Resistance Is
Futile!
Physics 2113
Jonathan Dowling
Physics 2113 Lecture 21: FRI 06 MAR
Current & Resistance III
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 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] =
V
and therefore : i =
R
and V = iR
Volt
≡ Ohm (abbr. Ω)
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
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:
ρ = or, as vectors, E = ρ J
J
Resistivity is associated
( resistance: R=V/I )
ρ ≡ [Ωm] = [Ohm⋅ meter]
with a material, resistance
with respect to a device
1
Conductivi
ty
:
σ
=
constructed with the material.
ρ
Example:
-
A
L
V
+
V
E= ,
L
i
J=
A
ρ=
V
L=RA
i
L
A
R=ρ
L
A
Longer → More resistance
Makes sense!
For a given material: Thicker → Less resistance
b
a
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? V
R≡
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!? ⎡V ⎤
Units : R = ⎢ ⎥ ≡ [Ω] = [Ohm]
⎣ A⎦
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
(c) P = V 2 / R → V 2 / ( 2R ) = 12 V 2 / 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 =
ρL
A
Rb =
ρ 3L / 2
ρL
=3
= 3Ra
A/2
A
Rc =
ρL / 2 ρL
=
= Ra
A/2
A
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.
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 Ω
•  i=V/R= 115V/10.6 Ω=10.8 A
•  Energy? P=dU/dt => ΔU=P×Δt = 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.
•