Nichrome Wire Resistivity: Zero Error, Averaging and Hidden Resistances
A Form 5 group is asked to determine the resistivity of a nichrome wire. Using a micrometer screw gauge they measure the diameter at five positions along the wire and record 0.51 mm, 0.52 mm, 0.52 mm, 0.53 mm and 0.52 mm. Before starting, they close the jaws fully and find the thimble scale sits 0.03 mm below the zero of the main scale. The length of the wire is measured with a metre rule as 0.750 m. Using a dry cell, an ammeter and a voltmeter connected across the wire, they obtain a current of 0.40 A and a potential difference of 2.4 V. The connecting leads are 1 m long, the crocodile clips feel warm after a minute, and the voltmeter is connected across both the wire and the ammeter. A friend says that taking five readings has removed all error and that quoting the diameter as 0.5150 mm would make the answer more accurate.
1. Correct the five diameter readings for zero error, justify the number of significant figures for the mean diameter, and calculate the resistivity of the nichrome wire.
Read the worked answer
Zero-error correction. With the jaws closed the gauge reads 0.03 mm below zero, so every reading is 0.03 mm too small. The correction is therefore added: 0.54, 0.55, 0.55, 0.56, 0.55 mm.
Mean. (d_{mean}=\frac{0.54+0.55+0.55+0.56+0.55}{5}=0.550\text{ mm})
The instrument resolves 0.01 mm, but averaging five readings justifies quoting one extra digit (0.001 mm), so 0.550 mm is defensible; 0.5150 mm would be invented precision because it implies resolution the gauge does not have.
Resistivity. (r = 2.75\times10^{-4}\text{ m}) (A = \pi r^2 = \pi(2.75\times10^{-4})^2 = 2.38\times10^{-7}\text{ m}^2) (R = V/I = 2.4/0.40 = 6.0\ \Omega) (\rho = RA/L = \frac{6.0\times2.38\times10^{-7}}{0.750} = 1.9\times10^{-6}\ \Omega\text{ m})
The answer is quoted to two significant figures, matching the two-figure resolution of the ammeter and voltmeter readings. Note that this value is an upper bound, because the measured R includes the ammeter and contact resistance (see below).
2. Assess the friend's two claims: that five readings remove all error, and that extra decimal places make the result more accurate.
Read the worked answer
The two claims confuse precision with accuracy and misunderstand the type of error present.
Claim 1 - five readings remove all error: false. Repeating removes random error only. The scatter of 0.51–0.53 mm is random and averaging shrinks its effect (uncertainty roughly half the range, about ±0.01 mm). The zero error, however, is systematic: every reading is shifted by the same −0.03 mm, so the mean is shifted by exactly the same amount. Averaging cannot cancel it; only calibrating or applying a correction can. Its size matters: 0.03/0.55 ≈ 5.5% in diameter, and because (A \propto d^2), the area - and hence ρ - would be about 11% too small if uncorrected.
Claim 2 - more decimal places mean more accuracy: false. Accuracy is limited by instrument resolution and by systematic effects, not by digits written down. Writing 0.5150 mm suggests a resolution of 0.0001 mm that the micrometer does not have; it is false precision.
A useful rule here: the percentage uncertainty of the diameter is doubled when it is squared for the area, so diameter dominates the uncertainty of ρ.
3. The leads are long, the clips are warm and the voltmeter spans the wire plus the ammeter. Explain the effect of each on the calculated resistivity and state one improvement.
Read the worked answer
Lead and contact resistance. The circuit gives (R_{measured}=V/I) for everything between the voltmeter terminals. If the voltmeter is connected across the wire and the ammeter, then (R_{measured}=R_{wire}+R_{ammeter}+R_{contacts}). Since ρ is calculated from this R, the value obtained is systematically too large. Warm clips confirm real power dissipation at the contacts, i.e. non-negligible contact resistance.
Heating. Current through the wire raises its temperature. For a metal, resistance rises with temperature, so the reading drifts upward during the experiment and the calculated ρ is too high; the warm clips simply show the same effect elsewhere in the circuit.
Improvement. Connect the voltmeter directly across the nichrome wire only (so the ammeter's resistance is excluded from the measured p.d.), and either keep the current small and switch off between readings, or use a metre bridge / potentiometer method where no current flows at balance and lead resistance is eliminated. Clean the clips and repeat quickly to limit heating.