Electric Circuit Analysis/Mesh Analysis/Answers

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Exercise 8: Answers

Model Answer

KVL arround abca loop:

 $I_1*R_1 + (I_1-I_3)*R_2 + (I_1-I_2)*R_3 = -V_s$

Therefore

 $I_1(R_1 +R_2 + R_3) - I_2(R_3) - I_3(R_2) = -V_s$    ...............   (1)

KVL arround acda loop:

  $(I_2-I_1)*R_3 + (I_2-I_3)*R_4 + I_2*R_5 = V_s$

Therefore

 $-I_1(R_3) + I_2(R_3 +R_4 + R_5) - I_3(R_4) = V_s$    ...............   (2)

KVL arround bcdb loop:

  $I_3*R_6 + (I_3-I_2)*R_4 + (I_3-I_1)*R_2 = 0$

Therefore

 $-I_1(R_2) - I_2(R_4) + I_3(R_2 +R_4 + R_6) = 0$    ...............   (3)

Now we can create a matrix with the above equations as follows:

 $\begin{bmatrix} (R_1 +R_2 + R_3) & (-R_3) & (-R_2) \\ (-R_3) & (R_3 +R_4 + R_5) & (-R_4) \\ (-R_2) & (-R_4) & (R_2 +R_4 + R_6)\end{bmatrix} . \begin{bmatrix} I_1\\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} -V_s \\ V_s \\ 0 \end{bmatrix}$

The following matrix is the above with values substituted:

$A.\vec{X} = \vec{Y}$ → $\begin{bmatrix} 10220& -10 000 & -20 \\ -10 000& 30 000 & -5000 \\ -20 & -5000 & 6020 \end{bmatrix} . \begin{bmatrix} I_1\\ I_2 \\ I_3 \end{bmatrix} = \begin{bmatrix} -9 \\ 9 \\ 0 \end{bmatrix}$

Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of the General matrix, and Determinants of alterations of the general matrix as follows:

Solving determinants of:

Matrix A : General matrix A from KVL equations

Matrix A1 : Genral Matrix A with Column 1 substituted by $\vec Y$ .

Matrix A2 : Genral Matrix A with Column 2 substituted by $\vec Y$ .

Matrix A3 : Genral Matrix A with Column 3 substituted by $\vec Y$ .

As follows:

 $det A = 9.86 \times 10^{11}$

 $det A1 = -857 700 000$

 $det A2 = 11 016 000$

 $det A3 = 6 300 000$

Now we can use the solved determinants to arrive at solutions for Mesh Currents $I_1; I_2 and I_3$ as follows:

1. $I_1 = \frac{det A1}{det A} = -0.00086968A$

2. $I_2 = \frac{det A2}{det A} = 0.00001117A$

3. $I_3 = \frac{det A3}{det A} = 0.000006388A$

Now we can solve for the current through $R_3$ as follows:

 $I_{R_3} = I_1 - I_2 = -0.881mA$

The negative sign means that $I_{R_3}$ is flowing in the direction of $I_2$ .

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