6. The Electric Battery Alessandro Volta Ranked the relative difference produced by combining pairs of metals and created the first battery using zinc and silver discs with an acidic solution

7. The Electric Battery  The metals of a simple battery (or electric cell) are called electrodes

8. The Electric Battery  The metals of a simple battery (or electric cell) are called electrodes  The medium in which they rest is called an electrolyte

9. The Electric Battery  The metals of a simple battery (or electric cell) are called electrodes  The medium in which they rest is called an electrolyte  The exposed parts of the metals are called terminals

10. The Electric Battery  In operation, the electrolyte dissolves one terminal, whose ions enter the electrolyte and leave electrons behind on the terminal, which is now negatively charged. These ions cause the electrolyte to become positively charged, and this causes electrons of the other terminal to flow into the electrolyte, leaving that terminal with a positive charge.

11. The Electric Battery  The difference in charges of the terminals causes a potential difference and so chemical energy is made available as electrical energy.  Voila……. Voltage

12. The Electric Battery Battery voltage is additive when batteries are arranged in series.

13. The Electric Battery When batteries are arranged in parallel……..

14. The Electric Battery When batteries are arranged in parallel…….. Their power is increased accordingly

15. The Electric Battery When batteries are arranged in parallel…….. Their power is increased accordingly (their current capacity)

16. Electric Current A conducting pathway between battery terminals creates a circuit, where charges flows to create an electric current.

17. Electric Current Conventional current flows from the positive terminal through the pathway to the negative terminal……

18. Electric Current Conventional current flows from the positive terminal through the pathway to the negative terminal…… THIS IS OPPOSITE ELECTRON FLOW

19. Batteries, Current Flow and Resistance

20. Charges moving through material will encounter an internal resistance indigenous to the material.

21. Batteries, Current Flow and Resistance This internal resistance also is found in batteries.

22. Batteries, Current Flow and Resistance The collection of charges within the battery due to the chemical reaction of the electrodes and electrolyte is known as the electromotive Force of the source device (battery). emf = ε (volts)

23. Batteries, Current Flow and Resistance Therefore, ideally the terminal voltage of the battery (outside measurement) should equal the emf.

24. Batteries, Current Flow and Resistance Therefore, ideally the terminal voltage of the battery (outside measurement) should equal the emf. And according to Ohms Law this should equal I x R.

25. Batteries, Current Flow and Resistance Therefore, ideally the terminal voltage of the battery (outside measurement) should equal the emf. And according to Ohms Law this should equal I x R. Which it doesn’t.

26. HUH !!!

27. Batteries, Current Flow and Resistance As stated earlier, all electrical devices possess some internal resistance. This internal resistance (r) is constant as is the current when a load is applied to the circuit. Therefore, the voltage will drop in accordance to Ohms Law.

28. Batteries, Current Flow and Resistance ε - Ir V ab = ε - Ir

29. Batteries, Current Flow and Resistance ε – Ir V ab = ε – Ir Read pages 955 to 957 for a complete detailed explanation of this concept do problem 25.35 & 25.41 on page 975

30. Stopped Thursday

31. ε – Ir V ab = ε – Ir Look at it this way…. V = I x R Due to losses: ε – V = Ir ε – V ab = Ir

32. ε – V = IR ε – V ab = IR Now internal resistance is difference between emf and terminal voltage divided by current.

33. Some Electrifying Concepts

34.  Charges in motion are called electric currents.

35. Some Electrifying Concepts  Charges in motion are called electric currents.  The force that puts the charges in motion is called voltage.

36. Some Electrifying Concepts  Charges in motion are called electric currents.  The force that puts the charges in motion is called voltage.  Once the charges are in motion, there will exist a resistance to them.

37. Some Electrifying Concepts  Charges in motion are called electric currents. I  The force that puts the charges in motion is called voltage. V  Once the charges are in motion, there exists a resistance to them. R

38. This is analogous to any force and object scenario. Once force has pushed the object it must remain applied, or some part thereof, to overcome the natural and/or intentional resistances to motion. I.e.: car in motion or hydraulic jack

39. Current is the amount of charge that passes per unit of time at a point along the pathway. I = ΔQ / Δt

40. Some more on current......

41. Current is the amount of charge that passes per unit of time at a point along the pathway. I = ΔQ / Δt …….

42. The units of current (I) are: Coulombs / second Coulombs / second or amperes (amps, A)

43. Ohm’s Law  The rule governing the flow of electrons through an electric circuit is that of Ohm’s Law  Where: V = IR

44. Ohm’s Law V = IR  The flow of electrons (electric current: “I” or amps) is proportional to the potential difference (“V” or volts) that causes it.  And is also proportional to the resistance (“Ω” or omega) encounter within it’s pathway (circuit). (“Ω” or omega) encounter within it’s pathway (circuit).  A resistor is a device used in a circuit to reduce the amount of current.

