1. Static Electricity vs. Electric Current Static electricity is a one time event. Electric current is a constant flow.

2. Electric current (I) is the rate at which charge passes a given point in a electric circuit. An electric circuit is a closed path along which charged particles move.

3. Current I = Δq t I = amps (A) An ammeter is a device used to measure current.

4. Determine the current for these situations: 1. A cross section of wire is isolated and 20 C of charge are determined to pass through it in 40 s. 20 C 40 s I = Δq t = 20 C 40 s =.5 A 2. A cross section of wire is isolated and 2 C of charge are determined to pass through it in.5 s. 2 C.5 s I = Δq t = 2 C.5 s = 4 A

5. Requirements necessary for an electric current: 1.There must be a closed conducting path which extends from the positive terminal to the negative terminal of a cell or battery (combination of cells).

6. Requirements necessary for an electric current: 2. There must be a difference in electric potential between the two end points in the circuit. The potential difference may be supplied by a cell or battery. In other words you need an energy source. The potential difference or voltage of a circuit can be measured using a voltmeter.

7. Batteries provide the voltage (electric pressure) necessary to move current through a circuit.

8. Light Bulbs Using only a light bulb, a wire, and a battery come up with 4 ways to create a complete circuit so that the light bulb lights up.

9. You need to create a closed circuit.

10. Conventional flow vs. THE TRUTH Ben Franklin envisioned positive charges as the carriers of charge. Because of this an early convention for the direction of an electric current was established to be in the direction which positive charges would move. The direction of an electric current is by convention the direction in which a positive charge would move. However, we now know that the positive charges (protons) are NOT the charged particles that are moving. IT’S THE ELECTRONS. And the negatively charged electrons would be moving in the opposite direction.

11. Alternating CurrentDirect Current AC - The electrons in alternating current flow in one direction, then in the opposite direction—over and over again. Electricity from a power plant is alternating current. DC - The electrons in direct current flow in one direction. The current produced by a battery is direct current.

12. In the United States, the current flow alternates 120 times per second. (In Europe it alternates 100 times per second.) The current supplied to your home by the local utility is alternating current.

13. War of Currents DC vs. AC Edison carried out a campaign to discourage the use of alternating current, including spreading disinformation on fatal AC accidents, publicly killing animals, and lobbying against the use of AC in state legislatures. Edison directed his technicians to preside over several AC-driven killings of animals, primarily stray cats and dogs but also unwanted cattle and horses. Acting on these directives, they were to demonstrate to the press that alternating current was more dangerous than Edison's system of direct current. EDISON WESTINGHOUSE TESLA

14. Warmup: On the circuit drawing below label conventional current flow and electron flow. Why is conventional current flow technically the “wrong” way?

15. How many electrons flow through a battery that delivers a current of 3 A for 12 s? (HINT: 1 C = 6.25 x 10 18 e) I = q t tI = qt t q = It q = (3A)(12s) q = 36 C #e = (36C)(6.25 x 10 18 e) #e = 2.25 x 10 20

16. Resistance (I) Current is the flow of electrons through a circuit. (R) Resistance discourages or controls the flow. It is caused by “things” that get in the way of a direct path. (V) Electric potential difference (voltage- like a battery) is necessary for current to flow. (Electric Pressure). The resistance of a conductor (like a wire) is the ratio of the potential difference applied to the circuit and the current that flows through it…

17. This is Ohm’s Law… R = V I R = Volts Amps R = ohm (Ω)

18. Resistors are “things” in a circuit that limit current flow. They can be controlled resistors or electrical devices like light bulbs or lamps.

