Report Sheet Lab 5 Factors Affecting Electrical Resistance - Answers - Jamie Assaf

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Jan 9, 2024

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Report Sheet FACTORS AFFECTING ELECTRICAL RESISTANCE (LAB 5) PROCEDURE Part 1 (Effect of Length) Length (m) Current (A) Resistance (Ω) 0.15 m 7.55 A 0.21 0.13 m 8.3 A 0.19 0.11 m 9.24 A 0.17 0.09 m 10.78 A 0.15 0.07 m 12.34 A 0.13 EVALUATION OF DATA Part 2 (Effect of Length) 1. Insert your Excel graph of the resistance (y-axis) versus the length (x-axis) here. 2. The slope of the best fit line = y = mx 3. What does the slope of the best-fit line signify? The slope of best fit line represents change in resistance with change in length in the slope = 1.375 ohm

4. Cross sectional area ( A = π r 2 ¿ = 15 mm^2. … 15*10^-6 mm^2 5. Resistivity (  ) of Play-Doh® = R*A/l for R = 0.21*l=0.15m  (0.21/15*10^-6) / (0.15) = 0.21*10^-4 ohm*m Part 3 (Effect of Cross-Sectional Area) Radius (m) Cross-Sectional Area (m 2 ) Current (A) Resistance (Ω) 0.017 m 15 mm 15.20 0.11 0.014 m 10 mm 12.40 0.13 0.013 m 15 mm 10.45 0.15 0.010 m 20 mm 8.50 0.19 Part 4 (Effect of Cross Section Area Analysis) 1. Insert your Excel plot of Resistance (y-axis) versus the inverse of the cross-sectional area (x-axis) here. 2. What relationship appears to exist between resistance and the cross-sectional area of the Play-Doh® resistors? The relationship between resistance and cross-sectional is in inverse, for example as the value of cross- section changes (increases) the resistance value decreases. 3. Slope of Resistance versus cross sectional are inverse = AB/AC = 0.06/1.7*10^4 = 3.52*10^-6 4. What does this slope represent? The slope represents varied resistance with 1/A, inverse. 5. Resistivity (  ) of Play-Doh® = 0.212*10^-4

Part 5 (Consolidation of your Findings) 1. ¿ ρ Step 2 − ρ Step 4 ∨ ¿ ρ Step 2 + ρ Step 4 2 ∗ 100% % Difference = ¿ = _____________________________________ 2. From the Resistance equation ( R = ρL A ), derive the units for this resistivity. POST-LAB QUESTION 1. If you are trying to power your electric lawnmower using a long extension cord, explain why it would be better to use 12-gauge (diameter = 2.053 mm) rather than 16-gauge (diameter = 1.291 mm) wire. 2. Use what you have learned about resistance and the equation for electrical power (power lost to joule heating is given by P = I 2 R ) to calculate the ratio of the power lost due to heating in these two wires. Assume the current in both wires is the same.

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