36 RESEARCHES ON II. 



R^Z^ + g-^, (1) 

 9 

 where p = the number of turns when the current was divided, 

 g = the resistance of the wire of the galvanometer, 

 r = the resistance of the dividing coil, 

 r^ := the number of turns developed when the current was not divided. 



de I'EUctricM of De la Rive. It will not, however, be quite useless to those who may not have these works 

 at hand to give a demonstration of it, which will be somewhat clearer than that which is in the latter journal. 

 Ohm's rule states that the intensity of a current F is in direct ratio of the electromotive force E of the pair, 

 and the inverse of the resistance R, so that we have 



R 



In the case where a galvanometer whose resistance is g, and a wire whose resistance is x, are introduced into 

 the circuit, we shall have 



F=. ^ (m) 



Let R'- =R+ p, p being as above any number of turns of the rheostat, and let the current be supposed to be 

 divided by means of two parallel conductors g and r: by this the resistance of the circuit itself is diminished; 

 now we must find the value of a wire whose reduced length %, is equal to the length of that portion of the 

 circuit which has been so modified. [Solution. — As in general the quantities of currents Q and Q^ which 

 pass through every one of the conductors r and g are in inverse ratio of their resistances, we shall have 



^=3_, from which ^+g^ = l±I- ; 



but by hypothesis % is such that it is capable of giving a free passage to Q + Q^, therefore 



Q+Q' _ 9 



from which, together with the preceding equation, we have 



gr 



- (»)] 



'• + , 



Now let the current running through the galvanometer before the division be expressed by 



F 



R' + g 



After the division by the coil r, we must consider three parts of the circuit : — 



1. That which was divided. , 



2. The galvanometer from which a portion of the current was taken. 



3. The wire which divides the current. 



The intensity in the first part of the circuit will be 



^'~ R^ + ^' 

 exactly the same as if instead of r and g parallel, we had substituted % ; representing the other two parts 

 by F^ and F^, we shall have 



—i = — , whence 3 — = = —2, 



^3 9 ^. + K 9 + r F: 



which gives 



9-\- r 

 for the portion of current passing through the galvanometer. This value, after the substitutions of F^ and %, 

 becomes 



R' {r + g) -{- r g 