19. Circuit Analogy The pipe is the counterpart of the wire in the electric circuit. The pump is the mechanical counterpart of the battery. The pressure generated by the pump, that drives the water through the pipe, is like the voltage generated by the battery to drive the electrons through the circuit. The seashells plug up the pipe and constrict the flow of the water creating a pressure difference from one end to the other. In a similar manner the resistance in the electric circuit resists the flow of electricity and creates a voltage drop from one end to the other. Energy is lost across the resistor and shows up as heat. R = V I

20. Try this… A student measures a current of.10 A flowing through a light bulb connected by short wires to a 12 V battery. What is the resistance of the light bulb? R = V I R = 12 V.10 A R = 120 Ω

21. Try this… A lamp with a resistance of 20 Ω is in a circuit that has a current of.05 A flowing through it. What is the potential difference across the lamp? R = V I I R = V I I V = I R V = (.05 A)(20 Ω) V = 1 V

22. Recall slope…What are the slopes for the following graphs? d t v t F x F a v = d t velocity acceleration a = v t spring constant k = F x m = F a mass

23. Guess what the slope is for this graph: V I Resistance (R) R = V I

24. Wires made of a certain types of metals are used in circuits because they are good conductors and allow the electrons within them to move relatively freely. Although wires allow current to flow through a circuit they also control the current or have some resistance.

25. Factors that affect the resistance of a wire: 1.LENGTH: Increasing the length (L) of the wire will increase the resistance of the wire. This is because the current (electrons) will now have further to travel and will encounter and collide with an increasing number of atoms.

26. Factors that affect the resistance of a wire: 2. AREA: Increasing the cross-sectional area (A) of a wire will decrease the wires resistance. This occurs because in making the wire thicker there is now more spaces between atoms through which the electrons can travel and thus flow easier. The wire isn’t resisting the flow as much. A = πr 2

27. Factors that affect the resistance of a wire: 3.RESISTIVITY (ρ) – this is a characteristic of a material that depends on its electronic structure and temperature. If a wire is made of a material that has a high resistivity then it will have a high resistance.

28. Combining all these factors gives us the following equation for the resistance of a wire: R = ρL A Short/thick/cold wires = low resistance (easy for electrons to flow) Long/thin/hot wires = high resistance (hard for electrons to flow) http://phet.colorado.edu/sims/resistance-in-a-wire/resistance-in-a-wire_en.html

29. Try this… Determine the resistance of a 4.00 m length of copper wire having a diameter of 2 mm. Assume the temperature of 20°C. 4 m d= 2 mm ρ copper = 1.72 x 10 -8 Ωm R = ρL A = (1.72 x 10 -8 Ωm)(4m) π(.001m) 2 =.0219 Ω

30. Determine the length of a copper wire that has a resistance of.172 Ω and cross-sectional area of 1 x 10 -4 m 2. The resistivity of copper is 1.72 x 10 -8 Ωm. AR = ρLA A AR = ρL ρ ρ L = AR ρ L = AR ρ L = (.001m 2 )(.172Ω) (1.72 x 10 -8 Ωm) L = 10,000 m

31. Which one of the five wires has the largest resistance? WireMaterialLengthDiameter Airon2 m6.4 x 10 -4 m Bcopper2 m6.4 x 10 -4 m Ccopper2 m1.2 x 10 -3 m Dcopper1 m1.2 x 10 -3 m EIron2 m1.2 x 10 -3 m ρ iron = 9.7 x 10 -8 Ωmρ copper = 1.7 x 10 -8 Ωm High resistance = long/thin/hot wires Wire A

32. WireMaterialLengthDiameter Airon2 m6.4 x 10 -4 m Bcopper2 m6.4 x 10 -4 m Ccopper2 m1.2 x 10 -3 m Dcopper1 m1.2 x 10 -3 m EIron2 m1.2 x 10 -3 m ρ iron = 9.7 x 10 -8 Ωmρ copper = 1.7 x 10 -8 Ωm Of the five wires, which one has the smallest resistance? Small resistance = short/thick/cold Wire D

33. WireMaterialLengthDiameter Airon2 m6.4 x 10 -4 m Bcopper2 m6.4 x 10 -4 m Ccopper2 m1.2 x 10 -3 m Dcopper1 m1.2 x 10 -3 m EIron2 m1.2 x 10 -3 m ρ iron = 9.7 x 10 -8 Ωmρ copper = 1.7 x 10 -8 Ωm Which of the wires carries the smallest current when they are connected to identical batteries? How does current (I) relate to resistance (R)? R = V I When current is low, resistance is high. High resistance = long/thin/hot wires Wire A

