=Vright

C. Vle^>Vright

(c) Now test your prediction. Connect the circuit in Figure 22-7. Use the voltmeter to measure the voltage across the battery and then use it to measure the voltage across the bulb.

Voltage across the Battery

Voltage across the bulb

What do you conclude about the voltage across the battery and the voltage across the bulb?

Now let's measure voltage and current in your circuit at the same time. To do this, connect a voltmeter and an ammeter so that you are measuring the voltage across the battery and the current entering the bulb at the same time. (See Figure 22-8.)

V

+-

A

+-

V

+ –

+

–

A

Figure 22-8: Meters connected to measure the voltage across the battery and the current through it. (The positive terminal of the battery is at the bottom.)

✍ Activity 22-5: Current and Voltage Measurements(a) Measure the voltage across the battery when the switch is closed and the

light is lit. Enter the value in the table below in Activity 22-6d

Page 22-12 Workshop Physics II Activity Guide SFU

© 1990-93 Dept. of Physics and Astronomy, Dickinson College Supported by FIPSE (U.S. Dept. of Ed.) and NSF. Modified at SFU by N. Alberding & S. Johnson, 2007,2014.



A

+-

+-

A+ -

V

+-

V

+-

V+

–

+

A

-

+ –A







1 bulb 2 bulbs

voltage

current



Tworesistors,onehavinghalftheresistanceoftheother,areconnectedtoabaWeryasshown.Whatisthevoltageacrossthebiggerresistor?

A)VB/2=0.75V

B)VB/3=0.50V

C)3VB/2=2.25V

D)2VB/3=1.00V

.

100 Ω

50 Ω1.5 V

Your stuff

➡ sowhichwaydoesDCcurrentflow?-_-➡ “Pleaseexplainhowelectricfieldiscalculatedinsidethecopperwire.”

➡ “SinceR=ρL/A,thegreaterthecrosssec8onalarea,thesmallertheresistance,butthegreaterthelengththehighertheresistance.Isthatwhylongcableshavetobeverythick?”

➡ “WhatifIputammeterrightbetween+and-?”➡ “thepartrela8ngtotheohm'slawandcurrentdensitystuffmakesnosensetome.”

New Symbols

“Howmanydifferentthingswillωsymbolize???”

A Big Idea Review

ElectricPoten8alScalarFunc8onthatcanbeusedtodetermineE

I

surface

~E · ~A = Qenclosed

✏0

Gauss’LawFluxthroughclosedsurfaceisalways

propor8onaltochargeenclosed

Gauss’LawCanbeusedtodetermineEfield

SpheresCylinders

InfinitePlanes


CapacitanceRelateschargeandpoten8alfortwoconductorsystem

ElectricFieldForceperunitcharge

ElectricFieldPropertyofSpaceCreatedbyCharges

Superposi8on

Coulomb’sLawForcelawbetween

pointchargesq2q1

ElectricPoten8alPoten8alenergyper

unitcharge�Va⇤b ⇥

�Ua⇤b

q= �Z b

a~E · d~l

A Note on Units

★ Forceisnewtons:N=kg•m/s2

★ ElectricField:newton/coulomb(N/C=V/m)★ Electricpoten8al:newton-meter/coulomb=volt

! kg•m2/s2C=V★ Capacitance:farad=coulomb/volt

! faradisbig,weusuallyuse• µF=10–6F• pF=10–12F(µµFinoldendays,“puffs”now)

• (nF=10–9notcustomaryinN.America)

ConductorsChargesfreetomove

SpheresCylinders

InfinitePlanes

Gauss’Law

FieldLines&Equipoten8als

Field

Lines

Equipoten8als Wa�b =bR

a

~F · d~l =bR

aq~E · d~l

WorkDoneByEField

�Va⇤b ⇥�Ua⇤b

q= �Z b

a~E · d~l

ChangeinPoten8alEnergy

CapacitorNetworks

Series:(1/C23) = (1/C2) + (1/C3)

ParallelC123 = C1 + C23

Applications of Big Ideas

WhatDeterminesHowTheyMove?

Theymoveun8lE = 0!

E = 0 inconductordetermineschargedensi8esonsurfaces


Today’sPlan:

1) Reviewofresistance&preflights

2) Workoutacircuitproblemindetail

KeyConcepts:

1) HowresistancedependsonA,L,σ,ρ

2) Howtocombineresistorsinseriesandparallel

3) Understandingresistorsincircuits


σ is conductivity here (not surface charge density)ρ is resistivity here (not volume charge density).

