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1 AP Physics 1 Lesson 12.a Electric Current and Circuits Outcomes 1. Determine the resistance of a resistor given length, cross-sectional area and length. 2. Relate the movement of charge to differences in potential. 3. Use Ohm s law to solve problems with simple direct current circuits. 4. Determine the effective resistance of a simple circuit with resistors in series, parallel or combination arrangements. 5. Use Kirchoff s Loop Rule to solve problems with simple direct current circuits. 6. Use Kirchoff s Junction Rule to solve problems with simple direct current circuits. 7. Determine the internal resistance of a battery in a circuit. 8. Determine the rate of energy dissipation in simple direct current circuits. 9. Link concepts of energy conservation and specific heat to simple circuits. 10. Determine the total effective resistance of a simple steady state circuit with resistors in series, parallel or combination arrangements. I= V R R= ρ L A R s =R +R 2 +R 3 1 = R p R 1 R 2 R 3 P=IV=I 2 R= V 2 R Name Date Period Engage 1. How does a light bulb work? 2. Draw arrows on the diagram or any other marks to help explain how the light bulb works. 3. Imagine that you are given a bulb, a battery and a wire segment. How would you connect the wire and the battery to get the light bulb to glow? Draw as many working set ups as you can. Explore The teacher will provide your group with a battery, bulb and wire segment. Experiment with these materials until you discover at least two different ways to arrange these materials to make the light bulb glow. 4. Carefully diagram successful arrangements. Explain 5. What general conditions have to be true for the light bulb to light? 1

2 Read Notes: A complete loop or path from one end of a battery to the other is referred to as a circuit. An open circuit is one that is not complete and has a gap or separation between one end of the battery and the other. A closed circuit is one that has a complete uninterrupted path from one end of the battery to the other. Explore II Examine the diagram on the right. There are two parallel metal plates. Plate A is covered with negatively charged particles. Plate B is covered with positively charged particles. Charge q ( a negative - charge) is in between the plates and is free to move. 6. Explain what will happen to charge q when it is released. 7. What happens to the velocity of charge A as it approaches plate B? Why? Explore III Examine the drawing on the right. There are two metal plates set up like before. This time there is a barrier preventing charges from directly passing from one plate to the other, but there is a wire connecting the plates. 8. What do you think will happen this time? Notes This arrangement describes the basic components of a simple electric circuit. When a large number of negative charges (or positive charges) are grouped together, there is a large amount of potential energy in the system. If a path is set up that will allow those charges to move to places where they are less concentrated, then they will spontaneously move along that path. That movement of charge along the path is called electric current. When the negative charges move from a state of high concentration to low concentration, the potential energy in the system decreases. Explain III 9. If large numbers of charges squeezed together have high potential energy, what does that potential energy become as the charges move? 2

3 10. What is being done with the electric potential energy in this circuit? Explore IV Read Let s examine charges moving in the wire. When there is no difference in the concentration of charged particles in a circuit, the electrons randomly move from one metal nucleus to another. If a difference in concentration of charge occurs between the ends of the wire, there is a net flow or drift of the electrons from the end with a higher concentration of charge toward the end with the lower concentration of charge. This movement decreases potential energy in the system. Let s examine this flow of charge with a model. The diagram below represents a wire segment. The white circles represent the positively charged nuclei of metal atoms in the wire. Place a coin or small piece of paper over nuclei 1, 2, 3, 4, 5 and 6. They represent the electrons. Now we will apply a voltage to the system. Draw a big (+) on the right hand side of the illustration, and a big (-) on the left hand side of the illustration. The penny covering nucleus 6 is now in a high potential energy state and is attracted to the big (+) charge. Pick it up and place it on the large positive charge you drew on the right hand side of the wire. The electron penny covering nucleus 5 now sees an empty space and will move to the right to cover nucleus #6. Move the electrons covering nucleus 5 in the manner you think appropriate. Repeat this process with the electron covering nucleus 4. 3

