Electrons in the Wire
Î
If the electrons move so slowly through the wire, why
does the light go on right away when we flip a switch?
1.
2.
3.
4.
Household wires have almost no resistance
The electric field inside the wire travels much faster
Light switches do not involve currents
None of the above
Think of what happens when you turn on a hose full of water. Water at
end of hose comes out immediately because of push by pressure wave.
PHY2054: Chapter 17
12
Electrons in the Wire, Part 2
Î
Okay, so the electric field in a wire travels quickly. But,
didn’t we just learn that E = 0 inside a conductor?
1.
2.
3.
4.
True, it can’t be the electric field after all!!
The electric field travels along the outside of the conductor
E = 0 inside the conductor applies only to static charges
None of the above
PHY2054: Chapter 17
13
Resistance and Ohm’s Law
ÎOhm’s
law is an empirical observation: for most materials
the current is proportional to the applied voltage
I ∝ ΔV
ÎWe
write the constant of proportionality as R and call it
the “resistance”, measured in ohms (Ω)
ΔV = IR
ÎExample,
‹R
120 V applied to a material gives I = 15 A.
= 120/15 = 8Ω
ÎMost
materials are “ohmic”, i.e. obey Ohm’s law over a
very wide range of applied voltages
‹ Common
“nonohmic” materials are semiconductors such as silicon
& germanium for which current rises exponentially with ΔV
PHY2054: Chapter 17
14
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