Chapter 27: Current & Resistance. HW For Chapter 27: 6, 18, 20, 30, 42, 48, 52, 56, 58, 62, 68

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Chapter 27: Current & Resistance HW For Chapter 27: 6, 18, 20, 30, 42, 48, 52, 56, 58, 62, 68

Positive Charges move from HI to LOW potential. HI V LOW V

Negative Charges move from LOW to HI potential. HI V LOW V

HOW FAST DO ELECTRONS MOVE IN A CURRENT CARRYING CONDUCTING WIRE??

Electron Speed is called the DRIFT Velocity. Drift velocity ~.001 m/s!!! Electric Fields travel at the speed of light!

Current I = dq dt I = Coulomb/second = Ampere Current flows from a higher potential to a lower potential (electrons flow the opposite way). Current carrying wires are neutral! DC current flows in one direction AC current oscillates back and forth Electrons have a drift velocity of.001m/s! Electric Fields travel at speed of light

Current is charge in motion Charge, e.g. electrons, exists in conductors with a number density, n e (n e approx 10 29 m -3 ) Somehow put that charge in motion: effective picture - all charges move with a velocity, v e real picture - a lot of random motion of charges with a small average equal to v e Current density, J, is given by J = q e n e v e = qnv unit of J is C/m 2 sec or A/m 2 (A Ampere) and 1A 1C/s current, I, is J times cross sectional area, I = J πr 2 for 10 Amp in 1mm x 1mm area, J = 10 +7 A/m 2, and v e is about 10-3 m/s (Yes, the average velocity is only 1mm/s!)

Atomic Vision of Ohm s Law E-field in conductor (resistor) provided by a battery Charges are put in motion, but scatter in a very short time from things that get in the way it s crowded inside that metal defects, lattice vibrations (phonons), etc Typical scattering time τ = 10-14 sec Charges ballistically accelerated for this time and then randomly scattered Average velocity attained in this time is v = Ft/m = qeτ/m Current density is J = qnv so current is proportional to E which is proportional to Voltage OHM s LAW J = (q 2 nτ/m)e or J = σ E σ = conductivity J E = σ σ = 2 qn τ m

Ohm s Law Demo: Vary applied voltage V. I R I Measure current I Does ratio remain I constant? V V R V I V slope = R I How to calculate the resistance? Include resistivity of material Include geometry of resistor

An ohmic device The resistance is constant over a wide range of voltages The relationship between current and voltage is linear The slope is related to the resistance Ohmic Material

Nonohmic Material Nonohmic materials are those whose resistance changes with voltage or current The current-voltage relationship is nonlinear A diode is a common example of a nonohmic device

Resistivity Property of bulk matter related to resistance of a sample is the resistivity (ρ) defined as: 1 m ρ = = σ qnτ ρ E 2 J where E = electric field and J = current density in conductor. I For uniform case: J, A = V = EL I ρl V = EL = ρjl = ρ L = I A A L V = IR where R = ρ A So, in fact, we can compute the resistance if we know a bit about the material, and YES, the property belongs to the material! e.g., for a copper wire, ρ ~ 10-8 Ω-m, 1mm radius, 1 m long, then R.01Ω; for glass, ρ ~ 10 +12 Ω-m; for semiconductors ρ ~ 1 Ω-m j J E = σ A σ = L E 2 qn τ m

Resistance: Resistivity R L ρ A 1 m ρ = = σ qnτ = 2 The LONGER the wire the GREATER the R The THINNER the wire the GREATER the R The HOTTER the wire the GREATER the R

Resistance Question Two cylindrical resistors, R 1 and R 2, are made of identical material. R 2 has twice the length of R 1 but half the radius of R 1. These resistors are then connected to a battery V as shown: V I 1 I 2 What is the relation between I 1, the current flowing in R 1, and I 2, the current flowing in R 2? (a) I 1 < I 2 (b) I 1 = I 2 (c) I 1 > I 2 The resistivity of both resistors is the same (ρ). Therefore the resistances are related as: L 2L L R = 2 1 1 2 = ρ = ρ = 8ρ 8 A2 ( A1 / 4) A1 The resistors have the same voltage across them; therefore V V 1 I 2 = = = I1 R 8R 8 2 1 R 1

Resistors

Radial Resistance of a Coaxial Assume the silicon between the conductors to be concentric elements of thickness dr The resistance of the hollow cylinder of silicon is dr = ρ dr 2πrL Cable: Leakage! R L ρ A 1 m ρ = = σ qnτ = 2 The total radial resistance is R b ρ b = dr = ln a 2πL a This is fairly high, which is desirable since you want the current to flow along the cable and not radially out of it

Resistance: Dependence on Temperature The HOTTER the wire the GREATER the R R = R 0 (1 + αδt) R = original resistance 0 α = Δ T = temperature coefficient of resistivity temperature change (<100 C) o

Resistivity Values

When are light bulbs more likely to blow? When hot or cold? The HOTTER the wire the GREATER the R! R = R 0 (1 + αδt) At lower Resistance, the bulb draws more current and it blows the filament!

