Download Lecture 4 Electric potential

Lecture 6 Current and Resistance Ch. 26
We are now leaving electrostatics.
• Cartoon -Invention of the battery and Voltaic Cell
• Topics
– What is current?
– Current density
– Conservation of Current
– Resistance
– Temperature dependence
– Ohms Law
– Batteries, terminal voltage, impedance matching
– Power dissipation
– Combination of resistors
• Demos
– Ohms Law demo on overhead projector
– T dependence of resistance
– Three 100 Watt light bulbs
• Elmo
– Puzzle - Resistor network figure out equivalent resistance
1
2
3
Loop of copper wire
Nothing moving;
electrostatic equilibrium
E 0
Now battery voltage forces
charge through the
conductor and we have a
field in the wire.
E0
4
What is Current I?
It is the amount of positive charge that moves past a certain point per unit time.
I
Q Coulomb

 Amp
t
sec ond
I
+ +
+
+
A
Copper wire with
voltage across it
+ +
+ +
L
vt  L
Drift
velocity of
charge
Q  charge per unit volume  volume
 nq  Avt
Q  nqAv t
Density of electrons
1.6 x 10-19 C
Divide both sides by Δt.
Q
I
 nqAv
t
5
What is the drift velocity vd ?
Example: What is the drift velocity for 1 Amp of current
flowing through a 14 gauge copper wire of radius 0.815
mm?
I
vd 
nqA
Drift velocity
No
n

vd 

8.4  10
22 atoms
cm3
I = 1 Amp
q = 1.6x10-19 C
= 8.4x1022 atoms/cm3
1amp
 1.6  10 19 C   (.0815cm)2
v d  3.5  10 5 m s
A = π(.0815 cm)2
 = 8.9 grams/cm 3
No = 6x1023 atoms/mole
M = 63.5 grams/mole
The higher the density
the smaller the drift
velocity
6
What is the current density J
?
Directions of current i is
defined as the direction
of positive charge. I is
not a vector.
i  nAqvd
r
r
J  nqvd
r r
i   J  dA
(Note positive charge moves in
direction of E) electron flow is
opposite E.
r
r
i
J  when J and A are parallel
7
A
Currents: Steady motion of charge and
conservation of current
i0  i1  i2 (Kirchoff's 2nd Rule)
Current is the same throughout
all sections in the diagram below;
it is continuous.
Current density J does vary
Higher J
8

Question: How does the drift speed compare to the
instantaneous speed?
v d  3.5  10 5 m s
Instantaneous speed ~ 106 m/s
v d  3.5  10 11v instant
(This tiny ratio is why Ohm’s Law works so well for
metals.)
Question: At this drift speed 3.5x10-5 m/s, it would take an electron 8
hours to go 1 meter. So why does the light come on immediately when
you turn on the light switch?
It’s like when the hose is full of water and you turn the faucet on,
it immediately comes out the ends. The charge in the wire is like
the water. A wave of electric field travels very rapidly down the
wire, causing the free charges to begin drifting.
9
Example: Recall typical TV tube,or CRT. The electron beam
has a speed 5x107 m/s. If the current is I = 100 microamps,
what is n?
n
I
qAv
Take A  1 mm 2
104 A
n
1.6  1019 C 106 m2  5  10 7 m s
 (10 3 m)2
 10 6 m 2
For CRT
n  1.2  1013
electrons
7 electrons

