Transcript Chapter 25 Current and Resistance

Chapter 25 Current and Resistance
Average Electric Current
Instantaneous Electric Current
Units:
Q
I
t
dQ
I
dt
Coulomb C
  Ampere  A
sec
s
• Scalar
• Sense determined by the movement of
the positive charge carrier
Microscopic view of current
A
+
+
E
J
+
+
+
+
Isn’t E = 0?
J 
I
A
Current density, j, is the electric current per unit cross sectional
area. Current density is a vector.
Charge carriers experience a force and accelerate. They collide
with the atoms in the metal and are slowed down eventually
reaching a terminal velocity called the drift velocity
Resistance model – Why doesn’t the
electron accelerate?
qE
eE
a

me
me
eE
vf  vi  at  vi 
t
me
eE
vf  vd  

me
Counting the charge flow
E
+
A
+
j
+
+
+
+
vd t
vd
Let:
n  number of charge carriers per unit volume
q  charge on a charge carrier (-e)
A  cross sectional area of the wire
Q  nA  v d t  q
Q nA  vd t  q
I

 nA vd q
t
t
Current density for electrons
A
-
-
-
-
E
-
-
vd
vd
J
E
I nA vd q
J  
 n vd q
A
A
J  nvd e
I J A
For non-uniform
current density
I   J dA
Ohm’s Law
+
+
+
+
+
E
J
+
JE
 eE 
ne2 
J  nvd e  n  
e 
E
me
 me 
J  E
E  J
ne2 

 conductivity
me
me
  2  resistivity
ne 
1


Scalar form of Ohm’s Law
+
+
b
E
J
b
+
+
+
+
E  J
a
I
Vba    E ds  E   J  
A
a



Vba  I     IR
R

A A
 A
Units of resistance, resistivity, and
conductivity

R

A A

RA
1
 
 RA
Ohms  
Ohms  meter     m
   m
1



1
mho siemens


m
m
m
Temperature effect on resistivity
(T)  o 1    T  To  
R

A
R(T)  R o 1    T  To  
1 

= temperature coefficient of resistivity
o T
Superconductors
• A class of materials and
compounds whose
resistances fall to
virtually zero below a
certain temperature, TC
– TC is called the
critical temperature
• The graph is the same
as a normal metal
above TC, but suddenly
drops to zero at TC
Superconductor Application
• An important
application of
superconductors is a
superconducting magnet
• The magnitude of the
magnetic field is about
10 times greater than a
normal electromagnet
• Used in MRI units
Electrical energy and power
Vba 
U b  Ua U

q
q
dU
V
dq
V
dU  Vdq
The power transformed in
an electric device is then:
Using Ohm’s Law
V  IR
dU dq
P

V  IV
dt
dt
P  IV  I  IR   I R
2
2
V
V
 
P  IV    V 
R
R
Example Problem
Given: Copper wire transmission line
length = 1500 m
diameter = 0.1cm
current = 50 A
cu = 1.67 x 10-6 -cm
Find: Power loss due to heating the wire
Example P27.6
The quantity of charge q (in coulombs)
that has passed through a surface of area
2.00 cm2 varies with time according to
the equation q = 4t3 + 5t + 6, where t is
in seconds.
(a) What is the instantaneous current
through the surface at t = 1.00 s?
(b) What is the value of the current
density?
I 1.00 s 


dq
 12t2  5
dt t1.00 s
t1.00 s
 17.0 A
J
I
17.0 A

A 2.00  104 m
2
 85.0 kA m
2
Example P27.15
A 0.900-V potential difference is maintained
across a 1.50-m length of tungsten wire that
has a cross-sectional area of 0.600 mm2.
What is the current in the wire?
A   0.600 m m
I
V A


2
2
1.00 m 
7
 1000 m m   6.00  10 m
 0.900 V   6.00  107 m 2 

 5.60  108   m  1.50 m 
I 6.43 A
2
Example P27.29
A certain lightbulb has a tungsten filament with a
resistance of 19.0 Ω when cold and 140 Ω when
hot. Assume that the resistivity of tungsten varies
linearly with temperature even over the large
temperature range involved here, and find the
temperature of the hot filament. Assume the
initial temperature is 20.0°C.


