LIXR CURRENT REGULATION 631 



This suggests solving for the value of R^ corresponding to the point 

 where h is a maximum by differentiating equation (2) with respect 

 to Ri and equating to 0. The nature of the equation shows also that 

 12 will have but one maximum. If the value of Ri corresponding to 

 maximum ii proves to be greater than 500 ohms, it will open up the 

 possibility of increasing the received current by adding the regulating 

 resistances at points (2), Fig. 1. 



In calculating the line and received currents for different values of 

 i?2 it is, of course, permissible to calculate separately corresponding 

 values of r^ and then substitute these values as constants in equation 

 (2). Obviously this procedure cannot be followed in finding the 

 derivative of ^2 with respect to Ri. The expression to be dealt with 

 in this differentiation is that which results from the substitution of 

 the right hand member of equation (8) for ^2 in equation (2). This 

 substitution gives the following explicit and rigorous equation for the 

 steady-state current in the receiving relays of a balanced symmetrical 

 bridge duplex telegraph circuit: 



,• -^^ (10) 



''~(ir+G)(2i?,+6)+i?2(i^2+6)+ 



^T{\T+G){2R2+hY+{2Ri-Yh)\R2T{R2^h)-VbRi{Ri-^2G)\ 



Equation (10) was found useful in calculating received currents as it 

 combines (2) and (8) and may be used instead of equations (1) and (7) 

 by changing G to Ri and R2 to a, but when it is differentiated and 

 equated to the resulting equation for R2 corresponding to maximum 

 received current is of an extremely impractical nature as it involves 

 various powers of Ro up to the sixth, together with an unusually large 

 number of terms. In this investigation, it was not necessary to solve 

 this equation for R-i as it was found that for values of R2 and T within 

 the practical ranges of 500 to 1750 ohms for R2 and 500 to 3000 

 ohms for T, ^2 is very nearly equal to 1/3 i?2 + 2^/7"+ 200. If this 

 expression be substituted for ^2 in equation (2) and the result differ- 

 entiated and equated to it leads to the following equation which 

 gives values of R2 corresponding fairly close to the point of maximum 

 received current: 



i?2 = V!6(r+2rH^+200) (ii) 



With a receiving relay of 400 ohms resistance and a battery tap re- 

 sistance of 120 ohms, as shown in Fig. 1, equation (11) becomes 



R2 = 10\/3(r+2ri+320) 



