sp lecture 20 - ac circuits lecture 20 - ac...electricity & magne?sm lecture 20, slide 22...
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ACCircuitsMaximumcurrents&voltages
Phasors:ASimpleTool
Electricity & MagnetismLecture 20
Today’sConcept:
Electricity&Magne?smLecture20,Slide1
Comments
doesn'tacapacitorblowupwhencurrentisthewrongway?
Howdoyouknowwhichwaythecurrentisgoingwhenitstartsorwhatorienta?ontodrawthevectorsatthestart?
Thepre-lecturesaresohardtofollowthattobequitehonesthemightaswellbespeakingklingon.Formulasflashbeforemyeyesasfastasmygradeswilldropinthisclass.Pleaseexplaintheseconceptsdifferentlythanthepre-lecture.
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MechanicalUniverse,ACCircuits
Resistors
R I = VR/R = Vmax/R sin(ωt)
Amplitude=Vmax/R
ε = Vmaxsin(ωt)
Electricity&Magne?smLecture20,Slide3
Capacitors
C I = VmaxωC cos(ωt)
whereXC=1/ωC islikethe“resistance” ofthecapacitorXCdependsonω
Amplitude=Vmax/XC
I = dQ/dt
Q = CV = CVmaxsin(ωt)
90o
ε = Vmaxsin(ωt)
Electricity&Magne?smLecture20,Slide4
Inductors
L I = − (Vmax/ωL) cos(ωt)
Amplitude = Vmax/XL
whereXL=ωL islikethe“resistance” oftheinductorXLdependsonω
dI/dt = VL = Vmaxsin(ωt)
90o
ε = Vmaxsin(ωt)
Electricity&Magne?smLecture20,Slide5
RL Clicker Question
AnRLcircuitisdrivenbyanAC generatorasshowninthefigure.
ForwhatdrivingfrequencyωofthegeneratorwillthecurrentthroughtheresistorbelargestA)ωlargeB)CurrentthroughRdoesn’tdependonω C)ωsmall
L
R
XL = ωL
Asω → 0, sodoesXL
As ω → 0, resistanceofcircuit→R currentgetsbigger
Electricity&Magne?smLecture20,Slide6
impedance
R Imax = Vmax/R VR inphasewithI
L Imax = Vmax/XL VL 90o aheadofIXL = ωL
C Imax = Vmax/XC VC 90o behind IXC = 1/ωC Currentcomesfirstsinceit
chargescapacitor
Becauseresistorsaresimple
Oppositeofcapacitor
Likeawireathighω
Likeawireatlowω
Summary
Electricity&Magne?smLecture20,Slide7
Vmax = Imax XL
Vmax = Imax XC
Vmax = Imax RV inphasewithI
V 90o behind I
V 90o aheadofI
MakessensetowriteeverythingintermsofIsincethisisthesameeverywhereinaone-loopcircuit: Imax XL
Imax XC
Imax R
Phasorsmakethis simpletosee
Alwayslooksthesame. Onlythelengthswillchange
εmax
“Doyouhaveanyfancy-schmancysimula?onsfortoshowme?”
Electricity&Magne?smLecture20,Slide8
AC Circuit Simulations
hfp://www2.epsd.us/robo?cs/phet/en/simula?on/circuit-construc?on-kit-ac.html
Imax XL
Imax XC
Imax R Imax XL
Imax R
Imax XC
εmax
Butnowweareaddingvectors:
Imax XL
Imax XC
Imax R
εmax
The Voltages still Add Up
Electricity&Magne?smLecture20,Slide9
Imax XL
Imax XC
Imax R
εmax
Make this Simpler
Imax XL
Imax XC
Imax R
Electricity&Magne?smLecture20,Slide10
Imax R
εmax = Imax ZImax(XL − XC)
Make this Simpler
Imax XL
Imax XC
Imax R
Electricity&Magne?smLecture20,Slide11
Imax R
εmax = Imax Z
Imax(XL − XC)
Make this Simpler
Electricity&Magne?smLecture20,Slide12
Imax R
εmax = Imax Z
Imax(XL − XC)
R(X
L − XC )
φ
φ
ImpedanceTriangle
Make this Simpler
Electricity&Magne?smLecture20,Slide13
φ
R (XL − X
C )
VCmax = Imax XC
VLmax = Imax XL
VRmax = Imax R
Summary
Imax = εmax / Z
εmax = Imax Z
Electricity&Magne?smLecture20,Slide14
Imax XL
Imax R
εmax
Example: RL Circuit Xc = 0
Electricity&Magne?smLecture20,Slide15
ABC
DrawVoltagePhasors
CheckPoint 2
Imax XL
Imax R
εmax
Electricity&Magne?smLecture20,Slide16
ABC
CheckPoint 4
DrawVoltagePhasorsImax XL
Imax R
εmax
Electricity&Magne?smLecture20,Slide17
TheCURRENTisTHECURRENT
ABCD
φ isthephasebetweengeneratorandcurrent
φ
CheckPoint 6
Imax XL
Imax R
εmax
Electricity&Magne?smLecture20,Slide18
CheckPoint 8
ABC
Whatdoesthevoltagephasordiagramlooklikewhenthecurrentisamaximum?
