134 ELECTRICAL ENGINEERING 



The total current 



EB 2 = EVG 2 + B 2 , 

 but, by equation 181, I = EY, and therefore 



Y = VG 2 + B 2 . ...... (187) 



(5) The conductance of a circuit is, from equation 184, 



n power component of current 



Or = : j -z - ... (loo) 



impressed e.m.f 



(6) The susceptance of a circuit is, from equation 185, 



wattless component of current 



-D = ; ; ;; 



impressed e.m.t. 



In the solution of series circuits it is not necessary to employ 

 the terms admittance, 'conductance and susceptance but the 

 solution of series-parallel circuits is very much simplified by their 

 use. 



84. Example. In the circuit in Fig. 102 determine the main 

 current and the currents in the two branches in magnitude and 

 phase relation with the impressed e.m.f. 



* L- n ^ 



IifRi Jaffa 



!X 2 J i 



FIG. 102. 

 (I) Using the constants R and X: 





from the vector diagram 



(3) 7 = /! 2 + 7 2 2 + 2 /i/ 2 cos <t> 

 and 



(4) tan^=f' sin ^ + 



/i cos fa + 



