Question

Q3: The switch in the has been open a long time before closing at t=0. Find il(t) for t> 0. t=0 2k-22 in 62.5H 12V 6 ΚΩ 2.5 UF 9 mA

Answer 1

page-1 * Given electric circuit is, Xe 2K Nw t=0 & 6ks 2.5 UF ama person 6-12v Case (1): Find initial conditions at t=0 * At t-o, inductor is short and capacitor is open as circuit will be in steady state. * circuit becomes, Ttv Evelo) 2 12v 2 kr MM =GMA illo) 2k Vc 10-) = 0 volt (as it is voltage across short circuit), Initial conditions are, ** 1110) = 1 (ot)=6mA Velo)= Velot) = 0 Volt

page-2 Case (2): convert the circuit in s-domain with initial conditions * circuit becomes, aka Gm CS am S 6k 362-55 al24 5 7 Source transformation To=GMA 6 kr 2k2 M OLIO + 12V 54 S 3 62.55 Gk on 2k 2 w bosts 0.375V 54 4X105 s 62.55

6652 zka page-3 MW 54 100375V •ux105 362.55 Source transformation I= 12 6m 2k 879 54 6m 4xios 56 6 0.3750 32k 362-55 parallel 4x1051/2k 4 x 10 x 2k 4X105 +2K 48x108 4x105 + aks 2 4x10 2x10²ts 4X105 S+800

page-4 * Circuit becomes, 6km Vx 54 6m • Ilis) 60.3750 Muxios 362055 S+200 Apply Nodal. Analysis at vi node. 6m Vx- 54 Vx+0.375 62055 + Va (4X105 + 5 +200 6k 54 6m 4 + + Vx ( ak +6255 $+200 4X105 0.375 62.55 .f 6ks S+200 + Vx (6x + 625 .) - hitta - 00315-6) 0-375 62.5 4x105 + S+200 4x105 Ś (9m – 6m-6m) vxl ak + 25 vel by Gorst UK TO ） S+200 = Ś( 9m-12m) Valor S+200 4X10S -3 m) S +

page-5 62.5(4x305)+ by (oxios)+ +200)(6162,5) 6X (62.5) (105) -3m s 2.5 115 2.5 1205 2 tot 4x103 vel 675x200 +(3x4 xyos) + (S+200)x3x6/is 6205x6x2x105 -3m S 12.5 2.5 (2.5x200) +(3x4x 4x103) + (S+200) x3 2.5x6x2x105 10² 200) *83:5). -3 Stor V₂ Vel 500+ 48x10 3 + (S+200)(7,5) (51200) (7.5))-- 3000 500+ 48000 + 7055+ 1500 o) = -2 3000 7.55 +50,0 3000 60.000) - 3x3000 S(7.55+50,000)

page 6 Vyc -9000 s(7.55+50,000) IL (s) = Vx+0.375 62-55 - 9000 s(7:55+50,000) .+0.375 62.55 —9000 +0.375 ( 7,55750,000) s* (7155+50,000)*6205 2.81255² + 187505-9000 (7.5 S+50,000)x6205 0.0455'+ 3005 – 144 S? ( 7.55+50,000) 0.0458 +3005 - 144 s7.5 (5+66666666) 6.x10 35² +405–1902 s? (5+666666666667)

page-7 IL(S) = 6x1035²7405-1902 S? ( 5 +6666-666667) A B + Il() (S+6666.666667) 6x10?s?+ 405 -1962 = As(s+66666666667) * put +B(8+6666-666667) +05 s=o in equation © 0+0 - 1962 = A(0) + B( 6666-666667)+O B- -19-2 6666666667 B- - 2.888103 * put s --6666.666667 in equation . 6x103 (-6666-666667) + 401–6666666667-) – 19.2= A(0)+B(0) +C(-6666-666667) -3 6x10 x 4 4 4 4 4 4 44.45-266666.6667-19-2 =c(444444444.45)

page-8 2666666667 – 266655.6667 -19.2 -6(41494444.45) - 19.2 4444 44 44-45 8 c = 43.199999815 c=0.43199999x10 * Equate sa coefficienta in equation , -3 6x10 Atc A = 6x103 mc 6x103 -0.63)99999x166 = 6000x106 -0.43199999x106 5,999.568 x106 A = 5.999 568 810 substitute A, B, C in equation o

5.999568x103 + (-2088x103) page-9 0043199999810 (S+6666666667 IL(S) = Apply Inverse. Laplace Transform, 6 Izlt) = 5.999568x10?-8-88x103)t +0.43199999xió - 6666.666667t -6666-6666674 Ame (+) = 5.999568x10?–12.28x1034 +0.43199999x10 * * *