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Electric Circuits - Key Vocabulary Term Electric Current The flow of electric charge. Electric Circuit Any complete path through which electricity travels. Closed Circuit Open Circuit Conductors Definition A circuit in which there is a complete path for electricity to flow. A circuit in which there is a break so current cannot flow. A material that easily carries electrical current. Insulators A material that poorly conducts electrical current or heat. Ohm’s Law Current Voltage Resistance A formula that says the current flowing in a circuit is the voltage divided by the resistance. V = IR Series Circuit A circuit that has only one path for current to flow through. Parallel Circuit A circuit that has multiple paths for the current to flow through. Power The flow of electric charge. Page | 1 Activity 1 – Which bulbs will light? Prediction: Point Bulb 1 Bulb 2 Lit / Unlit Lit / Unlit What will happen to bulbs 1 and 2 when you disconnect the wires at various points? Observation(s): Point Bulb 1 Bulb 2 Lit / Unlit Lit / Unlit A A Unlit Unlit B B Unlit Unlit C C Unlit Unlit D D Unlit Unlit E E Unlit Unlit F F Unlit Unlit Briefly explain your reasoning for point predictions B, D, and E. Conclusion: Consensus: (Your answer to the focus question in the “V” above.) New Terms Page | 2 Activity 2 – Which types of objects will allow bulbs to light? Prediction(s): Object Light? (Y or N) What type of object, when inserted into the loop, will allow the two test bulbs to light? Observation(s): Object Light? (Y or N) Foil Yes Cardboard Cardboard No Paperclip Paperclip Yes Shoestring Shoestring No Wax paper Wax paper No Foil Conclusion: Consensus: New Terms: Page | 3 PRACTICE SET 1: What is moving in the wires? Read the following: What’s moving? No one can see what moves through the wires, but something about the moving substance causes a compass needle to deflect. The property that enables the substance to do this is called CHARGE, from a Latin word that means “vehicle”. Particles that carry charge from one place to another are called “charge carriers”. The experiments you’ve done provide evidence that CHARGE is carried through wires, but they provide no evidence yet about the nature of the charge carriers. Which direction is it moving? The reversal of compass needle deflection when the battery orientation is reversed indicates a change in the direction of charge flow in the loop, but provides no information about which actual direction exists before or after the change. Scientists searched for hundred of years trying to determine which way the charge really moved, but were unable to do so until the late 1800’s. In the absence of any evidence, they decided to assume a direction for the motion. Such an assumption is “conventional” — that is, simply an “agreement” which isn’t necessarily right or wrong but is useful because it is necessary for communication. The international convention is that the charges circulating around a circuit leave the battery at the “positive” end (red spot), travel around the circuit and re-enter at the “negative” end (blue spot), and pass through the battery. In later Sections we will collect evidence to determine whether this “conventional” direction is accurate or not. 1. Study the two circuits below in which a paper clip has been inserted between wires in a circuit. Circuit A Circuit B Which of the following statements are true? _____________ (A) (B) (C) (D) The bulb will light more brightly in Circuit A . The bulb will light more brightly in Circuit B The bulb will be the same brightness in either case. The bulb will not light. Page | 4 2. Explain why you believe your answer to Question #1 is correct. Use the words “insulator” and “conductor” correctly as part of your explanation. The paperclip is a conductor so it does not matter where in the circuit it is placed; it will still allow the circuit to light up. An insulator would not allow the lights to turn on. 3. Write in your own words a definition of the word circuit which anyone could use to determine if a given set of connections is or is not a circuit. A complete path that allows current to flow through from one voltage to another. 4. We have observed in several activities that as soon as a very small gap is produced anywhere in the circuit, the bulbs go out. Would you classify air as a conductor or an insulator? Explain. Air is an insulator because it does not allow the charge to flow through easily. 