Transcript Resistors and Resistance

Lecture 08
Current & Resistance
October 3 or 5, 2005
New Topic

Current and
Resistance
Commercial Resistors
Color Coded
Conductors
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In the past we decided that in a
conductor
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The Electric Field is ZERO because we
postulated that no charges were to move.
Static Situation
Any electric field must be at the surface
That field must be normal to the surface
Let’s look at another situation.
Consider a conductor
V1
V2
Electric Field
Walla ….. a CIRCUIT
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The mobile electrons can “move” under the influence of
an electric field.
We then have a “current” (to be defined) flowing in the
wire.
But WAIT … how can we?????
Only if …
Symbol
From the Past
Q Flows and then stops.
Vo
+
-
A Different Situation
Vo
circuit
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Charge will begin to flow through the short.
Charge can flow back into the battery and discharge
it.
Wire can get warm, emit light or even burn our
(fuse).
The FLOW of charge is defined as a current.
Current is
a good thing
Or a
BAD thing
Franklin’s Impact on Physics!!!
ELECTRONS
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CURRENT
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Positive charge will leave the
battery from the positive
terminal and flow through an
external circuit to the negative
terminal.
Electrons will go the other
way.
Current is defined as the flow
of POSITIVE CHARGE.
+ charge does not normally
flow in a wire.
Definition
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Current is the total amount of charge
that flows through a “wire” in one
second.
Current is measured in Coulombs per
second.
A current of one coulomb per second is
defined as an AMPERE. (Amp.)
CONCISE DEFINITION: CURRENT
dq
i
dt
dq  idt
t2
q   i (t )dt
t1
Current will flow throughout the cross-section of the wire (usually).
Current through aa’ is the same as the current through bb’ and cc’.
What is DIFFERENT between aa’ and bb’ with respect to current?
What’s Different??
i=5 amps
A=.1 m2
A=0.05 m2
CURRENT DENSITY
current
j
unit  area
J=5 amps/.05 m2 = 100 amps/m2
J=5 amps / .1 m2 = 50 amps/m2
Question:
The current density in a wire of radius R is given by I0r.
What is the total current flowing through the wire?
Overhead Sol.
Current
Current can’t “pile up” at a point n a circuit.
Example
For BOTH
i0  i1  i2
First introduction to Kirchoff’s Node Equation
Consider
3A ^
5A  2A v
6A
8A
Question:
A 5 Amp current is set up in a circuit for 6 minutes
by a 6 Volt Battery. How much chemical energy is
provided by the battery?
Vo
+
CIRCUIT OF SOME SORT
WORK (Energy) per unit
Charge = qV
-
coul
60 sec
5amps  6 min  5
 6 min 
sec
min
 30  60coul  1800coulombs
qV  1800coul  6Volts  10 KJ
Definition
wire
i
V
High
Low
Current increases
with Potential
Difference (V)
V
R
i
Ohm’s Law
Volt
UNIT  1
 1 OHM
Ampere
SYMBOL  
Observations
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Wires and Resistors are made from conducting
materials.
These materials have some fundamental properties
associated with them.
Electrons are attached to atoms.
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Outer electrons weakly bound
Small Force (Applied Electric Field) can easily push
them.
They bump into things which retard their motion.
The more things that retard their motion, the more
difficult it is to push a current via an applied
potential difference.
 Thus, the resistance goes up.
What kinds of things cause
resistance?
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Sudden Constrictions in the
conductor including bends!
Underlying structure
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Amorphous
Crystalline
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Defects
Impurities
Thermal Collisions
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Temperature
Fundamental Property
RESISTANCE vs. RESISTIVITY
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Consider a wire made of some material.
Resistance is a property of the wire
itself … the material and the shape.
New Quantity:
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RESISTIVITY is a property of the material
itself regardless of its shape.
Consider our wire:
L
A
V
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What would happen to the current if we
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Increased the voltage?
Increased the Area?
Increased the length?
Summary
VA
i
L
or
L
V
i  iR
A
 is the resistivity
R is called the RESISTANCE
and is measured in
OHMS ()
Define CONDUCTIVITY s
s
1

