316 Steady Currents in Linear Conductors [CH. ix 



361. Suppose first that the whole system of currents in the network is 

 produced by a current X entering at P and leaving at Q, there being no 

 batteries in the network. Then all the E's vanish, and all the X's vanish 

 except X P and X Q , these being given by 



Xp = XQ X. 



Equation (292) now becomes 



so that V l -V 2 = (V 1 -V n )-(V 2 -V n ) 



Replacing 1, 2 by P, Q and P, Q by 1, 2, we find that if a current X 

 enters the network at 1 and leaves it at 2, the fall of potential from P 

 to Q is 



V P -V 9 = ~ (A 2P -A 2(? -A 1P + A ie ) (294), 



and since A^ = A gr , it is clear that the right-hand members of equations 

 (293) and (294) are identical. 



From this we have the theorem : 



The potential-fall from A to B when unit current traverses the network 

 from C to D, is the same as the potential-fall from C to D when unit current 

 traverses the network from A to B. 



362. Let it now be supposed that the whole flow of current in the 

 network is produced by a battery of electromotive force E placed in the 

 conductor PQ. We now take all the X's equal to zero in equation (292) 

 and all the E's equal to zero except E PQ which we put equal to E, and 

 E QP which we put equal to E. We then have 



J\^PQJ^PQ r \~ J\.QpLq P 



J\. f>f)Jls , 



Hence K- K = f (A Pl - A P2 - A Q1 + A Q2 ) ......... (295), 



and, by equation (289), the current flowing in the arm 12 is 



* = KE ( A P1 - A P2 - Aq, + A Q2 ) ............ (296). 



This expression remains unaltered if we replace 1, 2 by P, Q and P, Q by 

 1, 2. From this we deduce the theorem : 



