IRWIN 9e 9_7

IRWIN 9e 9_7 - Irwin Basic Engineering Circuit Analysis 9/E...

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Unformatted text preview: Irwin. Basic Engineering Circuit Analysis, 9/E 1 m 9.? It' {HI} = 0.5 cm; 2000: .-‘i.. 11ml the average rim-tr absorbed by each element in the circuit in Fig. FLT-'3'. 401'! 12Ufl Figure Fig.3:r SOLUTION: ’6} (t) = 0'5 2000f A if : 0-5 19 A L = 3”,“ : 3(2000)(6Om):3120.0, ' __—i.._. : i _-_- _.. It = 14‘” 312017, 1; f = “#533”... o ‘ 40+ fl;o+/20.j’+o) (0'5 L9) ‘ j = 0' 353 LII-50A j; : < 1410— £40 e , O! 0 f; = 0'853 1:45;; Chapter 9: Steady — State Power Analysis Problem 9.7 2 Irwin, Basic Engineering Circuit Analysis, 9/E .— V3 = (0' 3“ US°><'&0-i+°) \73= 4+. 65 136.513 P13 = V313 MG 2- "2.. {32% 1 48 w PQDJL : ; (0 353)2(4-O) '9- 2. FL 3 ON Va : ON _—————____________—__—..____—___J Problem 9.7 Chapter 9: Steady — State Power Analysis ...
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IRWIN 9e 9_7 - Irwin Basic Engineering Circuit Analysis 9/E...

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