# Carey Foster bridge

In electronics, the Carey Foster bridge is a bridge circuit used to measure medium resistances, or to measure small differences between two large resistances. It was invented by Carey Foster as a variant on the Wheatstone bridge. He first described it in his 1872 paper "On a Modified Form of Wheatstone's Bridge, and Methods of Measuring Small Resistances" (Telegraph Engineer's Journal, 1872–1873, 1, 196).

## Use

In the adjacent diagram, X and Y are resistances to be compared. P and Q are nearly equal resistances, forming the other half of the bridge. The bridge wire EF has a jockey contact D placed along it and is slid until the galvanometer G measures zero. The thick-bordered areas are thick copper busbars of very low resistance, to limit the influence on the measurement.
* Place a known resistance in position Y.
* Place the unknown resistance in position X.
* Adjust the contact D along the bridge wire EF so as to null the galvanometer. This position (as a percentage of distance from E to F) is ℓ$_1$.
* Swap X and Y. Adjust D to the new null point. This position is ℓ$_2$.
* If the resistance of the wire per percentage is σ, then the resistance difference is the resistance of the length of bridge wire between ℓ$_1$ and ℓ$_2$:
          $X-Y=\sigma (\ell _{2}-\ell _{1})\,$
To measure a low unknown resistance X, replace Y with a copper busbar that can be assumed to be of zero resistance.
In practical use, when the bridge is unbalanced, the galvanometer is shunted with a low resistance to avoid burning it out. It is only used at full sensitivity when the anticipated measurement is close to the null point.

### To measure σ

To measure the unit resistance of the bridge wire EF, put a known resistance (e.g., a standard 1 ohm resistance) that is less than that of the wire as X, and a copper busbar of assumed zero resistance as Y.

## Theory

Two resistances to be compared, X and Y, are connected in series with the bridge wire. Thus, considered as a Wheatstone bridge, the two resistances are X plus a length of bridge wire, and Y plus the remaining bridge wire. The two remaining arms are the nearly equal resistances P and Q, connected in the inner gaps of the bridge.
Let ℓ$_1$ be the null point D on the bridge wire EF in percent. α is the unknown left-side extra resistance EX and β is the unknown right-side extra resistance FY, and σ is the resistance per percent length of the bridge wire:
    ${P \over Q}={{X+\sigma (\ell _{1}+\alpha )} \over {Y+\sigma (100-\ell _{1}+\beta )}}$
and add 1 to each side:
    ${P \over Q}+1={{X+Y+\sigma (100+\alpha +\beta )} \over {Y+\sigma (100-\ell _{1}+\beta )}}$   (equation 1)
Now swap X and Y. ℓ$_2$ is the new null point reading in percent:
    ${P \over Q}={{Y+\sigma (\ell _{2}+\alpha )} \over {X+\sigma (100-\ell _{2}+\beta )}}$
and add 1 to each side:
    ${P \over Q}+1={{X+Y+\sigma (100+\alpha +\beta )} \over {X+\sigma (100-\ell _{2}+\beta )}}$   (equation 2)
Equations 1 and 2 have the same left-hand side and the same numerator on the right-hand side, meaning the denominator on the right-hand side must also be equal:
    ${\begin{aligned}&Y+\sigma (100-\ell _{1}+\beta )=X+\sigma (100-\ell _{2}+\beta )\\\Rightarrow&X-Y=\sigma (\ell _{2}-\ell _{1})\end{aligned}}$
Thus: the difference between X and Y is the resistance of the bridge wire between ℓ$_1$ and ℓ$_2$.
The bridge is most sensitive when P, Q, X and Y are all of comparable magnitude.