Wheatstone Bridge Calculator

The general four-arm bridge: the exact output voltage when it is out of balance, the balance condition R1·R4 = R2·R3 and the unknown resistance it gives you, the Thévenin resistance the detector sees, the sensitivity in volts per ohm, and what a real detector with its own resistance actually reads.

Last updated 24 Sep 2026

Wheatstone bridge

Four arms + excitation → output, balance, Rth

What are you doing?Deflection reads the imbalance and needs an accurate voltmeter and an accurate excitation. Null adjusts one arm until the output is zero and needs neither.

Excitation voltage across the bridge (V)More excitation gives proportionally more output and proportionally more self-heating, which is what really limits it.

Unit for the four arms

R1 — upper left arm

R2 — lower left arm (the output is taken between R1 and R2)

R3 — upper right arm

R4 — lower right armIn null mode this is the unknown: it is computed from the other three.

Detector resistance (Ω), 0 = ideal voltmeterA moving-coil galvanometer is tens to hundreds of ohms and loads the bridge badly; a modern instrumentation amplifier is megohms and does not. Always in ohms, whatever unit the arms use.

Two voltage dividers across one excitation supply, with the detector between their midpoints — the same four arm positions and the same names as the strain-gauge page, so the two drawings read alike. The dots show the current down each leg; the detector branch only carries current when you give it a resistance.

12.44mVExample

5 V excitation, R1 = R3 = R4 = 1,000 Ω and R2 = 1,010 Ω — one arm 1% high

The general bridge

V_out = V_ex × ( R2 ÷ (R1 + R2) − R4 ÷ (R3 + R4) ) = V_ex × (R2·R3 − R1·R4) ÷ ((R1 + R2)(R3 + R4))
balance: R1 · R4 = R2 · R3, so the unknown arm is R4 = R2 × R3 ÷ R1
detector Thévenin resistance: R_th = R1∥R2 + R3∥R4; a detector of R_D reads V_out × R_D ÷ (R_D + R_th)
sensitivity: dV_out ÷ dR2 = V_ex × R1 ÷ (R1 + R2)²

R1, R2: upper and lower arms of the left leg; the output’s positive terminal is between them
R3, R4: upper and lower arms of the right leg; the output’s negative terminal is between them
V ex: the excitation across the top and bottom rails, assumed to come from a source with no output resistance
R th: what the detector sees looking back into the output with the excitation source shorted

Worked example

5 V excitation, R1 = R3 = R4 = 1,000 Ω and R2 = 1,010 Ω — one arm 1% high
Left midpoint: 5 × 1,010 ÷ 2,010 = 2.512438 V
Right midpoint: 5 × 1,000 ÷ 2,000 = 2.500000 V
Output = the difference = 12.44 mV — note that a 1% change in one arm gives only 0.2488% of the excitation, which is the factor of four the bridge costs you and the reason the output needs amplifying
Balance would need R1 × R4 = R2 × R3: 1,000,000 against 1,010,000, so R4 would have to be 1,010 Ω
Thévenin resistance at the output = 1,000∥1,010 + 1,000∥1,000 = 1.002 kΩ
Sensitivity 1.2376 mV per ohm of R2, so 1% of R2 moves the output by 12.5 mV
The excitation supplies 24.94 mW into the two legs

What one arm out of balance is worth, at 5 V excitation

Computed by this page’s exact formula with 5 V excitation. The output is very nearly proportional to the imbalance at 0.1% and visibly is not at 10% — at a 10% imbalance the exact answer is about 4.8% below the linearised one. The last column is the Thévenin resistance with all arms at the nominal value, which is simply the arm resistance itself.
All four arms	R2 high by 0.1%	R2 high by 1%	R2 high by 10%	Detector sees
120 Ω (foil gauge)	1.2494 mV	12.4378 mV	119.0476 mV	120.0 Ω
350 Ω (foil gauge)	1.2494 mV	12.4378 mV	119.0476 mV	350.0 Ω
1 kΩ	1.2494 mV	12.4378 mV	119.0476 mV	1,000.0 Ω
100 Ω (Pt100 at 0 °C)	1.2494 mV	12.4378 mV	119.0476 mV	100.0 Ω

Balance, deflection, and which page you want

A Wheatstone bridge is two voltage dividers fed from one supply, with the detector across the two midpoints. Nothing more. Its value is that the detector reads a DIFFERENCE, so the large common part of the two divider voltages never reaches it: you can resolve a milliohm on top of a kilohm, which no ohmmeter will do for you. Balance happens when the two dividers have the same ratio, R2 ÷ (R1+R2) = R4 ÷ (R3+R4), which rearranges to the form worth memorising: R1 × R4 = R2 × R3. Opposite arms, multiplied.