 m
1
Resistivity – Ohm-Meters
Silver
Copper
Aluminum
Tungsten
Platinum
Silicon-Pure
Glass
Fused Quartz
1.62 x 10-8
1.69
2.65
5.25
10.6
2.5 x 103
1010 to 1014
~1016
WIRES
American Wire Gauge
Diameter - mm
30 gauge
0.255
24
0.511
18 (typical household)
1.024
14
1.628
12
2.05
BACK TO OHM
V=iR
i=V/R
Not EVERYTHING is a resistor
The Semiconductor Diode
Effect of Temperature
Temperature
(    0 )   0 (T  T0 )
   0   0 (T  T0 )   0 (1  T )
Linear over a limited temperature range.
QuestionFor an 18 gauge wire of length L to have a resistance
of 1 ohm what must L be if the material is copper?
L
R
A
R  1 .0 
L
A


d 2 / 4
1.69  10
8
d 2  1mm2  10 6 m 2
(d for 18 gauge is about 1 mm)
A current of 6.5 Amps exists in a 9 Ohm resistor for 5
minutes. How many coulombs and how many electrons
pass through the resistor in this time?
Coulombs: 6.5 COULOMBS per Second for 5 minutes
6.5C x 5 min X 60 sec/min = 1950 coulombs
Number of electrons = # coulombs / electron charge =
1950 / 1.6 x 10-19 = 1.22 x 10 +22
Microscopic Theory
J and E
i  JA
i
V
V
1V
J 


A RA
 L   L
A

 A
V 
J  s    sE
L
let A  0 and
J  sE
Consider a wire
N  n d tA
Ne
current 
 ne d A
t
J  ne d
Micro-View “Resistivity”
 depends on the material and is
the mean time between collisions
ease of motion – mobility
resistance to motion - scattering
eE
a a 
m
eE
 d  a  
m
ne 2
1
J  nevd 
E  sE  E
m

ne 2
s
m
Power
Battery supplies energy to the resistor
which, in turn, dissipates it in the form of
heat.
+
V
i
-
E
L
E
C
T
R
O
N
s
Work done on charge Q = Q x V
Work / time  POWER  P
QV Q
P
 V  iV  i iR   i 2 R
t
t
REMEMBER: P=iV and P=i2R
The Spectrum of Conductors
ENGINEERED MATERIALS!!!
Semiconductors
When 105 volts are applied across a wire that is 12 meters
long and has a 0.30 mm radius, the current density us 1.7 x
10 4 A/m2. What is the resistivity of the wire??
L
R
A
so that
RA

L
From the current density we can find the CURRENT.
i=JA = 1.7 x 104 amp/m2 X ( X 0.32) mm2 X (1m/1000mm)2
I = 4.8 ma
R= V/i = 105 volts / (4.8 x 10-3) amps = 2.18 x 104 ohms
 = 2.18 x 104 ohms x 3 x 10-7 m2 / 12m = 0.005 ohm-meters
The figure below gives the electrical potential V(x) along a copper
wire carrying a uniform current, from a point at higher potential
(x=0m) to a point at a lower potential (x=3m). The wire has a
radius of 2.45 mm. What is the current in the wire?
What does the graph tell us??
*The length of the wire is 3 meters.
*The potential difference across the
wire is 12 m volts.
*The wire is uniform.
Let’s get rid of the mm radius and
convert it to area in square meters:
A=r2 = 3.14159 x 2.452 x 10-6 m2
or
A=1.9 x 10-5 m 2
copper
12 volts
0 volts
Material is Copper so resistivity is (from table) = 1.69 x 10-8 ohm meters
We have all we need….
8
L 1.69 x10 ohm - m  3.0 m
R 
 2.67 m
5
A
1.9 x10
From Ohm' s Law :
6
V
12  10 volts
i 
 4.49 ma
3
R 2.67  10 ohms
i
i
Series Combinations
R1
V1
V1  iR1
V2
V
V2  iR2
and
V  V1  V2  iR  iR1  iR2
R  R1  R2
general :
R ( series )   Ri
i
R2
Parallel Combination??
R1, I1
V  iR
V V V
i  i1  i2  

R1 R2 R
R2, I2
V
so..
1
1
1


R1 R2 R
general
1
1

R
i Ri
What’s This???
In Fig. 28-39, find the
equivalent resistance
between points
(a) F and H and [2.5]
(b) F and G. [3.13]
The current density across a cylindrical conductor of radius R varies according to
the equation
J = J0(1 - (r/R)),
where r is the distance from the central axis. Thus the current density is a
maximum J0 at the axis (r = 0) and decreases linearly to zero at the surface (r =
R). Calculate the current in terms of J0 and the conductor's cross-sectional area A
= R2.
That’s it for Resistance
Next Time Electric Circuits