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Transcript
PHY 184
Spring 2007
Lecture 16
Title: Electric Current and Resistance
2/5/07
184 Lecture 16
1
Announcements
 Homework Set 4 is due tomorrow at 8:00 am.
 Midterm 1 will take place in class Thursday, February 8
• Will cover Chapters 16 - 19
• Homework Set 1 - 4
• You may bring one 8.5 x 11 inch sheet of equations, front and back,
prepared any way you prefer
• Bring a calculator
• Bring a No. 2 pencil
• Bring your MSU student ID card
 We will post Midterm 1 as Corrections Set 1 after the exam
• You can re-do all the problems in the Exam
• You will receive 30% credit for the problems you missed
• To get credit, you must do all the problems in Corrections Set 1, not just
the ones you missed
2/5/07
184 Lecture 16
2
Review
 Electric current i is the net charge passing a given point in a
given time
dq
i
dt
 The ampere is abbreviated as A and is given by
1C
1A
1s
 The current per unit area flowing through a conductor is the
current density J
 If the current is constant and perpendicular to a surface,
then and we can write an expression for the magnitude of
the current density
i
J
A
2/5/07
184 Lecture 16
3
Electron Drift Velocity
 In a conductor that is not carrying current, the
conduction electrons move randomly. (thermal motion)
 When current flows through the conductor, the
electrons have an additional coherent motion.
(drift velocity, vd )
 The magnitude of the velocity of random thermal
motion is on the order of 106 m/s while the
magnitude of the drift velocity is on the order of
10-4 m/s
 We can relate the current density J to the drift
velocity vd of the moving electrons.
2/5/07
184 Lecture 16
4
Electron Drift Velocity (2)
 Consider a conductor with cross sectional area A and
electric field E.
 Suppose that there are n electrons per unit volume.
 The negatively charged electrons will drift in a
direction opposite to the electric field.
 We assume that all the electrons have the same drift
velocity vd and that the current density J is uniform.
 In a time interval dt, each electron moves a distance
vddt .
 The volume that will pass through area A is then Avd dt;
the number of electrons is dn = nAvd dt .
2/5/07
184 Lecture 16
5
Electron Drift Velocity (3)
 Each electron has charge e so that the charge dq that flows
through the area A in time dt is
dq
i
 nevd A
dt
 So the current is
 … and the current density is
i
J   nevd
A
 The current density and the drift velocity are parallel
vectors, pointing in opposite directions. As vectors,


J  nev d
2/5/07
184 Lecture 16
6
Electron Drift Velocity (4)
 Consider a wire carrying a current
 The physical current carriers are negatively charged electrons.
 These electrons are moving to the left in this drawing.
 However, the electric field, current density and current are
directed to the right.
Comments
Electrons are negative charges!
On top of the coherent motion the electrons
have random (thermal) motion.
2/5/07
184 Lecture 16
7
Clicker Question
 The figure shows positive charge carriers that
drift at a speed vd to the left. In what
directions are J and E?
A) J and E point to the right
B) J points to the left, E to the right
C) J points to the right, E to the left
D) J and E point to the left
2/5/07
184 Lecture 16
8
Example - current through a wire (1)
 The current density in a cylindrical wire of radius R=2.0 mm is
uniform across a cross section of the wire and has the value 2.0
105 A/m2. What is the current i through the outer portion of the
wire between radial distances R/2 and R?
 J = current per unit area = di / dA
2/5/07
184 Lecture 16
R
9
Example - current through a wire (1)
 The current density in a cylindrical wire of radius R=2.0 mm is
uniform across a cross section of the wire and has the value 2.0
105 A/m2. What is the current i through the outer portion of the
wire between radial distances R/2 and R?
 J = current per unit area = di / dA
R
Area A’ (outer portion)
Current through A’
2/5/07
184 Lecture 16
10
Resistance and Resistivity
 Some materials conduct electricity better than others.
 If we apply a given voltage across a conductor, we get a
large current.
 If we apply the same voltage across an insulator, we get
very little current (ideal: none).
 The property of a material that describes its ability to
conduct electric currents is called the resistivity, 
 The property of a particular device or object that
describes it ability to conduct electric currents is
called the resistance, R
 Resistivity is a property of the material;
resistance is a property of a particular object made
from that material.
2/5/07
184 Lecture 16
11
Resistance and Resistivity (2)
 If we apply an electric potential difference V across a
conductor and measure the resulting current i in the
conductor, we define the resistance R of that conductor
as
V
R
i
 The unit of resistance is volt per ampere.
 In honor of George Simon Ohm (1789-1854) resistance
has been given the unit ohm, 
1V
1
1A
2/5/07
184 Lecture 16
12
Resistance and Resistivity (3)
 We will assume that the resistance of the device is
uniform for all directions of the current; e.g.,
uniform metals.
 The resistance R of a conductor depends on the
material from which the conductor is constructed
as well as the geometry of the conductor
 First we discuss the effects of the material and
then we will discuss the effects of geometry on
resistance.
2/5/07
184 Lecture 16
13
Resistivity
 The conducting properties of a material are characterized
in terms of its resistivity.
 We define the resistivity, , of a material by the ratio
E

