hw5_f08_soln - EECE 519 Homework 5 (Due Thursday, March 6)...

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Unformatted text preview: EECE 519 Homework 5 (Due Thursday, March 6) 1. Using mesh current analysis, determine the currents I1 and I2 . 100 + 15 0o V - I1 j750 -j 2000 I2 3 75 Mesh 1: 15 0o - 100I1 - (-j666.7)(I1 - I2 ) = 0 Mesh 2: -(-j666.7)(I2 - I1 ) - j750I2 - 75I2 = 0 Solving we get, I1 = 3.3 52.7o mA, I2 = 19.6 -85.3o mA 2. Solve for the voltages V1 and V2 using node-voltage analysis if is (t) = 40 cos 100tA. V1 is (t) 500F V2 40 0.2H 10 1 ZL = j(100)0.2 = j20, ZC = -j 10050010-6 = -j20 1 -V Node1: -40 0o + V1 + V-j202 = 0, (2 + j1)V1 - jV2 = 40 10 V2 2 -V Node2: V-j201 + V2 + j20 = 0, 2jV1 - V 2 = 0 40 Solving, V1 = 188.24-j47.06 = 194 -14.04o V, V2 = 96.12+j376.47 = 388 75.96o V v1 (t) = 194 cos(100t - 14.04o )V, v2 (t) = 388 cos(100t + 75.96o )V 3. Find the Thevenin and Norton equivalent circuits with respect to terminals a,b. 2 -j2 a j2 + 24 0o V - 2 90o A b 2-j2 ZT h = 2(2-j2) = 1.265 -18.4o = 2-j1 = (2-j2)(2+j1) = 6-j2 = 1.2 - j0.4 2+2-j2 (4+1) 5 -V ( V1 -24 - j2 + V1-j2T h = 0) (-j2), - jV1 + j24 - 4 + V1 - VT h = 0, 2 V1 (1 - j) - VT h = 4 - j24, and at the output node: h -V ( VT-j2 1 + VT h = 0) (-j2), VT h - V1 - jVT h = 0, V1 = (1 - j)VT h 2 2 VT h = 4-j24 -1-j2 = 8.8 + j6.4 = 10.88 36o V , IN = VT h ZT h = 10.88 36o 1.265 -18.4o = 8.6 54.4o A ...
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This homework help was uploaded on 04/07/2008 for the course ME 533 taught by Professor Bob during the Spring '08 term at Kansas State University.

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