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1-43. At node 1 in the circuit, applying KCL gives: 2Α 8 Ω 3 Ω Τ 12 V 6Ω 6Ω 4Ω LM 12-V₁ V1 = + (A) 2 + - 12-01 V1-V2 4 3 6 V2-V1 (Β) +2= ” + 3 4 12-V₁ 0-V1 V1-V2 (C) 2 + = + 3 6 4 12-01 0-V1 V2-V1 (D) +2 = 3 6 4 +

1-43. At node 1 in the circuit, applying KCL gives: 2Α 8 Ω 3 Ω Τ 12 V 6Ω 6Ω 4Ω LM 12-V₁ V1 = + (A) 2 + - 12-01 V1-V2 4 3 6 V2-V1 (Β) +2= ” + 3 4 12-V₁ 0-V1 V1-V2 (C) 2 + = + 3 6 4 12-01 0-V1 V2-V1 (D) +2 = 3 6 4 +

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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41-43. At node 1 in the circuit, applying KCL gives: 2Α 8 Ω 3Ω 12 V 6Ω Μ 4Ω 2 6Ω 12-01 3 +2 12-01 = 3 + 2 = (A) 2 + 12-01 (Β) 3 (C) 2 + (D) 12-01 3 = = + 6 V1 + 6 0-V1 6 0-01 6 01-02 4 V2-V1 4 + + V1-V2 4 V2-V1 4
Transcribed Image Text:41-43. At node 1 in the circuit, applying KCL gives: 2Α 8 Ω 3Ω 12 V 6Ω Μ 4Ω 2 6Ω 12-01 3 +2 12-01 = 3 + 2 = (A) 2 + 12-01 (Β) 3 (C) 2 + (D) 12-01 3 = = + 6 V1 + 6 0-V1 6 0-01 6 01-02 4 V2-V1 4 + + V1-V2 4 V2-V1 4
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