27 current and resistance 27-1 moving charges and electric currents electric currents——that is,...
TRANSCRIPT
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27 Current and Resist27 Current and Resistanceance
27-1 Moving Charges and Electric Currents
electric currents——that is,charges in motion.
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a. A loop of copper in electrostatic equilibrium.
b. This movement of charges is a current i.
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27-2 Electric Current27-2 Electric Current
The current I through the conductor has theSame value at planes aa`,bb`,cc`.
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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
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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
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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.
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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?
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Sample Problem 27-1Sample Problem 27-1
charge electrons moleculesi= per per per electron molecule second
Step one:Step one:
dt
dNei )10)((or
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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
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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
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27-3 Current Density27-3 Current Density
AdJi
AdJi
JAdAJJdAi JAdAJJdAi
A
iJ
A
iJ
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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
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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
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(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
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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:
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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
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Step four:Step four:
smrne
ivd /109.4 7
2
smrne
ivd /109.4 7
2
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27-4 Resistance and Resistivity27-4 Resistance and Resistivity
i
VR
i
VR
1 ohm =1Ω= 1 volt per ampere = 1V/A
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J
EJ
E
JE
JE
1
1 EJ
EJ
definition of ρ
definition of σ
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Calculating Resistance from ReCalculating Resistance from Resistivitysistivity
Resistance is a property of an object.Resistivity is a property of a material.
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LVE / LVE /
AiJ / AiJ /
Ai
LV
J
E
/
/
Ai
LV
J
E
/
/
A
LR
A
LR
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Variation with TemperatureVariation with Temperature
000 TTa 000 TTa
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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
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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.
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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.
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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
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m
eE
ne
Jvd
m
eE
ne
Jvd
Jne
mE
2 Jne
mE
2
ne
m2
ne
m2
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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
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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
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Step two:Step two:
nmm
ssmveff
40100.4
105.2/106.18
146
nmm
ssmveff
40100.4
105.2/106.18
146
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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
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RiP 2 RiP 2
R
VP
2
R
VP
2
or
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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:
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Step two:Step two: W
VP 400
36
120 2
W
VP 400
36
120 2
WPP 8002 WPP 8002
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27-8 Semiconductors27-8 Semiconductors
ne
m2
ne
m2
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27-9 Superconductors27-9 Superconductors
Superconductors are materials that lose allelectrical resistance at low temperatures.