306 Steady Currents in Linear Conductors [OH. ix 



From KirchhofFs Laws, we obtain the following equations: 



(Law I, point B) x l -x< 2 -x b = Q .................. (273), 



(Law I, point C) # 3 -# 4 + #6 = .................. (274), 



(Law II, circuit ABC) X& + x b E b - x 3 E s = Q .................. (275), 



(Law II, circuit BCD) x b R b + afjt 4 - x 2 R 2 = .................. (276). 



These four equations enable us to determine the ratios of the five currents 

 #n #2, #3> #4, #& We may begin by eliminating # a and # 4 from equations 

 (273), (274) and (276), and obtain 



X b (R b 4- R 2 4- -#4) + #3^4 ~ #1#2 = 0, 



and from this and equation (275), 



_ _ __ _ _ 



R 2 R 3 R^Ri R 1 (R b + R 2 4- R*) 4- R b R 9 R 3 (R b + R 2 + R*) 4- -R&K 4 



............ (277). 



The ratios of the other currents can be written down from symmetry. 



If the total current entering at A is denoted by X, we have X x^ + # 3 . 

 Thus if each of the fractions of equations (277) is denoted by 6, 



X = 6 {(R, 4- R 3 ) (R, 4- E.) + R b (E, + R, + R 3 + R 4 )} ...... (278), 



and this gives 6, and hence the actual values of the currents, in terms of the 

 total current entering at A. 



The fall of potential from A to D is given by 



and from equations (277) this is found to reduce to 



v A -v D =\e, 



where 



(R 3 4- .ft) + R b (R 1 R S + RA + ^ A 



so that X is the sum of the products of the five resistances taken three at 

 a time, omitting the two products of the three resistances which meet at the 

 points B and C. There is now a current X flowing through the network, and 

 having a fall of potential V A V D . Hence the equivalent resistance of the 

 network 



(R, + R 3 ) (R, + JB 4 ) 4- R 



by equation (278). 



