Solving Set of Mesh Equation of an Electrical Circuit

Kirchhoff’s Voltage Law (KVL) for Three Meshes

Step 1: Define Mesh Currents

We define the mesh currents as:

$ I_1 $ → Current in Mesh 1 (left loop)
$ I_2 $ → Current in Mesh 2 (right loop)
$ I_3 $ → Current in Mesh 3 (middle loop)

We assume all currents are in a clockwise direction.

Step 2: Write KVL Equations for Each Mesh

Mesh 1 (Left Loop)

Applying KVL:

\[V_1 - I_1 R_1 - I_1R_2 + I_2R_2 = 0\]

Rearrange:

\[I_1 (R_1 + R_2) + I_2(-R_2) = V_1\]

Mesh 2 (Middle Loop)

Applying KVL:

\[-I_2R_2 -I_2R_3 - I_2R_4 + I_3R_4 + I_2R_2 = 0\]

Rearrange:

\[I_1(-R_2) + I_2(R_2 - R_3 - R_4) + I_3R_4 = 0\]

Mesh 3 (Right Loop)

Applying KVL:

\[-I_3R_5 - V_2 + I_3R_4 - I_2R_4 = 0\]

Rearrange:

\[I_2(R_4) + I_3(R_4 - R_5) = V_2\]

Final KVL Equations for Three Meshes

\[I_1 (R_1 + R_2) + I_2(-R_2) = V_1\]
\[I_1(-R_2) + I_2(R_2 - R_3 - R_4) + I_3R_4 = 0\]
\[I_2(R_4) + I_3(R_4 - R_5) = V_2\]

These three equations can be solved simultaneously to find $ I_1 $, $ I_2 $, and $ I_3 $. The above equation can be written in matrix form as \begin{equation} \begin{pmatrix} (R_1 + R_2) & (-R_2) & 0 \\ (-R_2) & (R_2 - R_3 - R_4) & R_4 \\ (R_4) & (R_4 - R_5) & 0 \end{pmatrix} \begin{pmatrix} I_1\\ I_2 \\ I_3 \end{pmatrix}= \begin{pmatrix} V_1\\ 0 \\ V_2 \end{pmatrix} \end{equation}

[1]:

import numpy as np

[2]:

V1 = 130
V2 = 50
R1 = 10
R2 = 8
R3 = 5
R4 = 2
R5 = 5

[3]:

a00 = R1 + R2
a01 = -R2
a02 = 0
a10 = -R2
a11 = R2 - R3 - R4
a12 = R4
a20 = R4
a21 = R4 - R5
a22 = 0
A = np.array([[a00, a01, a02], [a10, a11, a12], [a20, a21, a22]], dtype=float)
b = np.array([V1, 0, V2], dtype=float)

Gauss Elimination Method

[4]:

# Gauss Elimination Method

def GaussElimination(A, b):
    x = np.copy(b)
    n = len(b)
    #Elimination phase
    for k in range(n-1):             # Fixed row/ Column no.
        for i in range(k+1, n):      # Row no. to be transformed
            if A[i, k] != 0:
                factor = A[i, k]/A[k, k]
                A[i, : ] = A[i, : ] - factor*A[k, : ]
                b[i] = b[i] - factor*b[k]

    # Back substitution
    for k in range(n-1, -1, -1):
        x[k] = (b[k] - np.dot(A[k, k+1:n], x[k+1:n]))/A[k,k]

    return x

[5]:

x = GaussElimination(A, b)
print(x)

[ -0.26315789 -16.84210526   7.36842105]

Gauss Seidal Method

[6]:

# Gauss Seidal Method

def GaussSeidel(A, b, maxiter=10, tol=1e-6):
    row, col = A.shape
    x = np.zeros_like(b)
    for iter in range(maxiter):
        for i in range(row):
            x_old = x.copy()
            j = i
            x[i] = (b[i] - np.sum(A[i, :j]*x[:j]) - np.sum(A[i, j+1:]*x[j+1:]))/A[i,i]
            err = np.abs(x_old -x)
            if all(errval < tol for errval in err) == True:
                break
    return x

[7]:

x = GaussSeidel(A, b, maxiter=10, tol=1e-6)
print(x)

[ -0.26315789 -16.84210526   7.36842105]