Page:Scientific Memoirs, Vol. 1 (1837).djvu/637

the length of the cylindrical armature; therefore the further increase of the number of convolutions can only be made by several series of convolutions placed one above another. Let the electromotive power of a series of convolutions which the length of the armature occupies ${}=\phi$; the length of the wire of all these convolutions, or, in this case, on account of the diameter of the wire being equal throughout, the resistance it offers ${}=\alpha$; let the length of the necessarily free ends of the wires together ${}=\beta$, the power therefore of the current of this first series of convolutions is

$\mu _{1}={\frac {\phi }{\alpha +\beta }};$

let $\gamma$ be the piece of the second series of convolutions by which its length, on account of its necessarily greater diameter, is greater than the length $\alpha$ of the first series, the power of the current from these two series is

$\mu _{2}={\frac {2\phi }{2\alpha +\gamma +\beta }},$

and in the same manner

$\mu _{3}={\frac {3\phi }{3\alpha +\gamma +\delta +\beta }},$

where $\delta$ designates the quantity by which the first series is surpassed in length by the third. If now the second series of convolutions does not add to the strength of the current, we put $\mu _{1}=\mu _{2}$, therefore

${\frac {\phi }{\alpha +\beta }}={\frac {2\phi }{2\alpha +\beta +\gamma }},$

whence we have

$\beta =\gamma ,$

i. e. as soon as the length of the free ends is only equal to the difference between the lengths of the second series of convolutions and those of the first, the second series would then add nothing to the strength of the current. In order to see what three series would do in this case, let us put $\beta =\gamma$ in the expression for $\mu$, and we obtain

$\mu _{3}={\frac {3\phi }{3\alpha +2\beta +\delta }};$

$\delta$ however is now greater than $\gamma$ or $\beta$, we therefore put $\delta =\beta +\mu$, where $\mu$ expresses a positive magnitude; we obtain by this

$\mu _{3}={\frac {3\phi }{3\alpha +3\beta +\mu }}={\frac {\phi }{(\alpha +\beta )+{\frac {\mu }{3}}}}.$

This last expression for $\mu _{3}$ is evidently smaller than ${\frac {\phi }{\alpha +\beta }}$, consequently three series of convolutions would only weaken the action of one or two series (which actions have been here assumed as equal).