Material and geometry set resistance
You will be able to: Use resistivity, length and cross-sectional area to compare uniform wires.
Why is a long thin wire harder to send current through?
Two wires made of the same metal need not have the same resistance. A longer route adds resistance; a wider cross section offers more conducting area. Keep material and temperature fixed when comparing geometry.
A useful starting point: Read a complete circuit path →
Words and symbols before equations
- Resistance R
- Potential difference divided by current for a component, in ohms (Ω).
- Resistivity ρ
- Material property in Ω·m; not the same quantity as R.
- Length L
- Conducting path length in m.
- Area A
- Cross-sectional area in m², perpendicular to charge flow; 1 mm²=10⁻⁶ m².
What this picture assumes
Uniform wire, fixed temperature and resistivity ρ=2.0×10⁻⁸ Ω·m. Sketch dimensions are schematic; use labeled values, not drawn width, for area.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- R=ρL/A=0.08 Ω. Area in SI is 5e-7 m².
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
A uniform conductor obeys R=ρL/A. Doubling L at fixed material and area doubles R. Doubling A halves R. Doubling a circular wire’s radius multiplies its area by four, so R falls to one quarter.
Resistivity describes the material at a specified temperature. Ordinary metallic conductors usually become more resistive as temperature rises. The simple fixed-ρ model deliberately holds temperature constant; an ideal ohmic model ignores heating-induced changes.
To investigate geometry, compare one material at one temperature and change only length or area. Plot R vertically against L horizontally: slope is ρ/A. Contact or lead resistance may shift the intercept, so do not confuse that offset with a change in material resistivity.
A worked example, step by step
A uniform wire has ρ=2.0×10⁻⁸ Ω·m, L=2.0 m and A=0.50 mm². Find R.
- Convert A=0.50×10⁻⁶ m².
- Substitute R=(2.0×10⁻⁸)(2.0)/(0.50×10⁻⁶).
- R=0.080 Ω.
- Doubling length with the same area and temperature would give 0.160 Ω.
Area depends on radius squared. Doubling radius is not the same as doubling area.
If both length and area double, what happens to R?
Compare with an explanation
R is unchanged because the two factors cancel in ρL/A.
Predict. Change one thing. Explain.
Keep material fixed. Double length, then separately double area. Explain each factor change using the labeled geometry and graph.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
R=ρL/A=0.08 Ω. Area in SI is 5e-7 m².
Uniform wire, fixed temperature and resistivity ρ=2.0×10⁻⁸ Ω·m. Sketch dimensions are schematic; use labeled values, not drawn width, for area.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, voltage or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionTwo same-material wires have equal temperature. Wire B has twice A’s length and three times A’s cross-sectional area. (a) State the governing relation. (b) Find R_B/R_A. (c) Compare resistivity. (d) Explain a useful controlled variable in a test.
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Compare with the answer and four-point rubric
- 1 point: R=ρL/A.
- 1 point: R_B/R_A=2/3.
- 1 point: Same resistivity at the same temperature.
- 1 point: Hold temperature and material fixed while varying one geometric factor.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Resistance versus resistivity?
Resistance belongs to a component; resistivity characterizes its material at stated conditions.
RECALL 2What is the slope of R versus L?
ρ/A for a uniform wire at fixed temperature.
RECALL 3How do metals usually respond to heating?
Their resistivity increases; the fixed-resistance model then becomes an approximation.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Material and geometry set resistance
- R=ρL/A for uniform material and geometry.
- Circular cross section: A=πr².
- 1 mm²=10⁻⁶ m²; compare wires at controlled temperature.
Remember: Area depends on radius squared. Doubling radius is not the same as doubling area.
Conditions: Uniform wire, fixed temperature and resistivity ρ=2.0×10⁻⁸ Ω·m. Sketch dimensions are schematic; use labeled values, not drawn width, for area.
Refresh Kid · AP Physics 2 Unit 3 (official Unit 11) · Objectives 11.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 11.3, objectives 11.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the third AP Physics 2 unit; College Board numbers it Unit 11; the first unit in this course is official Unit 9. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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