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Wheatstone bridge

The Wheatstone bridge, devised in 1843, provides an accurate method of determining the resistance of an unknown resistor. The circuit is shown in Figure 1. Four resistors are joined as shown, one of them being an unknown resistor whose resistance is to be measured and one a standard resistor. We will assume that R4 is the unknown. R3 can be varied and it is adjusted until no current flows in the galvanometer.

In this condition:

p.d. across AB = p.d. across AC and also
p.d. across BD = p.d. across CD

Therefore: I1R1 = I2R2 and I1R3 = I2R4

Balance condition: R1/R3 = R2/R4

The arms AB and BD are known as the ratio arms of the bridge.

It can be shown that for the highest sensitivity the galvanometer must be connected from the junction of the highest resistances to the junction of the lowest. The accuracy of the Wheatstone bridge should be of the order of 0.2%.

A switch is usually incorporated in the circuit in series with the cell, to prevent current flowing through the bridge at points other than the balance point and thus heating the components and changing their resistance.

Example problems
A short occurs in a telephone cable having a resistance of 0.45 Ω per metre. The circuit is tested with a Wheatstone bridge. The two resistors in the ratio arms of the Wheatstone bridge network have values of 100 Ω and 1110 Ω respectively. A balance condition is found when the variable resistor has a value of 400 Ω.

Calculate the distance down the cable where the short as occurred.
Let the total resistance of the two parts of the cable be R. 100/1110 = R/400 Therefore R = 36.04 Ω.
Therefore distance down the cable = 36.04/[2x0.45] = 40 m

© Keith Gibbs 2020