Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
Example: The unbalanced Y-load of Fig. below has balanced voltages of 100 V and the acb sequence. Calculate the line currents and the neutral current. Take ZA-152, ZB=(10+j5) 22 and Z (6-18) Solution: Is {= 100/120° 10 +j5 100/-120° 6-18 100/0⁰ 15 6.67/0° A 100/120° 11.18/26.56° 100/-120° 10/-53.13⁰ 8.94/93.44° A 10/-66.87° A = -(L₂ + Is + 1₂) = -(6.67 -0.54 +j8.92 +3.93-19.2) = -10.06 +0.28 10.06/178.4° A VAN B ar Zz N 20

Example: The unbalanced Y-load of Fig. below has balanced voltages of 100 V and the acb sequence. Calculate the line currents and the neutral current. Take ZA-152, ZB=(10+j5) 22 and Z (6-18) Solution: Is {= 100/120° 10 +j5 100/-120° 6-18 100/0⁰ 15 6.67/0° A 100/120° 11.18/26.56° 100/-120° 10/-53.13⁰ 8.94/93.44° A 10/-66.87° A = -(L₂ + Is + 1₂) = -(6.67 -0.54 +j8.92 +3.93-19.2) = -10.06 +0.28 10.06/178.4° A VAN B ar Zz N 20

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...
See similar textbooks
icon
Related questions
icon
Concept explainers
Question
Hello, I need the solution in multiple ways with correct steps, please.
L Example: The unbalanced Y-load of Fig. below has balanced voltages of 100 V and the acb sequence. Calculate the line currents and the neutral current. Take ZA=150, ZB=(10+j5) 2 and Zc (6-j8) Solution: 100/120⁰ 10 + j5 100/-120° 6-18 100/0⁰ 15 6.67/0° A 100/120° 11.18/26.56° 100/-120° 10/-53.13⁰ 8.94/93.44° A 10/-66.87° A I₂ VES I₂ = −(I + Is + 1) = -(6.67 -0.54 +18.92 +3.93 - 19.2) = -10.06 + 10.28 = 10.06/178.4° A B Zz ZA N Le
Transcribed Image Text:L Example: The unbalanced Y-load of Fig. below has balanced voltages of 100 V and the acb sequence. Calculate the line currents and the neutral current. Take ZA=150, ZB=(10+j5) 2 and Zc (6-j8) Solution: 100/120⁰ 10 + j5 100/-120° 6-18 100/0⁰ 15 6.67/0° A 100/120° 11.18/26.56° 100/-120° 10/-53.13⁰ 8.94/93.44° A 10/-66.87° A I₂ VES I₂ = −(I + Is + 1) = -(6.67 -0.54 +18.92 +3.93 - 19.2) = -10.06 + 10.28 = 10.06/178.4° A B Zz ZA N Le
Expert Solution
steps

Step by step

Solved in 4 steps with 14 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Types of Distribution in Power System
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,
SEE MORE TEXTBOOKS