HW5 Solution

# HW5 Solution - EE302 Homework#5 Chapter 3 Solution 36 4 i1...

This preview shows pages 1–5. Sign up to view the full content.

EE302 Homework #5 Chapter 3, Solution 36. Applying mesh analysis gives, 12 = 10I 1 – 6I 2 -10 = -6I 1 + 8I 2 or = 2 1 I I 4 3 3 5 5 6 , 11 4 3 3 5 = = Δ , 9 4 5 3 6 1 = = Δ 7 5 3 6 5 2 = = Δ , 11 9 I 1 1 = Δ Δ = 11 7 I 2 2 = Δ Δ = i 1 = -I 1 = -9/11 = -0.8181 A, i 2 = I 1 – I 2 = 10/11 = 1.4545 A. v o = 6i 2 = 6x1.4545 = 8.73 V . I 1 I 2 + 12 V + 10 V 4 Ω 6 Ω 2 Ω i 3 i 2 i 1

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

View Full Document
Chapter 3, Solution 39. For mesh 1, 0 6 10 2 10 2 1 = + I I I x But 2 1 I I I x = . Hence, 2 1 2 1 2 1 I 2 I 4 5 I 6 I 10 I 2 I 2 10 = ⎯→ + + = (1) For mesh 2, 2 1 1 2 4 3 6 0 6 8 12 I I I I = = + (2) Solving (1) and (2) leads to -0.900A A, 800 . 0 2 1 = = I I Chapter 3, Solution 40. Assume all currents are in mA and apply mesh analysis for mesh 1. 30 = 12i 1 – 6i 2 – 4i 3 15 = 6i 1 – 3i 2 – 2i 3 (1) for mesh 2, 0 = - 6i 1 + 14i 2 – 2i 3 0 = -3i 1 + 7i 2 – i 3 (2) for mesh 3, 0 = -4i 1 – 2i 2 + 10i 3 0 = -2i 1 – i 2 + 5i 3 (3) Solving (1), (2), and (3), we obtain, i o = i 1 = 4.29 mA . 4 k Ω + 30V i 1 i 3 6 k Ω 2 k Ω 4 k Ω 6 k Ω i 2 2 k Ω
Chapter 3, Solution 44. Loop 1 and 2 form a supermesh. For the supermesh, 6 i 1 + 4i 2 - 5i 3 + 12 = 0 (1) For loop 3, -i 1 – 4i 2 + 7i 3 + 6 = 0 (2) Also, i 2 = 3 + i 1 ( 3 ) Solving (1) to (3), i 1 = -3.067, i 3 = -1.3333; i o = i 1 – i 3 = -1.73 A + 12 V + 6 V 3 A i 1 i 2 i 3 4 Ω 5 Ω 2 Ω

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.

{[ snackBarMessage ]}

### Page1 / 5

HW5 Solution - EE302 Homework#5 Chapter 3 Solution 36 4 i1...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online