Physics 2102
Resistance Is
Futile!
Jonathan Dowling
Physics 2102
Lecture 10: TUE 23 FEB
Current & Resistance I
QuickTime™ and a
decompressor
are needed to see this picture.
QuickTime™ and a
decompressor
are needed to see this picture.
Georg Simon Ohm
(1789-1854)
EXAM I AVERAGE: 55/100
STANDARD DEVIATION: 15/100
APPROXIMATE EXAM I LETTER GRADE:
A=100–80
B=79–70
C=69–40
D=39–30
F=29–0
We do not assign an official letter grade for any midterm exam so the
above breakdown is an unofficial guide for use by students in our section
only; to give you an APPROXIMATE idea of where you stand.
The Exam I solutions have been posted at:
http://www.phys.lsu.edu/classes/spring2010/phys2102/exam1solutions.pdf
Exam I: Problem 2 — Integration with Cylindrical Shells
dr
Qenc    r dV
V
QuickTime™ and a
decompressor
are needed dV
to see this picture.
Qenc ≠ V unless  is a contant!
 = ar2 is NOT a constant!
L
V  r2L
dV  2 rdrL
r2

(a) Qtot  ar 2 rLdr  a r L / 2 
2
4
2
0
r1
(b)
Qenc
  ar 2 rLdr  a r L / 2  mC5
2
0
4
1
m4 m
1 1


C m4 m
m5 1 1


Resistance is NOT Futile!
Electrons are not “completely free to move” in a conductor. They move
erratically, colliding with the nuclei all the time: this is what we call
“resistance”.
The resistance is related to the potential we need to apply to a device to
drive a given current through it. The larger the resistance, the larger the
potential we need to drive the same current.
Ohm’s laws
V
R
i
Units : [R] 
and therefore : i 
V
and V  iR
R
Volt
 Ohm (abbr. )
Ampere
Georg Simon Ohm
(1789-1854)
"a professor who preaches such heresies
is unworthy to teach science.” Prussian
minister of education 1830
Devices specifically designed to have a constant value of R are called
resistors, and symbolized by
dq C
i
   Ampere   A
s 
dt
Vector :
Current Density and Drift


 
Speed
Same direction as E such that i   J  dA
J
The current is the flux of the current density!
If surface is perpendicular to a constant electric
field, then i=JA, or J=i/A
[J ] 
Units:
dA
J
Ampere
m2
E
i
Drift speed: vd :Velocity at which electrons move in order to establish a current.
Charge q in the length L of conductor: q  (n A L) e
L
n =density of electrons, e =electric charge
A
t
E
i
L
vd
i
q n ALe

 n A e vd
L
t
vd
i
J

n Ae n e


J  n e vd
vd 
Resistivity and resistance
Metal
“field lines”
These two devices could have the same resistance
R, when measured on the outgoing metal leads.
However, it is obvious that inside of them different
things go on.


E
resistivity:   or, as vectors, E   J
J
Resistivity is associated
( resistance: R=V/I )
  m  Ohm  meter 
with a material, resistance
with respect to a device
1
Conductivi
ty
:


constructed with the material.


Example:
A
-
L
V
+
V
E ,
L
i
J
A

Makes sense!
For a given material:
V
LRA
i
L
A
R
L
A
Longer  More resistance
Thicker  Less resistance
Resistivity and Temperature
Resistivity depends on
temperature:
 = 0(1+a(T-T0) )
• At what temperature would the resistance of a
copper conductor be double its resistance at
20.0°C?
• Does this same "doubling temperature" hold for
all copper conductors, regardless of shape or
size?
b
Power in electrical circuits
A battery “pumps” charges through the
resistor (or any device), by producing a
potential difference V between points a and
b. How much work does the battery do to
move a small amount of charge dq from b
to a?
a
dW = –dU = -dq•V = (dq/dt)•dt•V= iV•dt
The battery “power” is the work it does per unit time:
P = dW/dt = iV
P=iV is true for the battery pumping charges through any device. If the
device follows Ohm’s law (i.e., it is a resistor), then V=iR and
P = iV = i2R = V2/R
Ohm’s Law and Power in
Resistors
Watt? You Looking At!
V 
Units : R      Ohm
A
Ohm’s Law
V  iR
Power Dissipated
 by a Resistor:
P  iV  i R  V /R
2
2
 
