Module 8 Lab 8 Electric Circuits DHayes

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Northern Virginia Community College *

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MECHANICS

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Electrical Engineering

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Apr 3, 2024

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Circuits Name __ Donna Hayes ____ This Lab uses the PhET simulation Circuit Construction Kit AC . Each lab counts for 20 points and it is about 1.42% of the total grade, each little point counts, do not miss it! Introduction Current is a flow of charged particles, in metal wires the conduction electrons make up the flow of charges. The charges are moving because there is a potential difference. The current is measured in Amperes (A), the potential difference across a circuit element is called voltage and it is measured in Volts. A complete, closed path where electrons can flow is called a circuit. Current in the circuit passes through electrical devices, typical electrical devices are resistors, capacitors, inductors, diodes, LED, etc… In these activities you will work on circuits with resistors or traditional light bulbs. You will work with series and parallel connections of resistors or light bulbs. The last activity is dedicated to alternating current. Important formulae and assumptions for all activities. 1. When the elements in a circuit are connected in series, the same current passes through all elements which are in series. 2. When the elements in a circuit are connected in series, the sum of all voltages across each element adds up to the voltage provided by the power source in the circuit (battery or generator). 3. When the elements in a circuit are connected in parallel, the current is different for each element and the respective values add up to the total current drawn from the battery. 4. When the elements in a circuit are connected in parallel with each other, the voltage is the same for each element. 5. Power is the energy over time for a circuit element. The voltage, the current, and the resistance are linked by Ohm's law, which is Voltage = Resistanceequivalent ×current = ⇒⇒ V = RI Elaborating this equation we can also find the current as I = V R or the resistance as R = V I . If resistors are connected in series, the total resistance, also called equivalent resistance is 1/11

R equivalent = R 1 + R 2 + R 3 + R 4 + ... . If resistors are connected in parallel, the total resistance, also called equivalent resistance is R equivalent = ¿ The power is measured in SI units as Watts and it is equal to Power = voltage×current = VI where V is the potential difference across the circuit element and I is the current flow through it. Activity 1 Series circuit V ideo description of the available tools on the simulatio n Part A (4pts) 1. Open the simulation PhET simulation Circuit Construction Kit AC and double click on the tab labeled Lab. 2. On the right upper corner of the screen select "Show current, electrons", "Labels", and "Values". 3. From the options on the left side of the screen, drag one battery and three resistors. Using wires, build a circuit with one battery and 3 resistors in series. 4. By default the battery voltage is 9.0 V, change it into 10.0 V. (Click on the battery and a menu to change the value of the voltage will appear on the bottom of the screen). Keep the resistors values as 10 Ohms. 5. Using the Voltmeter, measure the voltage across each resistor. Report the values on Table 1. 2/11

6. Add a screenshot of your circuit diagram. [Screenshot] 7. Using the Ammeter with one probe, measure the current at the different resistors of the circuit. Is the current the same value? Yes/No. Report the values on Table 1. 8. Measure the total voltage across all resistors and report the value on the last column of Table 1. 9. Measure the current at the battery and report the value on the last column of Table 1. 10.Using the formula for resistors in series, calculate the equivalent resistance. Show all your work here: [ R Equivalent = R1 + R2 + R3 + R4 + R5 + etc… R Equivalent = 10.0 Ω + 10.0 Ω + 10.0 Ω = 30.0 Ω ] 11.Using Ohm's law, show that the current value obtained with I = V total R equivalent is the same that you measured in the simulation. Show all the steps here: [ I = V/R I = 10V/ 30 Ω = 0.33 A ] 3/11

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Table 1 Resistor 1 Resistor 2 Resistor 3 Total resistances, total voltage, and total current drawn from the battery. Resistance (Ω) 10.0 Ω 10.0 Ω 10.0 Ω 30.0 Ω Voltage (V) 3.33 V 3.33 V 3.33 V 9.99 V (10 V) Current (A) 0.33 A 0.33 A 0.33 A 0.33 A Part B (5pts) 1. Select all the resistors' elements in the circuit to edit and change their values to your own chosen values. Choose random numbers. 4/11

2. Include here a screenshot of your built circuit to show your chosen values. [Screenshot] 3. In the simulation, using the Voltmeter, measure the voltage across each resistor and report the values on Table 2. 4. Using the Ammeter, measure the current at each resistor and at the battery. The resistors are in series therefore the current should be the same for each one of them. Report the values in Table 2. 5. Show that the total voltage (i.e. The sum of all three voltages at the resistors) is equal to the battery's voltage. Show the work here: [ The total voltage at battery = 10.0 V The total voltage at battery = 0.72 V + 2.06 V + 7.22 V = 10.0 V ] 6. Calculate the resistance equivalent using the formula posted under the above section "Important formulae and assumptions for all activities". Show all the steps and calculations here: [ 7.0 Ω + 20.0 Ω + 16.0 Ω = 43.0 Ω ] 7. Using Ohm's law show that the resistance equivalent obtained at the step (6) is the same value as the one obtained from the equation R = V I . Show your work here: 5/11

