IRWIN 9e 8_120

# IRWIN 9e 8_120 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E -, 1 A 3.13) Determine To in the new-wk in Fig. PS. [EIII using h-iATLx-‘iB. 1253300 e Figure F8421: SOLUTION: Problem 8.120 Chapter 8: AC Steady—State Analysis 2 Irwin, Basic Engineering Circuit Analysis, 9/E “V1 ’Jf V3 : QLOO T4475 1' iZL 30" i Jiv) ‘i’ (vi-J1)VL1+Ji—\75 :0 ’Vl‘i jiU3 + UL, +JiV5—TO ,' {I V ~ ,._.~ . --« ~ g D J’ ) 1* ( My) \11 +Ji\/3 ,lz QO 4/) + Cl 2 610° \7, ._ 05 2 (1/.300 Problem 8.120 Chapter 8: AC Steady-State Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 6-9; 435-v6°v \712 2-0249w°v V3: 6-2.14iq'5i-I‘3V 0‘98 LQ3~ 25% 9/64 ~i57‘31°v 30: V5: 5'164-I57'320V 0/: Amwcw C09me 114370413413 O ‘f l-OOOOi' O O O —I'Oooo O +i-Qoooi O — i. ooooi «i-oooo+ Looooi 0 "Meoooi O ~i-oooo I'OOOO ' I'Oooo O 0 Chapter 8: AC Steady-State Analysis Problem 8.120 4 Irwin. Basic Engineering Circuit Analysis, 9/E x-I'oooo vz'OOOOi o-H'OOOOi I~oooO o—i Ivooooi C) O O O o ~poooo I : O O 0 “2.00003 6'0000 FONOO +6~Ooooi V: 5-6306 +wOH2i’ ~vo.n455 i 2.07g3r 5'35‘15 T 2'0783u' -—o-o5éo +'O'QX32£ —H'754H‘ wquyi VD: “LI-754W —i'95’??{ Problem 8.120 Chapter 8: AC Steady-State Analysis ...
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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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