Resistivity Lab

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Electrical Engineering

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Apr 3, 2024

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Copyright ©2021 Robert Hamby Resistance and Resistivity 1 q q q t q I    t q I    L R    R L   R T T R  Resistance and Resistivity Lab Online Purpose: To study the factors which affect Resistance. In a simple circuit there are three quantities, Current Voltage, and Resistance. Current is the movement of charge around the circuit where the current (I) is given by: The more charge (Δq) moving in the circuit, the greater will be the current. Also, if the charge is moving faster, then Δt will be smaller and again the greater will be the current. The Voltage is what “pushes” the charge around the circuit. The larger the Voltage, the greater will be the current. Resistance (R) is what impedes (tries to prevent) the movement of charge around the circuit. Since current and resistance are inversely proportional, if the current were to decrease, this would indicate a larger resistance. If the current were to increase, this would indicate a smaller resistance. There are four main factors which affect the Resistance of a device (conductor): 1) length (L) 2) cross-sectional area (A) 3) type of material (ρ) 4) temperature (T) Temperature: In most cases, the resistance of a device will increase as the temperature increases. This means that there will be more current moving through the device at lower temperatures. Remember, this is just the general rule, it will not apply for all materials or cases. That is: So as Length: Consider three wires which are all made of the same material and all have the same cross-sectional area (diameter). Let one wire be 1 meter long, the second wire be two meters long, and the third wire be three meters long. The same amount of charge (q) will move through each of the three wires and will travel the same speed through each wire. If the charges are all released at the same time, then the charge will get through and exit wire 1 (1 m long wire) first. Then the charge would get through and exit wire 2 (2 m long wire). And finally, the charge would exit the third wire. So Δt for wire 1 is the smallest and Δt for wire 3 is the largest. Since , wire 1 has the largest current and wire 3 has the smallest current. This means that wire 1 has the smallest resistance and wire 3 has the largest resistance. The longer the wire, the greater will be the resistance. So as In fact, the resistance of a wire is directly proportional to its length: If you double the length of the wire, the resistance will double. If you triple the length of the wire, the resistance would then triple.

Copyright ©2021 Robert Hamby Resistance and Resistivity 2 Wire 1 Wire 3 Wire 2 2 2 2         d r A   A R 1    R A t q I      R  L R    R   R A L R   L RA   A R 1  Cross-Sectional Area: For this discussion, the wires will be considered to be cylinders. This means that the cross-sectional area (A) will be the area of a circle. Where this area is Given by: Now consider three wires which are all made of the same material and all have the same length . But now each has a different cross-sectional area (diameter). The cross-sectional area of a wire will determine how many charges can move down the wire at the same time. Wire 1 has the smallest area and can move (in this analogy) 3 charges at a time. Wire 2 has twice the area can move two time more charge. So that would be 6 charges at a time. Finally, Wire 3 has the largest area and can move three times the charge (or 9 charges at a time). So Δq for wire 1 is the smallest and Δq for wire 3 is the largest. Since , wire 1 has the smallest current and wire 3 has the largest current. This means that wire 1 now has the largest resistance and wire 3 (which has the most current) has the smallest resistance. So for this case, the greater the cross-sectional area is, the smaller the resistance will be. So as Specifically, the resistance of a wire is inversely proportional to its cross-sectional area: If you double the area of the wire, the resistance will be one half (½) of its initial value. If you triple the area of the wire, the resistance would then be reduced to one third (1/3) of the original value. Type of Material: With all things being equal, the type of material will determine the resistance of the device. Some materials are very good a allowing charge to move through them (low resistance). These materials are called conductors. Other materials do not allow charges to move through them very well (high resistance). These materials are called insulators. All materials have a physical property called the Resistivity (ρ). This quantity determines how the material allows to charge to move through it. The larger the resistivity of the material, the more difficult it is for charges to move through the material. That is, the higher the resistivity, the greater will be the resistance of the material. Materials which have smaller resistivities would have a smaller resistance and would make for better conductors. So as In terms of type of material and resistance, resistance is directly proportional to the Resistivity. Putting all three relationships together: We get: which can be rearranged to to solve for the resistivity. Where the SI units for resistivity are Ώ-m