4 11. What is the net direction that the electrons are moving? Notes: We call this movement of electrons electron current. 12. What do you notice about the direction the circles or nuclei move? 13. So we have a net flow of negative charge to the right and a net flow of circles or positive charge states to the... Notes: We call this net movement of positive charge states to the left conventional current. Explain IV 14. Look at the picture on the right. The top of the battery has a high concentration of + charge. Draw the direction of electron current, and the direction of conventional current in the wire. 15. Why do charges (electrons) move in an electric circuit? 16. What is electron current? 17. What is conventional current? Elaborate Now we are going to investigate the how the presence of additional bulbs in a circuit affect the amount of current in the circuit. Prepare this circuit and note the brightness of the bulb. 18. Predict what will happen to the brightness of the bulb(s) if you add a second bulb as the diagram below illustrates. 19. Assemble the circuit and note the brightness. Are the bulbs brighter or dimmer? 20. Based on your observation, what do you believe happened to the amount of charge flowing through the closed circuit? 21. Predict what will happen to the brightness of the bulb(s) if you add a third bulb as the diagram below illustrates. 4

5 22. Assemble the circuit and note the brightness. Are the bulbs brighter or dimmer? 23. Based on your observation, what do you believe happened to the amount of charge flowing through the closed circuit? Explore V In this investigation, you will examine the relationship between the length of a nichrome wire segment and the amount of current in a circuit. Increasing the length of the nichrome wire simulates the effect of adding more light bulbs. 24. What do you predict will happen to the current in the circuit when the wire length L is increased? Read You may notice that the nichrome wire becomes warm while connected to the battery. The wire is a resistor. A resistor converts electric potential energy into heat. The ability of a material to resist the flow of charge is called resistance. Resistance is determined by the relationship R= ρ L A Where R= Resistance in Ohms, rho ρ is the resistivity of the material, L = length of the material, and A is the cross-sectional area of the material. 25. As the resistance in a circuit increases, what do you think happens to the current in the circuit? 5

6 Explore VI We will now make measurements of current in a circuit using an ammeter. An ammeter measures current in units of amperes. An ampere is a measure of the number of charges/second passing by the meter. Record and graph the ammeter measurements. It is not necessary to add the lightbulb to the circuit, but you can. Set Up Current Measurement (amperes) Graph Explain V 26. As the resistance increases, what happens to the current in the circuit? How does this compare to your predictions? You now have solid evidence that as the resistance in a circuit increases, the current decreases. This relationship is known as OHM s LAW. Ohm s law can be summarized by the following equation: Current = Voltage Resistance Amperes = Volts Ohms (Ω) I = V R Now you will connect our understanding of energy transformation with the ideas of current and resistance. 6

7 Explore VII 27. Prepare the following table. Make your current predictions. Check your predictions using the Electric Circuit DC Kit from the Phet website. The battery is 100V and each resistor is 10 Ohm. Set Up Current Prediction Actual Current 2D Model Sketch (the y-direction represents the voltage) Explain VI 28. As the resistance increases, what happens to the current in the circuit? 29. What happens to the electric potential energy of the charges in a circuit at each resistor? 30. What happens to the electric potential energy of the charges in a circuit at the battery? 31. How can you predict the current in a circuit? Notes Read The circuit you have worked with so far is called a series circuit, which consists of a single loop for the current to flow. The resistors are said to be in series because they are aligned one after the other. 7

8 Engage Record your responses. 32. What is the function of each of these items in an electric circuit? A. B. C. D. Battery Resistors Light bulb Electric Motor 33. Which item(s) increase(s) the electric potential energy of the charges in a circuit? 34. Which item(s) transform(s) electric potential energy into another form of energy? The current is measured for one resistor and two similar resistors. I. II. 35. Which line do predict would match set up I? Explain your thinking. 36. Which line do you predict would match set up II? Explain your thinking. Read Notes: Electric circuits transform energy from one form into another. Batteries convert chemical potential energy into electric potential energy. Resistors convert electric potential energy into heat. Light bulbs convert electric potential energy in to light and heat. Electric motors convert electric potential energy into mechanical energy (KE and or GPE). In the last section, you examined energy transformation with a nichrome wire. In this investigation, you will take a close look at energy transformation with resistors. Explore VIII We will now make measurements of current in a circuit using an ammeter and a voltmeter. An ammeter measures current in units of amperes. An ampere is a measure of the number of charges/second passing by the meter. A voltmeter measures volts or the difference in energy carried by each charge at two different points in the circuit. Record and graph the ammeter measurements. 37. As the voltage increases, what happens to the current in the circuit if the resistance is kept constant? 8

