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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 25 1

Current and Resistance

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

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!