} 



Electricity in a Network of Conductors. 279 



a battery of electromotive force E, the only electromotive 

 force in the system, and let the resistances and directions of 

 the currents in the various parts of the system be as indicated 

 in the diagram. Let r Q be the resistance of the battery and 

 the wires connecting it with A and B, and let y L , <y 2? & c - be 

 the strengths of the currents flowing in the resistances r 1} r l9 

 &c, respectively, in the directions indicated by the arrows. 

 By the principle of continuity we get 



73 = 71—75,-^ 



74=72 + 75,?- (1) 



76 = 71 + 72--' 

 Applying the second principle to the circuits BACB, ACDA, 

 CBDC, we obtain the three equations 



7i fa + n + n) + y 2 n — 75^3 = E , ■ 



Wx -72^2 + 75^5 =0, } 



71^3 - y^n - 75 fa + n + n) = 0, 



Eliminating y x and <y 2 , we find 



_ E(r 2 r B -r 1 r 4 ) . . 



7s- 5 9 W 



where 



D = r 5 r 6 (r , + r 2 + r 3 + r 4 ) + r 5 fa + r s ) fa + r 4 ) 



+ re(f^ + f^)<r 1 +r 4 )+-Vi(^'+r4)+f|gr 4 (r 1 +r I ). . (4) 



By substituting for 7 2 in the second and third of equations 

 (2) its value 7 6 — 7i> we get 



7](^l + ^2) + 75^5 -76^2 = 0,) ^ „. 



7l(V 3 + n) — 75 fa + ^4 + n) — 76^4 = 0. / 



From these we obtain, by eliminating y J} 



= 7efa^-^4) fg v 



nfa + ^2 + ^3 + ^4) + fa + ^2)fa + n)' 



By means of equations (3) and (6) we can very easily solve 

 the problem of finding the equivalent resistance of the system 

 of five resistances, r ly r 2 , &c, between A and B. For let R 

 be this equivalent resistance ; since y 6 is the current flowing 

 through the battery, we have 7 6 = E/(r 6 + R). Substituting 

 this value of y 6 in (6), equating the values of 75 given by (3) 

 and (6), and solving for R, we get 



R __ n fa + r 3 ) fa + n) + nr 3 (r 2 + r 4 ) + r 2 r 4 fa + r 3 ) 



n(n + ^2 + ^3 + n)+fa + ^2)fa + n) ^ ' 



The following theorem, which can be verified by experi- 

 ment, will be of use in what immediately follows. 