34. Resistance Variables Lab Coil #MetalLength (meters) Cross Sectional Area _____ Gauge V Volts (V) I Current (A) R Resistance (Ω) 1 Copper 10 m.325 mm 2 22 2 Copper 10 m.081 mm 2 28 3 Copper 20 m.325 mm 2 22 4 Copper 20 m.081 mm 2 22 5 Copper- Nickel 10 m.325 mm 2 22 First use R = ρL/A to predict the resistance of each circuit. ρ cop-nick = 4.9 x 10 -8 Ωmρ copper = 1.7 x 10 -8 Ωm

35. Schematic Circuit Diagrams There are many different ways to represent circuits. Below is an artist’s drawing and a schematic drawing. We will be using schematic drawings with symbols.

36. Types of Circuits: Series A series circuit can be constructed by connecting light bulbs in such a manner that there is ONE PATH for charge flow (CURRENT); the bulbs are added to the SAME LINE with no branching point- charge passes through every light bulb. Since there is only one current path in a series circuit, the current is the same through each resistor. I = I 1 = I 2 = I 3

37. Types of Circuits: Series I = I 1 = I 2 = I 3 If the current is the same everywhere within a series circuit then what is changing? V = I R http://phet.colorado.edu/sims/ohms-law/ohms-law_en.html

38. Types of Circuits: Series Let’s look at resistance: There are 3 resistors in this circuit. The total resistance of the entire circuit is called the equivalent resistance (R eq ). If we could replace all 3 resistors with just 1 resistor it would have the equivalent resistance of the entire circuit. R eq = R 1 + R 2 + R 3

39. Types of Circuits: Series Let’s look at voltage: The cell or battery of a circuit provides the maximum voltage allowed for the circuit. It you have a 12 V battery hooked up to your circuit then there is 12 V of “electrical pressure” pushing the current through the circuit. Current moves from an area of high pressure (12 V at the positive terminal of the battery) to an area of low pressure (0 V at the negative terminal of the battery). This is why you may often hear the term “voltage drop” across resistors. The voltage is decreasing as the current passes through resistors in the circuit because everything is together on one path. 12 V V = V 1 + V 2 + V 3

40. Series Summary: I = I 1 = I 2 = I 3 R eq = R 1 + R 2 + R 3 V = V 1 + V 2 + V 3

41. 30 Ω 10 Ω In this animation you should notice the following things: The battery or source is represented by an escalator which raises charges to a higher level of energy. As the charges move through the resistors (represented by the paddle wheels) they do work on the resistor and as a result, they lose electrical energy. The charges do more work (give up more electrical energy) as they pass through the larger resistor. By the time each charge makes it back to the battery, it has lost all the energy given to it by the battery. The total of the potential drops ( - potential difference) across the resistors is the same as the potential rise ( + potential difference) across the battery. This demonstrates that a charge can only do as much work as was done on it by the battery. The charges are positive so this is a representation of Conventional Current (the apparent flow of positive charges) The charges are only flowing in one direction so this would be considered direct current ( D.C. ).

42. Try this… 36 V 4 Ω 8 Ω 6 Ω Determine the (a) equivalent resistance of the entire circuit, (b) the current through each resistor, and (c) the potential drop across each resistor. I = I 1 = I 2 = I 3 R eq = R 1 + R 2 + R 3 V = V 1 + V 2 + V 3 a.R eq = R 1 + R 2 + R 3 R eq = 4 Ω + 6 Ω + 8 Ω R eq = 18 Ω b. I = I 1 = I 2 = I 3 I = V R eq I = 36 V 18 Ω I = 2 A

43. 36 V 4 Ω 8 Ω 6 Ω c. I = 2 A V = I R V 1 = I R 1 V 1 = (2A)(4 Ω) V 1 = 8 V V 2 = I R 2 V 2 = (2A)(6 Ω) V 2 = 12 V V 3 = I R 3 V 3 = (2A)(8 Ω) V 3 = 16 V V = V 1 + V 2 + V 3 V =8V +12V+ 16V V = 36 V

