How to Use a Wheatstone Bridge for Measuring Resistance

The instrument is connected as in Figure 116.

The unknown resistance or device to be measured is connected across the gap at B. One of the standard known coils is connected across the gap at A. A sensitive galvanometer or a telephone receiver and two cells of battery are also connected as shown.

If a telephone receiver is used, place it to the ear. If a galvanometer is used instead, watch the needle carefully. Then move the sharp edge of the knife-contact over the scale along the German-silver "slide wire" until a point is reached when there is no deflection of the needle or no sound in the telephone receiver.

If this point lies very far on one side or the other of the center division on the scale, substitute the next higher or lower known resistance spool until the point falls as near as possible to the center of the scale.

When this point is found, note the reading on the scale carefully. Now comes the hardest part. Almost all my readers have no doubt progressed far enough in arithmetic to be able to carry on the following simple calculation in proportion which must be made in order to find out the resistance of the unknown coil.

The unknown resistance, connected to B, bears the same ratio to the known coil, at A, that the number of divisions between the knife-contact and the right-hand end of the scale (lower row of figures) bears to the number of divisions between the knife-edge and the left-hand end of the scale (upper row of figures).

We will suppose that a 5-ohm coil was used at A in a test, and the needle of the galvanometer stopped swinging when the knife-contact rested on the 60th division from the left-hand end, or on the 40th from the right. Then, in order to find the value of the unknown resistance at B, it is simply necessary to multiply the standard resistance at A by the number of left-hand divisions and divide the product by the number of right-hand divisions. The answer will be the resistance of B in ohms.

The calculation in this case would be as follows:

5 X 40 = 200