M. E. Edlund on the Nature of Galvanic Resistance. 207 



found for the galvanic resistance. A galvanic current s divides 

 itself (see fig. 1), at a point a on the conduction-path, between 

 several similar conductors -p- ^ 



UfiJm &c -> wnicn with 

 the intensity 1 have the 



respective resistances r , r ls 

 r lt) &c, and all meet in a 

 point b. We have now to 

 determine the quantity of 

 the current that passes 

 through each of these con- 

 ductors. 



It is obvious that the current s must so distribute itself that 

 the resistances in all the conductors shall be perfectly equal — 

 that is, that the resistance undergone by each of these portions 

 of the current during its passage from the point a to the point 

 b shall be equally great. If the resistance in one of the con- 

 ductors were for a moment less than in the rest, the intensity 

 there would augment till the resistance became as great as in the 

 rest. Naming the respective currents s , s t , s u , &c, we obtain 

 the following, because the resistance is proportional to the in- 

 tensity : — 



Vo 



— s i r l —s u r il — ... _R. 



This signifies that the current-intensities in the respective 

 conductors are inversely proportional to the resistances with the 

 current -intensity 1, — a result which, as is well known, accords 

 with experiment. 



The conductors / , f p f ip &c. are now to be replaced by a 

 single conductor F, so that this alone shall cause the same re- 

 sistance as the conductors f , f t , / y/ , &c. together. 



The resistance is determined by the counterpressure, against 

 the propagation of the current, on the unit of surface of the cross 

 section of the conductor. This pressure was, when the current 

 passed simultaneously through / , //,///, &c, = R. If the re- 

 sistance sought (that of the conductor F for the current-inten- 

 sity 1) is denoted by sc 3 we have 



R=s#. 



In order to find R, we have, according to what was stated 

 above, 



R R R 



"o— T" *i 



■> s,i= —> 



and therefore 



r r t r n 



+ *, + *„+... =s=R (1 + 1 + 1 +;...). 



