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IRWIN 9e 8_123

IRWIN 9e 8_123 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 3.123 Us: bath a nodal analysis and a loop mmlysis, Each in conjunction with ME, to find la in the network in Fig. P3123. Figure P3423 SOLUTION: (7,471 + \7,—Y/L, +2[\7§l] = 140° I -JI J —jl(\7r‘_\72) + 91*UV‘JZUH:‘JI(240°) (1-3;)9’, +jl VfiC-PJUWI MAO“ KCL a1 @'— THJWLLZ ?5 WW +1 [21]: *5 I I j! Chapter 8: AC Steady-State Analysis Problem 8.123 2 Irwin. Basic Engineering Circuit Analysis, 9/E ——-—. —— Ji( vwvs) +j1vg : v5 jg \79 + (-i-Ji) {75 :0 KCLCut Sui/banodl5 1+:Tl:uo°+i4+i;+f3 -‘ cc.— 3 l firm + g—T/fl : ZLOO-i Var—05 +Yi‘i I ‘JI l I 1‘: (vi—v3) + \7. -— $14 = 210m ))vjl(\7q-\75) aim—j R73 -—v3 + TIC, =i1Lo° (i—jiiv, + ji‘vl’r('i-j2)fi,: u-qo‘ j3 Vq+ (—i—J’OUS 3:0 V2433 2:; O Problem 8.123 Chapter 8: AC Steady—State Analysis lrwin, Basic Engineering Circuit Analysis, 9/E 3 ”V3 4‘qu : llLOD ._ __ -- “1:7 +(—I+Jl)\7 — 'Iv - v, Jle‘lJ 3 w J 5-214700 V. :‘ ‘lvll'élHOflQV V2: 353?? L—l37«3°\x '3 :— nfil 1-4313“! '1,» : :2m4—qH-yw ‘5 = 155% 4—99's’°v Io : UL'\-73 l : 35-52 4—137? ~ I7I7L-lz7-3° I 10: 17.74'127-3°A Chapter 8: AC Steady-State Analysis Problem 8.123 Irwin, Basic Engineering Circuit Analysis, 9/E 5 :13 +i’/= T6 "1*" 1% ::i+2 i :rfii. ”Z I’+T3 8 i1: “1»: Problem 8.123 Chapter 8: AC Steady-State Analysis E Irwin, Basic Engineering Circuit Analysis, 9/E KVL: l(—1)+l(ix) :IZLo° {11—765 PF iris =7 i2Lo‘ -11 + Z :5 —i€' N: P2409] ,KVLT ICI") *J'C’I‘eHIC-3PO 7:6 - f3 +Ji I} ~( £5.31)“; #13 (15 Hzm‘i) 1:0 KCL ; f, + lid: .1 1:1 I, “—IL : —'2—LOD Chapter 8: AC Steady-State Analysis Problem 8.123 6 , Irwin, Basic Engineering Circuit Analysis, 9/E KVL‘ '(i) “J" TI-i I240°+ ;(-‘fo):o f-J‘Kfrfywzzo ~1+izd TH : 52'5'6 L'IZC?»6°A f3 :— 3541 Z'I13.75°A f6 2» 2. 55% 1429-55 “A Problem 8.123 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis. 9/E 7 j:0: ii”: f0: 5266 A—izq'6°v35'iiévi15i73° for. l7'7 L-‘i37'30 A MRTLHG SOWBN \/: [lei i 0 +2; o;oo o3,-~;_.';o i—2 (50;00'i I O; H A ¢i+li —|j 1: [an o; onzrzifl \/: iva’VKI lo: V(1) ' W3) /i 0/0 Emma: Y: Wm?» 'Hwoqgh 3 iroooo -/'Ooooi +i-0000i ~ 0 o O O O “0000 ”30000 O O ~i'oooo [-6000 O "i'OOOOi 0+ 2-0000] CeQumm HfiimqgfiS 'i'OOOO ~20000i O o +30000i "#0000— i'OODOi' O O l' 0000 O . ‘I-OoooJr'z-Ooooi O~b00cm Chapter 8: AC Steady-State Analysis Problem 8.123 8 Inn/in. Basic Engineering Circuit Analysis, 9/E I: O -2'00004 O O- IZ'OOOO O ’l'oooOi V: ‘*5*5000 +—H'SODOI ~2g-0000 " ZH'Ooooi -43-oooo " ’2'00006 —I'0000 ”12.00003 195000 —» ICI'Scaooi~ Io: , vBOmownome ‘__—_—___—_—_—._—_.———————————————"—_ E Problem 8.123 Chapter 8: AC Steady-State Analysis ...
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