Combining uncertainties
Choose when to add absolute uncertainties, add percentages or include powers.
Process wire measurements, find resistivity from a resistance–length graph and evaluate uncertainty.
The wire diameter is 0.50 mm. Find its resistivity from the supplied R–L line, excluding the contact/lead resistance. Give 3 significant figures.
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.
| L / m | V / V | I / A |
|---|---|---|
| 0.2 | 0.18 | 0.2 |
| 0.4 | 0.3 | 0.2 |
| 0.6 | 0.42 | 0.2 |
| 0.8 | 0.54 | 0.2 |
| 1 | 0.66 | 0.2 |
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.
| Point | Length / m | Resistance / Ω |
|---|---|---|
| 1 | 0.2 | 0.9 |
| 2 | 0.4 | 1.5 |
| 3 | 0.6 | 2.1 |
| 4 | 0.8 | 2.7 |
| 5 | 1 | 3.3 |
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.
Convert the wire diameter to metres. Give your answer to 3 significant figures.
Multiply the mm diameter by 10⁻³.
Answer: 5.00 × 10^-4 m (3 significant figures).
Find the radius in metres. Give your answer to 3 significant figures.
Radius is half the diameter. Keep unrounded values for the final calculation.
Answer: 2.50 × 10^-4 m (3 significant figures).
Calculate the circular cross-sectional area. Give your answer to 3 significant figures.
Use πd²/4 after converting diameter to metres.
Answer: 1.96 × 10^-7 m² (3 significant figures).
Find the resistance–length gradient using the first and last readings. Give your answer to 3 significant figures.
Subtract the resistances, then divide by the length difference. The fixed contact contribution cancels.
Answer: 3.00 Ω m⁻¹ (3 significant figures).
Now calculate the final answer to the main question.
Gradient = ρ/A, so ρ = gradient × πd²/4. Use diameter in metres.
Answer: 5.89 × 10^-7 Ω m (3 significant figures).
Gradient = ρ/A, so ρ = gradient × πd²/4. Use diameter in metres.
These tasks keep the same readings. Another question starts a new dataset. Your answers are checked separately.
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.
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.
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.
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.
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