Wheatstone Bridge Calculator

Enter the four bridge arm resistors and the supply voltage to find the output (galvanometer) voltage, the branch currents, and the resistance R4 needs for the bridge to balance.

Quick Facts

Balance condition
R1 × R4 = R2 × R3
Equivalent to R1/R2 = R3/R4; the output voltage is zero only when this holds.
Output voltage
Vg = Vs[R2/(R1+R2) − R4/(R3+R4)]
Differential voltage between the two midpoint nodes, assuming a high-impedance detector.
Unknown resistance
R4 = R2 × R3 ÷ R1
Solve for an unknown arm once the bridge is nulled (Vg = 0).
Common uses
Strain gauges, RTDs, load cells
Any sensor whose resistance shifts slightly can replace arm R4.

Your Results

Calculated
Output (Galvanometer) Voltage
-
Vg = Vs[R2/(R1+R2) − R4/(R3+R4)]
Left-Arm Current
-
I through R1 & R2 = Vs ÷ (R1+R2)
Right-Arm Current
-
I through R3 & R4 = Vs ÷ (R3+R4)
R4 Needed to Balance
-
R2 × R3 ÷ R1, for Vg = 0

Ready

Enter the four bridge resistors and the supply voltage, then press Calculate.

Formula and Method for the Wheatstone Bridge

A Wheatstone bridge is a diamond-shaped circuit of four resistors — R1 and R2 forming one voltage-divider leg, R3 and R4 forming a second leg — powered by the same supply voltage Vs. A galvanometer or high-impedance voltmeter is connected between the two midpoints (node B, between R1 and R2, and node D, between R3 and R4) to compare the two legs. The bridge is balanced, or "nulled," when both legs divide the supply voltage identically, which happens when R1 × R4 = R2 × R3. That single condition is why the circuit is used to measure an unknown resistance precisely: adjust one arm (traditionally R4) until the detector reads zero, then solve the balance equation for the unknown.

Deriving the output voltage

Each leg of the bridge is just a voltage divider off the same source. Node B sits at V_B = Vs × R2/(R1+R2), and node D sits at V_D = Vs × R4/(R3+R4). The detector reads the difference between them: Vg = V_B − V_D = Vs[R2/(R1+R2) − R4/(R3+R4)]. Setting Vg = 0 and solving gives the balance condition R1×R4 = R2×R3 (equivalently R1/R2 = R3/R4) — the two ratio pairs must match regardless of the actual supply voltage, which is why bridge measurements are largely insensitive to small drifts in Vs.

Common mistakes

  • Swapping arm positions: R1 and R2 must be the same physical leg (both connect to node B); mixing R1 with R3 or R2 with R4 in the formula gives the wrong balance condition.
  • Ignoring detector loading: the voltage-divider formula assumes a galvanometer or voltmeter that draws negligible current. A low-impedance meter pulls the bridge slightly off the ideal reading, especially far from balance.
  • Forgetting temperature drift: if R4 is a sensor (strain gauge, RTD, thermistor), R1, R2, and R3 should be stable, low-drift resistors, or their own temperature coefficients will show up as false signal.

Real-world applications

  • Unknown-resistance measurement: the classic use — null the bridge and solve R4 = R2×R3/R1 for a precise resistance reading without needing an accurately calibrated voltmeter.
  • Strain gauges and load cells: a strain gauge's resistance changes slightly under mechanical load; the bridge converts that tiny change into a measurable output voltage.
  • RTDs and thermistors: temperature sensors are wired as one bridge arm so a temperature change appears directly as a bridge output voltage.
  • Instrumentation amplifiers: the small differential output Vg is typically fed into an amplifier for data acquisition or control systems.

Frequently Asked Questions

What is the Wheatstone bridge balance condition?
A Wheatstone bridge is balanced when the products of opposite arms are equal: R1 × R4 = R2 × R3 (equivalently R1/R2 = R3/R4). At balance, the output voltage between the two midpoint nodes is zero regardless of the supply voltage.
How do I find an unknown resistance with a Wheatstone bridge?
Wire the unknown resistor in as R4, adjust it (or a known variable resistor in its place) until the detector reads zero, then solve the balance equation for R4: R4 = (R2 × R3) / R1. Because the result depends only on the ratio of known resistors, it does not require a precisely calibrated voltmeter.
What is the formula for the bridge's output voltage?
The output (galvanometer) voltage is Vg = Vs × [R2/(R1+R2) − R4/(R3+R4)], where Vs is the supply voltage. This is the difference between the two voltage-divider midpoints and is zero exactly when R1×R4 = R2×R3.
Why use a Wheatstone bridge instead of a simple voltage divider?
A bridge measures a null or a small differential signal rather than an absolute voltage, so it cancels out supply-voltage drift and common-mode errors that would otherwise swamp a tiny resistance change — which is why it's the standard circuit for strain gauges, RTDs, and other sensors with small resistance shifts.

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