Wheatstone's Bridge for measuring a given resistance. 119 



Multiply numerator and denominator by \/ = — — -, 



V de cd-\-dj+jc 



and we have 



u = 



2\/cde(c + d+e) {cd+ df+fc) + (c + d+ e) (cd+ df+fc) + cde 

 which has to be made a maximum. Differentiating and putting 



dc ' 



2\ / / cde(c + d+e)(cd + df+fc) + (c + d+ e) (cd+ df+fc) + cde 

 c 



\/cde (cd + df+fc) (c + d+e) 



X {cde{cd + df+fc)+cde(c + d+e)(d+f) + de(c + d-re)(cd + df+ft 



+ c{c + d+e)(d+f)+c(cd+df+fc)+cde. 



Therefore 



2^/cde(c + d+e)(cd + df+fc) + df(d+e)-c\d+f) 



_cde{c(cd+ df+fc) +c{c + d+e)(d+f) + {c + d + e){cd+df+fc)\ 

 y/cde [cd + df+fc) {c + d+J) 



Multiplying both sides of this equation by the denominator on 

 the right-hand side and reducing, we get 



{dfld + e)-c*{d+f)\ Vcde{c + d+e) (cd+df+fi) 

 = cde\c\d+f)-df(d+e)\, 

 which is satisfied by 



df(d+e)-c*(d+f) = 0, 

 which gives the required relation, 



-v< 



4rr <»> 



that is, c equals the square root of the product of the joint 

 resistance of the battery and the resistance to be measured 

 into the sum of the resistance of the galvanometer and the re- 

 sistance to be measured. Inserting this value of c n (6) and 

 (7), we find the values of a and b to be 



a^'ef, (!0) 



"We CU) 



In using the Wheatstone's bridge for measuring very high 

 resistances, as, for instance, the insulation resistances of (good) 



w 



