Electricity & MagnetismLecture 18

Today’sConcepts: A)Induc4on

B)RLCircuits

Electricity&Magne/smLecture18,Slide1

Stuff you said..

Willtherebemorelabac4vi4esinvolvingoscilloscopes?Unlikelabac4vi4esinvolvingmagnets,theoscilloscopesareforeign,new,andcuriouslyinteres4ngtoplay/learnwithduetotheabove?

WhataretheexactdifferencesbetweentheCapacitorandInductor.Inwhichsitua4onwouldIusetheCapacitorinsteadoftheInductororviceversa?

http://www.sfu.ca/phys/141/1154/Schedule141.pdf

Capacitors are yin — Inductors are yang

E = ��B

dt= �NA

dBdt

Today’s Experiment

Electricity&Magne/smLecture18,Slide2

Loop area A

E = ��B

dt= �A

dBdt

If there are more turns N in the loop,

Wrapawireintoacoiltomakean“inductor”…

dIdt

ε = −L

From the Prelecture: Self Inductance

Electricity&Magne/smLecture18,Slide2

currentI

L

dIdt

εL = −L

emfinducedacrossLtriestokeepIconstant.

Inductorspreventdiscon4nuouscurrentchanges!

It’slikeiner4a!

What this really means:

Electricity&Magne/smLecture18,Slide3

Twosolenoidsaremadewiththesamecrosssec/onalareaandtotalnumberofturns.InductorBistwiceaslongasinductorA

ComparetheinductanceofthetwosolenoidsA)LA = 4 LB B)LA = 2 LB C)LA = LB D)LA = (1/2) LB E)LA = (1/4) LB

(1/2)2 2

CheckPoint 2

Electricity&Magne/smLecture18,Slide4

RL

At t = 0:I = 0VL = VBATTVR = 0(L islikeagiantresistor)

VBATT

I = 0

At t >> L/R:VL = 0 VR = VBATTI = VBATT/R(Lislikeashortcircuit)

R

I = V/R

L

VBATT

Whennocurrentisflowingini4ally:

How to think about RL circuits Episode 1:

VL

Electricity&Magne/smLecture18,Slide5

WhatisthecurrentIthroughthever/calresistorimmediatelyaRertheswitchisclosed?

(+isinthedirec/onofthearrow)

A)I = V/R B)I = V/2R C)I = 0 D)I = −V/2R E)I = −V/R

Inthecircuit,theswitchhasbeenopenforalong/me,andthecurrentiszeroeverywhere.

At/met=0theswitchisclosed.

IL = 0

CheckPoint 4

Electricity&Magne/smLecture18,Slide6

WhatisthecurrentIthroughthever/calresistoraRertheswitchhasbeenclosedforalong/me?

(+isinthedirec/onofthearrow)

A)I=V/R B) I =V/2RC)I= 0D)I=−V/2R E)I= −V/R

RL Circuit (Long Time)

Electricity&Magne/smLecture18,Slide7

RL

VBATT

I = V/R

At t >> L/R:I = 0VL = 0VR = 0

R

I = 0

LR

I = VBATT/RVR = IR VL = VR

At t = 0:

When steady current is flowing initially:

How to Think about RL Circuits Episode 2:

VL

Electricity&Magne/smLecture18,Slide8

WhatisthecurrentIthroughthever/calresistorimmediatelyaRertheswitchisopened?

(+isinthedirec/onofthearrow)

A)I=V/RB)I=V/2RC)I=0 D)I= −V/2RE)I= −V/R

AReralong/me,theswitchisopened,abruptlydisconnec/ngthebaXeryfromthecircuit.

CheckPoint 6

Electricity&Magne/smLecture18,Slide9

R

I

V = IRdIdtV = L L

+

−

−

+

Why is there Exponential Behavior?

VL

Electricity&Magne/smLecture18,Slide10

where

Prelecture:

RL

I

Lecture:

VLVBATT

Didwemessup?

No:Theresistanceissimplytwiceasbiginonecase.

