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The meter bridge and the potentiometer are the two practicals where students most often get a number but cannot explain it in the viva. Both are "null" methods: you slide a jockey along a wire until a galvanometer reads zero. This post explains why that works, works through the standard calculations and goes through the errors that NCERT's lab manual asks you to think about.
Which exams include these instruments?
The JEE (Main) 2026 syllabus lists the Wheatstone bridge and metre bridge, and its experiment list includes the resistivity of a wire using a metre bridge. The JEE (Advanced) 2026 brochure includes the specific resistance of a wire using a meter bridge and post office box. The CBSE 2026-27 curriculum has two metre bridge experiments for Class 12, and an activity "to study the variation in potential drop with length of a wire for a steady current", which is the potentiometer principle.
The potentiometer itself is not named in either JEE syllabus or in the CBSE 2026-27 experiment list, though NCERT's lab manual still describes potentiometer experiments and some boards and olympiads use it. Check your own board's current list before deciding how much time to spend on it.
How does the meter bridge work?
The NCERT chapter Current Electricity derives the Wheatstone bridge balance condition: when the galvanometer shows no deflection, the ratios of resistances in the two arms are equal. It then notes that the meter bridge is a practical device using this principle.
In a meter bridge, two arms are a known resistance R and an unknown S, and the other two are the two parts of a uniform 100 cm wire. Because the wire is uniform, resistance is proportional to length, so at the null point l cm from the end on R's side:
R/S = l / (100 − l)
Example 1. With R = 4.0 Ω, the null point is at 40.0 cm. Then S = R(100 − l)/l = 4.0 × 60.0 / 40.0 = 6.0 Ω. If the wire under test is 0.50 m long with cross-section 1.0 × 10⁻⁷ m², its resistivity is ρ = SA/L = 6.0 × 1.0 × 10⁻⁷ / 0.50 = 1.2 × 10⁻⁶ Ω m.
Why should the null point be near the middle?
NCERT's lab manual for the metre bridge experiment tells you to choose R so the null point falls between 30 cm and 70 cm, and its discussion asks why. Here is the reason in numbers.
Suppose you can read l to within ±0.1 cm. Since S = R(100 − l)/l, the fractional error in S works out to Δl × 100 / [l(100 − l)].
At l = 50 cm: 0.1 × 100 / (50 × 50) = 0.004, or 0.4%.
At l = 30 cm: 0.1 × 100 / (30 × 70) ≈ 0.0048, or about 0.5%.
At l = 10 cm: 0.1 × 100 / (10 × 90) ≈ 0.011, or about 1.1%.
The product l(100 − l) is largest at 50 cm, so the error is smallest there.
What errors does the lab manual list?
The NCERT manual's sources of error for the metre bridge include:
A wire that is not of uniform cross-section, so resistance is not exactly proportional to length.
End resistances from the copper strips and connecting screws. The manual says the error can be reduced by interchanging the known and unknown resistances and repeating, then averaging.
A wire that is not taut and along the scale, which spoils the length readings.
A galvanometer that does not read zero with no current.
It also asks you to unplug the key between readings, because current heats the wire and changes its resistance, and to move the jockey gently, not scrape it along the wire.
How does the potentiometer work?
A driver cell drives a steady current through a long uniform wire, so the potential drops by the same amount per centimetre all along it. A cell connected against this drop through a galvanometer gives a null point where the two match. At that point no current flows through the cell, so the potentiometer measures its emf, not its terminal voltage. That is why it is better than a voltmeter for this job.
The potentiometer experiment in NCERT's lab manual compares two cells with E₁/E₂ = l₁/l₂.
Example 2. A Leclanche cell balances at 52.0 cm and a Daniell cell at 40.0 cm. E₁/E₂ = 52.0 / 40.0 = 1.30.
The companion experiment on internal resistance uses r = R(l₀ − l)/l, where l₀ is the balancing length with the cell on open circuit and l with a resistance R across it.
Example 3. l₀ = 60.0 cm, and with R = 10 Ω connected, l = 50.0 cm. r = 10 × (60.0 − 50.0) / 50.0 = 2.0 Ω.
Why can you not find a null point?
The manual's procedure gives the usual causes. If the galvanometer deflects the same way at both ends of the wire, connections may be loose, or the driver battery's emf may be less than that of the cells being tested. It also says the positive terminals of the driver battery and the cells must go to the same end of the wire, and that the rheostat should not be disturbed between the two balancing readings.
Our post on Kirchhoff's laws and sign conventions covers the circuit theory behind both instruments, and preparing for the practical exam and viva covers how to answer the examiner's questions. If you want to see how the null point moves as you change the resistances, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Day 1: Five meter bridge calculations, including one with interchanged resistors and one finding resistivity.
Day 2: Work out the percentage error at three different null points, as above.
Day 3: Four potentiometer problems: comparing emfs and finding internal resistance.
Day 4: Write two-line answers to ten viva questions from the error lists above.
Before your next practical, write down why the null point should be near the middle, in one sentence, and say it aloud.
