EE302Lecture13 - Lets take another look at this circuit and...

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Lets take another look at this circuit and solve it using Gauss Elimination and Cramer’s Rule for a 3x3 matrix: Chapter 3, Problem 20. For the circuit in Fig. 3.69, find v 1 , v 2 , and v 3 using nodal analysis. v 1 /4 v 2 /1 v 3 /4 Figure 3.69 We are free to draw a boundary (dotted line) across any part of the circuit and if the sum of currents across that boundary is zero, then we can write a KCL equation. For the above circuit, the equation is: v 1 /4 + v 2 /1 + v 3 /4 = 0. We can’t write any more KCL, but we can relate nodes 1 and 2 and nodes 1 and 3. Nodes 1 and 2 are related by the equation v 1 = 2 i + v 2 and nodes 1 and 3 are related by the equation v 1 = 12 + v 3 . We know that i = v 3 /4 so lets get rid of the i term in the second equation so that v 1 = 2( v 3 /4) + v 2 . We could find the solution by substitution of variables between the equations. However, lets look at a different method for finding the 1
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answer to this problem that is more organized than using what I would call the helter-skelter substitution method . Note that v 1 = 2( v 3 /4) + v 2 is the same as v 1 v 2 –2( v 3 /4) = 0 and v 1 = 12 + v 3 is the same as v 1 v 3 = 12. Getting the equations organized in a matrix form is important! The first step is to put all the v 1 terms in the first column and all the v 2 terms in the second column, and v 3 terms in the third column and the constants on the right hand side of the equation as shown below. The equations can be written in any order without affecting the answers. I will put the equation with the most terms as the last equation and the one with the lest terms first to make it easier: 1 v 1 + 0 v 2 1 v 3 = 12 1 v 1 – 1 v 2 0.5 v 3 = 0 .25 v 1 + 1 v 2 + .25 v 3 = 0 We can do the following math operations: 1) multiply an equation by any constant except 0, and 2) add any two equations together to form a new equation. The objective will be to perform operations so that all the off diagonal coefficients are 0 and the diagonal coefficients are 1. The answers will be on the right of the equal sign. This is the Gauss Elimination method for solving N equations with N unknowns. Lets make all the diagonal terms 1 by dividing each equation by its diagonal term: 1 v 1 + 0 v 2 1 v 3 = 12 – 1 v 1 + 1 v 2 + 0.5 v 3 = 0 1 v 1 + 4 v 2 + 1 v 3 = 0 2
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We need to combine the first equation with the second and third equations to get rid of the off diagonal terms in the first column. The first equation can be summed with the second equation and the
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This note was uploaded on 02/13/2010 for the course EE 302 taught by Professor Mccann during the Spring '06 term at University of Texas at Austin.

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EE302Lecture13 - Lets take another look at this circuit and...

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