Transcript PPT

Resistance Is
Futile!
Physics 2113
Jonathan Dowling
Physics 2102
Lecture 18: MON 05 OCT
Current & Resistance II
Georg Simon Ohm
(1789-1854)
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 mechanical analog is FRICTION.
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
Rº
i
Units : [R] =
V
and therefore : i =
and V = iR
R
Volt
º Ohm (abbr. W)
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 éCù
iº
= ê ú º [ Ampere] = [ A]
ësû
dt
Resistivity ρ vs. Resistance R
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.
resistivity:
Resistivity is associated
( resistance: R=V/I )
r º [Wm] = [Ohm× meter]
with a material, resistance
with respect to a device
1
Conductivity : s =
constructed with the material.
r
Example:
A
-
L
V
+
V
E= ,
L
i
J=
A
r=
Makes sense!
For a given material:
V
L=RA
i
L
A
L
R=r
A
Longer ® More resistance
Wider ® Less resistance
Fluid Flow: An Analogy for the PETEs!
• Amount of Water = Charge
• Pressure = Potential = Voltage
• Flow Rate = Current = Amps
The pressure at the end of the hose represents
voltage.
The amount of water in the tank represents
charge.
The rate of flow gallons/minute out the hose
represents current or amperage.
https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law
Fluid Flow: An Analogy for the PETEs!
R=r
• Amount of Water = Charge
• Pressure = Voltage
• Flow Rate = Current = Amps
Decrease hose width, decrease A, increase
resistance R.
Increase hose length L, increase resistance R.
Put pebbles in the hose, increase resistivity ρ,
increase resistance R.
https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law
L
A
26.4: Resistance and Resistivity:
The resistivity ρ of a resistor is defined as:
The SI unit for  is .m.
The conductivity  of a material is the reciprocal of its
resistivity:
Put pebbles or a filter in the hose, increase
resistivity ρ, increase resistance R.
26.4: Resistance and Resistivity, Calculating Resistance from Resistivity:
Think pumping water through a long hose. It is
easier if L is short and A is big (small R). It is
harder if L is long or A is small (big R).
If the streamlines representing the current
density are uniform throughout the wire,
the electric field E and the current density
J will be constant for all points within the
wire.
ICPP
The copper wire has radius r. What
happens to the Resistance R if you:
(a) Double the Length? R ® 2R
(b) Double the Area? R ® R / 2
(c) Double the Radius? R ® R / 4
A = pr2
What happens to the Resistivity ρ if you:
(a)Double the Length?
r®r
(b)Double the Area?
(c)Double the Radius?r ® r
r®r
Step I: The resistivity ρ is the same (all three are copper).
Find the Resistance R=ρL/A for each case:
Ra =
rL
A
Rb =
r ( 3L / 2 )
( A / 2)
=3
rL
A
Rc =
r ( L / 2)
( A / 2)
Rb > Ra = Rc
Step II: Rank the current using V=iR or i=V/R with V constant!
ia = ic > ib
Ranking is reversed since R is downstairs.
=
rL
A
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
Example, A material has resistivity, a block of the material has a resistance.:
26.4: Resistance and Resistivity, Variation with Temperature:
The relation between temperature and resistivity for copper—and for metals in general—
is fairly linear over a rather broad temperature range. For such linear relations we can
write an empirical approximation that is good enough for most engineering purposes:
Resistivity and Temperature
Resistivity depends on
temperature:
 =  0(1+ (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?
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 mechanical analog is FRICTION.
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
Rº
i
Units : [R] =
V
and therefore : i =
and V = iR
R
Volt
º Ohm (abbr. W)
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 éCù
iº
= ê ú º [ Ampere] = [ A]
ësû
dt
26.5: Ohm’s Law:
L
A=r2
A
Current Density: J=i/A
Units: [A/m2]
Resistance: R= L/A
Resitivity:  depends only on
Material and Temperature.