5. Indicate whether each of the following statements is True or False. Then state evidence which either supports or contradicts each statement. False [a] Charge moves out of each end of the battery into the loop. Evidence: They only move out of one end towards the other, because they flow from high to low energy. True [b] Light bulbs are non-directional devices. (Whichever way they are connected in the circuit, they behave the same way if you turn them around.) Evidence: If you disconnect the bulb and flip it around it still turns on with the same amount of brightness True [c] The battery determines the direction of flow of charge in a circuit. Evidence: Charges flow from positive to negative. True [d] A compass can be used to determine the exact direction that charge flows in a circuit. Evidence: After learning the future rules of electricity and magnetism it can. True [e] Metal substances are generally conductors. Evidence: Metal is a conductor and allows current to pass through. Page | 5 Activity 3 – And Then There Was Light! Focus Question: What are the essential components and configurations needed to make a bulb light? Equipment: One cell, one bulb, one wire. Directions: Make the bulb light using only the given equipment. Draw each successful circuit you create. There is more than one configuration that will be successful. You need to have at least two. Once you identified a successful configuration, you also need to trace the closed loop path around the circuit. Successful attempts (at least 2) Unsuccessful attempts (2) Can you light more than one bulb at a time? Try it. Conclusion: Consensus: Page | 6 Information: Current Current is the rate of charge flow past a given point in an electric circuit, measured in coulombs/second, which is named Amperes. Although it is electrons that are the mobile charge carriers that are responsible for electric current in conductors such as wires, it has long been the convention to take the direction of electric current as if it were the positive charges that are moving. The conventional current direction is the direction from high voltage to low voltage, high energy to low energy, and thus has some appeal in its parallel to the flow of water from high pressure to low (see the water analogy below). Critical Thinking Questions 1. What things can you think of that flow from place to place? What conditions are necessary for charges to flow? Water Electricity Wind Hair Blood 2. Is a current carrying wire electrically charged? No, current carrying wire allows the current to flow through it, but the wire is not charged to begin with, nor is it charged after turning the system off. Page | 7 Examine Picture A and Picture B closely. How are they similar? How are they different? How would each compare to the circuits below? Picture A Similarities They both start and finish at the same point. Differences Picture A only has one option for the skiers, while B has three. They both travel Picture B allows up the same more people to size ski-lift. go down the mountain at a time. Picture B They both have the same length to travel down the mountain. The riders have to split and take any path they want. Page | 8 Research Question What is the difference between the two circuits below? A B Procedure You will be assigned to build and test variables in one of the above circuits. You must determine a way to collect and organize the data/observations. You will put together your observations/data on a piece of poster paper. Include pictures and data tables on your poster. You will then present your findings to the class. Group A 1. Construct the circuit shown in Figure “A” above using three batteries, two bulbs, two bulb holders and wires. 2. Measure the voltage of a single battery in your circuit using the multimeter. (place the red lead by the positive terminal and the black lead by the negative terminal.) Record this voltage. Voltage of a single battery = 1.5 volts 3. Measure the voltage of all three batteries and record the total voltage in the circuit. Voltage of three batteries = 4.5 volts 4. Measure the voltage of each bulb. What observations can you make? Voltage of first bulb = ~4.5 volts 2 Bulb System Voltage of second bulb = ~4.5 volts 5. Add another bulb and again record the voltage of each bulb. What observations can you make? Voltage of first bulb = ~4.5 volts Voltage of second bulb = ~4.5 volts 3 Bulb System Voltage of third bulb = ~4.5 volts Page | 9 Group B 1. Construct the circuit shown in Figure “B”. The image below is for your 3 bulb system. 2. Measure the voltage of a single battery in your circuit using the multimeter. (place the red lead by the positive terminal and the black lead by the negative terminal.) Record this voltage. Voltage of a single battery = ~1.5 volts 3. Measure the total voltage of all three batteries in both circuits using a multimeter. Voltage of three batteries = ~4.5 volts 4. Measure the voltage of each bulb in each circuit and record the total voltage. Be sure to place the RED test probe on the POSITIVE terminal. What observations can you make? Voltage of first bulb = ~2.25 volts 2 Bulb System Voltage of second bulb = ~2.25 volts 5. Add another bulb and again record the voltage of each bulb. What observations can you make? Voltage of first bulb = ~1.5 volts Voltage of second bulb = ~1.5 volts 3 Bulb System Voltage of third bulb = ~1.5 volts Group C 1. Construct the circuit shown in Figure “A” above using three batteries, two bulbs, two bulb holders and wires. 