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Resistivity practical questions

Process wire measurements, find resistivity from a resistance–length graph and evaluate uncertainty.

Practise

Resistance–length gradient

Use the first and last positions on the supplied straight line to calculate its gradient in Ω m⁻¹. Give 3 significant figures.

Practical setup and assumptions

A uniform wire of circular cross-section is measured at a fixed temperature. The ammeter is in series and the voltmeter measures the wire plus a constant contact/lead resistance.

The graph follows R = (ρ/A)L + R₀. The contact and lead resistance R₀ is constant, so it affects the intercept rather than the gradient.

Readings for this dataset
Same-case electrical readings
L / mV / VI / A
0.20.280.2
0.40.480.2
0.60.680.2
0.80.880.2
11.080.2

The linked tasks use the same readings. These are generated practice data: treat them as exact unless an uncertainty is stated. Use the stated uncertainty rather than estimating it from scatter.

Resistance against wire lengthPractice readings with a given straight line; use the line shown for calculations. Horizontal axis: Length / m, 0 to 1.2. Vertical axis: Resistance / Ω, 0 to 10. Exact values are also listed below the graph.00.20.40.60.811.2012345678910Length / mResistance / Ω12345
Read the axes and units. The values are also available in the table below.
Larger graph
Resistance against wire lengthPractice readings with a given straight line; use the line shown for calculations. Horizontal axis: Length / m, 0 to 1.2. Vertical axis: Resistance / Ω, 0 to 10. Exact values are also listed below the graph.00.20.40.60.811.2012345678910Length / mResistance / Ω12345
Graph values
Coordinates
PointLength / mResistance / Ω
10.21.4
20.42.4
30.63.4
40.84.4
515.4
Work through the calculation

Check intermediate steps, then use your unrounded working for the final answer. Each numerical check uses 3 significant figures. These steps provide support; the main answer is checked separately.

Step 1

Find the resistance change between the first and last readings. Give your answer to 3 significant figures.

Show this step

Subtract first resistance from last resistance.

Answer: 4.00 Ω (3 significant figures).

Step 2

Now calculate the final answer to the main question.

Show this step

Subtract the two resistances and divide by the length difference. The constant intercept cancels.

Answer: 5.00 Ω m⁻¹ (3 significant figures).

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Subtract the two resistances and divide by the length difference. The constant intercept cancels.

Worked solution
  1. Gradient = (5.4 − 1.4)/(1.00 − 0.20) = 5 Ω m⁻¹.
  2. Answer: 5.00 Ω m⁻¹ (3 significant figures).
Next task in this dataset
Work through this dataset
  1. Read wire diameter
  2. Cross-sectional area
  3. Process V and I
  4. Choose graph axes
  5. Resistance–length gradient (current)
  6. Calculate resistivity
  7. Interpret the intercept
  8. Diameter and area uncertainty
  9. Propagate resistivity uncertainty
  10. Control wire heating

These tasks keep the same readings. Another question starts a new dataset. Your answers are checked separately.

Another question
Question type: Resistance–length gradient
Report a problem

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Calculate resistivity from wire measurements

A wire has resistance 2.0 Ω, length 1.0 m and diameter 0.40 mm. Radius is 0.20 mm = 2.0 × 10⁻⁴ m. Area = πr² = 1.26 × 10⁻⁷ m². Resistivity ρ = RA/L = 2.51 × 10⁻⁷ Ω m, keeping the area unrounded in the calculation.

The area formula uses radius, or πd²/4 if using diameter. Convert mm to metres before squaring. Resistivity depends on material and temperature; resistance also depends on length and area. A uniform-wire calculation assumes the stated diameter represents the tested length.

Practise the diameter measurement

From diameter to resistivity

Convert wire diameter d to metres, then A = πd²/4. Calculate each measured resistance from V/I. For a uniform wire at fixed temperature, Rwire = ρL/A.

Use the gradient when there is a series contribution

The supplied model includes a constant contact/lead contribution: Rmeasured = (ρ/A)L + R₀. Plot R vertically against L. Its gradient is ρ/A, so resistivity is gradient × area. A single Rmeasured/L ratio would include R₀/L.

Uncertainty and evaluation

Under the stated first-order maximum method, area’s percentage uncertainty is twice the diameter percentage. Add that to the gradient percentage for resistivity. Check zero error, measure diameter at several positions and orientations, and control heating. Repeats can reveal variation; they do not automatically remove systematic error.

These are synthetic ideal readings with a supplied line, not a full laboratory procedure or noisy best-fit exercise.

A-level relevance

AQA 7408: 3.5.1.3; required practical 5 (selected); OCR A H556: 4.2.4(a) (selected). Tasks cover the selected skills listed above.

AQA electricity specification · OCR A specification · AQA practical handbook · OCR practical skills handbook

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