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3) As shown in Fig. 1, a balanced three-phase, positive-sequence source with VAB = 480/0° V is applied to an unbalanced A load. Note that one leg of the A is open. Determine: a) the load currents IAB and IBC; b) the line currents IA, IB and IC which feed the A load; c) the zero-, positive, and negative-sequence components of the line currents Le LA V=480/0° V

3) As shown in Fig. 1, a balanced three-phase, positive-sequence source with VAB = 480/0° V is applied to an unbalanced A load. Note that one leg of the A is open. Determine: a) the load currents IAB and IBC; b) the line currents IA, IB and IC which feed the A load; c) the zero-, positive, and negative-sequence components of the line currents Le LA V=480/0° V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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**Problem Statement:** As shown in Fig. 1, a balanced three-phase, positive-sequence source with \( V_{AB} = 480 \angle 0^\circ \) V is applied to an unbalanced \( \Delta \) load. Note that one leg of the \( \Delta \) is open. Determine: a) the load currents \( I_{AB} \) and \( I_{BC} \); b) the line currents \( I_A \), \( I_B \), and \( I_C \) which feed the \( \Delta \) load; c) the zero-, positive-, and negative-sequence components of the line currents. **Diagram:** - The diagram shows a three-phase system with voltages and line currents labeled. - The open circuit in one leg of the delta load is indicated, influencing the current paths. - The system voltage \( V_{AB} \) is provided as 480 V at \( 0^\circ \). - Impedances are given as \( (j18 + j10) \Omega \). **Figure 1: Problem 3** **System Parameters:** *Generators:* - \( G_1 \): 500 MVA, 13.8 kV, \( x_d'' = x_2 = 0.20 \), \( x_0 = 0.10 \) pu - \( G_2 \): 750 MVA, 18 kV, \( x_d'' = x_2 = 0.18 \), \( x_0 = 0.09 \) pu - \( G_3 \): 1000 MVA, 20 kV, \( x_d'' = x_2 = 0.17 \), \( x_2 = 0.20 \), \( x_0 = 0.09 \) pu *Transformers:* - \( T_1 \): 500 MVA, 13.8/500 kV, \( \Delta \) – Y, \( x = 0.12 \) pu - \( T_2 \): 750 MVA, 18/500 kV, \( \Delta \) – Y, \( x = 0.10 \) pu - \( T_3 \): 1000 MVA, 20/500 kV, \( \Delta \) – Y, \(
Transcribed Image Text:**Problem Statement:** As shown in Fig. 1, a balanced three-phase, positive-sequence source with \( V_{AB} = 480 \angle 0^\circ \) V is applied to an unbalanced \( \Delta \) load. Note that one leg of the \( \Delta \) is open. Determine: a) the load currents \( I_{AB} \) and \( I_{BC} \); b) the line currents \( I_A \), \( I_B \), and \( I_C \) which feed the \( \Delta \) load; c) the zero-, positive-, and negative-sequence components of the line currents. **Diagram:** - The diagram shows a three-phase system with voltages and line currents labeled. - The open circuit in one leg of the delta load is indicated, influencing the current paths. - The system voltage \( V_{AB} \) is provided as 480 V at \( 0^\circ \). - Impedances are given as \( (j18 + j10) \Omega \). **Figure 1: Problem 3** **System Parameters:** *Generators:* - \( G_1 \): 500 MVA, 13.8 kV, \( x_d'' = x_2 = 0.20 \), \( x_0 = 0.10 \) pu - \( G_2 \): 750 MVA, 18 kV, \( x_d'' = x_2 = 0.18 \), \( x_0 = 0.09 \) pu - \( G_3 \): 1000 MVA, 20 kV, \( x_d'' = x_2 = 0.17 \), \( x_2 = 0.20 \), \( x_0 = 0.09 \) pu *Transformers:* - \( T_1 \): 500 MVA, 13.8/500 kV, \( \Delta \) – Y, \( x = 0.12 \) pu - \( T_2 \): 750 MVA, 18/500 kV, \( \Delta \) – Y, \( x = 0.10 \) pu - \( T_3 \): 1000 MVA, 20/500 kV, \( \Delta \) – Y, \(
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