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IRWIN 9e 5_64 - Inivin Basic Engineering Circuit Analysis...

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Unformatted text preview: Inivin, Basic Engineering Circuit Analysis. 9/E 1 5J5q USE Thévenjn’s tllem'em to find the pawer supplied by the 2—1? source in the circuit in Fig. P5.6«5l-. Figure P564 SOLUTION: KVL [Left Loop: 4—= ik1,+2\ll+il<1. \l = 1'41. imll ‘ I = I); 331%”; = o-ee4mA 3 Vac= in” 1L1, V0“ Lew Chapter 5: Additional Analysis Techniques Problem 5.64 ‘ Irwin. Basic Engineering Circuit Analysis, 9/E 1m. I: I" 1.96. I — I'+ 15c, I’= 1—1;G KVL AfijLE Loo]: 4:111, 1.2%; .TiltI V1: ikI 4U - 31: 13L- + KVL :efghtloopt 'ikl'l’iklsc‘ikls 0 HQ, ~ ikI +3LISc =0 KVL mm 4,051): 2v ~2kI .rikI 2kIl—3k12 dusk—,0 KVL gulf» L091): 4—: ikI'¢2l<I;+il<Igc It: 0'3qu I56." On Bryan RT"; 16? , 2.0811(1). O-Bnn Problem 5.64 Chapter 5: Additional Analysis Techniques Irwin, Basic Engineering Circuit Analysis, 9/E 3 L: 246? zoaamn 2.064k Ev=%; =ammg = 0‘ 32mN ______—_______———————-—-——-—-——_— Chapter 5: Additional Analysis Techniques Problem 5.64 " ...
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