Find the missing resistance that balances a Wheatstone bridge, or calculate its ideal differential output from all four arms and an excitation voltage. Use the resistor positions shown below because numbering conventions differ between diagrams.

Enter your values, then select Calculate.

Chosen layout: R1 upper left, R2 lower left, R4 upper right, R3 lower right. Excitation is top relative to bottom; output is left midpoint minus right midpoint. All resistances are positive; the detector is an ideal open circuit.

            Top: +Vexc
           /          \
       R1 upper     R4 upper
         left         right
          |             |
      Left midpoint   Right midpoint
          |             |
       R2 lower     R3 lower
         left         right
           \          /
             Bottom: 0 V
Output = left midpoint − right midpoint

Bridge formula

Vout = Vexc[R2/(R1 + R2) − R3/(R4 + R3)]. Balance requires R2R4 = R1R3. Thus R1 = R2R4/R3, R2 = R1R3/R4, R3 = R2R4/R1, or R4 = R1R3/R2.

Example

For R1 = 100 Ω, R2 = 200 Ω and R3 = 300 Ω, balance needs R4 = 100 × 300/200 = 150 Ω. Each midpoint is two-thirds of the excitation voltage, so their difference is zero.

Frequently asked questions

What does positive output mean?

The left midpoint is at a higher potential than the right for the chosen excitation direction. Reversing detector leads reverses the sign.

Does zero output always mean the bridge is balanced?

No. Zero excitation also gives zero output. Balance depends on matching resistance ratios.

Can this model include a loaded detector?

No. It assumes an infinite-impedance detector and ideal resistors. Detector loading or amplifier input currents require a fuller circuit analysis.

Reference: Analog Devices, Resistive Bridge Basics.