Combine length area and diameter changes for a metallic wire
Combine direct length and inverse area factors for one controlled sample while keeping circular geometry and contact accounts explicit.
Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Use direct length and inverse area scaling.
- Relate circular diameter to conducting area.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What area factor follows from twice a circular conducting diameter?
What you will learn
- Use R₂/R₁ = (L₂/L₁)(A₁/A₂).
- Check combined factors and resist unsupported rankings when controls change.
Combine independent factors
For the same uniform metallic material and temperature, combine the two dimension factors: R₂/R₁ = (L₂/L₁)(A₁/A₂). Doubling length and quadrupling area gives a resistance factor 2/4 = 1/2. Neither length alone nor area alone gives the final result.
If a circular wire's diameter doubles, its area becomes four times as large. Tripling length at that same time gives a total factor 3/4. An initial 8 Ω sample then becomes 6 Ω under the matched material and temperature conditions.
Check what is actually changed
The combined ratio applies to the conducting sample, with actual path length and conducting cross-section. A plastic coat's outside diameter is not the metal-core diameter. If a separately stated contact contribution remains fixed, add it after scaling the sample rather than applying the same ratio to every part of a measured total.
Temperature changes, different alloys and non-uniform sections can invalidate a simple two-factor prediction. If the given data change those controls without supplying a new material relation, explain what cannot be determined rather than forcing a convenient numerical factor.
Sources for this lesson
Pause and explain
Length triples and area quadruples. What resistance factor follows at matched conditions?
Worked example
An 8 Ω circular-core sample becomes three times as long and twice the conducting diameter. Material and temperature remain matched. Find R.
Show the worked solution
- Diameter factor 2 gives area factor 4.
- Resistance factor = 3/4.
- 8 × 3/4 = 6 Ω.
Answer 6 Ω.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Scale an 8 Ω wire to three times its length and twice its circular conducting diameter at matched conditions.
Give me a hint
Convert diameter to area, then combine the length and inverse-area factors.
Compare my reasoning
- Area factor = 2² = 4.
- Resistance factor = 3/4.
- R₂ = 8 × 3/4 = 6 Ω.
6 Ω.
Look for these in your work
- Square the diameter factor.
- Use both dimensions.
A 12 Ω sample has twice the length and 1.5 times the area at matched conditions. Find R.
Give me a hint
The combined resistance factor is the length factor divided by the area factor.
Compare my reasoning
- Factor = 2/1.5 = 4/3.
- R₂ = 12 × 4/3.
- R₂ = 16 Ω.
16 Ω.
Look for these in your work
- Use the inverse area factor.
- Keep the area factor distinct from diameter.
A wire doubles in length and conducting area but also changes to an unspecified alloy. Is unchanged R established?
Give me a hint
Check the lost same-material condition before using the cancelling dimension factors.
Compare my reasoning
- The dimension factors alone give 2/2 = 1.
- The material changed without a supplied relationship.
- The final resistance is not determined by those factors alone.
No; unchanged R requires matched material and temperature or additional material data.
Look for these in your work
- Identify the missing control.
- Do not force the dimensional factor onto a changed material.
Common mistakes
- Combine both dimension factors.
- Square a circular diameter factor.
- Do not scale fixed contacts or unknown material changes.
What can you explain now?
What resistance factor follows from twice the length and four times the area under matched conditions?
Compare with the explanation
One half.
The direct length factor and inverse area factor combine to 2/4.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Combine length and area scaling for a controlled wire sample, with circular diameter and boundary checks.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.