Chapter 8 Solutions - Soln0831

Transcript

Chapter 8, Solution 31. For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent circuit is shown in Figure (b). By KVL, v(0+) = v(0-) = 40, i(0+) = i(0-) = 1 By KCL, 2 = i(0+) + i 1 = 1 + i 1 which leads to i 1 = 1. By KVL, -v L + 40i 1 + v(0+) = 0 which leads to v L (0+) = 40x1 + 40 = 80 v L (0+) = 80 V, 40  i 1 40  10  + i v C (0+) = 40 V + + v 50V  +  (a) v vL  10  50V  0.5H (b) + 