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The Research Experience for Teachers Program Area(s): Physics, Physical ScienceComputer Science Tools: Scratch Activity Title: “Series & Parallel Circuit Visualization”Grade Level: 9 - 10Time Required: 60 minutesRecommended Group Size: 1 to 2 studentsSummary: Students should have a basic understanding of current, resistance, series circuit and parallel circuits. In this activity, students use Scratch to investigate concepts of brightness related to current flow and current flow with resistance in series and parallel circuits. Then Ohm’s law is introduced to extend the concepts to equivalent resistance and practice with calculating values for various puter Science Connection: Keywords: circuit, series, parallel, current, resistance, Ohm’s lawPre-Requisite Knowledge: Students should have an introductory understanding of basic circuits involving current & resistance.Learning Objective: Using Scratch to learn foundational concepts involving current flow in series and parallel circuits, and to practice calculations with various circuits.Materials List: Computer, paper, pencil or pen and calculatorPreparation: Open the webpage Read instructions on right. Click on green flag. Lab Activity: Series Answers (For Teachers):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: How many 10 ohm resistors are there? _____1____What is the total resistance of the circuit? ___10_ohms_____What is the current in the wire? ___0.9 amps____Press the “s” key for circuit s. Notice how fast it is moving. Fill in the blanks below:How many 10 ohm resistors are there? ___2____What is the total resistance of the circuit? ___20 ohms____What happened to the speed of the electron? ___slowed down____What is the current in the wire? ___0.45 amps____How much brighter is this one compared to circuit a? ____d____ (brighter, dimmer, no change)Press the “d” key for circuit d. Notice how fast it is moving. Fill in the blanks below:How many 10 ohm resistors are there? ___3____What is the total resistance of the circuit? __30 ohms_____What happened to the speed of the electron? ___slowed down____What is the current in the wire? ___0.3 amps___How much brighter is this one compared to circuit s? ____d____ (brighter, dimmer, no change)Main Series Concepts: based on what you saw, choose the best answer that finishes the sentences below. When resistors are added in series, the resistance of the circuit ___i____ (increases, decreases, stays the same). When resistors are added in series, the current of the circuit _____d___ (i, d, s).When current increases, the brightness of the bulb ___i___ (i, d, s)Lab Activity: ParallelPress 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:How many 10 ohm resistors are there? ___1___What is the total resistance of the circuit? ___10 ohms____What is the current in the wire? ____0.9 amps___ 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: How many 10 ohm resistors are there? ____2____What is the total resistance of the circuit? ___5 ohms____Compare the resistance in this one to circuit q. Resistance this one ___<__ circuit q. (<, >, =)Compare the speed of the electrons in the main wire to the branches. Branch speed __>__ Main wire speed. (<, >, =). Compare the speed of the electrons in the main wire in this one compared to circuit q. Speed of this one __>__ speed of circuit q. (<, >, =).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: How many 10 ohm resistors are there? ___3___What is the total resistance of the circuit? __30 ohms____Compare the resistance in this one to circuit w. Resistance this one ___<__ circuit w. (<, >, =)Compare the speed of the electrons in the main wire to the branches. Branch speed __>__ Main wire speed. (<, >, =). Compare the speed of the electrons in the main wire in this one compared to circuit q. Speed of this one __>__ speed of circuit q. (<, >, =).Main Parallel Concepts: based on what you saw, choose the best answer that finishes the sentences below.When resistors are added in parallel, the resistance of the circuit ___d__ (increases, decreases, stays the same).When resistors are added in parallel, the current in the main part of the circuit ___i___ (i, d, s)What happened to the current in the parallel branches as you added resistors in parallel? __s___ (i, d, s)Depending on your answer to #8c, what should happen to the brightness of the bulbs as you add 10 ohm resistors in series? ___s__ (i, d, s).SchematicsTo 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 & ConceptsTo 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: 1RTP=1R1+1R2+1R3+… An example is worked out below for a 2 ohm, 4 ohm and an 8 ohm resistor wired in parallel. 