44. Try this… 24 V 4 Ω 6 Ω A 2 A The diagram shows a circuit with three resistors. What is the resistance of resistor R 3 ? I = I 1 = I 2 = I 3 R eq = R 1 + R 2 + R 3 V = V 1 + V 2 + V 3 R eq = V I R eq = 24 V 2 A R eq = 12 Ω R 3 = R eq – R 1 – R 2 R eq = R 1 + R 2 + R 3 R 3 = 12 Ω – 4 Ω – 6 Ω R 3 = 2 Ω

45. Light bulbs in Series Remember how a light bulb works? What will happen to LB 2 if LB 1 blows out? LB 1LB 2 The circuit is no longer closed- LB 2 will also go out.

46. Christmas lights are connected in series. If one goes out they should all go out. But small Christmas light bulbs are designed so this doesn’t happen. Each bulb has a "shunt" of several turns of tiny wire inside the bulb near the bead. The shunt is intended to conduct current when the filament fails.

47. Electric Power (Watts): 3 different ways P = V I From Ohm ’ s Law V = I R P = I R I P = I 2 R P = V I I = V R P = V V R P = V 2 R Power is the rate at which electrical energy is supplied to a circuit or consumed by a load. Remember: 1 W = 1 J/s

48. Try this… A potential difference of 60 V is applied across a 15 Ω resistor. What is the power dissipated in the resistor? P = 240 W P = (60 V) 2 15 Ω P = 3600 V 2 15 Ω P = V 2 R

49. Try these… Determine the... a.... current in a 60-watt bulb plugged into a 120-volt outlet. b.... current in a 120-watt bulb plugged into a 120-volt outlet. c.... power of a saw that draws 12 amps of current when plugged into a 120-volt outlet. d.... power of a toaster that draws 6 amps of current when plugged into a 120-volt outlet. e.... current in a 1000-watt microwave when plugged into a 120-volt outlet I = P / V = (60 W) / (120 V) = 0.5 A I = P / V = (120 W) / (120 V) = 1.0 A P = V I = (120 V) (12 A) = 1440 W P = V I = (120 V) (6 A) = 720 W I = P / V = (1000 W) / (120 V) = 8 A

50. Electrical Energy (J) W = Pt P = I 2 R W = I 2 Rt W = V 2 t R or P = V 2 R Remember: 1 J = 1 Nm = 1 kg m 2 /s 2

51. Try this… A current of.4 A is measured in a 150 Ω resistor. How much energy is expended by the resistor in 30 seconds? W = I 2 Rt W = (.4) 2 (150 Ω)(30s) W = (.16)(150 Ω)(30s) W = 720 J

52. Try this… You are given 10 Ω, 20 Ω, and 30 Ω resistors, and a 3V battery. a.Draw a schematic diagram of these components in a series circuit. b.Determine the equivalent resistance of the circuit. c.Determine the current through each resistor. d.Determine the potential difference across each resistor. e.Determine the power dissipated by the 20 resistor.

53. Try this… Consider the following series circuit: a.What is the potential difference between: a and b, b and c, c and d, d and a. b.What is the current in the circuit? c.What power is dissipated by each resistor?

54. Types of Circuits: Parallel A parallel circuit can be constructed by connecting light bulbs in such a manner that there are SEVERAL PATHS for charge flow (CURRENT); the light bulbs are placed within a separate branch line, and a charge moving through the circuit will pass through only one of the branches during its path back to the low potential (negative) terminal of the battery. Since there is now several current paths, the total current of the circuit equals the sum of the current in each branch. I = I 1 + I 2 + I 3

55. Types of Circuits: Parallel Let’s take a closer look at how the current is flowing through each branch. 12 A 3 A 9 A 3 A 6 A 3 A 9 A 3 A 12 A Does this hold true? I = I 1 + I 2 + I 3

56. Types of Circuits: Parallel Let’s look at voltage: Each branch is hooked up to the same battery. Each branch has the same voltage (electric pressure). V = V 1 = V 2 = V 3 Outlets in a house are connected in parallel so you can use one appliance without having to turn them all on.

