## What is a Wheatstone bridge?

A **Wheatstone bridge** is an electrical circuit used to calculate an unknown resistance with the help of a bridge circuit. For this, the two legs of the bridge circuit are kept balanced and one leg of it includes the unknown resistance. The Wheatstone bridge principle is similar to the working of **potentiometer**. Slight modifications in the Wheatstone bridge can help in finding other quantities like capacitance and inductance as well. It also helps in finding the amount of a particular gas that is mixed among a sample. The Wheatstone bridge measurement is very accurate and the value of the unknown resistance is mostly found out in order to measure other physical values like temperature, force, pressure and so on. It can be used in all electronic circuits. The device was first invented by Samuel Hunter Christie in 1833. The concept was later modified and popularized by Sir Charles Wheatstone in the year 1843.

## Wheatstone Bridge Measurement

**Balanced Condition**

As shown in the circuit diagram, there are four resistances connected as a bridge circuit. The three resistors R1, R2 and R3 will have known values. The value of the resistance R_{X} will be unknown and has to be calculated. The value of resistance R2 is adjustable. A galvanometer has to be set between the points B and D.

**TAKE A LOOK : GALVANOMETERS**

The condition to be satisfied at the point of balance is given below.

If R2/R1 = R_{X}/R3, then V_{BD} = 0 and current through V_{G} = 0. To reach this condition, the adjustable resistor is varied. The direction of the current can be known from the value of the resistor R2.

As soon as the balance condition is obtained the value of the resistance R_{X }is obtained.

Thus, R_{X = }[R2/R1] x R3

This method is very accurate as the other values of resistors are of high precision.

#### Unbalanced condition

If the value of the resistors R1, R2 and R3 are fixed, then the value of the unknown resistor R_{X} has to be found out with the help of Kirchhoff’s voltage and current laws. For this, the bridge circuit will produce a voltage as well as a current as the circuit will not be balanced. This is the process mostly used in the measurement of strain gauges and resistance thermometers. This method is much easier to use as calculations can be done more easily with a voltage and current value than trying to make the circuit balanced.

To calculate the currents between the junctions B and D use Kirchhoff’s current law.

I3 – I_{X }+ I_{G }= 0

I1 – I2 – I_{G} = 0

This can be further simplified as

I3 = I_{X } and I1 = I2…………………………………..**[1]**

To calculate the voltage in the loops ABD and BCD use Kirchhoff’s voltage law.

[I3 R3] – [I_{G} R_{G}] – [I1 R1] = 0

[I_{X} R_{X}] – [I2 R2] + [I_{G} R_{G}] = 0

On re-arranging the above equations we get the value of unknown resistance as

R_{X }= [R2 I2 R3 I3]/ [R1 I1 I_{X}]…………………..**[2]**

Substituting value of **[1]** in **[2]**

R_{X} = [R3 R2]/ R1.

There may be cases when the values of all the resistors are known. If so, the value of the voltage between B and D [V_{G}] can be found out from the following equation.

V_{G }= V_{S }* ([R3 /{R3 + R_{X}}] – [R2/{R1 + R2}])

#### Applications of Wheatstone Bridge

Wheatstone bridge has certain useful applications. On its own, it can be used to determine unknown resistance quiet accurately. If you use an operational amplifier the same Wheatstone bridge can be used to measure temperature, strain etc. It can also read capacitance, impedance etc if used along with a varistor.

## 5 Comments

Hello,

firstly, great job on explaining how the wheatstone bridge functions! it is very detailed and i feel that it is very well done:)

however, as i was looking at the last equation which you have put up, i have recalculated it and i found out that it is wrong. This was the equation you have written:

VG = VS * ([R3 /{R3 + RX}] – [R2/{R1 + R2}])

if i assume the ratio R2/R1 = RX/R3 to be true, i will solve it in relations to RX and i would get RX=R2/R1*R3. If the value is correct, VG should be zero. However, the answer is

VG=VS*(R1-R2)/(R1+R2).

I did the equations again and the answer is supposed to be:

VG = VS * ([R2 /{R1 + R2}] – [RX/{R3 + RX}]).

Thank you so much for looking into this one more time. It would definitely help those who are studying this too. 🙂