Contents
1 Current and Resistance
1
1.1
Learning Objectives . . . . . . . . . . . . . . . . .
1
1.2
New Symbols . . . . . . . . . . . . . . . . . . . .
1
1.3
Current density . . . . . . . . . . . . . . . . . . .
2
1.4
Drift velocity . . . . . . . . . . . . . . . . . . . .
2
1.5
Electrical Resistance . . . . . . . . . . . . . . . .
4
1.6
Power . . . . . . . . . . . . . . . . . . . . . . . .
5
1.7
Summary . . . . . . . . . . . . . . . . . . . . . .
5
Copyright
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Reading Assignment
Knight Chapter 31
1
Current and Resistance
1.1
Learning Objectives
• To understand electric current and current density.
• To understand Ohm’s law that relates the current flowing
in a material to the potential difference across it.
• To understand the concept of resistance and also how it
relates to both material and geometrical parameters.
• To be able to calculate the power dissipated in a material
when current flows through it.
1.2
New Symbols
Table 1: New Symbols
I
J
σ
ρ
R
current
current density
conductivity
resistivity
resistance
A
Am−2
1/Ωm
Ωm
Ω
In a conducting wire, mobile charge carriers will move if there is
difference in potential between the two ends of the wire Vab . The
flow of charge is called a current.
Va
Vb
E
a
b
e-
F
A
I
conducting wire
I=
dQ
dt
Units : A (Ampères)
The current is the amount of mobile charge that flows through a
fixed surface (e.g. A on the Fig.) per unit time.
• In a metal the mobile charge carriers are electrons and they
will move to the left because Fx = −eEx .
• An electron moving to the left with charge −e has the same
effect as a positive charge +e moving to the right:
(−e)(−vx ) = (+e)(+vx ) = evx
and we defined current in terms of positive charge.
• You can think in terms of positive charge; +e flowing to
the right. This is called conventional current.
• Discuss faucets and water with negative mass.
2
1.3
Current density
If we normalize the current to the cross sectional area of the wire,
we have the current density:
J≡
1.4
I
A
Units : Am−2
Drift velocity
There are a lot of electrons in conductors and they do not all
travel at the same speed. They have a range of speeds and they
also don’t all move in the same direction. They move in both
directions (left and right).
The electrons that are moving most rapidly can be moving at ≈
c/100.
The effect of the electric field is to change the average electron
speed, or drift velocity
1 X
vi ,
vd =
N
from zero to some finite value.
Analogy: Traffic on the 401. Calculate the average speed by
counting vehicles that pass a fixed point for an entire day. Then
we will raise the 401 near Toronto by 5 km (e.g. add a gravitational potential) and calculate the average speed.
t=0
I
A
Q
t=t
vdt
E
Q
(Avd t)ne
=
= Anevd
t
t
Where n is the electron concentration. Therefore, the current
density is:
I=
J = nevd .
3
Example 1. Calculate the drift velocity for electrons in a copper
wire with a diameter of 1.0 mm if it is carrying a current of I =
1A. The concentration of electrons on copper - basically one per
atom - is n = 1.1 × 1029 m−3 .
1.5
Electrical Resistance
The ratio of J to E is a constant for many materials and it is
called the conductivity (σ). Materials with a high conductivity
conduct well. The inverse of the conductivity is called the resistivity (ρ).
1
J
=σ= .
E
ρ
Ohm0 s Law
Units: σ 1/Ωm; ρ Ωm.
Question: What can we say about the ratio V /I ?
Starting from Ohm’s law:
J
I L
1
=
=
E
AV
ρ
we have
V
L
= ρ ≡ R,
I
A
Second form of Ohm0 s Law
where R is the resistance of the material. Notice that the resistance contains both geometric (L and A) and materials (ρ)
parameters:
L
R=ρ .
A
Units : Ω (Ohms)
4
Separating material parameters from geometrical parameters is
a very smart thing to do.
• If we double the length of the wire, the resistance doubles.
• If we double the area of the wire, the resistance halves.
This is how our resistor would be represented in a circuit diagram:
V
R
I
1.6
A
Power
Considering the mobile charge carriers in the wire.
P =
U
I
U = I = I V.
e
e
Using the second form of Ohm’s Law V = IR, we have two
alternate forms
V2
P =I R=
.
R
2
Units: W (Watts) = Js−1 .
Example 2. A 12 V car starter motor draws I = 200 A. (a)
Calculate the amount of power dissipated and (b) calculate the
energy dissipated in a 5 s burst.
5
1.7
Summary
• Current is the amount of mobile charge that flows through
a fixed surface per unit time. (I = dQ/dt)
• Current density J = I/A
• Ohm’s Law: σ = J/E = 1/ρ where σ is the conductivity
and ρ is the resistivity.
• Resistance R = V /I = ρL//A
• Power P = IV = I 2 R = V 2 /R
6