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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
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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
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