Electricity&Magne/smLecture18,Slide11

ARerlong/meat0,movedto1 ARerlong/meat0,movedto2

AMerswitchmoved,whichcasehaslarger4meconstant?

A)Case1B)Case2C)Thesame

CheckPoint 8

Electricity&Magne/smLecture18,Slide12

A)Case1B)Case2C)Thesame

CheckPoint 10ARerlong/meat0,movedto1 ARerlong/meat0,movedto2

ImmediatelyaMerswitchmoved,inwhichcaseisthevoltageacrosstheinductorlarger?

Electricity&Magne/smLecture18,Slide13

I0 is V/R in both casesA) VL(0) = I0*2RB) VL(0) = I0*3R

A)Case1B)Case2C)Thesame

CheckPoint 12ARerlong/meat0,movedto1 ARerlong/meat0,movedto2

AMerswitchmovedforfinite4me,inwhichcaseisthe

currentthroughtheinductorlarger?

Electricity&Magne/smLecture18,Slide14

A) τ1 = L/2RB) τ2 = L/3R

VL

τ1τ2

ConceptualAnalysisOnceswitchisclosed,currentswillflowthroughthis2-loopcircuit.KVRandKCRcanbeusedtodeterminecurrentsasafunc4onof4me.

StrategicAnalysisDeterminecurrentsimmediatelyaMerswitchisclosed.DeterminevoltageacrossinductorimmediatelyaMerswitchisclosed.DeterminedIL/dtimmediatelyaMerswitchisclosed.

R1

LV

R2

R3

Calculation

Theswitchinthecircuitshownhasbeenopenforalong4me.Att = 0,theswitchisclosed.

WhatisdIL/dt,the4merateofchangeofthecurrentthroughtheinductorimmediatelyaMerswitchisclosed

Electricity&Magne/smLecture18,Slide15

WhatisIL,thecurrentintheinductor,immediatelyaRertheswitchisclosed?

A)IL = V/R1 up B)IL = V/R1 down C)IL = 0

IL = 0

Calculation

Theswitchinthecircuitshownhasbeenopenforalong4me.Att = 0,theswitchisclosed.

R1

LV

R2

R3

Electricity&Magne/smLecture18,Slide16

A)B)C)D)

Calculation

IL(t = 0 +) = 0

Theswitchinthecircuitshownhasbeenopenforalong4me.Att = 0,theswitchisclosed.

R1

LV

R2

R3

WhatisthemagnitudeofI2,thecurrentinR2,immediatelyaMertheswitchisclosed?

Electricity&Magne/smLecture18,Slide17

IL(t = 0 +) = 0 I2(t = 0 +) = V/(R1+ R2+ R3)

Calculation

Theswitchinthecircuitshownhasbeenopenforalong4me.Att = 0,theswitchisclosed.

R1

LV

R2

R3

I2

A)B)C)D)E)

WhatisthemagnitudeofVL,thevoltageacrosstheinductor,immediatelyaMertheswitchisclosed?

Electricity&Magne/smLecture18,Slide18

A)B)C)D)

VL(t = 0 +) = V(R2+ R3)/(R1+ R2+ R3)

Calculation

Theswitchinthecircuitshownhasbeenopenforalong4me.Att = 0,theswitchisclosed.

WhatisdIL/dt,the4merateofchangeofthecurrentthroughtheinductorimmediatelyaMerswitchisclosed

R1

LV

R2

R3

Electricity&Magne/smLecture18,Slide19

Follow Up

Theswitchinthecircuitshownhasbeenclosedforalong4me.

WhatisI2,thecurrentthroughR2?

(Posi4vevaluesindicatecurrentflowstotheright)

R1

LV

R2

R3

A)B)C)D)

Electricity&Magne/smLecture18,Slide20

Theswitchinthecircuitshownhasbeenclosedforalong4meatwhichpoint,theswitchisopened.

WhatisI2,thecurrentthroughR2immediatelyaMer

switchisopened?(Posi4vevaluesindicatecurrentflowstotheright)

Follow Up 2R1

LV

R2

R3

I2

Electricity&Magne/smLecture18,Slide21

A)B)C)D)E)