What it does
A Wheatstone bridge is four resistors in a diamond with a galvanometer across the middle. When the ratio of the two resistors on one side equals the ratio on the other side, no current flows through the galvanometer: the bridge is balanced. A metre bridge is the classroom version: two of the arms are the two parts of a uniform wire 1 metre long, split by a sliding contact called the jockey, so the ratio of those two arms is just the ratio of the two lengths.
This page gives you a metre bridge with a mystery resistor X in the left gap and a known resistor R in the right gap. Drag the jockey along the wire, or use the fine buttons, and watch the galvanometer needle. When it reads zero, the balance length l gives X = R l / (100 − l). Tap Show the answer to check yourself, or New mystery resistor for another one.
Under it, every current in the bridge is listed, worked out from Kirchhoff’s rules, and the junction and loop rules are written out with the live numbers. It follows the Class 12 Physics chapter on current electricity (Kirchhoff’s rules, the Wheatstone bridge, the metre bridge). If series and parallel resistors are new to you, start with the series and parallel builder; for Ohm’s law itself see the ohm's law lab.
How to use it
- Pick the known resistance R for the right gap.
- Drag the jockey (the triangle under the wire) or tap the wire to put it there. Use ◀ 1 cm, ◀ 0.1, 0.1 ▶ and 1 cm ▶ for fine steps.
- The line under the picture tells you which way to move: if the needle swings right, move the jockey left, and the other way round.
- When the needle reads zero the page says Balanced and shows X = R l / (100 − l) with your numbers.
- Tap Show the answer to see the mystery value, then New mystery resistor to try again.
Worked example
The page starts with X = 6 Ω (hidden) and R = 10 Ω. With the jockey at 20 cm the galvanometer carries about 5 mA and the needle swings left, so move right. At l = 37.5 cm the needle reads zero: X = 10 × 37.5 / 62.5 = 6 Ω. At that point the table shows the same current through X and R, and the same current along both parts of the wire, because none goes through the galvanometer. Change R to 5 Ω and the balance point moves to 100 × 6 / 11 = 54.5 cm; the answer is still 6 Ω.
Tips
- The result is most accurate when the balance point is near the middle of the wire, so choose R close to X.
- Swap X and R round: the balance length becomes 100 − l. Taking the average of both readings cancels small errors at the ends of the wire.
- The balance does not depend on the cell’s voltage or the galvanometer’s resistance: only the ratio of the arms matters.
FAQ
Why does a balanced bridge not care about the galvanometer? At balance its two ends are at the same voltage, so no current flows through it whatever its resistance.
Are the cell and wire values real? They are made-up teaching values: a 2 V cell with 0.5 Ω inside, a 2 Ω wire and a 50 Ω galvanometer.
Is anything saved? No. Everything is worked out on this page and nothing leaves your device.
Where the numbers come from
- Wheatstone bridge balance P/Q = R/S, the metre bridge X/R = l/(100 − l), and Kirchhoff's junction and loop rules: standard school physics results.
- Checked against an independent NumPy mesh-current solve of the full bridge (with the cell's internal resistance and the galvanometer's resistance) and the balance condition. The cell, wire and galvanometer values are made-up teaching values.
Standard scientific values, such as constants, atomic weights and melting points, are facts found in every textbook; we name where ours were checked. If you are an author or publisher and want a value, a credit or a table changed or removed, tell us and we will act quickly.
Last updated 2026-11-09.