IRWIN 9e 8_110

IRWIN 9e 8_110 - Irwin, Basic Engineering Circuit Analysis,...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E ____________________._.—————-——-———— Sam Fiml ‘1."_1E in the circuit in Fig. PH] 1|}I 1h;ng Nm'mn‘r: theorem. Hagar + 2mm u Figure Pam: SOLUTION: I 1'34‘450V 2409A J‘H_Q_ IOJ}. ISc ‘n J‘QJL lO-Q Is’c. Egg - —2£O°A Chapter 8: AC Steady—State Analysis Problem 8.110 2 Irwin, Basic Engineering Circuit Analysis, 9/E Ts! : Irv; m5“ : 2-25 445% JH is: +31; :—ZLO°+2'X31'LIS° ii’: ZLVQOOA Problem 8.110 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 3 Vx : —J’3(5-l2£~3q'39°) Vx : 1541 L ll‘i-HOV _____________________________————————-——- ‘ Chapter 8: AC Steady-State Analysis Problem 8.110 ...
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This note was uploaded on 11/15/2011 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.

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