Wheatstone Bridge Calculator
Balance a bridge to find an unknown resistance, or work out the resistor you need to balance it, with the ratio in between.
What it does
Measuring a resistance with a voltmeter and an ammeter is awkward, because the meters change the circuit they are measuring. There is a way round that problem which has been in use since 1843, and it works by making the answer come from a balance rather than from a reading.
The OpenStax College Physics text describes the arrangement. Two resistors R1 and R2 are known precisely, a third R3 can be varied and read, and the unknown resistor Rx sits in the other branch. A galvanometer joins the two branches, and R3 is adjusted until the galvanometer reads zero. At that point the two points it joins are at the same potential, so no current flows through it and the meter has no effect on the circuit at all.
The rest follows from the voltage drops. With no current in the galvanometer, the same current runs through each branch, and because the mid points are at the same potential the drop across R1 equals the drop across R3, and the drop across R2 equals the drop across Rx. Dividing one by the other cancels the currents and gives the unknown resistance as R3 times R2 over R1.
This page runs that balance in both directions. Give it the three known resistors and it returns the unknown resistance, or give it R1, R2 and the unknown resistance and it returns the setting for the variable resistor that would balance the bridge.
The limits are worth stating. The text is careful that the galvanometer current can never be exactly zero, and that the resistances of the wires and the connections, and the temperature of the resistors, both contribute uncertainty. The method measures a resistance, and nothing else.
How to use
- Choose whether you know the unknown resistance and want the balance setting, or the other way round.
- Enter R1 and R2, the two precisely known resistors.
- Enter R3, the variable resistor, when you are solving for the unknown resistance.
- Read the arm ratio, which is the number that multiplies R3 to give the answer.
- Read the unknown resistance, or the setting for R3 that balances the bridge.
- Keep the ratio of the two arms close to one, because that is what makes the answer least sensitive to errors in the resistors.
Formula
- Rx
- The unknown resistance being measured.
- R1
- First known resistor of one arm.
- R2
- Second known resistor of the same arm as R1.
- R3
- The variable resistor, set by the balance.
Because the balance condition is a ratio, the same bridge can be turned round to measure a known resistance in order to find an unknown one, which is how a standard resistor is used to check a reference. The three numbers that enter the answer each do so with a power of one, so a 1 percent error in any of them is a 1 percent error in the result.
Worked example
Example: a 1000 ohm arm, a 625 ohm arm and a 2500 ohm variable resistor
Defaults: solving for the unknown resistance, with R1 of 1000 ohms, R2 of 625 ohms and R3 of 2500 ohms.
- The arm ratio is R2 over R1, which is 625 over 1000 = 0.625000.
- At balance the unknown resistance is R3 times that ratio, which is 2500 x 0.625 = 1562.500000 ohms.
- The other way round, the variable resistor needed to balance a known 100 ohm resistance with R1 of 50 ohms and R2 of 175 ohms, is 100 x 50 over 175 = 28.571429 ohms.
- The measurement does not disturb the circuit, because with the galvanometer at zero no current flows through it.
- The accuracy comes from the resistors being precise, not from the galvanometer, which only has to show the difference between zero and something small.
- The sensitivity is the catch: a 1 percent error in R1, R2 or R3 moves the answer by 1 percent, so the resistors have to be good.
- The tool shows a ratio of 0.625000, an unknown resistance of 1562.500000 ohms, a balance resistor of 2500.000000 ohms, and the note that a 1 percent error in any arm gives a 1 percent error in the answer.
Reference table
| R3 | Arm ratio | Unknown resistance | Balance resistor for 100 ohm |
|---|---|---|---|
| 1000 ohm | 0.625000 | 625.000000 ohm | 160.000000 ohm |
| 2000 ohm | 0.625000 | 1250.000000 ohm | 160.000000 ohm |
| 2500 ohm | 0.625000 | 1562.500000 ohm | 160.000000 ohm |
| 4000 ohm | 0.625000 | 2500.000000 ohm | 160.000000 ohm |
Limits
- The galvanometer current can never be exactly zero, and the text says that limiting factor on the accuracy of the method.
- The resistance of the wires and the connections adds to the result, and it changes with time and with temperature.
- The method measures a resistance. Anything that depends on a resistance, such as a temperature or a strain, needs a separate way of turning it into a reading.
- The resistors themselves have a tolerance and a temperature coefficient, and those set the real limit on the precision.
Common mistakes
- Putting the ratio the wrong way round, so the answer is off by the square of the ratio.
- Mixing up the two arms, so the variable resistor and the unknown end up in the wrong branches.
- Ignoring the tolerance of the precision resistors, which is usually the whole of the error.
- Expecting a bridge to measure a capacitance or an inductance directly, which it does not do.
- Forgetting that the galvanometer resistance has to be high, so that the balance point is sharp and the reading is easy to find.
Try it yourself
Quick checks. Answers sit tucked under each question.
What is the balance setting for a 100 ohm unknown with R1 of 50 ohms and R2 of 175 ohms?
28.571429 ohms, because R3 is Rx times R1 over R2, which is 100 x 50 over 175. The arm ratio is 3.500000, so the answer is the unknown divided by that ratio.
Why does the order of the ratio matter?
Because the unknown resistance is R3 multiplied by the ratio of R2 to R1, and the arms have to sit in the branches the way the text describes. With the ratio the wrong way round the answer is out by the square of the ratio, which with a ratio of 0.625 is a factor of nearly three.
How accurate can the method be?
The text says it can be very accurate, often to four significant digits, and that the limit comes from the precision resistors and from the small currents the galvanometer cannot quite reach zero on. A 1 percent error in any of the three resistors is a 1 percent error in the result.
Questions
What is a Wheatstone bridge?
A circuit for measuring an unknown resistance by balancing it against three known ones. The OpenStax text calls it a null measurement device for calculating resistance by balancing potential drops in a circuit.
Why is it called a bridge?
Because the galvanometer forms a bridge between the two branches of the circuit, joining them at a point in each.
What is a null measurement?
A measurement where the instrument is arranged to read zero, so that it takes nothing from the circuit. The text notes that a voltmeter draws extra current and an ammeter reduces the flow, and that balancing avoids both.
What limits the accuracy?
The text names two things: the galvanometer cannot quite reach zero, and there are always uncertainties in R1, R2 and R3. The resistance of the wires and the connections adds to it as well.
What can the bridge be used for besides resistance?
Anything that can be turned into a resistance change, since the bridge output is a change in the balance. That covers strain gauges, thermistors, gas sensors and the load cell in a scale, but each needs its own way of reading the unbalance.
Does the order of the resistors matter?
Yes, because the balance condition is a ratio of the two arms. Swapping the arms inverts the ratio, and the answer is then out by the square of that ratio.
Sources
- OpenStax College Physics 2e, 21.5: null measurements and the Wheatstone bridge
- Physics LibreTexts, 21.5: null measurements, with the balance condition worked out