Copyright ©2021 Robert Hamby Resistance and Resistivity 3 A  A Slope   ) )( ( slope A   A L R   L A R         L Slope     A L R 1   ) ( L slope   L  A L Slope  From the overall equation, the previous relationships between the Resistance (R) and ρ, L, and A can all be seen together. In terms of Length and Resistance, since they are directly proportional, a graph of Resistance versus Length should produce a linear graph. The resistance will be the dependent variable (y) and the length will be the Independent variable (x). The slope of the best fit line will be equal to Since then: Then the slope could be used to find ρ: where A is the cross-sectional area. In terms of Cross-sectional Area and Resistance, they are inversely proportional. A graph of Resistance vs Cross-sectional Area (where the resistance will be the dependent variable (y) and the length will be the Independent variable (x)) will produce a hyperbolic graph as seen to the right. The graph shows that if the Area is increasing, then the Resistance will proportionally decrease. If however, a graph of Resistance vs 1/Area is made, a linear graph will result as seen in the graph at the right. This due to the Resistance being directly proportional to 1/A. The slope of the line will be equal to since The slope could be used to find ρ since So: where L is the Length. Finally, the relationship between type of material (ρ) and Resistance can be seen graphically as well. If a graph of Resistance vs Resistivity is made, the graph should be Linear showing the directly proportional relationship between R and ρ. For this, the length (L) and area (A) are constant. The slope of the best fit line would be The slope would be equal to the ratio of the constant Length divided by the constant area.

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Copyright ©2021 Robert Hamby Resistance and Resistivity 4 Resistance and Resistivity Procedure To start the Resistivity simulator, click on the following link or copy and paste it into your Browser. Resistance in a Wire (colorado.edu) After clicking (or pasting the link) your screen should appear as the figure below left. After the simulator opens, you should see the simulator as shown in the figure below right. This lab will be composed of three parts. Each part will examine the relationship between the Resistance of the wire and one of the three quantities which determine the Resistance of a conductor. In this lab you will examine the relationships between Resistance and Length, Resistance and Area, and Resistance and Resistivity. Part 1: Resistance and Length of the wire. Use the slider bar to make sure the Resistivity is set to 0.50 Ω cm (±0.02). Record this value in the space provided above Table 1. Then use the slider bar for Area to set the Area to 5.00 cm 2 (±0.05). Record this in provided space above Table 1.

Copyright ©2021 Robert Hamby Resistance and Resistivity 5 Finally, use the slider bar to set the Length to 2.00 cm (±0.08). See the figure below. Record this value for Trial 1 in table 1. Record the Resistance for this trial in Table 1. The value of the Resistance is located above the three slider bars. For this trial, the value is 0.200 Ω . Using R, A, and L, calculate and record the Resistivity of the wire. Then repeat this process for lengths of 4.00 cm, 6.00 cm, 8.00 cm, 10.00 cm, 12.00 cm, 14.00 cm, 16.00 cm, 18.00 cm, and 20.00 cm. After completing the table, calculate the average Resistivity and record the value in the provided space in Table 1. Show one sample calculation for the Resistivity below Table 1. Part 2: Resistance and Cross-Sectional Area of the wire. For this part, leave the Resistivity set to 0.50 Ω cm and record this value in the space provided above Table 2. Set the Length to 10.00 cm and record this value in the space provided above Table 2. Finally set the Area to 1.00 cm 2 (± 0.05) and record this value for trial 1 in Table 2. Record the value of the Resistance for this trial in Table 2. Using R, A, and L, calculate and record the Resistivity of the wire. Then repeat this process for cross-sectional areas of 2.00 cm 2 , 3.00 cm 2 , 4.00 cm 2 , 5.00 cm 2 , 6.00 cm 2 , 7.00 cm 2 , 8.00 cm 2 , 9.00 cm 2 , and 10.00 cm 2 . After completing the table, calculate the average Resistivity and record the value in the provided space in Table 2. Part 3: Resistance and Resistivity of the wire. For the last part of the lab, set the Length of the wire to 10.00 cm (±0.08). Record this value above Table 3. Then set the Cross-Sectional Area to 5.00 cm 2 (±0.05) and record its value above Table 3. Finally, set the Resistivity to 0.10 Ω cm (±0.02). Record the value of the Resistance for this trial in Table 3. Then repeat this process for Resistivities of 0.200 Ω cm, 0.300 Ω cm, 0.400 Ω cm, 0.500 Ω cm, 0.600 Ω cm, 0.700 Ω cm, 0.800 Ω cm, 0.900 Ω cm, 1.00 Ω cm.

