Scale metallic-wire resistance with cross-sectional area
Use inverse area proportionality at fixed material, length and temperature without substituting diameter for area.
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 conducting-area effect.
- Use inverse ratios and square units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is one square millimetre in square metres?
What you will learn
- Use R proportional to 1/A under matched conditions.
- Convert area units and recognise the quadratic diameter relationship for a circular section.
Use the inverse area factor
For the same metallic material, conducting length and temperature, resistance is inversely proportional to cross-sectional area: R₂/R₁ = A₁/A₂. If area doubles, sample resistance halves. A 12 Ω sample with area A has 4 Ω when the area becomes 3A under matched conditions.
Convert squared units consistently when required. One mm² is 10⁻⁶ m², not 10⁻³ m². Ratios of areas already in the same units need no unnecessary conversion, but both values must represent conducting cross-section rather than outside surface or a plastic coating.
Separate area and diameter
For a circular wire, area is πd²/4, so doubling diameter quadruples conducting area. At matched length, material and temperature this gives one-quarter of the original resistance, not one-half. A factor supplied for diameter must therefore be squared before using the inverse area relation.
The shape relation applies to the actual conducting circular core. An insulating coat can increase outside diameter without increasing that core area. The numerical scaling assumes a uniform wire and suitable controlled temperature; altered heating or composition needs additional information.
Pause and explain
A circular wire's conducting diameter doubles at matched conditions. What resistance factor follows?
Worked example
A 12 Ω uniform sample's conducting area becomes three times as large while length, material and temperature stay fixed. Find R.
Show the worked solution
- Area factor is 3.
- Resistance factor is its inverse, 1/3.
- 12/3 = 4 Ω.
Answer 4 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.
Find R when a 12 Ω wire's area becomes 3A at the same length, material and temperature.
Give me a hint
Invert the area factor before scaling resistance.
Compare my reasoning
- A₂/A₁ = 3.
- R₂/R₁ = 1/3.
- R₂ = 12/3 = 4 Ω.
4 Ω.
Look for these in your work
- Use the inverse factor.
- Retain the controls.
A wire is 8 Ω at area 0.50 mm². Find R at 2.0 mm² with all other conditions matched.
Give me a hint
The area values share units, so form their ratio directly.
Compare my reasoning
- Area factor = 2.0/0.50 = 4.
- Resistance factor = 1/4.
- R₂ = 8/4 = 2 Ω.
2 Ω.
Look for these in your work
- Use conducting areas.
- Do not apply a direct area factor.
A 16 Ω circular-core wire has twice its original conducting diameter at matched length, material and temperature. Find R and explain why 8 Ω is incorrect.
Give me a hint
Square the diameter factor to obtain the area factor.
Compare my reasoning
- Area factor = 2² = 4.
- Resistance factor = 1/4.
- R₂ = 16/4 = 4 Ω, rather than the diameter-only half.
4 Ω; doubling diameter quadruples area, so 8 Ω uses the wrong factor.
Look for these in your work
- Use circular core geometry.
- Distinguish diameter from area.
Common mistakes
- Use the inverse area factor.
- Area units are squared.
- Doubling circular diameter gives four times area.
What can you explain now?
What resistance factor follows when conducting area doubles under matched conditions?
Compare with the explanation
One half.
Inverse area proportionality divides the sample resistance by the area factor.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Use inverse conducting-area scaling with squared-unit and circular-diameter 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.