Wheatstone Bridge Balance
Condition for a balanced measurement bridge.
Formula
Variables
Example
Used to measure unknown R
How it works
A Wheatstone bridge is balanced when the two middle nodes are at the same electric potential, so no current flows through the detector branch. In that case, the ratio of the resistors in one side of the bridge must match the ratio on the other side, which gives R1/R2 = R3/R4. This comes directly from applying Kirchhoff’s laws to the two branches and demanding zero galvanometer current.
When to use it
Use this relation when analyzing or designing bridge circuits for precise resistance comparison, sensor readout, or calibration. It is especially useful when you need to determine an unknown resistor or tune a circuit so that a null detector reads zero.
Worked example, step by step
Given a bridge with R1 = 120 Ω, R2 = 300 Ω, and R3 = 180 Ω, find the value of R4 for balance. Start with the balance condition: R1/R2 = R3/R4 Substitute the known values: 120/300 = 180/R4 Simplify the left side: 0.4 = 180/R4 Cross-multiply: 0.4R4 = 180 Solve for R4: R4 = 180 / 0.4 = 450 Ω Check: 120/300 = 0.4 and 180/450 = 0.4, so the bridge is balanced. Final answer: R4 = 450 Ω
Common mistakes to avoid
- Swapping resistors from opposite arms and setting the wrong ratios equal
- Forgetting that the formula applies only at balance, when the detector current is zero
- Using the relation with an unbalanced bridge without checking whether the null condition is satisfied
Did You Know?
The Wheatstone bridge measures tiny resistance changes and sits inside the strain gauges weighing trucks and people.
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