Current and Resistance
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
1
Helproom hours
!  Strosacker learning center, BPS 1248
!
!
!
!
!
Mo: 10am – noon, 1pm – 9pm
Tue: noon – 6pm
We: noon – 2pm
Th: 10am – 1pm, 2pm – 7pm
Fri: noon to 2pm
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 21
2
Electric Current
!  Total charge is conserved, which means that charge flowing
in a conductor is never lost
!  The unit of current is Coulombs per second, which has been
given the unit Ampere (abbreviated A), named after French
physicist André-Marie Ampère, (1775-1836)
!  Some typical currents are
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
3
Current
!  Current is defined as the flow of positive charge
dq
= −nevd A
!  Current is i =
dt
!  Current flowing through a cross

section area A is i = J ⋅dA
∫
i
!  And the current density is J = = −nevd
A
!  A current that flows in only one direction, which does
not change with time, is called direct current
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
4
Current through portion of a wire
!  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 of J=2.0 105 A/m2.
PROBLEM What is the current i through the outer portion of
the wire between radial distances R/2 and R?
THINK
The current density is uniform
across the cross section, the
current density, J, the current i and
the cross section area A are related
i
by
J=
A
However, we only want the current
through the reduced cross sectional
are A’.
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
A’
5
Current through portion of a wire
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
of J=2.0 105 A/m2.
PROBLEM What is the current i through the outer portion of
the wire between radial distances R/2 and R?
CALCULATE
Cross sectional area A’ (outer portion)
2
3R
A′ = π R2 − π (R / 2 ) = π
= 9.424 ⋅10−6 m 2
4
2
Current through A’
(
)(
A’
)
i = JA′ = 2.0 ⋅10 5 A/m 2 9.424 ⋅10 −6 m 2 = 1.9 A
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
6
Resistivity and Resistance
!  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
!  The property of a material that describes its ability to
conduct electric currents is called the resistivity, ρ
!  The property of a particular device or object that describes
its 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
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
9
Resistance
!  If we apply an electric potential difference ΔV across a
conductor and measure the resulting current i in the
conductor, we define the resistance R of that conductor as
ΔV
R=
i
!  The unit of resistance is volt per ampere
!  In honor of Georg Simon Ohm (1789-1854), resistance has
been given the unit ohm, Ω
1V
1Ω=
1A
!  Rearrange the equation to get Ohm’s Law
ΔV
i=
or ΔV = iR
R
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
10
Resistivity
!  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 device depends on the material from
which the device is constructed as well as the geometry of
the device
!  The conducting properties of a material are characterized in
terms of its resistivity
!  We define the resistivity of a material by the ratio
E
ρ=
J
!  The units of resistivity are
E ] V/m V m
[
[ ρ ] = J = A/m2 = A = Ω m
[ ]
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
11
Typical Resistivities
!  The resistivities of some representative conductors at 20°C are listed
below (more in Table 25.1)
!  Typical values for the resistivity of metals used in wires are on the
order of 10-8 Ω m.
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
12
Resistance
!  Knowing the resistivity of the material, we can then
calculate the resistance of a conductor given its geometry
!  Consider a homogeneous wire of length L and constant
cross sectional area A, we can relate the electric field and
potential difference as
 
ΔV
ΔV = − ∫ E ⋅ds ⇒ E =
L
!  The magnitude of the current density is
i
J=
A
!  Combining our definitions gives us
E ΔV / L ΔV A iR A
A
L
ρ= =
=
=
=R
⇒ R=ρ
J
i/A
i L i L
L
A
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
13
Resistance of a Copper Wire
!  Standard wires that electricians put into residential housing
have fairly low resistance
PROBLEM
!  What is the resistance of a length of 100.0 m of standard
12-gauge copper wire, typically used in household wiring for
electrical outlets?
SOLUTION
!  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
!  See Table 25.2
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
16
Resistance of a Copper Wire
!  The formula to convert from the AWG size to the wire
diameter is
d = 0.127 ⋅92(36−AWG )/39 mm
!  So a 12-gauge copper wire has a diameter of 2.05 mm
!  Its cross sectional area is then
A = 14 π d 2 = 3.31 mm 2
!  Look up the resistivity of copper in Table 25.1
R=ρ
L
100.0 m
= (1.72⋅10-8 Ω m)
= 0.520 Ω
-6
2
A
3.31⋅10 m
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
17
Resistors
!  In many electronics applications one needs
a range of resistances in various parts of the
circuit
!  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 color-bands
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
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
20
Resistor Color Codes
Band 1 –Digit 1 - Brown – 1
Band 2 –Digit 2 - Green – 5
Band 3 – Power of 10 - Brown – 1
Band 4 – Tolerance - Gold
15·101 Ω = 150 Ω
5% tolerance
Color
Brown
Red
Gold
Silver
None
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
Tolerance
1%
2%
5%
10%
20%
21
Temperature Dependence of
Resistivity
!  The resistivity and resistance vary with temperature
!  For metals, this dependence on temperature is linear over a
broad range of temperatures
!  An empirical relationship for the temperature dependence
of the resistivity of metals is given by
ρ − ρ0 = ρ0α (T −T0 )
!  where
•  ρ is the resistivity at temperature T
•  ρ0 is the resistivity at temperature T0
•  α is the temperature coefficient of electric resistivity
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
22
Temperature Dependence of
Resistance
!  In everyday applications we are interested in the
temperature dependence of the resistance of various devices
!  The resistance of a device depends on the length and the
cross sectional area
!  These quantities depend on temperature
!  However, the temperature dependence of linear expansion
is much smaller than the temperature dependence of
resistivity of a particular conductor
!  The temperature dependence of the resistance of a
conductor is, to a good approximation,
R − R0 = R0α (T −T0 )
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
23
Temperature Dependence
!  Our equations for temperature dependence deal with
temperature differences so that one can use
°C as well as K (T-T0 in K is equal to T-T0 in °C)
!  Values of α for representative metals are shown below
(more can be found in Table 25.1)
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
24
Example: Which Metal is it?
!  A metal wire of 2mm diameter and length 300m has a
resistance of 1.6424 Ω at 20oC and 2.415 Ω at 150oC. Find
the values of α, R(0oC), ρ(0oC) and identify the metal!
!  Solution:
R(150oC)=2.415Ω=R(0oC)(1+α(150-0))
R(20oC)=1.6424Ω=R(0oC)(1+α(20-0))
First equation: R(0oC)=2.451Ω/(1+α(150-0)), substitute in
second equation, solve for α: α=3.9 10-3 oC-1
With α known, use any of the above equations to get
R(0oC)=1.5236Ω. Then use R(0oC)=ρ(0oC)L/A
ρ(0°C) ⋅ 300m
−8
1.5236Ω =
⇒
ρ
(0°C)
=
1.596
⋅10
Ωm
−3
2
0.25π (2 ⋅10 m)
ρ(20°C) = ρ(0°C)(1 + α ⋅ 20) = 1.72 ⋅10 −8 Ωm
Table 25.1: Copper!
February 11, 2014
Physics for Scientists & Engineers 2, Chapter 25
25