Kirchoff’s Law

Numerical Methods

Source inspiration: (Mathew 2000-2019).

Background

Kirchoff’s voltage law (KVL) states that the algebraic sum of voltage drops around any closed loop is zero. Using loop currents, resistor networks lead directly to linear systems.

NoteDefinition - KVL Loop Equation

For each loop, \[ \sum (R\,I) = \sum E, \tag{1} \] where \(R\,I\) are resistor drops and \(E\) are source voltages (with sign convention).

Collecting all loop equations gives

\[ A\mathbf{i}=\mathbf{b}, \tag{2} \]

with unknown loop-current vector \(\mathbf{i}\).

Network #1 (Two Loops)

From the module’s network model:

\[ \begin{aligned} (r_1+r_3+r_5)i_1-r_3i_2&=e_1,\\ -r_3i_1+(r_2+r_3+r_4+r_6)i_2&=0. \end{aligned} \tag{3} \]

Example 1

Given

\[ r_1=10,\; r_2=10,\; r_3=10,\; r_4=20,\; r_5=10,\; r_6=30,\; e_1=20. \]

Step 1: Substitute into (3):

\[ \begin{aligned} 30i_1-10i_2&=20,\\ -10i_1+70i_2&=0. \end{aligned} \]

Step 2: Matrix form and solve:

\[ \begin{pmatrix} 30 & -10\\ -10 & 70 \end{pmatrix} \begin{pmatrix} i_1\\ i_2 \end{pmatrix} = \begin{pmatrix} 20\\0 \end{pmatrix} \quad\Longrightarrow\quad \begin{pmatrix} i_1\\ i_2 \end{pmatrix} = \begin{pmatrix} \tfrac{7}{10}\\[2pt]\tfrac{1}{10} \end{pmatrix}. \]

So

\[ i_1=0.7,\qquad i_2=0.1. \]

Verification:

Loop equation Left side Right side
\(30i_1-10i_2=20\) \(30(0.7)-10(0.1)=20\) \(20\ \checkmark\)
\(-10i_1+70i_2=0\) \(-10(0.7)+70(0.1)=0\) \(0\ \checkmark\)

Network #2 (Three Loops, One Source)

From the module’s loop equations:

\[ \begin{aligned} (r_1+r_4+r_7)i_1-r_4i_2&=e_1,\\ -r_4i_1+(r_2+r_4+r_5+r_8)i_2-r_5i_3&=0,\\ -r_5i_2+(r_3+r_5+r_6+r_9)i_3&=0. \end{aligned} \tag{4} \]

Example 2

Given

\[ \begin{aligned} r_1=r_2=r_3=r_4=r_7=r_8=r_9&=10,\\ r_5=r_6&=20,\\ e_1&=36. \end{aligned} \]

Step 1: Substitute into (4):

\[ \begin{aligned} 30i_1-10i_2&=36,\\ -10i_1+50i_2-20i_3&=0,\\ -20i_2+60i_3&=0. \end{aligned} \]

Step 2: Solve

\[ \begin{pmatrix} i_1\\ i_2\\ i_3 \end{pmatrix} = \begin{pmatrix} \tfrac{13}{10}\\[2pt] \tfrac{3}{10}\\[2pt] \tfrac{1}{10} \end{pmatrix} = \begin{pmatrix} 1.3\\0.3\\0.1 \end{pmatrix}. \]

Verification:

Loop equation Left side Right side
\(30i_1-10i_2=36\) \(30(1.3)-10(0.3)=36\) \(36\ \checkmark\)
\(-10i_1+50i_2-20i_3=0\) \(-13+15-2=0\) \(0\ \checkmark\)
\(-20i_2+60i_3=0\) \(-6+6=0\) \(0\ \checkmark\)

Network #3 (Three Loops, Two Sources)

From the module’s loop equations:

\[ \begin{aligned} (r_1+r_3+r_4)i_1+r_3i_2+r_4i_3&=e_1,\\ r_3i_1+(r_2+r_3+r_5)i_2-r_5i_3&=e_2,\\ r_4i_1-r_5i_2+(r_4+r_5+r_6)i_3&=0. \end{aligned} \tag{5} \]

Example 3

Given

\[ r_1=r_2=r_4=1,\quad r_3=r_5=2,\quad r_6=4,\quad e_1=23,\quad e_2=29. \]

Step 1: Substitute into (5):

\[ \begin{aligned} 4i_1+2i_2+i_3&=23,\\ 2i_1+5i_2-2i_3&=29,\\ i_1-2i_2+7i_3&=0. \end{aligned} \]

Step 2: Solve

\[ \begin{pmatrix} i_1\\ i_2\\ i_3 \end{pmatrix} = \begin{pmatrix} 3\\5\\1 \end{pmatrix}. \]

Verification:

Loop equation Left side Right side
\(4i_1+2i_2+i_3=23\) \(4(3)+2(5)+1=23\) \(23\ \checkmark\)
\(2i_1+5i_2-2i_3=29\) \(2(3)+5(5)-2(1)=29\) \(29\ \checkmark\)
\(i_1-2i_2+7i_3=0\) \(3-10+7=0\) \(0\ \checkmark\)

Network solutions are only physically meaningful when resistor values are positive and the coefficient matrix is nonsingular. Near-singular systems can produce large sensitivity in current predictions.

Summary

Kirchoff loop modeling converts circuits into linear algebra:

  1. Write one KVL equation per independent loop.
  2. Assemble the system matrix and source vector.
  3. Solve for loop currents and verify each equation numerically.

This gives a direct bridge from circuit laws to matrix methods used throughout numerical analysis.

References

Mathew, John H. 2000-2019. Numerical Analysis - Numerical Methods Modules. https://web.archive.org/web/20190808102217/http://mathfaculty.fullerton.edu/mathews/n2003/NumericalUndergradMod.html.