140   19.0  1 4.50  103 C T 


T  1.42  103 C  T  20.0C
T  1.44  103 C
Resistor Values
• Values of resistors
are commonly
marked by colored
bands
The Lead Acid Electric Battery
I
Sulfuric Acid Electrolyte:
-
Terminals
+
Sulfuric
Acid
Solution
H2SO4
Spongy
Lead (Pb)
Lead Oxide
(PbO2)
Cell: 2 V
Battery: Multiple cells
H 2SO 4  2H 2O
2H 3O   SO 42
Oxidation at the Negative
Plate (Electrode:Anode):
Pb  SO42  PbSO4  2e
Reduction at the Positive
Plate (Electrode:Cathode):
PbO2  4H3O   SO 42  2e   PbSO 4  6H 2O
Batteries
Vab    Ir
  Electromotive Force
r  Batterey internal resistance
Kirchoff’s Rules
• Conservation of charge
• Junction (Node) Rule: At any junction point, the
sum of all currents entering the junction must
equal the sum of the currents leaving the junction.
• Conservation of energy
• Loop Rule: The some of the changes in potential
around any closed path of a circuit must be zero.
Energy in a circuit
Series Circuit
Apply the Loop Rule
Vac  Vab  Vbc  0
+
Vac
Vac  Vab  Vbc  IR1  IR 2  I  R1  R 2   IR eq
R eq  R1  R 2  .....
Parallel Circuits
I
I1
I3
+
Apply the
Junction Rule
I2
V
 1
V V V
1
1  V
I  I1  I2  I3 


 V 


R1 R 2 R 3
 R1 R 2 R 3  R eq
1
1
1
1



 ....
R eq R1 R 2 R 3
Rule Set – Problem Solving Strategy
• A resistor transversed in the direction of assumed
current is a negative voltage (potential drop)
• A resistors transversed in the opposite direction of
assumed current is a positive voltage (potential rise)
• A battery transversed from – to + is a positive voltage.
• A battery transversed from + to - is a negative voltage.
• Ohm’s Law applies for resistors.
• Both the loop rule and junction rule are normally
required to solve problems.
More about the Loop Rule
• Traveling around the loop from a
to b
• In (a), the resistor is traversed in
the direction of the current, the
potential across the resistor is – IR
• In (b), the resistor is traversed in
the direction opposite of the
current, the potential across the
resistor is is + IR
Loop Rule, final
• In (c), the source of emf is
traversed in the direction
of the emf (from – to +),
and the change in the
electric potential is +ε
• In (d), the source of emf is
traversed in the direction
opposite of the emf (from
+ to -), and the change in
the electric potential is -ε
Example Problem 1
Given:
R1  1690
R 3  1000
R 4  3000
V = 3 Volts
Find: current in each resistor
Example Problem 2
Given:
10V
5
10
20
20V
Find: current in the 20  resistor
Alternating Current
V  t   Vo
I  t   Io
V  t   Vo sin t
V  t  Vo
I(t) 

sin t  Io sin t
R
R
AC Power
P  I2 R  I2o R sin 2 t
1 2
1 Vo2
P  Io R 
2
2 R
T/2
1
1
2
sin  t dt  ?

