Transcript Current and Resistance

Current and
Resistance
February 18, 2008
February 18, 2008

Headlines:
 Today
we will finish capacitors and begin a
study of current, resistors and DC circuits.
 Watch for new WebAssign
 Quiz on Friday
 Regular Problem sessions on Wednesday
and Thursday
A Closer Look at this stuff..
q
Consider this virgin capacitor.
No dielectric experience.
Applied Voltage via a battery.
++++++++++++
V0
C0
-q
------------------
C0 
0 A
d
q  C0V0 
0 A
d
V0
Remove the Battery
q
++++++++++++
V0
The Voltage across the
capacitor remains V0
q remains the same as
well.
-q
-----------------The capacitor is fat (charged),
dumb and happy.
Slip in a Dielectric
Almost, but not quite, filling the space
Gaussian Surface
q
V0
-q
++++++++++++
- - - - - - - -
-q’
+ + + + + +
+q’
------------------
E0
E
E’ from induced
charges
in..small..gap
q
E

d
A


0
E0 
q   
  
0 A  0 
A little sheet
from the past..
-q’
+q’
- -q
-
+
q+
+
Esheet

q'


2 0 2 0 A
q'
0
2xEsheet
0
q'
Esheet / dialectric  2 

2 0 A  0 A
Some more sheet…
Edielectricch arg e
q
E 0
0 A
so
q  q'
E
0 A
 q'

0 A
Another look
Vo
+
-
Parallel  Plate
0 A
C0 
d
 0 AV0
Q0  C0V0 
d
Electric  Field
V0
E0 
d
Q0  0V0
0 

A
d
Add Dielectric to Capacitor
Vo
+

Original Structure

Disconnect Battery

Slip in Dielectric
-
+
V0
+
-
Note: Charge on plate does not change!
What happens?
o +
o -
i +
i 
E0
V0 1
E


d 
and
V
V  Ed  0

Potential Difference is REDUCED
by insertion of dielectric.
Q
Q
C

 C0
V0 /  
V
Charge on plate is Unchanged!
Capacitance increases by a factor of 
as we showed previously
SUMMARY OF RESULTS
V 
V0
E
E0

C  C0

New Topic:
Current and Resistance
Physical Resistors
What Happens?
“+”
REMEMBER, THE ELECTRONS
“+”
ARE ACTUALLY MOVING THE
OTHER WAY!
-
“+”
“+”
What’s Moving?
What is making the charged
move??
Battery
KEEP IN MIND


A wire is a conductor
We will assume that the conductor is essentially
an equi-potential
 It

really isn’t.
Electrons are moving in a conductor if a current
is flowing.
 This
means that there must be an electric field in the
conductor.
 This implies a difference in potential since E=DV/d
 We assume that the difference in potential is small
and that it can often be neglected.
 In this chapter, we will consider this difference and
what causes it.
DEFINITION

Current is the motion of POSITIVE
CHARGE through a circuit. Physically, it is
electrons that move but …
Conducting material
DQ,Dt
Conducting material
DQ,Dt
CURRENT
DQ
i
Dt
or
dq
i
dt
UNITS

A current of one coulomb per second is
defined as ONE AMPERE.
A charged belt, 30 cm wide, travels at 40 m/s
between a source of charge and a sphere. The belt
carries charge into the sphere at a rate
corresponding to 100 µA. Compute the surface
charge density on the belt.
[8.33e-06] C/m2
Comment on Current Flow
Question …..
A small sphere that carries a charge q is
whirled in a circle at the end of an
insulating string. The angular frequency
of rotation is ω. What average current
does this rotating charge represent?
An electric current is given by the expression I(t)
= 100 sin(120πt), where I is in amperes and t is in
seconds. What is the total charge carried by the
current from t = 0 to t = (1/240) s?
ANOTHER DEFINITION
current
I
J 

area
A
The figure represents a section of a circular
conductor of non-uniform diameter carrying a
current of 5.00 A. The radius of cross section A1 is
0.400 cm. (a) What is the magnitude of the current
density across A1? (b) If the current density across
A2 is one-fourth the value across A1, what is the
radius of the conductor at A2?
Ohm





