HW4 Solution

# HW4 Solution - Hence 4 4 1 4 3 2 1 3 2 1 = = V V V V V V(1...

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EE302 Homework #4 Chapter 3, Solution 6. i 1 + i 2 + i 3 = 0 0 2 10 6 4 12 0 0 0 = + + v v v or v 0 = 8.72 V Chapter 3, Solution 10. At node 1: 0 1 4 8 3 1 1 = + + V V V , also by Ohm’s law: 8 1 0 V I = At node 3: 0 1 4 2 1 3 3 0 = + + V V V I , since I o is known we have: 0 1 4 4 1 3 3 1 = + + V V V V Simplifying the equations yields Node 1: 4 125 . 1 3 1 = V V Node 3: 0 25 . 1 25 . 1 3 1 = + V V Thus V 1 = -32 V. Then by Ohm’s law 8 1 0 V I = , 8 32 0 = I A 00 . 4 0 = I

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Chapter 3, Solution 12. There are two unknown nodes, as shown in the circuit below. 10 Ω + _ 2 Ω 5 Ω 30 V 4 I x 1 Ω V o V 1 At node 1, 30 V 10 V 16 0 1 V V 2 0 V 10 30 V o 1 o 1 1 1 = = + + (1) At node o, 0 I 20 V 6 V 5 0 5 0 V I 4 1 V V x o 1 o x 1 o = + = + (2) But I x = V 1 /2. Substituting this in (2) leads to –15V 1 + 6V o = 0 or V 1 = 0.4V o (3) Substituting (3) into 1, 16(0.4V o ) – 10V o = 30 or V o = –8.33 V .
Chapter 3, Solution 17. 60 V + 60 V 4 Ω 10 Ω 2 Ω 8 Ω 3i 0 i 0 v 1 v 2 At node 1, 2 v v 8 v 4 v 60 2 1 1 1 + = 120 = 7v 1 - 4v 2 (1) At node 2, 3i 0 + 0 2 v v 10 v 60 2 1 2 = + But i 0 = . 4 v 60 1 Hence () 0 2 v v 10 v 60 4 v 60 3 2 1 2 1 = + + 1020 = 5v 1 + 12v 2 (2) Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 = = 4 v 60 1 1.73 A

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Chapter 3, Solution 20. Nodes 1 and 2 form a supernode; so do nodes 1 and 3.
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Unformatted text preview: Hence 4 4 1 4 3 2 1 3 2 1 = + + = + + V V V V V V (1) . V 1 . V 2 2 V 3 4 1 4 Between nodes 1 and 3, 12 12 1 3 3 1 = = + + V V V V (2) Similarly, between nodes 1 and 2, ( 3 ) i V V 2 2 1 + = But . Combining this with (2) and (3) gives 4 / 3 V i = . ( 4 ) 2 / 6 1 2 V V + = Solving (1), (2), and (4) leads to V . 15 V, 50 . 4 V, 00 . 3 3 2 1 = = = V V V Chapter 3, Solution 22. At node 1, 8 3 4 2 12 2 1 1 1 v v v v + + = 24 = 7v 1- v 2 (1) At node 2, 3 + 1 5 8 2 2 1 v v v v + = But, v = 12 - v 1 Hence, 24 + v 1- v 2 = 8 (v 2 + 60 + 5v 1 ) = 4 V 456 = 41v 1- 9v 2 ( 2 ) Solving (1) and (2), v 1 = - 10.9 V , v 2 = - 100.4 V...
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## This note was uploaded on 02/09/2010 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas.

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HW4 Solution - Hence 4 4 1 4 3 2 1 3 2 1 = = V V V V V V(1...

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