Scale metallic-wire resistance with length
Use direct proportionality to conducting length for a uniform wire of fixed material, cross-sectional area and temperature.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028. Use the stated source, operating conditions and meter ranges. Written practice is self-checked. Human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- State the qualitative wire-length effect.
- Use ratios and proportional changes.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the ratio of 1.20 m to 0.40 m?
What you will learn
- Use R proportional to L under matched conditions.
- Distinguish a sample length ratio from an unprovided contact contribution.
Use the direct length factor
For a uniform metallic wire at the same material, cross-sectional area and temperature, resistance is directly proportional to length. Thus R₂/R₁ = L₂/L₁. A 6 Ω sample at 0.40 m gives 18 Ω at 1.20 m under these matched conditions.
Use actual conducting length and consistent length units. If the two values are 20 cm and 80 cm, their ratio is four; converting both to metres gives the same factor. Do not mix 20 centimetres with a numeric 0.80 metre without conversion.
Apply the sample model, not every meter reading
The ratio describes the wire sample. A separately stated fixed contact resistance may add to a measured total and will not scale with sample length. For a 6 Ω sample plus a 1 Ω fixed contact account, tripling sample length gives 18 Ω sample and 19 Ω total, not 21 Ω total.
Heating, changed material or non-uniform area can break the simple matched comparison. Coiling a wire without changing its conducting length does not itself supply a length factor. State the controlled conditions whenever using a numerical direct proportion.
Pause and explain
A 6 Ω uniform sample's length triples under matched conditions. What is its new sample resistance?
Worked example
A uniform wire is 6 Ω at 0.40 m. Find its resistance at 1.20 m with matching material, area and temperature.
Show the worked solution
- Length factor = 1.20/0.40 = 3.
- R is directly proportional to length under the controls.
- R₂ = 6 × 3 = 18 Ω.
Answer 18 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 6 Ω at 0.40 m to 1.20 m under matched wire conditions.
Give me a hint
Find the new-to-old length ratio, then multiply sample resistance.
Compare my reasoning
- Factor = 1.20/0.40 = 3.
- R₂ = 6 × 3.
- R₂ = 18 Ω.
18 Ω.
Look for these in your work
- Use the direct factor.
- State the matched conditions.
A 12 Ω wire at 80 cm is shortened to 20 cm without changing material, area or temperature. Find its sample resistance.
Give me a hint
The new conducting length is one quarter of the old.
Compare my reasoning
- Factor = 20/80 = 1/4.
- R₂ = 12/4.
- R₂ = 3 Ω.
3 Ω.
Look for these in your work
- Use consistent centimetres.
- Scale the sample resistance directly.
A 6 Ω sample has an additional fixed 1 Ω contact contribution. Triple only its sample length under matched conditions and find the new measured total.
Give me a hint
Scale the sample, then add the unchanged contact contribution.
Compare my reasoning
- Sample R becomes 6 × 3 = 18 Ω.
- The contact remains 1 Ω.
- Total = 18 + 1 = 19 Ω.
19 Ω total, including 18 Ω sample and 1 Ω fixed contact.
Look for these in your work
- Keep the boundaries separate.
- Do not triple the fixed contact contribution.
Common mistakes
- Use conducting length and consistent units.
- Keep material, area and temperature fixed.
- Fixed contact resistance does not scale like the sample.
What can you explain now?
At matched conditions, what happens when a uniform wire's length triples?
Compare with the explanation
Its sample resistance triples.
Direct length proportionality applies to the controlled wire sample.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Use direct length ratios for a controlled wire sample and distinguish fixed contact contributions.
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.