1
.
2

10
m3
cm3
For Copper
n  8.5  1022
electrons
cm3
The lower the density the higher the speed.
10
Ohm’s Law
V  RI
R
V
I
Want to emphasize here that as long as we have current (charge moving)
due to an applied potential, the electric field is no longer zero inside the
conductor.
I
Potential difference
•
•
VB  VA  EL, where E is constant.
A
B
L
I  current  EL (Ohm' s law)
True for many materials – not all. Note that Ohms Law
is an experimental observation and is not a true law.
Constant of proportionality between V and I is known
as the resistance. The SI unit for resistance is called the
ohm.
V  RI
R
V
I
Demo: Show Ohm’s Law
Ohm 
Volt
amp
Best conductors
Silver
Copper – oxidizes
Gold – pretty inert
Non-ohmic materials
Diodes
Superconductors
11
A test of whether
or not a material
satisfies Ohm’s
Law
V  IR
V
I
R
1
 constant
R
Ohm' s law is satisfied
Slope 
Here the slope depends on
the potential difference.
Ohm's Law is violated for a
pn junction diode.
12
What is Resistance?
The collisions between the electrons and the atoms is the cause of
resistance and the cause for a very slow drift velocity of the electrons.
The higher the density, the more collisions you have.
field off
field on
extra distance electron
traveled
e-
The dashed lines represent the straight line tracks of electrons in between
collisions
----------Electric field is off.
----------Electric field is on. When the field is on, the electron
drifted further to B’...
13
Resistance: More on Resistance- Define
Resistivity Demo: Show temperature
• Depends on
shape, material, temperature.
dependence of resistance
• Most metals: R increases with increasing T
• Semi-conductors: R decreases with increasing T
Define a new constant which characterizes resistance of materials.
Resistivity
R
A
L
A
L
R
L

A
For materials ρ= 10-8 to 1015 ohms-meters
Example: What is the resistance of a 14 gauge Cu wire? Find the
resistance per unit length.
R cu
1.7  10 8 m


 8  10 3  m
3 2
L
A 3.14(.815  10 )
Build circuits with copper wire. We can neglect the resistance of
the wire. For short wires 1-2 m, this is a good approximation.
Note Conductivity = 1/Resistivity

14
Example Temperature variation of resistivity.
Calculate R at T= 20 C for
Fe – conductor
Si – insulator
-
  20 1   (T  20)
a long 6x106 m wire of area 1mm2.
a cube of Si 1 m on each side
Consider two examples of materials at T = 20o C, then
can be positive or negative
L
R 
A
ρ20 (Ω -m)
 (C-1)
L
Area
Fe
10 -7
0.005
6x106 m
1mm2(10- 600 KΩ
6m2)
Si
640
- 0.075
1m
1 m2
R (20oC)
0.640 KΩ
Question: You might ask is there a temperature where a conductor and
insulator are one and the same? Demo: Show temperature dependence of15resistance
Condition: RFe = RSi at what temperature?
Use
L
R 
A
RFe =
10-7
R  20 1   (T  20 C)
L
A
6 106 m
Ω-m [ 1 + .005 (T-20)]
106 m 2
1m
1m 2
Now, set RFe = RSi and solve for T
RSi = 640 Ω-m [ 1 - .075 (T-20)]
T = – 170.5 C or 102.6 K
(pretty low temperature)
T – 20 C = – 170.50 C
ρ20 (Ω -m)  (C-1)
L
Area
R (170.5oC)
Fe
10 -7
0.005
6x106 m
1mm2(106m2)
8.8 KΩ
600 KΩ
Si
640
- 0.075
1m
1 m2
8.8 KΩ
0.640 KΩ
R (20 C)
16
Resistance at Different Temperatures
T =293K
Cu
Nb
C
T = 77K (Liquid Nitrogen)
.1194 Ω
.0152 Ω
conductor
.0235Ω .0209 Ω
impure
.0553 Ω
.069 Ω
semiconductor
17
Power dissipation resistors
I
Potential energy decrease
U  Q( V )
U Q