Units: [•m]
V = iR
R = V / i = constant
Example
An electrical cable consists of 105 strands of fine wire, each
having r=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 ] 
Rd = 1.0x105
im = 1x10–3A
Rw = 1.5x103
im =
V1 = imRd
i1 = V1/Rw
V2 = imRw
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
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
V
Rº
i
V
and therefore : i =
R
Ohm’s laws
and V = iR
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 or i=V/R and
P = iV = i2R = V2/R
Ohm’s Law and Power in
Resistors
Watt? You Looking At!
éV ù
Units : R = ê ú º [W] = [Ohm]
ë Aû
Ohm’s Law
V = iR
Power Dissipated by a Resistor:
P = iV = i R = V /R
2
2
[ s ] = [W] = [Watt]
Units : P = J
Example, Rate of Energy Dissipation in a Wire Carrying Current:
ICPP: Why is this unwise???
P = iV ® 4P = (4i)V
Current i increases by 4.
House Circuit Breaker 3A.
P = iV = i R = V / R
2
2
i = P / V = 200W /120V = 1.67A
® 4P / V = 6.67A
P = iV
=i R
2
=V /R
2
Pa = Pb > Pd > Pc
(
( )
(
)
(a) P = V / R ® ( 2V ) / R = 4 V / R = 4P
2
2
(b) P = i R ® ( 2i ) R = 4 i 2 R = 4P
2
2
2
1
(c) P = V / R ® V / ( 2R ) = 2 V / R = 12 P
2
2
2
(d) P = i R ® i ( 2R ) = 2 ( i R ) = 2P
2
2
2
)
ICPP: The figure here shows three cylindrical copper
conductors along with their face areas and lengths.
Rank them according to the power dissipated by them,
greatest first, when the same potential difference V is
placed across their lengths.
P = iV = i R = V / R
2
2
Step I: The resistivity ρ is the same (all three are copper).
Find the Resistance R=ρL/A for each case:
Ra =
rL
A
Rb =
r 3L / 2
A/2
=3
rL
A
= 3Ra
Rc =
rL / 2
A/2
Ra = Rc < Rb
Step II: Rank the power using P=V2/R since V is same.
Pa = Pc > Pb
Ranking is reversed since R is downstairs.
=
rL
A
= Ra
Example
A P=1250 Watt 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=144Ω
• 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 => $4.32
• (d): Resistance should be the same, but it’s not: resistivity and
resistance increase with temperature. When the bulb is turned off,
it is colder than when it is turned on, so the resistance is lower.
i
V
P = iV
U = Pt
t in seconds
V
i
P = iV [J/s is Watt]
P = iV [Watt is J/s]
I’m switching to white
light LEDs!
Nobel Prize 2014!
My House Has Two Front Porch Lights.
Each Light Has a 100W Incandescent Bulb.
The Lights Come on at Dusk and Go Off at Dawn.
How Much Do these lights Cost Me Per Year?
Two 100W Bulbs @ 12 Hours Each = One 100W @ 24 Hours.
P = 100W = 0.1kW
T = 365 Days x 24 Hours/Day = 8670 Hours
Demco Rate: D = 0.1797$/(kW×Hour) (From My Bill!)
Cost = PxTxD = (0.1kW)x(8670 Hours)x(0.1797$/kW×Hour)
= $157.42
• Figure shows a person and a cow, each a radial distance D=60m from
the point where lightning of current i=10kA strikes the ground. The
current spreads through the ground uniformly over a hemisphere
centered on the strike point. The person's feet are separated by
radial distance Δrper=0.50m; the cow's front and rear hooves are
separated by radial distance Δrcow=1.50m. The resistivity of the
ground is ρgr=100 Ωm . The resistance both across the person, between
left and right feet, and across the cow, between front and rear
hooves, is R=4.00kΩ. (a) What is the current ip through the person? (b)
What is the current ic through the cow? Recall that a current ≥ 100mA
causes a heart attack and death.
•