Soln13 - EE 411 Spring 2008 HW #13 Solution (Q2) 13.24 (a)...

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EE 411 Spring 2008 HW #13 Solution (Q2) 13.24 (a) k = M/ L 1 L 2 = 1/ 4x2 = 0.3535 (b) ϖ = 4 1/4 F leads to 1/(j ϖ C) = –j/(4x0.25) = –j 1||(–j) = –j/(1 – j) = 0.5(1 – j) 1 H produces j ϖ M = j4 4 H produces j16 2 H becomes j8 12 = (2 + j16)I 1 + j4I 2 or 6 = (1 + j8)I 1 + j2I 2 (1) 0 = (j8 + 0.5 – j0.5)I 2 + j4I 1 or I 1 = (0.5 + j7.5)I 2 /(–j4) (2) Substituting (2) into (1), 24 = (–11.5 – j51.5)I 2 or I 2 = –24/(11.5 + j51.5) = –0.455 –77.41 ° V o = I 2 (0.5)(1 – j) = 0.3217 57.59 ° v o = 321.7cos(4t + 57.6 ° ) mV (c) From (2), I 1 = (0.5 + j7.5)I 2 /(–j4) = 0.855 –81.21 ° 0.5(1–j) 12 0 ° + - I 1 I 2 j4 2 j16 j8
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i 1 = 0.885cos(4t – 81.21 ° ) A, i 2 = –0.455cos(4t – 77.41 ° ) A At t = 2s, 4t = 8 rad = 98.37 ° i 1 = 0.885cos(98.37 ° – 81.21 ° ) = 0.8169 i 2 = –0.455cos(98.37 ° – 77.41 ° ) = –0.4249 w = 0.5L
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Soln13 - EE 411 Spring 2008 HW #13 Solution (Q2) 13.24 (a)...

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