Resistivity and Resistance
ÎResistance
depends on
‹ Fundamental
properties of the material (i.e., resistivity)
‹ Geometry, e.g. size and shape
ÎFor
materials with an “extruded” shape (constant cross
sectional area), we can write resistance as
L
R=ρ
A
‹ρ
= resistivity (property of material)
‹ A = cross sectional area
‹ L = length of material
ÎExample:
(
Find R of nichrome wire, r = 1mm, L = 4m
) (
)
R = 150 × 10−8 × 4 / 3.14 × 0.0012 = 6.0 Ω
PHY2054: Chapter 17
15
Resistivities of Common Materials
Material
ρ (Ω ⋅ m) @ 20° C
Temp. Coeff. / °C
Silver
1.59 x 10-8
0.0038
Copper
1.7 x 10-8
0.0039
Gold
2.44 x 10-8
0.0034
Aluminum
2.82 x 10-8
0.0034
Tungsten
5.6 x 10-8
0.0045
Iron
10 x 10-8
0.0050
Lead
22 x 10-8
0.0039
Nichrome
150 x 10-8
0.0004
Germanium
0.46
–0.048
Silicon
640
–0.075
Glass
1010 – 1014
Hard rubber
1013
Fuzed quartz
7.5 x 1017
PHY2054: Chapter 17
16
Dependence of R on Geometry
ÎExample:
Material has R = 2Ω for r = 2 mm, L = 3 m
‹ Reshape
same volume of material to r = 1 mm, L = 12 m
‹ (Area shrinks by x 4 so L increases by x 4 to maintain volume)
‹ Calculate new resistance (ρ cancels)
R
R'
2
R ′ ⎛ ρ L′ ⎞ ⎛ ρ L ⎞ L′ ⎛ r ⎞
2
/
4
2
=⎜
=
=
( ) = 16
⎜ ′⎟
⎟
⎜
⎟
2
2
R ⎝ π r′ ⎠ ⎝ π r ⎠ L ⎝ r ⎠
ÎSo
R’ = 2 x 16 = 32 Ω
PHY2054: Chapter 17
17
Dependence of ρ & R on Temperature
ρ increases with T because of extra thermal
motion of atoms in material
ÎGenerally,
increase per °C is called temperature coefficient α
‹ α is normally measured at T0 = 20 °C
‹ Fractional
ρ = ρ0 (1 + αΔT )
R = R0 (1 + αΔT )
ÎExample
involving copper wire
of Cu wire is 20 Ω at T = 20 °C
‹ Find R at T = 180 °C
‹ α = 0.0039/°C, ΔT = 180 – 20 = 160
‹ Resistance
R320 = 20 (1 + 0.0039 × 160 ) = 32.5 Ω
PHY2054: Chapter 17
Assumes α is constant
over this T range!
18
Electrical Energy and Power
ÎEnergy
delivered to circuit by EMF source of voltage V
ΔU = ΔQV
ÎPower
delivered
ΔU ΔQ
P=
=
V = IV
Δt
Δt
ÎPower
dissipated by resistor (heat)
‹ Voltage
drop is IR
ΔQ
PR =
× IR = I 2 R
Δt
ÎIn all circuits, power delivered = power dissipated
‹ EMF
source is needed to replenish power lost by resistors
‹ Dissipated power shows up as heat
PHY2054: Chapter 17
19
Power Example
ÎV
= 120, R = 15 Ω
‹I
= 120 / 15 = 8
‹ Delivered: P = IV = 8 × 120 = 960 W
‹ Dissipated: PR = I2R = 82 × 15 = 960 W
R
V
PHY2054: Chapter 17
20
Another Power Example
ÎV
= 120, R1 = 20 Ω, R2 = 40 Ω
‹I
= 120 / 60 = 2
‹ Delivered: P = IV = 2 × 120 = 240 W
‹ Dissipated R1: P1 = I2R1 = 22 × 20 = 80 W
‹ Dissipated R2: P2 = I2R2 = 22 × 40 = 160 W
‹ Total dissipated: 240 W
PHY2054: Chapter 17
21