ELECTRIC CURRENTS Topic 5.1 Electric potential difference, current and resistance.
Chapter 5 : Steady Electric Currents Static Electromagnetics, 2008 SpringProf. C.W. Baek & H. Kim...
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Transcript of Chapter 5 : Steady Electric Currents Static Electromagnetics, 2008 SpringProf. C.W. Baek & H. Kim...
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
•Types of electric currents– Convection currents result from motion of electrons and/or holes in a vacuum or rarefied gas. (electron beams in a CRT,
violent motion of charged particles in a thunder storm). Convection current , the result of hydrodynamic motion involving a
mass transport, are not governed by Ohm’s law.
– Electrolytic current are the result of migration of positive and negative ions.
Atoms of the conducting medium occupy regular positions in a crystalline structure and do not move. Electrons in the inner shells are tightly bound to the nuclei are not free to move away. Electrons in the outermost shells do not completely fill the shells; they are valence or conduction electrons and are
only very loosely bound to the nuclei. When an external E field is applied, an organized motion of the conduction electrons will result (e.g. electron
current in a metal wire). The average drift velocity of the electrons is very low (10 -4~10-3 m/s) because of collision with the atoms,
dissipating part of their kinetic energy as a heat.
Electrolysis chemical decomposition
+ion -ion
ElectrolyteUsually a diluted
salt solution
current
- Conduction currents result from drift motion of electrons and/or holes in conductors and semiconductors.
Chapter 5. Steady Electric Currents
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
• Steady current density
– Electric currents : motion of free charges
– Current density : current per unit area
Consider a tube of charge with volume charge density v
moving with a mean velocity uu along the axis of the tube. Over a period t, the charges move a distance l = u t. The amount of charge that crosses the tube's cross-sectional
surface s' in time t is therefore
If the charges are flowing through a surface s whose surface
normal is not necessarily parallel to uu,
' ' 'v v vq v l s u s t
n̂
vq u s t
v
qI u s J s
t
�������������������������� ��3 2 [C/m m/s=A/m ]J u
���������������������������� : (volume) current density
[A]S
I J d s ����������������������������
The total current flowing through an arbitrary surface S is
/ [A]=[C/s]I dQ dt2/ [A/m ]J dI ds
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
For metal,
For semiconductors : cf) resistivity :
Ohmic media : material following Ohm's law
e eJ E E ������������������������������������������
: point form of Ohm's law σ: conductivity (S/m)
( )e e h h e h 1 ( m)
12 12
1212
/
/
S
V E E V
I J d s JS J I S
V IJ E V I RI
S S
����������������������������
( )RS
The resistance of a material having a straight length , uniform
cross section area S, and conductivity :
Example 5-1.
In vacuum,
eF qE m a
u adt
– In conductors and semiconductors, electrons and/or holes can not be accelerated due to the collision.
– The drift velocity is proportional to the applied E field.
( : electron mobility)e eu E ��������������
3 2 [C/m m/s=A/m ]J u����������������������������
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
•Electromotive force (emf)Static (conservative) electric field :
This equation tells us that a steady current cannot be maintained in the same direction in a closed
circuit by an electrostatic field (Charge carriers collide with atoms and therefore dissipate energy
in the circuit).
This energy must come from a nonconservative field source for continuous current flow (e.g.
battery, generator, thermocouples, photovoltaic cells, fuel cells, etc.).
These energy sources, when connected in an electric circuit, provide a driving force to push a
current in a circuit : impressed electric field intensity EEi i .
– EMF of a battery : the line integral of the impressed field intensity EEi i from the negative to the
positive electrode inside the battery.
0C
E d ����������������������������
1
0C
J d
����������������������������
For an ohmic material :
1 1
2 2iE d E d= × =- ×ò ò
ur r ur rl lV
Insidethe source
2 1
1 20
CE d E d E d× = × + × =ò ò òur r ur r ur r
l l lÑOutside
the sourceInside
the source
1 1
12 1 22 2iE d E d V V V
�������������������������������������������������������� V
E��������������
2 1
1 20iE d E d× - × =ò ò
ur r ur rl l
1
2
1
2 (inside source, for )i
V E d
V E d emf
����������������������������
����������������������������
Current flows from (-) to (+) inside source!
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
• Kirchhoff's voltage law
When a resistor is connected between terminal 1 and 2 of the battery to complete
the circuit : the total electric field intensity (EE + EEii) must be used in the point form
of Ohm's law.