2. Add another bulb. What observations can you make about bulb brightness as you add a bulb? Add a fourth bulb and record any observations. 3. Unplug one bulb (you can select any bulb to unplug). What happens to the remaining bulbs? 4. Unplug a second bulb. What happens? 5. Come up with a hypothesis to explain your observations. 6. As you add additional bulbs do you think the resistance the current encounters increases or decreases? Why? 7. Do you think the electrical outlets in your home are connected like they are in Figure A or Figure B? Give two reasons why one type of circuit has an advantage over the other for connecting outlets. Page | 10 Group D 1. Construct the circuit shown in Figure “B” above using three batteries, two bulbs, two bulb holders and wires. 2. Add another bulb. What observations can you make about bulb brightness as you add a bulb? Add a fourth bulb and record any observations. 3. Unplug one bulb (you can select any bulb to unplug). What happens to the remaining bulbs? 4. Unplug a second bulb. What happens? 5. Come up with a hypothesis to explain your observations. 6. As you add additional bulbs do you think the resistance the current encounters increases or decreases? Why? 7. Do you think the electrical outlets in your home are connected like they are in Figure A or Figure B? Give two reasons why one type of circuit has an advantage over the other for connecting outlets. Group E 1. Build the circuit below using three batteries, three bulbs, three bulb holders and wires. 2. If Bulb B was unscrewed in the circuit below what would happen to (a) Bulb A (b) Bulb C? 3. Record the total voltage of the three batteries. 4. Record the voltage of bulb A, B and C. 5. What conclusions can you make about the voltage of bulbs A, B and C and the total voltage? Page | 11 Information: Circuit Building The principle circuit elements are: name variable unit source of emf V volts V resistor R ohms Ω ammeter amperes A voltmeter volts V symbol wires switch Series Circuits 1. In a series circuit there is only 1 path for electricity to flow. 2. Since there is only path for charges to flow, the current is the same everywhere. 3. In a series circuit, when there is a break in the circuit, the current stops flowing. Holiday lights used to be manufactured in series, meaning when one bulb burned out all of the bulbs would not light since the current stops. 4. Draw a picture of a series circuit below and label the following: a. Two Bulbs/Resistors b. 1 Battery c. 1 switch 5. As a class, build the series circuit from #4. a. What happens when you unscrew one bulb? When you unscrew a bulb, the other goes off. b. Add a third bulb and describe what happens to the overall brightness of the bulbs. When you add a third bulb the overall brightness decreases because current decreases. Page | 12 6. Resistors in series have the same current going through them. The voltage gets lower after each device that uses power which is known as the voltage drop. 7. The voltage drop is the drop in voltage across an electrical device that has current flowing through it. 8. This means the total of all the voltage drops must add up to the battery’s voltage. This rule is known as Kirchhoff’s Voltage Law. 9. Kirchhoff’s Voltage Law: The sum of all voltages in series resistors equals the total voltage from the battery. Vtotal = V1 + V2 + V3… Draw three 10 Ω resistors connected in series to a 6 volt battery: 10 10 10 To determine the total resistance in a series circuit you add: Rtotal = R1 + R2 + R3 + . . . Thus the total resistance of a circuit that has a 2 Ω , 4 Ω and 10 Ω resistor is: Rtotal = 2 + 4 + 10 Rtotal = 16 Ω Page | 13 Series Circuits Main Ideas Current Same throughout the circuit Voltage Is different for each bulb/resistor. Total Resistance Add the Resistance of Each Bulb. Adding a bulb Decreases the overall brightness Determine the total resistance in each circuit below. 3. 1. Rtotal = 1 + 2 = 3 Ω Rtotal = 1 + 1 = 2 Ω 2. 