Question

3. Find it) 61 422 244) I to sec LE 2.12 Zov 4.- Find it) two ways: a) in the time domain Transform the Laplace 6) using X- 10 cost Volts 112 tuo pac + lovelts 14

Answer 1

Now; i (0) - edull) (0)= (ot) = F Amps ( ( 10 10 - 2²6] 8 de PEG) *(-3.6e"}')) t Amps TCC 3 teosec PIC) X A 10 cost v --1 IDIONU 3IH domain To Time ou superposicion Theory o using lowost Ps luactive es Active 10 volts ⓇOJA X Plt :in =10 Amps t. iCoo) = 10 lov TH inductor Shoot coolul) sec 7- = 1 / 2 = tla -t Now) (t)= ((0)+ (() - Pe 10+ (5riode CS Scanned with CamScanner

3 :. f() = 10 - él Amps Active lo volt dc es 10 Cost es Now (1) then stale In stoody Inactile ACC) From Sprusordal sleody Analysis; 14 state lo lest MCA 10 20 (oms 5 max 2 کے Z tilj 1+; 10) Here wel (SwL=51) Plti = 10 Loº 52 re (1450 LUSO elt= Seto f(+7= 10 Case sthus کجام Alle 10 Jz ( wst cos 450+ sint sinuso) Plt) = (s wst + 55mnt) Amp CS Scanned with CamScanner

e(t)= 10-5étt sust - 5stat en Laplace Toans foom b Now los I(5) Laplace Tof of pocost = s²+1 in los los Laplace T of 10 = 10 3 10 S S Mesh Analysis From Now I(s) 10 16) Ce+s] # los 341 a I (s) eln los 5²41 (STI) los? 410-10s? I(S) = 5152+1)(5+1) 10 115) = 5(5+1)(51+0) CS Scanned with CamScanner

5 -55-5 Now; I(5) 10 3 sti s2) Tersfoom Laplace envease Now mps loutt)- se tuces e tucty - 5 Cost - 5 sint) CS Scanned with CamScanner

Sol: br X- M. (30% poust 2011 Thr opened too switch is Before te y (0-) =V (04)=0 Ampes Voits Chilial value So closed. es switch After t=0, Now, behaves open Capacitor sleady state, AL Circuit 4+2 or AM Vcl0O) = 20 x U4276 ar + uc division 204 From Voltage Rule -- 10 volts 2= ReqC =(6116) * (%) Vc (0) = 20x1 = 3/2 sec Velot (".CO) - Ve -t/2 (60) e VC= Now, -t/3/2 UCHE lot (0-1 lole - 2 t) volts V(t) = 10 10(1-e CS Scanned with CamScanner