5:10 PM Sat Jul 9 < Lab 07 Parallel and Series Circuits.docx PART IV: INTRODUCING THE IDEA OF 'POWER' Resistors heat up

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5 10 Pm Sat Jul 9 Lab 07 Parallel And Series Circuits Docx Part Iv Introducing The Idea Of Power Resistors Heat Up 1 (60.94 KiB) Viewed 128 times

5:10 PM Sat Jul 9 < Lab 07 Parallel and Series Circuits.docx PART IV: INTRODUCING THE IDEA OF 'POWER' Resistors heat up when current passes through them. The more resistance there is, the more they heat up. When this happens, we can discuss "How much power do they dissipate?", which just means, "How many Joules of energy are they releasing every second?". Power can be thought of as how quickly 'Q' moves. 'Q' is just a variable for heat energy, measured in Joules (like any other type of energy). In electric circuits, there are several different ways you can calculate power. Here's the easiest: POWER EQUALS CURRENT MES VOLTAGE OR P = IV So if I know how much current is passing through a resistor, and I know the voltage drop across that resistor, this can be used to easily find how much power each resistor dissipates. Your job is now to find which resistor out of all three sections above dissipates the least amount of power, and which one dissipates the most power. Show your work and your reasoning below: A
PART 1: RESISTORS IN SERIES Create a circuit using a 9V power supply and three resistors in series of DIFFERENT VALUES. Enter your resistor values below: R1 = 30 R2 = 20 R3-10 1. What is the Reff of the circuit? R1+R2+R3-3+2+1=60 2. What is the CALCULATED Itotal ? (You must show your calculations from here on. You can do it on this document, or you can hand write it and take a photo.) Itotal V/R 6=1.5 A 3. What are the CALCULATED voltage drops across each resistor? RI= Ilotal xR1=1.5x3=4.5 V R2=1.5 x2=3.0 V R3-1.5 x 1 1.5 V 4. Using PHET, build this circuit, insert an ammeter in series with the battery, and 3. voltmeters, (one across each resistor). Take a screenshot and insert it here: PART II: RESISTORS IN PARALLEL Create a circuit 9v power supply, and two resistors in parallel of DIFFERENT VALUES. Enter your resistor values below. R1 = 50 R2 = 100 1. What is the Reff of the circuit? (You must show your calculations from here on. You can do it on this document, or you can hand write it and take a photo.) 1/Ref 1/10 Reff 5x10/15 10/3 Reff 3.330 2. What is the CALCULATED total? Itotal 9/3.33-2.7 Amp 3. What is the voltage drop across the resistors? 150-10/15 x 2.7-1.8 Amp 100-5/15 x2.7 0.8 Ampl V=IR V50 1.8 x5-9V V100 0.9 x 10=9V V50-V100 because voltage drop is equal in parallel circuits 4. Calculate the current that should flow through EACH resistor below. 150 1.8 A 11002= 0.9Amp 5. Using PHET, build this circuit, insert an ammeter in series with the battery, plus 21 more in series with each resistor. Then add 2 voltmeters, (one across each resistor). Take a screenshot and insert it here: PART III: COMBINED CIRCUITS Create a circuit 12v power supply, and 4 resistors of DIFFERENT VALUES. Two must be in parallel, and the remaining two must be in series with these. Enter your resistor values below, (R1 and R2 will be the resistors in parallel): R1 = 100 R2 = 150 R3 = 200 R4=4022 6. What is the Reff of the circuit?[You must show your calculations from here on. You can do it on this document, or you can hand write it and take a photo.) Reff R1R2/R1+R2+R3+R4= 10x15/25+20+40-66 Reff-860 7. What is the CALCULATED Itotal? Itotal =v/Reff=12/86-0.182 A 8. What is the voltage drop across EACH of the resistors? V3-3.64 V-0.182 x20=3.64 V V4 R4 0.182x40-7.28 V V1 V2-R1 and R2-12-3.64-7.28-1.08 V V1 V2=1.08 V 9. Calculate the current that should flow through EACH resistor below. 12-10 x 0,182/10+15=0.0728 A 11-15 x 0.182/10+15=0,1092 A 13-14-0.182 A 10. Using PHET, build this circuit, insert an ammeter in series with the battery, plus 2 mare in series with R1 and R2. Then add 4 voltmeters. (one across each resistor). Take a screenshot and insert it here: