Resistivity Measurement with Systematic Errors
A student determines the resistivity of a constantan wire as part of a physics practical. They measure the diameter at five positions using a micrometer screw gauge and record: 0.42, 0.43, 0.41, 0.44, 0.42 mm. The micrometer reads 0.02 mm when fully closed. The wire length is measured with a metre rule as 0.500 m ± 0.001 m, but this includes 2 cm of connecting wire at each end. They connect the wire in series with a 6.0 V battery, an ammeter (resistance 0.10 Ω), and a rheostat. A voltmeter (resistance 1 MΩ) is connected across the wire only. With the rheostat adjusted to give a suitable current, the ammeter reads 0.48 A and the voltmeter reads 1.20 V. The experiment is at 20 °C. The accepted resistivity of constantan is 1.2 × 10^-6 Ω m.
1. Calculate the corrected mean diameter and its absolute uncertainty.
Read the worked answer
The mean of the raw readings is (0.42 + 0.43 + 0.41 + 0.44 + 0.42) / 5 = 0.424 mm. The micrometer has a zero error of +0.02 mm (it reads 0.02 mm when closed), so the corrected mean diameter is 0.424 − 0.02 = 0.404 mm = 4.04 × 10⁻⁴ m. The spread of the raw readings is 0.44 − 0.41 = 0.03 mm, so the half-range is ±0.015 mm. Since the zero error is a constant offset, it does not affect the spread. Thus the absolute uncertainty in the diameter is ±0.015 mm = ±1.5 × 10⁻⁵ m. The corrected diameter is therefore (4.04 ± 0.15) × 10⁻⁴ m.
2. Calculate the resistivity and its percentage uncertainty.
Read the worked answer
First, the resistance of the wire is R = V / I = 1.20 / 0.48 = 2.50 Ω. The measured length includes 2 cm of connecting wire at each end, so the length of constantan wire is L = 0.500 − 0.04 = 0.460 m. The cross-sectional area is A = π(d/2)² = π(4.04 × 10⁻⁴ / 2)² = 1.282 × 10⁻⁷ m². Therefore, ρ = R A / L = (2.50 × 1.282 × 10⁻⁷) / 0.460 = 6.97 × 10⁻⁷ Ω m.
Percentage uncertainty: ΔR/R = ΔV/V + ΔI/I = (0.01/1.20) + (0.01/0.48) = 0.0083 + 0.0208 = 0.0291 (2.9%). The diameter contributes 2Δd/d = 2(0.015/0.404) = 0.0743 (7.4%). The length uncertainty is ΔL/L = (0.001 + 0.005)/0.460 = 0.0130 (1.3%), assuming an additional ±0.005 m from the 2 cm correction. Total percentage uncertainty = 2.9% + 7.4% + 1.3% = 11.6%. Hence ρ = (7.0 ± 0.8) × 10⁻⁷ Ω m.
3. Evaluate whether the result is consistent with the accepted value, considering the effect of the ammeter resistance and the micrometer zero error.
Read the worked answer
The accepted resistivity is 1.2 × 10⁻⁶ Ω m. Our value, (7.0 ± 0.8) × 10⁻⁷ Ω m, is about 5.0 × 10⁻⁷ Ω m lower, which is roughly 6 times the uncertainty. Therefore, the result is not consistent with the accepted value.
Effect of micrometer zero error: We corrected for it. Had we not, the diameter would have been 0.424 mm, giving a larger area and a larger resistivity of about 7.7 × 10⁻⁷ Ω m-still too low.
Effect of ammeter resistance: The voltmeter is connected across the wire only, so it measures the true potential difference across the wire. The ammeter resistance does not affect this measurement. If the voltmeter had been across both wire and ammeter, the measured voltage would have included the ammeter drop (0.048 V), making R appear as 2.60 Ω and ρ = 7.25 × 10⁻⁷ Ω m-still too low.
The length correction (subtracting connecting wires) increased ρ from 6.4 × 10⁻⁷ to 7.0 × 10⁻⁷ Ω m. Without it, the value would be even lower. Thus, none of these systematic errors explain the discrepancy. The most likely reason is that the wire is not constantan (the accepted value may be for a different alloy), or the length measurement includes additional contact resistance. The student should question the material identification and check for other errors, such as the sign of the zero error or unintended heating of the wire.