J
E: magnitude of the applied field
J: magnitude of the current density
 The units of resistivity are
 V
  Vm
m
=
 m
 A A
 2 
m
2/5/07
184 Lecture 16
14
Typical Resistivities
 The resistivities of some representative conductors at
20° C are listed in the table below
Material
Silver
Copper
Gold
Aluminum
Nickel
Mercury
Resistivity  ( m )
1.5910-8
1.7210-8
2.4410-8
2.8210-8
6.8410-8
95.810-8
Resistivity  ((-cm)
ž  cm )
1.59
1.72
2.44
2.82
6.84
95.8
 As you can see, typical values for the resistivity of metals
used in wires are on the order of 10-8m.
2/5/07
184 Lecture 16
15
Resistance
 Knowing the resistivity of the material, we can then
calculate the resistance of a conductor given its geometry.
Derivation:
 Consider a homogeneous wire of length L and constant cross
sectional area A.
V
i
E
and J 
L
A
 … the resistance is
R
2/5/07
V EL  L


i JA A
184 Lecture 16
16
Resistance and resistivity
 For a wire,
L
R
A
2/5/07
184 Lecture 16
17
Clicker Question
 You have three cylindrical copper conductors. Rank them
according to the current through them, the greatest first,
when the same potential difference V is placed across their
lengths.
A: a, b, c
B: a and c tie, then b
C: b, a, c
D: a and b tie, then c
2/5/07
184 Lecture 16
18
Clicker Question
 You have three cylindrical copper conductors. Rank them
according to the current through them, the greatest first,
when the same potential difference V is placed across their
lengths.
B: a and c tie, then b
D: a and b tie, then c
2/5/07
184 Lecture 16
19
Example: Resistance of a Copper Wire
 Standard wires that electricians put into residential housing
have fairly low resistance.
 Question:
 What is the resistance of a length of 100 m of standard 12gauge copper wire, typically used in household wiring for
electrical outlets?
 Answer:
 The American Wire Gauge (AWG) size convention specifies
wire cross sectional area on a logarithmic scale.
 A lower gauge number corresponds to a thicker wire.
 Every reduction by 3 gauges doubles the cross-sectional
area.
2/5/07
184 Lecture 16
20
Example: Resistance of a Copper Wire (2)
 The formula to convert from the AWG size to the wire
diameter is
(36 AWG)/ 39
d  0.127  92
mm
 So a 12-gauge copper wire has a diameter of 2.05 mm
 Its cross sectional area is then
A   d  3.3 mm
1
4
2
2
 Look up the resistivity of copper in the table …
L
100 m
-8
R    (1.72 10 m)
 0.52 
-6
2
A
3.3 10 m
2/5/07
184 Lecture 16
21
Clicker Question
 A rectangular block of iron has dimensions 2.0cm x 2.0 cm x
10cm. A potential difference is to be applied to the block
between parallel sides. What is the ratio of the resistances
R(1)/R(2) of the block for the two arrangements
(1) and (2).
A)
B)
C)
2/5/07
(1)
2.0 cm
2.0 cm
184 Lecture 16
10 cm
(2)
22
Clicker Question
 A rectangular block of iron has dimensions 2.0cm x 2.0 cm x
10cm. A potential difference is to be applied to the block
between parallel sides. What is the ratio of the resistances
R(1)/R(2) of the block for the two arrangements
(1) and (2).
A)
(1)
2.0 cm
2.0 cm
10 cm
(2)
R1 L1 / A1 L1 A2 10 20

 
   25
R2 L2 / A2 L2 A1 2 4
2/5/07
184 Lecture 16
23
Resistors
 In many electronics applications one needs a range of
resistances in various parts of the circuits.
 For this purpose one can use commercially
available resistors.
 Resistors are commonly made from carbon,
inside a plastic cover with two wires sticking out at the
two ends for electrical connection.
 The value of the resistance is indicated by four colorbands on the plastic capsule.
 The first two bands are numbers for the mantissa, the
third is a power of ten, and the fourth is a tolerance for
the range of values.
2/5/07
184 Lecture 16
24
Resistors (2)
 The number associated with the colors are:
•
•
•
•
•
•
•
•
•
•
black = 0
brown = 1
red = 2
orange = 3
yellow = 4
green = 5
blue = 6
purple = 7
gray = 8
white = 9
For example, the single resistor shown here
has colors (top to bottom)
brown, green, brown and gold
Using our table, we can see that the
resistance is 15×101  = 150 
with a tolerance of 5%
 In the tolerance band
• gold means 5%
• silver means 10%
• no tolerance band means 20%
2/5/07
184 Lecture 16
25
Summary
J  nev d
L
R
A
2/5/07
.. speed of an electron
.. resistance to current
184 Lecture 16
26