Metre bridge: a drifting balance point and a mean that hides a bias
During the HSC Physics practical examination a candidate must find the resistance of a given wire with a metre bridge. The known resistance in the left gap is R = 2.0 Ω and the unknown wire is in the right gap. With the jockey tapped at successive points the candidate obtains balance at 38.6 cm, 40.2 cm and 41.9 cm from the left end, and records the mean balance length, 40.2 cm. The wire is then put in the left gap and R in the right gap; balance is now found at 58.4 cm from the left end. The answer sheet reports the unknown resistance as 2.98 Ω, obtained by substituting the mean length in S = R(100 − l)/l. The examiner notices that the copper strips and the wire ends are tarnished and that the candidate slid the jockey along the wire while pressing it down. Only ten minutes remain and the candidate asks whether the answer can be improved without repeating the whole experiment.
1. Compute S from each balance length and explain why the mean of the three is not the best estimate here.
Read the worked answer
S = R(100 − l)/l with R = 2.0 Ω.
- l = 38.6 cm: S = 2.0 × 61.4/38.6 = 3.18 Ω
- l = 40.2 cm: S = 2.0 × 59.8/40.2 = 2.98 Ω
- l = 41.9 cm: S = 2.0 × 58.1/41.9 = 2.77 Ω
The three values fall steadily, by about 0.21 Ω and then 0.21 Ω, instead of scattering above and below a mean. Random error scatters both ways; a one-sided drift is a systematic error, and averaging it does not remove the bias - it preserves it. The near-constant difference is the signature of an extra resistance added in series with the unknown, so S(measured) = S(true) + r with r ≈ 0.2 Ω. The reported 2.98 Ω is therefore about 0.2 Ω too high, and the reading taken at the largest balance length is the least contaminated.
2. State the value the candidate should report, and show how interchanging the gaps supports it.
Read the worked answer
The third reading gives 2.77 Ω, so report 2.8 Ω - two significant figures, matching the 2.0 Ω standard.
Interchanged gaps: the wire is now in the left gap and R = 2.0 Ω in the right, so S/R = l/(100 − l), giving S = 2.0 × 58.4/41.6 = 2.81 Ω.
Two arrangements, with the unknown in each gap in turn, agree at about 2.8 Ω, while the mean-based 2.98 Ω agrees with neither. The contact resistance does not vanish on interchange - it now adds to R instead of to S - but it changes the size of the bias, so agreement at 2.8 Ω is evidence that the drift is instrumental, not a property of the wire. The candidate should present 2.8 Ω, show the interchanged-gap calculation as a check, and note the likely contact/end resistance rather than silently averaging.
3. Explain the effect of sliding the jockey while pressing it, and state the correct procedure.
Read the worked answer
Sliding under pressure abrades the wire, thinning it locally and changing its resistance per unit length, and it varies the contact resistance from point to point. That produces exactly the kind of drift seen here, and it spoils the wire for later candidates; firm pressing also heats the contact, adding a thermoelectric effect.
The correct technique is to press or tap the jockey at a point, note the galvanometer deflection, lift, move, and press again - never slide while in contact. Contact strips should be cleaned with sandpaper before the experiment, and the jockey edge should be sharp.
In the examination itself, the candidate should state the probable contact/end resistance, use the interchanged-gap check, and report 2.8 Ω with a one-line remark on the bias, rather than averaging all three readings as if they were concordant.