Superconductivity 1911: H. K. Onnes, who had figured out how to make liquid helium, used it to cool mercury to 4.2 K and looked at its resistance: At low temperatures the resistance of some metals 0, measured to be less than 10-16 ρ conductor (i.e., ρ<10-24 Ωm)! Current can flow, even if E=0. Current in superconducting rings can flow for years with no decrease! 1957: Bardeen (UIUC!), Cooper, and Schrieffer ( BCS ) publish theoretical explanation, for which they get the Nobel prize in 1972. It was Bardeen s second Nobel prize (1956 transistor)

Superconductivity 1986: High temperature superconductors are discovered (Tc=77K) Important because liquid nitrogen (77 K) is much cheaper than liquid helium Highest critical temperature to date 138 K (-135 C = -211 F) Today: Superconducting loops are used to produce lossless electromagnets (only need to cool them, not fight dissipation of current) for particle physics. [Fermilab accelerator, IL] The Future: Smaller motors, lossless power transmission lines, magnetic levitation trains, quantum computers??...

Ohms Law: ΔV = IR L R = ρ A

Resistance QUESTION How much current will flow through a lamp that has a resistance of 60 Ohms when 12 Volts are impressed across it? USE OHMS LAW: ΔV = IR I ΔV 12V 12V = = = = R 60Ω 60 V / A.2A

Electrical Power As a charge moves from a to b, the electric potential energy of the system increases by QΔV The chemical energy in the battery must decrease by this same amount As the charge moves through the resistor (c to d), the system loses this electric potential energy during collisions of the electrons with the atoms of the resistor This energy is transformed into internal energy in the resistor as increased vibrational motion of the atoms in the resistor

Resistance What makes it glow?

Resistance The resistor is normally in contact with the air, so its increased temperature will result in a transfer of energy by heat into the air The resistor also emits thermal radiation After some time interval, the resistor reaches a constant temperature The rate at which the system loses potential energy as the charge passes through the resistor is equal to the rate at which the system gains internal energy in the resistor. The power is the rate at which the energy is delivered to the resistor

POWER P = I Δ V Energy time J s [ P] = = = Watt P = I Δ V = I ( IR ) = I R 2 ΔV P = I Δ V = V = R Δ V R 2 You pay for ENERGY not for ELECTRONS! Kilowatt-hour is the energy consumed in one hour: [kwh]=j NOT TIME! Power x Time

QUESTION The voltage and power on a light bulb read 120 V, 60 W How much current will flow through the bulb? USE: P = I ΔV I = P/ΔV = 60 W/120 V = 1/2 Amp

QUESTION The power and voltage on a light bulb read 120 V, 60 W What is the resistance of the filament? (I =.5 A) Hint: USE OHMS LAW: V = IR R = V/I = 120 V/.5 A = 240 Ω

QUESTION The power rating for two light bulbs read 30W and 60W. Which bulb has the greatest resistance at 120V? 2 2 P = V / R R= V / P R V W 2 = (120 ) / 30 = 480Ω R V W 2 = (120 ) / 60 = 240Ω (fyi: Power ratings are for bulbs in parallel only!)

Quick Quiz 27.8 For the two lightbulbs shown in this figure, rank the current values at the points, from greatest to least. I a = I b > I c = I d > I e = I f. The 60 W bulb has the lowest resistance and therefore draws the most current! Which light burns the brightest?

If V = 120V, What is I? USE P = IV=> I = P/V Appliance _ Power Current (A) Hair Dryer Electric Iron TV Computer 1600 Watts 1200 Watts 100 Watts 45 Watts

If V = 120V, What is I? USE P = IV=> I = P/V Appliance _ Power Current (A) Hair Dryer 1600 Watts 13.3 A Electric Iron 1200 Watts 10 A TV 100 Watts.83 A Computer 45 Watts.38 A

Electric Bill: Cost to run for 1 hr @ $.05 per 1 kw-hr? Cost = Power x Time x Rate Appliance _ Power Cost Hair Dryer Electric Iron TV Computer 1600 Watts 1200 Watts 100 Watts 45 Watts

Electric Bill: Cost to run for 1 hr @ $.05 per 1 kw-hr? Cost = Power x Time x Rate Appliance _ Power Cost Hair Dryer 1600 Watts $0.08 Electric Iron 1200 Watts $0.06 TV 100 Watts $0.005 Computer 45 Watts $0.003

Why is Power Transmitted with AC instead of DC? Viva La Resistance! (Modern transmission grids use AC voltages up to 765,000 volts.)

Limitations of DC Transmission Large currents in wires produce heat and energy losses by P = I 2 R. Large expensive conductors would be needed or else very high voltage drops (and efficiency losses) would result. High loads of direct current could rarely be transmitted for distances greater than one mile without introducing excessive voltage drops. Direct current can not easily be changed to higher or lower voltages. Separate electrical lines are needed to distribute power to appliances that used different voltages, for example, lighting and electric motors.

Advantages of AC Transmission Alternating Current can be transformed to step the voltage up or down with transformers. Power is transmitted at great distances at HIGH voltages and LOW currents and then stepped down to low voltages for use in homes (240V) and industry (440V). Convert AC to DC with a rectifier in appliances. AC is more efficient for Transmission & Distribution of electrical power than DC!

In The Future. Long Distance AC Power Transmission may not be needed!!!