Chapter 8 Solutions - Soln0804 110224

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Chapter 8, Solution 4. (a) At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a). i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V. i(0+) = i(0-) = 5A Hence, v(0+) = v(0-) = 25V 3 i 40V + +  v 5  (a) 3 i 40V 0.25 H + vL  iC +  0.1F iR 4A 5 (b) (b) i C = Cdv/dt or dv(0+)/dt = i C (0+)/C For t = 0+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b). Since i and v cannot change abruptly, i R = v/5 = 25/5 = 5A, i(0+) + 4 =i C (0+) + i R (0+) 5 + 4 = i C (0+) + 5 which leads to i C (0+) = 4 dv(0+)/dt = 4/0.1 = 40 V/s Similarly, v L = Ldi/dt which leads to di(0+)/dt = v L (0+)/L 3i(0+) + v L (0+) + v(0+) = 0 15 + v L (0+) + 25 = 0 or v L (0+) = -40 di(0+)/dt = -40/0.25 = –160 A/s (c) As t approaches infinity, we have the equivalent circuit in Figure (c). 3 i + 4A v 5  (c) i() = -5(4)/(3 + 5) = –2.5 A v() = 5(4 – 2.5) = 7.5 V