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A balanced 3-phase Y-connected source 866V, 60Hz, feeds a A-connected loads (Zph=177- j2460) via a line. Each phase of the line is 1+j20. Phase-sequence is positive. Determine: 1) line and phase current; 2) power absorbed by the load; 3) power dissipated by the line. 50020V 500/120 V 500-120° 192 ww 1Q 152 j20 voo -553.13" A j2 Q2 vor -52-66.87" A B j20 000 -5 ZIZA: A TAB 177 02 --j246 2 -246 02 177 02 www 177 Ω -f246 Q HE -TAC VBC VCA *

A balanced 3-phase Y-connected source 866V, 60Hz, feeds a A-connected loads (Zph=177- j2460) via a line. Each phase of the line is 1+j20. Phase-sequence is positive. Determine: 1) line and phase current; 2) power absorbed by the load; 3) power dissipated by the line. 50020V 500/120 V 500-120° 192 ww 1Q 152 j20 voo -553.13" A j2 Q2 vor -52-66.87" A B j20 000 -5 ZIZA: A TAB 177 02 --j246 2 -246 02 177 02 www 177 Ω -f246 Q HE -TAC VBC VCA *

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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### Three-Phase Power Problem Analysis **Problem Statement:** A balanced 3-phase Y-connected source with a voltage of 866V and frequency of 60Hz feeds a Δ-connected load (Zph = 177 - j246Ω) via a line. Each phase of the line has an impedance of 1 + j2Ω. The phase sequence is positive. We need to determine the following: 1. Line and phase current 2. Power absorbed by the load 3. Power dissipated by the line **Circuit Diagram Explanation:** The provided circuit diagram illustrates the connections and components of the three-phase system: 1. **Source:** - The source has three lines labeled \( a, b, \) and \( c \), each providing a voltage of 500∠0° V, 500∠120° V, and 500∠-120° V respectively. 2. **Line Impedances:** - Each phase of the line has an impedance of \(1 + j2Ω\). 3. **Load:** - The load is Δ-connected with a phase impedance of \(177 - j246Ω\). The voltages across the load are denoted as \( \vec{V}_{AB} \), \( \vec{V}_{BC} \), and \( \vec{V}_{CA} \). - The current through each phase of the load is denoted as \( I_{AB} \), \( I_{BC} \), and \( I_{CA} \). 4. **Points of Interest:** - Voltages and currents are annotated with both their magnitudes and phase angles at various points in the circuit. - The notation \( \vec{I}_{na}, \vec{I}_{nb}, \) and \( \vec{I}_{nc} \) represents the line currents at points \( a, b, \) and \( c \). ### Solution Steps: 1. **Line and Phase Current:** - Calculate the line currents using Kirchhoff's laws, Ohm's Law, and the impedance of the line and load. 2. **Power Absorbed by the Load:** - Use the formula \( P = \sqrt{3} \times V_{LL} \times I_{L} \times \cos{φ} \) to calculate the power absorbed by the load, where \( φ \) is the phase
Transcribed Image Text:### Three-Phase Power Problem Analysis **Problem Statement:** A balanced 3-phase Y-connected source with a voltage of 866V and frequency of 60Hz feeds a Δ-connected load (Zph = 177 - j246Ω) via a line. Each phase of the line has an impedance of 1 + j2Ω. The phase sequence is positive. We need to determine the following: 1. Line and phase current 2. Power absorbed by the load 3. Power dissipated by the line **Circuit Diagram Explanation:** The provided circuit diagram illustrates the connections and components of the three-phase system: 1. **Source:** - The source has three lines labeled \( a, b, \) and \( c \), each providing a voltage of 500∠0° V, 500∠120° V, and 500∠-120° V respectively. 2. **Line Impedances:** - Each phase of the line has an impedance of \(1 + j2Ω\). 3. **Load:** - The load is Δ-connected with a phase impedance of \(177 - j246Ω\). The voltages across the load are denoted as \( \vec{V}_{AB} \), \( \vec{V}_{BC} \), and \( \vec{V}_{CA} \). - The current through each phase of the load is denoted as \( I_{AB} \), \( I_{BC} \), and \( I_{CA} \). 4. **Points of Interest:** - Voltages and currents are annotated with both their magnitudes and phase angles at various points in the circuit. - The notation \( \vec{I}_{na}, \vec{I}_{nb}, \) and \( \vec{I}_{nc} \) represents the line currents at points \( a, b, \) and \( c \). ### Solution Steps: 1. **Line and Phase Current:** - Calculate the line currents using Kirchhoff's laws, Ohm's Law, and the impedance of the line and load. 2. **Power Absorbed by the Load:** - Use the formula \( P = \sqrt{3} \times V_{LL} \times I_{L} \times \cos{φ} \) to calculate the power absorbed by the load, where \( φ \) is the phase
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