sp lecture 20 - ac circuits lecture 20 - ac...electricity & magne?sm lecture 20, slide 22...

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