General Physics (PHY 2140)

Transcription

1 General Physics (PHY 2140) Lecture 9 Electrodynamics Electric current temperature variation of resistance electrical energy and power Chapter

2 Department of Physics and Astronomy announces the Fall 2003 opening of The Physics Resource Center on Monday, September 22 in Room 172 of Physics Research Building. Hours of operation: Mondays, Tuesdays, Wednesdays Thursdays and Fridays 11 AM to 6 PM 11 AM to 3 PM Undergraduate students taking PHY will be able to get assistance in this Center with their homework, labwork and other issues related to their physics course. The Center will be open: Monday, September 22 to Wednesday, December 10,

3 Lightning Review Last lecture: 1. Current and resistance Q = t Current and drift speed Resistance and Ohm s law is proportional to Resistivity material property = = ρ = nqv A R d RA l a Review Problem: Consider two resistors wired one after another. f there is an electric current moving through the combination, the current in the second resistor is a. equal to b. half c. smaller, but not necessarily half the current through the first resistor. R 1 R 2 b c 3

4 17.4 Resistivity - Example (a) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius m. Cross section: ( ) 2 A= πr = π m = m Resistivity (Table): 1.5 x 10 6 Ωm. Resistance/unit length: 6 R ρ Ωm = = = 7 2 l A m 4.6 Ω m 4

5 17.4 Resistivity - Example (b) f a potential difference of 10.0 is maintained across a 1.0-m length of the nichrome wire, what is the current? 10.0 = = = R 4.6Ω 2.2A 5

6 17.4 Temperature ariation of Resistance -ntro The resistivity of a metal depends on many (environmental) factors. The most important factor is the temperature. For most metals, the resistivity increases with increasing temperature. The increased resistivity arises because of larger friction caused by the more violent motion of the atoms of the metal. 6

7 For most metals, resistivity increases approx. linearly with temperature. ρ ρ= ρ 1 ( ) o + α T To T ρ is the resistivity at temperature T (measured in Celsius). ρ ο is the reference resistivity at the reference temperature T ο (usually taken to be 20 o C). α is a parameter called temperature coefficient of resistivity. For a conductor with fixed cross section. ρ Metallic Conductor R = Ro 1 + α T T ( ) o T Superconductor 7

8 17.5 Temperature ariation of Resistance - Example Platinum Resistance Thermometer A resistance thermometer, which measures temperature by measuring g the change in the resistance of a conductor, is made of platinum and has a resistance of 50.0 Ω at 20 o C. When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 Ω.. Find the melting point of ndium. Solution: Using α=3.92x10-3 ( o C) -1 from table

9 Platinum Resistance Thermometer A resistance thermometer, which measures temperature by measuring g the change in the resistance of a conductor, is made of platinum and has a resistance of 50.0 Ω at 20 o C. When the device is immersed in a vessel containing melting indium, its resistance increases to 76.8 Ω.. Find the melting point of ndium. Solution: Using α=3.92x10-3 ( o C) -1 from table R o =50.0 Ω. T o =20 o C. R=76.8 Ω.. T R R 76.8Ω 50.0Ω o To = = 1 αr 3 o T = 137 o = 157 o C C ( o C) [ ] Ω 9

10 Appendix: 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 (i.e., ρ<10-24 Ωm)! 16 ρ conductor Current can flow, even if E=0. Current in superconducting rings can flow for years with no decrease! 1957: Bardeen (UUC! UUC!), Cooper, and Schrieffer ( BCS ) publish theoretical explanation, for which they get the Nobel prize in t was Bardeen s second Nobel prize (1956 transistor) 10

11 17.7 Electrical energy and power n any circuit, battery is used to induce electrical current chemical energy of the battery is transformed into kinetic energy of mobile charge carriers (electrical energy gain) Any device that possesses resistance (resistor) present in the circuit will transform electrical energy into heat kinetic energy of charge carriers is transformed into heat via collisions with atoms in a conductor (electrical energy loss) = R C D +- B A 11

12 Electrical energy Consider circuit on the right in detail AB: charge gains electrical energy form the battery E = Q B C (battery looses chemical energy) CD: electrical energy lost (transferred into heat) Back to A: same potential energy (zero) as before Gained electrical energy = lost electrical energy on the resistor A D 12

13 Power Compute rate of energy loss (power dissipated on the resistor) Use Ohm s law E Q P= = = t t P= = R= ( ) 2 2 R Units of power: S: watt delivered energy: kilowatt-hours ( )( ) kwh = 10 W 3600 s = J 13

14 Example Power Transmission line A high-voltage transmission line with resistance of 0.31 Ω/km carries 1000A, starting at 700 k, for a distance of 160 km. What is the power loss due to resistance in the wire? Given: = ρ=0.31 Ω/km L=160 km =1000 A Find: P=? Observations: 1. Given resistance/length, compute total resistance 2. Given resistance and current, compute power loss ( )( ) R = ρl = 0.31Ω km 160 km = 49.6 Ω Now compute power 2 ( ) ( ) 2 6 P = R= 1000 A 49.6 Ω = W 14

15 Mini-quiz Why do the old light bulbs usually fail just after you turn them on? When the light bulb is off, its filament is cold, so its resistance is large. Once the switch it thrown, current passes through the filament heating it up, thus increasing the resistance, ( ) R= Ro 1 + α T T This leads to decreased amount of power delivered to the light bulb, as P= 2 R Thus, there is a power spike just after the switch is thrown, which burns the light bulb. Resume: electrical devices are better be turned off if there is a power loss o 15

16 Direct Current Circuits 16

17 18.1 Sources of EMF Steady current (constant in magnitude and direction) requires a complete circuit path cannot be only resistance cannot be only potential drops in direction of current flow Electromotive Force (EMF) provides increase in potential E converts some external form of energy into electrical energy Single emf and a single resistor: emf can be thought of as a charge pump = R +- = R = E E 17

18 EMF Each real battery has some internal resistance AB: potential increases by E on the source of EMF, then decreases by r (because of the internal resistance) Thus, terminal voltage on the battery is = E r Note: E is the same as the terminal voltage when the current is zero (open circuit) B r E A C D R 18

19 EMF (continued) Now add a load resistance R Since it is connected by a conducting wire to the battery terminal voltage is the same as the potential difference across the load resistance = E r = R, or E = r + R Thus, the current in the circuit is = r E B A Power output: E 2 2 R + r E = r+ R C D R Note: we ll assume r negligible unless otherwise is stated 19

20 Measurements oltmeters measure Potential Difference (or voltage) across a device by being placed in parallel with the device. Ammeters measure current through a device by being placed in series with the device. A 20

21 Direct Current Circuits Two Basic Principles: Conservation of Charge Conservation of Energy Resistance Networks R ab eq = R ab eq a R eq b 21

22 Resistors in series Conservation of Charge = 1 = 2 = 3 Conservation of Energy ab = R eq ab = = + + = + + R = R + R + R eq oltage Divider: ab = 1 R 1 1 R eq b a R 1 1 = 1 R 1 R 2 2 = 2 R 2 R 3 3 = 3 R 3 22

23 Resistors in parallel Conservation of Charge = Conservation of Energy ab = 1 = 2 = = R eq ab ab = + + R R R R eq R 1 1 = 1 R = + + = ab ab ab a R 2 2 = 2 R 2 R 3 3 = 3 R 3 b Current Divider: 1 = R R eq 1 23

24 Example: Determine the equivalent resistance of the circuit as shown. Determine the voltage across and current through each resistor. Determine the power dissipated in each resistor Determine the power delivered by the battery =18 R 1 =4Ω R 2 =3Ω R 3 =6Ω 24