4. Rtotal = 1 + 1 + 1 = 3 Ω Rtotal = 2 + 3 + 1 = 6 Ω Page | 14 Information: Resistance and Ohm’s Law The electrical resistance of a circuit component or device is defined as the ratio of the voltage applied to the electric current which flows through it: V V = IR or R Triangle: I If the resistance is constant over a considerable range of voltage, then Ohm's law, V = IR, can be used to predict the behavior of the material. Classwork Practice Problems 1. A light bulb is plugged into a wall outlet (120 V). It uses 16 A. What is the light bulbs resistance? Looking For Given Resistance I = 16 Amps V = 120 Volts Relationship Solution R = 7.5 Ω 2. A flash light bulb is labeled to uses 1.77 A. Its resistance is 3.20 Ω. What voltage is the light bulb rated for? Looking For Given Relationship Solution Voltage I = 1.77 Amps R = 3.2 Ω V = IR V = (1.77)(3.2) V = 5.66 Volts 3. A stereo speaker has a resistance of 16.00 Ω. When it is operating at full power (exactly 200 watts) it uses 50 volts of electricity. What is the current drawn by the speaker? Looking For Given Current R = 16 Ω V = 50 Volts Relationship Solution I = 3.125 Amps 4. A toaster plugged into the wall, (120 volts), uses 24 amps of electricity. What is the resistance of the toaster? Looking For Given Resistance I = 24 Amps V = 120 Volts Relationship Solution R=5Ω Information: Power The electric power in watts (W) is associated with a complete electric circuit. A circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. For a resistor in a circuit the power is given by the product of applied voltage and the electric current: or or Triangle: Classwork Practice Problems 5. Find the current drawn from a 1200 W hair dryer connected to a 120 V source. Find the resistance of the hair dryer. Looking For Given Current P = 1,200 Watts V = 120 Volts Looking For Given Resistance I = 10 Amps V = 120 Volts Relationship Solution I = 10 Amps Relationship Solution R = 12 Ω Page | 15 6. A car lighter has a resistance of 4 Ω. If it draws from a 12 V battery, what is the power dissipated? Looking For Given Relationship Solution Current R=4Ω V = 12 Volts Looking For Given Relationship Solution Power I = 3 Amps V = 12 Volts P = IV P = (3)(12) P = 36 Watts I = 3 Amps Series Circuit Problem Solving 1. A series circuit contains a 12-V battery and three bulbs with a resistance of 4 Ω, 6 Ω, and 8 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 4Ω 0.667 A 2.667 V 1.779 W Bulb 2 6Ω 0.667 A 4V 2.668 W Bulb 3 8Ω 0.667 A 5.336 V 3.56 W Total 18 Ω 0.667 A 12 V 8.004 W 2. A series circuit contains a 120-V battery and three bulbs with a resistance of 12 Ω, 20 Ω, and 60 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 12 Ω 1.3 A 16 V 2.08 W Bulb 2 20 Ω 1.3 A 26 V 3.38 W Bulb 3 60 Ω 1.3 A 78 V 10.1 W Total 92 Ω 1.3 A 120 V 15.6 W Page | 16 3. A series circuit contains a 6-V battery and three bulbs with a resistance of 2 Ω, 4 Ω, and 6 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 2Ω 0.5 A 1V 0.5 W Bulb 2 4Ω 0.5 A 2V 1W Bulb 3 6Ω 0.5 A 3V 1.5 W Total 12 Ω 0.5 A 6V 3W Series Circuits - Homework 1. A series circuit contains a 9-V battery and three bulbs with a resistance of 1 Ω, 3 Ω, and 6 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 1Ω 0.9 A .9 V .81 W Bulb 2 3Ω 0.9 A 2.7 V 2.43 W Bulb 3 6Ω 0.9 A 5.4 V 4.86 W Total 10 Ω 0.9 A 9V 8.1 W 2. A series circuit contains a 60-V battery and three bulbs with a resistance of 5 Ω, 10 Ω, and 15 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 5Ω 2A 10 V 20 W Bulb 2 10 Ω 2A 20 V 40 W Bulb 3 15 Ω 2A 30 V 60 W Total 30 Ω 2A 60 V 120 W Page | 17 Parallel Circuits 1. In a parallel circuit there is more than 1 path for electricity to flow. 2. The voltage is the same across each branch of a parallel circuit. 3. Draw a picture of a parallel circuit below and label the following: a. Two Bulbs/Resistors b. 1 Battery c. 1 switch 4. As a class, build the parallel circuit above. Use the picture below for help. a. What happens when you unscrew one bulb? When you unscrew a bulb, the other bulbs stay on because there is more than one path for the current to follow. b. Add a third bulb and describe what happens to the overall brightness of the bulbs. When you add a third bulb, the overall brightness of the bulbs remains the same. 5. One main advantage parallel circuits have over series circuits is that each device in the circuit may be turned off independently without stopping the current in the other devices in the circuit. For example, parallel circuits allow you to turn off one light in your home without all of the other lights in your home going out. 6. Because each branch in a parallel circuit has the same voltage, the current in each branch is different. Page | 18 Parallel Circuits Main Ideas Current Varies down paths based on resistor present. Voltage Same down parallel paths Total Resistance Found by taking Adding a bulb Does not affect bulb brightness Draw three 10 Ω resistors connected in parallel: R1, R2, R3 = 10 Ω Summary Information: Series and Parallel A Series Circuit (Figure 1) is a circuit in which resistors are arranged in a chain, so the current has only one path to take. The current is the same through each resistor. The