Copyright ©2021 Robert Hamby Resistance and Resistivity 6 Resistance and Resistivity Part 1 Table 1 Resistivity: 0.50 (Ώcm) Cross-Sectional Area: 5.02(cm^2) Trial Length (cm) Resistance (Ώ) Resistivity (Ώ cm) 1 1.98 0.197 0.499 2 3.96 0.394 0.499 3 6.05 0.603 0.500 4 8.03 0.800 0.500 5 10.01 0.997 0.499 6 11.99 1.19 0.498 7 13.97 1.39 0.499 8 16.06 1.60 0.500 9 18.04 1.80 0.501 10 20.00 1.99 0.499 Average ρ 0.500 Sample Calculation for resistivity (ρ): (10 points) 𝑅 = 𝜌 𝐿 𝐴 𝜌 = 𝑅𝐴 𝐿 𝜌 = (0.197)(5.02) 1.98 𝜌 = 0.499 Ώ cm

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Copyright ©2021 Robert Hamby Resistance and Resistivity 7 Part 2 Table 2: Resistivity: 0.50 Length: 10.00 Trial Area (cm 2 ) Resistance (Ώ) Resistivity (Ώ cm) 1 0.97 5.15 0.500 2 2.04 2.45 0.500 3 3.03 1.65 0.500 4 4.02 1.24 0.498 5 5.02 0.996 0.500 6 6.01 0.832 0.500 7 7.00 0.714 0.500 8 8.00 0.625 0.5 9 8.99 0.556 0.500 10 9.99 0.501 0.500 Average ρ 0.500 Sample Calculation for resistivity (ρ): (10 points) 𝑅 = 𝜌 𝐿 𝐴 𝜌 = 𝑅𝐴 𝐿 𝜌 = (0.97)(5.15) 10.00 𝜌 = 0.500 Ώ cm Part 3 Length: 10.00 cm Cross-Sectional Area: 4.97 cm^2 Ratio of Length to Area (L/A) : 2.01 Table 3: Trial Resistivity (Ώ cm) Resistance (Ώ) 1 0.1 0.201 2 0.2 0.402 3 0.3 0.604

Copyright ©2021 Robert Hamby Resistance and Resistivity 8 4 0.4 0.805 5 0.50 1.01 6 0.60 1.21 7 0.70 1.41 8 0.80 1.61 9 0.90 1.81 10 1.00 2.01 (10 points) Analysis of Resistance and Resistivity Lab Online Name: Wyatt Collier Course/Section: PHY-1631-007 Instructor: Dale Bobar 1a) Using Excel, or some other graphing program, and the data in Table 1, make a graph of measured Resistance versus Length. Display the trendline on the graph. Using dimensional analysis, determine the units of the slope of this graph. (10 points) 𝑦 = 𝑚𝑥 + 𝑏 𝑦 = 0.0996𝑥 − 0.0004 𝑚 = 0.0996𝑥 [𝑚] = 𝑅 𝐿 = 𝛺 𝑐𝑚 𝑚 = 0 .0996 𝛺 𝑐𝑚 1b) Use the slope of the line to calculate the Resistivity (ρ). (5 points) 𝑚 = 𝑦 ଶ − 𝑦 ଵ 𝑥 ଶ − 𝑥 ଵ 𝑚 = 1.6 − 0.603 16.06 − 6.05 y = 0.0996x - 0.0004 0 0.5 1 1.5 2 2.5 0 5 10 15 20 25 Resistance (Ohms) Length (cm) Length vs. Resistance

