Given100 10°SAV₁I9V₂USE10 23035345Applying Kez @ node 1 (V₁):V₁ - 1000V₁V₁-V2 = 0++53+j4V₁ [ = + 3+ 4+ = √5] + V² [ ]V₂=100 40°5Equation Ⓒ5Applying kel@ hode 2 (V₂):V+=10 2305-jsVz=•[=] + v₂ [ + + + + ]· 10/230° → liquation @25SOLVE FORVI and V₂ :Answer should be :V₁.53. 914 L25.5°-V₂ = 53. 988 429.7"7

The given node equation can be solved by using the cramer's rule. It depends on the determinant of…

Answered: VI and V₂ : V₁ = 53. 914 L25.5° V₂ =…

Engineering

Electrical Engineering

VI and V₂ : V₁ = 53. 914 L25.5° V₂ = 53. 988 4²9.7" SOLVE FOR Answer should be :

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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Concept explainers

Nodal Analysis

In the analysis of any electric circuit, nodal analysis is the method of determining the potential difference or voltage between different nodes in an electric circuit. A node is a point where two or more branches connect each other. In the nodal analysis, the voltages are determined in the terms of branch currents. The nodal analysis is also known as the node voltage method or branch current method. The nodal analysis method is one of the several applications of Kirchhoff's Current Law (KCL).

Question

Given
100 10°
SA
V₁
I
9
V₂
USE
10 230
35
345
Applying Kez @ node 1 (V₁):
V₁ - 1000
V₁
V₁-V2 = 0
+
+
5
3+j4
V₁ [ = + 3+ 4+ = √5] + V² [ ]
V₂
=
100 40°
5
Equation Ⓒ
5
Applying kel@ hode 2 (V₂):
V
+
=
10 230
5
-js
Vz
=
•[=] + v₂ [ + + + + ]· 10/230° → liquation @2
5
SOLVE FOR
VI and V₂ :
Answer should be :
V₁.
53. 914 L25.5°
-
V₂ = 53. 988 429.7"
7

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