Electric Currents in Networks of Conductors. 

 Removing the common factor rp, we have to evaluate 



237 





■ 7T 



-it— 6 -l 



9 





—ir-6 tt + 2 -2 





-6 -2 nr + 2 



This is very easily reduced to 





o e 77 





nr 



-112 



5 



which is equal 

 and therefore 



to 



-0 2 t 





d n = ry27r(7r + 07r-6 2 ). 



The value of the minor of the leading element of the network 

 determinant, viz. d n _i, is 



ry{(7r + 2) 2 -4}=rV7r(7r + 4). 



Hence the resistance of the network between A and B, = R, is 



d n -i r 7T(7T + 4) 



=rp 



27r + 27rfl-2fl 2 



7T— 4 



We can check this in the extreme cases when = or 6= 



7T 



When = 0, the network-resistance is simply that of three 



conductors whose resistances are 2pr, irpr, and irpr joined in 



multiple arc, as in Plate VII. fig. 12, because PQ now coin- 



2tt 

 cides with AB ; and this is simply rp , . It is seen at 



once that the above value for R becomes this when 9 is put 



equal to zero. Now, when 0=-^, the diameter PQ joins 



points at equal potential (fig. 13), and is not traversed by any 

 current at all ; and hence its removal will not affect the 

 resistance between the points A and B. 



Hence the resistance of the network simply reduces to that 

 of a circle measured at the ends of a diameter, or to two con- 

 ductors of resistance irpr joined in multiple arc, and this is 



equal to pr^-. By putting 0=y in the general solution 



for R above, we get it reduced to rp y ; and accordingly this 

 Phil. Mag. S. 5. Vol. 20. No. 124. Sept. 1885. S 



