Mathematics · AC Distribution Systems
A single-phase ring distributor ABC is fed at point A. The loads at B and C are: At point B: 20 A at 0.8 power factor lagging At point C: 15 A at 0.6 power factor lagging The impedances of the sections are: AB = (1 + j1) Ω BC = (1 + j2) Ω CA = (1 + j3) Ω If the current in section AB is (16 - j12) A and the current in section AC is (9 - j12) A, what is the total current fed at point A?
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 1(26 - j22) A
- 2(34 - j30) A
- 3(25 - j24) A✓
- 4(27 - j20) A
Correct answer
(25 - j24) A
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