Electric Circuit Analysis/Mesh Analysis/Answers

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 $$I_2$$.