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The Research Experience for Teachers Program http://www.cs.appstate.edu/ret Introduction/Motivation: The behavior of the current in a circuit depends on how much resistance the current experiences, and when the circuit is more complex the change in current is not intuitive. To understand the concepts behind current behavior is important as more complicated calculations depend on a firm understanding of concepts. Materials List: Computer with internet access, paper, pen or pencil and calculator Background: You should have a basic understanding of current, resistance, series circuit and parallel circuit. You will look at the behavior of current in two circuits as they go from one resistor to three resistors keeping a constant voltage. Note: the resistors are representing light bulbs and the yellow and orange lines coming from the electron(s) should actually be emanating from the resistors representing light produced from electrons transferring energy to the resistors. Note: electron speed = current. Procedure: Open the webpage https://scratch.mit.edu/projects/68919222/ Read instructions on right. Click on green flag. Lab Activity: Series 1. Press the “a” key for circuit a, you should see a red letter e moving left to right. Notice how fast it is moving, and notice the length of the orange and yellow lines indicating brightness. Fill in the blanks below: a. How many 10 ohm resistors are there? _________ b. What is the total resistance of the circuit? _________ c. What is the current in the wire? _______ 2. Press the “s” key for circuit s. Notice how fast it is moving. Fill in the blanks below: a. How many 10 ohm resistors are there? _______ b. What is the total resistance of the circuit? _______ c. What happened to the speed of the electron? _______ d. What is the current in the wire? _______ e. How much brighter is this one compared to circuit a? ________ (brighter, dimmer, no change) 3. Press the “d” key for circuit d. Notice how fast it is moving. Fill in the blanks below: a. How many 10 ohm resistors are there? _______ b. What is the total resistance of the circuit? _______ c. What happened to the speed of the electron? _______ d. What is the current in the wire? ______ e. How much brighter is this one compared to circuit s? ________ (brighter, dimmer, no change) 4. Main Series Concepts: based on what you saw, choose the best answer that finishes the sentences below. a. When resistors are added in series, the resistance of the circuit ________ (increases, decreases, stays the same). b. When resistors are added in series, the current of the circuit ________ (i, d, s). c. When current increases, the brightness of the bulb ______ (i, d, s) Lab Activity: Parallel 5. Press the “q” key for circuit q. Notice how fast it is moving and notice the orange and yellow lines indicating brightness. Fill in the blanks below: a. How many 10 ohm resistors are there? _______ b. What is the total resistance of the circuit? _______ c. What is the current in the wire? _______ 6. Press the “w” key for circuit w. Notice how fast they are moving at the beginning, and then in the parallel branches. Fill in the blanks below: a. How many 10 ohm resistors are there? ________ b. What is the total resistance of the circuit? _______ c. Compare the resistance in this one to circuit q. i. Resistance this one _____ circuit q. (<, >, =) d. Compare the speed of the electrons in the main wire to the branches. i. Branch speed ____ Main wire speed. (<, >, =). e. Compare the speed of the electrons in the main wire in this one compared to circuit q. i. Speed of this one ____ speed of circuit q. (<, >, =). 7. Press the “e” key for circuit e. Notice how fast they are moving at the beginning, and then in the parallel branches. Fill in the blanks below: a. How many 10 ohm resistors are there? ______ b. What is the total resistance of the circuit? ______ c. Compare the resistance in this one to circuit w. i. Resistance this one _____ circuit w. (<, >, =) d. Compare the speed of the electrons in the main wire to the branches. i. Branch speed ____ Main wire speed. (<, >, =). e. Compare the speed of the electrons in the main wire in this one compared to circuit q. i. Speed of this one ____ speed of circuit q. (<, >, =). 8. Main Parallel Concepts: based on what you saw, choose the best answer that finishes the sentences below. a. When resistors are added in parallel, the resistance of the circuit _____ (increases, decreases, stays the same). b. When resistors are added in parallel, the current in the main part of the circuit ________ (i, d, s) c. What happened to the current in the parallel branches as you added resistors in parallel? _____ (i, d, s) d. Depending on your answer to #8c, what should happen to the brightness of the bulbs as you add 10 ohm resistors in series? _____ (i, d, s). Schematics To understand circuits, you must understand how objects (wires, batteries, resistors, lightbulbs, etc.) can be represented in introductory schematic diagrams. (Note: lightbulbs are resistors that can glow) Wire Lightbulb/Resistor Resistor Battery Calculations & Concepts To find total resistance in series (RTS) add all the resistors RTS = R1 + R2 + R3 + … To find total resistance in parallel (RTP) use the following relationship: 1 𝑅𝑇𝑃 = 1 𝑅1 + 1 𝑅2 + 1 𝑅3 worked out below for a 2 ohm, 4 ohm and an 8 ohm resistor wired in parallel. 