1. Electric Potential

2. Let’s review... 10 kg 12 m KE + PE g = E

3. Force by Hand? 10 kg 12 m KE + PE g = E W E,b = 100N N R,b = 100N a = 0 Work ext = F ext  x = 100N  m  =  J Work ext ?

4. Which buckets were filled? 10 kg 12 m KE + PE g = E WORK  PE g = mgh

5. Energy sloshes 10 kg 12 m KE + PE g = E 10 kg

6. B A X q test =+1 C

7. Like Gravitational potential energy a charged object can have potential energy due to it’s location in an electric field. Just like work is required to lift an object, work is required to push a charged particle against the electric field of a charged body.

8. Electrical potential = electric potential energy divided by charge V = PE/q 1 volt = 1 Joule/Coulomb PE = q V

9. Current, voltage and resistance

10. Consider a flashlight ….. battery

11. Wiring Diagram or a Schematic Battery Resistor Switch

12. Predict the relative brightness

13. CURRENT: Measure of the rate of flow of charge. I =  q/  t Measured in coulombs/second, which we define as the Ampere (A).

14. How do we measure current? use an ammeter How do we insert it? I = ? A What drives the current?

15. Electromotive Force (EMF) The battery acts like a (charge) pump. EMF is the energy that one coulomb of charge gains in passing through a battery. Measured in J/C or volts. We’ll just call this the “battery voltage” Also given the symbol  batt 

16. EMF and Voltage Loses 12 J/C  batt = 12 volts A B  V AB =  J/C D C  V CD =  J/C

17. Loses 12 J/C  = 12 volts

18. Loses 12 J/C  = 12 volts

19. Loses 12 J/C  = 12 volts

20. Loses 12 J/C  = 12 volts

21. Loses 12 J/C  = 12 volts

22. Loses 12 J/C  = 12 volts

23. Loses 12 J/C  = 12 volts

24. Loses 12 J/C  = 12 volts

25. Loses 12 J/C  = 12 volts

26. Loses 12 J/C  = 12 volts

27. How do we measure voltage? Voltage is a measure of the electric potential. use a voltmeter How do we insert it? V 10.000 A.001 A9.999 A  V A,B A B

28. Consider a simple circuit What if we increase the push? more push means more flow What if we change the bulb? not all identical pushes produce identical flows

29. Resistance... …is futile

30. Resistance Represents the ability of a circuit element to impede the flow of current. R =  V/I volt/ampere = ohm  Georg S. Ohm (1789-1854) V = I R

31. Which has more resistance? steel wood

32. Which has more resistance? steel

33. Which has more resistance? steel

34. Which has more resistance? Steel at 0 o C Steel at 25 o C

35. More flow more glow! AC Delco -+ adjustable battery

36. R R V batt = I batt R circuit

37. AB indicator bulb Which box contains the greater resistance?

38. AB Which is box A and which is box B? R R R R B A

39. AB The bigger the R the smaller the current R R R R B A through the battery. of the circuit,

41. Junction

42. R R V batt = I batt R circuit

43. AB Which is box A and which is box B? R R A B

44. Adding in Series What happens to I batt if a bulb or network of bulbs is inserted between these two bulbs? In all cases, it goes down. Why? Clog up an existing path, increase the resistance of the circuit, and decrease the current through the battery. A B

45. Summary When a bulb or network of bulbs is added in series, the resistance of the circuit increases. The important thing is not what is added but how it is added. If you have to break the circuit, you are adding something in series. A B

46. Adding in Parallel What happens to I batt when a bulb (or network) is added in parallel to bulb B? A B A It goes up. Why? Add a new path (opportunity), decrease the resistance of the circuit, and increase the current through the battery.

47. Summary When a bulb or network of bulbs is added in parallel, the resistance of the circuit decreases. Again, the important thing is not what is added but how it is added. If you don’t have to break the circuit, you are adding something in parallel. A B A

48. AB Which is box A and which is box B? R R A B

49. AB R R B A

50. Rank all the bulbs below for brightness. Explain your reasoning. A CB A>B=C A gets all of the current while B and C each get half of total the current. B and C share equally since each branch has equal resistance. Be the river

51. Rank the bulbs Rank all the bulbs below for brightness. Explain your reasoning. Batteries and bulbs are all the same. A CB F ED A=F>B=C=D=E Both circuits have the same current. Which comes first does not matter.

52. Current favors the branch of least resistance. C<D Less current will take the left branch since it has more resistance than the right branch. Which is brighter, C or D? Explain. A C B D Ranking: A>D>B=C Be the river

53. A C B D E E>A=B>C=D Be the river

54. Short Circuits A B What happens to the brightness of each bulb when the switch is closed? SP 117

55. Rank the Resistances A C R R

56. B C A R R R

57. The need for more A CB Closed:A>B=C Open:A=B, C out What happens to B as switch is opened?

58. Review of Current Model  More flow more glow.  The current through the battery depends on the resistance of the circuit.  It’s not what you add but how you add it.  Current favors the path of least resistance.  Branches connected directly to a battery are independent.

59. Add a bulb in series What about the currents now?

60. 12 J/C

61. 6 J/C What do we mean by sharing?

62.  = 12 volts

74.  = 12 volts

75. Start Here

77.  PE/person = 75 kg x 10 N/kg x 1000m  PE/person = 750,000 J  PE/person = ?

78. Start Here  PE/person = 75 kg x 10 N/kg x 1000m  PE/person = 750,000 J  PE/person = ? R R I I

79. What if there is more than one path? How bright are the bulbs? 12 J/C

80. What if there is more than one path? 12 J/C

81. What if there is more than one path? 6 J/C What about the currents?

82. Voltage Model Voltage: how much energy each coulomb of charge gains (battery) or loses (bulb) in going through an element. More voltage, more glow. More volt, more jolt. What goes up, must come down! voltage rises and drops must be equal around any loop.

83. Applying The Voltage Model  =12 V Rank the bulbs.

84. Applying The Voltage Model  =12 V Rank the bulbs. VaVa >6 volts =6 volts <6 volts VbVb VcVc V a + V b + V c = 12 volts -or- V a + V b = 12 volts

85. Applying The Voltage Model  =12 V Which meter reads more? VbVb VcVc VaVa

86. Applying The Voltage Model  =12 V Which meter reads more? VbVb VcVc VaVa VV VV Which network has the greater resistance? R R

87. Applying The Voltage Model  =12 V For series networks, the voltage drop is greater across the larger R. VbVb VcVc VaVa VV VV R R

88. Voltage Model Voltage: how much energy each coulomb of charge gains (battery) or loses (bulb) in going through an element. More voltage, more glow. More volt, more jolt. What goes up, must come down! voltage rises and drops must be equal around any loop.

89. Voltage Model The voltage for series elements DIVIDES. The voltage divides equally for identical Rs. Otherwise, the voltage drop is greater across the greater/lesser R. A B  R.

90. Be careful with your words... 2  4  12 volts 200  4 volts 8 volts The greater the resistance,...the greater the voltage. the share of 6 volts

91. ParallelSeries I I/2  I I    parallel series currentvoltage splitssame divides