9 Now you will investigate the relationship between increasing voltage and current in a circuit. In the previous activities we kept the voltage (power source) constant and changed the resistance. Set Up Graph Use a single 25 Ohm resistor Current Prediction (amperes) Current Measurement (amperes) Voltage (volts) Explain VII 38. Compare your predictions to the evidence. Were your ideas supported by the evidence or has your thinking changed? 39. As the voltage increases, what happens to the current in the circuit? 40. Find the slope of the line. How does this slope compare to the resistance? 41. If you were to use two 25 ohm resistors, what do you predict the slope of the line would be? If time permits, check your prediction. Engage Record your responses. 42. What is happening in these pictures? What are these pictures about? 43. What would happen to the rate at which cars could move through the toll booths if some of the lanes were closed? Examine the circuits below. I. II. 44. Which circuit provides more paths for charge to flow through the light bulbs? 45. Which circuit will allow more charge to flow through the light bulbs? 9

10 Explore IX In the last lessons, you examined energy transformation in series circuits. A series circuit has a single loop or path for charge to flow through. In this lesson, you will investigate parallel circuits. A parallel circuit has multiple paths for charge to flow through. Set Up Brightness Prediction Rank the Brightness of Each Set Up Brightness Observation Explain IIX 46. Compare your predictions to the evidence. Were your ideas supported by the evidence or has your thinking changed? 47. Why do you think the bulbs in the parallel circuit were brighter than the bulbs in the series circuit? 48. Why do you think the bulbs in the parallel circuit were about the same brightness as the single bulb? Compare the current in the parallel bulbs to the single series bulb. 49. Which circuit do you think will remove energy from the battery quicker? Why? 50. As the number of paths in a circuit increases, what do you think happens to the current in the circuit? 51. If the current increases as the number of paths increase, what must happen to the total resistance of the circuit? 10

11 Explore X Now you will take a closer look at the characteristics of series and parallel circuits. I am glad you noticed, in the third and fourth set-ups the single wire at the bottom of each picture should connect to the right side of the base at Y and not the middle of the wire that connects X and Y. Set Up Prediction Observation What will happen if bulb X is removed? Simply unscrew the bulb. What will happen if bulb X is removed? Simply unscrew the bulb. What will happen if bulb X is removed? Simply unscrew the bulb. What will happen if bulb X and Y are removed? Simply unscrew the bulbs. Explain IX 52. Compare your predictions to the evidence. Were your ideas supported by the evidence or has your thinking changed? 53. What happens to a series circuit if a bulb is removed? 54. What happens to a parallel circuit if a bulb is removed? 55. Summarize the differences between series and parallel circuits. 11

12 Explore XI Now you will make measurements of current in series and parallel circuits using an ammeter and a voltmeter. An ammeter measures current in units of amperes. An ampere is a measure of the number of charges/second passing by the meter. A voltmeter measures volts or the difference in energy carried by each charge at two different points in the circuit. Record and graph the ammeter measurements. Plot both data sets on the same grid Set Up I (series) Graph Use two 25 Ohm resistors Current Prediction (amperes) Current Measurement (amperes) Voltage (volts) Set Up II (parallel) Use two 25 Ohm resistors Current Prediction (amperes) Current Measurement (amperes) Voltage (volts) Explain X 56. Compare your predictions to the evidence. Were your ideas supported by the evidence or has your thinking changed? 57. As the voltage increases, what happens to the current in the circuit? 58. Compare the slopes for each set of data. Which circuit has a greater resistance? Explore XII 59. In this activity you and your group will draw a circuit diagram for a black box made up of lights, wires, batteries, and switches. You will use the Phet applets to create the circuits following teacher s example. It is called a black box because all of the wires and connections will be hidden. Other students can see the voltage source, the bulbs, and the switches. By unscrewing bulbs and flipping switches they need to draw possible circuit diagrams that would match their observations. We will do a walk- around where you and your group will have 3 minutes to explore each of the other groups circuits and to draw a circuit diagram you think will match the circuit your observed. After the big reveal we will see which group has the most circuits identified. What do you win? Bragging rights. Rules: No more than 5 bulbs of equal resistance. One battery No more than 5 switches. All bulbs, switches, and the battery are visible and in the top half of your circuit. 12

