Resistivity
The resistance of a conductor that obeys Ohm’s law
depends upon three factors:
1. The material of which it is composed;
the ability to carry an electric current
varies more than almost any other
physical property of matter.
2. Its length (L): The longer the conductor,
the greater its resistance.
3. Its cross-sectional area (A). The thicker the
conductor the less its resistance.
It’s much easier to travel on an
open highway than a rocky
trail.
Resistance
R =!
Also, the longer
the trip, the
harder it is.
It’s much easier to travel on an
open highway than a rocky
trail.
Resistance is
futile.
Resistance
R =!
L
A
Where ρ = resistivity of the material. This is
a characteristic property of the material and
it varies with temperature
L
A
L = length (m)
A = cross sectional area (m2)
ρ = resistivity of the material. This is a
characteristic property of the material
and it varies with temperature
ρ = “roe”
Calculate the resistance of a piece of aluminum that is
10.0 cm long and has a cross sectional area of 1 x 10-4m2.
(ρ = 2.82x10-5Ωm)
R =!
L
A
= (2.82 "10 !5 )
0.1m
10 ! 4 m 2
= 2.82 x 10-2 Ω
1
Repeat the calculation for a piece of hard rubber…
ρ = 1.0 x 1013Ωm
R =!
L
A
ΔR = αRΔT
= (1"1013 )
0.1m
10 ! 4 m 2
An alternative formula, used to
calculate the resistance of a material at
any temperature is given by….
ρ = ρo [1 + α(T-To)]
In this equation, resistivity is referenced to
some standard resistivity ρo at some standard
temperature To
In terms of resistance …
R = Ro [1 + α(T – To)]
Playing with Ohm’s Law…
Volts x amps =
Amps = Coulombs/second
Joules/Coulomb x Coulombs/second
= Joules/second
Power = Volts x Amps
Where α is the temperature coefficient
of resistivity of the material. This is a
characteristic property of the material.
A negative coefficient indicates the
resistivity decreases with increasing
temperature.
= 1.00 x 1016 Ω
Volts = Joules/Coulomb
The resistivities of nearly all substances vary with
temperature. In general, metals increase in resistivity
with an increase in temperature while non metals
decrease in resistivity.
= Watts
Wait for the
better equations
Example: A metal wire has a resistance of 10.00 Ω at a
temperature of 20oC. If the same wire has a resistance of
10.55 Ω at 90oC, what is the resistance of the wire when
its temperature is -20oC?
Step 1: find the temp. coefficient of resistivity, α
R = Ro [1 + α(T – To )]
10.00 = 10.55[1 + α(20-90)]
α = 7.857x10-4
Step 2: find the resistance at -20oC
R = Ro [1 + α(T – To )]
R = 10[1 + 7.857x10-4(-20-20)] = 9.68 Ω
POWER
P = VI
But V = IR
so….
P = I2 R
P = VI
2
What is the current flow to a 1000W
hairdryer plugged into a 120V socket?
1) 8.3A
2) 10A
3) 12.2A
Ohm’s Law Equations:
4) 15.8A
5) 20A
V = IR
P = IV
P = (V/R)V = V2/R
P = V2/R
Lets say you want to protect your new
speakers by putting a fuse in line with the
speakers. Assume you have 100W speakers
and that they are 8 ohms. What size (amps)
fuse should you use so that it just blows at
100 Watts?
P = I2R
100W =
Each mini-group needs:
1 battery
I2
8
I2 = 100/8 = 12.5
I = 3.5 amp fuse
1 battery holder
2 wires
1 VERY small pinch of steel
wool
A fuse is used to
prevent an appliance
from receiving too
much current. If the
current goes through
a fuse first and is too
much for the fuse to
handle, the fuse will
burn up, protecting
the appliance.
Two light bulbs operate from 120V, but one has a
power rating of 25W, and the other has a power rating of
100W. Which bulb has the higher resistance?
1) 25W bulb
2) 100W bulb
4) Depends on the bulb
3) Both the same
3