1. Electricity Chapter 6

4. Fiat lux! (Then there was light!) The generator, battery, or power pack supplies the energy…

5. Fiat lux! (Then there was light!) …the wire carries a charge…

6. Fiat lux! (Then there was light!) …the electrical potential energy is converted into heat and light…

7. Fiat lux! (Then there was light!) …and the charge must be allowed to return to its source.

8. Fiat lux! (Then there was light!) The potential energy is measured in Volts.

9. Fiat lux! (Then there was light!) The charge is measured in Coulombs

10. Fiat lux! (Then there was light!) The power of the bulb is measured in Watts.

11. Fiat lux! (Then there was light!) The flow of charge is measured in Amperes.

12. How did you get your bulb to light? You made a circuit You provided the work. The generator converted it into electrical energy. Electrons (- charge) moved through the wires, heated up the filament in the bulb, and returned to the generator.

13. Things to measure: Charge Voltage Current Resistance Power Energy

14. Things to measure: Charge(q) Voltage(V) Current(I) Resistance(R) Power(P) Energy(E)

15. Things to measure: Charge(q) in Coulombs Voltage(V)in volts Current(I)in amps Resistance(R) in ohms Power(P) in Watts Energy(E) in Joules

16. Things to measure: Charge(q) in Coulombs(C) Voltage(V)in volts(V) Current(I)in amps(A) Resistance(R) in ohms(  ) Power(P) in Watts(W) Energy(E) in Joules(J)

17. Measure the current and voltage V=___ VI=____ A A V

18. Resistance --the resistance to the flow of charge Resistance=Voltage/Current

19. Resistance --the resistance to the flow of charge R=V/I

20. Resistance Measure the current and voltage. V=___ VI=____ A R=V/I=____ 

21. Warning! Not all bulbs are identical. Their resistance varies with brightness!

22. Try a different bulb. Measure the current and voltage. V=___ VI=____ A R=V/I=____ 

23. Make a table Circuit #VoltageCurrentResistance 1 2 3 4 5

24. A V A V A V A V A V

25. ?R V=6.0 V I=2.0 A

26. ?R V=12.0 V I=2.0 A V=6.0 V I=3.0 A V=12.0 V I=4.0 A V=9.0 V I=2.0 A

27. ?I V=6.0 V R=2.0  V=12.0 V R=2.0  V=6.0 V R=12.0  V=6.0 V R=4.0 

28. ?V I=6.0 A R=2.0  R=12.0  R=4.0  I=.50 A I=2.0 A I=.60 A

29. A V A V A V A V A V

30. Series and Parallel Measure the current and voltage of each circuit A V A V

31. Series and Parallel Which is which?

33. Series circuit Voltage is split between the bulbs. Current is the same Add the resistances

34. Parallel circuit Voltage is the same for each leg Current splits into each leg. Resistance goes down

35. Voltage is the same for each leg Current splits into each leg. Resistance goes down Voltage is split between the bulbs. Current is the same Add the resistances R Total =R 1 +R 2 1/R Total =1/R 1 +1/R 2

42. Power P=IV = I 2 R =V 2 /R

43. Power P=IV=(C/s)(J/C) = I 2 R=(C 2 /s 2 )(J/C)/(C/s) =V 2 /R=(J 2 /C 2 )/((J/C)/(C/s))

44. Power P=IV=(C/s)(J/C)=J/s = I 2 R=(C 2 /s 2 )(J/C)/(C/s)=J/s =V 2 /R=(J 2 /C 2 )/((J/C)/(C/s))=J/s

45. Power P=IV=(C/s)(J/C)=J/s=W = I 2 R=(C 2 /s 2 )(J/C)/(C/s)=J/s=W =V 2 /R=(J 2 /C 2 )/((J/C)/(C/s))=J/s=W

46. Power P=IV=(C/s)(J/C)=J/s=W = I 2 R=(C 2 /s 2 )(J/C)/(C/s)=J/s=W =V 2 /R=(J 2 /C 2 )/((J/C)/(C/s))=J/s=W The units of power are Watts

47. Home wiring A panel of circuit breakers allows only up to some maximum current through each circuit in the house. If you plug in too many things, it breaks the circuit.

48. A typical house: Houses are wired with 240 V. This is stepped down with a transformer to 120 V for most circuits. The highest power appliances (stove, AC, dryer, water heater) might use 240 V circuits.

49. A typical house: The US average electrical use in a home is about 1200 W. What causes variation in load?

50. A typical house: The US average electrical use in a home is about 1200 W. What causes variation in load? Most homes can carry over 100 A safely.

