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CH3 Solutions

# CH3 Solutions - 3.1 3.2 3.3 3.9 3.11 3.14 3.15 3.17 3.21...

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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 .

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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
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

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CH3 Solutions - 3.1 3.2 3.3 3.9 3.11 3.14 3.15 3.17 3.21...

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