1. Copyright © 2009 Pearson Education, Inc. Chapter 25 Electric Currents and Resistance

2. Copyright © 2009 Pearson Education, Inc. 25-4 Resistivity Example 25-5: Speaker wires. Suppose you want to connect your stereo to remote speakers. (a) If each wire must be 20 m long, what diameter copper wire should you use to keep the resistance less than 0.10 Ω per wire? (b) If the current to each speaker is 4.0 A, what is the potential difference, or voltage drop, across each wire?

3. Copyright © 2009 Pearson Education, Inc. For any given material, the resistivity increases with temperature: Semiconductors are complex materials, and may have resistivities that decrease with temperature. 25-4 Resistivity

4. Copyright © 2009 Pearson Education, Inc. 25-4 Resistivity Example 25-7: Resistance thermometer. The variation in electrical resistance with temperature can be used to make precise temperature measurements. Platinum is commonly used since it is relatively free from corrosive effects and has a high melting point. Suppose at 20.0°C the resistance of a platinum resistance thermometer is 164.2 Ω. When placed in a particular solution, the resistance is 187.4 Ω. What is the temperature of this solution?

5. ConcepTest 25.3aWires I Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? d A = 4d B 1) d A = 4d B d A = 2d B 2) d A = 2d B d A = d B 3) d A = d B 4) d A = 1/2d B 5) d A = 1/4d B

6. area is less arearadiussquared diameter of A must be two times less than the diameter of B The resistance of wire A is greater because its area is less than wire B. Since area is related to radius (or diameter) squared, the diameter of A must be two times less than the diameter of B. ConcepTest 25.3aWires I Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? d A = 4d B 1) d A = 4d B d A = 2d B 2) d A = 2d B d A = d B 3) d A = d B 4) d A = 1/2d B 5) d A = 1/4d B

7. Copyright © 2009 Pearson Education, Inc. Power, as in kinematics, is the energy transformed by a device per unit time: 25-5 Electric Power or

8. Copyright © 2009 Pearson Education, Inc. The unit of power is the watt, W. For ohmic devices, we can make the substitutions: 25-5 Electric Power

9. Copyright © 2009 Pearson Education, Inc. 25-5 Electric Power Example 25-8: Headlights. Calculate the resistance of a 40-W automobile headlight designed for 12 V.

10. Copyright © 2009 Pearson Education, Inc. What you pay for on your electric bill is not power, but energy – the power consumption multiplied by the time. We have been measuring energy in joules, but the electric company measures it in kilowatt-hours, kWh: 1 kWh = (1000 W)(3600 s) = 3.60 x 10 6 J. 25-5 Electric Power

11. Copyright © 2009 Pearson Education, Inc. 25-5 Electric Power Example 25-9: Electric heater. An electric heater draws a steady 15.0 A on a 120-V line. How much power does it require and how much does it cost per month (30 days) if it operates 3.0 h per day and the electric company charges 9.2 cents per kWh?

12. Copyright © 2009 Pearson Education, Inc. Current from a battery flows steadily in one direction (direct current, DC). Current from a power plant varies sinusoidally (alternating current, AC). 25-7 Alternating Current

13. Copyright © 2009 Pearson Education, Inc. The voltage varies sinusoidally with time: as does the current: 25-7 Alternating Current,,

14. Copyright © 2009 Pearson Education, Inc. Multiplying the current and the voltage gives the power: 25-7 Alternating Current

15. Copyright © 2009 Pearson Education, Inc. Usually we are interested in the average power: 25-7 Alternating Current.

16. Copyright © 2009 Pearson Education, Inc. The current and voltage both have average values of zero, so we square them, take the average, then take the square root, yielding the root-mean-square (rms) value: 25-7 Alternating Current

17. Copyright © 2009 Pearson Education, Inc. 25-7 Alternating Current Example 25-13: Hair dryer. (a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain?