If the resistor has a conductivity , length , and uniform cross section S, J = I / S.
– Kirchhoff's voltage law : Around a closed path in an electric circuit, the algebraic
sum of the emf’s (voltage rises) is equal to the algebraic sum of the voltage drops
across the resistances.
i iJ
J E E E E
��������������
���������������������������������������������������������������������� R
1 1C
IJ d RI
S
���������������������������� V
1
2
1
i iC C C C
i
J d E E d E d E d
E d
������������������������������������������������������������������������������������������������������������������������������
����������������������������
V
(V)j k kj k
R I V
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
• Equation of continuity and Kirchhoff's current law
– Principle of conservation of charge : If a net current I flows across the surface out of (into) the
region, the charge in the volume must decrease (increase) at a rate that equals the current.
– Kirchhoff's current law : Algebraic sum of all the dc currents flowing out of (into) a junction
in an electric circuit is zero.
VVVS
dvt
dvJdvdt
d
dt
dQsdJI
)(A/m 3
tJ
: Equation of continuity
For steady currents, and therefore =0t
0J
��������������0 0 (A)jS
j
J d s I ����������������������������
( )
/ 0
J E Et
Et
������������������������������������������
��������������( / ) 3
0 (C/m )te
: relaxation time
decay to 1/e (36.8% value)
For a good conductor(e.g. copper), = 1.5210-19 [s]
Charge relaxation
For ac currents, and therefore 0t
0J
��������������0 0 (A)jS
j
J d s I ����������������������������
Really?
Quasi-static case (at low frequency =0)
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
• Power dissipation and Joule's law
– Power dissipated in a conducting medium in the presence of an electrostatic field EE Microscopically, electrons in the conducting medium moving under the influence of an
electric field collide with atoms or lattice sites Energy is thus transmitted from the
electric field to the atoms in thermal vibration.
The work W done by an electric field EE in moving a charge q a distance is
For a given volume V, the total electric power converted into heat is
In a conductor of a constant cross section, , with measured in the direction JJ.
Since V = RI,
0
limt
ww qE p qE u
t
��������������������������������������������������������
: )(W/m 3JEdv
dP Power density under steady-current conditions
Joule’s law: (W) dvJEPV
dvJEdvuqNEpdPi
iiii
i
dsddv d
L SP Ed Jds VI
2 (W)P I R
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
• Boundary conditions for current density
– For steady current density J J in the absence of nonconservative energy sources
(1) Normal component : the normal component of a divergenceless vector field is
continuous.
(2) Tangential component : the tangential component of a curl-free vector field is
continuous across an interface.
01
0
0 0
dJJ
sdJJ
C
S
Differential form Integral form
)(A/m 221 nn JJ
2
1
2
1
t
t
J
J The ratio of the tangential components of J J at two sides of an interface is equal to the ratio of the conductivities.
[HW] Solve Example 5-3.
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
• Resistance calculations
We have calculated the resistance of a conducting medium having a straight length
, uniform cross section area S, and conductivity .
This equation can not be used if the S of the conductor is not uniform How can
we calculate the resistance?
– Procedures for resistance calculation
(1) Choose an appropriate coordinate system for the given geometry.
(2) Assume a potential difference V0 between the conductor material.
(3) Find EE within the conductor (by solving Laplace's equation and taking ).
(5) Find resistance R by taking the ratio V0 / I.
( )RS
S
L
S
L
sdE
dE
sdJ
dE
I
VR
SS
sdEsdJI (4) Find the total current I from
E V ��������������
cf) S
L
E d sQC
V E d
����������������������������
����������������������������
RC
Chapter 5 : Steady Electric Currents
Static Electromagnetics, 2008 Spring Prof. C.W. Baek & H. Kim
• Example 5-6 : Resistance of a conducting flat circular washer
Sol.
(1) Choose a coordinate system : CCS
(2) Assume a potential difference V0.
(3) Find E E .
Boundary conditions are :
(4) Find the total current I.
(5) Find R.
V0+
-
2
1 220
d VV c c
d
0 02 2ˆ ˆ, V VV
V E Vr r
�������������� 0
0 at 0
at / 2
V
V V
0 0 02 2 2ˆ lnb
S a
V V hVdr bJ E I J d s h
r r a
��������������������������������������������������������
(at = /2 surface)
ab
hI
VR
ln2
0