Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/449

with a mercury cup, into which one electrode of the galvanometer may be plunged.

The rest of the apparatus is arranged, as in Wheatstone's Bridge, with resistance coils or boxes $A$ and $C$, and a wire $PR$ with a sliding contact piece $Q$, to which the other electrode of the galvanometer is connected.

Now let the galvanometer be connected to $S$ and $Q$, and let $A_{1}$ and $C_{1}$ be so arranged, and the position of $Q$ so determined, that there is no current in the galvanometer wire.

Then we know that

${\frac {XS}{SY}}={\frac {A_{1}+PQ}{C_{1}+QR}}$

where $XS,\,PQ,$ &c. stand for the resistances in these conductors.

From this we get

${\frac {XS}{XY}}={\frac {A_{1}+PQ_{1}}{A_{1}+C_{1}+PR}}.$

Now let the electrode of the galvanometer be connected to $S'$, and let resistance be transferred from $C$ to $A$ (by carrying resistance coils from one side to the other) till electric equilibrium of the galvanometer wire can be obtained by placing $Q$ at some point of the wire, say $Q_{2}$. Let the values of $C$ and $A$ be now $C_{2}$ and $A_{2}$, and let

$A_{2}+C_{2}+PR=A_{1}+C_{1}+PR=R.$

Then we have, as before,

${\frac {XS'}{XY}}={\frac {A_{2}+PQ_{2}}{R}}.$

Whence

${\frac {SS'}{XY}}={\frac {A_{2}-A_{1}+Q_{1}Q_{2}}{R}}.$

In the same way, placing the apparatus on the second conductor at $TT'$ and again transferring resistance, we get, when the electrode is in $T'$,

${\frac {XT'}{XY}}={\frac {A_{3}+PQ_{3}}{R}},$

and when it is in $T$,

${\frac {XT}{XY}}={\frac {A_{4}+PQ_{4}}{R}}.$

Whence

${\frac {T'T}{XY}}={\frac {A_{4}-A_{3}+Q_{3}Q_{4}}{R}}.$

We can now deduce the ratio of the resistances $SS'$ and $T'T$, for

${\frac {SS'}{T'T}}={\frac {A_{2}-A_{1}+Q_{1}Q_{2}}{A_{4}-A_{3}+Q_{3}Q_{4}}}.$