Homework Help is Here – Start Your Trial Now!
Engineering
Electrical Engineering
I would like to know the steps to solve the problem to get Equation 10.1.2 in detail, thank you in advance for the answer.

I would like to know the steps to solve the problem to get Equation 10.1.2 in detail, thank you in advance for the answer.

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
Question
I would like to know the steps to solve the problem to get Equation 10.1.2 in detail, thank you in advance for the answer.
Thus, the frequency-domain equivalent circuit is as shown in Fig. 10.2. or 20/0° V Figure 10.2 At node 2, 10 92 www Applying KCL at node 1, 20 - V₁ 10 By simplifying, we get A₁ = 20 0 Frequency-domain equivalent of the circuit in Fig. 10.1. But Ix = V₁/-j2.5. Substituting 2V₁ -j2.5 V₁ We obtain the determinants as Δ= V₁ = j2.5 15 V₂ = 21x + (1 + j1.5)V₁ + j2.5V₂ = 20 -j2.5 Q Δι A 1+j1.5 j2.5 11 = 300, 42 A V₁ -j2.5 11V₁ +15V₂ = 0 Equations (10.1.1) and (10.1.2) can be put in matrix form as 25][]-[20] + j4 92 m V₁ - V₂ V₂ j4 j2 this gives V₁ - V₂ j4 15 1+j1.5 11 + j2.5 15 300 15 - j5 21₁ -220 15 - j5 Δ2 = = V₂2₂ V₁ - V₂ j4 V₂ j2 = 15 - j5 1+j1.5 20 11 0 18.97/18.43° V = ¹ + j2 92 13.91/198.3° V (10.1.1) (10.1.2) = -220
Transcribed Image Text:Thus, the frequency-domain equivalent circuit is as shown in Fig. 10.2. or 20/0° V Figure 10.2 At node 2, 10 92 www Applying KCL at node 1, 20 - V₁ 10 By simplifying, we get A₁ = 20 0 Frequency-domain equivalent of the circuit in Fig. 10.1. But Ix = V₁/-j2.5. Substituting 2V₁ -j2.5 V₁ We obtain the determinants as Δ= V₁ = j2.5 15 V₂ = 21x + (1 + j1.5)V₁ + j2.5V₂ = 20 -j2.5 Q Δι A 1+j1.5 j2.5 11 = 300, 42 A V₁ -j2.5 11V₁ +15V₂ = 0 Equations (10.1.1) and (10.1.2) can be put in matrix form as 25][]-[20] + j4 92 m V₁ - V₂ V₂ j4 j2 this gives V₁ - V₂ j4 15 1+j1.5 11 + j2.5 15 300 15 - j5 21₁ -220 15 - j5 Δ2 = = V₂2₂ V₁ - V₂ j4 V₂ j2 = 15 - j5 1+j1.5 20 11 0 18.97/18.43° V = ¹ + j2 92 13.91/198.3° V (10.1.1) (10.1.2) = -220
Expert Solution
steps

Step by step

Solved in 4 steps with 5 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Analysis of Multiple Order System and Steady State Error
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