A particular object will
resist the flow of current.
It is found that for any
conducting object, the
current is proportional to
the applied voltage.
STATEMENT: DV=IR
R is called the resistance
of the object.
An object that allows a
current flow of one
ampere when one volt is
applied to it has a
resistance of one OHM.
Ohm’s Law
DV  IR
Graph
DV  IR
A DIODE
Resistance Varies with Applied Voltage
(actually with current)
Let’s look at the atomic level ..
Conduction is via electrons.
 They are weak and small and don’t
exercise much.
 Positive charge is big and strong and
doesn’t intimidate easily.
 It’s an ugly situation … something like
……

+
-
Vb  Va
E
l
The Current
Electrons are going the opposite way from
the current. (WHY?)
 They probably follow a path like …

Average “drift”
speed - vd
IN
OUT
Notation
vd average drift velocity of the electron
 n number of electrons (mobile) per unit
volume.
 Dt interval of time
 Dx average distance the electron moves
in time Dt.
 Q total amount of CHARGE that goes
through a surface of the conductor in time
Dt.

The Diagram
DQ  (nAvd Dt )e
DQ
I avg 
 nAvd e
Dt
I avg
J
 nevd
A
J  nev d
Often a Vector
We return to the diagram …..





Consider an electron.
Assume that whenever it
“bumps” into something it
loses its momentum and
comes to rest.
It’s velocity therefore starts
at zero, the electric field
accelerates it until it has
another debilitating collision
with something else.
During the time it
accelerates, its velocity
increases linearly .
The average distance that
the electron travels between
collisions is called the
“mean free path”.
Starting when the electron is at
We showed two slides ago::
rest:
v  v0 + at  at
F eE
a

m
m
eE
v  vd 

m
Let n= number of charge carriers
per unit volume (mobile electrons)
eE
J  nqvd  nevd  ne 
m
or
ne 2 E
J
  E
m
so
ne 2

m
1


Finally
  vd
Reference
The average drift velocity of an
electron is about 10-4 m/s
Ponder
How can a current go through
a resistor and generate heat
(Power) without decreasing
the current itself?
Loses Energy
Gets it back
Exit
Conductivity
In metals, the bigger the electric field at a
point, the bigger the current density.
J  E
 is the conductivity of the material.
=(1/) is the resistivity of the material
  0 1 +  (T  T0 )
A conductor of uniform radius 1.20 cm carries a
current of 3.00 A produced by an electric field of
120 V/m. What is the resistivity of the material?
The rod in the figure is made of two materials. The
figure is not drawn to scale. Each conductor has a
square cross section 3.00 mm on a side. The first
material has a resistivity of 4.00 × 10–3 Ω · m and is
25.0 cm long, while the second material has a
resistivity of 6.00 × 10–3 Ω · m and is 40.0 cm long.
What is the resistance between the ends of the rod?
Going to the usual limit …
dI
J 
dA
and
I   JdA
Example
A cylindrical conductor of radius R has
a current density given by
(a) J0 (constant)
(b) gr
Find the total current in each case.
Range of  and 
Ye old RESISTANCE
DV  El
1 DV
1 DV
I
J  E 
E

 El
 l
A
l
DV 
I
A
l
R
A
DV  V  IR
REMEMBER
R
L
A
DV  IR
Temperature
Effect
D
DT
  0 (1 + DT )
A closed circuit
Power
In time Dt, a charge DQ is pushed through
the resistor by the battery. The amount of work
done by the battery is :
DW  VDQ
Power :
DW
DQ
V
 VI
Dt
Dt
Power  P  IV  I IR   I 2 R
2
E
P  I 2 R  IV 
R