The null method. Put the unknown in one arm, make one of the others adjustable, and turn it until the detector reads zero. The unknown is then R4 = R2 × R3 ÷ R1, and look at what is NOT in that expression: the excitation voltage, the detector’s gain, the detector’s linearity, and any offset in the wiring that affects both legs equally. The answer is a ratio of three resistances, so the accuracy is the accuracy of those three and nothing else, and the detector only has to be able to tell zero from not-zero — which is the one thing a cheap instrument does well. That is why the null method was the basis of precision resistance metrology for a century, and why a decade box and a galvanometer still beat a four-digit multimeter on a good day.

The deflection method. Leave the bridge fixed and read the imbalance as a voltage. Now the excitation’s accuracy and stability go straight into the answer, and so does the detector’s. The usual repair is to make the measurement ratiometric: feed the analog-to-digital converter’s reference from the same supply that excites the bridge, so a drift in the excitation cancels out of the ratio. Texas Instruments’ bridge-measurement guide devotes a section to exactly this. Deflection is what every sensor bridge does, because you cannot re-balance a bridge fast enough to follow a load cell, and it is why sensor outputs are quoted in millivolts per volt.

The output is not a quarter of the imbalance. With one arm changed by a fraction x and the other three equal, the exact output is V·x ÷ (2(2+x)), not V·x ÷ 4. The difference is a factor of 1/(1 + x/2): negligible at 0.1%, 0.5% low at a 1% imbalance, and 5% low at 10%. This page uses the exact form everywhere, and so should you whenever the imbalance is more than a fraction of a per cent.

What the detector sees. Looking back into the output with the excitation shorted, the bridge is R1∥R2 in series with R3∥R4 — for four equal arms, just the arm resistance. A detector whose own resistance is comparable to that drags the reading down by R_D ÷ (R_D + R_th), which is why a 50 Ω galvanometer on a 350 Ω bridge reads under a seventh of the open-circuit voltage — and why it does not matter at all when you are only looking for zero.

Which page you want. This one is the general bridge: any four arms, any excitation, the exact output, the balance condition and the unknown resistance. If your bridge is a strain gauge — if you are thinking in terms of a gauge factor, microstrain, a quarter, half or full arrangement, lead-wire compensation or the amplifier gain to fill a converter — go to the strain gauge calculator instead. It uses the same four arm names in the same four positions, so the two drawings read identically; it simply starts from strain rather than from resistance. For a platinum sensor in a bridge see the Pt100 RTD calculator, and for the amplifier that follows any of them the non-inverting amplifier calculator. A bridge with a load across the middle is the classic network that no series or parallel reduction will simplify; the delta-wye transformation is how you get at it.

Frequently asked questions

What is the Wheatstone bridge balance condition?

R1 × R4 = R2 × R3, with R1 and R2 the upper and lower arms of one leg and R3 and R4 the upper and lower arms of the other. Equivalently, the two legs must have the same divider ratio. The unknown arm is then R4 = R2 × R3 ÷ R1.

What is the output voltage of an unbalanced bridge?

V_out = V_ex × (R2·R3 − R1·R4) ÷ ((R1+R2)(R3+R4)). The familiar V·ΔR/4R is the first term of that expanded about balance with one arm active, and it is about ΔR/2R low — half a per cent at a 1% imbalance.

Why is a null measurement more accurate than a deflection measurement?

Because at balance the answer is a ratio of three resistances and nothing else. The excitation voltage, the detector’s gain and its calibration all drop out — the detector only has to distinguish zero from not-zero. In deflection mode every one of those goes straight into the result.

What resistance does the detector see?

R1∥R2 in series with R3∥R4, with the excitation source treated as a short. For four equal arms of R that is just R. A detector of resistance R_D reads R_D ÷ (R_D + R_th) of the open-circuit voltage.

How much excitation should I use?

As much as the arms can dissipate without warming measurably, and no more. The output is proportional to the excitation but the self-heating goes as its square, and a resistance that drifts with its own dissipation is indistinguishable from the signal. 5 V and 10 V are the usual figures for 350 Ω arms.

Is this the same as a strain gauge calculator?

No. This page is the general bridge — four arbitrary arms, an exact output, and the balance condition. The strain gauge page on this site is the transducer case: gauge factor, microstrain, quarter/half/full arrangements, lead-wire effects and amplifier gain. Both use the same arm names in the same positions.

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References

Nilsson JW, Riedel SA. Electric Circuits, 11th ed. Pearson, 2019. §3.6 “Measuring Resistance — The Wheatstone Bridge”, and §3.7 for the delta-wye transformation that a loaded bridge needs. Section titles confirmed against Pearson’s own published front matter for the eleventh edition.
Texas Instruments. A Basic Guide to Bridge Measurements, application report SBAA532A, revised March 2024 (B. Lizon, J. Wu). Bridge topologies, the deflection output of a resistive bridge, and §3.1 on ratiometric measurement: driving the converter’s reference from the bridge excitation so that excitation drift cancels.
Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Chapter 1 and Chapter 5: bridges, Thévenin equivalents and the instrumentation amplifier that usually follows one.