3.15 METER BRIDGE
A meter bridge.
Wire \( AC \) is \( 1 m \) long. \( R \) is a resistance to be measured and \( S \) is a standard resistance.
The meter bridge is shown in Fig. 3.27. 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 \( R_{cm}l \), where \( R_{cm} \) is the resistance of the wire per unit centimetre. The portion \( DC \) of the wire similarly has a resistance \( R_{cm}(100-l ) \).
The four arms \(AB\), \(BC\), \(DA\) and \(CD\) [with resistances \(R\), \(S\), \(R_{cm}l\) and \(R_{cm}(100-l )\)] 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 \(l= l_1\). The four resistances of the bridge at the balance point then are \(R\), \(S\), \(R_{cm}l_1\) and \(R_{cm}(100–l_1)\). The balance condition, Eq. [3.83(a)] gives
\[ \frac{R}{S}=\frac{R_{c m} l_{1}}{R_{c m}\left(100-l_{1}\right)}=\frac{l_{1}}{100-l_{1}} \](3.85)
Thus, once we have found out \( l_1 \), the unknown resistance \( R \) is known in terms of the standard known resistance \( S \) by
\[ R=S \frac{l_{1}}{100-l_{1}} \](3.86)
By choosing various values of \(S\), we would get various values of \(l_{1}\), and calculate \(R\) each time. An error in measurement of \(l_{1}\) 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 \(l_{1}\) is close to \(50 cm\). ( This requires a suitable choice of \(S\).)
Example 3.9
In a meter bridge (Fig. 3.27), the null point is found at a distance of \( 33.7 cm \) from \( A \). If now a resistance of \( 12\Omega \) is connected in parallel with \( S \), the null point occurs at \( 51.9 cm \). Determine the values of \( R \) and \( S \).
From the first balance point, we get
\[ \frac{R}{S}=\frac{33.7}{66.3} \](3.87)
After \( S \) is connected in parallel with a resistance of \( 12\Omega \), the resistance across the gap changes from \( S \) to \( S_{eq} \), where \[ S_{eq}=\frac{12S}{S+12} \]
and hence the new balance condition now gives
\[ \frac{51.9}{48.1}=\frac{R}{S_{e q}}=\frac{R(S+12)}{12 S} \](3.88)
Substituting the value of \( R/S \) from Eq. (3.87), we get \[ \frac{51.9}{48.1}=\frac{S+12}{12} \cdot \frac{33.7}{66.3} \] which gives \( S=13.5\Omega \). Using the value of \( R/S \) above, we get \( R=6.86\Omega \).