[R Equivalent = V/I R Equivlant = 10 V/ 1.10 A = 9.09 Ω] Table 2 Resistor 1 Resistor 2 Resistor 3 Battery Resistance (Ω) 7.0 Ω 20.0 Ω 16.0 Ω none Voltage (V) 0.72 V 2.06 V 7.22 V 10.0 V Current (A) 0.10 A 0.10 A 0.10 A 0.10 A Activity 2 Parallel circuit (4pts) 1. With the same three resistors from Activity 1B, build a circuit with three resistors in parallel with the battery. 6/11

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2. Include a screenshot of your built circuit here. Make sure to show the values on the image. [Screenshot] 3. In the simulation, using the Voltmeter, measure the voltage across each resistor and report the values on the table below. Is the voltage the same for each resistor? Yes/No? 4. Using the Ammeter, measure the current of each resistor. Report the values on Table 3. 5. Using the Ammeter, measure the total current from the battery. Report the value of the current in on Table 3. Table 3 Resistor 1 Resistor 2 Resistor 3 Battery Resistance (Ω) 25.0 Ω 100.0 Ω 35.0 Ω zero Voltage (V) 1.56 V 6.25 V 2.19 V 10 V Current (A) 0.06 A 0.06 A 0.06 A 0.06 A 7/11

6. Calculate the resistance equivalent using the formula posted under the section "Important formulae and assumptions for all activities". Show all the steps here: [ 25.0 Ω + 100.0 Ω + 35.0 Ω = 160.0 Ω ] 7. Show that the current from the battery is equal to the sum of the currents taken at the three resistors. Show all the steps here: [ 10 V + 0.06 A = 10.06 V ] Activity 3 Electric Power and compound circuit (5.5pts) In section 8.9 of our book, there is a Screencast called "Car battery power" explaining about a compound circuit with a car battery and four light bulbs. In this activity, you will build a similar circuit. Remember that power is measured in watts and that it is equal to Power = voltage×current = VI 1. From the options on the left of the screen, drag four light bulbs and build the following circuits with the battery. ALT: The circuit for this activity is represented in the image. In the image, it is shown one battery connected in parallel with two light bulbs, and further in parallel with a combination of two light bulbs linked in series. 8/11 Light Bulb 1 Light Bulb 2 Light Bulb 3 Light Bulb 4 Bat ter y

The battery voltage is 12 Volts. Each light bulb has a resistance of 20 Ohms. The light bulbs are emitting light. 2. Use a 12 V battery and four 20 ohm light bulbs. 3. Using the Ammeter, find the current for each branch, a.k.a. find the current for each light bulb. Report the values on Table 4. 4. Using the Ammeter, find the current at the battery. Using the Voltmeter, find the voltage for each light bulb. Report the values on Table 4. 5. Using the formula for power, calculate the power for each light bulb. Show all the steps and report the values on Table 4. 6. Using the same formula for power, determine the power at the battery. Show all the steps on Table 4. 7. What can you say about the brightness of the four light bulbs? Do they all have the same brightness, (Yes/No)? Provide an explanation to your answer. [ Bulbs 3 & 4 are less brighter than bulbs 1 & 2. I believe it’s because bulbs 3 & 4 are connected thru the same wire so they are sharing power where bulbs 1 & 2 have their own wires. ] 8. Determine the total power dissipated through the light bulbs. Show all the steps here: [ 12.60 W + 12.60 W + 6.30 W + 6.30 W = 37.8 W ] 9/11

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9. Is this value obtained at step (8) equal to the power provided by the battery? Yes/No? Light bulb 1 Light bulb 2 Light bulb 3 Light bulb 4 Battery Current (A) 0.60 A 0.60 A 0.30 A 0.30 A 1.50 A Voltage (V) 12.0 V 12.0 V 6.0 V 6.0 V 12.0 V Power (W) 12.60 W 12.60 W 6.30 W 6.30 W 13.50 W Table 4 Activity 4 Alternating current circuit (1.5pt) A lternating C urrent circuits differ from D irect C urrent circuits in that the direction of their current changes with time. In the classroom, the lights are actually turning on and off 60 times each second, much faster than our eyes can see, so the lights appear to be constantly on. The simulation slows down the current so we can observe it. 10/11

1. Build a series circuit with one light bulb, and instead of the battery use the "AC Voltage" power source. Include the screenshot here. [Screenshot] 2. Observe the flow of electrons. Describe the actions of the electrons in your AC circuit. [ The electrons teeter back and forth like a seesaw. ] 3. Click on the " Current chart" tool and place the sensor on your wire. Observe the graph of current vs time and the brightness of the light bulb. What do you see? Please provide an explanation to what you see. [ The brightness and the electrical current are proportionate to each other. As the electrical current increases, the brightness of the bulb increases. As the electrical current declines, the brightness of the bulb diminishes as well. ] 11/11

Module 8 Lab 8 Electric Circuits DHayes

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