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http://www.nearingzero.net (nz105.jpg)
Comment on exam scores…
Titan Quest screenshot, just after escaping a monster
that ate a slow-footed companion.
It just goes to show, you do not
have to be faster than the
monsters… you just have to be
faster than your slowest companion.
Comment on exam scores…
 Exam 1 grade distribution (regrades not included)…
I will provide the Exam 1 grade distribution during this lecture.
Today’s agenda:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s Law and Resistance.
You must be able to use Ohm’s Law and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
Electric Current
Definition of Electric Current
The average current that passes any point in a conductor
during a time t is defined as
Q
Iav 
t
where Q is the amount of charge passing the point.
dQ
The instantaneous current is I =
.
dt
1C
.
One ampere of current is one coulomb per second: 1A =
1s
Currents in battery-operated devices are often in the milliamp
range: 1 mA = 10-3 A.
“m” for milli—another abbreviation to remember!
Here’s a really simple circuit:
+-
current
Don’t try that at home! (Why not?)
The current is in the direction of flow of positive charge…
…opposite to the flow of electrons, which are usually the charge
carriers.
+-
current
electrons
An electron flowing from – to + gives rise to the same
“conventional current” as a proton flowing from + to -.
“Conventional” refers to our convention, which is always to
consider the effect of + charges (for example, electric field
direction is defined relative to + charges).
“Hey, that figure you just showed me is confusing. Why don’t
electrons flow like this?”
+-
current
Good question.
electrons
+-
current
electrons
Electrons “want” to get away from - and go to +.
Chemical reactions (or whatever energy mechanism the battery
uses) “force” electrons to the negative terminal. The battery
won’t “let” electrons flow the wrong way inside it. So electrons
pick the easiest path—through the external wires towards the +
terminal.
Of course, real electrons don’t “want” anything.
Note!
Current is a scalar quantity, and it has a sign associated with it.
In diagrams, assume that a current indicated by a
symbol and an arrow is the conventional current.
I1
If your calculation produces a negative value for the current,
that means the conventional current actually flows opposite to
the direction indicated by the arrow.
Example: 3.8x1021 electrons pass through a point in a wire in 4
minutes. What was the average current?
Q Ne
Iav 

t
t
Iav 
21
19
3.8

10
1.6

10



 4  60 
Iav  2.53A
Today’s agenda:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s Law and Resistance.
You must be able to use Ohm’s Law and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
Current Density
When we study details of charge transport, we use the concept
of current density.
Current density is the amount of charge that flows across a unit
of area in a unit of time.
Current density: charge per area per time (current / area).
A current density J flowing through an infinitesimal area dA
produces an infinitesimal current dI.
dA
J
dI  J  dA
Current density is a vector. Its
direction is the direction of the
velocity of positive charge carriers.
The total current passing through A is just
I

