Question

3. Find it) vo 412 w X 2/4) t=0 sec LE 320 zov 2 4.- (25) Find itt) two ways: a) in the time domain 6) using the Laplace Transform t=0 sec et ilt) + 10 cost Volts 12 mm lovolts MI

Answer 1

3. The given circuit 62 WWW 42 mw t-o sec. icts 201 22 trosea The switch is open for a Long time and it is closed at For evaluation of initial conditions: Circuit at tco and t=0 capacitor behaves open circuit at steady state 402 W + Uclo) 201 to Ucco) vollage across capacitor at Velo ) =ov allow the capacitor does not sudden change in vollage lucrot ) Nc66) = OV

circuit at too 6ur Vect) 42 ict) + + 20v vect) - 2 F aut Let vollage across capacitor Vcct) Node equation. Vect) -20 6 + ict) + Vect) 6 ict) с dvect) dt current across Capacitor Vect) [ ó + 5] + t duelt) dt 20 6 «ct) { $ ] + + dvelt) dt 20 6 2 Vilt) + 3 duelt) dt 20. solving above differential equation 3 dvc ct) 20 - 2 Velt) dt ducit) (10 - Vect)] dt 3 dvcct) (10 - Nelt) 2 at

Integrating On both sides - ducit) 2 dt ماننا df10 - vect)] 2 ttc -in [10 - Volt)] - Ž Im (10 - Vect) - 2 + + Co 3 both sides Apply Exponential on 10 - Velt) - 22€ c к Vect) - 2/3 € To к (1) د where is constant apply initial values to (i) Ni(t) It:ot 10 K - 430) 10 к 1 K =10 O to - K V tro Velt) = 10 C1-e2gt) ] The current through capacitus ict) e duc lt) = a al lock-e ezt) la eft - 22€ 2 x 10 (-1) (-2/3) è 2/3 € olm А i(t)= 10 22/35 tzo. 3

The electrical circuit trosec ict) + 10cost v leuw lov 1H a ict) the time domain Source DC contains and time varging In given circuit the voltages. Here I am using Supperpo super position theorem. (casei) only DC lov is present ct) case i only 1200) lo cost is present Total ilt) = response ict) + izlt). casei);- t=osec 1,ct). I + lov TH to Source inductor ico) = oA - No when to

www 1,(t) 31H tov It circuit Looks like RL simple in the circuit given by The current response ikAD = Il final + [zfinal lo - ct) I - ldii(t) L dict) dt VLCE). - 0 dt 10 - iict) - diilt) = 0 dt dict) dt 10 - iet) diilt) 1o-i,ct) =dt both sides Integrate on S dict) 10-1(4) = sat - Im Cio-i, ct)] = ttc 10 -1i(t) -t K. e ict) = 10 -tk е K 210 i, (f) =0 03 10 - 1 K lict) = 10 C1-ét) А too

caseii. w izct) locost TH izcot) =O A To cost 2ct) diz(t) = 0 dt diect) + 2(t) = locost. — (i) dt Equation is represent first order lineal differrential Equation. standard form of dylt) dt こ P(x) dy ct) + Q(x) solution to above Equation s pcx)da yct) facx) ef {pcxdx e da tc eSP(x) dx Integrating factor. PCL) - 1 QCE) = locOSE SPLE) dt izct), e - facts & Spitsat dt to Sidt Sidt iz(t) e S 100 lo cost e dt tc iz It) et So cost. to dttc

t iz Lt) e cost e at tc E I I I cost e etat Apply OV method U= cost du=etat Sudu UN s du dr I = cost. et cost. et - Sesint. et at cost et + fsint état 1 cost et + Simket - s cost et dt 1 I 2 I cost et + sintet 1 et (cost +Simt] 2. iz(t) et LO et [cost + sint tc 2 P2 (t) = 10 (cast + simt] - -t te C izlot) =0 10 (Coso + sino] 11 të c 2 5tc C-5 lilt) to 6A 5 ( cost + sint] - 51-5

ict) = {(t) + {{Ct) = locinét) + 5 [(cost + simt) -ět] ict) 15 + 5 COSE +56int А too using Laplace domain To 7 ilt) t=osec locost um (H + LOV . 10S Laplace Transform 10 10 cost S s²+ 12 5²+1 Domain 10 10 S Lit LOT 1 S - I(S) los 5"+1 S 10

los I(S) (1+s] + $+1 10 Tos | ILS) = + S st1 (s+1] 10 TOS + 1 SCS+I) (s+ 1) (5+1) 10 +10 [Goo.com] S(541) (5+1) (541) 10 + 10 [ St 1 (541) S(5+1) (S+ 1) (8 +1) 10 + 10 (541) (5+1) (5+1) (5 +1) st) & 10 ICS) = S(5+1) srt 1 partial fraction 1 (5+1) (5+1) l partial fraction 2 = PF2 = PE 1 10 A B + PFI = s(s+1) S St I 10 A = PFI sls=0 =10 (5+1) Is=0 PFI. (5+1) Is=-1 10 B = - Is=-1 -10 S 10 PFI = 10 S (5+1)

- 10 Pf= A BS + c f (5+1) (5+1) (5+1) st - 10 = A (5+1) + (8 + c)(s+1) (5+1) (5+1) (52+1) (5+1) -10 5[A + 8] +s ce+c] + A+ c) By compasing coefficients A+ C =-10 2 A+B=0 B+c=0 A = - B B=0 Atc=-10 - Btc = -10 A=c B=-C C+C =-10 =-5 B=+5 - 5 A=-5 B = +5 c =-5 PF2 - 5 5S - 5 + (sti) - 5 S 5 - 5 S+1 (571) (5+1) 10 10 I (S) 5 S + 5 5 sti st (5'+1) +1) (5+1) - 10 S 15 S+1 + 5 (5²+1) + 5 S (5+1) I(s) 15 5 + 5S + 10 S S+) $1 s'+1

to Ics) By applying Inverse Laplace transform S - 10 I(S) 15 5 + 5 ols + S+) st set L.T Sint <> (lt) = [10 - 15 et 5 st1 15e + 5 cost + 5 sint] L.T -t sti sti i(t) = -t 10 - 15e -15e + 5 cost +5 sint eest osaint A to.
Part A and Part B answers are same.