57. Types of Circuits: Parallel Let’s look at RESISTANCE: The actual amount of current always varies inversely with the amount of overall resistance. I = V R eq 1 = 1 + 1 + 1 R eq R 1 R 2 R 3

58. 1 = 1 + 1 + 1 R eq R 1 R 2 R 3 The total resistance of a parallel circuit is a fraction of all of the resistors added together!!!! Types of Circuits: Parallel

59. I = V R eq So the more resistors you add in parallel the lower the equivalent resistance will be. And because R eq  the I  sometimes to a potentially dangerous level. http://phet.colorado.edu/sims/ohms-law/ohms-law_en.html This is why we have circuit breakers or fuses inserted into the main line of our homes. When we are trying to use too many electrical devices (resistors) in parallel the current gets to be too much and a fuse in blown.

60. Inside the fuse is a small piece of metal, across which current must pass. During normal flow of current, the fuse allows the current to pass unobstructed. But during an unsafe overload, the small piece of metal melts, stopping the flow of current. Circuit breakers are switches that are tripped when the current flow passes a unsafe limit. The excess of current typically triggers an electromagnet, which trips the circuit breaker when an unsafe limit is reached. Once tripped, the switches simply turn off. That stops the flow of electricity, which will remain off until the switch is reset.

61. Parallel Summary: I = I 1 + I 2 + I 3 V = V 1 = V 2 = V 3 1 = 1 + 1 + 1 R eq R 1 R 2 R 3

62. In this animation you should notice the following things: More current flows through the smaller resistance. (More charges take the easiest path.) The battery or source is represented by an escalator which raises charges to a higher level of energy. As the charges move through the resistors (represented by the paddle wheels) they do work on the resistor and as a result, they lose electrical energy. By the time each charge makes it back to the battery, it has lost all the electrical energy given to it by the battery. The total of the potential drops ( - potential difference) of each "branch" or path is the same as the potential rise ( + potential difference) across the battery. This demonstrates that a charge can only do as much work as was done on it by the battery. The charges are positive so this is a representation of conventional current (the apparent flow of positive charges) The charges are only flowing in one direction so this would be considered direct current ( D.C. ). 30 Ω10 Ω

63. Series vs. Parallel SeriesParallel I V R I = I 1 + I 2 + I 3 V = V 1 = V 2 = V 3 1 = 1 + 1 + 1 R eq R 1 R 2 R 3 I = I 1 = I 2 = I 3 R eq = R 1 + R 2 + R 3 V = V 1 + V 2 + V 3 The more resistors you have the less the equivalent resistance is, and current increases. Voltage never changes. The more resistors you have the more the equivalent resistance is, and current never changes. The voltage across each resistor adds up to the total voltage of the source (battery).

64. Hooking up Devices Properly Ammeters, because of the way they are built, have very little resistance. So they can be placed in series with other devices in a circuit and not disrupt the current. A Voltmeters on the other hand have a very high resistance and when placed in series would disrupt the current so they are placed in parallel within the circuit- separate branch. V

65. Try this… Calculate (a) the equivalent resistance, (b) the potential difference across each resistor, and (c) the current through each resistor. 12 V 4 Ω6 Ω12 Ω I = I 1 + I 2 + I 3 V = V 1 = V 2 = V 3 1 = 1 + 1 + 1 R eq R 1 R 2 R 3 a. 1 = 1 + 1 + 1 R eq R 1 R 2 R 3 1 = 3 + 2 + 1 R eq 12 Ω 12 Ω 12 Ω Need a common denominator 1 = 1 + 1 + 1 R eq 4 Ω 6 Ω 12 Ω 1 = 6 R eq 12 Ω R eq = 12 Ω 1 6 = 2 Ω NOTE: Notice how the R eq is less than any of the resistors in the circuit.

