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IRWIN 9e 9_23

# IRWIN 9e 9_23 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 9:; Determine the average pawer supplied 10 the Hem-wk in Fig. PLEA. 11H 1n 1:1 I'm Figure Fig.2; SOLUTION: ﬁﬂ 1Q Chapter 9: Steady — State Power Analysis Problem 9.23 2 Irwin, Basic Engineering Circuit Analysis. 9/E KCL Vat MW g'ﬂOdgj. ’ +3:— : 1&0 y: + x9. ; w i '11 ii \7’ + : i 31V,er = 150°) (31\$); — leLTQ-IT31)\73=O ..]2.\7l '1' (i+j2) 1—1—13: 219°1\7m‘\73 = 180° V, ; 0‘75’ Law-51W \7} s i L‘li'ﬂlov (N<\ = 0' 5? L42.2.¥°v PI = VzI _vco4(<9v;@1) Q. P; z \$9.2 w(—/¥-79°-—0°.) 2. PI = 4—16.22 mN Problem 9.23 Chapter 9: Steady — State Power Analysis Irwin, Basic Engineering Circuit Analysis, 9/E 3 77 A 2(3-1-3-;2i_|+lj—;2il+§l-l;lol] A z 3' 0000 'l’ Looool O —§.ooool - l~0000 ‘l' Lemon; 0 —c3.oooor Loooo.+ 3.0000? -—|.oooo 0+ l-OOOol O 7) 5;; [01,95 5] |.0000 0.6191 T 04523”? 0466;103077; 049,331+ 03346? 27 A: [i431 l—i ~11: —:; l+§i—-l 3? ol] Chapter 9: Steady — State Power Analysis Problem 9.23 E Irwin, Basic Engineering Circuit Analysis, 9/E 1.0000 + 3.0000! lioooo —loooooi -—I-oooo O ‘ 010000; 0 'f’ bacon; >> A\b ‘oooao +5~0000i mLoaoo Q i ‘ 00630 On: ~— 0‘6’5‘4 “r 0~ £43305 0‘36’5 ~ 0‘3077; 0‘493) 'i' 0.384“. 7) m Problem 9.23 Chapter 9: Steady - State Power Analysis ...
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IRWIN 9e 9_23 - Irwin Basic Engineering Circuit Analysis...

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