Phy-112 lab 3 resitance Kyle Pfannkuch

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Grand Canyon University *

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112

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Electrical Engineering

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Dec 6, 2023

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PHY 112 L Name: Kyle Pfannkuch Class: PHY-112L, Monday 3:15-5:45 Professor: Lee Brown Lab 3: Resistance of a Wire 1. Testable Question: How is the length of the wire related to the resistance of the wire? 2. Hypothesis: As the length of the wire increases the resistance of the wire as well because the current will have a further distance to travel and therefore have more obstacles in it’s path to overcome. 3. Variables: Control(s): Brass Wire, Aluminum Wire, Current Independent: Length (L) (m) Dependent: Resistance (R) (ohms) 4. Experimental Design: i L (m) V (V) R (ohms) 1 – 8 1 – 8 5. Materials: Brass wire Battery Ammeter Voltmeter
PHY 112 L Micrometer Power Supply 6. Procedure: Metal ρ (μΩ∙cm) Brass 7.29 Steel 79.1 Aluminum 4.96 Copper 1.78 Nichrome 108 1. Obtain a brass wire and measure the diameter to find the area using a micrometer, mark measurement in the table. 2. Clamp wire into resistance apparatus and connect circuits like schematic above.
PHY 112 L 3. Turn on power supply and adjust knobs accordingly. 4. Record reading on ammeter and continues increasing length of wire. 5. Grab another wire and repeat steps 1-4. 6. Equation being investigated: R = ρ/A*L 7. Data Table: Data for brass wire rho= 7.29 microohms/cm L (cm) V (V) I (Amps) R (microohms) 3 2.8 1.056 2302.105 A = 0.0088cm^2 6 5.7 1.056 4604.211 9 8.6 1.056 6906.316 Diameter = 1.097 mm 12 11.5 1.056 9208.421 15 14.4 1.056 11510.53 Rho for copper = 7.29 microohms/cm 18 17.2 1.056 13812.63 21 20.1 1.056 16114.74 24 23 1.056 18416.84 Data for Aluminum L (cm) V (V) I (Amps) R (microohms) 3 2.2 1.055 875.2941 A = 0.017cm^2 6 4.4 1.055 1750.588 9 6.6 1.055 2625.882 Diameter = 1.475 mm 12 8.9 1.055 3501.176 15 11.1 1.055 4376.471 Rho for aluminum = 4.96 microohms/ohm 18 13.2 1.055 5251.765 21 15.4 1.055 6127.059 24 17.6 1.055 7002.353 8. Analysis:
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PHY 112 L 0 5 10 15 20 25 30 0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 f(x) = 291.76 x + 0 R² = 1 f(x) = 767.37 x − 0 R² = 1 L (cm) vs R (µΩ) Linear () Linear () L (cm) R (µΩ) MS(1) = 767.37 microohms TS(1) = 7.29 microohms/cm / 0.0088 cm^2 = 828.4 microohms % error (1) = 828.4-767.37/828.4 *100 = 7.4 % error MS(2) = 291.76 microohms TS(2) = 4.96 microohms/cm / 0.017 cm^2 = 291.8 microohms % error (2) = 291.8-291.76 / 291.8 = 0.014 % error 9. Conclusion: The length is proportional to the resistance which is supported by the equations R = 767.37L - 1E-11 and R = 291.76L + 2E-12 10. Evaluation: The hypothesis in the beginning is supported because as the length increased the current had to travel a longer distance and overcome more obstacles therefore increasing resistance proportionally to the length. The level of accuracy for the copper wire was 7.4% error and for the aluminum it was 0.014% error. Possible sources of systematic error could have been
PHY 112 L loose connections in the wire, or the power supply not being hooked up correctly. The R^2 value for both wires was 1.0. Possible sources of random error could be the power supply giving wrong readings of the current or not having the sliding end of the apparatus exactly lined up on the measurement.