Chapter 8 Solutions - Soln0828 110224

Transcript

Chapter 8, Solution 28. The characteristic equation is 1 Ls2  Rs   0   C s1,2  1 2 1 s  4s  0 2 0.2   s2  8 s  10  0 8  64  40 –6.45 and –1.5505  0.838, 7.162 2 i( t )  i s  Ae 6.45t  Be 1.5505t But [i s /C] = 10 or i s = 0.2x10 = 2 i (t )  2  Ae 6.45t  Be 1.5505t i(0) = 1 = 2 + A + B or A + B = –1 or A = –1–B (1) di (t )  6.45 Ae 6.45t  1.5505Be 1.5505t dt di (0) but  0  6.45 A  1.5505B dt (2) Solving (1) and (2) gives –6.45(–1–B) – 1.5505B = 0 or (6.45–1.5505)B = –6.45 B = –6.45/(4.9) = –1.3163 and A = –1–1.3163 = –2.3163 A= –2.3163, B= –1.3163 Hence, i(t) = [2–2.3163e–6.45t –1.3163e–1.5505t] A.