Carey Foster bridge

The Carey Foster bridge. The thick-edged areas are busbars of almost zero resistance.

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.

1. Place a known resistance in position Y.
2. Place the unknown resistance in position X.
3. 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 ℓ₁.
4. Swap X and Y. Adjust D to the new null point. This position is ℓ₂.
5. If the resistance of the wire per percentage is σ, then the resistance difference is the resistance of the length of bridge wire between ℓ₁ and ℓ₂:

${\displaystyle X-Y=\sigma (\ell _{2}-\ell _{1})\,}$

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.

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.

A standard Wheatstone bridge for comparison. Points A, B, C and D in both circuit diagrams correspond. X and Y correspond to R₁ and R₂, P and Q correspond to R₃ and R_X. Note that with the Carey Foster bridge, we are measuring R₁ rather than R_X.

Let ℓ₁ 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:

${\displaystyle {P \over Q}={{X+\sigma (\ell _{1}+\alpha )} \over {Y+\sigma (100-\ell _{1}+\beta )}}}$

and add 1 to each side:

${\displaystyle {P \over Q}+1={{X+Y+\sigma (100+\alpha +\beta )} \over {Y+\sigma (100-\ell _{1}+\beta )}}}$       (equation 1)

Now swap X and Y. ℓ₂ is the new null point reading in percent:

${\displaystyle {P \over Q}={{Y+\sigma (\ell _{2}+\alpha )} \over {X+\sigma (100-\ell _{2}+\beta )}}}$

and add 1 to each side:

${\displaystyle {P \over Q}+1={{X+Y+\sigma (100+\alpha +\beta )} \over {X+\sigma (100-\ell _{2}+\beta )}}}$       (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:

${\displaystyle {\begin{aligned}&Y+\sigma (100-\ell _{1}+\beta )=X+\sigma (100-\ell _{2}+\beta )\\\Rightarrow {}&X-Y=\sigma (\ell _{2}-\ell _{1})\end{aligned}}}$

Thus: the difference between X and Y is the resistance of the bridge wire between ℓ₁ and ℓ₂.

The bridge is most sensitive when P, Q, X and Y are all of comparable magnitude.

• A. H. Fison (1919). "Obituary notices: .... George Carey Foster, 1835–1919; ..." J. Chem. Soc., Trans. 115: 412–427. doi:10.1039/CT9191500408.