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EE240-A2-2010 Assignment #3 SOLUTION 1. Nodal Analysis Problem: Find the matrix equations for V 1 , V 2 , and V 4 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 10 1 2/5 -1 V2 = 8 -1/8 29/40 -7/10 V4 -6 2. Mesh Analysis Problem : Find I 1 , I 2 , I 3 and I 4 by mesh analysis. ANS: Define the currents I 5 and I 6 as in the diagram. I A = I 6- 5. KVL around loop of I 5 : 5I A = 5(I 6- 5) = 3I 5 + 12(I 5 +2) + 8 (I 5-I 6 ) (1) KVL around loop of I 6 : 0 = 5I 6 + 8 (I 6-I 5 ) + 9 + 4(I 6-5) (2) Simplify: 23I 5- 13I 6 = -49 (1)' -8I 5 +17I 6 = +11 (2)' Solve: I 5 = -2.404 A I 6 = -0.484 A Now convert back to original currents: I 1 = I 5 = -2.404 A I 2 = -2 A I 3 = I 6- 5 = -5.484 I 4 = I 6 = -0.484 A 5A 2A I 5 I 6... View Full Document