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IRWIN 9e 14_18

# IRWIN 9e 14_18 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 1.5.43 Liar: 11112511 analysis [011ml Z-‘DIIIII. .r D. in the network in Fig. Plnltli. Figure P1443 SOLUTION: 11(9): 4‘s KCL: Iéa>+1xts)=1,(s) in“) z.- I‘(s)—I3(\$) ~ 1,3) : gar) — 13(5) 92 029(5)—.I.<s)—I5(s) lM’S) : — KCL'. 179+ 152(5) =13) I’M) :, 13(1) —— 13(3) Chapter 14: Application of the Laplace Transform To Circuit Analysis Problem 14.18 2 _ Irwin, Basic Engineering Circuit Analysis, 9/E KVL" SC“ Ixt‘91i< I’m)» 103(9) =0 —-S (13(5) 413(g))+ 131;) ——Ia(\$) 43(3) =0 ~Sliés) 41(91— (am) gar—o \Iois): i(I5(\$)) —8 (436))» 415 + («913(9): o (615 1a) I50): ﬂs 138): § 51(34—1) 13: o? ‘3 ) SGML) \/O(s)= 2 SQH) .3...“ 3 A ELM-i) 7* 3% 0?: A(S+I)+8(s) “13:0 °?=A M4; .1 0?: - B B:-2. Vo(\$) :: i — a? 3 5+: V003):- ‘RO— U?) M(t)\/ Problem 14.18 Chapter 14: Application of the Laplace Transform To Circuit Analysis ...
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• Fall '10
• MicahStickel
• Circuit Analysis, Basic Engineering Circuit, Engineering Circuit Analysis, Laplace Transform To Circuit Analysis

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IRWIN 9e 14_18 - Irwin Basic Engineering Circuit Analysis...

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