Question

15 3 N Find ilt) 612 412 2(t) IF 320 t=0 sec I OV I (25) Find ilt) two ways:

voltage Is 40 volts

Answer 1

un Me 4 Flat t=o see LE 22 UCE) yov i(t) using time domain Analysis Before switch close capacity is in steadystate. Replaced by open cirent. u Initial value of Capacitor voltage volo) 2 Vcló) = ov Adler Scaitch closed 6 Time constant 4 u € REG 2 REG=6/6 yov 2 VCUT IT 2 K T = R(= 3x2=1.5 sec Capacitor is actas openarent At Stecody state (422) 6. V((*) 40 Final vale of Cap Voltage VC(O) = yo (voltage divider 6+12 ел.) عه VCCOO) = 20v CapacAr Vo Hase Eguchon Valt)* V(H) = fonas +(vioimal - Vtinad)e 20+ (0-200ety = 200-e 1/5) V

V(t)= 20/16 - 1/5)v corrent througth Capacitor icts = care at TCL) = ( (2001-ething) - 10 [o-ething -4415 i(t) = 20e 3 i(t) osing leplace Tkanstorm iccol= ov &((0) = 20v Y = 15 sex After Scitch operation Replace Capaaher with initial conditions 6 42 (5) AAN (5) 2/3 22

Apply Nodal Aralyns at UCS) W(S)-40 s vs 2/5 + VCS) ५० Vcs (9) [+++421 J- N(S) +] 40 65 VCS) 2+35 yo 6S 1 VCS) 40 S(35+2) 40 2 3 35+2 VCS) -2°[$ - ...) Apply Inverse Laplace It V (t) = 20 [l-e V15) V ICS) 20 VOS) 45 SS+2 2/5 35+2 10 (1-33 5+2 =10 [-(-23) [52] LT 10 (2) 3542 Sta Apply Inverse Laplace T/ icts = 20 let/15]