PRINCIPLES OF ALTERNATING CURRENT 225 



Problem. Suppose that we have an inductance of 

 0.01 henry and 50 ohms resistance in series with a capacity 

 of 40 microfarads on a 60-cycle, 110-volt circuit. How 

 much current will flow and what will be the power factor 

 of the circuit? 



110 



110 110 



V50 2 + (37.7 -66.4) 2 



110 



= ^-^ = 1.73 amperes. 

 5/.6 



tancj> = ^-=0.575; 



<]) = 29 55' (leading current); 

 or 



cos <b = ^ = 0.870 or <b = 29 55'. 



Parallel Circuits. When there are two paths in parallel 

 for the current to flow through, the calculation of the 

 current becomes more difficult. It is generally easier to 

 solve parallel circuits by the vector addition of the currents 

 in the different branches. Suppose a circuit as given in 

 Fig. 135. The magnitude and phase of the current in each 

 branch is first determined and plotted on cross-section 

 paper as in Fig. 136. The current in the inductive path is 

 calculated to be 01 1, lagging behind the voltage OE by the 

 angle <j>i. The current in the capacity branch is 01 2, 

 leading the impressed force by the angle fa. Now the 

 line current must be the vector sum of the two branch 



