phy112l.lab3resistancewire3

PHY 112 L Lab 3: Resistance of a Wire 1. Testable Question: A. How is the resistance of a wire related to the length of the wire with a constant cross- sectional area? B. How is the resistance of a wire related to the cross sectional area with a constant length? 2. Hypothesis: A. As the length of the wire increases the resistance of the wire increases because there will be more collisions causing it to clash more with the atoms. B. As the cross sectional area increases the resistance will decreases because there will be less collisions between the electrons and the atoms. 3. Variables: A. Control(s): resistivity of material (ρ), cross sectional area (A), Current (I) Independent: length of the wire (L) Dependent: resistance of the wire (R) B. Control(s): resistivity of material (ρ), current (I), length of wire (L) Independent: cross sectional area (A) Dependent: resistance of the wire (R) 4. Experimental Design: ρ = __ (μΩ∙cm ) I = __ (A) A1= ___ (mm^2) A2= ___ (mm^2) i L(cm) R1 (m Ohm) R2 (m Ohm) 1 – 8 A1 L 1-8 R 1-8 1 – 8 A2 L 1-8 R 1-8 5. Materials:  micrometer  set of wires of different metals and cross sectional areas  power supply  wire resistance operator  digital multimeter 6. Procedure: Setup:

PHY 112 L  The resistivity, ρ, of a number of resistance wires: Metal ρ (μΩ∙cm) Brass 7.29 Steel 79.1 Aluminum 4.96 Copper 1.78 Nichrome 108 1. Select 2 wires from tube container. 2. Separate them by size obtaining a thick and thin wire. 3. Measure the diameter of both wires using a micrometer and record. 4. Set up as shown above. 5. Place the thick wire into the wire resistance at 4cm. 6. Increase the distance by 2 cm for 8 trails. 7. Record the voltage of the wire using the digital multimeter. 8. Repeat steps 4-7 with the thin wire. 9. Calculate resistance of the wire by using R= ρL/A 7. Data Table: ρ = 7.29(μΩ∙cm) I = 1.000 (A) A1= .541(mm^2) A2 = 1.25 (mm^2)

PHY 112 L T.S= 1000x7.29x10^-6/(1.25x0.01) =0.585 % diff = abs(T.S-M.S)/avg(T.S,M.S) x 100 R1: =abs(1.38-1.37)/avg(1.38,1,37) x 100 = 0.385 % R2 : =abs(0.585-0.582)/avg(0.585,0.582) x 100 = 0.00437% 9. Conclusion: A. The resistance of a wire is directly related to the length of the wire with a constant cross-sectional area. This is supported by: R=(1.37)L B. The resistance of a wire is directly related to the cross sectional are with a constant length. This is supported by: R= (0.582)L 10. Evaluation: A. The hypothesis was supported. As the length. Of the wire increases the resistance of the wire increases. The accuracy was excellent with a 0.385% error that may have come from the micrometer not starting exactly at 0. The prediction R^2 was excellent at a 0.999 that may have come from not having the resistance wire tight enough. B. The hypothesis was supported. As the cross sectional area increases the resistance will decrease. The accuracy was excellent with a 0.00437% error that may have come from the micrometer not starting exactly at 0. The prediction R^2 was excellent at a 1.00 but errors still occur that may have come from not having the resistance wire tight enough.