Question

Q1

In the circuit shown below, the switch was closed for a very long time and opens at t = 0. Calculate: 1. The initial value of inductor current, i_(0*), 2. The initial value of capacitor voltage, vc(0*), 3. The minimum value of the capacitor voltage, V cm 4. The time, tm, when the minimum in p.3 occurs. Try command executed. The exact answers are displayed next to your entries. Your score: 0 (Fail). Your score is not recorded for assessment purpose. New Problem Help Try Examine Topic Menu 1:1 (0*) = 12 A 545.5258 mA Vh 15 vc + 8.8 ICC 2: vc(0*) = 12 1.8002 V 1.6K 3: Vcm = 12 -10.5069 V 4: tm = 12 246.0715 us

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Answer 1

Sult- Switch was closed for very long time and opens at too Let us draw the circuit and find the values of eco-) and Veco-) L isa + 303n C Tv 2 16oor . Ei=3.nH jou =6.8uF Now we know, that in a dc circuit, in steady state capacitor behaves as open circuit and inductor behaves as short circuit ie 1 ise 216001 ve Now 5 0.54(65A I = lo 15+ 1600x3.3 - 1000+ 3x3 current divisjon And from rule - 1600 E ILS Trou +3.3 0.5455258A= 545.5258 ma

and nd Vc = "9030043.3) Vc = 1600 x 1x2.3 1600x I, = 1.8002 Volls Now, we know that current and voltage are continaious in inductor and capacitors it. sudden change in them cannot happen, hence nee é, co-) = { ecot) . and co-) = V(ot) Now e, cot) = 545.525 8 mal aus and Ivecot) = 169002 volts) Aug After switch is opend circuit will look like 3.3.0 ILC0+)-545.5258 ml + anavalt = 6.8uf Z16oon & 3.3m4

R2=3.32 éict) + Vilt) Rakoon (=6.80/v. C=6.84 t {Rahoon fr & ve +) = 1.8002 Volts Now for capacitor éstia edvelt ...d and for inductor V lt) = L dill Now applying mesh analysis in given circuit. - valt) of R, (2, - 12) = 0 =) - Rec-R, &, -Velmi - 1,--éc & I₂ = ? R, (3,-1) + R, I, +VL = 0 an =) Now 3) and R, ic +(R,+R,)in = -V - @ putting sic = cover and V, audio -Ricove -Ridea Ve -- (V) Riedle + CR1+4,722 = -L dd ---Qiy dt dt

Now we will apply taplace transform in equis) (v, f (V) 5) -R,C(SV_15) – Vecot)) - R, IL(5) z Ve63)...(vii) and R,C (SVECS) - VECOT)) +(1+R2)7715) = -2(57,69-7,6) . -- (viii) Now to find Veit) and i(t), we will first find Vecs) and I (s) then take inverse Laplace transform from equ (vii) Valsy (1 + Rics) = R, CV, (ot) - R, IC's) R. IL15) or Vecs) = RicVelot) - R. I, is) (Itri (5) (1x) I putting Value of Vees in equ vili) and solving for IL(S) LIL (0) IL(S) = RitR & SL (RRCS AR, Cvelet) (ITQ,CS) putting this value of Iq (s) in equ (ix)

ILCS) ♡ (S) = P, C Vecot) - Ri (1 + R, C5) (i+R, (S) 7) Vecs) - Roc Vecot), Rik I. (ot) (1+R,CS) PCRitR2751)(1+R,CS)-(a?cs +Rievetot)) 2) Vecs) = 19.5861767103 (1+0.010895) 2-880376224 (3.5904x1085?*$*$$45g + 39.204760?5-1603,28) transform Now taking inverse Laptace = Velt)= 1.8002 e-91.91176€ -S4s.gssst - 11.96543e . sinn .: (6708.67705) fince go expression for Va(t) is very complex ; hence I cannot find its minima.