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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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