Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/443

349.]
401
WHEATSTONE'S BRIDGE.

Instead of the whole conductor being a uniform wire, we may make the part near ${\displaystyle O}$ of such a wire, and the parts on each side may be coils of any form, the resistance of which is accurately known.

We shall now use a different notation instead of the symmetrical notation with which we commenced.

Let the whole resistance of ${\displaystyle BAC}$ be ${\displaystyle R}$.

Let ${\displaystyle c=mR}$ and ${\displaystyle b=(1-m)R}$.

Let the whole resistance of ${\displaystyle BOC}$ be ${\displaystyle S}$.

Let ${\displaystyle \beta =nS}$ and ${\displaystyle \gamma =(1-n)S}$.

The value of ${\displaystyle n}$ is read off directly, and that of ${\displaystyle m}$ is deduced from it when there is no sensible deviation of the galvanometer.

Let the resistance of the battery and its connexions be ${\displaystyle B}$, and that of the galvanometer and its connexions ${\displaystyle G}$.

We find as before

 {\displaystyle {\begin{aligned}D=G\{BR+BS+RS\}&+m(1-m)R^{2}(B+S)+n(1-n)S^{2}(B+R)\\&+(m+n-2mn)BRS,\end{aligned}}}

and if ${\displaystyle \xi }$ is the current in the galvanometer wire

 ${\displaystyle \xi ={\frac {ERS}{D}}(n-m)}$.

In order to obtain the most accurate results we must make the deviation of the needle as great as possible compared with the value of ${\displaystyle (n-m)}$. This may be done by properly choosing the dimensions of the galvanometer and the standard resistance wire.

It will be shewn, when we come to Galvanometry, Art. 716, that when the form of a galvanometer wire is changed while its mass remains constant, the deviation of the needle for unit current is proportional to the length, but the resistance increases as the square of the length. Hence the maximum deflexion is shewn to occur when the resistance of the galvanometer wire is equal to the constant resistance of the rest of the circuit.

In the present case, if ${\displaystyle \delta }$ is the deviation,

 ${\displaystyle \delta =C{\sqrt {G}}\xi }$,

where ${\displaystyle C}$ is some constant, and ${\displaystyle G}$ is the galvanometer resistance which varies as the square of the length of the wire. Hence we find that in the value of ${\displaystyle D}$, when ${\displaystyle \delta }$ is a maximum, the part involving ${\displaystyle G}$ must be made equal to the rest of the expression.

If we also put ${\displaystyle m=n}$, as is the case if we have made a correct observation, we find the best value of ${\displaystyle G}$ to be

 ${\displaystyle G=n(1-n)(R+S)}$.