Metre Bridge with End Errors and a Null Near the End
In a metre-bridge experiment, a student connects an unknown resistor R in the left gap and a 2.0 Ω standard resistor S in the right gap. The galvanometer shows a null at 89.5 cm from the left end. On interchanging R and S, the null is at 10.8 cm from the left end. The wire is 1.000 m long, but its soldered ends and the copper strips add small end resistances. Pressing the jockey harder shifts the null by 1–2 mm, and the wire is slightly non-uniform near its ends. The student must report R with a sensible uncertainty and decide whether to average the two null points, apply an end correction, or repeat with a different S. The laboratory has standard resistors of 1 Ω, 5 Ω, 10 Ω, 20 Ω and 50 Ω, all 1% tolerance, and a metre bridge wire of uniform cross-section only over its central 90 cm.
1. Using the ideal metre-bridge formula, calculate R from each null point. Why do the two values differ, and which one would you trust less?
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With R in the left gap, R/S = l/(100−l), so R₁ = 2.0×89.5/10.5 = 17.05 Ω. After interchange, S/R = l/(100−l), so R₂ = 2.0×(100−10.8)/10.8 = 2.0×89.2/10.8 = 16.52 Ω. The ideal formula assumes zero end resistance, a uniform wire and perfect contacts. The two values differ by about 0.53 Ω, or 3%, which is much larger than the 1% tolerance of S. Both nulls are near the ends, where a small shift in the jockey produces a large fractional change in the short side of the bridge. The 10.8 cm null is especially sensitive because the small length appears directly in the denominator; the 89.5 cm null is also sensitive because 100−l is small. Neither value alone is trustworthy.
2. Estimate R and its uncertainty by combining the two observations as equivalent null positions, taking the jockey reproducibility as ±2 mm.
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The second observation, with the gaps interchanged, is equivalent to a null at 100−10.8 = 89.2 cm for the original arrangement. Average the two equivalent positions: l_avg = (89.5+89.2)/2 = 89.35 cm. Then R = S l/(100−l) = 2.0×89.35/10.65 = 16.78 Ω, so R ≈ 16.8 Ω. For an uncertainty of Δl = 0.2 cm, use the fractional sensitivity ΔR/R ≈ Δl/l + Δl/(100−l) = 0.2/89.35 + 0.2/10.65 = 0.0022+0.0188 = 0.021. Thus ΔR ≈ 0.021×16.8 = 0.35 Ω. Report R = 16.8 ± 0.4 Ω from the random error alone. This average reduces the effect of a symmetrical end error, but it does not remove a one-sided contact resistance or non-uniformity, so the systematic uncertainty may still be larger.
3. Which standard resistor would you choose for a repeat measurement, and what procedure would you use to reduce end and non-uniformity errors?
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Choose S close to R so that the null falls near the middle of the wire. With R ≈ 16.8 Ω, a 20 Ω standard gives l = 100R/(R+S) = 100×16.8/36.8 = 45.7 cm, well inside the central uniform 90 cm and far from both ends. A 10 Ω standard would give about 62.7 cm, also acceptable. Avoid the 1 Ω, 5 Ω and 50 Ω choices because they push the null toward an end. For the repeat, clean the copper strips and binding posts, keep the jockey pressure light and consistent, and measure both with R left and with R right. Average the equivalent positions and state the random uncertainty from the jockey shifts. This procedure reduces end effects and keeps the measurement on the uniform part of the wire.