Chapter 25: Current, Resistance and Electromotive Force
Charge carrier motion in a conductor in two parts
Constant Acceleration
F = ma = qE
Randomizing Collisions (momentum, energy)
=>Resulting Motion
http://phys23p.sl.psu.edu/phys_anim/EM/random_walk.avi
Average motion = Drift Velocity = vd ~10-4 m/s
Typical speeds ~ 106 m/s
p212c25: 1
Current Flow = net motion of charges
= Charge carriers
charge q
speed vd
vd
I
I = Current = rate at which charge passes the area
I≡
dQ
dt
units:
1C
= 1 Amp = 1 A
1S
vd
I
Negative charge carriers move in opposite direction of
conventional current.
p212c25: 2
Connection with microscopic picture:
E
q
A
vd
ℓ = v d dt
dQ = charge that passes through A
= number that pass through A × charge on each
= (n A vd dt) q,
n = number density = number/volume
dQ
= nqv d A
dt
Current Density:
I
J ≡ = nqvd
A
J = nqvd
I=
(works for negative charge carriers, multiple types of charge carriers as well)
J = n1q 1vd 1 + n2 q 2 v d 2 +…
p212c25: 3
Example: 18 gauge copper wire (~1.02 mm in diameter)
-constant current of 2A
-n = 8.5x1028 m-3 (property of copper)
find J, vd
p212c25: 4
Current as a response to an applied electric field
J = J( E)
=σE
σ = conductivity
E 1
ρ ≡ = = resistivity
J σ
V m V
units =
= m = Ωm
A m2 A
p212c25: 5
J
J
slope = 1/ρ
E
J
linear response
Ohm’s Law
E
nonlinear response
E
diode
nonlinear response
direction dependence!
ρ depends upon
•material
•E
•Temperature
If ρ does not depend on E, the material is said to obey
“Ohm’s Law”
p212c25: 6
For a cylindrical conductor
E = ρJ
Vab = V = Eℓ
= ρJℓ = I
E
J
ρℓ
I
a
A
V ρℓ
R≡ =
= resistance
I
A
V = IR
b
see also example 28-3 re: alternate geometries
Example: 50 meter length of 18 gauge copper wire (~1.02 mm in diameter)
constant current of 2A
ρ = 1.72x10-8 Ω.m
find E,V, R
p212c25: 7
Temperature Dependence of ρ
ρ
ρ
T
Metallic Conductor
ρ
T
Semiconductor
T
Superconductor
For small changes in temperature:
ρ T = ρ o [1 + α ( T − To )]
α = temperature coefficient of resistivity
RT = Ro [1 + α (T − To )]
p212c25: 8
Resistor Color Codes
Color number
black 0
brown 1
red
2
orange 3
yellow 4
green 5
blue 6
violet 7
gray 8
white 9
color range
none ±20%
silver ±10%
gold ±5%
value: n1n2x10n3±x%
10x102±5%
p212c25: 9
Electromotive Force and Circuits
Steady current
requires a complete circuit
path cannot be only resistance
cannot be only potential drops in direction of current
flow
Electromotive Force (EMF)
provides increase in potential E
converts some external form of energy into electrical energy
Single emf and a single resistor:
V = IR
V = IR = E
I
+ E
p212c25: 10
Measurements
Voltmeters measure Potential Difference (or voltage)
across a device by being placed in parallel with the
device.
V
Ammeters measure current through a device by being
placed in series with the device.
A
p212c25: 11
Real Sources and Internal Resistance
E
a
r
b
Ideal emf E
determined by how energy is converted into electrical energy
Internal Resistance r
unavoidable “internal” losses
aging batteries => increasing internal resistance
p212c25: 12
V
Open Circuit
I=0
Vr=0
Vab= E
r
E
a
A
b
V
Short Circuit
Vr = Ir = E
Vab= 0
r
E
a
b
A
p212c25: 13
V
Complete Circuit
E = Ir + IR
E
I=
R+r
Vab = E − Ir
r
E
a
b
A
I
R
Charging Battery
V
Vab = E + Ir
I
A
E
a
r
b
p212c25: 14
Energy and
Power
dW = dQVab = Vab Idt
P=
a
I +
dW
= IVab
dt
-
b
a
V=IR
+ -
b
V= E
Ideal emf
P = IE
Power deliverd by ( I out of +)
or to ( I into of +) battery
Resistance
V2
P = IV = I R =
R
2
p212c25: 15
Power and Real Sources
E
Discharging Battery
Vab = E − Ir
I
r
a
b
I
b
I
Power delivered externally : IVab = IE − I 2 r
E
Charging Battery
I
a
r
Vab = E + Ir
Power delivered by external source : IVab = IE + I 2 r
p212c25: 16
Real Battery with Load
I=
V
E
R+r
E
ER
E
VR = IR =
=
R + r 1+ r R
PE = IE =
E2
R+r
r
a
b
A
I
R
2
 E 
Pr = I r = 
 r
R+r 
2
max PR
2
 E 
PR = I R = 
 R
R+r
P
1
eff = R =
PE 1 + r R
2
dPR
= 0⇒ R = r
dR
"Impedance Matching"
p212c25: 17
Complete Circuit Example
V
I=
Vab =
PE =
E=12V
r = 2Ω
a
b
A
Pr =
PR =
PVab =
R = 4Ω
p212c25: 18
Theory of Metallic Conduction
Constant Acceleration between randomizing
collisions
(momentum, velocity randomized)
E
ρ=
J
J = nqvd
E
F qE
a= =
m m
τ = mean time between
v = v o + aτ
collisions
qE
τ = vd
vavg = a τ =
ℓ = mean free path
m
= (v o ) avg ⋅ τ
nq 2 τ J = nqvd =
E
m
m
ρ= 2
nq τ
p212c25: 19
Example: What is the mean time between collisions and the
mean free path for conduction electrons in copper?
ρ = 127
. × 10 −8 Ω ⋅ m
n = 8.5 × 10 28 m −3
m = 9.1 × 10 −31 kg
q = e = 16
. × 10 −19 C
m
v o ≈ 1x10 6
s
p212c25: 20
Physiological Effects of Current
Nerve action involves electrical pulses
currents can interfere with nervous system
~.1A can interfere with essential functions
(heartbeat, e.g.)
currents can cause involuntary convulsive muscle action
~.01 A
Joule Heating (I2R)
With skin resistance
dry skin: R ~ 500kΩ
wet skin: R ~ 1000Ω
p212c25: 21