EE 240 - HOMEWORK - Solution 3
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EE 240 - HOMEWORK - Solution 3

Course: E E 240, Spring 2011

School: University of Alberta

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EE240-A2-2010 Assignment #3 SOLUTION 1. Nodal Analysis Problem: Find the matrix equations for V1 , V2 , and V4 by NODAL analysis. ANS: (1) At the supernode: 6A = V1/4 + (V1-V2)/8 + (V4-V2)/2 + (V4-20)/5 (2) Constraint: V1-V4 = 4 Ix = 4(20-V2)/10 (3) V2 Node: 2Vx = 2(V4-20)/5 = (V2-V1)/8 + (V2-V4)/2 + (V2-20)/10 (1)' V1(1/4+1/8) + V2(-1/8-1/2) + V4(1/2+1/5) = 10 (2)' V1 + V2(4/10) + V4(-1) = 8 (3)' V1(-1/8) +...

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#3 SOLUTION 1. EE240-A2-2010 Assignment Nodal Analysis Problem: Find the matrix equations for V1 , V2 , and V4 by NODAL analysis. ANS: (1) At the supernode: 6A = V1/4 + (V1-V2)/8 + (V4-V2)/2 + (V4-20)/5 (2) Constraint: V1-V4 = 4 Ix = 4(20-V2)/10 (3) V2 Node: 2Vx = 2(V4-20)/5 = (V2-V1)/8 + (V2-V4)/2 + (V2-20)/10 (1)' V1(1/4+1/8) + V2(-1/8-1/2) + V4(1/2+1/5) = 10 (2)' V1 + V2(4/10) + V4(-1) = 8 (3)' V1(-1/8) + V2(1/8+1/2+1/10) + V4(-1/2-2/5) = -8+2 = -6 Matrix form: 3/8 -5/8 7/10 V1 1 2/5 -1 V2 -1/8 29/40 -7/10 V4 = 10 8 -6 I5 2A 5A Mesh I6 2. Analysis Problem: Find I1 , I2 , I3 and I4 by mesh analysis. ANS: Define the currents I5 and I6 as in the diagram. IA = I6 - 5. KVL around loop of I5: KVL around loop of I6: Simplify: Solve: 5IA = 5(I6 - 5) = 3I5 + 12(I5+2) + 8 (I5-I6) 0 = 5I6 + 8 (I6-I5) + 9 + 4(I6-5) 23I5 - 13I6 = -49 -8I5 +17I6 = +11 I5 = -2.404 A I6 = -0.484 A Now convert back to original currents: I1 = I5 = -2.404 A I2 = -2 A I3 = I6 - 5 = -5.484 I4 = I6 = -0.484 A (1) (2) (1)' (2)'

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