Electric Current
No longer static equilibrium
Charges move in conductors
No longer E=0 in conductor
Free charges should experience force and
therefore accelerate in conductor
It is seen that free charges move, but not
accelerate. How can this be?
Conduction in metals
–
E
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
The path of an electron
through a metal.
Path is almost completely
random, but in presence of an
electric field there is a drift
τ = “collision time” = mean (typical) time bet ween collisions
1 !
1
!
!a = F = (−eE)
m
m
!
eE
!vd = !aτ = −
τ
m
acceleration (bet ween collisions)
“drift velocity” is velocity of average
motion of electrons
Current density, J:
Count how many charge carriers cross a surface
per unit area (of the surface) per unit time.
We really want how much charge, rather than how
many carriers. So multiply by charge of carrier.
q · (number of carriers crossing dA in time dt)
!
J≡
dA · dt
direction of vector is direction of motion of charge carriers
Well now:
Recall?
+
A
V
E
A
ρ
J
E
E
(a)
(a)
+
A
V
A
90°
E
90°
ρ
J
E
E
(b)
(b)
+
V
A
and we see that the
charge crossing the
A
surface per unit time
is
!
dI = J!–· dA
We will return to
A
this, but let’s mention
this is the
“electric current”
–
A
θ
θ
ρ
E
J
E
(c)
–
E
(c)
I=
!
!
J!A· dA
vd
n Charges/unit volume
+
+
+
+
+
+
+
+
+
+
+
(COB)
J! = nq!vd
A
+
Ohm’s Law
Put together t wo previous results, for metals (q=-e):
J! = nq!vd
&
!
eE
!vd = −
τ
m
2
ne
τ
!
J! =
E
m
⇒
Current density is proportional to applied
field
Not a “law” but an empirical relation for
how conductors behave in some conditions
Counterpart for current later
!
J! = σ E
⇒
σ = “conductivity”
ρ = “resistivity”
ne2 τ
σ=
m
1
ρ=
σ
! = ρJ!
E
Material
Resistivity (V.m/A)
Copper
1.68 x 10-8
Blood, human
0.70
Ceramics, glass, rubber
1010 — 1017
other conduction mechanisms
(see text for some of these, but won’t cover here)
semiconductors
ionic solutions
plasmas
superconductors
Example: Copper at room temperature contains
1.1 x 1029 free electrons per cubic meter. What is the collision
time for these electrons?
Ans: 1.9 x 10-14 s
need m=9.11 x 10-31 kg, e=1.6 x 10-19 C
and conductivity for table, and
m
ρ= 2
ne τ
I=
!
Current
! = charge crossing surface S per unit time
J! · dA
S
dq
=
Unit:
dt
Ampere = Coulomb/sec
Surface S usually cross section
of wire:
+
+
+
+
+
+
–
–
–
–
–
–
(a)
I
we need to supply a
direction for I:
the direction of I is the
direction of motion of
positive charge carriers
Analogy: water
flow in a pipe
charge
conser vation
These four wires are made of the same metal. Rank in order,
from largest to smallest, the electron currents ia to id.
1.
2.
3.
4.
5.
id > ia > ib > ic
ib = id > ia = ic
ic > ia = ib > id
ib = ic > ia = id
ic > ib > ia > id
These four wires are made of the same metal. Rank in order,
from largest to smallest, the electron currents ia to id.
1.
2.
3.
4.
5.
id > ia > ib > ic
ib = id > ia = ic
ic > ia = ib > id
ib = ic > ia = id
ic > ib > ia > id
Rank in order, from largest to smallest, the current
densities Ja to Jd in these four wires.
1.
2.
3.
4.
5.
Jb = Jd > Ja > Jc
Jb > Ja > Jc > Jd
Jb > Ja = Jd > Jc
Jc > Jb > Ja > Jd
Jc > Jb > Ja = Jd
Rank in order, from largest to smallest, the current
densities Ja to Jd in these four wires.
1.
2.
3.
4.
5.
Jb = Jd > Ja > Jc
Jb > Ja > Jc > Jd
Jb > Ja = Jd > Jc
Jc > Jb > Ja > Jd
Jc > Jb > Ja = Jd
Ohm’s Law, again
+
V
–
E
ρ
A
J
recall:
V
I=
R
1!
!
J= E
ρ
(COB)
ρ"
R=
A
units of R: ohm (Ω) = Volt/Ampere
Resistance
Example: What should be the diameter of an aluminum wire
that carrier 15 A when the voltage across 1.0 m of the wire
is 0.25 V?
Ans: 1.4 mm
(need resistivity of aluminum: 2.65 x 10-8 Ω.m)
Ohm’s Law
You double the voltage across 1) Ohm’s law is obeyed since the
current still increases when V
a certain conductor and you
increases
observe the current increases
2) Ohm’s law is not obeyed
three times. What can you
conclude?
3) This has nothing to do with Ohm’s
law
Ohm’s Law
You double the voltage across 1) Ohm’s law is obeyed since the
current still increases when V
a certain conductor and you
increases
observe the current increases
2) Ohm’s law is not obeyed
three times. What can you
conclude?
3) This has nothing to do with Ohm’s
law
Ohm’s law, V = I R, states that the
relationship between voltage and current
is linear. Thus for a conductor that
obeys Ohm’s Law, the current must
double when you double the voltage.
Follow-up: Where could this situation occur?
Wires I
Two wires, A and B, are made of the
1) dA = 4 dB
same metal and have equal length,
but the resistance of wire A is four
2) dA = 2 dB
times the resistance of wire B. How
do their diameters compare?
3) dA = dB
4) dA = 1/2 dB
5) dA = 1/4 dB
Wires I
Two wires, A and B, are made of the
1) dA = 4 dB
same metal and have equal length,
but the resistance of wire A is four
2) dA = 2 dB
times the resistance of wire B. How
do their diameters compare?
3) dA = dB
4) dA = 1/2 dB
5) dA = 1/4 dB
The resistance of wire A is greater because its area is less than
wire B. Since area is related to radius (or diameter) squared, the
diameter of A must be two times less than B.
Wires II
A wire of resistance R is
1) it decreases by a factor 4
stretched uniformly (keeping its
volume constant) until it is twice
2) it decreases by a factor 2
3) it stays the same
its original length. What happens
to the resistance?
4) it increases by a factor 2
5) it increases by a factor 4
Wires II
A wire of resistance R is
1) it decreases by a factor 4
stretched uniformly (keeping its
volume constant) until it is twice
2) it decreases by a factor 2
3) it stays the same
its original length. What happens
to the resistance?
4) it increases by a factor 2
5) it increases by a factor 4
Keeping the volume (= area x length) constant means
that if the length is doubled, the area is halved.
Since
, this increases the resistance by four.
Resistors: devices that are pieces of conductors
made to have specific resistance
Notation for
circuit diagrams:
we’ll mostly use A
and sometimes D
Electric Power
Electrons loose energy in collisions with
atoms in lattice
This energy goes into lattice vibrations,
raising the temperature of the material
To maintain current energy has to be
provided externally at a fixed rate P=dE/dt
This energy is usually from a battery or
some other such source
Power: general device
V
P =IV
I
derivation:
dW = dq V
⇒
dW
dq
P =
=
V
dt
dt
Power dissipated by resistor
2
V
P = IV = I 2 R =
R
Example: A power line has 0.20 ohms resistance per
kilometer of length. It carries 300 A of current. Find:
(a) the voltage across 1.0 km of the wire, and
(b) the power dissipated in each kilometer of the wire
Ans: (a) 60 V, (b) 18 kW