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TRANSCRIBE THE FOLLOWING SOLUTION IN DIGITAL FORMAT

TRANSCRIBE THE FOLLOWING SOLUTION IN DIGITAL FORMAT

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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TRANSCRIBE THE FOLLOWING SOLUTION IN DIGITAL FORMAT

Gven Vin Vm cos(w+-$) - 365 (10³¹4-30) : Vin - 34-3° (polar w=b²³ and = -30 Goouf 47 500μF >> JXcz -m → n 2mH S Guit becomes HH XC, ⇒ 4 Xc₁= wc - 103³*600x106 Xcz= Yax - 103x s00% ба CUL = 10³ * 2 *10³ -j1.667 MH (00.2 Nodal equahon @ Node VI Y-3430 100+ (-j1-667) V₁ form) Nodal equation @Node V2 V₂-VI + V₂ +5√2₂ = 0 -ja J२ +5 <= 1.666752 13V₂ V₂ [ 5<0] = V/12. V₂ = 0·11 v₁ V2 = 01190° *[**] --2 -ja. - Q5 5V₂ F -ja - 3V₂ + V-V2 - 5V₂ = O -J2 V₂ ja • V₁ = 10 V₂<-90°
Transcribed Image Text:Gven Vin Vm cos(w+-$) - 365 (10³¹4-30) : Vin - 34-3° (polar w=b²³ and = -30 Goouf 47 500μF >> JXcz -m → n 2mH S Guit becomes HH XC, ⇒ 4 Xc₁= wc - 103³*600x106 Xcz= Yax - 103x s00% ба CUL = 10³ * 2 *10³ -j1.667 MH (00.2 Nodal equahon @ Node VI Y-3430 100+ (-j1-667) V₁ form) Nodal equation @Node V2 V₂-VI + V₂ +5√2₂ = 0 -ja J२ +5 <= 1.666752 13V₂ V₂ [ 5<0] = V/12. V₂ = 0·11 v₁ V2 = 01190° *[**] --2 -ja. - Q5 5V₂ F -ja - 3V₂ + V-V2 - 5V₂ = O -J2 V₂ ja • V₁ = 10 V₂<-90°
Putting & in i 10V₂4-900-34-80 100-j1-667 V₂ [102-900 1₂ [100 -3 + 10V₂<-90-V2 -ja -3 + 10490 -ja 1100-j1-667 ---5) = -ja - SV₂ = 0 34-30 100-j1-66-7. 100-j1.667. 2[3.05777 <-168-68] > 3<3° *103 O 1/2 = 9.80 97018 <166. 639 V₂ = 9.8097 Cos (10³ +●+16639°) mV
Transcribed Image Text:Putting & in i 10V₂4-900-34-80 100-j1-667 V₂ [102-900 1₂ [100 -3 + 10V₂<-90-V2 -ja -3 + 10490 -ja 1100-j1-667 ---5) = -ja - SV₂ = 0 34-30 100-j1-66-7. 100-j1.667. 2[3.05777 <-168-68] > 3<3° *103 O 1/2 = 9.80 97018 <166. 639 V₂ = 9.8097 Cos (10³ +●+16639°) mV
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