Copyright ©2021 Robert Hamby Resistance and Resistivity 9 𝑚 = 0.0996 𝑅 = 𝜌 𝐿 𝐴 𝜌 = 𝑅 𝐿 (𝐴) 𝜌 = 𝑚(𝐴) 𝜌 = (0.0996)(5.02) 𝜌 = 0 .500 Ώ cm 2) What was the relationship between the length of the wire and the measured Resistance ? Was this the expected result? Explain. (5 points) As the length of the wire increases, so does the resistance. The expected result from the graph makes sense because as the length increases, the electrons in the wire experience more collision as they pass through, thus, creating more resistance. 3) Calculate the percent difference between the average value of Resistivity in Table 1 and the value for Resistivity found using the slope of line (in question 1b). (3 points) % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = ተ ቀ 𝐿 𝐴 ቁ ௠ − ቀ 𝐿 𝐴 ቁ ௧௔௕௟௘ (ቀ 𝐿 𝐴 ቁ ௠ + ቀ 𝐿 𝐴 ቁ ௧௔௕௟௘ )/2 ተ 𝑥100% % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = ฬ 0.500 − 0.500 (0.500 + 0.500)/2 ฬ 𝑥100% % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0% 4a) Using Excel, or some other graphing program, and the data in Table 2, make a graph of measured Resistance versus 1/Area. Display the trendline and the trendline equation on the graph. Using dimensional analysis, determine the units of the slope of this graph. (10 points) 𝑦 = 𝑚𝑥 + 𝑏 𝑦 = 4.9953𝑥 + 0.0004 𝑚 = 4.9953𝑥 [𝑚] = 𝑅 ቀ 1 𝐴 ቁ = 𝛺 ቀ 1 𝑐𝑚 ଶ ቁ = 𝛺𝑐𝑚 y = 4.9953x + 0.0004 0 1 2 3 4 5 6 0 0.2 0.4 0.6 0.8 1 1.2 Resistance (Ohms) Inverse Area (cm^2) Resistance vs. Inverse Area

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Copyright ©2021 Robert Hamby Resistance and Resistivity 10 4b) Use the slope of the line to calculate the Resistivity. (5 points) 𝑚 = 4.9953 𝑅 = 𝜌 𝐿 𝐴 𝜌 = 𝑅 𝐿 (𝐴) 𝜌 = ቀ 𝑚 𝐿 ቁ = 4.9953𝛺𝑐𝑚 ଶ 10𝑐𝑚 = 0 .49953𝛺𝑐𝑚 5) What was the relationship between the Cross-sectional Area of the wire and the measured Resistance ? Was this the expected result? Explain. (5 points) The cross-sectional area of the wire has an inverse relationship with the measured resistance. This was the expected result because as they decreased, the resistance increased and vice versa. 6) Calculate the percent difference between the average value of Resistivity in Table 2 and the value for Resistivity found using the slope of line (in question 4b). (3 points) % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = ተ ቀ 𝐿 𝐴 ቁ ௠ − ቀ 𝐿 𝐴 ቁ ௧௔௕௟௘ (ቀ 𝐿 𝐴 ቁ ௠ + ቀ 𝐿 𝐴 ቁ ௧௔௕௟௘ )/2 ተ 𝑥100% % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = ฬ 0.49953 − 0.500 (0.49953 + 0.500)/2 ฬ 𝑥100% % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0 .94% 7a) Using Excel, or some other graphing program, and the data in Table 3, make a graph of measured Resistance versus Resistivity. Display the trendline and the trendline equation on the graph. Using dimensional analysis determine the units of the slope of this graph. (10 points) 𝑅 𝜌 = 𝛺 𝛺𝑐𝑚 = 1 𝑐𝑚 7b) What does the slope of the line represent? (5 points) The slope of the line is in units of ଵ ௖௠ , so the slope represents ଵ ௟௘௡௚௧௛ y = 2.011x + 0.0011 0 0.5 1 1.5 2 2.5 0 0.2 0.4 0.6 0.8 1 1.2 Resistance Resistivity Resistance vs. Resistivity

Copyright ©2021 Robert Hamby Resistance and Resistivity 11 8) What was the relationship between the Resistivity of the wire and the measured Resistance ? Was this the expected result? Explain. (5 points) 𝜌 = 𝑅𝐴 𝐿 𝑅 = 𝜌 𝐿 𝐴 … … (1) 𝑅 𝜌 = 𝐿 𝐴 … … (2) Equation (1) demonstrates that resistance has a linear relationship with resistivity. The ratio of the resistivity in equation (2) represents the constant value of the ratio of the length to the area. This verifies the linear relationship between the resistivity and the resistance. 9) Calculate the percent difference between the slope found in question 7a and the ratio of the Length to Area calculated above Table 3. (4 points) % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = ተ ቀ 𝐿 𝐴 ቁ ௠ − ቀ 𝐿 𝐴 ቁ ௧௔௕௟௘ (ቀ 𝐿 𝐴 ቁ ௠ + ቀ 𝐿 𝐴 ቁ ௧௔௕௟௘ )/2 ተ 𝑥100% % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = ฬ 2.011 − 2.01 (2.011 + 2.01)/2 ฬ 𝑥100% % 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 0 .05%