Combining uncertainties
Choose when to add absolute uncertainties, add percentages or include powers.
Resistivity practical questions
Process wire measurements, find resistivity from a resistance–length graph and evaluate uncertainty.
Practise
Propagate resistivity uncertaintyThe R–L gradient has quoted uncertainty 2%. Diameter 0.55 mm has uncertainty ±0.01 mm. Use the approximate maximum-uncertainty method, treating the input uncertainties as independent, for ρ = gradient × πd²/4. Find percentage uncertainty in ρ. Give 2 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
| L / m | V / V | I / A |
|---|---|---|
| 0.2 | 0.15 | 0.15 |
| 0.4 | 0.27 | 0.15 |
| 0.6 | 0.39 | 0.15 |
| 0.8 | 0.51 | 0.15 |
| 1 | 0.63 | 0.15 |
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.
Hint
Add the gradient percentage and twice the diameter percentage. π and the factor 4 are exact.
Worked solution
- Δρ/ρ × 100% ≈ 2% + 2 × (0.01/0.55 × 100%).
- This is stated first-order maximum propagation, not a statistical standard uncertainty.
- Answer: 5.6 % (2 significant figures).
Work through this dataset
- Read wire diameter
- Cross-sectional area
- Process V and I
- Choose graph axes
- Resistance–length gradient
- Calculate resistivity
- Interpret the intercept
- Diameter and area uncertainty
- Propagate resistivity uncertainty (current)
- Control wire heating
These tasks keep the same readings. Another question starts a new dataset. Your answers are checked separately.
Question type: Propagate resistivity uncertainty
Report a problem
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.
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