Metre Bridge Readings That Do Not Add Up to 100 cm
In a practical revision session a diploma student measures an unknown coil with a metre bridge. The resistance box in the right gap is 6.0 ohm and the coil is in the left gap. Pressing the jockey near 40 cm gives a null at 40.0 cm, and the student records 4.00 ohm. A second student repeats the run after interchanging the two gaps and finds the null at 57.6 cm from the left end, which works out to 4.41 ohm for the same coil. The bridge wire is 1 m of 26 SWG manganin on a wooden base; the copper strips are tarnished and two terminal screws are only finger-tight. The record book quotes the accepted value as 4.2 ohm with a tolerance of plus or minus 2 per cent. The students must decide what single value goes into the record, whether the two readings may simply be averaged, and what must be done before the examiner signs the page.
1. Compute the unknown resistance from both balance points and show why the geometric mean, not either single value, is the number to report.
Read the worked answer
For the coil in the left gap: X/R = l/(100 − l), so X = 6.0 × 40.0/60.0 = 4.00 ohm.
After interchange the 6.0 ohm box is on the left and the coil on the right: R/X = l/(100 − l), so 6.0/X = 57.6/42.4 = 1.358, giving X = 4.42 ohm (4.41 with the recorded figures).
The two balance points do not add to 100 cm: 40.0 + 57.6 = 97.6. In an ideal bridge the second null would sit at 100 − 40.0 = 60.0 cm. The 2.4 cm shortfall is a systematic end or contact resistance in series with the gaps, which behaves like a shift of the effective wire ends.
If the fault inflates one reading by (1 + d), swapping the gaps deflates the other by 1/(1 + d), so X1 × X2 = X² to first order: X = sqrt(4.00 × 4.41) = sqrt(17.64) = 4.20 ohm.
That combined value lies inside the plus or minus 2 per cent band (4.116 to 4.284 ohm), whereas 4.00 (−4.8 per cent) and 4.41 (+5.0 per cent) each fall outside it.
2. The student proposes reporting the arithmetic mean of 4.00 ohm and 4.41 ohm. Is that defensible here, and what physical fault explains the non-complementary balance points?
Read the worked answer
Arithmetic mean = (4.00 + 4.41)/2 = 4.205, i.e. 4.20 ohm to three figures - numerically the same as the geometric mean, because d is only about 5 per cent and the two means differ by well under 0.1 per cent. The averaging formula is therefore not the real issue; either mean lands on the same recorded number.
What is at stake is whether averaging is legitimate at all. It is legitimate here only because the interchange makes the two errors equal and opposite by construction, not because averaging is generally safe. Both single readings sit outside the plus or minus 2 per cent tolerance, and a symmetric plus or minus 5 per cent spread about one accepted value is the signature of a systematic fault, not random scatter. Random scatter would shrink on repetition; this fault will not.
The physics: tarnished copper strips and finger-tight terminals put roughly 0.1 to 0.5 ohm of contact resistance in series with the gaps. Its relative effect is larger in the 4 ohm coil gap than in the 6 ohm box gap, so the two configurations cannot give complementary balance points. Fix the cause before trusting either number.
3. List the corrective actions in order of cost, and predict what the readings should do once they are taken.
Read the worked answer
Actions, cheapest first:
- Clean the copper strips and the wire ends with fine emery and dilute nitric acid, then dry them; oxide films a few micrometres thick can add tenths of an ohm.
- Tighten every terminal properly - finger-tight screws are the commonest cause of this asymmetry - and use short, thick connecting leads.
- Press the jockey firmly and lift it between trials; never drag it along the wire, which thins the wire and changes its resistance per centimetre.
- Keep the resistance-box plugs clean and firmly seated, and do not handle the coil, so its temperature stays at ambient.
Prediction after correction: contact resistance becomes negligible beside both 4 ohm and 6 ohm, the relative error in the two gaps becomes nearly equal, and the balance points become complementary, summing to about 100 cm (roughly 40.8 cm and 59.2 cm). The resistance computed from either gap then agrees to within about 0.5 per cent and the plain l/(100 − l) formula is valid without an interchange correction.
Record 4.20 ohm with an uncertainty of about 0.04 ohm, note that a correction was applied, and repeat the measurement rather than presenting a corrected but unverified figure.