How this calculation works
A Wheatstone bridge compares two divider midpoints, so its output is a signed differential voltage. Equal arm ratios balance those midpoints even when the individual resistor values are not all equal.
Open output = Vs[R2/(R1+R2) − R4/(R3+R4)]; differential Rth = R1||R2 + R3||R4; loaded output = open × load/(Rth + load)A worked example
With five volts and equal 1000-ohm left resistors, midpoint A is 2.5 V. A right arm of 1000 and 1100 ohms places B at about 2.61905 V, so open A-minus-B output is about minus 0.119048 V. The differential Thevenin resistance is about 1023.81 ohms. A 100000-ohm load slightly reduces the signed output magnitude, showing why loading matters in an otherwise balanced-looking measurement circuit.
Good to know
Ideal source and resistors only; no sensor sensitivity, tolerance, self-heating, amplifier limits or safe excitation rating. Output sign depends on node order. Do not interpret a negative value as invalid merely because the load current runs from B toward A.
Reference: Analog Devices — resistive bridge basics (opens in a new tab) ↗
What this calculation answers
A Wheatstone bridge compares two divider midpoints, so its output is a signed differential voltage. Equal arm ratios balance those midpoints even when the individual resistor values are not all equal.
A finite load between the midpoint nodes changes the differential output. This page uses the bridge’s Thevenin equivalent seen across those nodes, instead of incorrectly treating the load as two separate resistors to ground.
The Loaded voltage divider calculator examines one midpoint relative to a reference. A differential bridge requires both divider arms and the load path between them.
Use the right input conventions
R1 and R2 are the top and bottom left arm; R3 and R4 are top and bottom right. A positive output means left midpoint A is above right midpoint B. Keep that node convention when comparing a meter or amplifier.
The excitation source is ideal. The differential load represents an actual resistive connection between A and B; it is not a grounded oscilloscope input unless the measurement arrangement is equivalent.
Interpret the result
With the ideal voltage source set to zero for resistance calculation, each vertical arm contributes a parallel resistance to the common supply node. Those two midpoint-to-common resistances add in the differential Thevenin path.
Increasing differential load resistance reduces loading and moves the output toward its open-circuit value. This does not establish instrument input common-mode limits, sensor calibration or safe excitation power.
Work through the example
With five volts and equal 1000-ohm left resistors, midpoint A is 2.5 V. A right arm of 1000 and 1100 ohms places B at about 2.61905 V, so open A-minus-B output is about minus 0.119048 V.
The differential Thevenin resistance is about 1023.81 ohms. A 100000-ohm load slightly reduces the signed output magnitude, showing why loading matters in an otherwise balanced-looking measurement circuit.
Where the model stops
Ideal source and resistors only; no sensor sensitivity, tolerance, self-heating, amplifier limits or safe excitation rating.
Output sign depends on node order. Do not interpret a negative value as invalid merely because the load current runs from B toward A.
A few questions, answered
Why can output be negative?
The chosen output is A minus B. Reversing which midpoint is higher reverses that sign.
Is the load connected to ground?
No. It is a differential resistor directly between the two bridge midpoint nodes.