Back to: DISCRETE COMPONENTS

### introduction

The Wheatstone bridge is a circuit that is used to find the value of unknown resistance, to a high level of accuracy. This is achieved by balancing two legs of a circuit. One of these legs includes the unknown resistance and the other contains a variable resistor. Between the midpoints of each leg is a galvanometer, to measure current. The idea is that the variable resistor is altered until the galvanometer reads 0A. At this point, the two legs must be balanced and the unknown resistance can be calculated.

The Wheatstone bridge was invented by a British physicist and mathematician, Samuel Hunter Christie in 1833 and was improved and popularised by Sir Charles Wheatstone in 1843.

### how it works

An example is shown in figure 1. R4 is an unknown resistance. R1, R2 and R3 are resistors with known values. R2 is adjustable. In operation, R2 is adjusted until the bridge is balanced and no current flows through the galvanometer, g. At this point, the voltage difference between points g+ and g- is zero. Therefore, the ratio of the resistors in the known leg (R2 / R1) is equal to the ratio of the two resistances, which includes the unknown resistance (R4 / R3). If the bridge is unbalanced, the direction of the current indicates if R2 is too low or too high.

A Wheatstone bridge can also be made with no variable resistance. Here the voltage across the meter can be used to calculate the value of R4 by using Kirchoff’s voltage law. Similarly, the current through the meter can be used to calculate the value of R4, by using Kirchoff’s current law.

### resistors in series

I think that the arrangement of the Wheatstone bridge often makes it hard for people to understand. The important concept to realise and understand is that the Wheatstone bridge is composed of 2 legs, each containing 2 resistors in series, with the 2 legs in parallel! When you look at it like this, you can then approach each leg as a resistor in series problem, making calculations to solve the unknown resistance! I have rearranged the conventional Wheatstone bridge to display this concept, see figure 2.

We know that when 2 resistors are in series, the same current will flow through both of them (Kirchoff’s current law) and there will be a voltage drop across each one.

Therefore we can calculate the current flowing through R1 and R2 by:

I = V / R

I = 5 / (R1 + R2)

I = 5 / (20Ω + 30Ω)

I = 5 / 50

I = 0.1A

We can calculate the voltage at point A by calculating the voltage drop across R2:

VR2 = I X R2

VR2 = 0.1A x 30Ω

VR2 = 3V

Because we know the voltage of the source (VS) and we know the voltage drop across R2, we can calculate the voltage drop across R1:

VR1 = VS – VR2

VR1 = 5V – 3V

VR1 = 2V

Now, look at figure 3. Here we have the same 2 resistors in series but in parallel with a leg with the same values, in series. Now, because the value of R3 and R4 are the same as that of R1 and R2, the voltage drops will be the same.

You may have wondered about the galvanometer, if it has any effect on the circuit, that should be taken into account! The answer is no. Devices such as galvanometers and ammeters have very little resistance and therefore have no meaningful impact on our circuit. Also, the reason that a galvanometer is often used is that it can show the direction of the current, which is useful for determining which way the bridge needs to be balanced. You could use an ammeter but not all of them can show the direction!

Because of this, if we measured the voltage at points A and B, the voltage would be zero! This is because they are at the same potential, relative to ground. Therefore there is no difference between them. This is the principle of the Wheatstone bridge – **when balanced the difference at these points is zero**.