66. 12 V 4 Ω6 Ω12 Ω b. V = V 1 = V 2 = V 3 V = 12 V c.I = V R I 1 = V R 1 I 1 = 12 V 4 Ω I 1 = 3 A I 2 = V R 2 I 2 = 12 V 6 Ω I 2 = 2 A I 3 = V R 3 I 3 = 12 V 4 Ω I 3 = 3 A

67. Try this… Find the magnitude of current that is flowing through all 3 ammeters. I = I 1 + I 2 + I 3 V = V 1 = V 2 = V 3 1 = 1 + 1 + 1 R eq R 1 R 2 R 3 A1A2 A3 12 V 10 Ω15 Ω A1 = I 1 = V R 1 I 1 = 12 V 10 Ω I 1 = 1.2 A A2 = I 2 = V R 2 I 2 = 12 V 15 Ω I 2 =.8 A A3 = I 1 + I 2 A3 =1.2A +.8A A3 = 2 A

68. Which two of the resistor arrangements below have the same equivalent resistance? A B D C 1 Ω 8 Ω 2 Ω

69. Which circuit below would have the lowest voltmeter reading? 6 V 20 Ω 40 Ω V 6 V 20 Ω 40 Ω V 6 V 20 Ω 40 Ω V 6 V 20 Ω 40 Ω V A BD C

70. Arrange the schematic diagrams below in order of increasing equivalent resistance. 1234

71. Find the resistance of R 3. R 1 = 6 ΩR 2 = 6 ΩR 3 = ? R eq = 2 Ω 1 = 1 + 1 + 1 R eq R 1 R 2 R 3

72. Simplifying Resistors in Combination Circuits 3Ω3Ω 11Ω

73. 4 Ω 18 Ω

74. 4 Ω 18 Ω

75. Combo Circuits Quiz

76. Now we can analyze other aspects of the circuit…in order to do this we must first simplify. R 1 = 1 Ω R 2 = 8 Ω R 3 = 8 Ω 1 = 1 + 1 R 2,3 R 2 R 3 1 = 1 + 1 = 2 = 4 Ω R 2,3 8 Ω 8 Ω 8 Ω R 1 = 1 ΩR 2,3 = 4 Ω 100 V 0 V 100 V0 V

77. Next simplified the circuit down to one resistor. R 1 = 1 ΩR 2,3 = 4 Ω 100 V0 V R T = R 1 + R 2,3 R T = 1 Ω + 4 Ω R T = 5 Ω 100 V0 V

78. We went from R 1 = 1 Ω R 2 = 8 Ω R 3 = 8 Ω 100 V 0 V to R 1 = 1 ΩR 2,3 = 4 Ω 100 V0 V to R T = 5 Ω 100 V0 V

79. Next find the total current flowing through the simplified circuit. R T = 5 Ω 100 V0 V I T = V T R T I T = 100 V 5 Ω I T = 20 A Now we can go back and un-simplify the circuit and find the current and voltage through specific resistors…

80. Find the current and voltage through R 1. Hint: Always find current first, then voltage. R 1 = 1 ΩR 2,3 = 4 Ω 100 V0 V R T = 5 Ω 100 V0 V We already found the current through R T and since this is a series circuit I T = I 1 = I 2,3 so I 1 = 20 A. Next find voltage…

81. R 1 = 1 ΩR 2,3 = 4 Ω 100 V0 V I 1 = 20 AI 2,3 = 20 A V 1 = I 1 R 1 In a series circuit V T = V 1 + V 2,3 so the voltages across both of these resistors will add up to 100 V. V 1 = (20A)( 1 Ω) V 1 = 20 V V 2,3 = I 2,3 R 2,3 V 2,3 = (20A)( 4 Ω) V 2,3 = 80 V Let’s check to see if the voltages make sense: V T = V 1 + V 2,3 100 V = 20 V + 80 V

82. Now we can find the separate currents and voltages through R 2 and R 3. R 1 = 1 ΩR 2,3 = 4 Ω 100 V0 V I 1 = 20 AI 2,3 = 20 A R 1 = 1 Ω R 2 = 8 Ω R 3 = 8 Ω 100 V 0 V I 1 = 20 A I 2,3 = 20 A

