Physics 9702 · Paper 2

Resistivity

Resistance belongs to an object. Cut a wire in half and its resistance halves. belongs to the material: copper has one number, and that number is the same whatever shape you cut it into.

R = ρL/A

Double a wire's length and you have two resistors in series: R doubles. Double its area and you have two in parallel: R halves. So $R \propto L/A$ , and the material fixes the constant.

Same material throughout. Double the length and the current must fight through two wires, one after the other: R doubles. Double the thickness and the current has two wires to share: R halves.

R = \frac{\rho L}{A}

Symbols

$R$ = resistance of the wire (\Omega)
$\rho$ = resistivity of the material (\Omega\,\text{m})
$L$ = length (m)
$A$ = cross-sectional area (\text{m}^2)

The unit of ρ is the ohm metre (Ω m), from rearranging: $\rho = RA/L$ gives $\Omega \cdot \text{m}^2 / \text{m} = \Omega\,\text{m}$ . It is not ohms per metre.
The numbers span an enormous range: metals are near $10^{-8}$ Ω m, while insulators reach $10^{16}$ Ω m. That gap, 24 powers of ten, is why a wire conducts and its plastic coating does not.
Resistivity rises with temperature in a metal, for the same vibrating-ion reason as resistance.

Your turn— tap to reveal the worked answer (9702/11/M/J/24 Q31)

What is the unit of resistivity: Ω m⁻², Ω m⁻¹, Ω, or Ω m?

Answer: Ω m. Work it from $\rho = RA/L$ : ohms times square metres over metres leaves ohm metres. (9702/11/M/J/24 Q31)

Using the formula

Worked example

Resistance of a copper wire

A copper wire is 2.0 m long with diameter 0.50 mm. Copper's resistivity is $1.7 \times 10^{-8}$ Ω m. Find its resistance.

Area first, in metres: $A = \pi r^2 = \pi (0.25 \times 10^{-3})^2 = 2.0 \times 10^{-7}$ m².
Then $R = \dfrac{\rho L}{A} = \dfrac{1.7 \times 10^{-8} \times 2.0}{2.0 \times 10^{-7}}$ .
This gives $R \approx 0.17$ Ω. Copper wire has very small resistance, which is exactly why wires are made from copper.

Answer

R \approx 0.17 \ \Omega

Common mistake

The question gives a diameter in mm. Two traps in one: halve it to get the radius, and convert mm to m before squaring. Most lost marks on resistivity come from this step.

The stretched-wire trap

Stretch a wire to twice its length and its volume stays the same, so its area halves. Length doubled and area halved: $R \propto L/A$ goes up by $2 \div \tfrac{1}{2} = 4$ .
So stretching to double length makes the resistance four times bigger, not twice. The step that questions test is this: the volume stays constant.

Your turn— tap to reveal the worked answer (9702-style)

A wire of resistance 10 Ω is stretched to twice its original length. What is its new resistance?

Answer: 40 Ω. Length ×2 and, because the volume is fixed, area ×½. So R changes by a factor of 4. (9702-style)

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