ELECTRIC CIRCUIT THEORY 347 



by (259), (260) and (261) in terms of the indicial admittances of the 

 component networks. 



A simple example will now be worked out illustrating the method 

 of solution just discussed. Suppose that a unit e.m.f. is impressed 

 on a transmission line (infinitely long) of distributed constants R,L,C, 

 through a terminal resistance Ro. Required the terminal line voltage 

 V. 



The operational equation of this problem is gotten in the usual 

 manner. The current entering the line is 



V 



V 



Cp 



Lp+R 



It is also obviously equal to ^- (1— F) : equating the two expressions, 



Ro 



and rearranging we get : — 



I'LP+R 



v= ' ^P 



^o+^l 



. , .\LP+R 



cp 



Writing i?/2L = p and setting Ro = V'L/C, this becomes 



V= Vl + V^ ■ . (262) 



l + Vl + 2p/^ 



This operational equation can, of course, be solved in a number of 

 ways, though, as a matter of fact, its numerical solution is quite 

 troublesome. This point will be returned to later: we shall first 

 formulate the problem in accordance with the method just discussed. 



The indicial admittance of the line is know n ; it is 



^^e-'^'Io{pt)=A{t). 



Consequently the current entering the line is explicitly 



But the current is also equal to -^ (l— V{t)); equating, we get 



Ro 



V{t) =l-Ro^^f V{t)A (/ - T)dr. 



