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1) The load on a 460 V, 60 Hz, Y-connected, balanced three-phase source consists of three equal Y-connected impedances of 100 + j100 № and three equal A-connected impedances of 300-j300 2, Draw the circuit diagram, and compute the line current, power and the power factor for the total load.

1) The load on a 460 V, 60 Hz, Y-connected, balanced three-phase source consists of three equal Y-connected impedances of 100 + j100 № and three equal A-connected impedances of 300-j300 2, Draw the circuit diagram, and compute the line current, power and the power factor for the total load.

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
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### Three-Phase Balanced Load Analysis #### Problem Description 1) The load on a 460 V, 60 Hz, Y-connected, balanced three-phase source consists of: - Three equal Y-connected impedances of \(100 + j100 \, \Omega\) - Three equal Δ-connected impedances of \(300 - j300 \, \Omega\) Draw the circuit diagram, and compute the line current, power, and the power factor for the total load. #### Solution Outline 1. **Circuit Diagram** - Draw the Y-connected and Δ-connected loads connected to the three-phase source. 2. **Impedance Calculations** - Calculate the total equivalent impedance for the Y-connected and Δ-connected loads. 3. **Line Current Calculation** - Apply Ohm’s Law to find the line current. 4. **Power Calculation** - Calculate the total real, reactive, and apparent power. 5. **Power Factor Calculation** - Compute the overall power factor for the combined load. #### Step-by-Step Solution 1. **Draw the Circuit Diagram** - The Y-connected load will have three impedances of \(100 + j100 \, \Omega\) connected to each phase. - The Δ-connected load will have three impedances of \(300 - j300 \, \Omega\) connected phase-to-phase. 2. **Calculate Equivalent Impedance** - Y-connected impedance: \[ Z_Y = 100 + j100 \, \Omega \] - For Δ-connected impedance \(Z_\Delta = 300 - j300 \, \Omega\), convert to an equivalent Y-connected impedance \(Z'_\Delta\): \[ Z'_\Delta = \frac{Z_\Delta}{3} = \frac{300 - j300}{3} = 100 - j100 \, \Omega \] 3. **Total Equivalent Y-Impedance** - The total equivalent impedance per phase \(Z_{eq}\) (sum of Y and equivalent Y of Δ): \[ Z_{eq} = Z_Y + Z'_\Delta = (100 + j100) + (100 - j100) = 200 \, \Omega \] 4. **Line Current** - Line-to-neutral voltage \(V_{LN}\): \[ V_{LN} = \frac{
Transcribed Image Text:### Three-Phase Balanced Load Analysis #### Problem Description 1) The load on a 460 V, 60 Hz, Y-connected, balanced three-phase source consists of: - Three equal Y-connected impedances of \(100 + j100 \, \Omega\) - Three equal Δ-connected impedances of \(300 - j300 \, \Omega\) Draw the circuit diagram, and compute the line current, power, and the power factor for the total load. #### Solution Outline 1. **Circuit Diagram** - Draw the Y-connected and Δ-connected loads connected to the three-phase source. 2. **Impedance Calculations** - Calculate the total equivalent impedance for the Y-connected and Δ-connected loads. 3. **Line Current Calculation** - Apply Ohm’s Law to find the line current. 4. **Power Calculation** - Calculate the total real, reactive, and apparent power. 5. **Power Factor Calculation** - Compute the overall power factor for the combined load. #### Step-by-Step Solution 1. **Draw the Circuit Diagram** - The Y-connected load will have three impedances of \(100 + j100 \, \Omega\) connected to each phase. - The Δ-connected load will have three impedances of \(300 - j300 \, \Omega\) connected phase-to-phase. 2. **Calculate Equivalent Impedance** - Y-connected impedance: \[ Z_Y = 100 + j100 \, \Omega \] - For Δ-connected impedance \(Z_\Delta = 300 - j300 \, \Omega\), convert to an equivalent Y-connected impedance \(Z'_\Delta\): \[ Z'_\Delta = \frac{Z_\Delta}{3} = \frac{300 - j300}{3} = 100 - j100 \, \Omega \] 3. **Total Equivalent Y-Impedance** - The total equivalent impedance per phase \(Z_{eq}\) (sum of Y and equivalent Y of Δ): \[ Z_{eq} = Z_Y + Z'_\Delta = (100 + j100) + (100 - j100) = 200 \, \Omega \] 4. **Line Current** - Line-to-neutral voltage \(V_{LN}\): \[ V_{LN} = \frac{
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