Question

Problem 4 (20 pts.): In the following circuit, the switch was open for a long time. At i = 0, the switch closes. Determine the quantities given in the following table. 0.1 pF Xt=0" + + Ve(1) - 7.5 mA (1 v(t) 320k12 30 k 12 a Ve(0) b Vc(0) с v(O) d VcSS e T f ve(t) for t> 0. g v(t) for 1>0. h (0) Vss

Answer 1

Solution: For t<0 condition- the capacitor element branch gets disconnected from the circuit therefore the circuit is in initially relaxed condition and acts as a short circuit element such that the voltage across capacitor is zero volt and the voltage across 20 kilo -ohm resistor is equal to 150 volt (using basic form of Ohm's law).

For t>0 condition:The remaining quantities can be derived as shown below with detailed explanation -

Solution : For the circuit shown in the given figure- Olet tão tuelt 7.SMA VIH220kn SBokn (a) Drawing the circuit for t<0 condition : The switch is kept in open position such that the network to right contains the capacitor element gets disonnected from the circuit such that capacitor element is initially relaxed so that, at this condition the capacitor element is treated as the short circuited. Taking 'O' as ground node - icojao >30K72 7.SMA VO) 20K 0 the (b) Here vro] = o volt Now initial voltage across the 20k2 resistor is equal to - 1103 = 7.5mA x 2062 = 4.5 x103 x 20 x103 vots 150 volt At t= 0+ condition: The switch gets closed, since the voltage across the capacitor element cannot change instantly therefore- verot) = 10) = 0 out

of steady circuited d) Calculation state volnage across Capacitor :- At steady state condition, capereitar acts as an open liloo)=0 330k2 ? Vous 7.SMA O.C the refore (مناجا + Vloo) 2012 te (0) The wimage doop acones 2018 resistor V10) = Y.SMA X 2062 Vioo) = 1 sovelt acons Capacitor Now using KVL to obtain the voltage =0 Vel00) - V100) + V2!) 30kn the element As voltage doop across 30kiz resister is 'ovelt' since no current flowing through V100) - 150 +0= 4, 100) - · 150 volt

therefore Vass - voltage at steady state across capacitor | Vass = V10) = 150 volt e) calculation of Time constant (r): (Andready state) : calculation ou equirment residence Crea) seen from the capacitor terminals : Replace the independent current source by their internal resistances wje; 7:50A Duc current source acts as open aircuited now equivalent resistance Req is given as Rear an J 2062 230k2 (d) Req > 2012 and 30 KR resister in series o DD Req = 20k2 + 30KR Req= 50 kr ReqxC second. Also, Time constant (e) 50x10° x 0.1X10.6 5x103 sec [T = 5 milliseconds]

f) Redrawing the circuit for tro: pruf { Boke Isle + 7.SMA V Velt 70KL Зок Vogo. T17 e) Source Transformation to Spain the weetage source equiment model eving Visma Sovete in parallel with 2012 revistar Applying O'lt 20K2 + viit 1500t iety S30kr (f) Using KUL in the mesh 5) 20000lt) + 2.674 +30000 ich = 150 As current flowing through capacitor istl= couilts Ampere at 0.1X106 duelty Ampere dt of Pht the cause of ich in egr or (abonox 01X10") d Volt + V (4 + 30000X 0-1X10°d vec H = 150 (it) = list dt dt =150 > (5x109) UCH + Vesty

duelt + Velt 150 at (5x169) 5x103 Event +200 Ve 1+) =30,000 d at Let g =D at ; (D+200)ųCH = 30,000 Here egn (2) represents the differential equarion for netthage across Capacitor - com solution and The fotae seletion of differential equatim (2) coneita mplementary form particular integral solucion V CH = V CHI + Vect) (f) ie; (PD) RMS >0 =0 4 for complementary form of seletion! Maping 1D + 200) Ult) forming Ausillion equation : Replace's by im' we get- AE = m +200 20 ROOL of auxiliary equation a m =-200 As roots ove real therefore, welhoge reopone makes form of - Velti Gent volt OF

UCH - 2005 -ce CF for Particular solution in Velth CRD) (0+200) 1 - X 30,000 ane - e if flad to fin) Put Dea in denominato gires - YP.. flos 4 f(a)=0 then Yp.2 2.Ie flas ar if f'conto and soon Delt ot - 30000 e PI (D+200) Replease 'b' by 'o'we get Velt 30x10 3 (P.1) Velt PL = 150 volt Now, Total volt votrenge response uit > Velha V (+) + VIH PᏛ CC - 2005 Velt) = 150 + Ge At t=0 U(0= 150 + Géo Veo) = 150+61 Vild) = o volt 192-150 s'hce,

-200t volt Anself) at 2000 -200€ mill. Ampere ; txo+ Put the name of 'ci in ear we get The complete voltage response across the capaciter Velt = 150-150 € i to now, current flowing through capacitors ict) = cave(t ampere = 0.1X100 (150) - histē =-0.1X106x150% (-200) e -200t [{e1) Now, from figure le) The veerage VH) across 2012 resister is equal to - Velt) + Vookra volt Verty + (30x10%) lét -2005 = 150-150 e + Box102)x(3x109) -200t - 2005 = 150-150e tgoe = 3e g ) O'INF VH) - f ANCH Wit Zgona 30kn Voorn -200t - 200t = 150-60e volt Vet) ; vaid for to

A) voltage vioty i equal to O Al +=ot ar vest from eq? 0 150 -60 e go volt VIO) Ans. W ů At it to the steady State werage ves across 20k2 resistor Vico) - VSS -00 150 -600 vou Uss = 150-0 Uss 150 volt Anseil 150 voet o final Answers obtained - uloj = Ovolt veletes =0 volt voj = 150 volt Vess e = 5 milli-seconds - 200t = 150 - 150 e volt for t>o - 200 t VGH = 150 - 60 e volt for to veots go volt 150 volt Velt) Answer,