Question

Q4: The switch in the has been open a long time before closing at t=0. Find iz(t) for t> 0. t = 0 2.5 k 2 il 1362.5H 15V 2.5 UF 9 mA

Answer 1

2.5 krz at o wm- txo ( 10 ) Į 9mA osv i 10+) = of> tro] T2.5 HF 162.5t ii +7o GMA T Sistema We Conver Vis) ei qi Into Laptave domain we Apply KCL at 1, +ILS) mA 103 SL Sc VⓇ) so VG) +1515 2.5KR 3 MA ILS VS) SL 15 VS) SL + SMA VIS) SC + (s) 2.5kr (Sx2.5 kr 08 + vesz[4542_57 S 6 - OY 5125)ma ves) [ sc + 2.5 kr + 1] v«> [(S?lcx2-5k) + SL +2.5k] 3 mA - colo SL-2.5K S 3 <3 3*2'5*10 X10 XBL 8 ves[8Lc.2.5K) +SL +2^5«] =

08 [8**62.5 x2-5 x10 6x2.5x180) HS 62-5) +2:51] 3x2.5x62.5 OY Vis) VS) ICS) VCS) SL (3x 2.5x62:5) (saxo EX0-390625) +(562.5) +2.5K) 468-75 SO-390625) +(62'55) 4(2.5K) 468.75 S4025 [(52x0-390625) + 562.5 +2:5K] IS) = 7.5 (5) (50.390625 +562.5 +2.5K) SB + s20.390625 +562.5+2 5k Calculation for A, ommit s from LHS, sin Remaining we put s=0 А 34103 7.5 25k Calculation for BPG, we multi s on bothsides. multibly (B5+)5 7.5 = A + sx0.390625 +562.5 +2.5K we put s=0 s²xo. Exo.390625 +562-542.5K 0 = A + (6 + 6/2754 (S)[0-390625 +62,5 4.2511

A B or B AXO:390625 0.390625 B=-3x10 300.390625 ) = 1-171875x153 Now we But 3=1 on both sidest + Btc 0.390625 +625 +2.5K = (3x103) 0.390625 +62-5 +2.5K 7.5 = (3x10x16-390625 +625 +2.5K) +B+C 7.5 +(-6171875 tio) +6 = -0.1868 7.688 C IS) - 3x103 (+-174875 xios) – 0.1865 (33x0.390625) 462.55 +2.5K S 0.478208 11 Es 3x103 S 35 S2 + 5160 + 4400 Sa+100s +6400 ! 3x102) 35 (S+80)2 0.478208 (S +8032 465) = (3x103) 3(5 +80 -80) (51 803² 0.478208 (5+80) 2 19 3x10-3 S 3 S480 +240 (S+8072 0.478208 (9+802

14eg Snvоn. 8. ХаЫats“ ты" 14 -44) = (3x to')ца - зе. 5 -2ҷо t e° ) got. dit) -(зх16°) - go t о. 41s20g tе. ult) + 2 39-52 11te №t]au) go 1 —3e