PROBLEM OF WHEATSTONE's BRIDGE. 343 



The numerators are directly deduced by the known rule. Putting 

 we have 



o = E 



^ 



If /> is the resistance of the whole set of conductors, 0, b, a', b', 

 and r, comprised between the points A and B, the intensity I of the 

 total current may be written 



E 



~R+V 



from this follows 



A r(a + b) (a' + b') + ab(a' + V) + a'b'(a + b) 



_=R+ ___ 



or 



_ r(a + b) (a' + b') + ab (a' + b') + a'b' (a + b) 

 P ~ r(a + a' + b + b') + (a + a')(b + b') ' 



If, for shortness sake, we represent by (a, b) the resistance 



i ab 



of the two branches a and b arranged in parallel, this 



- - 

 r i a + b 



a ~b 



value of p may then be written 



(,)(*,*) 



