Lesson 12
(1) Electric Current
Imagine a piece of surface in space. If an amount of charge Dq goes through the
surface in time Dt , the current through the surface is defined by
I=
Dq
Dt
The unit C/s for current is called the Ampere A .
Electric current is due to the movement of charged particles, called the carriers.
Consider a beam of identical charged particles all with charge q and all travelling
with the same velocity u in the same direction. Let the cross-sectional area of the
beam be A , and the number density of particles in the beam be n . To calculate the
beam current, we focus our attention on a particular cross-section of the beam, and
count the number the charged particles going through that cross-section in a time
Dt . Since each particle travels a distance uDt during this time, all those particles
upstream of the cross-section and within the distance uDt from the cross-section
will go through the cross-section during the time Dt . These particles are to be found
in a volume uDtA . Therefore the number of such particles is nuDtA , and they carry
the total charge Dq = nqu ADt . The beam current is therefore
I = nqu A
The current density j is defined as current per unit surface area. It is a vector in the
direction of the beam velocity if the carriers are positive, opposite if the carriers are
negative. We thus have
j = nqu
In the case of current in a metallic wire, the carriers are electrons. The number
density n is that of the free electrons. Instead of all moving with the same velocity,
the free electrons have random velocity , with an average value called the drift
velocity u d . The magnitude of the current density is
j = neud
and the current I = jA is in the direction opposite to the drift velocity of the
electrons.
Example: A copper wire with a circular cross-sectional of diameter 1.0mm carries a
current of 10.0A. Assuming each copper atom contributes 2.0 free electrons, find the
drift velocity. (Density of copper = 9000 kg/m3, atomic mass = 64)
Solution:
number density of atoms =
mass density
9000
=
= 8.4 ´10 28 m-3
-27
mass of one atom 64 ´1.67´10
n = 2 ´ number densityof atoms =1.68´1029 m-3
j=
I
10
=
=1.27 ´10 7 A / m 2
2
-3
A p ( 0.5´10 )
ud =
j
1.27´10 7
=
= 4.7 ´10-4 m / s = 0.47mm / s
29
-19
ne 1.68´10 ´1.6 ´10
This is a remarkably small velocity.
(2) Ohm’s Law
When an electric potential difference exists between two ends of a metal wire, a
current flows from the end with higher potential to that with the lower potential.
Denoting by a and b the ends with higher and lower potential, and Va ,Vb the
potentials there, we define the potential difference
V = Va -Vb
Ohm’s law states the current I is proportional to V so that the ratio V I is a
constant. The constant is called the electrical resistance R of the wire. We therefore
have
V = IR
The unit Volt Ampere is called the Ohm ( W). A resistor is symbolically drawn as
shown:
The electrical resistance is proportional to the length L and inversely proportional
to the cross-sectional area A of the wire. Introducing the constant of
proportionality r called the electrical resistivity, we can write
L
R=r
A
Example: Calculate the resistance at 20C of a 1.0m long copper wire with circular
cross-section of diameter 1.0mm, given that the resistivity of copper is
1.7´10-8 W× m . Calculate the potential difference to produce a current of 10A in this
wire.
Solution: R =1.7 ´10-8 ´
1.0
p ( 0.5´10-3 )
2
= 0.022W
V = IR =10 ´ 0.022 = 0.22V
The potential difference produces an electric field inside the wire, pointing from a to
b . The electric field is related to the potential difference by
V = EL
Using this in Ohm’s law and expressing R in terms of resistivity, also introducing the
E
current density j = I A , we find j =
r
In general, in a conducting medium of any shape, at a point where the electric field is
E , a current density
j=
E
r
arises. This equation is the microscopic form of Ohm’s law.
Electrical resistivity is a property of the material forming the conducting medium. It
varies over a great range for different materials, and is used to classify materials
into conductor, insulator, semi-conductors, and super-conductors. It also depends
on temperature.
(3) Power Dissipation of Resistor
This electric field in a metallic wire exerts a force on the free electrons in the
opposite direction. As the electrons move, they collide with the ions forming the
crystal lattice and suffer a friction force as a result. At equilibrium, when the electric
force is balanced by the friction force, the electrons will have a steady drift velocity,
which accounts for the current.
Even though electrons are the carriers, it is advantageous to think of the current as
due to the movement of positive charge in the direction of the current. When a
quantity of charge Dq moves from one end a of the wire to the other end b ,
Work done by electric field = DW = Dq (Va -Vb ) = DqV
Power delivered by electric field is P =
DW Dq
= V = IV
Dt
Dt
Because of friction, this power is converted into heat, and we can consider the
power as the power dissipated in the resistor. Using Ohm’s law, we can have
different ways of calculating the power dissipated:
P = IV = I 2 R =
V2
R
Although we have been concerned with DC (direct current), it turns out that for AC
(alternating current) , both Ohm’s law and the formulas for power ( which is now
taken to be average power over one cycle of oscillation of AC) are valid provided the
current and the voltage refer to their root-mean-square values. The rms voltage we
get from household electrical outlets is 120V. The rated power of an appliance is the
power dissipated when connected to a 120V rms voltage.
Example: How much current does a 100-W light bulb draw? What is the bulb’s
resistance?
Solution: From P = IV ,
From P =
I=
P 100
=
= 0.83A
V 120
V2
V 2 120 2
, R=
=
=144W
R
P 100
Note that the higher the power rated for an appliance, the lower the resistance, and
the higher the current the appliance draws.
Example: How much does it cost to run a 100-W light bulb for a month if the cost of
electricity is $0.25 per kWh?
Solution: 1.0 kWh =103 ´ 3600 J = 3.6 ´106 J
EnergyinkWh = PowerinkW ´Timeinhours = 0.1´30 ´ 24 = 72kWh
Cost = 72 ´$0.2 = $14.4
Energy used in kWh= power in kilowatts time in hours
(4) Mechanisms of Electrical Conduction
1. Metal
Microscopic theory; Temperature dependence , breakdown of classical theory
2. Semiconductors
P and N types; P-N junction; transistor
3. Superconductors
4. Electrolytes
5. Ionized gases (plasmas)