total resistance of the circuit is found by simply adding up the resistance values of the individual resistors: RT = R1 + R2 + R3 + ... A Parallel Circuit (Figure 2 & 3) is a circuit in which the resistors are arranged with their heads connected together, and their tails connected together [Figures 2 and 3]. The current in a parallel circuit breaks up, with some flowing along each parallel branch and re-combining when the branches meet again. The voltage across each resistor in parallel is the same. The total resistance of a set of resistors in parallel is found by adding up the reciprocals of the resistance values, and then taking the reciprocal of the total: 1/RT = 1/R1 + 1/R2 + 1/R3 +... Page | 19 Figure 1 Resistors connected in series. Figure 2 Example of a circuit containing three resistors connected in parallel Figure3 Circuit containing resistors in parallel, equivalent to Figure 2 If you add a resistor in: equivalent resistance Series increases Parallel Decreases total current in circuit decreases Increases current through each device the same depends on its resistance voltage across each device depends on its resistance the same Many circuits have a combination of series and parallel resistors. Generally, the total resistance in a circuit like this is found by reducing the different series and parallel combinations step-by-step to end up with a single equivalent resistance for the circuit. This allows the current to be determined easily. The current flowing through each resistor can then be found by undoing the reduction process. General rules for doing the reduction process include: 1. Two (or more) resistors with their heads directly connected together and their tails directly connected together are in parallel, and they can be reduced to one resistor using the equivalent resistance equation for resistors in parallel. 2. Two resistors connected together so that the tail of one is connected to the head of the next, with no other path for the current to take along the line connecting them, are in series and can be reduced to one equivalent resistor. Finally, remember that for resistors in series, the current is the same for each resistor, and for resistors in parallel, the voltage is the same for each one. CHALLENGE SAMPLE PROBLEM: What is the equivalent resistance of this circuit? NOTE: It has both series and parallel! R1 = 2 Ω R2 = 4 Ω R3 = 5 Ω R4 = 10 Ω First look at R3 and R4. These two resistors are in parallel because any current flow has a choice of which resistor to go through. Thus, Rp = (1/R3+1/R4)-1 = (1/5 + 1/10)-1 = 3.33 Ω And we have that the starting circuit is equivalent to one that looks like the following. Now we can easily see that R1,R2, and Rp are in series because any current must flow through each resistor (or resistor branch). RT = 2 + 4 + 3.33 = 9.33 Ω Page | 20 1. In a flashlight TWO 1.5 volt batteries are connected in series to TWO 60 bulbs. a. Draw the circuit. b. Calculate the current flowing through the bulbs. Looking For Given Current R1 = 60 Ω R2 = 60 Ω Vtotal = 3 Volts Relationship Solution I = 0.25 Amps c. Calculate the power in the circuit. Looking For Given Relationship Solution Power I = 0.25 Amps V = 3 Volts P = IV P = (0.25)(3) P = 0.75 Watts Critical Thinking Questions 2. To connect a pair of resistors so that your combined resistance would increase they should be connected how? To see the resistance decrease they should be connected how? i. ii. To have the combined resistance increase, you should add another resistor in series. To have the combined resistance decrease, you should add another resistor in parallel. 3. Why is there a difference in total resistance if you connect three 60 Ω resistors in series versus three 60 Ω resistors in parallel? There is a different total resistance because the three resistors added in series becomes 180 Ω of resistance. When added in parallel it becomes 20 Ω because there is more than one pathway to travel down for the current. 4. Find the combined resistance of a 12, 4 and 6 Ω resistors in series and then again in parallel. Series: Req = 12 + 4 + 6 = 22 Ω Parallel: (FLIP) Ω 5. Why do Christmas lights turn off if one light bulb is bad? How are they wired? If the lights turn off, it means that the bulbs are wired in series. The other situation would be if the remaining lights stayed on, then it would be parallel. Page | 21 Parallel Circuit Problem Solving 1. A parallel circuit contains a 12-V battery and three bulbs with a resistance of 4 Ω, 6 Ω, and 8 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 4Ω 3A 12 V 36 W Bulb 2 6Ω 2A 12 V 24 W Bulb 3 8Ω 1.5 A 12 V 18 W Total 1.85 Ω 6.5 A 12 V 78 W 2. A parallel circuit contains a 120-V battery and three bulbs with a resistance of 20 Ω, 30 Ω, and 60 