Physics 2113 ���Lecture 21: FRI 06 MAR

Current & Resistance III

Physics 2113 Jonathan Dowling

Georg Simon Ohm (1789-1854)

Resistance Is Futile!

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.

) (abbr. OhmAmpere

Volt [R] :Units Ω≡=

Georg Simon Ohm (1789-1854)

"a professor who preaches such heresies is unworthy to teach science.” Prussian minister of education 1830

iVR ≡ iRV

RVi == and : thereforeand

Ohm’s laws

Devices specifically designed to have a constant value of R are called resistors, and symbolized by

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”.

Resistance is NOT Futile!

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i ≡ dqdt

= Cs

⎡ ⎣ ⎢

⎤ ⎦ ⎥ ≡ Ampere[ ] = A[ ]

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.

Metal “field lines”

resistivity: JEJE !!

ρρ == vectors,as or,

Resistivity is associated with a material, resistance with respect to a device constructed with the material. ρ

σ 1 :tyConductivi =

Example: A

L V

+ -

AiJ

LVE == ,

LAR

AiL

V==ρ

ALR ρ=

Makes sense! For a given material: resistance LessThicker

resistance MoreLonger →→

Resistivity and Resistance

( resistance: R=V/I )

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ρ ≡ Ωm[ ] = Ohm ⋅meter[ ]

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?

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

iVR ≡ iRV

RVi == and : thereforeand

Ohm’s laws

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Units : R = VA⎡ ⎣ ⎢

⎤ ⎦ ⎥ ≡ Ω[ ] = Ohm[ ]

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V = iR

Ohm’s Law and Power in Resistors

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P = iV = i2R =V 2 /R

Units : P = Js[ ] = W[ ] = Watt[ ]

Power Dissipated by a Resistor:

Ohm’s Law

Watt You Looking At!?

Example, Rate of Energy Dissipation in a Wire Carrying Current:

ICPP: Why is this unwise??? P = iV = i2R =V 2 / RP = iV → 4P = (4i)VCurrent i increases by 4. House Circuit Breaker 3A.

i = P /V = 200W /120V = 1.67A→ 4P /V = 6.67A

P = iV= i2R=V 2 / R

(b) P = i2R→ 2i( )2 R = 4 i2R( ) = 4P(c) P =V 2 / R→V 2 / 2R( ) = 1

2 V2 / R( ) = 1

2 P(d) P = i2R→ i2 2R( ) = 2 i2R( ) = 2P

Pa = Pb > Pd > Pc(a) P =V 2 / R→ 2V( )2 / R = 4 V 2 / R( ) = 4P

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 = i2R =V 2 / R

Ra =ρLA

Rb =ρ3L / 2A / 2

= 3ρLA

= 3Ra Rc =ρL / 2A / 2

= ρLA

= Ra

Ra = Rc < Rb

Step I: The resistivity ρ is the same (all three are copper). Find the Resistance R=ρL/A for each case:

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.

P = iV

U = Pt ���t in seconds

P = iV [J/s is Watt]

P = iV [Watt is J/s]

i V

V i

���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

I’m switching to white light LEDs! Nobel Prize 2014!

•  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.

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