Page:Scientific Memoirs, Vol. 1 (1837).djvu/625

In order now to find the resistance which the electric current suffers in its passage through the different wires, I first reduce their lengths all to one diagonal, and indeed to that of the wire of the multiplier, on the principle that two wires of the same metal offer then the same resistance to conduction when their lengths are in the same proportion as their diagonals (See Ohm's Galvanic Chain). In this case therefore the reduced lengths of the wires express their resistance to conduction: to have therefore a general idea of the problem, I suppose the multiplier, the conducting wires, and the electromotive spirals (with their free ends) to have the three reduced lengths, $L,\,l,$ and $\lambda$, and the electromotive power produced in the spirals to be represented by $x$, then ${\frac {x}{L+l+\lambda }}$ will be in effect the current which takes place, and we therefore have

${\frac {x}{L+l+\lambda }}=p\sin .{\frac {1}{2}}\alpha$

$x=(L+l+\lambda )\cdot p\cdot \sin .{\frac {1}{2}}\alpha$

(A.)

If we now consider the electromotive power in a convolution of the wire as unity, representing the unknown deviation produced by a convolution by $\xi$, and its reduced length by ($\lambda$); then granting the probable hypothesis, that at one and the same distance of the convolutions the electromotive force is directly as the number of convolutions, the following relation will take place for the number $n$, and for the reduced lengths $\lambda _{n}$ belonging to it (this is not necessarily $n\,\lambda$, because the free ends of the spirals need not increase in the same ratio for every number of convolutions)

${\frac {1}{n}}={\frac {(L+l+(\lambda ))p\cdot \sin .{\frac {1}{2}}\xi }{(L+l+\lambda _{n})p\cdot \sin .{\frac {1}{2}}\alpha }}$

therefore

$\sin .{\frac {1}{2}}\alpha =n\cdot {\frac {L+l+(\lambda )}{L+l+\lambda _{n}}}\cdot \sin .{\frac {1}{2}}\xi$

(B.)

In the experiments just mentioned $l+\lambda$ continued of the same magnitude for every number of convolutions, as the conducting and spiral wire consisted of one piece, besides $L$ remains the same, we therefore have $L+l+(\lambda )=L+l+\lambda _{n}$ and the equation B becomes changed into the following:

$\sin .{\frac {1}{2}}\alpha =n\sin .{\frac {1}{2}}\xi$

(C.)

If we now put instead of ${\frac {1}{2}}\alpha$ the values contained in the last column of our table of experiments, we obtain eleven equations, from which after the method of the least square, we shall be able to determine $\xi$, and if we bring this value of $\xi$ into the equation (C), we shall find the deviations $\alpha$ belonging to the number $n$ of convolutions, and the differences between this and the observed values will show whether the assumed