Transcript Lecture 4 Electric potential

Lecture 6 Current and Resistance Chp. 27
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•
•
•
•
Cartoon -Invention of the battery and Voltaic Cell
Opening Demo - Lemon Battery
Warm-up problem
Physlet
Topics
•
Demos
– Lemon Battery estimate internal resistance
– Ohms Law demo on overhead projector
– T dependence of resistance
– Three 100 Watt light bulbs
Puzzles
•
– Resistor network figure out equivalent resistance
Loop of copper wire
Nothing moving;
electrostatic equilibrium
E0
Now battery forces charge
through the conductor. We
have a field in the wire.
E0
What is Current?
It is the amount of positive charge that moves past a certain point per
unit time.
Q Coulomb
I

 Amp
t
sec ond
I
Copper wire with
voltage across it
+ +
+
+
A
+ +
+ +
L
v t   L
Drift
velocity of
charge
Density of electrons
Q = charge per unit volume x volume
nq x Avt
Q = nqAv t
1.6 x 10-19 C
Divide both sides by t.
I
Q
 nqAv
t
What causes charges to move in the wire?
Question
How many charges are available to move?
Example
What is the drift velocity for 1 Amp of current flowing through
a 14 gauge copper wire of radius 0.815 mm?
Drift velocity
I
vd 
nqA
I = 1 Amp
q = 1.6x10-19 C
A = (.0815 cm)2
No
= 8.4x1022 atoms/cm3
n
 = 8.9 grams/cm 3


vd 
1
8.4 10 22 1.6 1019   (.0815) 2
5 m
v d  3.5 10
s
No = 6x1023 atoms/mole
M = 63.5 grams/mole
The higher the density
the smaller the drift
velocity
Drift speed of electrons and current
density
Directions of current i is
defined as the direction
of positive charge.
i  nAqvd
i
J
A
J  nqvd
(Note positive charge moves in
direction of E) electron flow is
opposite E.
Currents: Steady motion of charge and
conservation of current
i = 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.
Question: How does the drift speed compare to the instantaneous speed?
Instantaneous speed  106 m/s
vd  3.5x10-11 • vinstant
(This tiny ratio is why Ohm’s Law works so well for metals.)
At this drift speed 3.5x10-5 m/s, it would take an electron 8 hours to go 1 meter.
Question: 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.
Example: Recall typical TV tube, CRT, or PC monitor. The electron beam has a speed 5x10 7 m/s. If
the current is I = 100 microamps, what is n?
I
104 A
n

qAv 1.6 1019 C 106  5 107 m s
Take A = 1mm2
= (10-3)2
= 10-6 m2
For CRT
n = 1.2x1013 e/m3 = 1.2x107 e/cm3
For Copper
n = 8.5x1022 e/cm3
The lower the density the
higher the speed.
What is Resistance?
The collisions between the electrons and the atoms is the cause of resistance and a very
slow drift velocity of the electrons. The higher density, the more collisions.
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 traveled drifted further to B I.
Ohm’s Law
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
•
•
A
B
Potential difference
VB - VA = E L
Constant E
I = current  E L (Ohm’s Law)
L
True for many materials – not all. Note that this 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
Ohm = volt/amp
Best conductors
Silver – w/ sulpher
Copper – oxidizes
Gold – pretty inert
Non-ohmic materials
Demo: Show Ohm’s Law
Diodes
Superconductors
A test of whether or
not a material
satisfies Ohm’s Law
V = IR
I = V/R
Slope = 1/R = constant
Ohm’s Law is satisfied.
Here the slope depends on
the potential difference.
Ohm’s Law is violated.
Resistance: What is it? Denote it by R
• Depends on shape, material, temperature.
• Most metals: R increases with increasing T
• Semi-conductors: R decreases with increasing T
Define a new constant which characterizes materials.
Resistivity
A
R
L
A
R
L
L

A
Demo: Show temperature dependence of resistance
For materials  = 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
3 



8

10
m
L
A 3.14(.815103 ) 2
Build circuits with copper wire. We can neglect the resistance of
the wire. For short wires 1-2 m, this is a good approximation.
Example Temperature variation of resistivity.
 = 20 [ 1 +  (T-20) ]
L
R 
can be positive or negative
A
Consider two examples of materials at T = 20oC.
(-m)
(k-1)
L
Area
R (20oC)
Fe
10-7
.005
6x106 m
1mm2(10-6m2)
60,000 
Si
640
-.075
1m
1 m2
640 
Fe – conductor
-
a long 6x106 m wire.
Si – insulator
-
a cube of Si 1 m on each side
Question: You might ask is there a temperature where a conductor and insulator are one
and the same?
Condition: RFe = RSi at what temperature?
Set RFe = RSi and solve for T
Use
R
L

