Resistance and Resistivity
Inquiry Lab Kit for AP® Physics 1
Materials Included In Kit
Nichrome wire,16-gauge, 15 m Nichrome wire, 18-gauge, 15 m Nichrome wire, 22-gauge, 15 m
Electrical tape Multimeter
Additional Materials Required
Meter sticks, 12 Multimeters, 12 Vernier calipers, 12
Wire cutters* *for Prelab Preparation
Prelab Preparation
- Measure a 1.2-meter piece of 18-gauge nichrome wire. Cut with wire cutters. Repeat for total number of lab pairs.
- Repeat step 1 for 16-gauge nichrome wire.
- Repeat step 1 for 22-gauge nichrome wire.
Safety Precautions
Be cautious of the ends of the wires. Wear safety glasses when performing this experiment. Please follow all normal laboratory safety guidelines.
Disposal
All materials may be saved and stored for future use.
Lab Hints
- This laboratory activity can be completed in two 50-minute class periods. It is important to allow time between the Introductory Activity and the Guided-Inquiry Activity for students to discuss and design the guided-inquiry procedures. Also, all student-designed procedures must be approved for safety before students are allowed to implement them in the lab. Prelab Questions may be completed before lab begins the first day, and analysis of the results may be completed the day after the lab or as homework. An additional lab period would be needed for students to complete an optional inquiry investigation (see Opportunities for Inquiry in the Further Extensions section).
- The sample data was collected using a student multimeter (Flinn Catalog No. AP4639). If other mutlimeters will be used, it is recommended that you test the usability of the device prior to the lab.
- The students should maintain one of the multimeter probes at the 0-m mark when conducting the lab. If the probes are shifted on the wire, the resistance will not be measured through the same expanse of the wire and the same defects in the wire will not be equally encountered.
- It is important to remind students to “zero” the multimeter by pressing the probe tips together and recording the resistance. The probes have a certain amount of resistance that contribute the measurements of the wires. This value needs to be subtracted in order to attain the true resistance of the wires.
Teacher Tips
- Nichrome metal is an alloy with a nominal percent composition of 79% nickel and 21% chromium.
- Nano-circuitry is a burgeoning field of electronics. Nano-circuits are made with component parts on the scale of 10–9 meters, a billionth of a meter. The dimensions of a wire in such a circuit are typically between 1 nm to 100 nm, in both length and thickness (diameter). At this scale, there is a significant probability of electrons encountering defects within the metal crystal lattice. This means resistance in nano-wires can be great. Therefore, single-walled carbon nanotubes (SWCNT) are a viable replacement for metallic and semi-metallic wires in nano-circuitry.
- As the temperature of most conductors increases, the resistance through the conductor increases as well. A few conductors deviate from this generalization. The change in resistance based on temperature can be calculated using the following equation
{13796_Tips_Equation_2}
where ΔR is the change in resistance, R is the original resistance, and ΔT is the change in temperature. The proportionality, α, is called temperature coefficient of resistance.
Further Extensions
Opportunities for Inquiry
The resistance of a wire is also temperature dependent. As the temperature of a wire increases, there is greater random movement of its mobile electrons. As the temperature decreases, the mobile electrons slow down. Write a detailed procedure to determine the effect temperature has on resistance.
Alignment to the Curriculum Framework for AP® Physics 1 Enduring Understandings and Essential Knowledge
Materials have many macroscopic properties that result from the arrangement and interactions of the atoms and molecules that make up the material. (1E) 1E2: Matter has a property called resistivity. Learning Objectives 1E2.1: The student is able to choose and justify the selection of data needed to determine resistivity for a given material. Science Practices 3.1 The student can pose scientific questions. 3.2 The student can refine scientific questions. 3.3 The student can evaluate scientific questions. 4.1 The student can justify the selection of the kind of data needed to answer a particular scientific question. 4.2 The student can design a plan for collecting data to answer a particular scientific question. 4.3 The student can collect data to answer a particular scientific question. 4.4 The student can evaluate sources of data to answer a particular scientific question. 5.1 The student can analyze data to identify patterns or relationships. 5.3 The student can evaluate the evidence provided by data sets in a particular scientific question. 6.1 The student can justify claims with evidence.