T T / 2
2
Root Mean Square (rms)
V  t   Vo sin t
I(t)  Io sin t
2
V
V2  o
2
Vrms
2
I
I2  o
2
2
V
Vo
2
o
 V 

2
2
I rms 
1 2
2
P  I o R  I rms R
2
2
1 Vo2 Vrms
P

2 R
R
2
I
Io
2
o
I 

2
2
The Wheatstone bridge
a simple Ohmmeter
I3R 3  I1R1
I3R x  I1R 2
R2
Rx 
R3
R1
Charging a capacitor in an RC circuit
Same
Symbol

At t = 0, Qo = 0 and I o 
R
Solving the charging differential
equation
Kirchoff’s loop rule
Q
  IR   0
C
Convert to a simple equation
in Current by taking the first
derivative w.r.t. time
dI 1 dQ
R 
0
dt C dt
dI
1
R  I
dt
C
Separate variables
dI
1

dt
I
RC
Integrate the results
I t 

Io
t
dI
1
 
dt
I
RC
0
t
ln  I  t    ln  I o   
RC
It
t
ln 

RC
 Io 
t

It
 e RC
Io
I  t   Io e

t
RC
Charge buildup
dQ
It 
 Io e
dt
dQ  Io e
Q t 

0

t
RC
t
dQ  Io  e

t
RC
dt

t
RC
dt
0

Q  t   Io  RC  e

t

RC
t
t




RC

I
RC
1

e



o
0


Discharging the capacitor in an RC
circuit
At t = 0, Q = Qo
Solving the discharging differential
equation
Q
 IR  0
C
dQ 1
R
 Q
dt C
Kirchoff’s loop rule
dQ
I
since charge is decreasing
dt
dQ
1

dt
Q
RC
Separate variables
Q t 
Integrate

Qo
t
dQ
1
 
dt
Q
RC
0
Charge and current decay
t
ln Q  t    ln  Q o   
RC
Qt
t
ln 

RC
 Qo 
t

Qt
 e RC
Qo
Q  t   Qo e

t
RC
Charge and current decay
dQ
d
I
  Qo e
dt
dt
Qo  RCt
I
e
RC
I  Io e

t
RC

t
RC
More on microscopic picture of
conduction (See Ch 38)
qE
eE
a

me
me
eE
vf  vd  

me
 eE 
ne2 
J  nvd e  n  
e 
E
me
 me 
J  E
E  J
ne2 

 conductivity
me
me
  2  resistivity
ne 
1


Mean Free Path
me
 2
ne 
  vav 
path length
vt
1



# of collisions n ions Avt n ions A
me vav
 2
ne 
3
1
2
kT  m e v av
2
2
What is vav for an electron? (M-B)


3 1.38x1023 J / K 300K
3kT
5
vav 


1.17x10
m/s
31
me
9.11x10 kg
Evaluation of classical conduction theory
m e v av


2
ne 
me
3kT
me
ne 2 
•Resistivity does not depend on the electric field.
•Temperature dependence of resistivity does not agree with
laboratory measurement.
•Calculated resistivities calculated from M-B average velocities and
path lengths are 6x greater than measured.
Quantum theory of free electrons
• Electrons are fermions not bosons
• There can be at most 2 electrons with
the same set of values for their
spatial quantum numbers
• The energy level of the last filled
(highest) energy level at T=0 K is
found from quantum mechanics
2
h
Ef 
8me
3
2
 3N 
N

  0.3646 eV nm  
 V 
V
3
2
Fermi Energies
Fermi-Dirac Distribution
k=8.62x10-5 eV/K
kT
At room temperature ~ 300K
kT=.026 eV
Only a few electrons can be
taken to higher energy levels
at room temperature.
If the temperature is very high
Tf 
Ef
k
More electrons might be in
higher energy states, but for
typical temperatures below
this Fermi Temperature, the
electrons have energy like the
T=0 K case
Correcting the classical picture
m e vav m e u f

 2
2
ne 
ne 
Use Fermi Speed
Correct the mean path length
uf 
2E f
me

1
n ions A
Area is not physical size of lattice ion
but vibration amplitude of oscillation
E~r2~A~T
Electrical Safety
• Current kills, not voltage (70 mA)
• Normal body resistance = 105 
But could be less than 1000 
• Take advantage of insulators, remove
conductors
• Work with one hand at a time
• Shipboard is more dangerous
• Electrical safety is an officer responsibility