A
L
= 20 [ 1 +  (T-20) ] A
6  106 m
RFe =
1 + .005 (T-20)]
106 m 2
1m
RSi = 640 -m [ 1 + .075 (T-20)] 1m 2
10-7 -m [
T – 20 = -196C
T = -176C
(pretty low temperature)
Resistance at different Temperatures
Cu
.1194 
.0152 
conductor
Nb
.0235 
.0209 
impure
C
.0553 
.069 
semiconductor
T
= 300 K
= 77 K
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,
P = I2R or V2/R
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
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 = I
 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.
Terminal Voltage decreases quickly.
How do you visualize this?
What is terminal
voltage?
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.
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 battery
rR
P = I2R = power dissipated in load
P
P
2
E
R
( r  R) 2
dP
0
dR
Solve for R
R=r
You get max. power when load resistor equals
internal resistance of battery.
(battery doesn’t last long)
?
R
Demo show lemon or apple batteries
Estimate internal resistance by adjusting R until the terminal
voltage is half of the open circuit voltage. r = R
For lemons rlemons  3600 .
 - Ir = V
open circuit
voltage
Note: When V 
terminal
voltage

2
, r=R
Combination of resistors
Resistors in series
V = R1I + R2I = (R1 + R2)I
Requiv = R1 + R2
Resistors in parallel
Voltages are the same, currents add.
I = I1 + I2
V/R = V/R1 + V/R2

1/R = 1/R1 + 1/R2
Requiv = R1R2 /(R1 + R2)
Demo: 3 lightbulb resistor puzzle
Equivalence of two versions of Ohm’s Law
E = J

V = RI
LE = L J
V = L J
R =  (L/A)
L = AR
V = ARJ = RJA
I
 V = RI
Warm-up Set 6
1. HRW6 27.TB.08. [119812] Current is a measure of:
amount of charge that moves past a point per unit time
force that moves a charge past a point
energy used to move a charge past a point
speed with which a charge moves past a point
resistance to the movement of a charge past a point
2. HRW6 27.TB.14. [119818] In a conductor carrying a current we expect the electron drift speed to be:
about the same as the average electron speed
much less than the average electron speed
less than the electron speed at high temperature and greater than the electron speed at low temperature
less than the electron speed at low temperature and greater than the electron speed at high temperature
much greater than the average electron speed
3. HRW6 27.TB.49. [119853] You buy a "75 W" light bulb. The label means that:
the bulb is expected to "burn out" after you use up its 75 watts
none of these
no matter how you use the bulb, the power will be 75 W
the bulb was filled with 75 W at the factory
the actual power dissipated will be much higher than 75 W since most of the power appears as heat
What is the electric field in a sphere of uniform
distribution of positive charge. (nucleus of protons)

R
r

E




Q
4 3
R
3
EdA 
qenc
0
4 r 3

2
3
E 4 r 
0
r
Q
E

r
3
30 4 0 R
Cq
V
V  E 0S
S

E0 
0
 qA
E0 
q
0A
E
E0
V 
E 0S
V


V0

V
C
qS
0A
0A
S
Cq
C
V
q
V0
C  C 0
Find the capacitance of a ordinary piece of coaxial cable (TV cable)
2k
Er 
r
For a long wire we found that
where r is radial to the wire.
a
• r
a
Va  Vb    E. ds  2k 
b
b

E. ds  Edscos180 Eds Edr
dr
 2k ln r
r
outer insulator
metal braid
with - q
•
a
radius b
b
ds = - dr because path of integration is radially inward
Va Vb  2k ln
or
b
V  2k ln
a

Q
b
V
ln
20L a
Va is higher than Vb

 QL
k
1

4 0  air

a = 0.5 mm
C 2 0
 b
L ln a
a
b

Q2

0
L
C  QV 
Qln ba
C
2 0 L
ln ba
signal
wire
radius a
with + q
Insulator
(dielectric )
b = 2.0 mm
2
C 6 1011 6 1011


L
ln 4
1.38
C
 43 PF m
L
C
 86 PF m
L
0 (air)
=2
Capacitance of two concentric spherical shells
-q
Integration path
E
+q
a
a
a
b
b
Va  Vb    E. ds    Edr
E. ds  Edscos180 Eds Edr
b
ds = - dr
a
a
a
b
b
b
2
 Va  Vb    Edr    kq/r dr  kq 
1
1 1
b  a)
V  kq  kq(  )  kq(
rb
a b
ab
a


ab
ab
C  q /V 
 4 0
k(b  a)
ba
dr
r2
Model of coaxial cable for calculation of capacitance
Outer metal braid
Signal wire