?? 13.10? ???? ?? Vo? ???. j1Ω 4Ω + 200∠45^∘...

Transcript

00:01 Given system of equations if we try to put it in ax is equals to b form or ax is equals to b form we will have the a matrix as the coefficient matrix that is 5, 1, 2, 1, 4, 3, 2, minus 2 and 8 and b matrix will be equals to the constant values that is 19, minus 2 and 39 whereas x matrix will be the terms that is x1, x2 and x3.

00:27 So, x1, x2 and x3 these are the terms.

00:30 So matrix a we can divide it into upper matrix, lower matrix and the diagonal matrix where lower matrix will be equals to 0, 0, 0, 1, 0, 0, 2, 3 and 0 only the lower part upper matrix will only be the upper triangular matrix so that would be 0, 1, 2, 0, 0, minus 2 and 0, 0, 0 so that is the upper matrix and diagonal matrix will be equals to only the diagonal elements of the matrix that is 5, 4 and 8 so diagonal would be 5, 4 and 8 rest terms would be 0.

01:06 So these are the matrices.

01:08 Now the question has asked us to compare two methods one is jacob method and the other is the gaussian method.

01:14 So, so the jacobi method the formula for calculating x of k is equals to determinant negative of b minus l plus u matrix into x of k minus 1 matrix.

01:33 Let this be equation number 1 and the question has asked us to treat x of 0 the first one as a matrix of 1, 1, 1 that is transpose of the matrix.

01:43 So if we put all of these equations and try to find x of 1 here we can write x of 1 is equals to d inverse of b minus l plus u of x of 0.

01:55 So x of 0 we already have once we put on all these values and find the calculations we will have 3 .2 minus 0 .25 and 4 .25 these are the values for x of 1.

02:07 Similarly for x of 2 it would be d inverse of b minus l plus u of x of 1.

02:13 So putting those values we will have x of 1, x of 1 we already calculated x of 2 based on x of 1 will come out to be 2 .15, 0 .825 and 4 .1688.

02:32 So we can see that we are already getting closer.

02:35 So now we can calculate x of 3 because the question has asked us to do a 5 iterations.

02:39 So finding it similarly we will have 1 .9675, 1 .0469 and 4 .0281.

02:51 Then x of 4 will become with the same formula 1 .9794, 1 .0222 and 3 .9905...

Question

?? 13.10? ???? ?? $V_o$? ???. $j1\Omega$ $4\Omega$ $+$ $200\angle45^\circ V$ $+$ $I_1$ $j8\Omega$ $j5\Omega$ $I_2$ $10\Omega$ $V_o$ $-$ ?? 13.10 ???? 13.1.

Question

?? 13.10? ???? ?? $V_o$? ???. $j1\Omega$ $4\Omega$ $+$ $200\angle45^\circ V$ $+$ $I_1$ $j8\Omega$ $j5\Omega$ $I_2$ $10\Omega$ $V_o$ $-$ ?? 13.10 ???? 13.1.

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