IRWIN 9e 9_22

# IRWIN 9e 9_22 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 9.2: Given the Hem-xiii; in Fig. PLl-EE. ﬁml the average power supplied and the 10ml average pai‘ﬂr’m' ahmrhcd. Figure P922 SOLUTION: j2 n Chapter 9: Steady — State Power Analysis Problem 9.22 L 2 Inivin. Basic Engineering Circuit Analysis. 9/E f‘ = 5.3,?L knee:va f2; 5‘3¥[;26.5—7A 2' 2— PZ—Q’ 5 234 g: fax“?! .— lé" 5‘3?k&6.5¥0—5.37'L:I;6.§;b 1: 3' '7'6 LL51 436A m ; vs 1.: '2. KS - Eyck/J) M43)”— @455) ' "naryﬁu... 2.. I35- 43. 26M le) v Problem 9.22 Chapter 9: Steady — State Power Analysis lrwin, Basic Engineering Circuit Analysis, 9/E 3 >>A =L—Qi —l 1.2!} l I -3—i .2 —I+Ql I-l 2+: -p2-Qi}0l0-IJ A 2 o — {zvooool —'\.oooo‘ l-oooo (21.0000 \.0000 I 0+ l-Ooool ~3°oooo — Looool .3. 0000 —-/:oooo +Q-OOOOI l'°°°° 'I~O°°0l 07-0000 '1'“ l- Doool -2-oooo—§-Ooo°l 7 = ' ‘ . « . - W >A [‘3"l lglj'll~3-123—I-l-ml-iJ-l-iaz—ﬂljolo ~l] A : l a O ——D?. 0000' "' 0000 " 0000 O + 3,0000! ‘~ 0000 l'OOOOl -— 3.0000 —— I. 0000; 2.0000 —- [.0000 +5 . 0000! \noooo -- 1. cool Q-oooo + Loooo'r —°?~oooo 320000] 0 [10000 o “labooo 77 E= Loy-)oyﬂ lo: 0 O '2 7) AH; - Wmvxivxa ', Moulle )1; close. {0 [Sinaulodz ‘ovs baAly SCoJeJ 0 Chapter 9: Steady — State Power Analysis Problem 9.22 4 lrwin, Basic Engineering Circuit Analysis, 9/E Regalia “may be {naccw‘scdlc- RCOND = 33783811 e-ols. Oh; = ‘?~6000 * 977'20003 0'0000 " 32-00003 “9-6000 —97.:oooi - '2' 0000 —-33. 0000! >> A 2 . 0 Jr 3.0000! —— I-oooo l- 0000 O + 3.000; l'oooo 0+ 1.00007 -— 3-0000 —I.oooo'l Q-oooo "loOOOO 1— 3.000; I. 0000 ... Looob; 3'0000 + LOOOO; -Q.oooo “10000; O Ivoooo o _l.oooo 7> A\b OMS : —-ou0000 ~f 0.0000; 9' 6000 ~4-‘3000; O 0 19.4000... 4.3000: 7? 1 Problem 9.22 Chapter 9: Steady — State Power Analysis ...
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## This note was uploaded on 11/17/2011 for the course EE 2120 taught by Professor Aravena during the Fall '08 term at LSU.

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IRWIN 9e 9_22 - Irwin Basic Engineering Circuit Analysis...

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