IRWIN 9e 8_39

IRWIN 9e 8_39 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 3.39 find 139“) and in“) in [hit network in Fig. P839. Figure P339 SOLUTION: 2L : j(20)(o*4) :J'SJz '2'" : l :: j), Z'SJL ‘* Jl'zoxo-oz) - _ r : —J‘ 512 2&2. ‘ J'Czo) (0'01) i ff. 3 1310" 2" 510°“ 5 i = [C5‘j5)llvfl‘5]"i8l/5 UV Chapter 8: AC Steadetate Analysis Problem 8.39 2 Irwin, Basic Engineering Circuit Analysis. 9/E 2?: [{5‘j52g-j2-5) IIJ'E HS 3—J’5—j2-5 — - , - . ° ' l5 2%. (1%: 7277 1m)! _ - . _. 5‘og°l|5 ‘7‘, : 108- L 51-921 66v is: 2540‘” : 12m; 5w°A 2-OgL-5I'H° KCL= j: 1 j: + T0 f0 2 T5 ' iI :70 =9-71A75'IOA qaflt) : OH). (0% (101+ 75.10) A _" _—__—_————-———_—-——-——'—'"_—“‘_’ __ 1, Problem 8.39 - Chapter 8: AC Steady-State Analysis ...
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