I A

V

L

σ

V=EL I =JA

Observables:

R= LσA

Ohm’sLaw:J=σ E

Conduc8vity–highforgoodconductors.

I/A=σ V/L I=V/(L/σ A)

I=V/RR=Resistanceρ=1/σ


R = L σA

IislikeflowrateofwaterVislikepressureRishowharditisforwatertoflowinapipe

This is just like Plumbing!

TomakeRbig,makeLlongorAsmall

TomakeRsmall,makeLshortorAbig


A B C A B C

Samecurrentthroughbothresistors

Comparevoltagesacrossresistors

1 CheckPoint: Two Resistors 2



CheckPoint: Current DensityTheSAMEamountofcurrentIpassesthroughthreedifferentresistors.R2hastwicethecross-sec8onalareaandthesamelengthasR1,andR3isthree8mesaslongasR1buthasthesamecross-sec8onalareaasR1.InwhichcaseistheCURRENTDENSITYthroughtheresistorthesmallest?

SameCurrent

σ 2σ σ

Voltage

Current

Resistance

Series Parallel

Differentforeachresistor.Vtotal =V1 +V2

IncreasesReq =R1 +R2

SameforeachresistorItotal=I1= I2

Sameforeachresistor.Vtotal=V1=V2

Decreases1/Req=1/R1+1/R2

WiringEachresistoronthesamewire.

Eachresistoronadifferentwire.

DifferentforeachresistorItotal =I1 + I2

R1 R2

R1

R2

Resistor Summary


Voltage Divider

Current Divider

Symbols★ Resistorsymbol(ANSI)

! N.America,Japan,China(?)

★ Alternateresistorsymbol(DIN)! Europe,MiddleEast,Aus/NZ,Africa(?)

4.7 k = 4700 ohm 1.8 Ω = 1.8 ohm

4k7 =4700 ohm 1R8= 1.8 ohm

★ VoltageSource

★ ElectrochemicalCell(“baWery”)some8mesusedforvoltagesource

R2inserieswithR3

CheckPoint: Resistor Network 1


CurrentthroughR2andR3isthesame

ThreeresistorsareconnectedtoabaWerywithemfVasshown.Theresistancesoftheresistorsareallthesame,i.e.R1=R2=R3=R.

ComparethecurrentthroughR2withthecurrentthroughR3:

A.I2>I3B.I2=I3C.I2<I3

R1 = R2 = R3 = R

CheckPoint2ComparethecurrentthroughR1withthecurrentthroughR2

I1 I2

CheckPoint3ComparethevoltageacrossR2withthevoltageacross R3

V2 V3

CheckPoint4Comparethevoltageacross R1withthevoltageacrossR2

V1 V2





ComparethecurrentthroughR1withthecurrentthroughR2:

A.I1/I23=1/2B.I1/I23=1/3C.I1=I23D.I1/I23=2E.I1/I23=3

Weknow:

I23

Similarly:

I1




ComparethevoltageacrossR2withthevoltageacrossR3:

A.V2>V3

B.V2=V3=VC.V2=V3<VD.V2<V3

Considerloop

V2 V3

V23




ComparethevoltageacrossR1withthevoltageacrossR2.

A.V1=V2=VB.V1=1/2V2=VC.V1=2V2=VD.V1=1/2V2=1/5VE.V1=1/2V2=1/2V

R1inparallelwithseriescombina8onofR2andR3

V1

V23

Calculation

Inthecircuitshown:V = 18V,

R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.

WhatisV2,thevoltageacrossR2?

ConceptualAnalysis:Ohm’sLaw:whencurrentIflowsthroughresistanceR,thepoten8aldropVisgivenby:

V = IR.Resistancesarecombinedinseriesandparallelcombina8ons

Rseries = Ra + Rb

(1/Rparallel) = (1/Ra) + (1/Rb)

StrategicAnalysis:CombineresistancestoformequivalentresistancesEvaluatevoltagesorcurrentsfromOhm’sLawExpandcircuitbackusingknowledgeofvoltagesandcurrents

V

R1 R2

R4

R3


Calculation

CombineResistances:

A)inseriesB)inparallelC)neitherinseriesnorinparallel

R1andR2areconnected:

Parallel:Canmakealoopthatcontainsonlythosetworesistors

ParallelCombina8on

Ra

Rb

SeriesCombina8on

Series:Everyloopwithresistor1alsohasresistor2.

Ra Rb

V

R1 R2

R4

R3

Inthecircuitshown:V = 18V,R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.



