Chapter 27 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
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
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