IRWIN 9e 13_32

# IRWIN 9e 13_32 - lnNin Basic Engineering Circuit Analysis...

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Unformatted text preview: lnNin, Basic Engineering Circuit Analysis, 9/E 1 ________._________.___.__—_._— 13.32 Giwn the i'nJJim-ing i'unutinm FM}. find flit]. 3(5 + 6) (3 + ms? + 63 +13} [5 + 4H: + 8] 5(32 + E: + 32} [a] H5] = [b] H5) = SOLUTION: (a) PG) : 3(§+6) (\$4902 +65%?) (5+3)CS)+63-H3) 3+3 “3.53 H3153 3/06) '5 A(Sl+55+'8) +8 (3+3) (H3403) +8 "(s+3)(5+3~j3) LeiL s: -3 "3(-3-ié) : A (-31.; €53) +157) 9A: *q AZ—J Let 5: “3+J3 (—zﬂ'l) (' 3+}; +6) = 8('—3+j3 +3)(-5+j3+3+33) (3)0608: (~3+J5)(3~U3) _______________________—.———————- . Chapter 13: The Laplace Transform Problem 13.32 2 Irwin, Basic Engineering Circuit Analysis. 91E FG): ~I +_.___L'___.. +4.... 5+3 5+3'j3 g+3+j3 FOE) : [(53% 26,3): ((34 3f] LLLt) (6) H3) : (3+3) (8+8) 5(81495-r31) Fm: in“ 12 + 8* 5 “WW swan {SM} (5+2) = L+____E_____. 1‘ 121,. §(.s‘1+2.s+32) 5 Shy-JV 9%er (5N) {3+2}: M31”: +3z)4ﬁ(s) {WWW} + 8* “MﬂH-Jw) 1.61530 W): M32) A=l La: 9: “LIN” Problem 13.32 Chapter 13: The Laplace Transform lrwin, Basic Engineering Circuit Analysis. 9/E 3 (—Hﬁwq) (*quﬁ") = AE—qu) Q+X("4+J") +32] + 8 (www) U8) 8 (“9+ij (38) 1' (JV) (may) 8: 0'5 4- 90° 8* I 054 90° [3(5): __i_,+ 0.54110" + 05130" 5 S‘HJ'J“! sw—Uq PH): [1+ 6% C023 (Lit—90°) ] ULC‘U _______________________________ 5 Chapter 13: The Laplace Transform Problem 13.32 ...
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IRWIN 9e 13_32 - lnNin Basic Engineering Circuit Analysis...

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