Add the resistance of thin slices
You will be able to: Integrate a position-dependent resistivity over constant cross-section.
How do you calculate resistance when the material varies along a wire?
Suppose one end of a composite resistive strip is more resistive than the other. Current must pass through every slice in sequence, so slice resistances add.
A useful starting point: Separate material resistivity from object resistance →
Words and symbols before equations
- Slice dx
- A small length along the current path.
- Local resistivity ρ(x)
- Material resistivity at position x.
- Series addition
- The same current passes through successive slices, so their voltage drops add.
What this picture assumes
Constant area; prescribed ρ(x) = ρ₀(1+βx/L). Slices are in series. Resistivity is positive throughout the controls; temperature is not calculated.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Average ρ = 4 μΩ·m; total R = 4 Ω. Integrate graph area and divide by A = 1 mm².
- 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 thin slice of uniform area A contributes dR = ρ(x)dx/A. Adding slices gives R = (1/A)∫₀ᴸρ(x)dx.
For ρ(x) = ρ₀(1+βx/L), the integral gives R = ρ₀L(1+β/2)/A. The average resistivity for this linear profile is the mean of its endpoint values.
This model specifies constant area and a prescribed positive resistivity profile. It does not infer temperature from current or solve heat flow. If area varied too, it would have to remain inside the integral.
A worked example, step by step
Let ρ₀ = 2×10⁻⁶ Ω·m, β = 2, L = 1 m and A = 1 mm². Find R.
- Write R = (ρ₀/A)∫₀ᴸ(1+2x/L)dx.
- The bracket integrates to L + L = 2L.
- R = 2ρ₀L/A = 4 Ω.
- The end resistivities are 2 and 6 μΩ·m; their average is 4 μΩ·m.
Do not substitute only the high-resistivity endpoint for the entire wire.
Does β = 0 recover the uniform-wire formula?
Compare with an explanation
Yes. The integral becomes ρ₀L/A.
Predict. Change one thing. Explain.
Change the gradient β while holding ρ₀, L and A fixed. Compare the resistivity graph area with the total resistance.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Average ρ = 4 μΩ·m; total R = 4 Ω. Integrate graph area and divide by A = 1 mm².
Constant area; prescribed ρ(x) = ρ₀(1+βx/L). Slices are in series. Resistivity is positive throughout the controls; temperature is not calculated.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, circuit topology, charge or energy conservation, or RC 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 questionDerive R for the linear profile and explain the units of the area under its ρ-versus-x graph.
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Compare with the answer and four-point rubric
- 1 point: dR = ρ(x)dx/A.
- 1 point: Integrate to ρ₀[L+βL/2]/A.
- 1 point: The graph area has units Ω·m².
- 1 point: Dividing by cross-sectional m² gives resistance in Ω.
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 1Why add dR instead of d(1/R)?
The slices form a series path.
RECALL 2What graph area matters?
The resistivity-versus-position area.
RECALL 3What if area also changes?
Use ∫ρ(x)/A(x) dx instead of taking A outside.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Add the resistance of thin slices
- dR = ρ(x)dx/A.
- R = (1/A)∫₀ᴸρ(x)dx.
- Linear profile: R = ρ₀L(1+β/2)/A.
Remember: Do not substitute only the high-resistivity endpoint for the entire wire.
Conditions: Constant area; prescribed ρ(x) = ρ₀(1+βx/L). Slices are in series. Resistivity is positive throughout the controls; temperature is not calculated.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (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. This is E&M Unit 4: Electric Circuits, numbered Unit 11 in the official combined Physics C sequence. Topics 11.1–11.8 retain their official identifiers. Models include signed charge flow, prescribed current-density and resistivity integrals, DC resistor networks, nonideal batteries and meters, capacitor combinations and finite-resistance RC transients. Circuit schematics use conventional-current references and explicit node connectivity. Initial capacitor voltage and positive time constants are stated. Unequal ideal sources are never directly wired in parallel. The optional spatial wire view supplements a complete 2D current-density explanation. Checked with the Fall 2026 clarifications. 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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