surface
J  dA
No OSE’s on this page.
Simpler, less-general
OSE on next page.
Current density: charge per area per time.
J
A
If J is constant and parallel to dA (like in a wire), then
I
I   J  dA  J  dA  JA  J 
A
surface
surface
Now let’s take a “microscopic” view of current and calculate J.
vt
q
v
A
If n is the number of charges
per volume, then the number of
charges that pass through a
surface A in a time t is
number
 volume   n  vt A 
volume
The total amount of charge passing through A is the number of
charges times the charge of each.
vt
q
v
Q  nqvt A
A
Divide by t to get the current…
Q
I
 nqv A
t
…and by A to get J:
J  nqv .
To account for the vector nature of the current density,
J  nqv
Not quite
“official” yet.
and if the charge carriers are electrons, q=-e so that
Je  n e v.
Not quite
“official” yet.
The – sign demonstrates that the velocity of the electrons is
antiparallel to the conventional current direction.
Currents in Materials
Metals are conductors because they have “free” electrons,
which are not bound to metal atoms.
In a cubic meter of a typical conductor there roughly 1028 free
electrons, moving with typical speeds of 1,000,000 m/s…
…but the electrons move in random directions, and there is no
net flow of charge, until you apply an electric field.
Thanks to Dr. Yew San Hor for this slide.
E
electron “drift” velocity
inside a
conductor
just one
electron
shown, for
simplicity
The electric field accelerates the electron, but only until the
electron collides with a “scattering center.” Then the electron’s
velocity is randomized and the acceleration begins again.
Some predictions based on this model are off by a factor or 10
or so, but with the inclusion of some quantum mechanics it
becomes accurate. The “scattering” idea is useful.
A greatly oversimplified model, but the “idea” is useful.
Even though some details of the model on the previous slide
are wrong, it points us in the right direction, and works when
you take quantum mechanics into account.
In particular, the velocity that should be used in
J  n q v.
is not the charge carrier’s velocity (electrons in this example).
Instead, we should the use net velocity of the collection of
electrons, the net velocity caused by the electric field.
Quantum mechanics shows us how to deal
correctly with the collection of electrons.
This “net velocity” is like the terminal velocity of a parachutist;
we call it the “drift velocity.”
J  n q vd .
It’s the drift velocity that we should use in our equations for
current and current density in conductors:
J  n q vd
I  nqvd A
I
vd 
nqA
Example: the 12-gauge copper wire in a home has a crosssectional area of 3.31x10-6 m2 and carries a current of 10 A.
The conduction electron density in copper is 8.49x1028
electrons/m3. Calculate the drift speed of the electrons.
I
vd 
nqA
I
vd 
neA
10 C/s
vd 
(8.49 1028 m -3 )(1.60 1019 C)(3.31106 m 2 )
vd  2.22 104 m/s
Today’s agenda:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s Law and Resistance.
You must be able to use Ohm’s Law and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
Resistance
The resistance of a material is a measure of how easily a
charge flows through it.
Resistance: how much “push” is needed to
get a given current to flow.
V
R
I
1V
.
The unit of resistance is the ohm: 1  
1A
Resistances of kilohms and megohms are common:
1 k  103 , 1 M=106.
Every circuit component has resistance.
This is the symbol we use for a “resistor:”
All wires have resistance. Obviously, for efficiency in carrying a
current, we want a wire having a low resistance. In idealized
problems, we will consider wire resistance to be zero.
Lamps, batteries, and other devices in circuits have resistance.
Resistors are often intentionally used in
circuits. The picture shows a strip of five
resistors (you tear off the paper and
solder the resistors into circuits).
The little bands of color on the resistors have meaning. Here
are a couple of handy web links:
1. http://www.dannyg.com/examples/res2/resistor.htm
2. http://www.digikey.com/en/resources/conversioncalculators/conversion-calculator-resistor-color-code-4-band
Ohm’s Law
In some materials, the resistance is constant over a wide range
of voltages.
For such materials, we write V  IR, and call the equation
“Ohm’s Law.”
In fact, Ohm’s Law is not a “Law” in the same sense as
Newton’s Laws…
… and in advanced Physics classes you will write something
other than V=IR when you write Ohm’s Law.
Newton’s Laws demand; Ohm’s Law suggests.
I
Materials that follow Ohm’s Law are called
“ohmic” materials, and have linear I vs. V
graphs.
Materials that do not follow Ohm’s Law are
called “nonohmic” materials, and have
curved I vs. V graphs.
slope=1/R
V
I
V
I
Materials that follow Ohm’s Law are called
“ohmic” materials, and have linear I vs. V
graphs.
Materials that do not follow Ohm’s Law are
called “nonohmic” materials, and have
curved I vs. V graphs.
slope=1/R
V
I
V
Today’s agenda:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s Law and Resistance.
You must be able to use Ohm’s Law and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
Resistivity
It is also experimentally observed (and justified by quantum
mechanics) that the resistance of a metal wire is well-described
by
L
R
,
A
where  is a “constant” called the resistivity of the wire
material, L is the wire length, and A its cross-sectional area.
This makes sense: a longer wire or higher-resistivity wire should
have a greater resistance. A larger area means more “space”
for electrons to get through, hence lower resistance.
R = L / A,
units of 
are m

A
L
The longer a wire, the “harder” it is to push electrons through
it.
The greater the resistivity, the “harder” it is to push electrons
through it.
The greater the cross-sectional area, the “easier” it is to push
electrons through it.
Resistivity is a useful tool in physics because it depends on the
properties of the wire material, and not the geometry.
Resistivities range from roughly 10-8 ·m for copper wire to
1015 ·m for hard rubber. That’s an incredible range of 23
orders of magnitude, and doesn’t even include superconductors
(we might talk about them some time).
Example (will not be worked in class): Suppose you want to
connect your stereo to remote speakers.
(a) If each wire must be 20 m long, what diameter copper wire
should you use to make the resistance 0.10  per wire.
R = L / A
A = L / R
A =  (d/2)2
 (d/2)2 = L / R
geometry!
(d/2)2 = L / R
d/2= ( L / R )½
don’t skip steps!
d = 2 ( L / R )½
d = 2 [ (1.68x10-8) (20) /  (0.1) ]½
d = 0.0021 m = 2.1 mm
(b) If the current to each speaker is 4.0 A, what is the voltage
drop across each wire?
V=IR
V = (4.0) (0.10)
V = 0.4 V
Homework hint you can look up the resistivity of copper in a
table in your text.
Ohm’s Law Revisited
The equation for resistivity I introduced five slides back is a
semi-empirical one. Here’s almost how we define resistivity:
E
 .
J
NOT an official
starting equation!
Our equation relating R and  follows from the above equation.
We define conductivity  as the inverse of the resistivity:
1
1
  , or   .


