27 Current and Resist27 Current and Resistanceance

27-1 Moving Charges and Electric Currents

electric currents——that is,charges in motion.

a. A loop of copper in electrostatic equilibrium.

b. This movement of charges is a current i.

27-2 Electric Current27-2 Electric Current

The current I through the conductor has theSame value at planes aa`,bb`,cc`.

An electric current I in a conductor is defined by:

dt

dqi dt

dqi

We can fine the charge that passes through theplane in a time interval extending from 0 to t byIntegration:

t

idtdqq0

t

idtdqq0

The SI unit for current is the coulomb per Second,also called the ampere(A):

1 ampere=1 A=1 coulomb per second=1C/s1 ampere=1 A=1 coulomb per second=1C/s

210 iii 210 iii

The Direction of CurrentsThe Direction of Currents

A current arrow is drawn in the direction in which positive charge carriers would move,Even if the actual charge carriers are negative and move in the opposite direction.

CHECKPOINT 1CHECKPOINT 1

What are the magnitude and direction of the current I in the lower right-hand wire?What are the magnitude and direction of the current I in the lower right-hand wire?

Sample Problem 27-1Sample Problem 27-1

charge electrons moleculesi= per per per electron molecule second

Step one:Step one:

dt

dNei )10)((or

Step two:Step two:

molecules molecules moles mass volume per = per per unit per unit per second mole mass volume second

Step three:Step three:

dt

dV

M

N

dt

dV

MN

dt

dN massAmassa

1

dt

dV

M

N

dt

dV

MN

dt

dN massAmassa

1

Step four:Step four:

dt

dVMeNi massA 110

dt

dVMeNi massA 110

Step five:Step five:

MA

smmkgmolkg

molCi

1.24

/10450/1000/018.0

1002.6106.1103631

12319

MA

smmkgmolkg

molCi

1.24

/10450/1000/018.0

1002.6106.1103631

12319

27-3 Current Density27-3 Current Density

AdJi

AdJi

JAdAJJdAi JAdAJJdAi

A

iJ

A

iJ

Drift SpeedDrift Speed

enALq enALq

dv

Lt

dv

Lt

dd

nAevvL

nALe

t

qi

/ dd

nAevvL

nALe

t

qi

/

ne

J

nAe

ivd

ne

J

nAe

ivd dvneJ

dvneJ

Sample Problem 27-2Sample Problem 27-2

Step one:Step one:

262

222

10424.9002.04

3

4

3

2

mm

RRRA

262

222

10424.9002.04

3

4

3

2

mm

RRRA

Step two:Step two:

A

mmAAJi

9.1

10424.9/100.2 2625

A

mmAAJi

9.1

10424.9/100.2 2625

(b)

JdAJdAAdJ 0cos

JdAJdAAdJ 0cosStep one:Step one:

Step two:Step two:

AmmA

aRR

Rar

a

drrardrarJdAAdJi

R

R

R

R

R

R

1.7002.0/100.332

15

32

15

16242

22

4411

44

4

2/

4

2/

3

2/

2

AmmA

aRR

Rar

a

drrardrarJdAAdJi

R

R

R

R

R

R

1.7002.0/100.332

15

32

15

16242

22

4411

44

4

2/

4

2/

3

2/

2

Sample Problem 27-3Sample Problem 27-3

atoms atoms moles massn= per unit = per per unit per unit volume mole mass volume

Step one:Step one:

Step two:Step two:

328

3

33123

1049.8

/1054.63

/1096.81002.6

m

molkg

mkgmoln

328

3

33123

1049.8

/1054.63

/1096.81002.6

m

molkg

mkgmoln

Step three:Step three:

dnevA

i

dnevA

i

Step four:Step four:

smrne

ivd /109.4 7

2

smrne

ivd /109.4 7

2

27-4 Resistance and Resistivity27-4 Resistance and Resistivity

i

VR

i

VR

1 ohm =1Ω＝ 1 volt per ampere = 1V/A

J

EJ

E

JE

JE

1

1 EJ

EJ

definition of ρ

definition of σ

Calculating Resistance from ReCalculating Resistance from Resistivitysistivity

Resistance is a property of an object.Resistivity is a property of a material.

LVE / LVE /

AiJ / AiJ /

Ai

LV

J

E

/

/

Ai

LV

J

E

/

/

A

LR

A

LR

Variation with TemperatureVariation with Temperature

000 TTa 000 TTa

Sample Problem 27-4Sample Problem 27-4

Step one:Step one: 242 1044.12.1 mcmA 242 1044.12.1 mcmA

Step two:Step two: 4100.1A

LR

4100.1

A

LR

Step three:Step three: 7105.6A

LR

7105.6

A

LR

27-5 Ohm’s Law27-5 Ohm’s Law

Ohm’s law is an assertion that the current through a device is always directly proportionalto the potential different applied to the device.

A conducting device obeys Ohm’s law whenthe resistance of the device is independentof the magnitude and polarity of the applieddifference.

A conducting material obeys Ohm’s law whenthe resistivity of the material is independentof the magnitude and direction of the appliedelectric field.

27-6 A Microscopic View of O27-6 A Microscopic View of Ohm’s Lawhm’s Law

m

eE

m

Fa

m

eE

m

Fa

m

eEavd

m

eEavd

m

eE

ne

Jvd

m

eE

ne

Jvd

Jne

mE

2 Jne

mE

2

ne

m2

ne

m2

Sample Problem 27-5Sample Problem 27-5

Step one:Step one:

2ne

m

2ne

m

skgmc

mcm

/1067.3/1067.3

1069.1106.11049.8172217

8219328

skgmc

mcm

/1067.3/1067.3

1069.1106.11049.8172217

8219328

s

kg

sm

smkg

sCm

CJC

Am

VC

m

C

/

/

/

/2

22

2

2

2

2

2

2

s

kg

sm

smkg

sCm

CJC

Am

VC

m

C

/

/

/

/2

22

2

2

2

2

2

2

sskg

kg 1417

31

105.2/1067.3

101.9

sskg

kg 1417

31

105.2/1067.3

101.9

Step two:Step two:

nmm

ssmveff

40100.4

105.2/106.18

146

nmm

ssmveff

40100.4

105.2/106.18

146

27-7 Power in Electric Circuits27-7 Power in Electric Circuits

iVP iVP

Ws

J

s

C

C

JAV 11111

W

s

J

s

C

C

JAV 11111

RiP 2 RiP 2

R

VP

2

R

VP

2

or

Sample Problem 27-6Sample Problem 27-6

W

V

R

VP 200

72

120 22

W

V

R

VP 200

72

120 22

Step one:Step one:

Step two:Step two: W

VP 400

36

120 2

W

VP 400

36

120 2

WPP 8002 WPP 8002

27-8 Semiconductors27-8 Semiconductors

ne

m2

ne

m2

27-9 Superconductors27-9 Superconductors

Superconductors are materials that lose allelectrical resistance at low temperatures.