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A 1-phase ring distributor ABC is fed at A. The loads at B and C are 20 A at 0.8 p.f. lagging and 15 A at 0.6 p.f. lagging respectively, both expressed with reference to voltage at A. The total impedances of the sections AB, BC and CA are (1 + j1), (1 + j2) and (1 + j3) ohm respectively. Find the total current fed at A and the current in each section.

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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.

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