Thevenin’s theorem will be used to solve this problem. The ring distributor is shown in Fig.(a). Imagine feeder BC to be removed [Fig. (b)].
Current in A B = 20(0.8 − j0.6) = (16 − j12) A Current in section AC = 15(0.6 − j0.8) = (9 − j12)A
Drop over AB = (16 − j12)(1 + j1) = (28 + j 4) V Drop over AC = (9 − j12)(1 + j 3) = (45 + j15)V
Obviously, point C is at a lower potential as compared to point B.
p.d. between B and C = (45 + j15) − (28 + j 4) = (17 + j11) V
Impedance of the network as looked into from points B and C is = (1 + j1) + (1 + j 3) = (2 + j4)Ω..
The equivalent Thevenin’s source is shown in Fig.(c) with feeder BC connected across it.
Current in A B = (16 − j12) + (2.6 − j1.53) = 18.6 − j 13.53 = 23 ∠− 36° A
Current in BC = (9 − j12) − (2.6 − j1.53) = 6.4 − j11.5 = 13.2 ∠− 60.9° A
Total current fed at point A = (16 − j12) + (9 − j12) = 25 − j24 = 34.6 ∠− 43.8° A.