HW2%20solutions2 - MAE 3780 MECHATRONICS FALL 2008 H O M E...

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MAE 3780 MECHATRONICS FALL 2008 1 HOMEWORK 2 SOLUTIONS 1. An equation for the upper left corner node does not need to be written because its voltage is known to be 3V. An equation for the upper right corner node does not need to be written because the rightmost two resistors can be combined in series. KCL at node 1: 0 2 1 = + + i i i 0 25 . 0 5 . 0 5 . 0 3 2 1 1 1 = + + v v v v 6 4 8 2 1 = v v KCL at node 2: 0 5 . 0 3 2 = + + i i 0 25 . 0 5 . 0 5 . 0 25 . 0 2 2 1 = + + + v v v 5 . 0 333 . 5 4 2 1 = + v v Using the following Matlab code, we get: A = [8 -4; ... % KCL node 1 -4 5.333333]; % KCL node 2 b = [6 -.5]'; % righthand side x = A\b % solve V 75 . 0 , V 125 . 1 2 1 = = v v Therefore,
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FALL 2008 2 A 75 . 3 5 . 0 125 . 1 3 5 . 0 3 1 = = = v i 2. Before writing any equations, it is helpful to label positive and negative sides to all circuit components in order to mitigate sign errors. In the following equations, a voltage from the negative to the positive side of a component is considered positive. This choice is arbitrary, but as long as it is consistent, it will lead to the right answer. KVL for loop 1: () 0 1 2 2 2 1 1 = I I I , or 2 3 2 1 = + I I KVL for loop 2: () 0 3 1 2 2 1 = A V I I I , or 0 4 2 1 = A V I I KVL for loop 3: 0 2 3 3 3 = I I V A , or 0 5 3 = + A V I constraint: 2 3 2 = + I I Using the following Matlab code, we get A = [-3 1 0 0; ... % KVL loop 1 1 -4 0 -1; ... % KVL loop 2 0 0 -5 1; ... % KVL loop 3 0 -1 1 0]; % constraint
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This note was uploaded on 10/02/2010 for the course MAE 3780 taught by Professor Wickenhieser during the Fall '08 term at Cornell University (Engineering School).

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HW2%20solutions2 - MAE 3780 MECHATRONICS FALL 2008 H O M E...

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