Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/455

strength of current, so as to determine how the strength of current effects the resistance.

If $y$ is the current in the galvanometer, the actual current through the battery is $x_{0}$ with the key down and $x_{1}$ with the key up, where

$x_{0}=y\left(1+{\frac {b}{\alpha +\gamma }}\right),\qquad x_{1}=y\left(1+{\frac {b}{\gamma }}+{\frac {\alpha c}{\gamma (\alpha +c)}}\right),$

the resistance of the battery is

$a={\frac {c\gamma }{\alpha }}$,

and the electromotive force of the battery is

$E=y\left(b+c+{\frac {c}{\alpha }}(b+\gamma )\right)$.

The method of Art. 356 for finding the resistance of the galvanometer differs from this only in making and breaking contact between $O$ and $A$ instead of between $O$ and $B$, and by exchanging $\alpha$ and $\beta$ we obtain for this case

${\frac {y_{0}-y_{1}}{y}}={\frac {\beta }{\gamma }}{\frac {c\gamma -b\beta }{(c+\beta )(\beta +\gamma )}}$.

On the Comparison of Electromotive Forces.

358.] The following method of comparing the electromotive forces of voltaic and thermoelectric arrangements, when no current passes through them, requires only a set of resistance coils and a constant battery.

Let the electromotive force $E$ of the battery be greater than that of either of the electromotors to be compared, then, if a sufficient

Fig. 38.

resistance, $R_{1}$, be interposed between the points $A_{1},\,B_{1}$ of the primary circuit $EB_{1}A_{1}E$, the electromotive force from $B_{1}$ to $A_{1}$