CH3 Solutions - 3.1, 3.2, 3.3, 3.9, 3.11 3.14, 3.15, 3.17,...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 3.1, 3.2, 3.3, 3.9, 3.11 3.14, 3.15, 3.17, 3.21, 3.26, 3.30 3.40, 3.41, 3.42, 3.51, 3.53, 3.73, 3.74 Solution 3.1. Applying KCL: 0 4 1 3 2 1 1 = + - - V V V 1. 0 1 2 2 1 2 2 2 = + + - V V V V 2. 4 3 4 2 1 = - V V 0 2 2 1 = + V -V Solving the equations, V . V 8 4 1 = and V . V 4 2 2 = Solution 3.2. Applying KCL at each node, we can obtain: 0 10 20 30 20 2 1 1 1 = + + - v v v - v 0 10 30 30 1 2 2 2 = + + - v v v v Simplify the equations, 20 3 5 5 2 1 = v - v . 0 5 3 2 1 = + v v - Solving the two equations, V 41 5 1 . v = and V 24 3 2 . v = Solution 3.3. node 1: 2 5 0 1 3 1 2 1 = + . v v v v node 2: 3 25 0 1 2 1 2 = + . v v v node 3: 3 33 0 25 0 3 1 3 = + . v . v v Solving for V 34 . 0 2 = v therefore, V 34 . 0 = v .
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 Solution 3.9. At node 1: v 1 200 + v 1 v 2 5 + v 1 v 3 100 = 0 At node 2: v 2 v 1 5 + i + 0.2 = At node 3: i + v 3 v 1 100 + v 3 50 = For the voltage source , we have: v 3 v 2 = 50 V Solving the system: v 1 =− 45.53 V, v 2 48.69 V, v 3 = 1.31 V and, finally, i = 491 mA . Solution 3.11. At node 1: v 1 3 0.5 + v 1 0.5 + v 1 v 2 0.25 = At node 2: v 2 v 1 0.25 + v 2 0.75 + 0.5 = Solving the system: v 1 = 1.125 V, v 2 = 0.75 V i = 3 v 1 0.5 = 3.75 A. Solution 3.14. For mesh #1: ( ) ( ) 1 3 3 1 2 1 = + + i i For mesh #2: ( ) ( ) 2 2 3 3 2 1 = + + i i Solving, A 455 . 0 A 091 . 0 2 1 = = i i
Background image of page 2
3 Solution 3.15. Mesh #1 () 0 10 10 15 20 2 1 = + + I - I Mesh #2 50 10 10 40 10 1 2 = + + I - I Therefore, A 1923 0 1 . I = and A 865 0 2 . I = , V 727 6 10 1 2 10 . - i i v = = Solution 3.17. Mesh #1 (on the left-hand side) 0 ) ( 3 2 2 2 1 1 = I I I If treat mesh #2 (middle) and mesh #3 (on the right-hand side) as a single loop containing the four resistors (but not the current source), we can write 0 ) ( 3 2 3 1 1 2 3 3 2 = I I I I I From the current source: 2 2 3 = I I Solving the system, I 1 = - 0.333 A I 2 = - 1.222 A I 3 = 0.778 A Chapter 3, Solution 21. a) Specify the nodes (e.g., A on the upper left corner of the circuit in Figure P3.10, and B on the right corner). Choose one node as the reference or ground node. If possible, ground one of the sources in the circuit. Note that this is possible here. When using KCL, assume all unknown
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/19/2008 for the course ECE 3413 taught by Professor Bruce during the Spring '08 term at Mississippi State.

Page1 / 7

CH3 Solutions - 3.1, 3.2, 3.3, 3.9, 3.11 3.14, 3.15, 3.17,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online