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node v₁: 3 (V₁-V₂) SA-3.1052 SA SA- 3(V₁-V3) 3052 SA- - 3V₁-3V3 V₁ V₂ 305 4V. +V₂+3V₂. 3052 -4V₁ + V₂ + 3V₂ = -150A (V₁-V₂) V₁-V₂ node v₂: -(V-V.). (v.₂-K). V+₂=0 102 40-52 3022 node V3: VI-V + VIOVI + VI SO V3-V₂ 105 403 -12 (V₂-v₁-3(V₂ -V₂ ) -4V₂ = 0 -1913 +3V₂ +12yY₁ = 0 2- 80. V₂-U 40 40 -12V3+12V₁-3V3+3V₂-4V3=0 =O =O (V₂-V₂) (V₂ -v₁) 40 - 80-V₂-V3- 40 U₂U INI VV 30 =O 240-3(V₂-V3-4(V₂-V₁) = 0 -3V₂ +3V3-4v₂ +4v₁ = -240 3V3-7V₂+4v₁ = -240

node v₁: 3 (V₁-V₂) SA-3.1052 SA SA- 3(V₁-V3) 3052 SA- - 3V₁-3V3 V₁ V₂ 305 4V. +V₂+3V₂. 3052 -4V₁ + V₂ + 3V₂ = -150A (V₁-V₂) V₁-V₂ node v₂: -(V-V.). (v.₂-K). V+₂=0 102 40-52 3022 node V3: VI-V + VIOVI + VI SO V3-V₂ 105 403 -12 (V₂-v₁-3(V₂ -V₂ ) -4V₂ = 0 -1913 +3V₂ +12yY₁ = 0 2- 80. V₂-U 40 40 -12V3+12V₁-3V3+3V₂-4V3=0 =O =O (V₂-V₂) (V₂ -v₁) 40 - 80-V₂-V3- 40 U₂U INI VV 30 =O 240-3(V₂-V3-4(V₂-V₁) = 0 -3V₂ +3V3-4v₂ +4v₁ = -240 3V3-7V₂+4v₁ = -240

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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node ₁: 3 (V₁-V₂) SA-3.1052 SA- ● 3(V₁-V3) 3052 34-3V3 305 node v₂: 103 V3-V₁ SA- 5A - = 4V. +V₂+3V₂ 3052 1052 2- -4v₁ + V₂+3V₂ =-150A (V₁-V₂) 40 3052 0 node V3: V₁-V₂ V₂ = 0 305 (V₂-V₂> 140 V₁-V₂ V₂ 40 30V₂-V3 (U₂ - V₂) - 12 = 0 40-52 3022 -12 (V3-v₁-3(V₂-V₂ ) -4V₂=0 30 O VI-VI + V1=0 V3-V₂ 405 30. -12V₂+12V,-3V3+3V₂-4V3=0 -19v3 +3V₂ +12V₁ = 0 4-U₂) - (U₂-V1) 30 V₂-V₁ 30 80-V-V3 - V2-V 40 30 =O 240-3(V₂-V₂-4 (V₂-V₁) = 0 -3v₂ +3V3-4v₂ +4v₁ = -240 3V3-7 V₂ +4v₁ = -240
Transcribed Image Text:node ₁: 3 (V₁-V₂) SA-3.1052 SA- ● 3(V₁-V3) 3052 34-3V3 305 node v₂: 103 V3-V₁ SA- 5A - = 4V. +V₂+3V₂ 3052 1052 2- -4v₁ + V₂+3V₂ =-150A (V₁-V₂) 40 3052 0 node V3: V₁-V₂ V₂ = 0 305 (V₂-V₂> 140 V₁-V₂ V₂ 40 30V₂-V3 (U₂ - V₂) - 12 = 0 40-52 3022 -12 (V3-v₁-3(V₂-V₂ ) -4V₂=0 30 O VI-VI + V1=0 V3-V₂ 405 30. -12V₂+12V,-3V3+3V₂-4V3=0 -19v3 +3V₂ +12V₁ = 0 4-U₂) - (U₂-V1) 30 V₂-V₁ 30 80-V-V3 - V2-V 40 30 =O 240-3(V₂-V₂-4 (V₂-V₁) = 0 -3v₂ +3V3-4v₂ +4v₁ = -240 3V3-7 V₂ +4v₁ = -240
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