Physics 2102 Lecture 10: TUE 23 FEB. Physics 2102

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Resistance Is Futile! Physics 2102 Jonathan Dowling Physics 2102 Lecture 10: TUE 23 FEB Current & Resistance I 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

dr Exam I: Problem 2 Integration with Cylindrical Shells dv L Q enc = ( ) "! r dv V Q enc ρv unless ρ is a contant! ρ = ar 2 is NOT a constant! V =!r 2 L dv = 2!rdrL (a) r 2" Q tot = ar 2 2!rLdr = a!r 4 C 2 L / 2! # $ 0 m 4 m 5 1 m 1 % & (b) r 1 Q enc = " ar 2 2!rLdr = a!r 4 C 1 L / 2! # $ 0 m 4 m 5 1 m 1 % &

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. V R! i Resistance is NOT Futile! and therefore : Volt Units :[R] = " Ohm (abbr.!) Ampere Ohm s laws V i = and V = ir R 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 i! dq " = C % dt # $ s & '! [ Ampere ] = [ A]

Current Density and Drift Speed Vector : J! Same direction as E!!! such that i =! J " da 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 da J Units: [ J ] = Ampere 2 m E i Drift speed: v d :Velocity at which electrons move in order to establish a current. L Charge q in the length L of conductor: q = ( n AL) e A E i n =density of electrons, e =electric charge t = L v d q n ALe i = = = t L v d n Aev d i v d = = n Ae!! J = n ev d J n e

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. Resistivity is associated with a material, resistance with respect to a device constructed with the material. resistivity: E!! = or, as vectors, E =! J! J ( resistance: R=V/I )! " [#m] = [ Ohm$ meter] Conductivi ty : " = 1! Example: A V E =, J = L i A! = V i L A = R A L L R =! A - L V + Makes sense! For a given material: Longer! Thicker! More resistance Less resistance

Resistivity and Temperature Resistivity depends on temperature: ρ = ρ 0 (1+α (T-T 0 ) ) 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 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 and P = iv = i 2 R = V 2 /R

Ohm s s Law and Power in Watt? You Looking At! Ohm s Law V = ir Resistors! Units : R = V $ " # A% & ' ( [ ] = [ Ohm] Power Dissipated by a Resistor: P = iv = i 2 R = V 2 /R Units : P = [ J ] = [ W] = [ Watt] s

L A=πr 2 A Current Density: J=i/A Units: [A/m 2 ] 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 R A /R B, measured between their ends? A R=ρL/A B A A =π r 2 A B =π ((2r) 2 r 2 )=3πr 2 R A /R B = A B /A A = 3 L A =L B =L Cancels

Integration with Cylindrical Shells dr da i L i enc =! J! d A! = J dacos" A i enc J A unless J is constant! J(r) = ar 2 is NOT a constant! A =!r 2 da = 2!rdr! A r 2" i tot = ar 2 2!rdr = a!r 4 C /s 2 / 2! # $ 0 m 4 m 4 1 % & r 1 i enc = " ar 2 2!rdr = a!r 4 C /s 1 / 2! # $ 0 m 4 m 4 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=i 2 R=V 2 /R; V=iR Know P, V; need R to calculate current! P=1250W; V=115V => R=V 2 /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) (b) (c) (d) What is the resistance of the bulb? What is the current in the bulb? How much does it cost per month to leave the light turned on continuously? Assume electric energy costs 6 /kw h. Is the resistance different when the bulb is turned off? Resistance: same as before, R=V 2 /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]

R d = 1.0x10 5 Ω R w = 1.5x10 3 Ω i m = 1x10 3 A i m = V 1 = i m R d i 1 = V 1 /R w V 2 = i m R w

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!