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IRWIN 9e 5_60 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin. Basic Engineering Circuit Analysis. 9/E 1 5.15": US: Norton’s tllflfll'ficl'fl [u find P; in the umwurk in Fig. Pififl. 2000 II a m 4 m Figure P545»: SOLUTION: ‘é:<6k+2k>Ix L: —6 z—O'Vs’mn 8k V“ = 200012 +2: It Vet: 2000 C’ o, 75m)+ 2k(_o-’15m) \/o H I (N < ___________________._______.._.___————-— Chapter 5: Additional Analysis Techniques Problem 5.60 2 lrwin, Basic Engineering Circuit Analysis. 9/E 9.00011 1- 5kg Zmll 4“ 3k!)— 4L9. \lo ____—___—___—.__._—___——————————_——-——-——-———-— Problem 5.60 Chapter 5: Additional Analysis Techniques Inlvin, Basic Engineering Circuit Analysis, 9/E 3 Io: ( (2m) 8M k4k+ 4I< I) 0' 545mA \/6 = (o- 545:0(40 2‘ 18V — ' ________—______—.._—_———————————--— Chapter 5: Additional Analysis Techniques Problem 5.60 I ...
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