lab4solution

lab4solution - 10/6/2007 Page 1 of 6 ECE220 Lab4 Solution...

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ECE220 Lab4 Solution Created by Rong Guo 6.35. Part.1. Use KCL at nodes , and use KVL at Loop1, Loop2, and Loop3, we can get six equations. I 1 -I 4 -I 5 =0 I 2 +I 5 -I 6 =0 I 3 +I 6 +I 4 =0 7*I 4 -5*I 6 -8*I 5 =0 4*I 2 -8*I 5 -2*I 1 =20 4*I 2 +5*I 6 -3*I 3 =10 Part.2. >>A=[1 0 0 -1 -1 0;0 1 0 0 1 -1;0 0 1 1 0 1;0 0 0 7 -8 -5;-2 4 0 0 -8 0;0 4 -3 0 0 5] 10/6/2007 Page 1 of 6
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A = 1 0 0 -1 -1 0 0 1 0 0 1 -1 0 0 1 1 0 1 0 0 0 7 -8 -5 -2 4 0 0 -8 0 0 4 -3 0 0 5 >> b=[0;0;0;0;20;10] b = 0 0 0 0 20 10 Part.3. >> I=inv(A)*b I = -1.9351 1.7853 0.1498 -0.8115 -1.1236 0.6617 Part.4. >> R=[2;4;3;7;8;5] R = 2 4 3 10/6/2007 Page 2 of 6
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7 8 5 >> V=I.*R V = -3.8702 7.1411 0.4494 -5.6804 -8.9888 3.3084 Part.5. >> syms rx % define symbol rx to represent the 7 Ohm resistor. >> A=[1 0 0 -1 -1 0;0 1 0 0 1 -1;0 0 1 1 0 1;0 0 0 rx -8 -5;-2 4 0 0 -8 0;0 4 -3 0 0 5] A = [ 1, 0, 0, -1, -1, 0] [ 0, 1, 0,
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This note was uploaded on 03/30/2009 for the course ECE 220 taught by Professor Nilson during the Spring '08 term at N.C. State.

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lab4solution - 10/6/2007 Page 1 of 6 ECE220 Lab4 Solution...

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