Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/445

Let us suppose that $\beta$ is a standard resistance coil, and that we wish to adjust $\gamma$ to be equal in resistance to $\beta$.

Two other coils, $b$ and $c$, are prepared which are equal or nearly equal to each other, and the four coils are placed with their electrodes in mercury cups so that the current of the battery is divided between two branches, one consisting of $\beta$ and $\gamma$ and the other of $b$ and $c$. The coils $b$ and $c$ are connected by a wire $PR$, as uniform in its resistance as possible, and furnished with a scale of equal parts.

The galvanometer wire connects the junction of $\beta$ and $\gamma$ with a point $Q$ of the wire $PR$, and the point of contact at $Q$ is made to vary till on closing first the battery circuit and then the galvanometer circuit, no deflexion of the galvanometer needle is observed.

The coils $\beta$ and $\gamma$ are then made to change places, and a new position is found for $Q$. If this new position is the same as the old one, then we know that the exchange of $\beta$ and $\gamma$ has produced no change in the proportions of the resistances, and therefore $\gamma$ is rightly adjusted. If $Q$ has to be moved, the direction and amount of the change will indicate the nature and amount of the alteration of the length of the wire of $\gamma$, which will make its resistance equal to that of $\beta$.

If the resistances of the coils $b$ and $c$, each including part of the wire $PR$ up to its zero reading, are equal to that of $b$ and $c$ divisions of the wire respectively, then, if $x$ is the scale reading of $Q$ in the first case, and $y$ that in the second,

${\frac {c+x}{b-x}}={\frac {\beta }{\gamma }},\qquad {\frac {c+y}{b-y}}={\frac {\gamma }{\beta }},$

whence

${\frac {\gamma ^{2}}{\beta ^{2}}}=1+{\frac {(b+c)(y-x)}{(c+x)(b-y)}}$

Since $b-y$ is nearly equal to $c+x$, and both are great with respect to $x$ or $y$, we may write this

${\frac {\gamma ^{2}}{\beta ^{2}}}=1+4{\frac {y-x}{b+c}},$

and

$\gamma =\beta \left(1+2{\frac {y-x}{b+c}}\right).$

When $\gamma$ is adjusted as well as we can, we substitute for $b$ and $c$ other coils of (say) ten times greater resistance.

The remaining difference between $\beta$ and $\gamma$ will now produce a ten times greater difference in the position of $Q$ than with the