Wheatstone Bridge Calculator

Calculate unknown resistance using the Wheatstone bridge calculator.

Clear
Bridge output (difference)-12.4378 mVunbalanced
Output voltage2.5 V5 V × 1,000 ÷ (1,000 + 1,000)
Division ratio0.550% of the input
Current through the divider2.5 mAthe quiescent current this divider wastes continuously
Power wasted in the divider12.5 mW
Dropped across R12.5 V
Thévenin resistance500 ΩR1 in parallel with R2 — the output impedance, and what decides how much a load disturbs the divider
As a current divider, through R150% of the totalnote the ratio is the OPPOSITE way round from the voltage divider — current prefers the lower resistance
As a current divider, through R250% of the total
Second arm output2.5124 V
Balance condition1 against 0.990099a bridge balances when these ratios match, which is why it measures a resistance ratio without needing an accurate supply
R4 that would balance the bridge1 kΩ

The formula

V_out = V_in × R₂ ÷ (R₁+R₂); a current divider uses the opposite ratio

The ratio flips between voltage and current

A voltage divider gives the lower resistor's share: V_out = V_in × R₂/(R₁+R₂). A current divider gives the opposite ratio, because current takes the path of least resistance — the branch with R₁ carries the fraction R₂/(R₁+R₂). Both are shown above precisely because swapping them is such an easy mistake.

The Thévenin resistance is the figure that decides whether a divider is usable. It is R₁ in parallel with R₂, and any load comparable to it will drag the output down noticeably. A rule of thumb is that the load should be at least ten times the Thévenin resistance, or the divider needs a buffer amplifier.

There is a genuine trade-off in choosing the resistor values. Low values give a stiff output that tolerates loading but waste current continuously; high values waste almost nothing but cannot drive anything and pick up noise. A Wheatstone bridge sidesteps the supply-accuracy problem entirely: it balances on a ratio, so the reading is independent of the supply voltage.