IRWIN 9e 8_108

# IRWIN 9e 8_108 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin. Basic Engineering Circuit Analysis, 9/E 3.1!:3 Find the Thévcniu’s equivalent fur the [remark in Fig. PEJDE at termilmls A—B. Figure Pad-:3 SOLUTION: KCL 0-1'@: T3+I,:: IVA Chapter 8: AC Steady-State Ana|ysis Problem 8.108 2 . Irwin, Basic Engineering Circuit Analysis, 9/E ﬂ. _— VX : VOL " V2— voc ”V?— + VII—<7). : 1EVOC—V1] 'Jl l vac ' V; - JI( r451) : —jz[VoC-\7J ”J iv; +(""J')VL l” (PUD—(70¢, :0 Kat, a1®t :3": qLo° \70c ’V1 3 9100 vi ~47): + \10c 5 WA—QO" (ml)? *J'Vz =2 LIA—«90° vim, +(~I'Jl)\71 +(:+jz)\75¢=: 0 ~47}, ”Wm. “’4‘90‘: I V. : XLWOOV "\7l :2 WM Alleog ° v VOL: 5'66 Al35°V Problem 8.108 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E A 4 0°Ao :1- Sc [19 (-1“), o 8 --‘f, +(Hji) f1 "'JITScS'HA'qO" nu... KCL‘ 2R7} +116: I1 \7} : "J! (Hzoi‘fgc) “ *J2( H400 —T\$c) + it. : I). Chapter 8: AC Steady-State Analysis Problem 8.108 4 |rwin, Basic Engineering Circuit Analysis, 9/E —: .+< rm) EC :ma" ‘1', : L440°A (mi) E «j: T};- 5661445" —Il+( ma) 1 : New T1: 1'53 A7Ir6° A R : 2‘53 LIQ'H3OA ‘2}H = 3c : 5~ee LL25" ISC 2‘53 L'8'L'36 7TH : 2.31 4H6 57°—Q. _______________________—.——————— Problem 8.108 Chapter 8: AC Steady-State Analysis ...
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