Constantan Resistivity: Zero Error, Uncertainty and a Doubtful Agreement
A student is asked to determine the resistivity of a sample of constantan wire about 0.70 m long. With the jaws of a micrometer screw gauge closed, the thimble reads −0.03 mm; the student does not notice this. Five readings of the diameter along the wire, uncorrected, are 0.48, 0.49, 0.47, 0.48 and 0.48 mm. The length is measured with a metre rule from the 2.0 cm mark to the 72.0 cm mark; the rule has 1 mm divisions. A digital ohmmeter, specified as ±(1% of reading + 2 in the last digit), reads 2.4 Ω. The accepted resistivity of constantan at room temperature is 4.9 × 10⁻⁷ Ω m. The wire has a slight kink near one end and is taped to the bench. The student calculates a value and writes: 'The result agrees with the accepted value within experimental error.'
1. Correct the mean diameter for the zero error and estimate its absolute uncertainty. State the corrected diameter in metres and explain the sign of the correction.
Read the worked answer
The zero error is −0.03 mm. When the jaws are closed the gauge reads below zero, so every reading is too small by 0.03 mm. The correction is therefore add 0.03 mm to each reading.
Mean uncorrected diameter:
[ \bar d = \frac{0.48+0.49+0.47+0.48+0.48}{5}=0.48\ \text{mm} ]
Corrected diameter:
[ d = 0.48 + 0.03 = 0.51\ \text{mm} = 5.1\times10^{-4}\ \text{m} ]
Uncertainty: the five readings span 0.47–0.49 mm, so the scatter is about ±0.01 mm. The resolution is 0.01 mm and the zero-error setting is also uncertain by about 0.01 mm. A reasonable absolute uncertainty is therefore ±0.01 mm, i.e. about 2%. Hence (d=(5.1\pm0.1)\times10^{-4}\ \text{m}).
2. Calculate the resistivity and its percentage uncertainty. State the final result with a sensible number of significant figures.
Read the worked answer
Length:
[ L = 72.0-2.0 = 70.0\ \text{cm} = 0.700\ \text{m} ]
Cross-sectional area:
[ A=\frac{\pi d^2}{4}=\frac{\pi(5.1\times10^{-4})^2}{4}=2.04\times10^{-7}\ \text{m}^2 ]
Resistivity:
[ \rho=\frac{RA}{L}=\frac{(2.4)(2.04\times10^{-7})}{0.700}=7.0\times10^{-7}\ \Omega\ \text{m} ]
Percentage uncertainties:
- diameter: about 2%, so area: about 4%;
- resistance: ±(1% + 0.2 Ω) at 2.4 Ω, i.e. about 9%;
- length: about 0.1%.
Combining in quadrature:
[ \frac{\Delta\rho}{\rho}\approx\sqrt{4^2+9^2+0.1^2}\approx10% ]
So (\rho=(7.0\pm0.7)\times10^{-7}\ \Omega\ \text{m}). The uncertainty is dominated by the ohmmeter specification, so quoting more than two significant figures for the resistivity would be misleading.
3. Evaluate the student's claim that the result agrees with the accepted value within experimental error. Identify the most likely systematic error and state whether repeating the diameter readings would remove it.
Read the worked answer
Accepted value: (4.9\times10^{-7}\ \Omega\ \text{m}). Student value: (7.0\times10^{-7}\ \Omega\ \text{m}).
Difference:
[ \frac{7.0-4.9}{4.9}\times100%\approx43% ]
The experimental uncertainty is only about 10%, so the discrepancy is roughly four times the uncertainty. The claim of agreement is therefore not justified; the result is significantly high.
If the zero error had been ignored, the diameter would have been taken as 0.48 mm, giving about (6.2\times10^{-7}\ \Omega\ \text{m}), still about 27% high. The remaining discrepancy is most likely a systematic resistance in the leads and contacts of the ohmmeter. To obtain (4.9\times10^{-7}\ \Omega\ \text{m}) with the corrected diameter and length, the resistance would need to be about 1.7 Ω rather than 2.4 Ω - an excess of about 0.7 Ω, which is plausible for poor contact resistance.
Repeating the diameter readings would reduce the random scatter, but it would not remove a zero error that has already been corrected nor a contact-resistance error. A better test is to measure the resistance with a four-terminal method or to subtract the lead resistance by short-circuiting the probes.