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Determine the equivalent impedance across 'a' and 'b'. j120 5.2 W 4.2 j9n be ww 4.0 j130 j112 j20Ω 8.2. MM ΖΩ j8n

Determine the equivalent impedance across 'a' and 'b'. j120 5.2 W 4.2 j9n be ww 4.0 j130 j112 j20Ω 8.2. MM ΖΩ j8n

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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Basic Electrical Engineering (UPVOTE WILL BE GIVEN. PLEASE WRITE THE COMPLETE SOLUTIONS LEGIBLY. NO LONG EXPLANATION NEEDED. ANSWER IN 4 DECIMAL PLACES. BOX THE FINAL ANSWERS.

Determine the equivalent impedance across 'a' and 'b'. j12Ω 5Ω (00) Μ 4Ω j9n 2 ww 4Ω j13Ω j112 @j20Ω 8Ω. WM ΖΩ j8Ω
Transcribed Image Text:Determine the equivalent impedance across 'a' and 'b'. j12Ω 5Ω (00) Μ 4Ω j9n 2 ww 4Ω j13Ω j112 @j20Ω 8Ω. WM ΖΩ j8Ω
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Please resolve the problem above. The same givens/values above but follow the way of solving in the attached photo. 

1. Determine the equivalent impedance across 'a' and 'b'. home m M 1.22 j6Ω M 2.522 j12 j10Ω 4Ω W W jon j152 3.2 Ans: Zab Zab = 17.4966/74.44° : 4.6932 + j16.8554 Ω 5.2 j7 b Gotham (+)+(5) 4+jlo 2+ vize 3.5 ed ut Aller зјол Zeb 1541 Zcd 5 ja Zbo LUM a bilo 45 زودتر Zabe Zb Zeg Z = imaginary па Zeg 2 j 12 & 3 jq py d Zd Zbd= =35.8799/82.2727 с bzc Zeq Zad Zeg = 6,3238+113.4691 cd= Zc Zbt Zc Zdt Z bZd Zb zb 2db = (1+; 4) + (216 + 2)² Zcd= 8.6+j27.8=29 07972/100912 Zab Zob Zeg Zcd= 8.6+j27.8=29.0998/12:001/72 db = ²c2b + 2c 2d + 2b zd Zcb = -1 +j46 = 46.0109 291.2454.52 |Zab=4.6932 +j16.8555 Zod=48243 +j35.5541 Zab=17.4967 274.44075 Zd Zc Zb+ ZcZd + Zb Zd Zc لموان دل -Tak A=Zc=2 tj 12 B=Zd=3+;9 c=²6=j15 R -|F +
Transcribed Image Text:1. Determine the equivalent impedance across 'a' and 'b'. home m M 1.22 j6Ω M 2.522 j12 j10Ω 4Ω W W jon j152 3.2 Ans: Zab Zab = 17.4966/74.44° : 4.6932 + j16.8554 Ω 5.2 j7 b Gotham (+)+(5) 4+jlo 2+ vize 3.5 ed ut Aller зјол Zeb 1541 Zcd 5 ja Zbo LUM a bilo 45 زودتر Zabe Zb Zeg Z = imaginary па Zeg 2 j 12 & 3 jq py d Zd Zbd= =35.8799/82.2727 с bzc Zeq Zad Zeg = 6,3238+113.4691 cd= Zc Zbt Zc Zdt Z bZd Zb zb 2db = (1+; 4) + (216 + 2)² Zcd= 8.6+j27.8=29 07972/100912 Zab Zob Zeg Zcd= 8.6+j27.8=29.0998/12:001/72 db = ²c2b + 2c 2d + 2b zd Zcb = -1 +j46 = 46.0109 291.2454.52 |Zab=4.6932 +j16.8555 Zod=48243 +j35.5541 Zab=17.4967 274.44075 Zd Zc Zb+ ZcZd + Zb Zd Zc لموان دل -Tak A=Zc=2 tj 12 B=Zd=3+;9 c=²6=j15 R -|F +
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