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Q5 (Large Single Phase Resistive load connected to 3 phase supply) Single Phase Resistive Load eg Electric Arc Furnace Additional C and L to give balanced LINE currents VRN Vey V Xe = 3.R XL = - T3.R In this example: R=0.250 Therefore: Zc=-j0.43 Zz=j0.43 Calculate the load phase currents: 4.4 cm 44030° VRY IRY = ZRY = 1760Z30° 0.2520° A 2.56cm V YB IYB = ZYB 440Z - 90° = 102320" 0.43Z-90° 2 om 440Z150° VBR IBR = ZBR = 1023260° 0.43290° V BR VBN -lyB IRY VRY IBR [36 120 + VRN Talvo 360 Scale: 1cm=400A -IRY VRB VYN VYB

Q5 (Large Single Phase Resistive load connected to 3 phase supply) Single Phase Resistive Load eg Electric Arc Furnace Additional C and L to give balanced LINE currents VRN Vey V Xe = 3.R XL = - T3.R In this example: R=0.250 Therefore: Zc=-j0.43 Zz=j0.43 Calculate the load phase currents: 4.4 cm 44030° VRY IRY = ZRY = 1760Z30° 0.2520° A 2.56cm V YB IYB = ZYB 440Z - 90° = 102320" 0.43Z-90° 2 om 440Z150° VBR IBR = ZBR = 1023260° 0.43290° V BR VBN -lyB IRY VRY IBR [36 120 + VRN Talvo 360 Scale: 1cm=400A -IRY VRB VYN VYB

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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How did they get Zc=j0.43 and Zl=j0.43? How did they get the diagram break down simply?
Q5 (Large Single Phase Resistive load connected to 3 phase supply) Single Phase Resistive Load eg Electric Arc Furnace Additional C and L to give balanced LINE currents R VRN Vey Xe = 3.R C. XL = -3.R In this example: R=0.250 Therefore: Zc=-j0.43 Zz=j0.43 Calculate the load phase currents: 4.4 cm 440430° VRY IRY = ZRY = 1760Z30° 0.2520° A 2.56cm V YB IYR = ZYB 440Z - 90° = 102320" 0.43Z - 90° 2 cm 440Z150° VBR IBR = ZBR = 1023Z60° 0.43290° VBR VRY VBN -lyB IBR IRY [36 120 360 + VRN Scale: -IRY 1cm=400A VRB VYN VYB
Transcribed Image Text:Q5 (Large Single Phase Resistive load connected to 3 phase supply) Single Phase Resistive Load eg Electric Arc Furnace Additional C and L to give balanced LINE currents R VRN Vey Xe = 3.R C. XL = -3.R In this example: R=0.250 Therefore: Zc=-j0.43 Zz=j0.43 Calculate the load phase currents: 4.4 cm 440430° VRY IRY = ZRY = 1760Z30° 0.2520° A 2.56cm V YB IYR = ZYB 440Z - 90° = 102320" 0.43Z - 90° 2 cm 440Z150° VBR IBR = ZBR = 1023Z60° 0.43290° VBR VRY VBN -lyB IBR IRY [36 120 360 + VRN Scale: -IRY 1cm=400A VRB VYN VYB
A large single phase load (represented by a resistance of 0.25Q) is to be connected to a three phase 440V/50HZ system. Determine he necessary (delta connected) Capacitance and Inductance reactances to give balanced line currents. Calculate the resultant load phase currents and from a scaled phasor diagram plot and estimate the three line currents, and the total power (W) taken from the supply. Q5
Transcribed Image Text:A large single phase load (represented by a resistance of 0.25Q) is to be connected to a three phase 440V/50HZ system. Determine he necessary (delta connected) Capacitance and Inductance reactances to give balanced line currents. Calculate the resultant load phase currents and from a scaled phasor diagram plot and estimate the three line currents, and the total power (W) taken from the supply. Q5
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