Question

Q4: The switch in the has been open a long time before closing at t=0. Find iz(t) for t>0. 2.5 k92 ina 262.5H 15V 2.5HF 9 mA

Answer 1

tao is For long For s/co open so for loo time capacitor open $. induchor Shooted 2.5K2 + +)15.V Velo) i Lloy From Pigi HYL (0) IS 2.515 -6mai Veconj -ov -inductos shosted For tjo Slc is closed with initial condition Vae 25oo hs' +! & gma 15 s is @LÅLCO) va Kal at - gxio3 15 Use A Lillo) .hs. Vee cs & Vact s + 2500 gxió + Vse s LS test 2Soo & illo) + 15 20 2500s

9x103 Lillo) IS Vse Puth It les s ک to bo]. 25oos hs 9x103 (-6x103) 3 Soos use (2500 2500+2 sooles 3765 s 2500 Ls 1. SX103 6x103 Esco+2500LCS2+w5 - Vre b 2500 Ls gxio3 s soot3.9062552+62-55 *2500LS avizelec 62587 (3.90625(52418 (527165+ $40 Vze (5241657699) 1406.25= Voe 11 360 Ve= 52165 +640 From Rig i, (s) = Voethico) 25 m? 360 52168 +690 +62.5(-6x103) 62-55. 360 +52 +1654640) (0.375) 62.55 (524165+640) 360-0.37552_65 -240 5162-5534 /0005240000 Pl(s).

i; (s) -0.37552-68 +120 s(62:55?+ 1000s +40000) By postal Fraction -0.37552-65+120 5(62.5524/000's +40000) A BSIC + 62-5521000$+40000 ی -037552-65+1203 62.5AS+ 1000 ASHG0000 A +BS2cs. -0.37552-65+/20 = (62.5A+B) 52+(1000 A+C)5+90000 A. s coefficient compare 120 = 40000 A. A= 3x103 - 6 = 1000 A+ c= - 600 A = LO -6-1000x3x103 C = -9 -0.75 - 62-SA +3 13 -0.375 - 62.5X3x103 -0.5625. PL(S) = 3x103 -0.56255 S 62.552710oos HG0000 i ((S) = 3x10 0:56255 ģ t 62.5 625 3x102 5271657640 -gx1035 -0.149 (572x85+64 +576) + 6

s . 3x103 3x10 6.072 X 2 4 IL(+) =: 3x103-9x1o e cos 240.-3X102. ērt sin24t i lis) 3X103 gxios 0199 (5+3)²+(24)² (s+8)2+(242 9c69) 3x103 gx103 (5+8-8) o.144 (5+8) 3+ (24)? (s+8)? +242 9x103(5+8) + gx16 28-0.144. (s+ 2)2+(2472 (st82 24² 9x183(3+3) (S+8² +242 gxio3 sto - 3x103 24 (S+8) 2+ 262 (5+2)3724? Talde inverse laplace | ((t) = (3-g gestaszet-zeetsin 2017 mA. 3x102 s -8% t70