TRANSIENT SOLUTIONS OF ELECTRICAL NETWORKS 

 Similarly, when we have the expression 



119 



1 



we can replace it by 

 2R, 



G -\- jcoC 



1 - g-2i 



where 



R, 



-p= G + jc^C """"^ i?o->0, P-^0. (31) 



The structure which gives the impedance in (31) is an open circuited 

 line in series with a resistance Rq. The impedance of the combination 

 is 



1 + e-2^ 



Ro + Ro coth P = Ro 



1 + 



1 - e~^P 



1 2J?o 



J 1 - e-2/ 



If then the series impedances of the line approach zero, Rq 

 the impedance of the combination approaches 



1 



and 



G -\- jcoC 



F. Solution for a Resistance and Capacity in Series 

 As a second example let us obtain the solution of a resistance and 

 capacity in series. To obtain the solution we solve the case of a 



Ro— O P— o 



o/p-y. 



Fig. 3 — Open circuited line and series resistance. 



resistance in series with an open circuited line as shown on Fig. 3. 

 The steady state solution for the current in this circuit is 



i? + 



joiC 



Replacing -^— =;, by , 



jwC I — e 



?3^ , and substituting in the above equation 



there results 

 E 



t — 



R + 



7? 1 



7r^5 — - when i?o -> and P -^ 0, and -^ ->-^— ?> 

 zKq 1 JUL 



1 - e-2^ 

 in accordance with equation (18). 