Calculation

WefirstwillcombineresistancesR2+R3+R4:

Whichofthefollowingistrue?

A)R2,R3andR4areconnectedinseries

B)R2,R3,andR4areconnectedinparallel

C)R3andR4areconnectedinseries(R34) whichisconnectedinparallelwithR2

D)R2andR4areconnectedinseries(R24) whichisconnectedinparallelwithR3

E)R2andR4areconnectedinparallel(R24) whichisconnectedinparallelwithR3

V

R1 R2

R4

R3




Calculation

RedrawthecircuitusingtheequivalentresistorR24=seriescombina8onofR2andR4.

R2andR4areconnectedinseries(R24) whichisconnectedinparallelwithR3

V

R1 R2

R4

R3



(A)(B)(C)

V

R1

R3

R24

V

R1

R3

R24

V

R1

R3 R24


Calculation

CombineResistances:R2andR4areconnectedinseries=R24

R3andR24areconnectedinparallel=R234

A)R234 = 1 Ω B)R234 = 2 Ω C)R234 = 4 Ω D)R234 = 6 Ω

WhatisthevalueofR234?

(1/Rparallel) = (1/Ra) + (1/Rb)

R2 and R4 in series



V

R1

R3 R24

R24 = R2 + R4 = 2Ω + 4Ω = 6Ω

1/R234 = (1/3) + (1/6) = (3/6) Ω -1 R234 = 2 Ω


Calculation

R234

R1 and R234 areinseries.R1234 = 1 + 2 = 3 Ω

Ohm’sLawtellsus:I1234 = V/R1234

= 18 / 3

= 6 Amps

I1 = I234

Ournexttaskistocalculatethetotalcurrentinthecircuit


R24 = 6Ω R234 = 2Ω


V

R1

R234

= I1234

V

R1234


a

b

Calculation

WhatisVab,thevoltageacrossR234 ?

A)Vab = 1 V B)Vab = 2 V C)Vab = 9 V D)Vab = 12 V E)Vab = 16 V

R234

I234 = I1234 Since R1 inserieswith R234

V234 = I234 R234

= 6 x 2

= 12 Volts


R24 = 6Ω R234 = 2Ω I1234 = 6 A


= I1234

V

R1234

V

R1

R234I1 = I234


Calculation

V = 18V R1 = 1Ω

R2 = 2Ω

R3 = 3Ω

R4 = 4ΩR24 = 6Ω

R234 = 2Ω

I1234 = 6 Amps

I234 = 6 Amps

V234 = 12V

WhatisV2?

V

R1

R234

V

R1

R24R3

Whichofthefollowingaretrue?

A)V234 = V24 B)I234 = I24 C)BothA+B D)None

Ohm’sLaw

I24 = V24 / R24

= 12 / 6

= 2 Amps

R3 and R24 werecombinedinparalleltogetR234 Voltagesaresame!


Calculation

V = 18V R1 = 1Ω

R2 = 2Ω

R3 = 3Ω

R4 = 4Ω.

R24 = 6Ω

R234 = 2Ω

I1234 = 6 Amps I234 = 6 Amps

V234 = 12V V24 = 12V

I24 = 2 Amps

WhatisV2?

V

R1

R24R3

Whichofthefollowingaretrue?

A)V24 = V2B)I24 = I2 C)BothA+BD)None

V

R1 R2

R4

R3

I1234

Ohm’sLaw

V2 = I2 R2

= 2 x 2

= 4 Volts!

TheProblemCanNowBeSolved!

R2 andR4wherecombinedinseriestogetR24Currentsaresame!

I24


Quick Follow-Ups

WhatisI3?

A)I3 = 2 A B)I3 = 3 A C)I3 = 4 A

V

R1

R234

a

b

V

R1 R2

R4

R3=

V3 = V234 = 12V

WhatisI1?

WeknowI1 = I1234 = 6 A

NOTE: I2 = V2/R2 = 4/2 = 2 A

V = 18V R1 = 1Ω

R2 = 2Ω R3 = 3Ω

R4 = 4ΩR24 = 6Ω

R234 = 2ΩV234= 12V V2 = 4V I1234 = 6 AmpsI3 = V3/R3 = 12V/3Ω = 4A

I1 = I2 + I3 MakeSense?

I1 I2I3


V0

r

R VL

r

V0

+

VLR

Usuallycan’tsupplytoomuchcurrenttotheloadwithoutvoltage“sagging”

Model for Real Battery: Internal Resistance


electricity magnetism lecture 9: electric current lecture 09...electric current electricity &...

Documents