RESISTANCE AND CAPACITANCE IN SERIES 

 any time t after opening the switch 



R 



dt. 



Jo Jo ^ 



The solution is Vji=^ Vq = IR{1 — e'^'^^) 



1 



But iji = I— ic 



Figure 3.37 



If the generator and switch are then disconnected, the capacitance discharges 

 again, through the resistance, and Vji = Vq= y^-tiRC^ where V is the 

 potential difference to which the capacitance was charged. If the original 

 charging process was substantially completed, V is, of course, equal to IR. 



Resistance and capacitance in parallel, connected to constant-voltage alternator 

 (Figure 3.38) 



Figure 3.38 



In this case, we have : 

 Current in the resistance = 



V 

 R 



V _jV 



J'Xq Xq 



Current in capacitance = 



Therefore, total current = '^l 15 + y~ I 



V 

 So the impedance = 



1 



(-; 



+ 



Xc) 



R^ Xc 



Compare this with the expression for two resistances in parallel. 



If we take the expression for the impedance of the parallel combination 

 and rewrite it in the form 



Z = 



-jXc . R 

 R-jXc 

 41 



