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  1. G. Planté, Comptes Rendus,  85, pp. 794–6; 1877.
  2. 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.
  3. 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=∞
  4. 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.
  5. Clearly such a device is a transformer rather than a generator, since no work is done on the charges coming from the original source.

1877 (1)

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

Planté, G.

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

Comptes Rendus (1)

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

Other (4)

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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Figures (2)

Figs. 1–5
Figs. 1–5

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

Fig. 6
Fig. 6

Diagram of a direct current potential transformer using four condensers.

Equations (6)

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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
d F d x = d d x ( 1 + x ) 1 / x = ( 1 + x ) 1 / x x 2 ( 1 + x ) [ x - x log e ( 1 + x ) - log e ( 1 + x ) ]
F = ( 1 + 1 ) 1 / 1 = 2
T = 2 n
x=0         x=0         z=