Correlation to Next Generation Science Standards (NGSS)†
Science & Engineering Practices
Developing and using models Planning and carrying out investigations Analyzing and interpreting data Using mathematics and computational thinking Constructing explanations and designing solutions Obtaining, evaluation, and communicating information
Disciplinary Core Ideas
HS-PS3.A: Definitions of Energy HS-PS4.A: Wave Properties HS-PS3.D: Energy in Chemical Processes HS-PS3.C: Relationship between Energy and Forces HS-PS1.A: Structure and Properties of Matter
Crosscutting Concepts
Patterns Cause and effect Systems and system models Energy and matter Structure and function Stability and change
Performance Expectations
HS-PS4-1. Use mathematical representations to support a claim regarding relationships among the frequency, wavelength, and speed of waves traveling in various media. HS-PS3-5. Develop and use a model of two objects interacting through electric or magnetic fields to illustrate the forces between objects and the changes in energy of the objects due to the interaction. HS-PS2-6. Communicate scientific and technical information about why the molecular-level structure is important in the functioning of designed materials.
Answers to Prelab Questions
- For the following materials, provide a qualitative assessment of the resistivity. Explain how you made your determination.
- Conductor
The resistivity value of a conductor will be low. A conductor may be defined as a material through which electricity readily flows. In order for electricity to flow, there needs to be low resistivity.
- Insulator
The resistivity value of an insulator will be high, when compared to that of a conductor. An insulator may be defined as a material through which electricity does not readily flow. In order to impede the flow of electricity, there must be high resistivity.
- Semi-conductor
The resistivity value of a semi-conductor will lie somewhere in between the values of a conductor and an insulator. Semi-conductors conduct electricity better than an insulator, but worse than a conductor. The resistivity value of a semi-conductor will be higher than that of a conductor and lower than that of an insulator.
- Calculate the resistance of a 1.00-m piece of copper wire that is 0.750 mm in diameter.
{13796_PreLabAnswers_Equation_3}
R = 0.0380 Ω
- Calculate the expected resistance if the copper wire piece were cut to 0.500 m. Explain any difference in resistance giving respect to resistivity.
The resistance of the wire increased. The length of the wire decreased by a factor of two, and the resistance increased by a factor of two. At shorter lengths, there is an increased chance the electrons will collide with imperfections in the metal crystal lattice, increasing resistance.
- Calculate the expected resistance of the copper wire if its diameter were increased to 1.50 mm and the length remained 1.00 m. Explain any difference in resistance giving respect to resistivity.
The resistance of the wire decreased. The radius of the wire doubled, which quadrupled the cross-sectional area of the wire. With a wider area for electrons to pass through, the likelihood of the electrons colliding with imperfections in the metal crystal lattice decrease, decreasing the resistance of the wire.
Sample Data
Introductory Activity
{13796_Data_Table_1}
Guided-Inquiry Activity
{13796_Data_Table_2}
Overall Average Resistivity
{13796_Data_Equation_4}
Procedure AnalysisThe percent error value, 12%, indicates the procedure, as written, is suitable for determining the resistivity of nichrome wire. Likely sources of error may have resulted from a lack of precision in the multimeter readings. If the multimeter were able to measure to more decimal places, the resistance values would be more precise (greater significant figures), giving greater precision to the calculated resistivity values. To confirm the reliability of this procedure, the experiment should be repeated with other multimeters or ohmmeters with greater sensitivity.
Answers to Questions
Guided-Inquiry Discussion Questions
- In step 5 of the Introductory Activity, the probes of the multimeter were pressed together and the resistance was measured. Why is this a necessary step?
The probes of the multimeter were pressed together to determine if there was resistance within the probes and circuit of the multimeter. This is an important step in order to accurately measure the resistance and determine the resistivity of nichrome wire. The resistance of the multimeter and probes should be subtracted from the value displayed to attain the true resistance value of the wire.
- Identify a trend in your data between resistance and the separation distance of the probes. Does this trend agree with Equation 1 found in the Background section?
As the separation of the probes on the wire increased, the measured resistance increased as well. This agrees with the direct relationship established by Equation 1 between resistance and the length of the wire.
- Calculate the average diameter of the wire. Then calculate the cross-sectional area of the wire in square-meters (m2).
{13796_Answers_Equation_5}
- Rearrange Equation 1 from the Background to solve for resistivity (ρ).
{13796_Answers_Equation_6}
- Calculate the resistivity of nichrome wire for each separation distance. Determine the average resistivity. Record these values in an appropriate data table.
