electricity magnetism lecture 9: electric current lecture 09...electric current electricity &...
TRANSCRIPT
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Electricity & Magnetism Lecture 9: Electric Current
Today’sConcept:
ElectricCurrent
Electricity&Magne8smLecture9,Slide1
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Midterm
! B9201:6:30pmto8:00pm! Coversuptotoday’slecturematerial(Unit9)andnextweek’stutorials
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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
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(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
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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.
Workshop Physics II: Unit 22 – Batteries, Bulbs, & Current Flow Page 22–11Authors: 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.
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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.
(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.
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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.
(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.
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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
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Your stuff
➡ sowhichwaydoesDCcurrentflow?-_-➡ “Pleaseexplainhowelectricfieldiscalculatedinsidethecopperwire.”
➡ “SinceR=ρL/A,thegreaterthecrosssec8onalarea,thesmallertheresistance,butthegreaterthelengththehighertheresistance.Isthatwhylongcableshavetobeverythick?”
➡ “WhatifIputammeterrightbetween+and-?”➡ “thepartrela8ngtotheohm'slawandcurrentdensitystuffmakesnosensetome.”
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New Symbols
“Howmanydifferentthingswillωsymbolize???”
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A Big Idea Review
ElectricPoten8alScalarFunc8onthatcanbeusedtodetermineE
I
surface
~E · ~A = Qenclosed
✏0
Gauss’LawFluxthroughclosedsurfaceisalways
propor8onaltochargeenclosed
Gauss’LawCanbeusedtodetermineEfield
SpheresCylinders
InfinitePlanes
Electricity&Magne8smLecture9,Slide2
CapacitanceRelateschargeandpoten8alfortwoconductorsystem
ElectricFieldForceperunitcharge
ElectricFieldPropertyofSpaceCreatedbyCharges
Superposi8on
Coulomb’sLawForcelawbetween
pointchargesq2q1
ElectricPoten8alPoten8alenergyper
unitcharge�Va⇤b ⇥
�Ua⇤b
q= �Z b
a~E · d~l
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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)
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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
Electricity&Magne8smLecture9,Slide3
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Today’sPlan:
1) Reviewofresistance&preflights
2) Workoutacircuitproblemindetail
KeyConcepts:
1) HowresistancedependsonA,L,σ,ρ
2) Howtocombineresistorsinseriesandparallel
3) Understandingresistorsincircuits
Electricity&Magne8smLecture9,Slide4
σ is conductivity here (not surface charge density)ρ is resistivity here (not volume charge density).
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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
Electricity&Magne8smLecture9,Slide5
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R = L σA
IislikeflowrateofwaterVislikepressureRishowharditisforwatertoflowinapipe
This is just like Plumbing!
TomakeRbig,makeLlongorAsmall
TomakeRsmall,makeLshortorAbig
Electricity&Magne8smLecture9,Slide6
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A B C A B C
Samecurrentthroughbothresistors
Comparevoltagesacrossresistors
1 CheckPoint: Two Resistors 2
Electricity&Magne8smLecture9,Slide7
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Electricity&Magne8smLecture9,Slide8
CheckPoint: Current DensityTheSAMEamountofcurrentIpassesthroughthreedifferentresistors.R2hastwicethecross-sec8onalareaandthesamelengthasR1,andR3isthree8mesaslongasR1buthasthesamecross-sec8onalareaasR1.InwhichcaseistheCURRENTDENSITYthroughtheresistorthesmallest?
SameCurrent
σ 2σ σ
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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
Electricity&Magne8smLecture9,Slide9
Voltage Divider
Current Divider
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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
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R2inserieswithR3
CheckPoint: Resistor Network 1
Electricity&Magne8smLecture9,Slide10
CurrentthroughR2andR3isthesame
ThreeresistorsareconnectedtoabaWerywithemfVasshown.Theresistancesoftheresistorsareallthesame,i.e.R1=R2=R3=R.
