Chapter 8 Solutions - Soln0833 101029

Transcript

Chapter 8, Solution 33. We may transform the current sources to voltage sources. For t = 0-, the equivalent circuit is shown in Figure (a). 10  i i 1H + 30V +  v + 5  5 v 20V 4F +   (a) (b) i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V For t > 0, we have a series RLC circuit, shown in (b).  = R/(2L) = 5/2 = 2.5  o  1 / LC  1 / 4 = 0.5, clearly  >  o (overdamped response) s 1,2 =     2   2o  2.5  6.25  0.25 = –4.949, –0.0505 v(t) = V s + [A 1 e–4.949t + A 2 e–0.0505t], V s = 20. v(0) = 10 = 20 + A 1 + A 2 or A 2 = –10 – A 1 (1) i(0) = Cdv(0)/dt or dv(0)/dt = –2/4 = –1/2 Hence, –0.5 = – 4.949A 1 – 0.0505A 2 From (1) and (2), –0.5 = –4.949A 1 + 0.0505(10 + A 1 ) or –4.898A 1 = –0.5–0.505 = –1.005 A 1 = 0.2052, A 2 = –10.205 v(t) = [20 + 0.2052e–4.949t – 10.205e-0.05t] V. (2)