Electric Circuit Analysis/Mesh Analysis/Answers

Template:Robelbox Template:Image 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:

 $[\begin{matrix} (R_{1} + R_{2} + R_{3}) & (- R_{3}) & (- R_{2}) \\ (- R_{3}) & (R_{3} + R_{4} + R_{5}) & (- R_{4}) \\ (- R_{2}) & (- R_{4}) & (R_{2} + R_{4} + R_{6}) \end{matrix}] . [\begin{matrix} I_{1} \\ I_{2} \\ I_{3} \end{matrix}] = [\begin{matrix} - V_{s} \\ V_{s} \\ 0 \end{matrix}]$

The following matrix is the above with values substituted:

$A . \vec{X} = \vec{Y}$ → $[\begin{matrix} 10220 & - 10000 & - 20 \\ - 10000 & 30000 & - 5000 \\ - 20 & - 5000 & 6020 \end{matrix}] . [\begin{matrix} I_{1} \\ I_{2} \\ I_{3} \end{matrix}] = [\begin{matrix} - 9 \\ 9 \\ 0 \end{matrix}]$

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:

 $d e t A = 9.86 \times 1 0^{11}$

 $d e t A 1 = - 857700000$

 $d e t A 2 = 11016000$

 $d e t A 3 = 6300000$

Now we can use the solved determinants to arrive at solutions for Mesh Currents $I_{1}; I_{2} a n d I_{3}$ as follows:

1. $I_{1} = \frac{d e t A 1}{d e t A} = - 0.00086968 A$

2. $I_{2} = \frac{d e t A 2}{d e t A} = 0.00001117 A$

3. $I_{3} = \frac{d e t A 3}{d e t A} = 0.000006388 A$

Now we can solve for the current through $R_{3}$ as follows:

 $I_{R_{3}} = I_{1} - I_{2} = - 0.881 m A$

The negative sign means that $I_{R_{3}}$ is flowing in the direction of $I_{2}$ .

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Electric Circuit Analysis/Mesh Analysis/Answers

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