A 3 -phase,3-wire, 100 V system supplies a balanced delta connected load of
z = 2 + j7 ohm .The line currents when one of the lines is opened is
A. 20.6 < 74.05, 20.6 < 74.05, 0
B. 20.6 < 74.05, 0, 0
C. 41.2 < 90, 0, 0
D. none of these
Answers
Answer:
a
Explanation:
Answer:
Line current IA = 5∠ − 45° − 5 ∠ 75° = 8.66 ∠ − 75°
Opposition of every delta branch = 20 cos 45° = 14.14 Ω
Complete power consumed = 3 I2R = 3 × 52 × 14.14 = 1060.6 W
Explanation:
Let VAB = 100∠0°. Since stage grouping is ABC, VBC = 100 ∠-120° and VCA = 100° ∠ 120
Stage current IAB = VAB/ZAB = 100∠0°/20∠45°
IBC = VBC/ZBC = (100 ∠ - 120°)/20∠45°
= 5 ∠ - 165°, ICA = VCA/ZCA = 100 ∠120°/20∠45° = 5 ∠75°
Applying KCL to intersection A, we have
IA + ICA - IAB = 0 IA = IAB - ICA
∴ Line current IA = 5∠ − 45° − 5 ∠ 75° = 8.66 ∠ − 75°
Since the framework is adjusted, IB will slack IA by 120° and IC will slack IA by 240°.
∵ IB = (8.66 ∠ 120°) = 8.66 ∠ − 195° IC= 8.66∠( − 75° − 240°) = 8.66∠ − 315° = 8.66∠ 45°
W1 will be W1 = VAC IC cos ϕ . Phasor VAC is the opposite of phasor VCA. Consequently, VAC is the converse of phasor VCA.
Consequently, VAC slacks the reference vector by 60° though IA slacks by 75°. Consequently, stage contrast between the two is (75° - 60°) = 15°
∴ W1 = 100 × 8.66 × cos 15° = 836.5 W
Likewise W2 = VBC IB cos φ = 100 x 8 .66 x cos 75° = 224.1W
∴ W1 + W2 = 836.5 + 224.1 = 1060.6 W
Opposition of every delta branch = 20 cos 45° = 14.14 Ω
Complete power consumed = 3 I2R = 3 × 52 × 14.14 = 1060.6 W
Consequently, it demonstrates that the amount of the two watt meter readings gives the complete power consumed
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