22 



ALTERNATING CURRENTS 



inductive and non-inductive, and condensers. The magnitudes of 

 these are marked in the diagram. An alternating p.d. of 500 volts, 

 having a frequency of 50, is applied across the extreme terminals 



r-20 



rx/WVHE 



L- 



TTrmnr 



r-/0i L--IZ 

 FIG. 16. Circuit containing Resistances, Inductances, and Capacities. 



L and N. The problem is to find the currents in the various 

 branches. 



We begin by finding the joint impedance of the branches a and /3 

 between L and M, and the joint impedance of the branches y, 8, and 

 between M and N. 



The frequency being 50, we have p = 2?r X 50 = 314, say. 



Hence the capacity reactance in the branch a is -^ = 9 .. . .- 1 6 



= 80 '4 ohms ; and since the resistance is 20, the impedance of the 

 branch a is v / 20 r +~80'4 a ~ = 82 '8 ohms, and its phase angle is 



82'8 1 



tan- 1 = 76 25'. The admittance of this branch is ~ 



20 o*'o 



= 0-0121, say. 



Considering next the branch /3, we have for its inductive reactance 

 pL = 314 x 0-12 = 37-7. The impedance is \/10 2 + 37'7 2 = 39, 



1 37'7 



the admittance = 0'0256, and the phase angle tan" 1 -JQ- 



= + 75 8'. 



Applying the rule for the composition of parallel admittances, 

 we have for the joint admittance of a and /3 



\/(0-0121 cos 76 25' + 0'0~256 cos 75 b 8') 2 + (- 0-0121 sin 76 25' 

 + 0-0256 sin 75 8') 2 = \/0-00945 2 + 0'013 2 = 0'01605 



Hence the joint impedance between L and M is 62'3 ohms, and 



. 0-013 

 the joint phase angle is tan- 1 n . nn ^ i~ = + 



Taking next the branches 7, 

 proceeding as before, we find 



and t between M and 



and 



