Resistance
Is
Futile!
Physics 2102
Spring 2007
Jonathan Dowling
Lecture 14: FRI 13 FEB
Current and Resistance
Ch26.3–7
Georg Simon Ohm
(1789-1854)
Watt?
Ohm’s Law and Power in
Resistors
Ohm’s Law
"V %
Units : R = $ ' ( [)] = [Ohm]
# A&
V = iR
Power Dissipated
! by a Resistor:
2
2
P = iV = i R = V /R
[ ]
Units : P = J s = [W ] = [Watt ]
L
A=πr2
A
Current Density: J=i/A
Units: [A/m2]
Resistance: R=ρL/A
Resitivity: ρ depends only on
Material and Temperature.
Units: [Ω•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=1500Ω, 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=ρL/A
B
AA=π r2
AB=π ((2r)2−r2)=3πr2
RA/RB= AB/AA= 3
LA=LB=L Cancels
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 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=144 Ω
• 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 => $4.32
• (d): Resistance should be the same, but it’s not: resistivity and
resistance increase with temperature. When the bulb is turned off,
it is colder than when it is turned on, so the resistance is lower.
Example
An electrical cable consists of 105 strands of fine wire, each
having 2.35 Ω resistance. The same potential difference is
applied between the ends of all the strands and results in a
total current of 0.720 A.
(a) What is the current in each strand?
i=I/105=0.720A/105=[0.00686] A
(b) What is the applied potential difference?
V=iR=[0.016121] V
(c) What is the resistance of the cable?
R=V/I=[.0224 ] Ω
i
V
P = iV
U = Pt
t in seconds
V
i
P = iV [J/s is Watt]
P = iV [Watt is J/s]
Rd = 1.0x105Ω
im = 1x10–3A
Rw = 1.5x103Ω
im =
V1 = imRd i1 = V1/Rw V2 = imRw My House Has Two Front Porch Lights.
Each Light Has a 100W Bulb.
The Lights Come on at Dusk and Go Off at Dawn.
How Much Does this Cost Me Per Year?
Two 100W Bulbs @ 12 Hours Each = One 100W @ 24 Hours.
P = 100W = 0.1kW
T = 365 Days x 24 Hours/Day = 8670 Hours
Demco Rate: D = 0.1797$/kW•Hour (From My Bill!)
Cost = PxTxD = (0.1kW)x(8670 Hours)x(0.1797$/kW•Hour)
= $157.42 = 13 shots of Goldschläger!