With the above definitions,
E   J,
Think of this as our
definition of resistivity.
J  E.
In anisotropic materials, 
and  are tensors. A tensor
is like a matrix, only worse.
The “official” Ohm’s law, valid for non-ohmic materials.
Cautions!
In this context:
 is not volume density!
 is not surface density!
Example: the 12-gauge copper wire in a home has a crosssectional area of 3.31x10-6 m2 and carries a current of 10 A.
Calculate the magnitude of the electric field in the wire.
I
E  J  
A
(1.72 108   m) 10 C/s 
E
(3.31106 m 2 )
E  5.20 102 V/m
Question: are we still assuming the electrostatic case?
Homework hint (not needed in this particular example): in this chapter it is safe to use V=Ed.
Today’s agenda:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s Law and Resistance.
You must be able to use Ohm’s Law and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
Temperature Dependence of Resistivity
Many materials have resistivities that depend on temperature.
We can model* this temperature dependence by an equation
of the form
  0 1    T  T0   ,
where 0 is the resistivity at temperature T0, and  is the
temperature coefficient of resistivity.
*T0 is a reference temperature, often taken to be 0 °C or 20 °C. This approximation
can be used if the temperature range is “not too great;” i.e. 100 °C or so.
Resistance thermometers made of carbon (inexpensive) and
platinum (expensive) are widely used to measure very low
temperatures.
Example: a carbon resistance thermometer in the shape of a
cylinder 1 cm long and 4 mm in diameter is attached to a
sample. The thermometer has a resistance of 0.030 . What is
the temperature of the sample?
This is the starting equation:
  0 1    T  T0  
We can look up the resistivity of carbon at 20 C.
We use the thermometer dimensions to calculate the resistivity
when the resistance is 0.03 , and use the above equation
directly.
Or we can rewrite the equation in terms or R. Let’s first do the
calculation using resistivity.
Example: a carbon resistance thermometer in the shape of a
cylinder 1 cm long and 4 mm in diameter is attached to a
sample. The thermometer has a resistance of 0.030 . What is
the temperature of the sample?
The resistivity of carbon at 20 C is
0  3.519 105   m
L
R
A
RA
(R) 
L
(R  0.03) 
2
0.03


0.002
 

 0.01
 3.7699 105   m
Example: a carbon resistance thermometer in the shape of a
cylinder 1 cm long and 4 mm in diameter is attached to a
sample. The thermometer has a resistance of 0.030 . What is
the temperature of the sample?
  0 1    T  T0  
  0.0005 C-1

  T  T0    1
0

1 
T  T0    1
  0 
 3.7699 105 
1
T  20 
 1  122.6 C

5
0.0005  3.519 10

Example: a carbon resistance thermometer in the shape of a
cylinder 1 cm long and 4 mm in diameter is attached to a
sample. The thermometer has a resistance of 0.030 . What is
the temperature of the sample?
Alternatively, we can use the resistivity of carbon at 20 C to
calculate the resistance at 20 C.
T0  20C
0  3.519 105   m
L = 0.01 m
0 L
R 0  2  0.02800 
r
This is the resistance at 20 C.
r = 0.002 m
Example: a carbon resistance thermometer in the shape of a
cylinder 1 cm long and 4 mm in diameter is attached to a
sample. The thermometer has a resistance of 0.030 . What is
the temperature of the sample?
  0 1    T  T0  
RA R 0 A 0

1    T  T0  
L
L0
If we assume A/L = A0/L0, then
R  R 0 1    T  T0  
Example: a carbon resistance thermometer in the shape of a
cylinder 1 cm long and 4 mm in diameter is attached to a
sample. The thermometer has a resistance of 0.030 . What is
the temperature of the sample?
R  R 0 1    T  T0  
R
  T  T0  
1
R0

1 R
T  T0  
 1
  R0 
1
 .030 
T  20 
 1  122.9 C

0.0005  .028 
The result is very sensitive to significant figures in resistivity and .