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IRWIN 9e 13_53

# IRWIN 9e 13_53 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 13.53 The switch in the eireuii in Fig. P135?- Jms been eJnsed ['m' a 10113.:- time and is nj‘ieneil at I = U. Fiml {III} the I D, using Laplace transforms. E F 411 _|. Cl) Figure P1353 SOLUTION: Chapter 13: The Laplace Transform ProbIem 13.53 Irwin, Basic Engineering Circuit Analysis. 91E vc (0') : 6+ 2H +q (1(6)): 30+ L403) VCCO‘) 3 13V 6v 2/gH ﬁf1(i)di+wi(;t)+ g: curt) : 6 3 3 out 1: 1(3) + H Imﬁg 516:) :2; m0“): .6, 3 S 3 3 5 I6 I“) + ILIC3)+ 251(5) ~2L(O') ‘13. 5 S I(:)~[_Ié_ T mu] : U? + 21m“) 3 1(3): [16+ mi 15"] 3 '57 *6 S 13>er£1£ ‘S S Problem 13.53 Chapter 13: The Laplace Transform IMin, Basic Engineering Circuit Analysis. 9/E 3 I63) : '65 7‘ I? mi...” 231+ ILS'HS S Ics) : 2.035%) 2(\$1+65+8) I3) 2 -3s+9 (3+k4)(3+1) 1(5) 2' A ’F B \$+q 3+1 ~33+ﬁ : A T 8 (SW) C 3+1) 3+” 5‘” —33+q: A(3+2)’/’E[s+q) Lei s: “2 -302) +9 : B(~ UV) IS: 23 8:15. L Let S: A! Chapter 13: The Laplace Transform Problem 13.53 4 lwvin. Basic Engineering Circuit Analysis, 9/E -3(«u) M = Aﬁwz) -2A= z; “:21. ’2. IQ): 2.2.]. 4 «LS—— 1 l s+~4 5+“L _ ~)J: w) : ~__2_L.€ ”* +1.5. e ]u(t)A 2. z _________________________._.__———————— 5 Problem 13.53 Chapter 13: The Laplace Transform ...
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