* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
[Type the company name] Resistance vs. CrossSectional Area IB Physics HL Adam Pfoertsch [Pick the date] Abstract As a student, one is expected to know mathematical formulae by heart, how to take them apart, rearrange them and know what changes what. However, when put into practice, certain equations might not fit perfectly anymore, which is why nothing beats experimenting and putting models to the test. In this experiment I wanted to figure out if the equation for finding resistivity in a material is precise compared to a real life counterpart. By putting a current through several pieces of aluminum wire and then comparing the results to each other. I expected the results to show that the larger the cross-sectional are of a wire, the less resistant it is, therefore letting more voltage flow through it. Expectations What I expected was that as the cross-sectional are of the wire increases, the resistivity of it 𝐿 decreases. According to 𝑅 = 𝑝 𝐴 with p being the resistivity constant of the material, L being the length in meters, and A being the cross-sectional area in cm2. Since A is inversely proportional to R, when A increases, this means that R should decrease. Also, if A decreases, R should proportionately increase. With my data however, I used a different formula to find R, instead I used 𝑉 = 𝐼𝑅 but since I am finding 𝑉 𝐼 R I simply moved the I over to make𝑅 = . With this, I can now find R because I collected V and I in my data collecting. Variables Current (Amps) – Independent Voltage – Dependent Material/Length of wire – controlled Power Supply strength - controlled Materials A power supply that can independently control Voltage and Amps 2 crocodile clips with preferably close to zero resistance Aluminum wire A Multimeter Procedure 1. Make sure the power supply is off before connecting any wires to it 2. Cut three pieces of Aluminum wire to 30 cm 3. Connect the crocodile clips to each a plus and negative port and connect them to the aluminum wire making sure the circuit is closed properly 4. Set the multimeter to Voltage and make sure the cables are in the right ports 5. Turn on the power supply with both knobs at zero 6. Connect the ends of the multimeter to the crocodile clips at the aluminum wire. 7. Turn the voltage knob up all the way. The displays should still show zero 8. Slowly turn the current knob up, watching the voltage on both the supply meter and the voltmeter 9. Measure each readable value (Amps, Volts on Power supply/Voltmeter) at .5 Amp steps 10. Combine two pieces of aluminum wire together and repeat steps 8 and 9 11. Repeat with three pieces of aluminum wire Figure 1: Schematic of Experiment Power supply Crocodile Clip Aluminium Wire Data Table 1: Measurements with 1 (one) piece of Aluminum Wire Current (Amps) Volts Volts Resistance (Power (Multimeter) (Ω) Supply) 0.50 0.1 0.0110 0.0220 1.00 0.2 0.0245 0.0245 1.51 0.3 0.0366 0.0242 2.00 0.4 0.0516 0.0258 2.51 0.5 0.0645 0.0257 3.00 0.5 0.0531 0.0177 Table 2: Measurements with 2 (two) pieces of Aluminum wire Current (Amps) 0.51 1.00 1.50 2.00 2.49 3.00 Volts (Power Supply) 0.1 0.2 0.3 0.4 0.5 0.6 Volts Resistance (Multimeter) (Ω) 0.0097 0.0203 0.0310 0.0431 0.0566 0.0686 0.0190 0.0203 0.0207 0.0216 0.0227 0.0229 Diameter: 1.20mm Diameter: 4.38mm Table 3: Measurements with 3 (three) pieces of Aluminum Wire Current (Amps) 0.50 1.00 1.50 2.00 2.50 3.00 Volts (Power Supply) 0.1 0.2 0.3 0.4 0.5 0.6 Volts Resistance (Multimeter) (Ω) 0.0108 0.0212 0.0316 0.0427 0.0533 0.0638 0.0216 0.0212 0.0211 0.0214 0.0213 0.0213 Diameter: 7.5mm Figure 2: First Trial 0.0800 0.0700 Voltage 0.0600 0.0500 0.0400 0.0300 0.0200 0.0100 0.0000 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 2.50 3.00 3.50 Current (Amps) Figure 3: Second Trial 0.0800 0.0700 Voltage 0.0600 0.0500 0.0400 0.0300 0.0200 0.0100 0.0000 0.00 0.50 1.00 1.50 2.00 Current (Amps) Figure 4: Third Trial 0.08 0.07 Voltage 0.06 0.05 0.04 0.03 0.02 0.01 0 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 Current (Amps) Figure 5: Combined Graphs 0.0800 0.0700 0.0600 1 wire Voltage 0.0500 2 wires 0.0400 3 wires 0.0300 Linear (1 wire) Linear (2 wires) 0.0200 Linear (3 wires) 0.0100 0.0000 0.00 0.50 1.00 1.50 2.00 2.50 Current (Amps) Calculations: 𝑅= 𝑅= .0110 .50 = .0220Ω, 𝑅 = .0097 .51 = .0190Ω, etc. 𝑉 𝐼 3.00 3.50 Analysis While at first the data may seem inconclusive, upon further inspection one can see that the resistance does change depending on the three wires. While only one wire has the highest resistance of the three, the second trial and third trial are both lower than each, fully confirming my expectations. While they are only very minimal differences, this is also expected since a piece of aluminum wire 30 cm long cannot possibly have a high amount of resistance, compared to maybe an inch thick steel wire. However, the first trial contains one data point that does not match up to the rest, which also throws off the trend line by a lot. This last point should be ignored because it is most likely either a recording error or measuring error. Otherwise, the two other data lines seem probably and realistic. Evaluation This experiment could have been improved in several ways to enhance it. First, while a sufficient number of data was collected, more could have always been better. While we do get to see that there is a difference, more data points would have enabled a more precise way to distinguish them. For example, instead of collecting data every .5 amps, data could be collected every .25 or even every tenth of an amp. This way, a change would be discoverable more easily than with current data. Additionally, instead of only using up to three pieces of aluminum wire, more could have definitely been used. With another 3 thicknesses of wires, more change could have been noticeable and would have strengthened the idea of thickness decreasing resistance. However, only so many wires fit into one crocodile clip, so another way of connecting the two would have to be conceived, which might change some of the variables, and change the data. Uncertainties, as with every experiment, also remains a slight problem here. While the power supply can only display voltage to a tenth, this problem was solved by hooking up the multimeter to the circuit, which can display up the a thousandth of a volt. If we had instead relied on the power supply read out, we might have gotten completely different data, data that might not have confirmed the model, or even showed information that is not expected in any way. Conclusion In conclusion, this experiment fully confirmed the initial hypothesis and even the model given to students in the textbook. By having tested that cross-sectional area does influence resistance in an inverse way, we are now more confident when using the formula in solving theoretical problems that it will confirm if we did it in practice.