electrical network

We shall treat an electrical network as a weighted graph $G$ where

vertices are labelled $1$ through $n$ with $1$ being the voltage source and $n$ is the ground (so there is a battery connected to them completing the circuit)
the weight of each edge is $\frac{1}{R_{e}}$ , where $R_{e}$ is the resistance of the resistor at that edge
At each edge, we specify an arbitrary direction and keep track of the information
- $V_{e}$ the voltage,
  - note that this quantity relies on the direction of the arrows, and if we reverse the arrows, the voltage would be the negation of the previous.
- $R_{e}$ the resistance,
- and $I_{e}$ the current
  - similarly, if we reverse the direction, then the current would be negated

That satisfies Kirchoff's laws: (see potential function for another interpretation of these independent on direction of the graph)

Kirchoff's first law: for any $v \in G$ , the current coming in is equal to the current coming out:

\sum_{i \overset{e}{\to} v} I_{e} = \sum_{v \overset{e^{'}}{\to} i} I_{e^{'}}

Kirchoff's second law: For all undirected cycles in $G$ , the signed sum of the voltages in this cycle is 0.
Ohm's law: for all $e \in E$ ,

V_{e} = I_{e} \cdot R_{e}

Properties

Associated to a network we have

We are concerned with finding the effective resistance of $G$ : if we assume $U_{n} = 0$ and $I_{b a t} = 1$ , then

R_{e f f} = \frac{det \tilde{\tilde{K}}}{det \tilde{K}}

where $\tilde{\tilde{K}}$ is obtained by removing the 1st and $n$ th rows and columns of $K$ , and $\tilde{K}$ is obtained by removing the $n$ th row and column of $K$ .

One more trick for electrical networks: Y-delta transform

Examples

First, some examples from classical circuits:

Series connection

Suppose we have a series connection

\usepackage{circuitikz}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document}
\begin{circuitikz}  
\draw  
(0,0) node[left]{}  
to[short, o-] (1,0)  
to[R=$R_1$] (3,0)  
to[R=$R_2$] (5,0)  
to[short, -o] (6,0) node[right]{};  
\end{circuitikz}
\end{document}

which corresponds to the graph $G$

\usepackage{tikz-cd}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document}
\begin{tikzcd}
1 \arrow[r, no head, "\frac{1}{R_1}"] & 2 \arrow[r, no head, "\frac{1}{R_2}"] & 3 
\end{tikzcd} 
\end{document}

and if we glue together 1 and 3, we get the graph $G^{'}$

\usepackage{tikz-cd}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document}
\begin{tikzcd}
1 \arrow[r, bend left, "\frac{1}{R_1}"] \arrow[r, bend right, "\frac{1}{R_2}"] & 2
\end{tikzcd} 
\end{document}

Thus,

R_{e f f} = \frac{\frac{1}{R_{1}} + \frac{1}{R_{2}}}{\frac{1}{R_{1}} \cdot \frac{1}{R_{2}}}

Parallel connection

Suppose we have the parallel connection

\usepackage{circuitikz}

\begin{document}
\begin{circuitikz}
\draw (0,0) node[left]{} to[short, o-] (0,1)  
to[R=$R_1$] (4,1)  
to[short, -o] (4,0) node[right]{};  
  
\draw (0,0) to[short] (0,-1)  
to[R=$R_2$] (4,-1)  
to[short] (4,0);
\end{circuitikz} 
\end{document}

Then $G$ is given by

\usepackage{tikz-cd}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document}
\begin{tikzcd}
1 \arrow[r, bend left, "\frac{1}{R_1}"]\arrow[r, bend right, "\frac{1}{R_2}"] & 2
\end{tikzcd} 
\end{document}

and $G^{'}$ is given by

\usepackage{tikz-cd}
\usepackage{amsmath}
\usepackage{amsfonts}
\begin{document}
\begin{tikzcd}
1 \arrow[loop up, "\frac{1}{R_1}"] \arrow[loop down, "\frac{1}{R_2}"]
\end{tikzcd} 
\end{document}

R_{e f f} = \frac{1}{\frac{1}{R_{1}} + \frac{1}{R_{2}}}

Wheatstone bridge

Note that this circuit is not the composition of series and parallel networks!
Pasted image 20260430132840.png|300
It has Kirchoff matrix

K = [\begin{matrix} R_{1}^{- 1} + R_{2}^{- 1} & - R_{1}^{- 1} & - R_{2}^{- 1} & 0 \\ - R_{1}^{- 1} & R_{1}^{- 1} + R_{3}^{- 1} + R_{4}^{- 1} & - R_{3}^{- 1} & - R_{4}^{- 1} \\ - R_{2}^{- 1} & - R_{3}^{- 1} & R_{2}^{- 1} + R_{3}^{- 1} + R_{5}^{- 1} & - R_{5}^{- 1} \\ 0 & - R_{4}^{- 1} & - R_{5}^{- 1} & R_{4}^{- 1} + R_{5}^{- 1} \end{matrix}]

\tilde{K} = [\begin{matrix} R_{1}^{- 1} + R_{2}^{- 1} & - R_{1}^{- 1} & - R_{2}^{- 1} \\ - R_{1}^{- 1} & R_{1}^{- 1} + R_{3}^{- 1} + R_{4}^{- 1} & - R_{3}^{- 1} \\ - R_{2}^{- 1} & - R_{3}^{- 1} & R_{2}^{- 1} + R_{3}^{- 1} + R_{5}^{- 1} \end{matrix}]

and

\tilde{\tilde{K}} = [\begin{matrix} R_{1}^{- 1} + R_{3}^{- 1} + R_{4}^{- 1} & - R_{3}^{- 1} \\ - R_{3}^{- 1} & R_{2}^{- 1} + R_{3}^{- 1} + R_{5}^{- 1} \end{matrix}]

And with these you can calculate the effective resistance of the graph (though it is a little bit tedious)