February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 1
Current and Resistance
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February 11, 2014 Physics for Scientists & Engineers 2, Chapter 21 2
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 3
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
Current ! Current is defined as the flow of positive charge ! Current is
! Current flowing through a cross section area A is
! And the current density is
! 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
i =J ⋅dA∫
i = dq
dt= −nevd A
J = i
A= −nevd
Current through portion of a wire
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 5
! 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 by However, we only want the current
through the reduced cross sectional are A’.
A’ J = i
A
Current through portion of a wire
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 6
Cross sectional area A’ (outer portion)
′A = πR2 −π R / 2( )2 = π 3R2
4= 9.424 ⋅10−6 m2
i = J ′A = 2.0 ⋅105 A/m2( ) 9.424 ⋅10−6 m2( )=1.9 A
A’ Current through A’
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
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 9
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 10
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
! The unit of resistance is volt per ampere ! In honor of Georg Simon Ohm (1789-1854), resistance has
been given the unit ohm, Ω ! Rearrange the equation to get Ohm’s Law
R = ΔV
i
1 Ω = 1 V
1 A
i = ΔV
R or ΔV = iR
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 11
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
! The units of resistivity are ρ = E
J
ρ[ ]= E[ ]
J[ ] =V/mA/m2 =
V mA
=Ω m
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 12
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.
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
! The magnitude of the current density is
! Combining our definitions gives us
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 13
ΔV = −
E ⋅ds∫ ⇒ E = ΔV
L
J = i
A
ρ = E
J= ΔV / L
i / A= ΔV
iAL= iR
iAL= R A
L ⇒ R = ρ L
A
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 16
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 17
Resistance of a Copper Wire ! The formula to convert from the AWG size to the wire
diameter is
! So a 12-gauge copper wire has a diameter of 2.05 mm ! Its cross sectional area is then ! Look up the resistivity of copper in Table 25.1
d = 0.127 ⋅92(36−AWG )/39 mm
A = 14πd2 = 3.31 mm2
R = ρ L
A= (1.72 ⋅10-8 Ω m) 100.0 m
3.31⋅10-6 m2 = 0.520 Ω
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 20
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
Resistor Color Codes
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 21
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 Tolerance Brown 1%
Red 2% Gold 5% Silver 10% None 20%
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 22
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
! where • ρ is the resistivity at temperature T • ρ0 is the resistivity at temperature T0
• α is the temperature coefficient of electric resistivity
ρ− ρ0 = ρ0α T −T0( )
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 23
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 24
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)
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
February 11, 2014 Physics for Scientists & Engineers 2, Chapter 25 25
1.5236Ω =ρ(0°C) ⋅ 300m0.25π (2 ⋅10−3m)2
⇒ ρ(0°C) = 1.596 ⋅10−8Ωm
ρ(20°C) = ρ(0°C)(1+α ⋅20) = 1.72 ⋅10−8Ωm Table 25.1: Copper!