1 𝑅𝑇𝑃 1 𝑅𝑇𝑃 1 1 1 = 2Ω + 4Ω + 8Ω = 4+2+1 8Ω 7 = 8Ω 1 𝑅𝑇𝑃 𝑅𝑇𝑃 1 4 2 1 = 8Ω + 8Ω + 8Ω = 8Ω 7 𝑜𝑟 𝑅𝑇𝑃 = 8Ω 7 It’s important to remember “electricity will take the path of least resistance”. Brightness depends on current flow: increase current flow increase brightness. or 1.14Ω +… An example is Practice 1. Find the total resistance of the following circuits: Circuit A - 4Ω, 8Ω & 12Ω resistors in parallel. Circuit B - 4Ω, 8Ω & 12Ω resistors in series. 4Ω 4Ω 8Ω 12Ω 8Ω 12Ω 12 Volts A 12 Volts For numbers 2 through 6, answer the questions qualitatively (no numbers or equations needed). 2. Circuit B, explain how and why current changes as you add resistors. 3. Circuit A, explain how and why the current changes at point A as you add the resistors. 4. Circuit A, which branch has the most current moving through it, why? 5. Circuit B, what would happen to the brightness of the 8Ω lightbulb as the 12Ω & 4Ω resistors are removed, why? 6. Circuit A, what would happen to the brightness of the 8Ω lightbulb as the 12Ω & 4Ω resistors are removed, why? Ohm’s Law – Fancy definition and then easier definition...and then explanation Fancy: The current through a conductor between two points is directly proportional to the potential difference across the two points. (potential difference = voltage drop) Easy: The speed of stuff flowing through a metal is affected directly by the voltage drop across that metal. Variables & units: Variable I V R Current Voltage Resistance Unit Ampere Volt Ohm Unit Abbreviation A V Ω Graphical representation: The graph shows how voltage and current is related, but where is resistance? It’s the slope of the graph! Voltage (V) Explanation: The speed of the current out of the battery depends on the total resistance of the circuit and the strength of the battery pushing the current around. Building an equation: If the slope of the graph is resistance, we can create an equation for Ohm’s law that can be used to calculate values for circuits. Current (A) The slope of a linear graph is the change in vertical (voltage, V) divided by change in horizontal (current, I). Slope = Resistance 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑆𝑙𝑜𝑝𝑒 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝐶𝑢𝑟𝑟𝑒𝑛𝑡 𝑅= 𝑉 𝐼 Solve for V and Ohm’s law is commonly displayed as V = IR Calculation Examples 1Ω 2Ω 3Ω For a series circuit, in order to calculate the current out of the battery at point A or to find the voltage drop across a specific resistor you must: 1. first find total resistance, a. Add in series RTS = 1Ω + 2Ω + 3Ω RTS = 6Ω A 6 Volts 2. second use Ohm’s law equation to solve for current at point A using RTS = 6Ω, a. 𝑉 = 𝐼𝑅 𝑉 𝐼=𝑅 6𝑉 𝐼 = 6Ω 𝑰 = 𝟏𝑨 b. The current out of the battery is 1 ampere. i. Since the circuit is series, there’s only 1 path for the current to take. So the current through each resistor is the same. 3. third to find the voltage drop across each resistor use V = IR with each resistor & the current through the circuit. a. For the 1Ω resistor V = IR V = (1A)(1Ω) V = 1V b. For the 2Ω resistor V = IR V = (1A)(2Ω) V = 2V c. For the 3Ω resistor V = IR V = (1A)(3Ω) V = 3V i. Notice: if you add all the voltage drops it equals the voltage of the battery. For a parallel circuit, in order to calculate the current out of the battery at point A or to find the current through a specific resistor, you must: 1Ω 2Ω 1. first find total resistance, a. Use the relationship: i. ii. 1 𝑅𝑇𝑃 1 𝑅𝑇𝑃 = = iii. 𝑅𝑇𝑃 = 1 + 1Ω 6 + 6Ω 6Ω 11 1 𝑅𝑇𝑃 = 1 + 2Ω 3 + 6Ω 1 3Ω 2 6Ω 1 𝑅1 + 1 𝑅2 + 1 𝑅𝑇𝑃 1 𝑅3 = 3Ω 6+3+2 6Ω 1 𝑅𝑇𝑃 11 = 6Ω A 𝑹𝑻𝑷 = 𝟎. 𝟓𝟓𝛀 6 Volts 2. second use Ohm’s law equation to solve for current at point A using 𝑹𝑻𝑷 = 𝟎. 𝟓𝟓𝛀, 𝑉 6𝑉 a. 𝑉 = 𝐼𝑅 𝐼=𝑅 𝐼 = 0.55Ω 𝑰 = 𝟏𝟏𝑨 b. The current out of the battery is 11 amperes. i. Since the circuit is parallel, there’s 3 paths for the current to take. So the current through each resistor is different if the resistances are different. The current will be the same if the resistors are the same. c. The voltage drop across each resistor, in a parallel circuit, is easy to obtain as it is equal to the voltage drop across the battery. Concept – voltage drops across parallel circuits are equal for each branch. This means that the voltage drop across: i. the 1Ω resistor = 6V ii. the 2Ω resistor = 6V iii. the 3Ω resistor = 6V 3. third, to find the currents through each resistor use V = IR with each individual resistor and the voltage of the battery. The current through: 𝑉 6𝑉 a. The 1Ω resistor: 𝑉 = 𝐼𝑅 𝐼=𝑅 𝐼 = 1Ω 𝑰 = 𝟔𝑨 lowest resistance, highest current b. The 2Ω resistor: 𝑉 = 𝐼𝑅 𝐼= c. The 3Ω resistor: 𝑉 = 𝐼𝑅 𝐼= 𝑉 𝑅 𝑉 𝑅 𝐼= 𝐼= 6𝑉 2Ω 6𝑉 3Ω 𝑰 = 𝟑𝑨 𝑰 = 𝟐𝑨 highest resistance, lowest current i. Electricity follows the path of least resistance. 2Ω Practice 3Ω 1. For the parallel circuit, find the total resistance, current at point A and current through each branch. 4Ω A 12 Volts 2Ω 3Ω 2. Find the total resistance, current and voltage drop across each resistor. 12 Volts 4Ω