14 Power The rate of energy dissipation by a resistor in a simple circuit is defined by the following relationships (combining Ohm s Law with the basic power equation P=IV): P=IV=I 2 R= V 2 R The units used to measure the rate of energy dissipation are given below: Power = Joules second =Watts Another useful relationship in determining a rate of energy transformation is determined through: Energy Power =Force velocity= N m s = Joules second =Watts Since power is the change in energy over time, energy is the product of power and time. Energy = Power x time A bulbs brightness or any other electric appliance s output is based on the power dissipated by this particular resistor. Ex. A 20W bulb is brighter than a 10W bulb or a 1500W vacuum cleaner is stronger than a 500W vacuum cleaner. The cost of using electricity is based on the power of the appliance and how long it is being used. Ex. Using a 1000W microwave oven for 2 minutes uses twice the energy of using it for 1 minute and therefore also costs you twice as much. Problem Demonstrations: Resistance Resistance is the ability of a medium to restrict the flow of charge in a circuit. A resistor converts electric potential energy into heat. The physical properties of the medium, the temperature of the medium, and the physical dimensions of the medium define its resistance. Resistivity is the material specific constant used in the determination of resistance at a specific temperature. Table of Resistivities. PROBLEM The five resistors shown below have lengths and cross sectional areas indicated and are made of material with the same resistivity. Which has the greatest resistance? The relationship between resistance and the physical dimensions of the conductive medium is defined: R = ρ l A (at a specified temperature) where l represents the length of the conductor, A is its cross-sectional area, and (rho) is the resistivity of the material. PROBLEM Calculate the resistance at 20 C of an aluminum wire that is meter long and has a cross-sectional area of 1.00 x 10-3 square meter. SOLUTION We know from the table given above that aluminum has a resistivity of 2.65 x 10-8 Ω m. Therefore: R = ρ l A (at a specified temperature) =2.65x 10-8 Ωm 0.200m /1.00 x 10-3 m2 =5.30x 10-6 Ω 14

15 Current The SI unit of current is the ampere. current = q 1 coulomb 1 ampere = t second PROBLEM What is the electric current in a conductor if 240 coulombs of charge pass through it in 1.0 minute? SOLUTION We solve the problem by using the relation I = Qt and remembering that time must be measured in seconds. I= Q/t = 240 C/60. s = 4.0A Ohm s Law Ohm s law can be summarized by the following equation: Current = Voltage Resistance = Amperes = Volts Ohms(Ω) = I = V R This relationship is only valid for ohmic resistors that do not change resistance over a range of current values. In reality, as a resistor heats up, its resistance changes. PROBLEM Determine the resistance of resistor A and resistor B. R A =V/I = 5V/2.5a=2Ω R B =V/I = 5V/5a=1Ω If resistor A was connected in series with a 6 volt battery, what current would be measured in the resistor? I=V/R=6B/2Ω=3a Power The rate of energy dissipation by a resistor in a simple circuit is defined by the following relationships: P=IV=I 2 R= V 2 R The units used to measure the rate of energy dissipation are given below: Power = Joules second =Watts Another useful relationship in determining a rate of energy transformation is determined through: Power =Force velocity= N m s = Joules second =Watts PROBLEM 1989B3. A series circuit consists of a battery of negligible internal resistance, a variable resistor, and an electric motor of negligible resistance. The current in the circuit is 2 amperes when the resistance in the circuit is adjusted to 10 ohms. Under these conditions the motor lifts a l-kilogram mass vertically at a constant speed of 2 meters per second. a. Determine the electrical power that is i. dissipated in the resistor P=I 2 R=(2a) 2 (10Ω)=40watts ii. used by the motor in lifting the mass P=Fv=(1kg)(10m/s 2 )(2m/s)=20 watts iii. supplied by the battery 40 watts + 20 watts = 60 watts b. Determine the potential difference across i. the resistor P=IV V=P/I =(40watts)(2a)=20V ii. the motor P=IV V=P/I =(20watts)(2a)=10V iii. the battery V=Vdrop R + Vdrop Motor = 30V The resistor is now adjusted until the mass rises vertically at a constant speed of 3 meters per second. The voltage drop across the motor is proportional to the speed of the motor, and the current remains constant. c. Determine the voltage drop across the motor. V 3m/s = 1.5(10V) = 15 V d. Determine the new resistance in the circuit. 30V-15V=15V (15V)/(2a)=7.5Ω 15