ComparethecurrentthroughR2withthecurrentthroughR3:
A.I2>I3B.I2=I3C.I2<I3
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R1 = R2 = R3 = R
CheckPoint2ComparethecurrentthroughR1withthecurrentthroughR2
I1 I2
CheckPoint3ComparethevoltageacrossR2withthevoltageacross R3
V2 V3
CheckPoint4Comparethevoltageacross R1withthevoltageacrossR2
V1 V2
Electricity&Magne8smLecture9,Slide11
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CheckPoint: Resistor Network 2
Electricity&Magne8smLecture9,Slide12
ThreeresistorsareconnectedtoabaWerywithemfVasshown.Theresistancesoftheresistorsareallthesame,i.e.R1=R2=R3=R.
ComparethecurrentthroughR1withthecurrentthroughR2:
A.I1/I23=1/2B.I1/I23=1/3C.I1=I23D.I1/I23=2E.I1/I23=3
Weknow:
I23
Similarly:
I1
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CheckPoint: Resistor Network 3
Electricity&Magne8smLecture9,Slide13
ThreeresistorsareconnectedtoabaWerywithemfVasshown.Theresistancesoftheresistorsareallthesame,i.e.R1=R2=R3=R.
ComparethevoltageacrossR2withthevoltageacrossR3:
A.V2>V3
B.V2=V3=VC.V2=V3<VD.V2<V3
Considerloop
V2 V3
V23
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CheckPoint: Resistor Network 4
Electricity&Magne8smLecture9,Slide14
ThreeresistorsareconnectedtoabaWerywithemfVasshown.Theresistancesoftheresistorsareallthesame,i.e.R1=R2=R3=R.
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
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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
Electricity&Magne8smLecture9,Slide15
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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.
WhatisV2,thevoltageacrossR2?
Electricity&Magne8smLecture9,Slide16
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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
Inthecircuitshown:V = 18V,R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.
WhatisV2,thevoltageacrossR2?
Electricity&Magne8smLecture9,Slide17
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Calculation
RedrawthecircuitusingtheequivalentresistorR24=seriescombina8onofR2andR4.
R2andR4areconnectedinseries(R24) whichisconnectedinparallelwithR3
V
R1 R2
R4
R3
Inthecircuitshown:V = 18V,R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.
WhatisV2,thevoltageacrossR2?
(A)(B)(C)
V
R1
R3
R24
V
R1
R3
R24
V
R1
R3 R24
Electricity&Magne8smLecture9,Slide18
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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
Inthecircuitshown:V = 18V,R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.
WhatisV2,thevoltageacrossR2?
V
R1
R3 R24
R24 = R2 + R4 = 2Ω + 4Ω = 6Ω
1/R234 = (1/3) + (1/6) = (3/6) Ω -1 R234 = 2 Ω
Electricity&Magne8smLecture9,Slide19
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Calculation
R234
R1 and R234 areinseries.R1234 = 1 + 2 = 3 Ω
Ohm’sLawtellsus:I1234 = V/R1234
= 18 / 3
= 6 Amps
I1 = I234
Ournexttaskistocalculatethetotalcurrentinthecircuit
Inthecircuitshown:V = 18V,R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.
R24 = 6Ω R234 = 2Ω
WhatisV2,thevoltageacrossR2?
V
R1
R234
= I1234
V
R1234
Electricity&Magne8smLecture9,Slide20
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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
Inthecircuitshown:V = 18V,R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.
R24 = 6Ω R234 = 2Ω I1234 = 6 A
WhatisV2,thevoltageacrossR2?
= I1234
V
R1234
V
R1
R234I1 = I234
Electricity&Magne8smLecture9,Slide21
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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!
Electricity&Magne8smLecture9,Slide22
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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
Electricity&Magne8smLecture9,Slide23
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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
Electricity&Magne8smLecture9,Slide24
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V0
r
R VL
r
V0
+
VLR
Usuallycan’tsupplytoomuchcurrenttotheloadwithoutvoltage“sagging”
Model for Real Battery: Internal Resistance
Electricity&Magne8smLecture10,Slide21