18. Copyright © 2009 Pearson Education, Inc. Electrons in a conductor have large, random speeds just due to their temperature. When a potential difference is applied, the electrons also acquire an average drift velocity, which is generally considerably smaller than the thermal velocity. 25-8 Microscopic View of Electric Current: Current Density and Drift Velocity

19. Copyright © 2009 Pearson Education, Inc. 25-8 Microscopic View of Electric Current: Current Density and Drift Velocity We define the current density (current per unit area) – this is a convenient concept for relating the microscopic motions of electrons to the macroscopic current: If the current is not uniform:. (Remember the water in the pipe)

20. Copyright © 2009 Pearson Education, Inc. This drift speed is related to the current in the wire, and also to the number of electrons per unit volume: 25-8 Microscopic View of Electric Current: Current Density and Drift Velocity and

21. Copyright © 2009 Pearson Education, Inc. 25-8 Microscopic View of Electric Current: Current Density and Drift Velocity Example 25-14: Electron speeds in a wire. A copper wire 3.2 mm in diameter carries a 5.0- A current. Determine (a) the current density in the wire, and (b) the drift velocity of the free electrons. (c) Estimate the rms speed of electrons assuming they behave like an ideal gas at 20°C. Assume that one electron per Cu atom is free to move (the others remain bound to the atom).

22. Copyright © 2009 Pearson Education, Inc. 25-8 Microscopic View of Electric Current: Current Density and Drift Velocity The electric field inside a current-carrying wire can be found from the relationship between the current, voltage, and resistance. Writing R = ρ l /A, I = jA, and V = E l, and substituting in Ohm’s law gives:

23. Copyright © 2009 Pearson Education, Inc. 25-8 Microscopic View of Electric Current: Current Density and Drift Velocity Example 25-15: Electric field inside a wire. What is the electric field inside the wire of the earlier example? (The current density was found to be 6.2 x 10 5 A/m 2.)

24. Copyright © 2009 Pearson Education, Inc. In general, resistivity decreases as temperature decreases. Some materials, however, have resistivity that falls abruptly to zero at a very low temperature, called the critical temperature, T C. 25-9 Superconductivity Purely quantum mechanical; CANNOT be explained using classical physics.

25. Copyright © 2009 Pearson Education, Inc. Experiments have shown that currents, once started, can flow through these materials for years without decreasing even without a potential difference. Critical temperatures are low; for many years no material was found to be superconducting above 23 K. Since 1987, new materials have been found that are superconducting below 90 K, and work on higher temperature superconductors is continuing. 25-9 Superconductivity

26. Copyright © 2009 Pearson Education, Inc. A battery is a source of constant potential difference. Electric current is the rate of flow of electric charge. Conventional current is in the direction that positive charge would flow. Resistance is the ratio of voltage to current: Summary of Chapter 25

27. Copyright © 2009 Pearson Education, Inc. Ohmic materials have constant resistance, independent of voltage. Resistance is determined by shape and material: ρ is the resistivity. Summary of Chapter 25

28. Copyright © 2009 Pearson Education, Inc. Power in an electric circuit: Direct current is constant. Alternating current varies sinusoidally: Summary of Chapter 25

29. Copyright © 2009 Pearson Education, Inc. The average (rms) current and voltage: Relation between drift speed and current: Summary of Chapter 25

30. Copyright © 2009 Pearson Education, Inc. Chapter 26 DC Circuits

31. Copyright © 2009 Pearson Education, Inc. Electric circuit needs battery or generator to produce current – these are called sources of emf. Battery is a nearly constant voltage source, but does have a small internal resistance, which reduces the actual voltage from the ideal emf: 26-1 EMF and Terminal Voltage

32. Copyright © 2009 Pearson Education, Inc. This resistance behaves as though it were in series with the emf. 26-1 EMF and Terminal Voltage

33. Copyright © 2009 Pearson Education, Inc. 26-1 EMF and Terminal Voltage Example 26-1: Battery with internal resistance. A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, V ab, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.

34. Copyright © 2009 Pearson Education, Inc. A series connection has a single path from the battery, through each circuit element in turn, then back to the battery. 26-2 Resistors in Series and in Parallel

35. Copyright © 2009 Pearson Education, Inc. The current through each resistor is the same; the voltage depends on the resistance. The sum of the voltage drops across the resistors equals the battery voltage: 26-2 Resistors in Series

36. Copyright © 2009 Pearson Education, Inc. From this we get the equivalent resistance (that single resistance that gives the same current in the circuit): 26-2 Resistors in Series Unless an internal resistance r is specified assume V constant.

