IRWIN 9e 14_11 - Irwin. Basic Engineering Circuit Analysis,...

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Unformatted text preview: Irwin. Basic Engineering Circuit Analysis, 9/E 1 mm Find 1'0“). r 0.111 the new-uric in Fig. P]-—L.l l. Figure Fm+11 SOLUTION: KCL: I¢s)+IILS)=Is(S) I8) '— 135%" I4“) KCL: 1'55) : ITS) + Ids) t 1'29): 13(3)’II“) \«CL: Ig(g)+I//(S) : IMLS) I”{§) = 14(5) ‘14“) KVL" ‘3‘ LC‘)‘T 1(1 I'CS))+-.L( 19(5)) : o “2'? C13“) “149] + I [:IHG) — 13(5):] + 14:) = a #10263) - Ks 1‘36) + (a: + '/s) LU) =0 ___________________.———-———-———————-———— E Chapter 14: Application of the Laplace Transform To Circuit Analysis Problem 14.11 2 Irwin, Basic Engineering Circuit Analysis, 9/E I315): %_ 1| (9) : 2'3 _% 13(5)+(o1+ ifs) L‘cfiiéz KVL: 313(3) + (I’M) + 1/3 Ia) :0 513(3) 4— 1(136) —1.(:)) + VS (13(5) Jud) : 0 419+ (8+ '/s+‘)13(5) - '/g 1.48) -_- 0 (3+ y5+1)13(s) — 131.1“): 4/3 -:j§-,‘.[3(s): if? — (01+ 1%.)1‘13) I3CS)=-S[i'3- (.21. 1?) L161] 13“): 1+ 3 (4+ lid-Tau) GT llSHJC‘l T Caruipfllqw> _ l/S 1H“) 2 2/5 (511' 3H) (—1 1- [23 +I]Q(s)>’% L16): 1’3 $2.9 2- +; \9 1+! I'H(S)_.i/_g 327,1.” V s (o‘eflr 43+ 21+ $2+J+I-—/)I (g): slum 3 '* ”' .5 <43 5 1- 3&L+3J)L1(S): disfi S ' & 1 g): lira §x3+3§+33 Vo(§): 111(9) (l) Hr Problem 14.11 Chapter 14: Application of the Laplace Transform To Circuit Analysis lrwin. Basic Engineering Circuit Analysis, 9/E 3 V06); & L+ 3+3 o‘esjr 3&2'1'33 \/o($) ; .i'2—(SL1'J-f'3) ____________________________._. V05) _ A I3 5* - _- + 7' 3+3/if-] J; JT3/4TJJZ—j f 13($)(5+ 3/4-+j J; +CCS)(5+ 3/4-] Alt 3:0 %= A (3/41 “§)(%*]/§ ’3: 0'408Ll24-75° V0 (‘9): [L r Most (q: 5 + 0.1. 76?] “WV Chapter 14: Application of the Laplace Transform To Circuit Analysis Problem 14.11 ...
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IRWIN 9e 14_11 - Irwin. Basic Engineering Circuit Analysis,...

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