An introduction to electronics
An introduction to electronics

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An introduction to electronics

3.2 The Wheatstone bridge

Originally developed in the nineteenth century, a Wheatstone bridge provided an accurate way of measuring resistances without being able to measure current or voltage values, but only being able to detect the presence or absence of a current. A simple galvanometer, as illustrated in Figure 13, could show the absence of a current through the Wheatstone bridge in either direction. The long needle visible in the centre of the galvanometer would deflect to one side or the other if any current was detected, but show no deflection in the absence of a current.

Described image
Figure 13  An early galvanometer showing magnet and rotating coil

Figure 14(a) shows a circuit made of four resistors forming a Wheatstone bridge. Its purpose here is to show whether there is any current flowing between cap v sub left and cap v sub right. Figure 14(b) shows an equivalent way of drawing the circuit.

Described image
Figure 14  Equivalent examples of a Wheatstone bridge

The bridge is said to be balanced (that is, no current flows through the bridge and the needle of the galvanometer shows no deflection) if the voltages cap v sub left and cap v sub right are equal. It can be shown that the bridge is balanced if, and only if, equation left hand side cap r sub one divided by cap r sub two equals right hand side cap r sub three divided by cap r sub four, as follows.

When cap v sub left minus cap v sub right equals zero then equation left hand side cap v sub left equals right hand side cap v sub right. Then the Wheatstone bridge can be viewed as two voltage dividers, cap r sub one and cap r sub two on the left and cap r sub three and cap r sub four on the right. Applying the voltage divider equation gives equation left hand side cap v sub left equals right hand side cap r sub two divided by open cap r sub one plus cap r sub two close times cap v sub cap s and equation left hand side cap v sub right equals right hand side cap r sub four divided by open cap r sub three plus cap r sub four close times cap v sub cap s.

So

equation left hand side cap r sub two divided by open cap r sub one plus cap r sub two close equals right hand side cap r sub four divided by open cap r sub three plus cap r sub four close

and

equation left hand side cap r sub two times open cap r sub three plus cap r sub four close equals right hand side cap r sub four times open cap r sub one plus cap r sub two close

Multiplying out the brackets gives

equation left hand side cap r sub two times cap r sub three plus cap r sub two times cap r sub four equals right hand side cap r sub four times cap r sub one plus cap r sub four times cap r sub two

which simplifies to

equation left hand side cap r sub two times cap r sub three equals right hand side cap r sub four times cap r sub one

and

equation left hand side cap r sub three divided by cap r sub four equals right hand side cap r sub one divided by cap r sub two

So, if cap r sub four were unknown, cap r sub one, cap r sub two and cap r sub three could be chosen so that the needle of a galvanometer showed no deflection due to the current. Then

equation left hand side cap r sub four equals right hand side cap r sub two multiplication cap r sub three divided by cap r sub one

SAQ 4

Assume the Wheatstone bridge shown in Figure 14 is balanced. If equation left hand side cap r sub one equals right hand side 1000 postfix times normal cap omega, equation left hand side cap r sub two equals right hand side 10 postfix times k normal cap omega and equation left hand side cap r sub three equals right hand side 50 postfix times normal cap omega, what is the resistance of cap r sub four?

Answer

By the formula given in the text,

equation sequence cap r sub four equals cap r sub two multiplication cap r sub three divided by cap r sub one equals 10 postfix times 000 multiplication 50 divided by 1000 postfix times normal cap omega equals 500 postfix times normal cap omega
T212_1

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