Units : P  J s  W   Watt 
L
A=r2
A
Current Density: J=i/A
Units: [A/m2]
Resistance: R=L/A
QuickTime™ and a
decompressor
are needed to see this picture.
Resitivity:  depends only on
Material and Temperature.
Units: [•m]
Example
A human being can be electrocuted if a
current as small as i=100 mA passes near
the heart. An electrician working with
sweaty hands makes good contact with
the two conductors he is holding. If his
resistance is R=1500, what might the
fatal voltage be?
(Ans: 150 V) Use: V=iR
Example
Two conductors are made of the same material and have the
same length. Conductor A is a solid wire of diameter r=1.0mm.
Conductor B is a hollow tube of outside diameter 2r=2.0mm and
inside diameter r=1.0mm. What is the resistance ratio RA/RB,
measured between their ends?
A
R=L/A
B
AA= r2
AB= ((2r)2-r2)=3r2
RA/RB= AB/AA= 3
LA=LB=L Cancels
Integration with Cylindrical Shells
dr
ienc
r
r
  J • dA   J dA cos
A
QuickTime™ and a
decompressor
i picture.
are needed to see this
A
ienc ≠ J•A unless J is constant!
J(r) = ar2 is NOT a constant!
L
A  r2
dA  2 rdr
dA
r2
itot   ar 2 rdr  a r / 2 
2
4
2
0
r1
ienc
  ar 2 rdr  a r / 2  Cm/4s
2
0
4
1
m4
1


C / s m4
m4 1


Example
A P=1250Watt radiant heater is constructed to operate at
V=115Volts.
(a) What will be the current in the heater?
(b) What is the resistance of the heating coil?
(c) How much thermal energy is produced in 1.0 hr by the
heater?
• Formulas: P=i2R=V2/R; V=iR
• Know P, V; need R to calculate current!
• P=1250W; V=115V => R=V2/P=(115V)2/1250W=10.6 
• i=V/R= 115V/10.6 =10.8 A
• Energy? P=dU/dt => U=P•t = 1250W 3600 sec= 4.5 MJ
= 1.250kW•hr
Example
A 100 W lightbulb is plugged into a standard 120 V outlet.
(a) What is the resistance of the bulb?
(b) What is the current in the bulb?
(c) How much does it cost per month to leave the light turned on
continuously? Assume electric energy costs 6¢/kW·h.
(d) Is the resistance different when the bulb is turned off?
• Resistance: same as before, R=V2/P=144 
• Current, same as before, i=V/R=0.83 A
• We pay for energy used (kW h):
U=Pt=0.1kW  (30 24) h = 72 kW h => $4.32
• (d): Resistance should be the same, but it’s not: resistivity and
resistance increase with temperature. When the bulb is turned off,
it is colder than when it is turned on, so the resistance is lower.
Example
An electrical cable consists of 105 strands of fine wire, each
having 2.35  resistance. The same potential difference is
applied between the ends of all the strands and results in a
total current of 0.720 A.
(a) What is the current in each strand?
i=I/105=0.720A/105=[0.00686] A
(b) What is the applied potential difference?
V=iR=[0.016121] V
(c) What is the resistance of the cable?
R=V/I=[.0224 ] 
i
V
QuickTi me™ and a
decompressor
are needed to see thi s pi ctur e.
P = iV
QuickTime™ and a
decompressor
are needed to see this picture.
V
U = Pt
t in seconds
i
QuickTime™ and a
decompressor
are needed to see this picture.
P = iV [J/s is Watt]
P = iV [Watt is J/s]
Quic kTime™ and a
dec ompres sor
are needed to see this pic ture.
Rd = 1.0x105
im = 1x10–3A
Rw = 1.5x103
im =
QuickTime™ and a
decompressor
are needed to see this picture.
V1 = imRd
i1 = V1/Rw
V2 = imRw
Quic kT i me™ and a
dec om pres s or
are needed t o s ee thi s pi c ture.
QuickTime™ and a
decompressor
are needed to see this picture.
Quic kT i me™ and a
dec om pres s or
are needed t o s ee thi s pi c ture.
My House Has Two Front Porch Lights.
Each Light Has a 100W Bulb.
The Lights Come on at Dusk and Go Off at Dawn.
How Much Does this Cost Me Per Year?
Two 100W Bulbs @ 12 Hours Each = One 100W @ 24 Hours.
P = 100W = 0.1kW
T = 365 Days x 24 Hours/Day = 8670 Hours
Demco Rate: D = 0.1797$/kW•Hour (From My Bill!)
Cost = PxTxD = (0.1kW)x(8670 Hours)x(0.1797$/kW•Hour)
= $157.42 = 13 shots of Goldschläger!
QuickTime™ and a
decompressor
are needed to see this picture.