# Soln6 - EE411 Spring 2008 HW#6 Solution(Q4 8.60 At t = 0 4u(t = 0 so that i1(0 = 0 = i2(0 Applying nodal analysis 4 = 0.5di1/dt i1 i2(2(3(4(1 Also

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EE411 Spring 2008 HW#6 Solution (Q4) 8.60 At t = 0-, 4u(t) = 0 so that i 1 (0) = 0 = i 2 (0) (1) Applying nodal analysis, 4 = 0.5di 1 /dt + i 1 + i 2 ( 2 ) Also, i 2 = [1di 1 /dt – 1di 2 /dt]/3 or 3i 2 = di 1 /dt – di 2 /dt (3) Taking the derivative of (2), 0 = d 2 i 1 /dt 2 + 2di 1 /dt + 2di 2 /dt (4) From (2) and (3), di 2 /dt = di 1 /dt – 3i 2 = di 1 /dt – 3(4 – i 1 – 0.5di 1 /dt) = di 1 /dt – 12 + 3i 1 + 1.5di 1 /dt Substituting this into (4), d 2 i 1 /dt 2 + 7di 1 /dt + 6i 1 = 24 which gives s 2 + 7s + 6 = 0 = (s + 1)(s + 6) Thus, i 1 (t) = I s + [Ae -t + Be -6t ], 6I s = 24 or I s = 4 i 1 (t) = 4 + [Ae -t + Be -6t ] and i 1 (0) = 4 + [A + B] (5) i 2 = 4 – i 1 – 0.5di 1 /dt = i 1 (t) = 4 + -4 - [Ae -t + Be -6t ] – [-Ae -t - 6Be -6t ] = [-0.5Ae -t + 2Be -6t ] and i 2 (0) = 0 = -0.5A + 2B (6) From (5) and (6), A = -3.2 and B = -0.8 i 1 (t) = {4 + [-3.2e

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## This note was uploaded on 10/26/2009 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas at Austin.

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Soln6 - EE411 Spring 2008 HW#6 Solution(Q4 8.60 At t = 0 4u(t = 0 so that i1(0 = 0 = i2(0 Applying nodal analysis 4 = 0.5di1/dt i1 i2(2(3(4(1 Also

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