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19 A balanced three-phase voltage source, Van = 200/0° V rms, positive phase sequence, is connected to an unbalanced A load: ZAB= 60 + 10, Zgc = 30 + 130. ZCA 30-130 2. Find: (a) the three phasor line currents; (b) the power de- livered to each phase of the load; (c) the power provided by each phase of the 'BC source.

19 A balanced three-phase voltage source, Van = 200/0° V rms, positive phase sequence, is connected to an unbalanced A load: ZAB= 60 + 10, Zgc = 30 + 130. ZCA 30-130 2. Find: (a) the three phasor line currents; (b) the power de- livered to each phase of the load; (c) the power provided by each phase of the 'BC source.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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19 A balanced three-phase voltage source, Van = 200/0° V rms, positive phase sequence, is connected to an unbalanced A load: ZAB= 60 + 10, Zgc = 30 + 130, ZCA 30 130 2. Find: (a) the three phasor line currents; (b) the power de- livered to each phase of the load; (c) the power provided by each phase of the 'BC source.
Transcribed Image Text:19 A balanced three-phase voltage source, Van = 200/0° V rms, positive phase sequence, is connected to an unbalanced A load: ZAB= 60 + 10, Zgc = 30 + 130, ZCA 30 130 2. Find: (a) the three phasor line currents; (b) the power de- livered to each phase of the load; (c) the power provided by each phase of the 'BC source.
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How do you get the value of Vab and Vac here? (YELLOW HIGHLIGHTED PART)

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Given Van = 200/0° Vrms Unbalanced delta cormected load ZBC = 30+330 ZAB = 60+JO (a) & df line cursents = ? ab = √3 Van / +30° Vab = √3X200 (0+30° 34641+30 V Vạc - 84641290 V Vca = 346-41 -210° Vab ZAB IBC = V₂c ZBC TAB = ) ICA = Vea ZCA = 34641 490 30+J30 346-41 L-210 30-J30 3 phasor line currents IA = IAB - ICA IB = IBC-IAB Ic = / Ic = ZCA = 30-53012 1-сл ICA fto C 346-414-30° => AB = 3.7735/430° A 60 > IBC = 8.1649 -135 A ⇒ICA 8-1649 -165 A LAB ZBE TA = 13-823/21-2° ATB = 13.923-141-2°A ZAB 5.7735/30-81649-165° = 13-823/21-2 9-1649-135-5-7735/30° = 13-823/-141-2 ICA-IBC = 8-1649-165-8-1649 C-135 = 4.226/120* IB Ic= 4.226 120 A
Transcribed Image Text:Given Van = 200/0° Vrms Unbalanced delta cormected load ZBC = 30+330 ZAB = 60+JO (a) & df line cursents = ? ab = √3 Van / +30° Vab = √3X200 (0+30° 34641+30 V Vạc - 84641290 V Vca = 346-41 -210° Vab ZAB IBC = V₂c ZBC TAB = ) ICA = Vea ZCA = 34641 490 30+J30 346-41 L-210 30-J30 3 phasor line currents IA = IAB - ICA IB = IBC-IAB Ic = / Ic = ZCA = 30-53012 1-сл ICA fto C 346-414-30° => AB = 3.7735/430° A 60 > IBC = 8.1649 -135 A ⇒ICA 8-1649 -165 A LAB ZBE TA = 13-823/21-2° ATB = 13.923-141-2°A ZAB 5.7735/30-81649-165° = 13-823/21-2 9-1649-135-5-7735/30° = 13-823/-141-2 ICA-IBC = 8-1649-165-8-1649 C-135 = 4.226/120* IB Ic= 4.226 120 A
19 A balanced three-phase voltage source, Van = 200/0° V rms, positive phase sequence, is connected to an unbalanced A load: ZAB= 60 + 10, Zgc = 30 + 130, ZCA 30 130 2. Find: (a) the three phasor line currents; (b) the power de- livered to each phase of the load; (c) the power provided by each phase of the 'BC source.
Transcribed Image Text:19 A balanced three-phase voltage source, Van = 200/0° V rms, positive phase sequence, is connected to an unbalanced A load: ZAB= 60 + 10, Zgc = 30 + 130, ZCA 30 130 2. Find: (a) the three phasor line currents; (b) the power de- livered to each phase of the load; (c) the power provided by each phase of the 'BC source.
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