(V )
t
t
P  IV
(drop the minus sign)
Rate of potential energy decreases equals rate of thermal energy increases in resistor.
Called Joule heating
• good for stove and electric oven
• nuisance in a PC – need a fan to cool computer
Also since V = IR,
V2
P  I R or
R
2
All are equivalent.
Example: How much power is dissipated when I = 2A flows through the Fe resistor of
R = 10,000 .
P = I2R = 22x104  = 40,000 Watts
18
Batteries
A device that stores chemical energy and converts it to electrical energy.
Emf of a battery is the amount of increase of electrical potential of the charge when it
flows from negative to positive in the battery. (Emf stands for electromotive force.)
Carbon-zinc = Emf = 1.5V
Lead-acid in car = Emf = 2V per cell
(large areas of cells give lots of current)
Car battery has 6 cells or 12 volts.
Power of a battery = P
P = εI
ε is the Emf
Batteries are rated by their energy content. Normally they give an equivalent measure
such as the charge content in mA-Hrs
milliamp-Hours
Internal Resistance
Charge = (coulomb/seconds) x seconds
As the battery runs out of chemical energy the internal resistance increases.
What is terminal
voltage?
Terminal Voltage decreases quickly.
How do you visualize this?
19
What is the relationship between Emf,
resistance, current, and terminal voltage?
Circuit model looks like this:
I
r
•
R

Terminal voltage = V
V = IR (decrease in PE)
•
  Ir  IR
  Ir  V  IR
  I (r  R )

I
(r  R)
The terminal voltage decrease =  - Ir as the internal resistance r increases or
when I increases.
20
Example: This is called impedance matching. The question is
what value of load resistor R do you want to maximize power
transfer from the battery to the load.
I
E
=current from the battery
rR
P = I2R = power dissipated in load
P
P
2
E
R
( r  R) 2
dP
0
dR
?
R
Solve for R
R=r
You get max. power when load resistor equals
internal resistance of battery.
(battery doesn’t last long)
21
Combination of resistors
Resistors in series
Current is the same in both the resistors
V  R1I  R2I  (R1  R2 )I
Reqiv  R1  R2
Resistors in parallel
Voltages are the same, currents add.
I  I1  I 2
V V
V


R R1 R2
1
1
1


R R1 R2
RR
Requiv  1 2
R1  R2
So,
22
Equivalent Resistance
R eq  (R  R ) (R  R )
 2R 2R

4R 2
R eq 
4R
R eq  R
R eq  R (R  R )

 R 2R
2R 2
R eq 
3R
R eq 
2
R
3
23
Resistance cube
I
I
The figure above shows 12 identical resistors of value R attached to form a cube.
Find the equivalent resistance of this network as measured across the body
diagonal---that is, between points A and B. (Hint: Imagine a voltage V is applied
between A and B, causing a total current I to flow. Use the symmetry arguments
to determine the current that would flow in branches AD, DC, and CB.)
24
Resistance Cube cont.
I
I
3
I
6
I
3
I
3
I
6
I
I 3
6
I
6
I
3
I
3
V  R eq I
V  VAD  VDC  VCB
I
6
I
6
I
Because the resistors are
identical, the current
divides uniformly at each
junction.
I
I
I
R eq I  R  R  R
3
6
3
5
R eq I  RI
6
5
R eq  R
6
25
Chapter 26 Problem 21
A wire with a resistance of 3.0 is drawn out through a die so that
its new length is three times its original length. Find the
resistance of the longer wire, assuming that the resistivity and
density of the material are unchanged.
26
Chapter 26 Problem 23
Two conductors are made of the same material and have the same
length. Conductor A is a solid wire of diameter 0.5 mm.
Conductor B is a hollow tube of outside diameter 3.0 mm and
inside diameter 1.5 mm. What is the resistance ratio RA/RB,
measured between their ends?
27
Chapter 26 Problem 25
A common flashlight bulb is rated at 0.30 A and 2.9 V (the values
of the current and voltage under operating conditions). If the
resistance of the bulb filament at room temperature (20°C) is 1.8 ,
what is the temperature of the filament when the bulb is on? The
filament is made of tungsten.
28
Chapter 26 Problem 27
When 125 V is applied across a wire that is 12 m long and has a
0.30 mm radius, the current density is 1.2 multiplied by 104 A/m2.
Find the resistivity of the wire.
29
30
31
32
33
34
35
36