A 3- phase, 4 – wire, 380 V supply is connected to an unbalanced load having phase impedances of ZR = (8 + j6) Q, ZY = (8 - j6) Q and ZB = 5 Q. Impedance ofthe neutral wire is ZN = (0.5 +j1) Q. Ignoring the impedances of line wires and intemal %3D impedances of the emfsources, find the phase currents and voltages of the load.

Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
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DEGREE: ELECTRICAL ENGINEERING 

SUBJECT/COURSE: AC CIRCUITS

 

NOTE: Please solve in this way. 

 

1. Please have a good handwriting, some of the answers are not readable. Thank you!

2. GIVEN.(include symbols and units)

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5. STEP-by-STEP SOLUTION with Formulas and Symbols. No Shortcut, No skipping, and detailed as possible

6. FINAL ANSWERS must be rounded up to three decimal places.

A 3– phase, 4 – wire, 380 V supply is connected to an unbalanced load having phase impedances of ZR = (8 + j6) Q, Zy
= (8 – j6) Q and ZB = 5 Q. Impedance of the neutral wire is ZN = (0.5 +j1) Q. Ignoring the impedances of line wires and intemal
%3D
impedances of the emfsources, find the phase currents and voltages of the load.
Transcribed Image Text:A 3– phase, 4 – wire, 380 V supply is connected to an unbalanced load having phase impedances of ZR = (8 + j6) Q, Zy = (8 – j6) Q and ZB = 5 Q. Impedance of the neutral wire is ZN = (0.5 +j1) Q. Ignoring the impedances of line wires and intemal %3D impedances of the emfsources, find the phase currents and voltages of the load.
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