1RTP=12Ω+14Ω+18Ω 1RTP=48Ω+28Ω+18Ω 1RTP=4+2+18Ω=78Ω RTP1=8Ω7 or RTP=8Ω7 or 1.14Ω It’s important to remember “electricity will take the path of least resistance”.Brightness depends on current flow: increase current flow increase brightness.Practice 1. Find the total resistance of the following circuits: 416560030226012 Volts4Ω 8Ω 12Ω12 Volts4Ω 8Ω 12Ω412750240030A12 Volts12Ω8Ω4ΩA12 Volts12Ω8Ω4Ω Circuit A - 4Ω, 8Ω & 12Ω resistors in parallel. Circuit B - 4Ω, 8Ω & 12Ω resistors in series. RT = 2.18? RT = 24?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.The current decreases because there is more resistance to current flow.3. Circuit A, explain how and why the current changes at point A as you add the resistors.The current increases at point A because there are more paths for the current to take thereby decreasing total resistance of the circuit. This is like drinking with more than one straw side by side. 4. Circuit A, which branch has the most current moving through it, why? The 4? branch because more current flows through the path of least resistance5. Circuit B, what would happen to the brightness of the 8? lightbulb as the 12? & 4? resistors are removed, why?It would get brighter as less resistance provides more current flow, and higher current flow means brighter bulb6. Circuit A, what would happen to the brightness of the 8? lightbulb as the 12? & 4? resistors are removed, why?There would be no change in the brightness. Current (in parallel circuits) depends on the voltage drop divided by the resistance. Voltage drop across resistors in parallel is the same for all, and since the voltage and resistance is constant the current through the resistor is also constant. I = V/ROhm’s Law – Fancy definition and then easier definition...and then explanationFancy: 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. 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.VariableUnitUnit AbbreviationCurrentIAmpereAVoltageVVoltVResistanceROhm?Variables & units: 5496560130810Current (A)Voltage (V)00Current (A)Voltage (V)Graphical representation: The graph shows how voltage and current is related, but where is resistance? It’s the slope of the graph!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.The slope of a linear graph is the change in vertical (voltage, V) divided by change in horizontal (current, I). Slope = Resistance Slope=change in verticalchange in horizontal Resistance= VoltageCurrent R=VI Solve for V and Ohm’s law is commonly displayed as V = IR4513580140970A6 Volts1Ω 2Ω 3ΩA6 Volts1Ω 2Ω 3ΩCalculation Examples 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:first find total resistance, Add in series RTS = 1? + 2? + 3? RTS = 6?second use Ohm’s law equation to solve for current at point A using RTS = 6?, V=IR I=VR I=6V6Ω I=1A The current out of the battery is 1 ampere. Since the circuit is series, there’s only 1 path for the current to take. So the current through each resistor is the same. third to find the voltage drop across each resistor use V = IR with each resistor & the current through the circuit. For the 1? resistor V = IR V = (1A)(1?) V = 1VFor the 2? resistor V = IR V = (1A)(2?) V = 2VFor the 3? resistor V = IR V = (1A)(3?) V = 3VNotice: if you add all the voltage drops it equals the voltage of the battery. 506857019050A6 Volts3Ω2Ω1ΩA6 Volts3Ω2Ω1ΩFor 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: first find total resistance, Use the relationship: 1RTP=1R1+1R2+1R31RTP=11?+12?+13?1RTP=66?+36?+26? 1RTP=6+3+26?1RTP=116?RTP= 6Ω11 RTP= 0.55Ω second use Ohm’s law equation to solve for current at point A using RTP= 0.55Ω, V=IR I=VR I=6V0.55Ω I=11A The current out of the battery is 11 amperes. 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. 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:the 1? resistor = 6V the 2? resistor = 6V the 3? resistor = 6V third, to find the currents through each resistor use V = IR with each individual resistor and the voltage of the battery. The current through:The 1? resistor: V=IR I=VR I=6V1Ω I=6A lowest resistance, highest currentThe 2? resistor: V=IR I=VR I=6V2Ω I=3A The 3? resistor: V=IR I=VR I=6V3Ω I=2A highest resistance, lowest currentElectricity follows the path of least resistance. 514985090805A12 Volts4Ω3Ω2ΩA12 Volts4Ω3Ω2ΩPracticeFor the parallel circuit, find the total resistance, current at point A and current through each branch. RT = 0.92? Current at point A = 13 ampsCurrent through the 4? resistor = 3 ampsCurrent through the 3? resistor = 4 ampsCurrent through the 2? resistor = 6 amps454025010858512 Volts2Ω 3Ω 4Ω12 Volts2Ω 3Ω 4ΩFind the total resistance, current and voltage drop across each resistor. RT = 9?Current out of the battery = 1.3 amps Voltage drop across the 4? resistor = 5.2 voltsVoltage drop across the 3? resistor = 3.9 voltsVoltage drop across the 2? resistor = 2.6 volts ................
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