Question

S2* (8 pts) In the circuit to the right, the DC source, V = 2 volts for all time. The switch has been open a long time. Also, the capacitor is charged to 1 volt, so ve(0-) =1. At t = 0, the switch closes. Assume: C = "4 F, = 1 H, R = 10 ohms. a) Find the current through the resistor for t> 0. (hint: at t= 0+, di/dt = .04 amps per second. *) Plot your result using MATLAB or EXCEL. b) At t= (- what is the energy stored in the capacitor in joules, and what is the energy stored in the inductor? c) After the switch is closed for a long time, what is the energy stored in the capacitor and in the inductor? (*Finding this initial condition is tricky. See if you can do it. If not, just use the 04 value and move on.)

Answer 1

1 (0-) when txo true son Po co w Boco.za is a ve (oo)= 1v [given] Eclo-) = Cros = 2 x 22 x 1² = 235 1. (b) = 0.1953 El60-) = L cos a * 1X(0.2) = 0,04 = 0.025 (2 Ve(a) t>o and at t=d ILCO 1,60 e 50-24 v da Velo): Vs = 2V . Ec (a) = 1 x cx le cas = 2*4X(2)= -0.5 Jocs ما اداء (ع) XIX (0.2) = 1X0.00 = 0.02 jouls > (a B002 when txo Vs-ve = L.die ogla Toscot ) 2 VCS) = (5 149)- 012) 1469) + 0,3 - vecs) ca i-ie in bed -O applying replace Toanstorm

ti (svecs)-1) = 9c9-ves á svęcs + versy - = I4cs) ve(s)(+ b)-Mattias Vegas VEC9 [atóts] = VE (S) (55 25420) = 40 +43°4 55" VEU = 5s'tus tuo 1 + 0 SJ 10 2052 S (554+25 +20) - by Vecs) = $70.85+8 Sesto:us+4) taking inverse Laplace Transform vect) = 2- [603 (1.998) -0.1005 SOC11997)] v iect), Vente -0.21 lt)0.2-0.le I cos (1.99+)-011005 sin (1.99t)]

>> syms s; V = (s^2+0.8*9+8)/(* (s^2+0.4*3+4)); Vt = ilaplace (V); >> Vt = vpa (Vt, 4) Vt = 2.0 - 1.0*exp(-0.2*t) * (COS (1.99*t) - 0.1005*sin(1.99*t)) >> t = 0:0.0001:35; figure; plot (t, subs (Vt, t), 'linewidth',2):grid on; grid minor; xlabel('t');ylabel ('Amps');title('i R(t)');

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