Wheatstone Bridge Calculator
Enter four resistances and input voltage to compute the balanced unknown R_x=R₂R₃/R₁ and the output voltage V_out=V_in·(R_x/(R_x+R₃)−R₂/(R₁+R₂)). R₁=100,R₂=200,R₃=100,R_x=200 → balanced, V_out=0.
Input Data
Results
At a glance:The Wheatstone bridge (Wheatstone bridge): four resistors R₁, R₂, R₃, R_x form a diamond; the supply V_in is across one diagonal and a galvanometer (or voltmeter) across the other. At balance V_out=0, giving the balance condition R_x/R₃=R₂/R₁, i.e. R_x=R₂·R₃/R₁. Off balance, V_out=V_in·(R_x/(R_x+R₃) − R₂/(R₁+R₂)). Invented by Samuel Hunter Christie (1833) and improved/popularised by Charles Wheatstone (1843) to measure unknown resistances precisely (accuracy to 10⁻⁴ Ω). Classic example: R₁=100 Ω, R₂=200 Ω, R₃=100 Ω, R_x=200 Ω → balanced (R_x=R₂·R₃/R₁=200×100/100=200 Ω), V_out=0. Off-balance example: with R_x=250 Ω, V_out>0 (R_x above balance), current flows from R₃ toward R₁. Applications: (1) precise resistance measurement — set three known resistors, adjust R_x until the galvanometer reads zero, then back-calculate R_x; (2) strain gauges — R_x is a gauge whose resistance changes slightly under deformation (ΔR/R≈10⁻⁶), so V_out reveals strain; (3) temperature sensing (platinum Pt100/Pt1000) — R_x is a platinum resistor whose value tracks temperature, V_out infers the temperature; (4) pressure sensing and load cells — four gauges form a bridge measuring force and weight (core of electronic scales); (5) AC bridges compare capacitance/inductance. History: Christie first built it in 1833; Wheatstone demonstrated it to the Royal Society in 1843, and it became named after him. Galvanometer sensitivity sets the bridge accuracy; modern designs use op-amps. Notes: (1) larger V_in raises sensitivity but increases power; (2) R₁–R₄ typically 100 Ω–10 kΩ — too low wastes power, too high is noise-sensitive.
Formula
Balance condition: R_x/R₃ = R₂/R₁
Balanced resistance: R_x = R₂·R₃/R₁
Output voltage: V_out = V_in·(R_x/(R_x+R₃) − R₂/(R₁+R₂))
Current direction: R_x > balance ⇒ positive; R_x < balance ⇒ negative
$$\frac{R_x}{R_3} = \frac{R_2}{R_1}, \quad R_x = \frac{R_2 R_3}{R_1}$$$$V_{out} = V_{in} \left( \frac{R_x}{R_x + R_3} - \frac{R_2}{R_1 + R_2} \right)$$How to Use
- Enter the four resistances R₁, R₂, R₃, R_x (Ω) and the bridge input voltage V_in (V).
- The tool computes the balanced R_x*, the output voltage V_out and whether it is balanced.
- Balance gives V_out=0 and R_x=R₂R₃/R₁.
Case Studies
Measuring an unknown resistor
Set R₁=100, R₂=200, R₃=100 Ω, adjust R_x until the galvanometer reads zero.
At balance R_x=R₂·R₃/R₁=200×100/100=200 Ω.
This yields the unknown to ~10⁻⁴ Ω precision without an ammeter.
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.