A sample calculation is provided for the 0.20-m separation distance.
{13796_Answers_Equation_7}
- How would the measured resistance of a 1.0-m nichrome wire be affected if the radius of the wire were doubled? Halved? Qualitatively predict the factor by which the resistance would change. Hint: Use Equation 1 from the Background section.
If the radius of the wire were doubled, then the bottom area term would be increased by a factor of four. Therefore, the resistance of the wire would decrease because the ratio of length to area will be smaller. If the radius of the wire were halved, then the bottom area term would be decreased by a fourth. The ratio of the wire length to area would increase, thereby increasing the resistance of the wire.
- Write a detailed procedure to determine the effect the thickness of a wire has on its resistance.
- Remove the 18-gauge wire from the meter stick and place the 16-gauge wire on the meter stick. Tape the ends to secure the wire in place.
- Turn the multimeter on and set it to measure resistance.
- Press the measuring probes together and record the resistance of the probes in an appropriate data table.
- Place one probe at the 0-cm mark and the other at the 20-cm mark. Record the resistance.
- Repeat step 4 at the 40-, 50-, 60-, 80- and 100-cm marks, maintaining the probe at the 0-cm mark.
- Turn the multimeter off.
- Using calipers, measure and record the diameter of the wire in five places.
- Repeat steps a–g, replacing the 16-gauge wire with the 22-gauge wire.
Review Questions AP® Physics 1
- A 30.4-gram sample of titanium metal is spun into a cylindrical wire of length, L, and cross-sectional radius, r. The density of titanium is 4.50 g/cm3 and its resistivity is 4.20 x 10–7 Ω∙m. Determine the values of L and r if the resistance of the wire is measured to be 1.87 Ω.
The sample of titanium metal can only be made into a wire of a certain volume based on the density of the material. The wire is assumed to be a cylinder with a cross-section of a circle.
{13796_Answers_Equation_8}
- A wire of length L has a measured resistance of 3.50 Ω. The wire is then stretched to a new length five times that of the original. The density and resistivity of the wire are unaffected. Find the resistance of the elongated wire.
The density of the wire is unaffected, so the mass and volume of the wire remain constant after being stretched. The resistivity value of the metal is also unchanged. The shape of the wire is assumed to be a cylinder.
{13796_Answers_Figure_1}
- Two wires made of the same material are drawn to be equal lengths. The resistance of the wires are measured with a multimeter to be 14 Ω and 20 Ω, respectively. The diameter of the first wire is 1.2 mm. What is the diameter of the second wire?
The resistivity, ρ, is the same for the two wires.
The length, L, can be any positive value as long as the same value is used for both wires. The length of the wires will be assumed to be 1 meter.
{13796_Answers_Figure_2}
- Data were collected for various gauges (thicknesses) of platinum wire. The data are summarized in the following chart. The equation for the regression is: y = (1.34 x 10–7)x–2.
{13796_Answers_Figure_3}
- Based on the previous results, explain why the regression is proportional to x–2 or 1/x2.
The diameter of the wire is accounted for in the cross-sectional area, A, of the wire in the following equation: R = ρ x L/A, where R is resistance, ρ is resistivity, L is the length, and A is the cross-sectional area. The crosssection of a cylindrical wire is a circle, where Area = π x r2 = π x (d/2)2. Rearranging the first equation gives R/L = ρ x 4π/d2. This leads to the inverse-square function based on the diameter.
- Can the resistivity of platinum be determined from the data and regression? Explain.
The resistance per unit length is proportional to the inverse-square of the diameter, as shown in this equation: R/L = ρ x 4π/d2. The proportionality is the product of four times the resistivity divided by pi, π. Based on the regression, the proportionality equals 1.34 x 10–7. Therefore, 1.34 x 10–7 = 4/π x ρ, ρ = 1.0 x 10–7 Ω∙m. This is a fair approximation of the resistivity value of platinum, which has a literature value of 1.05 x 10–7 Ω∙m (CRC Handbook of Chemistry and Physics, 94th ed.).
References
AP® Physics 1: Algebra-Based and Physics 2: Algebra-Based Curriculum Framework; The College Board: New York, NY, 2014.
CRC Handbook of Chemistry and Physics. 94th ed. CRC Press: Boca Raton, FL, 2013–2014; pp 15–37.
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