Mr. 0. Heaviside on Duplex Telegraphy. 35 



ever, a simple solution may be obtained in terms of f;g, and^ 

 the external resistance. Thus, differentiating (2) with respect 



to b and then solving for ~, we obtain 



dx _ &Kc+f) + bc(c + g)} 2 + cx(bc -fgj 



db 



\w(g + b)(c+f) + bc(g + b + c+f)} 2 + bc(bc-.fgy' 



(5) 



and since the right-hand side of (2) is unaltered by inter- 

 changing/and g, and b and c, we can find -j- by making these 

 changes in (5). Thus 



dx_ \f<9 + h) + bc(b + f)¥ + bx(bc-fgy 



do ~ ^{ g + h){c+f) + bc(g + b + c+f)Y + bc{bc-fgY \ K } 



Equations (4), (5), and (6) must now be manipulated to 

 obtain b and c in terms of /, g, and x. After going through 

 the usual algebraical drudgery, which it is unnecessary to give 

 here, we obtain 



whence I " 



a=zVfg. _ J 



It will be found that these values of a, b, and c make the 

 received current a maximum with any given battery, receiving 

 instrument, and line. The strength of the received current is 

 then 



G= " ' '' //./■ /— " v 00 



or, which is the same, 



G= E 



a + b + c+f+g + x + by/l+c\/S, 



J J 



It will be observed that in the above solution (?) one of the 

 resistances (a) is independent of x, and is the same for all lines 

 with the same receiving instrument and battery. Since ba- 

 lance is always obtained in practice by adjusting one. of the 

 resistances, say c, it follows that only the other two, a and b, 

 need be calculated. This is readily done for a ; but for b it is 

 more difficult, since x cannot be simply expressed in terms of 



D2 



