Lab+4+Assignment+esponse 2

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Jun 19, 2024

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5CL Lab 4 Assignment (Remote) Viktoriya Reshetnyak 05/09/2024 When finished adding your responses on each slide, save this as a PDF file and upload it to CANVAS

Activity 1: Hand Draw the circuit diagram for one of your simple circuit of a light bulb and a power supply, include the voltmeter and ammeter connection in the diagram. Label all circuit elements along with the voltage and current measurements made. Provide evidence that the voltage drops across ALL elements in your circuit sum to zero. 2 when we measure the voltage so that it hits the red cable first (+), the + 1.61 V. If we move the wires closer to the battery, switching the locations of the red and black wires, the voltage change is -1.61 V. Therefore, it is evident that the sum is Zero. The battery had the same magnitude, only opposite in sign. V battery voltage- lightbulb voltage=0

Activity 1: Graph 3 V vs. I from #1 and #2 measurement. Include legend showing the resistance of the lightbulb. I vs. R from #3 and #4 measurement. Include legend showing the voltage of the battery.

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Activity 1: From the observation of the graphs, draw some conclusions from activity 1. ● From the provided graphs, it's evident that as the resistance increases, the voltage also increases proportionally while keeping the current constant. This relationship is in line with Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them, given constant temperature and other physical conditions. 4

Member name: ______________________________ Activity 2: (Series circuit 1) Screenshot and Hand Draw the circuit diagram for your light bulbs in the series circuit. Include in the diagram all the voltmeters and ammeters you place to measure the voltage and current for each circuit element. Label all circuit elements along with voltage and current measurements. With the bulbs in series, how much current flows through the bulbs compared to the current flowing through the power supply? Provide evidence that current is conserved through the elements in your circuit. 5 ● when measuring the current coming out of the buttery, the current was 0.87 A. I then measured the current coming out of each lightbulb / coming in to each lightbulb, which was 0.87A. 3 measurements were needed, since I have 3 elements ( two lightbulb and one battery) Because the current is the same throughout , the current is conserved throughout the single loop.

Member name: ______________________________ Activity 2: (Series circuit 2) Screenshot and Hand Draw the circuit diagram for your light bulbs in the series circuit. Include in the diagram all the voltmeters and ammeters you place to measure the voltage and current for each circuit element. Label all circuit elements along with voltage and current measurements. With the bulbs in series, how much current flows through the bulbs compared to the current flowing through the power supply? Provide evidence that current is conserved through the elements in your circuit. 6 ● when measuring the current coming out of the buttery, the current was 0.71 A. I then measured the current coming out of each lightbulb / coming in to each lightbulb, which was 0.71A. 3 measurements were needed, since I have 3 elements ( two lightbulb and one battery) Because the current is the same throughout , the current is conserved throughout the single loop.

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Member name: ______________________________ Activity 3: (Parallel Circuit 1) Screenshot and Hand Draw the circuit diagram for your light bulbs in the parallel circuit. Include in the diagram all the voltmeters and ammeters you place to measure the voltage and current for each circuit element. Label all circuit elements along with voltage and current measurements. With the bulbs in parallel, how much current flows through each bulb compared to the current flowing through the power supply? Provide evidence that current is conserved through the elements in your circuit. 7 ● I=I(2)+I(1) ● I=9.47 A; I(2)=8.50A; I(1)=0.97A ● When measuring the current, the total I is 9.47 A. It makes sense that the current coming out of each battery is 8.50A and 0.97 A . This shows that I=I(2)+I(1), in other words the sum of the currents entering the junction is equal to the sum of the currents leaving the junction

Member name: ______________________________ Activity 3: (Parallel Circuit 2) Screenshot and Hand Draw the circuit diagram for your light bulbs in the parallel circuit. Include in the diagram all the voltmeters and ammeters you place to measure the voltage and current for each circuit element. Label all circuit elements along with voltage and current measurements. With the bulbs in parallel, how much current flows through each bulb compared to the current flowing through the power supply? Provide evidence that current is conserved through the elements in your circuit. 8 ● I =I(2)+I(1) ● I=3.40 A; I(2)=2.40A; I(1)=1.00A ● When measuring the current, the total I is 3.40 A. It makes sense that the current coming out of each battery is 2.40A and 1.00 A . This shows that I=I(2)+I(1), in other words the sum of the currents entering the junction is equal to the sum of the currents leaving the junction

Activity 3: In your parallel circuit, there are 3 loops. Use one of the parallel circuit and d raw each of these loops separately and label their voltage and current measurements. In each loop, show that the voltages of all elements sum to zero. 9 when measuring the voltage drop across elements in my parallel circuit, I get +85V, and -85 v (when switching the cables). When measuring them for the battery and for each of the two lightbulb, we see the same results. This shows that the voltage changes across the elements sun to zero. (Thay have the same voltage drop across them.)

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Compare the voltage drops on the lightbulbs in parallel to those for the light bulbs in series. Compare the brightness of the lightbulbs in the series circuit to the brightness of the lightbulbs in the parallel circuit. Bulb brightness is proportional to power (P = IV) through the bulb. Which configuration (series or parallel) consumes more power (show your work) , and which configuration is robust to one of the bulbs breaking? How would you like your car headlights to be wired? How would you wire up a city or town and why? 10 The lightbulbs in the parallel circuit was brighter compared to the lightbulbs in the series circuit. The parallel configuration consumes more power and is more robust to one of the bulbs breaking, since more charge is being stored in the lightbulb from the battery. This is because in parallel, the equivalent capacitor has greater current and can hold more charge, and thus power. Using our current and power values for the parallel circuit, I got a power of about P=(0.97 A)* (85 V) = 82W (for Lightbulb 88.0 Ohms) or P=V^2/R= 85^2/88=82W. P=(8.50 A)* (85 V) = 722.5 W ( for Lightbulb 10.0 Ohms) or P=V^2/R=85^2?10=722.5 W. For the series circuit, I got a power of P=(0.87 A)*(76.32V) = 66W (for Lightbulb 88.0 Ohms) and P=(0.87 A)*(8.67 V)=7.5W . In the series circuit, the total capacitance to hold charge is lower than any of the individual lightbulbs. For car headlights, a parallel configuration would help produce a brighter light. For an entire city or town, it is better to use a series configuration to prevent breaking.