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 20 Ω 6A 120 V 720 W Bulb 2 30 Ω 4A 120 V 480 W Bulb 3 60 Ω 2A 120 V 240 W Total 10 Ω 12 A 120 V 1,440 W 3. A parallel circuit contains a 6-V battery and three bulbs with a resistance of 2 Ω, 4 Ω, and 8 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 2Ω 3A 6V 18 W Bulb 2 4Ω 1.5 A 6V 9W Bulb 3 8Ω 0.75 A 6V 4.5 W Total 1.14 Ω 5.25 A 6V 31.5 W Page | 22 Parallel Circuit Homework 1. A parallel circuit contains a 20-V battery and three bulbs with a resistance of 2 Ω, 4 Ω, and 6 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 2Ω 10 A 20 V 200 W Bulb 2 4Ω 5A 20 V 100 W Bulb 3 6Ω 3.33 A 20 V 66.6 W Total 1.09 Ω 18.33 A 20 V 366.6 W 2. A parallel circuit contains a 60-V battery and three bulbs with a resistance of 5 Ω, 10 Ω, and 15 Ω. What is the total resistance in the circuit? What is the voltage drop of each bulb? What is the current that each bulb uses? What is the power in each bulb? Current (Amps) Voltage (Volts) Power (Watts) Bulb 1 Resistance (Ohms) 5Ω 12 A 60 V 720 W Bulb 2 10 Ω 6A 60 V 360 W Bulb 3 15 Ω 4A 60 V 240 W Total 2.73 Ω 22 A 60 V 1,320 W Page | 23 The Tollway Analogy The flow of charge through the wires of a circuit can be compared to the flow of cars along a toll way system in a very crowded metropolitan area. The main source of resistance on a toll way system are the tollbooths. Stopping cars and forcing them to pay a toll at a tollbooth not only slows the cars down, but in a highly trafficked area, will also cause a bottleneck with a backup for miles. The rate at which cars flow past a point on that toll way system is reduced significantly by the presence of a tollbooth. Clearly, tollbooths are the main resistor to car flow. This can also be related to thinking about walking through the hallways at school. If there were multiple pathways to walk down, then it would be much easier to walk to the front office. If the class were restricted to only one hallway for the trip, then it would take longer and would be more crowded with people pushing their way through. Come up with your own scenario that shows a relationship between multiple paths compared to one in order to explain series vs. parallel circuits. Page | 24 Resistivity The resistance of a wire is proportional to the length of the wire and inversely proportional to the thickness of the wire. The temperature and the material of the wire will also play a role in how fast or slow a current can flow. R = Resistance ρ = Resistivity l = length of wire A = Area of wire R l A Note: This equation is just for background information and is not necessary for you to know. You can think of how resistance changes within a circuit, and relate it to how people travel through hallways. A Thick Wire (large cross-sectional area) is similar to a wide hallway, while a thin wire would be a smaller hallway. Humans would rather walk down the wide hallway because travel is easier, similar to how current travels down the path with least resistance. A Long Wire would be similar to a long hallway to walk down, while a short wire would be like a short hallway. Humans would rather walk down the shorter hallway because they are able to reach their destination quicker and easier. A Hot Wire would be similar to a hallway on a summer day with excited students moving about, while a cold hallway would be found in the winter time with students sitting bundled up and not moving around much. If a person walked down the hot hallway, they would have more people in their way to say “Hi” to, causing them to take longer to make it through the hallway. Critical Thinking Questions 1. Why are thick wires rather than thin wires usually used to carry large current? Thick wires have less resistance so you can have more current travel through the wire. The reason there is less resistance is because there is a lot of room for the current to travel through since there is a larger cross section. 2. Could the same length of a copper and aluminum wire have the same resistance? Explain. Yes, the same length of copper and aluminum could have the same resistance because you could vary the thickness of the wire to make the overall resistance come out to the same answer. 3. Compare the Resistances of the following: a. Thick Wire vs. Thin Wire The thin wire has more resistance because it has less room for the current to travel through. b. Long Wire vs. Short Wire The long wire has more resistance because it has more distance for the current to travel through. c. Hot Wire vs. Cool Wire The hot wire has more resistance because the particles inside are moving around more compared to the cold wire, causing more collisions. d. Series Circuit vs. Parallel Circuit The Resistance in a series circuit is higher than a parallel circuit that uses the same resistors. Page | 25