Nichrome Wire and the Vanishing Resistance Trend
Students test how the length of nichrome wire affects current. They cut 20, 40, 60, and 80 cm lengths, connect each to a 3 V battery and an ammeter in series, and record current after 5 seconds. Results: 20 cm -> 0.62 A; 40 cm -> 0.48 A; 60 cm -> 0.41 A; 80 cm -> 0.40 A. A student concludes 'resistance increases with length only up to 60 cm, then stops changing.' The wire becomes warm, and the alligator clips are taped to the wire with masking tape. The teacher notes the ammeter was zeroed once at the start.
1. Critique the student's conclusion. Identify at least two measurement artifacts that could produce the plateau.
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The conclusion is an artifact, not a property of the wire.
Compute resistance from R = V/I:
- 20 cm: 3 / 0.62 = 4.84 ohm
- 40 cm: 3 / 0.48 = 6.25 ohm
- 60 cm: 3 / 0.41 = 7.32 ohm
- 80 cm: 3 / 0.40 = 7.50 ohm
A uniform wire should give R proportional to length. The flattening at 80 cm is suspicious.
Artifact 1: fixed contact resistance. The clips, tape, and contacts add a roughly constant resistance R_c in series. Then I = V / (R_wire + R_c). As the wire gets longer, R_wire grows but R_c does not, so the current changes less and less. At long lengths the fixed contact resistance dominates and the curve appears to plateau.
Artifact 2: heating. Nichrome has a positive temperature coefficient. After 5 s the wire is not at steady state, and shorter wires carry more current and heat more. Resistance drifts during the measurement, so 'current after 5 seconds' is not a stable value.
Other contributors: battery voltage sag under load, ammeter internal resistance, poor contact on the 80 cm sample, and a single zeroing with no calibration check.
2. What should the graph of resistance vs length look like for a uniform wire, and how would you calculate resistance from the measurements? Show one calculation.
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Expected relationship: R = ρL/A, so R vs L should be a straight line through the origin for a wire of constant diameter and temperature.
Calculation: use the voltage actually across the wire, not the battery's nominal 3 V. If the measured voltage across the 40 cm wire were 3.0 V and the current 0.48 A, then R = 3.0 / 0.48 = 6.25 ohm.
Using the 20 cm and 40 cm points gives a slope of (6.25 - 4.84) / 0.20 m = 7.05 ohm/m. That predicts about 7.66 ohm at 60 cm and 9.9 ohm at 80 cm. The measured 80 cm value of 7.50 ohm falls far below the prediction, which is evidence of a series artifact rather than a real plateau. A correct analysis would plot R vs L, fit a line, and examine residuals for curvature.
3. Redesign the procedure to test the relationship. Include how to handle heating and contact resistance.
Read the worked answer
Redesign features:
- Use a four-wire measurement: drive a small constant current through the wire and measure the voltage across a fixed central span with a separate voltmeter. This removes clip and contact resistance from the measurement.
- Keep the current low (for example 0.1 A) and pulse it briefly, or use a multimeter in ohmmeter mode with the battery disconnected, so the wire stays near room temperature.
- Sand the wire ends and use the same clips for every trial; do not use masking tape as an electrical contact.
- Measure the wire diameter with a micrometer and compute resistivity from R = ρL/A.
- Repeat each length three times, allow cooling between trials, and record temperature.
- Plot R vs L and check linearity and residuals.
A water bath at fixed temperature would further control the temperature coefficient. The key judgment is that the original setup measured the sum of wire, contact, and battery resistances, not the wire alone.