EE 240 - HOMEWORK - Solution 3

# EE 240 - HOMEWORK - Solution 3 - 2 Mesh Analysis Problem...

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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

{[ snackBarMessage ]}

### Page1 / 2

EE 240 - HOMEWORK - Solution 3 - 2 Mesh Analysis Problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online