Circuit Analysis & Fundamentals
Wheatstone Bridge Calculator
Balance condition and unknown resistance, or the bridge output (galvanometer) voltage of an unbalanced bridge — with the bridge schematic and the Vg-vs-Rx curve.
About this calculator
A Wheatstone bridge is two voltage dividers driven by the same source, with a meter connected between their midpoints. It is one of the oldest and most precise ways to measure resistance, and the front end of countless sensor circuits: strain gauges, RTDs, and pressure transducers all read out as a bridge.
The bridge is balanced when the two dividers produce equal midpoint voltages, so the meter reads zero. That happens when the resistor ratios match: R1/R3 = R2/Rx, which rearranges to the balance condition Rx = R2·R3 / R1. In balance mode this calculator returns the unknown resistor Rx from the three known arms — the measurement principle behind a classic resistance bridge, where balance is found by adjusting a known arm until the meter nulls.
When the bridge is not balanced, the midpoints differ and a bridge (galvanometer) voltage appears: V_g = V_s·[R3/(R1+R3) − Rx/(R2+Rx)]. In output mode the tool computes that voltage from all four arms and the excitation, along with each midpoint voltage and how far Rx sits from balance. It plots V_g against Rx so you can see the null at balance and the polarity change as Rx moves through it — the small, roughly linear signal around balance that sensor bridges exploit. Use it to size a resistance bridge, to estimate a sensor bridge's output, or to teach why the null is independent of the supply voltage while the unbalanced output is proportional to it.
Design notes & common mistakes
- Balance condition: R_x = R_2·R_3 / R_1. The null depends only on ratios, so it is independent of the excitation voltage and its drift.
- Unbalanced output: V_g = V_s·[R3/(R1+R3) − Rx/(R2+Rx)]. It scales with V_s, so sensor bridges are rated in mV per volt.
- The sign of V_g flips as Rx passes through balance — the bridge tells you which way the unknown deviates, not just by how much.
- This model assumes an ideal (infinite-impedance) meter; a real galvanometer or amplifier input loads the bridge slightly.
Assumptions
- Four resistive arms across an ideal DC excitation source; the detector between the midpoints is ideal (draws no current).
- Linear resistors; balance and output depend only on the four arm values and the excitation.
- Lead and contact resistances are neglected (they matter in precise low-resistance measurement).
When to use this calculator
Appropriate for
- Finding an unknown resistance from a balanced bridge's three known arms
- Estimating the output voltage of an unbalanced or sensor bridge
- Teaching bridge balance and why the null is independent of the excitation
Not suitable for
- AC impedance bridges or bridges with reactive arms
- Precise low-resistance work where lead and contact resistance dominate (use Kelvin/4-wire methods)
- Modelling a real sensor's drift, self-heating, linearity, or detector loading
What this calculator does not cover
- Ideal detector assumed — a real meter or amplifier input loads the bridge and shifts the reading slightly.
- DC resistive bridge only; it does not model AC (impedance) bridges or reactive arms.
- Neglects lead, contact, and wiring resistance, which dominate error in low-value resistance measurement.
- Reports the static balance or output, not the temperature drift, self-heating, or noise of a real sensor bridge.
- As with every calculator on this site: results are preliminary and educational, are not verified for any specific installation, and must be reviewed against the applicable code edition and stamped by a licensed Professional Engineer before real-world use.
Frequently asked questions
What is the Wheatstone bridge balance condition?
The bridge is balanced — the meter reads zero — when the two divider ratios are equal: R1/R3 = R2/Rx. Solving for the unknown arm gives Rx = R2·R3 / R1. Because it depends only on the resistance ratios, the balance point is independent of the excitation voltage.
How do I calculate the output voltage of an unbalanced bridge?
Treat each side as a voltage divider and subtract the midpoint voltages: V_g = V_s·[R3/(R1+R3) − Rx/(R2+Rx)]. The result is positive or negative depending on which way Rx deviates from balance, and it scales directly with the excitation voltage V_s.
Why are Wheatstone bridges used in sensors?
Because a bridge converts a small fractional change in resistance — from a strain gauge, RTD, or pressure element — into a differential voltage that is zero at rest and roughly linear near balance. Working around a null rejects common-mode effects and lets a small signal be amplified cleanly, which is why sensor bridges are specified in millivolts of output per volt of excitation.
References
- Irwin, J. D., Basic Engineering Circuit Analysis, 11th ed. (the Wheatstone bridge)
- Nilsson, J. & Riedel, S., Electric Circuits, 11th ed. (bridge circuits and balance)
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