45. Power Electric power can be converted into mechanical work, thermal energy or light.

46. Power  The power transferred by an electrical device is equal to the rate of energy transferred.  In an ideal device: energy in = energy out energy in = energy out

47. Power  Rather than Joules the unit of electrical power is equivalently expressed in Watts.  In practice, kilowatts are often used

48. Power  Power relationships: For current and voltageFor current and voltage P = I V

49. Power  Power relationships: For current and resistanceFor current and resistance P = I 2 R

50. Power  Power relationships: For voltage and resistanceFor voltage and resistance P = V 2 / R

51. Power A problem dealing with power

52. A Power Problem  Problem: An electric kettle contains 2 liters of water which it heats from 20 o c to boiling in 5 minutes. The supply voltage is 200 V and a kWh (kilowatt-hour) unit costs 2 cents.

53. A Power Problem  Problem: An electric kettle contains 2 liters of water which it heats from 20 o c to boiling in 5 minutes. The supply voltage is 200 V and a kWh (kilowatt-hour) unit costs 2 cents. Calculate: 1.the power consumed (assume that heat loss is negligible) 2.the cost of using the kettle under these conditions six times 3.the resistance of the heating element 4.the current in the element

54. A Power Problem Consider:  heat gained: H = mc (t2 – t1) [ where c is the specific heat of water.] H = mc (t2 – t1) [ where c is the specific heat of water.]  The specific heat of water is: 4.18 J / g * C  Density of water is 1

55. A Power Problem Solution part a: Since heat losses are neglected, conservation of energy mandates that electrical energy supplied to the kettle equals heat generated by the kettle, which just so happens to be “H” in this case.

56. A Power Problem Solution part a: Since heat losses are neglected, conservation of energy mandates that electrical energy supplied to the kettle equals heat generated by the kettle, which just so happens to be “H” in this case. H = mc (t2 – t1) H = m x ( 4.18 J / g C) x ((100 – 20) o C)

57. A Power Problem Solution part a: H = mc (t2 – t1) H = m x ( 4.18 J / g C) x ((100 – 20) o C) mass = ?? mass = ??

58. A Power Problem Solution part a: Density of water = 1 g/ml 2 liter = 2000 ml = 2000 g mass = 2000 g mass = 2000 g

59. A Power Problem Solution part a: Density of water = 1 g/ml 2 liter = 2000 ml = 2000 g mass = 2000 g mass = 2000 g H = mc (t2 – t1) H = 2000g x ( 4.18 J / g C) x ((100 – 20) o C) [ notice how the units cancel ]

60. A Power Problem Solution part a: Density of water = 1 g/ml 2 liter = 2000 ml = 2000 g mass = 2000 g mass = 2000 g H = mc (t2 – t1) H = 2000g x ( 4.18 J / g C) x ((100 – 20) o C) [ notice how the units cancel ] therefore: H = 6.69 x 10 3 Joules this is energy

61. A Power Problem Solution part a: H = 6.69 x 10 3 Joules this is energy The power is the energy consumed per second, which is:

62. A Power Problem Solution part a: H = 6.69 x 10 3 Joules this is energy The power is the energy consumed per second, which is: P = H / t P = 6.69 x 10 3 Joules / (5 x 60 seconds)

63. A Power Problem Solution part a: H = 6.69 x 10 3 Joules this is energy The power is the energy consumed per second, which is: P = H / t P = 6.69 x 10 3 Joules / (5 x 60 seconds) P = 2.23 x 10 3 J/s or 2.23 kilowatts

64. A Power Problem Solution part b: P = 2.23 kW The kettle uses 2.23 kW for 5 minutes each time it is boiled. When it is used 6 times, 2.23 kW is used for 30 minutes or ½ hour. The cost is thus:

65. A Power Problem Solution part b: P = 2.23 kW The kettle uses 2.23 kW for 5 minutes each time it is boiled. When it is used 6 times, 2.23 kW is used for 30 minutes or ½ hour. The cost is thus: 2.23 kW x ½ hour x 2 cents / kWh = 2.23 cents cost = 2.23 cents

66. A Power Problem Solution part c: P = 2.23 kW cost = 2.23 cents the power consumed is 2.23 cents and the supply voltage is 200 V. But P = V 2 / R where R is the kettles heating element

67. A Power Problem Solution part c: P = 2.23 kW cost = 2.23 cents the power consumed is 2.23 cents and the supply voltage is 200 V. But P = V 2 / R where R is the kettles heating element R = V 2 / P R = (200v) 2 / 2.23 x 10 3 w R = 17.9 Ω

68. A Power Problem Solution part d: P = 2.23 kW cost = 2.23 cents R = 17.9 Ω I = 11.2 A Power is also I x V. Therefore, the current is:

69. A Power Problem Solution part d: P = 2.23 kW cost = 2.23 cents R = 17.9 Ω I = 11.2 A Power is also I x V. Therefore, the current is: I = P / V I = 2.23 x 10 3 w / 200 V I = 11.2 A

70. A Power Problem Solutions: P = 2.23 kW cost = 2.23 cents R = 17.9 Ω I = 11.2 A End of Problem

71. Deriving the Power Formulas AP  Given Ohms Law and that P = I * V, derive the two power formulas you were just given.

72. Deriving the Power Formulas AP Numero Uno 1. V=IR and P=IV 2. First: V=IR 3. Sub: P=IxIxR 4. Thus: P=I 2 R

73. Deriving the Power Formulas AP Numero Dos 1. V=IR and P=IV 2. First: I=V/E 3. Sub: P=VxV/R 4. Thus: P=V 2 /R

74. End of PowerPoint