Download Steady Electric Currents Electric Current and Current Density Ohm`s

UNIVERSITY OF TECHNOLOGY, SYDNEY
FACULTY OF ENGINEERING
48531 Electromechanical Systems
Steady Electric Currents
Topics to cover:
1. Electric Current and Current Density
2. Ohm’s Law
3. Electric Resistance
4. Electric Power Dissipation and Joule’s Law
Ohm’s Law
• Point Form
J =σE
where σ is the conductivity. The SI unit for conductivity is
ampere per volt-meter (A/Vm) or siemens per meter (S/m). The
reciprocal of conductivity is known as resistivity in ohm-meters
(Ωm).
• Circuital form
I=
V
R
where V is the voltage across the conductor and R is the
resistance of the conductor. The unit of resistance is ohms (Ω).
Electric Current and
Current Density
• Electric charges in motion constitutes an electric current. The unit
of current is ampere or A, defined as the rate of movement of
charge crossing a given plane in coulomb per second, or
dQ
I =
dt
• Electric current density is defined as the current per unit cross
sectional area:
∆I
∴ I = J • ds
J = lim
n
∆s → 0 ∆ s
S
∫
where n is the normal direction of the plane. The unit of J is A/m2.
• Current density is a vector while current is not.
Conductivity
• Conduction currents in conductors and semiconductors are caused by
drift motion of conduction electrons and/or holes.
• Consider the steady motion of one
kind of charge carriers, each of
charge q (which is negative for
electrons), across an element of
surface ∆s with a velocity u. If N is
the number of charge carriers per
unit volume, then in time ∆t each
charge carrier moves a distance u∆t,
and the amount of charge carrier
passing through the surface ∆s is
∆Q = Nqu • a n ∆s∆t
Conductivity - cont.
Electric Resistance
Since current is the time rate of change of charge, we have
∆I =
∆Q
= Nqu • an ∆s = J • ∆s
∆t
where J=Nqu (A/m2) is the volume current density or simply current
density. It can be justified analytically that for most conducting
materials the average drift velocity is directly proportional to the applied
external electric field strength, or u=−µeE, where µe is the mobility of
electrons in a conductor. Therefore, we obtain the point form of Ohm's
law:
J = −NqµeE =σE
where σ is conductivity. The SI unit for conductivity is ampere per voltmeter (A/Vm) or siemens per meter (S/m). The reciprocal of conductivity
is known as resistivity in ohm-meters (Ωm).
Electric Resistance - cont.
• The reciprocal of resistance is defined as conductance or G=1/R.
The unit for conductance is Siemens (S) or (Ω-1).
• Resistance Calculation
or
R1
R1
R2
R2
R s = R1 + R2
Rs
Rp
• The voltage-current relationship
Consider a piece of homogeneous material of conductivity σ, length l,
and uniform cross section A, as shown below. Within the conductor,
J=σE, where both J and E are in the direction of current flow. The
potential difference or voltage between terminals 1 and 2 is
V12 = El
E = V12 l
or
and the total current is
I = ∫ J • dA = JA = σEA =
A
or
I=
V12
R
where
or
G p = G1 + G 2
l
σA
(Ω)
is the electric resistance between the terminals.
Electric Resistance - Example
A metal hemisphere of radius Re, buried with its flat face lying in the surface of the ground, is used as an earthing electrode. It may be assumed
that a current flowing to earth spreads out uniformly and radially from the
electrode for a great distance. Show that, as the distance for which this is
true tends to infinity, the resistance between the electrode and earth tends
to the limiting value ρ/2πRe, where ρ is the resistivity of the earth.
metallic hemisphere
1
1
1
=
+
R p R1 R2
R=
σA
V
l 12
ground
hemispherical cap with
resistance dR, thickness dr
and cross-sectional area A.
Electric Resistance - Example
Solution:
To determine the total resistance between the metallic cap and earth (at
∞) we can sum the incremental resistances of the thin hemispherical caps
(extending from Re to ∞). First, choose a hemispherical cap of thichness
dr, and the incremental resistance of the cap is
dR = ρ
dr
,
A
A = 2 πr 2
where
Therefore, the total resistance
ρ
R=
2π
∞
ρ
dr
=
∫ r 2 2π
Re

 −
Power Dissipation and
Joule's Law
• Under the influence of an electric field, conduction electrons in a
conductor undergo a drift motion macroscopically. Microscopically,
these electrons collide with atoms on lattice sites. Energy is thus
transmitted from the electric field to the atoms in thermal vibration.
• The Joule's law states that for a given volume Vc the total electric
power converted into heat is
∞
1
ρ
=
r  Re 2 πRe
P = ∫ E • Jdv
(W)
Vc
as required.
Power Dissipation and
Joule's Law - cont.
• In a conductor of uniform cross section, dv=dAdl, with dl measured
in the direction of J. The above equation becomes
P = ∫ Edl ∫ JdA = VI
l
Summary
I = ∫ J • ds
or
I=
• Ohm’s Law
J =σE
• Resistance
l
R=
σA
1
1
1
=
+
R p R1 R2
R s = R1 + R2
G p = G1 + G 2
P = I2R
This is an expression for power dissipation in a resistor of resistance
R.
and
I =
A
where I is the current in the conductor. Since V=RI, we have
dQ
dt
• Electrical Current
• Joule’s Law
P = VI = I 2 R
S
V
R