70 Mr. 0. Heaviside on the 



and consequently,, on making Z 1 Z 4 = Z 2 Z 3 , we make A/B in- 

 dependent of Z 6 . Hence, the current in the bridge-wire is 

 independent of branch ti altogether when the general condition 

 of an induction-balance is satisfied, making branches 5 and 6 

 conjugate. 



But, as is known to all who have had occasion to work out 

 problems concerning the steady distribution of current in a 

 network, there is a great deal of labour involved, which, when 

 it is the special state involved in a resistance-balance, is wholly 

 unnecessary. This remark applies with immensely greater 

 force when the balance is to be a universal one, for transient 

 as well as permanent currents ; so that the proper course is 

 either to assume the existence of the property required at the 

 beginning, and so avoid the reductions from the complex 

 general to the simple special state, or else to purposely arrange 

 so that the reductions shall be of the simplest character. 

 Thus, to show that C 5 is independent of branch 6, when there 

 is an. impressed force in (say) side 1, making no assumptions 

 concerning the nature of branch 6, we may ask this question, 

 Under what circumstances is C 5 independent of C 6 ? And, to 

 answer it, solve for C 5 in terms of e l and C 6 , and equate the 

 coefficient of C 6 to zero. 



Thus, writing down the equations of E..M.F. in the circuits 

 ABiBgA and BiCB^ in the above figure, we have 



«i=Z 1 C 1 + Z 5 C 5 — Z 2 C«j,) (\§d\ 



o=z,c,+ZA-z,c,J • <■ ' 



when there is no mutual induction between different branches, 

 but not restricting Z to a particular form ; and now putting 



C4 = C 6 — Ci + C 5 , C 3 = Ci — C 5 , C 2 = C 6 — Ci, . (16d) 

 we obtain 



ei + Z 2 C 6 =(Z 1 + Z 2 )C 1 + Z 5 C 6 , 



Z 4 C 6 =(Z3 + Z 4 )C 1 -(Z 3 + Z !l + Z 5 )C 5 ;. 

 which give 



(Z3 + Z>, + (Z 2 Z 3 -Z,Z 4 )C 6 



^-(Z 1 +z 2 )(Z 8 +z 4 )+z 6 (Z 1 +z 2 +z 3 +z 4 y ■ (lbd) 



making C 5 independent of C 6 when the condition of conjugacy 

 of branches 5 and (5 is satisfied. 



If there are impressed forces in all four sides of the quadri- 

 lateral, then (I8d) obviously becomes 



(Z, + Z«)fo- f ,WZ , + Z,Kg.-g4)-'-(Z 1 ,Z,-Z,Z < )C. , 10A 

 ° 5 ~ (Z 1 + Z 2 )(Z 3 + Z 4 ) + (Z 1 + Z 2 + Z 3 + Z 4 )Z 5 -> V w > 



(lid) 