51. Typical home electrical power use

53. 67 ft 24 ft 47 ft 24 ft

54. AppliancePowerHours /week Total energy use

55. For example: Lighting Fluorescent bulbs: 15 @ 23 W each Incandescent bulbs:17 @ 60 W each and 10 @ 40 W each (DOE estimates 8 hrs/bulb/day, an immense overestimate for me) Average ~15 hrs/week

56. For example: Lighting 15 x 23 W = 345 W 17 x 60 W = 1020 W= 1765 W total 10 x 40 W = 400 W (Average ~15 hrs/week) 1765 W x 15 h= 27000 Wh= 27 kWh/week

57. Let us all estimate: Heat You will need about.25 kWh/ square foot/ week in the winter, half of that spring and fall. For me,.25 kWh / sq. ft x 2700 sq. ft = 680 kWh/week

58. Let us all estimate: Hot water You will need about 8 kWh/ adult / week, 4 kWh/ child / week, For me, 3 adult x 8 kWh/ adult / week= 24 kWh/wk 2 children x 4 kWh/ child / week= 8 kWh/wk 32 kWh/wk

59. By my estimates: My home uses about: Heating:340 kWh/week Lighting:27 kWh/week Hot water:32kWh/week Appliances:60kWh/week 500 kWh/week

60. Adjust your estimates based on choices: Use of heating / cooling Insulation / windows Use of lighting Use of appliances Choices of appliances Location of home Water conservation Multi-family buildings Passive or active solar

61. …and, a few words about heat.

62. Conduction ColdHot Objects in contact

63. Conduction ColdHot Not so cold Not so hot

64. Conduction ColdHot Not so cold Not so hot Hot (fast) particles collide with cool (slow) particles. The fast ones slow down while the slow ones speed up

65. Conduction ColdHot Not so cold Not so hot Fast (hot) particles collide with slow (cool) particles. The hot ones cool down while the cool ones warm up

66. Let’s try to warm up a cup of cold coffee. Step 1: Add heat.

67. Let’s try to warm up a cup of cold coffee. Step 1: Add heat. Well, that was easy.

68. Let’s try to warm up a cup of cold coffee. What if you add half as much heat?

69. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) b) c)

70. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) c)

71. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) Use half as much coffee (and cup) c)

72. Let’s try to warm up a cup of cold coffee. What if you add half as much heat? a) Raise the temperature only half as much. b) Use half as much coffee (and cup) c) Use a different substance

73. The effect of heat, q! When something warms up: The heat, q, depends on: The mass of the sample (m) The change in temperature (  T) The nature of the sample (C)

74. The effect of heat (q) When something warms up: The heat, q, depends on: The mass of the sample (m) The change in temperature (  T) The nature of the sample (C) C is the specific heat capacity for a given substance. Its units are (J/g o C)

75. q=mC  T q – heat, in Joules m –mass, in grams C –specific heat capacity, in J/g o C  T—change in temperature (T final -T initial )

76. C water =4.184 J/g o C Every substance has its own specific heat capacity

77. How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T

78. How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T = 50.g x 4.18 J/g o C x (80. o C-15 o C)

79. How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T = 50.g x 4.18 J/g o C x (80. o C-15 o C) = 50.g x 4.18 J/g o C x (65 o C)

80. How much heat? How much heat does it take to raise 50.g water from 15 o C to 80. o C? q=mC  T = 50.g x 4.18 J/g o C x (80. o C-15 o C) = 50.g x 4.18 J/g o C x (65 o C) =14000 J (14 kJ)

81. What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC

82. What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C )

83. What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C ) = 31 o C

84. What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C ) = 31 o C If the temperature starts at 25 o C, it will heat up to …

85. What is the change in temperature? If you add 1550 J to 12 g water, how much will it heat up?  T =q/mC  1550 J / (12 g x 4.18 J/g o C ) = 31 o C If the temperature starts at 25 o C, it will heat up to 56 o C

86. Calorimetry --the measurement of heat.

87. Calorimetry --the measurement of heat. If one thing gains heat…

88. Calorimetry --the measurement of heat. If one thing gains heat… …something else lost it.

89. If 75 g of a metal at 96 o C is placed in 58 g of water at 21 o C and the final temperature reaches 35 o C, what is the specific heat capacity of the metal?

90. Step 1 How much heat did the water gain?

91. Step 1 How much heat did the water gain? q=mC  T Mass of water, in grams Specific heat of water, 4.18 J/g o C Change in the temperature of water, in o C

92. Step 2 How much heat did the metal lose?

93. Step 2 How much heat did the metal lose? Heat lost = - heat gained q lost =-q gained

94. Step 3 What is the specific heat capacity of the metal?

95. Step 3 What is the specific heat capacity of the metal? C=q/m  T Mass of metal, in grams Specific heat of metal, in J/g o C Change in the temperature of metal, in o C Heat lost by metal

96. If 75 g of a metal at 96 o C is placed in 58 g of water at 21 o C and the final temperature reaches 35 o C, what is the specific heat capacity of the metal?.74 J/g o C

97. If 250 g of a metal at 96 o C is placed in 120 g of water at 26 o C and the final temperature reaches 33 o C, what is the specific heat capacity of the metal?

98. Efficiency Efficiency= Work accomplished Energy expended = Work out Work in Usually expressed as a %

99. ResistanceVoltageCurrentPower Total 6V 12 

100. ResistanceVoltageCurrentPower 1 2 Total 10  20  18V

101. ResistanceVoltageCurrentPower 1 2 Total 10  30  12V

102. ResistanceVoltageCurrentPower 1 2 3 Total 10  20  40  120V