Electricity Magnetism Lecture 9: Electric Current Lecture 09 - Electric Current.pdf · After you...
Transcript of Electricity Magnetism Lecture 9: Electric Current Lecture 09 - Electric Current.pdf · After you...
Electricity & MagnetismLecture 9: Electric Current
Today’sConcept:
ElectricCurrent
Electricity&Magne8smLecture9,Slide1
Battery and BulbWhichcurrentflowmodeliscorrect?
30 minDeveloping a Model for Current Flow in a CircuitSeveral models for current flow in the circuit might be proposed. Four are diagrammed below.
After you have discussed the various ideas you will be asked to figure out how to use one or more ammeters in your circuit to measure current and test your model.
Model AThere will be no electric current left to flow in the wire attachedto the base of the battery
Model BThe electric current willtravel in a direction towardthe bulb in both wires
Model CThe direction of the current willbe in the direction shown, but there will be less current in the return wire
Model DThe direction of the current willbe as shown, and it will be the same in both wires
Figure 22-5: Four alternative models for current flow
Measuring Current with an Ammeter The ammeter is a device that measures current and displays it. It will allow you to explore the current flowing at different locations in an electric circuit.
Current is typically measured in amperes (A) or milliamperes (mA). (1 ampere = 1000 milliamperes.) Usually we just refer to current as "amps" or "milliamps".
To measure the current flowing through a part of the circuit, you must "insert" the ammeter at the point of interest. Disconnect the circuit, put in the ammeter, and reconnect with
Workshop Physics II: Unit 22 – Batteries, Bulbs, & Current Flow Page 22–7Authors: 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.
(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.
Activity 22-5
Whentheswitchisclosed,andbulbsareiden8cal
A)Topbulbisbrighter
B)BoHombulbisbrighter
C)Bothareequallybright
Activity 22-4
Whentheswitchisclosed
A)VbaHery=Vbulb
B)VbaHery<Vbulb
C)VbaHery>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.
Howdowthecurrentsmeasuredcompare?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.
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.
Your stuff
➡ sowhichwaydoesDCcurrentflow?-_-➡ “Howmanydifferentthingswillsigmasymbolize???”➡ “SinceR=ρL/A,thegreaterthecrosssec8onalarea,thesmallertheresistance,butthegreaterthelengththehighertheresistance.Isthatwhylongcableshavetobeverythick?”
➡ “WhatifIputammeterrightbetween+and-?”➡ “thepartrela8ngtotheohm'slawandcurrentdensitystuffmakesnosensetome.”
A Note on Units
★ Forceisnewtons:N=kg•m/s2
★ ElectricField:newton/coulomb(N/C=V/m)★ Electricpoten>al:newton-meter/coulomb=volt
Ø kg•m2/s2C=V★ Capacitance:farad=coulomb/volt
Ø faradisbig,weusuallyuse• µF=10–6F• pF=10–12F(µµFinoldendays,“puffs”now)• (nF=10–9notcustomaryinN.America)
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).
I A
V
L
s
V=ELI =JA
Observables:
R=LσA
Ohm’sLaw:J=σ E
Conduc>vity–highforgoodconductors.
I/A=σ V/L I=V/(L/σ A)
I=V/RR=Resistanceρ=1/σ
Electricity&Magne8smLecture9,Slide5
R = L σA
IislikeflowrateofwaterVislikepressureRishowharditisforwatertoflowinapipe
This is just like Plumbing!
TomakeRbig,makeLlongorAsmall
TomakeRsmall,makeLshortorAbig
Electricity&Magne8smLecture9,Slide6
Samecurrentthroughbothresistors
Comparevoltagesacrossresistors
1 CheckPoint: Two Resistors 2
Electricity&Magne8smLecture9,Slide7
A B C A B C
Electricity&Magne8smLecture9,Slide8
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
Electricity&Magne8smLecture9,Slide9
Voltage Divider
Current Divider
Symbols★ Resistorsymbol(ANSI)
Ø N.America,Japan,China(?)
★ Alternateresistorsymbol(DIN)Ø Europe,MiddleEast,Aus/NZ,Africa(?)
4.7 k = 4700 ohm1.8 Ω = 1.8 ohm
4k7 =4700 ohm1R8= 1.8 ohm
★ VoltageSource
★ ElectrochemicalCell(“baHery”)some>mesusedforvoltagesource
R2inserieswithR3
CheckPoint: Resistor Network 1
Electricity&Magne8smLecture9,Slide10
CurrentthroughR2andR3isthesame
ThreeresistorsareconnectedtoabaHerywithemfVasshown.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
Electricity&Magne8smLecture9,Slide11
CheckPoint: Resistor Network 2
Electricity&Magne8smLecture9,Slide12
ThreeresistorsareconnectedtoabaHerywithemfVasshown.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
CheckPoint: Resistor Network 3
Electricity&Magne8smLecture9,Slide13
ThreeresistorsareconnectedtoabaHerywithemfVasshown.Theresistancesoftheresistorsareallthesame,i.e.R1=R2=R3=R.
ComparethevoltageacrossR2withthevoltageacrossR3:
A.V2>V3
B.V2=V3=VC.V2=V3<VD.V2<V3
Considerloop
V2 V3
V23
CheckPoint: Resistor Network 4
Electricity&Magne8smLecture9,Slide14
ThreeresistorsareconnectedtoabaHerywithemfVasshown.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
R1inparallelwithseriescombina>onofR2andR3
V1
V23
Calculation
Inthecircuitshown:V = 18V,
R1=1Ω, R2=2Ω, R3=3Ω, and R4=4Ω.
WhatisV2,thevoltageacrossR2?
ConceptualAnalysis:Ohm’sLaw:whencurrentIflowsthroughresistanceR,thepoten>aldropVisgivenby:
V = IR.Resistancesarecombinedinseriesandparallelcombina>ons
Rseries = Ra + Rb
(1/Rparallel) = (1/Ra) + (1/Rb)
StrategicAnalysis:CombineresistancestoformequivalentresistancesEvaluatevoltagesorcurrentsfromOhm’sLawExpandcircuitbackusingknowledgeofvoltagesandcurrents
V
R1 R2
R4
R3
Electricity&Magne8smLecture9,Slide15
Calculation
CombineResistances:
A)inseriesB)inparallelC)neitherinseriesnorinparallel
R1andR2areconnected:
Parallel:Canmakealoopthatcontainsonlythosetworesistors
ParallelCombina>on
Ra
Rb
SeriesCombina>on
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
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
Calculation
RedrawthecircuitusingtheequivalentresistorR24=seriescombina>onofR2andR4.
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
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
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
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
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
Calculation
V = 18V R1 = 1Ω
R2 = 2Ω
R3 = 3Ω
R4 = 4Ω.
R24 = 6Ω
R234 = 2Ω
I1234 = 6 AmpsI234 = 6 Amps
V234 = 12VV24 = 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
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= 12VV2 = 4VI1234 = 6 AmpsI3 = V3/R3 = 12V/3Ω = 4A
I1 = I2 + I3 MakeSense?
I1 I2I3
Electricity&Magne8smLecture9,Slide24