Physics 2 Lab 3

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Temple University *

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Electrical Engineering

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Jan 9, 2024

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Lab 3- Ohm’s Law Purpose: To practice Ohm’s Law and understand the application and non application of it. We also see how current and voltage are interrelated with each other. We also learn more about power supplies and how to use them in a circuit. Graph and interpret I to V data. Apparatus: Computer with PASCO 850 interface, DC power supply, Capstone and Excel software, Current and voltage probes,light bulbs, wires, various resistors Procedure: Setup ● Insert the blue current probe into a Pasport input and voltage probe into an analog input. Open Capstone and choose “voltage sensor” in the Hardware Setup menu. ● Set up a graph and assign the voltage to the y-axis and current to the x-axis. Also open digits display for both current and voltage. ● Before turning on the power supply, connect the power supply, current probe and a 68 resistor in series with each other to make a complete circuit. Afterwards, connect the voltage circuit in parallel across the resistor. Attach the positive red to the positive end of the circuit. ● Switch the control on the DC power supply to 0V and turn on the power supply.

● Slowly increase the voltage to 5.0V and monitor the digits meter to see whether your voltage reading corresponds to the power supply’s output and current is flowing through circuit. Data Collection ● Set the power supply to 0V. Open Capstone and switch the data collection from continuous mode to keep mode and click preview to begin data collection. ● Increase the voltage in 0.5V increments up to 5V, pressing keep sample after each 0.5V increment to obtain data for I-V for voltages from 0 to 5V. ● Click stop and set the power supply back to 0V. ● Click the curve fitting button at the top of the I-V graph and select the appropriate type of fitting function. Record the slope and y-intercept of the fit line with their units. ● Replace the resistor with a light bulb and repeat the previous steps. Use small increments to capture more data for the lower voltages because of the light bulb filaments. ● Fit to a quadratic function using the curve fit tool in the software. ● Save the results. Precautions: ● Assign the voltage to the y-axis and the current to the x-axis. ● Attach the positive red to the positive ends of the circuit. ● Set the power supply to 0V. ● Increase the voltage in 0.5V increments and ensure to hit “keep sample” after each time.

Errors: ● Ensure the power supply is off before connecting the power supply to the probe. ● The current not flowing in your circuit because of the power supply’s output. ● The power supply not being set to 0V. Data: 68 Ohms resistor graph

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Light bulb linear graph Light bulb as Quadratic graph

Conclusion: In today’s lab we were able to see how current resistance and voltage are interrelated. In the first graph which we obtained from measuring the current in our loop and the voltage from the 68 ohms resistor by increasing the voltage by 0.5V, keeping track of each, up to 5V. We found the slope to be 67.7Ohms which is almost 68ohms and the graph was linear. This is accurate because we used a 68 ohms resistor. In the second graph which we obtained from measuring the current in our loop and the voltage from a 6v light bulb by increasing in small voltage increments. We found the voltage to be way significantly less than that of the 68 ohms resistor used earlier since the voltage in the light bulb used was smaller than the resistor. This was also linear. I got a 17.7 slope which should have been smaller than that had I highlighted the main points from what I needed from the graph, I would have gotten a smaller number which would have been more accurate than the 17.7 slope. In the last graph, we plotted the relationship between current and voltage of a non ohmic circuit at a low current from which we got a quadratic equation of 44.4 and 0.482 and an RMSE of 0.0782. This equation was smaller than our 67.7ohms we got from our 68 ohm resistor. Questions: Question 1. If a resistor follows Ohm’s Law, what does the slope represent when you plot V as a function of I (with V on the y-axis)? What about when you plot I as a function of V (with I on the y-axis)? When you plot V as a function of I, the slope represents the resistance, contrary to when you plot I as a function of V, the slope would represent 1/R resistance.

Question 2. Based on your experimental data and fitting, did your 68 resistor follow Ohm’s law? Explain. Yes it followed Ohm’s Law because we had a 67.7 slope when we calculated that and it was a linear slope too. Question 3. Describe what happened to the current through the light bulb as the potential increased: · Was the response linear in any region? · Which fitting method was closer to the data (use the RMSE root mean square error value to back up your answer)? · Other observations? In light of this, is the light bulb filament an Ohmic circuit component? (i.e. does it follow Ohm’s Law?) As always, base your answer on your experimental data. As the potential increased, so did the current of the light bulb, which increased linearly too. As the resistance was increasing, so was the current at a smaller rate. It follows Ohm’s Law because current and resistance are inversely proportional. Question 4. Hypothesize as to why the data is linear at low voltages/currents, and then deviates at higher values. (Hint: our eyes tell us that the light intensity is higher for these higher voltages/currents. Assuming the temperature of the bulb filament corresponds to the light intensity, what can we say about the how the resistance of the filament changes with temperature? A simple qualitative answer is all that is sought here.)

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The data is linear because at lower voltages/currents and deviates at higher values because as the temperature increases, so does the resistance. The more the resistance increases, the more there is a decrease in current. Question 5. Why isn’t it a good idea to power light bulbs using constant current rather than constant voltage? To help you answer this question note that the heating of the filament is what causes the light production. The heat energy “burned up” in the bulb every second is the power: P = IV = I 2 R = V 2 /R. Use these equations and your answer to Question 4 to help you answer this question. It is a bad idea to power light bulbs using constant current rather than constant voltage because if you increase the power by increasing the voltage and keep the current constant, it will prevent the light bulb from burning out whereas if you did viceversa, it would burn out the bulb.