16 Resistors in Series When resistors are linked together in series in a circuit, the total effective resistance is the sum of the individual resistors. R s =R 1 +R 2 +R 3 Kirchoff s Loop Rule The sum of all of the changes in potential difference in a circuit is equal to zero. The voltage source (battery) provides an increase in potential, while the other circuit elements (resistors, motors, light bulbs) lower the potential. PROBLEM Examine the circuit below. a) Determine the total effective resistance of the circuit. R s=r 1+R 2+R 3 = 3Ω+6Ω+9Ω = 18Ω b) Determine the current in the circuit. I=V/R = 24V/18Ω=1.33a c) Determine the voltage drop across each resistor. V=IR V 12=(1.33a)(3Ω)=4V V 23=(1.33a)(6Ω)=8V V 34=(1.33a)(9Ω)=12V 0 = V source + V drop1 + V drop2 + V drop3 Resistors in Parallel When resistors are linked together in parallel, the total effective resistance is determined through the following relationship 1 = R p R 1 R 2 R 3 Kirchoff s Junction Rule The sum of all of the currents entering a junction equals the sum of the currents leaving a junction. PROBLEM a) Determine the total effective resistance of the circuit. 1/Rs=1/R 1+1/R 2+1/R 3 =[1/3Ω+1/6Ω+1/9Ω] -1 = 1.64Ω b) Determine the current in the circuit. I=V/R = 24V/1.64Ω=14.67a c) Determine the current in each resistor. I=V/R 24V/3Ω=8a 24V/6Ω=4a 24V/9Ω=2.67a (8a+4a+2.67a=14.6a) V source = V 1 = V 2 = V 3 Internal Resistance Real batteries have a certain resistance as a result of the physical and chemical properties of their construction. In certain problems this resistance must be included in the problem solving process. The internal resistance is treated as a normal resistor in series with the battery. The terminal voltage is the actual voltage the external circuit sees from the battery. The emf (electromotive force) is the voltage across the battery that has not been adjusted for the internal resistance. Problem A battery with an emf of 24 volts and an internal resistance of 1 ohm is connected to an external circuit as shown on the right. Determine each of the following: a) the equivalent resistance of the combination of the 4 ohm, 8 ohm, and 12 ohm resistors. [1/R 1+(1/(R 2+R 3))]-1 RP=[1/12Ω+(1/(4Ω+8Ω))] -1 R P=6Ω b)the current in the 5 ohm resistor I = V/R I=24V/(12Ω) = 2a c)the terminal voltage VAC, of the battery V AC = emf-ir internal = 24V-(2a)(1Ω)=22V d) the rate at which energy is dissipated in the 12 ohm resistor P=I 2 R P=(1a) 2 (12Ω)=12watts e) the magnitude of the potential difference VBC V BC=V terminal-ir 5Ω=22V-(2a)(5Ω)=12V f) the power delivered by the battery to the external circuit P=IV = (2a)(22V) = 44 watts 16

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