Chapter 8, Solution 50.
For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A.
For t > 0, we have a parallel RLC circuit.
i
Is = 3 + 6 = 9A and R = 10||40 = 8 ohms
= 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25
o = 1/ LC = 1/ 01.0x4 = 5
Since > o, we have a overdamped response.
s1,2 = 2
o
2 -10, -2.5
Thus, i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9
i(0) = 3 = 9 + A + B or A + B = -6
di/dt = [-10Ae-10t] + [-2.5Be-2.5t],
v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A
Thus, A = 2 and B = -8
Clearly, i(t) = { 9 + [2e-10t] + [-8e-2.5t]} A
10
+
v
10 H
10 mF
3A 6A 40