IXc
IRε
IXL
IXc
IR
εIXL
Electricity&Magne?smLecture20,Slide19
Whatdoesthevoltagephasordiagramlooklikewhenthecapacitorisfullycharged?
ABC
IXc
IRε
IXL
IXc
IR
εIXL
CheckPoint 10
Electricity&Magne?smLecture20,Slide20
CheckPoint 12
Whatdoesthevoltagephasordiagramlooklikewhenthevoltageacrosscapacitorisatitsposi?vemaximum?
A B C
IXc
IRε
IXL
IXc
IR
εIXL
Electricity&Magne?smLecture20,Slide21
ConceptualAnalysisThemaximumvoltageforeachcomponentisrelatedtoitsreactanceandtothe
maximumcurrent.Theimpedancetriangledeterminestherela?onshipbetweenthemaximum
voltagesforthecomponentsStrategicAnalysis
UseVmaxandImaxtodetermineZ UseimpedancetriangletodetermineRUseVCmaxandimpedancetriangletodetermineXL
C
RLV
Calculation
Electricity&Magne?smLecture20,Slide22
ConsidertheharmonicallydrivenseriesLCRcircuitshown.Vmax = 100 V Imax = 2 mA VCmax = 113 V Thecurrentleadsgeneratorvoltageby45o
LandRareunknown.
WhatisXL,thereactanceoftheinductor,atthisfrequency?
~
CompareXLandXCatthisfrequency:
A)XL < XC B)XL = XC C)XL > XC D)Notenoughinforma?on
Thisinforma?onisdeterminedfromthephaseCurrentleadsvoltage
VL = ImaxXL VC = ImaxXC
VR (phaseofcurrent)
VL
VC V leads
IR
V
45ο
Calculation
C
RLV
Electricity&Magne?smLecture20,Slide23
ConsidertheharmonicallydrivenseriesLCRcircuitshown.Vmax = 100 V Imax = 2 mA VCmax = 113 V Thecurrentleadsgeneratorvoltageby45o
LandRareunknown.
WhatisXL,thereactanceoftheinductor,atthisfrequency?
~
WhatisZ,thetotalimpedanceofthecircuit?
A)B)C)D)35.4 kΩ50 kΩ 21.1 kΩ70.7 kΩ
Calculation
C
RLV
Electricity&Magne?smLecture20,Slide24
ConsidertheharmonicallydrivenseriesLCRcircuitshown.Vmax = 100 V Imax = 2 mA VCmax = 113 V Thecurrentleadsgeneratorvoltageby45o
LandRareunknown.
WhatisXL,thereactanceoftheinductor,atthisfrequency?
~
WhatisR?A)B)C)D)
Z = 50 kΩ
sin(45°) =0.707
cos(45°) =0.70735.4 kΩ50 kΩ 21.1 kΩ70.7 kΩ
= 50 kΩ × 0.707
= 35.4 kΩ
Calculation
C
RLV
DeterminedfromimpedancetriangleR
Z = 50kΩ
(XC − XL)45ο R = Z cos(45o)
Electricity&Magne?smLecture20,Slide25
ConsidertheharmonicallydrivenseriesLCRcircuitshown.Vmax = 100 V Imax = 2 mA VCmax = 113 V Thecurrentleadsgeneratorvoltageby45o
LandRareunknown.
WhatisXL,thereactanceoftheinductor,atthisfrequency?
~
A)B)C)D)
Z = 50 kΩ
R = 35.4kΩ
WhatisXC?
35.4 kΩ50 kΩ 21.1 kΩ70.7 kΩ
VCmax = ImaxXC
Calculation
ConsidertheharmonicallydrivenseriesLCRcircuitshown.Vmax = 100 V Imax = 2 mA VCmax = 113 V Thecurrentleadsgeneratorvoltageby45o
LandRareunknown.
WhatisXL,thereactanceoftheinductor,atthisfrequency?
~
C
RLV
XL = XC − R
XL = 56.5 kΩ − 35.4 kΩ
R
Z(XC − XL)
45ο
Westartwiththeimpedancetriangle:
Electricity&Magne?smLecture20,Slide26