3.15 Meter Bridge

NCERT Class 12 physics for blind and visually impaired students.

The meter bridge is shown in Fig. 3.27.

FIGURE 3.27 A meter bridge. Wire AC is 1 m long. R is a resistance to be measured and S is a standard resistance.

It consists of a wire of length 1m and of uniform cross sectional area stretched taut and clamped between two thick metallic strips bent at right angles, as shown. The metallic strip has two gaps across which resistors can be connected. The end points where the wire is clamped are connected to a cell through a key. One end of a galvanometer is connected to the metallic strip midway between the two gaps. The other end of the galvanometer is connected to a ‘jockey’. The jockey is essentially a metallic rod whose one end has a knife-edge which can slide over the wire to make electrical connection.

R is an unknown resistance whose value we want to determine. It is connected across one of the gaps. Across the other gap, we connect a standard known resistance S. The jockey is connected to some point D on the wire, a distance l cm from the end A. The jockey can be moved along the wire. The portion AD of the wire has a resistance Rcml, where Rcm is the resistance of the wire per unit centimetre. The portion DC of the wire similarly has a resistance Rcm (100 − l).

The four arms AB, BC, DA and CD [with resistances R, S, Rcm l and Rcm(100 − l)] obviously form a Wheatstone bridge with AC as the battery arm and BD the galvanometer arm. If the jockey is moved along the wire, then there will be one position where the galvanometer will show no current. Let the distance of the jockey from the end A at the balance point be l = l1. The four resistances of the bridge at the balance point then are R, S, Rcm l1 and Rcm(100 − l1). The balance condition, Equation [3.83(a)] gives

R/S = Rcm l1/{Rcm (100 − l1)} = l1/(100 − l1) .. .. (3.85)

Thus, once we have found out l1, the unknown resistance R is known in terms of the standard known resistance S by

R = S l1/(100 − l1) .. .. (3.86)

By choosing various values of S, we would get various values of l1, and calculate R each time. An error in measurement of l1 would naturally result in an error in R. It can be shown that the percentage error in R can be minimised by adjusting the balance point near the middle of the bridge, i.e., when l1 is close to 50 cm. (This requires a suitable choice of S.)

Example 3.9

In a metre bridge (Fig. 3.27), the null point is found at a distance of 33.7 cm from A. If now a resistance of 12 Ω is connected in parallel with S, the null point occurs at 51.9 cm. Determine the values of R and S.

Solution:


From the first balance point, we get


R/S = 33.7/66.3 .. .. (3.87)

After S is connected in parallel with a resistance of 12Ω, the resistance across the gap changes from S to S eq, where

S eq= 12S/(S + 12)


and hence the new balance condition now gives


51.9/48.1 = R /Seq = R(S + 12) /12S (3.88)

Substituting the value of R/S from Equation (3.87), we get


51.9/48.1 = (S + 12) /12 • 33.7/66.3

which gives S = 13.5 Ω. Using the value of R/S above, we get R = 6.86 Ω.