37. ConcepTest 26.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R

38. equal evenly 3 V drop Since the resistors are all equal, the voltage will drop evenly across the 3 resistors, with 1/3 of 9 V across each one. So we get a 3 V drop across each. ConcepTest 26.1aSeries Resistors I 9 V Assume that the voltage of the battery is 9 V and that the three resistors are identical. What is the potential difference across each resistor? 1) 12 V 2) zero 3) 3 V 4) 4 V 5) you need to know the actual value of R

39. ConcepTest 26.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V

40. ConcepTest 26.1bSeries Resistors II 12 V R 1 = 4  R 2 = 2  In the circuit below, what is the voltage across ? In the circuit below, what is the voltage across R 1 ? 1) 12 V 2) zero 3) 6 V 4) 8 V 5) 4 V The voltage drop across R 1 has to be twice as big as the drop across R 2.V 1 = 8 V The voltage drop across R 1 has to be twice as big as the drop across R 2. This means that V 1 = 8 V and V 2 = 4 V. Or else you could find the current I = V/R = (12 V)/(6  = 2 A, and then use Ohm’s law to get voltages. Follow-up: What happens if the voltage is 24 V?

41. Copyright © 2009 Pearson Education, Inc. A parallel connection splits the current; the voltage across each resistor is the same: 26-2 Resistors in Parallel

42. Copyright © 2009 Pearson Education, Inc. The total current is the sum of the currents across each resistor: 26-2 Resistors in Parallel,

43. Copyright © 2009 Pearson Education, Inc. This gives the reciprocal of the equivalent resistance: 26-2 Resistors in Parallel

44. Copyright © 2009 Pearson Education, Inc. An analogy using water may be helpful in visualizing parallel circuits. The water (current) splits into two streams; each falls the same height, and the total current is the sum of the two currents. With two pipes open, the resistance to water flow is half what it is with one pipe open. 26-2 Resistors in Parallel

45. Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Conceptual Example 26-2: Series or parallel? (a) The lightbulbs in the figure are identical. Which configuration produces more light? (b) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current.

46. ConcepTest 26.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5  R 2 = 2  1) 10 A 2) zero 3) 5 A 4) 2 A 5) 7 A

47. voltagesame V 1 = I 1 R 1 I 1 = 2 A The voltage is the same (10 V) across each resistor because they are in parallel. Thus, we can use Ohm’s law, V 1 = I 1 R 1 to find the current I 1 = 2 A. ConcepTest 26.2aParallel Resistors I In the circuit below, what is the current through ? In the circuit below, what is the current through R 1 ? 10 V R 1 = 5  R 2 = 2  1) 10 A 2) zero 3) 5 A 4) 2 A 5) 7 A Follow-up: What is the total current through the battery?

48. ConcepTest 26.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit?

49. ConcepTest 26.2bParallel Resistors II 1) increases 2) remains the same 3) decreases 4) drops to zero resistance of the circuit drops resistance decreasescurrent must increase As we add parallel resistors, the overall resistance of the circuit drops. Since V = IR, and V is held constant by the battery, when resistance decreases, the current must increase. Points P and Q are connected to a battery of fixed voltage. As more resistors R are added to the parallel circuit, what happens to the total current in the circuit? Follow-up: What happens to the current through each resistor?

50. Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Conceptual Example 26-3: An illuminating surprise. A 100-W, 120-V lightbulb and a 60-W, 120-V lightbulb are connected in two different ways as shown. In each case, which bulb glows more brightly? Ignore change of filament resistance with current (and temperature).

51. Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Example: Current in one branch. What is the current through the 500-Ω resistor shown?

52. Copyright © 2009 Pearson Education, Inc. 26-2 Resistors in Series and in Parallel Example 26-8: Analyzing a circuit. A 9.0-V battery whose internal resistance r is 0.50 Ω is connected in the circuit shown. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? Note: slight error in figure and text