Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/442

in $BC$. the resistance of $BC$ being still $a$, then the value of $D$ would remain the same, and the current in $BC$ due to the electromotive force $E$ acting along $OA$ would be equal to the current in $OA$ due to the electromotive force $E$ acting in $BC$.

But if we simply disconnect the battery and the galvanometer, and without altering their respective resistances connect the battery to $O$ and $A$ and the galvanometer to $B$ and $C$, then in the value of $D$ we must exchange the values of $B$ and $G$. If $D'$ be the value of $D$ after this exchange, we find $${\begin{aligned}D'-D&=(G-B)\{(b+c)(\beta +\gamma )-(b+\gamma )(\beta +c)\},\\&=(B-G)\{(b-\beta )(c-\gamma \}.\end{aligned}}$$

Let us suppose that the resistance of the galvanometer is greater than that of the battery.

Let us also suppose that in its original position the galvanometer connects the junction of the two conductors of least resistance $\beta$, $\gamma$ with the junction of the two conductors of greatest resistance $b$, $c$, or, in other words, we shall suppose that if the quantities $b$, $c$, $\gamma$, $\beta$ are arranged in order of magnitude, $b$ and $c$ stand together, and $\gamma$ and $\beta$ stand together. Hence the quantities $b-\beta$ and $c-\gamma$ are of the same sign, so that their product is positive, and therefore $D'-D$ is of the same sign as $B-G$.

If therefore the galvanometer is made to connect the junction of the two greatest resistances with that of the two least, and if the galvanometer resistance is greater than that of the battery, then the value of $D$ will be less, and the value of the deflexion of the galvanometer greater, than if the connexions are exchanged.

The rule therefore for obtaining the greatest galvanometer deflexion in a given system is as follows:

Of the two resistances, that of the battery and that of the galvanometer, connect the greater resistance so as to join the two greatest to the two least of the four other resistances.

349.] We shall suppose that we have to determine the ratio of the resistances of the conductors $AB$ and $AC$, and that this is to be done by finding a point $O$ on the conductor $BOC$, such that when the points $A$ and $O$ are connected by a wire, in the course of which a galvanometer is inserted, no sensible deflexion of the galvanometer needle occurs when the battery is made to act between $B$ and $C$.

The conductor $BOC$ may be supposed to be a wire of uniform resistance divided into equal parts, so that the ratio of the resistances of $BO$ and $OC$ may be read off at once.