For wires of the same material at the same temperature, resistance increases with length and decreases with cross-sectional area. When a circular wire’s diameter is halved, its cross-sectional area falls to one quarter. That area change alone makes the wire’s resistance four times as large.
The 2027 SEC G3 Physics K323 syllabus covers the relationship between resistance, length and cross-sectional area. This example changes both dimensions so you can see how their effects combine.
- Check that the material and temperature match.
- Write the length multiplier.
- Square the diameter multiplier to get the area multiplier.
- Divide the length multiplier by the area multiplier.
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Separate the two multipliers
Cross-sectional area for a round wire is π(d/2)², where d is the diameter. If the diameter is multiplied by ½, the area is multiplied by (½)² = ¼. With the material and temperature held fixed, resistance is proportional to length divided by cross-sectional area.
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Worked example | Twice as long and half as thick
Wire A has a resistance of 3.0 Ω. Wire B is made from the same material. It is twice as long as A and has half its diameter. The length factor is 2 and the area factor is ¼, so the resistance factor is 2 ÷ ¼ = 8. The resistance of B is 8 × 3.0 = 24 Ω.
Predict the direction before calculating. The extra length and smaller area both increase resistance, so the result should exceed 3.0 Ω.
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Try a case where both dimensions double
A third wire has twice the length and twice the diameter of A. Its area is four times as large, so its resistance factor is 2 ÷ 4 = ½. Its resistance is 1.5 Ω. Here the increase in cross-sectional area outweighs the increase in length.
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Frequently asked questions
Can I use radius instead of diameter?
Yes. A diameter ratio gives the same radius ratio. Square either ratio to find the area factor.
What if the wires are made from different metals?
The material affects resistance. You need the relevant material information before comparing them from dimensions alone.