Chapter 26: Current, Resistance and Electromotive Force
Charge carrier motion in a conductor in two parts
Constant Acceleration



F  ma  qE
Randomizing Collisions (momentum, energy)
=>Resulting Motion
Average motion = Drift Velocity = vd ~10-4 m/s
Typical speeds ~ 106 m/s
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Current Flow = net motion of charges
= Charge carriers
charge q
speed vd
vd
I
I = Current = rate at which charge passes the area
dQ
1C
I
units:
 1 Amp  1 A
dt
1S
vd
I
Negative charge carriers move in opposite direction of
conventional current.
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Connection with microscopic picture:
E
q
A
vd
  vd dt
dQ = charge that passes through A
= number that pass through A  charge on each
= (n A vd dt) q,
n = number density = number/volume
dQ
I
 nqv d A
dt
Current Density:
I
J   nqv d
A


J  nqv d
(works for negative charge carriers, multiple types of charge carriers as well)



J  n1q1v d 1  n2 q 2 v d 2 
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Example: 18 gauge copper wire (~1.02 mm in diameter)
-constant current of 2A
-n = 8.5x1028 m-3 (property of copper)
find J, vd
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Current as a response to an applied electric field
  
J  J(E)

 E
 = conductivity
E 1
    resistivit y
J 
V m V
units =
= m  m
2
A
Am
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J
J
slope = 1/
E
J
linear response
Ohm’s Law
 depends upon
E
nonlinear response
E
diode
nonlinear response
direction dependence!
•material
•E
•Temperature
If  does not depend on E, the material is said to obey
“Ohm’s Law”
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For a cylindrical conductor
E  J
E
J
Vab  V  E
 J  I

A
V 
R 
 resistance
I
A
V  IR
I
a
b
see also example 28-3 re: alternate geometries
Example: 50 meter length of 18 gauge copper wire (~1.02 mm in diameter)
constant current of 2A
 = 1.72x10-8 .m
find E,V, R
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Temperature Dependence of 


T
Metallic Conductor

T
Semiconductor
T
Superconductor
For small changes in temperature:
 T   o [1   (T  To )]
  temperature coefficient of resistivity
RT  Ro [1   (T  To )]
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Resistor Color Codes
Color number
black 0
brown 1
red
2
orange 3
yellow 4
green 5
blue 6
violet 7
gray 8
white 9
color range
none ±20%
silver ±10%
gold ±5%
value: n1n2x10n3±x%
10x102±5%
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Electromotive Force and Circuits
Steady current
requires a complete circuit
path cannot be only resistance
cannot be only potential drops in direction of current flow
Electromotive Force (EMF)
provides increase in potential E
converts some external form of energy into electrical energy
Single emf and a single resistor:
V = IR
I
+ -
V = IR = E
E
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Measurements
Voltmeters measure Potential Difference (or voltage)
across a device by being placed in parallel with the
device.
V
Ammeters measure current through a device by being
placed in series with the device.
A
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Real Sources and Internal Resistance
r
a E
b
Ideal emf E
determined by how energy is converted into electrical energy
Internal Resistance r
unavoidable “internal” losses
aging batteries => increasing internal resistance
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Open Circuit
I=0
Vr=0
Vab= E
V
E
r
a
A
b
V
Short Circuit
Vr = Ir = E
Vab= 0
r
a
E
b
A
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V
Complete Circuit
E Ir  IR
E
I
Rr
Vab  E Ir
E
r
a
b
A
I
R
Charging Battery
V
Vab  E Ir
E
I
A
a
r
b
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Energy and
Power
dW  dQVab  Vab Idt
dW
P
 IVab
dt
a
I +
-
b
V=IR
+ -
b
V= E
Resistance
2
V
P  IV  I R 
R
2
a
Ideal emf
P  IE
Power deliverd by ( I out of +)
or to ( I into of +) battery
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Power and Real Sources
E
Discharging Battery
I
r
a
b
I
b
I
Vab  E Ir
Power delivered externally : IVab  IE I 2 r
E
Charging Battery
I
a
r
Vab  E Ir
Power delivered by external source : IVab  IE I 2 r
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Real Battery with Load
V
E
I
Rr
E
ER
E
VR  IR 

R  r 1 r R
E2
PE  IE
Rr
r
a
b
A
I
R
2
E 

2
Pr  I r  
 r
Rr
2
 E 
PR  I 2 R  
 R
Rr
PR
1
eff 

PE 1  r R
max PR
dPR
 0 R r
dR
"Impedance Matching"
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Complete Circuit Example
V
I
Vab 
PE 
E =12V r = 2
a
b
A
Pr 
PR 
PVab 
R = 4
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Theory of Metallic Conduction
Constant Acceleration between randomizing collisions
(momentum, velocity randomized)
E

J


J  nqv d



E
 F qE
a 
m m
 

  mean time between
v  v o  a

collisions
qE



v avg  a 
  vd
  mean free path
m
= (v o ) avg  

nq 2  

J  nqv d 
E
m
m
 2
nq 
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Example: What is the mean time between collisions and the
mean free path for conduction electrons in copper?
  127
.  10 8   m
n  8.5  10 28 m 3
m  9.1  10 31 kg
q  e  16
.  10 19 C
6 m
v o  1x10
s
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Physiological Effects of Current
Nerve action involves electrical pulses
currents can interfere with nervous system
~.1A can interfere with essential functions
(heartbeat, e.g.)
currents can cause involuntary convulsive muscle action
~.01 A
Joule Heating (I2R)
With skin resistance
dry skin: R ~ 500k
wet skin: R ~ 1000
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