How this instrument works
A Wheatstone bridge is four resistors wired into a diamond, split into two parallel branches across the same supply: one branch holds the two known resistors, R1 and R2, in series; the other holds the variable resistor R3 in series with the unknown, Rx. A galvanometer bridges the midpoint of each branch. Turn R3 until the galvanometer reads exactly zero — the bridge is ‘balanced’ — and the two midpoints sit at the same potential, so no current crosses between them. At that instant the branches divide the supply voltage identically, which forces R1·Rx = R2·R3, and rearranging for the one true unknown gives Rx = R2·R3 ⁄ R1.
The shape of the formula is what makes the bridge useful: balance depends only on the ratio between resistors, never on the supply voltage or on how accurately the galvanometer's scale is calibrated. A battery that has sagged from 6.0 V to 5.4 V, or a galvanometer whose markings are only roughly right, still finds the same null point, because both branches are thrown off by the same factor and the ratio cancels it out. That is why bridge measurements were trusted for precision resistance work for more than a century before digital multimeters existed — the instrument only has to detect zero, a job a cheap galvanometer does well, rather than measure an absolute value accurately.
The formula Rx = R2·R3 ⁄ R1 is only true at that null point; read R3 before the galvanometer has settled to zero and the number means nothing, since the derivation assumes zero current through the detector branch. R1 also cannot be zero, since it sits in the denominator — a shorted ratio arm gives no meaningful ratio at all, only a division failure. And the bridge modeled here is a DC, purely resistive circuit: extend the same idea to AC excitation with reactive components in the arms and the null condition becomes complex-valued, the territory of the related but distinct impedance bridge.
- Enter the fixed values of the ratio arm into Known resistor, R1 and Known resistor, R2 — these stay constant during a measurement.
- Wire the unknown component into the Rx position and adjust Variable (balancing) resistor, R3 until the detector reads zero current.
- Confirm the null: the galvanometer must show no deflection at all, not merely a small one, before R3 is treated as final.
- Read the result in Unknown resistance, Rx — it is computed automatically from the three resistor values at that balance point.
- Switch any resistor field to kΩ if the values run into the thousands of ohms.
Worked example — nulling a bridge at R3 = 150 Ω
Two known resistors go into the ratio arm: R1 = 100 Ω and R2 = 200 Ω, a common pairing on a decade resistance box. An unknown resistor — a strain-gauge lead, say, or a suspect wire-wound resistor pulled from a scrap bin — sits in series with the variable arm, R3. Turning the R3 dial slowly, the galvanometer's needle swings less and less until, at R3 = 150 Ω, it sits dead still: the bridge is balanced. Reading the formula off that null gives Rx = 200 × 150 ⁄ 100 = 300 Ω.
The cross-check confirms it without a calculator: R1·Rx should equal R2·R3, and indeed 100 × 300 = 30,000 while 200 × 150 = 30,000 too, the two products matching exactly as balance requires. That equality is the whole reason a nineteenth-century technician trusted this circuit over a direct meter reading — the arithmetic, not the needle's precision, is what pins the number to exactly 300 Ω.
Questions
What does it mean for the bridge to be ‘balanced’?
It means the galvanometer connecting the two branch midpoints reads exactly zero current, which happens only when both midpoints sit at the same electrical potential. At that single point the ratio of resistors in one branch matches the ratio in the other, R1 ⁄ R3 = R2 ⁄ Rx, and rearranging gives Rx = R2·R3 ⁄ R1. Away from that null the formula does not apply — the bridge has to actually be balanced first, usually by turning R3.
Why measure resistance this way instead of with an ohmmeter?
Because balance depends only on the ratio between resistors, not on the accuracy of the supply voltage or the galvanometer's calibration — the meter only has to detect zero, which a cheap galvanometer does reliably. That made the bridge the standard for precision resistance work long before digital multimeters existed, and it is still how strain-gauge load cells and platinum resistance thermometers get read today, since those sensors report a tiny resistance change that a bridge turns into an easily measured voltage.
What happens if Known resistor, R1 is set to zero?
The formula divides by R1, so a zero value makes Rx undefined rather than merely large — the calculator flags it instead of returning a number. Physically, a zero-ohm R1 shorts that arm of the bridge, which removes the ratio the whole measurement depends on; every real bridge circuit needs R1 to be a genuine, nonzero resistor.
Does the formula still work if the bridge isn't perfectly balanced?
No — Rx = R2·R3 ⁄ R1 is derived on the assumption that zero current crosses the galvanometer branch, and it only holds exactly at that null. An unbalanced bridge still produces a small detector voltage, but reading Rx from that voltage needs a different, non-null relation involving the supply voltage and the detector's own resistance, not this simple ratio.
Who actually uses a Wheatstone bridge today?
Anyone measuring a small resistance change precisely: strain-gauge load cells in a kitchen scale or a crane's overload sensor, platinum RTDs in a lab thermometer, and technicians sorting precision resistors by comparing an unknown against a known standard. The bridge topology survives inside modern instruments even where a digital readout has replaced the galvanometer and dial.
What's the most common mistake when using this formula?
Putting a resistor in the wrong arm. R1 and R2 must be the fixed ratio arm and R3 the one turned to find balance — swap R3 into the R1 or R2 field and the calculator still returns a number, just not the resistance actually connected in the circuit. Since Rx = R2·R3 ⁄ R1 is not symmetric in its three inputs, mislabeling any one of them changes the answer.