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Full Article | PDF Article**Journal of the Optical Society of America**- Vol. 14,
- Issue 4,
- pp. 323-327
- (1927)
- •doi: 10.1364/JOSA.14.000323

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- G. Planté, Comptes Rendus, 85, pp. 794–6; 1877.

- This cycle of connections will have to be repeated many times in order to charge the second group to the voltage (1+x) V, which is approached as an asymptote.

- It is worthy of notice that Lim F=Lim (1+x)1/x=Lim (1+1/z)z=Napierian base, e.x=0 x=0 z=∞

- The expression in brackets is 0 when x=0. Its derivative is −(1+x), which is negative for positive values of x. The factor in brackets is therefore negative when x is positive. The factor in front of the brackets is positive for positive values of F and x. Hence the product of the two factors is negative when x is positive.

- Clearly such a device is a transformer rather than a generator, since no work is done on the charges coming from the original source.

G. Planté, Comptes Rendus, 85, pp. 794–6; 1877.

G. Planté, Comptes Rendus, 85, pp. 794–6; 1877.

This cycle of connections will have to be repeated many times in order to charge the second group to the voltage (1+x) V, which is approached as an asymptote.

It is worthy of notice that Lim F=Lim (1+x)1/x=Lim (1+1/z)z=Napierian base, e.x=0 x=0 z=∞

The expression in brackets is 0 when x=0. Its derivative is −(1+x), which is negative for positive values of x. The factor in brackets is therefore negative when x is positive. The factor in front of the brackets is positive for positive values of F and x. Hence the product of the two factors is negative when x is positive.

Clearly such a device is a transformer rather than a generator, since no work is done on the charges coming from the original source.

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Connections showing the application of the cumulative and group principles to four condensers.

Diagram of a direct current potential transformer using four condensers.

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$$(1+x)V+x(1+x)V={(1+x)}^{2}V$$

$$T={(1+x)}^{y}={(1+x)}^{n/x}=\{{(1+x)}^{1/x}{\}}^{n}={F}^{n}$$

$$\frac{dF}{dx}=\frac{d}{dx}{(1+x)}^{1/x}=\frac{{(1+x)}^{1/x}}{{x}^{2}(1+x)}[x-x\hspace{0.17em}{\text{log}}_{e}(1+x)-{\text{log}}_{e}(1+x)]$$

$$F={(1+1)}^{1/1}=2$$

$$T={2}^{n}$$

$$\text{x}=0\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\text{x}=0\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\text{z}=\infty $$

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