83. R 1 = 1 Ω R 2 = 8 Ω R 3 = 8 Ω 100 V 0 V I 1 = 20 A I 2,3 = 20 A R 2 and R 3 are in parallel so we know that V 2,3 = V 2 = V 3 therefore V 2 and V 3 are 80 V. Now we can find I 2 and I 3. I 2 = V 2,3 R 2 I 2 = 80 V 8 Ω I 2 = 10 A I 3 = V 2,3 R 3 I 3 = 80 V 8 Ω I 3 = 10 A V 1 = 20 V V 2,3 = 80 V

84. Let’s try another one… R 3 = 6 Ω R 1 = 4 ΩR 2 = 2 Ω Resistors R 1 and R 2 are in series so R 1,2 = R 1 + R 2 R 1,2 = R 1 + R 2 R 1,2 = 4 Ω + 2 Ω = 6 Ω R 3 = 6 Ω R 1,2 = 6 Ω 100 V0 V 100 V0 V

85. R 3 = 6 Ω R 1,2 = 6 Ω 100 V0 V 1 = 1 + 1 R T R 1,2 R 3 1 = 1 + 1 = 2 = 3 Ω R T 6 Ω 6 Ω 6 Ω R T = 3 Ω 100 V0 V

86. R 3 = 6 Ω R 1 = 4 ΩR 2 = 2 Ω 100 V0 V R 3 = 6 Ω R 1,2 = 6 Ω 100 V0 V R T = 3 Ω 100 V0 V

87. What do we do next? R T = 3 Ω 100 V0 V I T = V T R T I T = 100 V 3 Ω I T = 33.33 A Next we un-simplify the circuit and find the rest…

88. R 3 = 6 Ω R 1,2 = 6 Ω 100 V0 V I T = 33.33 A Since R 1,2 and R 3 are in parallel they have voltage in common: V T = V 1,2 = V 3 therefore V 1,2 and V 3 are both 100 V. We can now find each resistors individual current flow, I T = I 1,2 + I 3. I 1,2 = V 1,2 R 1,2 I 1,2 = 100 V 6 Ω I 1,2 = 16.67 A I 3 = V 3 R 3 I 3 = 100 V 6 Ω I 3 = 16.67 A

89. Now let’s separate R 1 and R 2 … R 3 = 6 Ω R 1,2 = 6 Ω 100 V0 V R 3 = 6 Ω R 1 = 4 ΩR 2 = 2 Ω 100 V0 V I 3 = 16.67 A I 1,2 = 16.67 A

90. R 3 = 6 Ω R 1 = 4 ΩR 2 = 2 Ω 100 V0 V I 3 = 16.67 A I 1,2 = 16.67 A R 1 and R 2 are in series with each other so they have the same amount of current flowing through them: I 1,2 = I 1 = I 2 therefore the current flowing through both of them is 16.67 A which will help us find the voltage drop across each resistor. V 1 = I 1,2 R 1 V 1 = (16.67A)( 4 Ω) V 1 = 66.68 V V 2 = I 1,2 R 2 V 2 = (16.67A)( 2 Ω) V 2 = 33.34 V

91. More practice… a)Which letter shows the graph of voltage vs. current for the smallest resistance? b) Which letter shows the graph of voltage vs. current for the largest resistance?

92. a)What is the total resistance of this circuit? b) What is the total current of this circuit? c) What is the amount of current running through each resistor?

93. a) What is the total resistance of this circuit? b) What is the total current of this circuit?

94. a) Find the equivalent resistance. b) Find the current (I T ) going through this circuit. c) Find potential drop across R 1 & R 2

95. a) Find combined resistance (R T ). b) Find the current in R 1. c) Find I 3. d) Find R 2. e) Find value of the second resistor.

96. a)If the voltage drop across the 3 ohm resistor is 4 volts, then what would the voltage drop be across the 6 ohm resistor? b) Find the total voltage in this series circuit. c) Find combined resistance in this circuit. d) Find the total current in this circuit.

97. a) Current in this circuit? b) Potential difference in 20 ohm resistor? c) Equivalent resistance in the circuit?

98. Find R 2.

99. What happens to the brightness of the light bulbs as we add more and more bulbs into the circuit? http://phet.colorado.edu/sims/ohms-law/ohms-law_en.html http://phet.colorado.edu/simulations/sims.php?sim=Circuit_Construction_Kit_DC_Only